Fibonacci sequence: Difference between revisions
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<lang tailspin> |
<lang tailspin> |
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templates nthFibonacci |
templates nthFibonacci |
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<=0|=1> $ ! |
when <=0|=1> do $ ! |
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otherwise ($ - 1 -> #) + ($ - 2 -> #) ! |
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end nthFibonacci |
end nthFibonacci |
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</lang> |
</lang> |
Revision as of 15:56, 6 August 2020
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
The Fibonacci sequence is a sequence Fn of natural numbers defined recursively:
F0 = 0 F1 = 1 Fn = Fn-1 + Fn-2, if n>1
- Task
Write a function to generate the nth Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:
Fn = Fn+2 - Fn+1, if n<0
support for negative n in the solution is optional.
- Related tasks
- References
- Wikipedia, Fibonacci number
- Wikipedia, Lucas number
- MathWorld, Fibonacci Number
- Some identities for r-Fibonacci numbers
- OEIS Fibonacci numbers
- OEIS Lucas numbers
0815
<lang 0815> %<:0D:>~$<:01:~%>=<:a94fad42221f2702:>~> }:_s:{x{={~$x+%{=>~>x~-x<:0D:~>~>~^:_s:? </lang>
11l
<lang 11l>F fib_iter(n)
I n < 2 R n V fib_prev = 1 V fib = 1 L 2 .< n (fib_prev, fib) = (fib, fib + fib_prev) R fib
L(i) 1..20
print(fib_iter(i), end' ‘ ’)
print()</lang>
- Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
360 Assembly
For maximum compatibility, programs use only the basic instruction set.
using fullword integers
<lang 360asm>* Fibonacci sequence 05/11/2014
- integer (31 bits) = 10 decimals -> max fibo(46)
FIBONACC CSECT
USING FIBONACC,R12 base register
SAVEAREA B STM-SAVEAREA(R15) skip savearea
DC 17F'0' savearea DC CL8'FIBONACC' eyecatcher
STM STM R14,R12,12(R13) save previous context
ST R13,4(R15) link backward ST R15,8(R13) link forward LR R12,R15 set addressability
- ----
LA R1,0 f(n-2)=0 LA R2,1 f(n-1)=1 LA R4,2 n=2 LA R6,1 step LH R7,NN limit
LOOP EQU * for n=2 to nn
LR R3,R2 f(n)=f(n-1) AR R3,R1 f(n)=f(n-1)+f(n-2) CVD R4,PW n convert binary to packed (PL8) UNPK ZW,PW packed (PL8) to zoned (ZL16) MVC CW,ZW zoned (ZL16) to char (CL16) OI CW+L'CW-1,X'F0' zap sign MVC WTOBUF+5(2),CW+14 output CVD R3,PW f(n) binary to packed decimal (PL8) MVC ZN,EM load mask ED ZN,PW packed dec (PL8) to char (CL20) MVC WTOBUF+9(14),ZN+6 output WTO MF=(E,WTOMSG) write buffer LR R1,R2 f(n-2)=f(n-1) LR R2,R3 f(n-1)=f(n) BXLE R4,R6,LOOP endfor n
- ----
LM R14,R12,12(R13) restore previous savearea pointer XR R15,R15 return code set to 0 BR R14 return to caller
- ---- DATA
NN DC H'46' nn max n PW DS PL8 15num ZW DS ZL16 CW DS CL16 ZN DS CL20
- ' b 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0' 15num
EM DC XL20'402020206B2020206B2020206B2020206B202120' mask WTOMSG DS 0F
DC H'80',XL2'0000'
- fibo(46)=1836311903
WTOBUF DC CL80'fibo(12)=1234567890'
REGEQU END FIBONACC</lang>
- Output:
... fibo(41)= 165,580,141 fibo(42)= 267,914,296 fibo(43)= 433,494,437 fibo(44)= 701,408,733 fibo(45)= 1,134,903,170 fibo(46)= 1,836,311,903
using packed decimals
<lang 360asm>* Fibonacci sequence 31/07/2018
- packed dec (PL8) = 15 decimals => max fibo(73)
FIBOWTOP CSECT
USING FIBOWTOP,R13 base register B 72(R15) skip savearea DC 17F'0' savearea SAVE (14,12) save previous context ST R13,4(R15) link backward ST R15,8(R13) link forward LR R13,R15 set addressability
- ----
ZAP FNM2,=P'0' f(0)=0 ZAP FNM1,=P'1' f(1)=1 LA R4,2 n=2 LA R6,1 step LH R7,NN limit
LOOP EQU * for n=2 to nn
ZAP FN,FNM1 f(n)=f(n-2) AP FN,FNM2 f(n)=f(n-1)+f(n-2) CVD R4,PW n MVC ZN,EM load mask ED ZN,PW packed dec (PL8) to char (CL16) MVC WTOBUF+5(2),ZN+L'ZN-2 output MVC ZN,EM load mask ED ZN,FN packed dec (PL8) to char (CL16) MVC WTOBUF+9(L'ZN),ZN output WTO MF=(E,WTOMSG) write buffer ZAP FNM2,FNM1 f(n-2)=f(n-1) ZAP FNM1,FN f(n-1)=f(n) BXLE R4,R6,LOOP endfor n
- ----
L R13,4(0,R13) restore previous savearea pointer RETURN (14,12),RC=0 restore registers from calling sav
- ---- DATA
NN DC H'73' nn FNM2 DS PL8 f(n-2) FNM1 DS PL8 f(n-1) FN DS PL8 f(n) PW DS PL8 15num ZN DS CL20
- ' b 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0' 15num
EM DC XL20'402020206B2020206B2020206B2020206B202120' mask WTOMSG DS 0F
DC H'80',XL2'0000'
- fibo(73)=806515533049393
WTOBUF DC CL80'fibo(12)=123456789012345 '
REGEQU END FIBOWTOP</lang>
- Output:
... fibo(68)= 72,723,460,248,141 fibo(69)= 117,669,030,460,994 fibo(70)= 190,392,490,709,135 fibo(71)= 308,061,521,170,129 fibo(72)= 498,454,011,879,264 fibo(73)= 806,515,533,049,393
6502 Assembly
This subroutine stores the first n—by default the first ten—Fibonacci numbers in memory, beginning (because, why not?) at address 3867 decimal = F1B hex. Intermediate results are stored in three sequential addresses within the low 256 bytes of memory, which are the most economical to access.
The results are calculated and stored, but are not output to the screen or any other physical device: how to do that would depend on the hardware and the operating system. <lang 6502asm> LDA #0
STA $F0 ; LOWER NUMBER LDA #1 STA $F1 ; HIGHER NUMBER LDX #0
LOOP: LDA $F1
STA $0F1B,X STA $F2 ; OLD HIGHER NUMBER ADC $F0 STA $F1 ; NEW HIGHER NUMBER LDA $F2 STA $F0 ; NEW LOWER NUMBER INX CPX #$0A ; STOP AT FIB(10) BMI LOOP RTS ; RETURN FROM SUBROUTINE</lang>
8080 Assembly
This subroutine expects to be called with the value of in register A, and returns also in A. You may want to take steps to save the previous contents of B, C, and D. The routine only works with fairly small values of . <lang 8080asm>FIBNCI: MOV C, A ; C will store the counter
DCR C ; decrement, because we know f(1) already MVI A, 1 MVI B, 0
LOOP: MOV D, A
ADD B ; A := A + B MOV B, D DCR C JNZ LOOP ; jump if not zero RET ; return from subroutine</lang>
8th
An iterative solution: <lang forth>
- fibon \ n -- fib(n)
>r 0 1 ( tuck n:+ ) \ fib(n-2) fib(n-1) -- fib(n-1) fib(n) r> n:1- times ;
- fib \ n -- fib(n)
dup 1 n:= if 1 ;; then fibon nip ;
</lang>
ABAP
Iterative
<lang ABAP>FORM fibonacci_iter USING index TYPE i
CHANGING number_fib TYPE i. DATA: lv_old type i, lv_cur type i. Do index times. If sy-index = 1 or sy-index = 2. lv_cur = 1. lv_old = 0. endif. number_fib = lv_cur + lv_old. lv_old = lv_cur. lv_cur = number_fib. enddo.
ENDFORM.</lang>
Impure Functional
<lang ABAP>cl_demo_output=>display( REDUCE #( INIT fibnm = VALUE stringtab( ( |0| ) ( |1| ) )
n TYPE string x = `0` y = `1` FOR i = 1 WHILE i <= 100 NEXT n = ( x + y ) fibnm = VALUE #( BASE fibnm ( n ) ) x = y y = n ) ).</lang>
ACL2
Fast, tail recursive solution: <lang Lisp>(defun fast-fib-r (n a b)
(if (or (zp n) (zp (1- n))) b (fast-fib-r (1- n) b (+ a b))))
(defun fast-fib (n)
(fast-fib-r n 1 1))
(defun first-fibs-r (n i)
(declare (xargs :measure (nfix (- n i)))) (if (zp (- n i)) nil (cons (fast-fib i) (first-fibs-r n (1+ i)))))
(defun first-fibs (n)
(first-fibs-r n 0))</lang>
- Output:
>(first-fibs 20) (1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765)
ActionScript
<lang actionscript>public function fib(n:uint):uint {
if (n < 2) return n; return fib(n - 1) + fib(n - 2);
}</lang>
Ada
Recursive
<lang Ada>with Ada.Text_IO, Ada.Command_Line;
procedure Fib is
X: Positive := Positive'Value(Ada.Command_Line.Argument(1));
function Fib(P: Positive) return Positive is begin if P <= 2 then return 1; else return Fib(P-1) + Fib(P-2); end if; end Fib;
begin
Ada.Text_IO.Put("Fibonacci(" & Integer'Image(X) & " ) = "); Ada.Text_IO.Put_Line(Integer'Image(Fib(X)));
end Fib;</lang>
Iterative, build-in integers
<lang ada>with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Fibonacci is
function Fibonacci (N : Natural) return Natural is This : Natural := 0; That : Natural := 1; Sum : Natural; begin for I in 1..N loop Sum := This + That; That := This; This := Sum; end loop; return This; end Fibonacci;
begin
for N in 0..10 loop Put_Line (Positive'Image (Fibonacci (N))); end loop;
end Test_Fibonacci;</lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55
Iterative, long integers
Using the big integer implementation from a cryptographic library [1].
<lang Ada>with Ada.Text_IO, Ada.Command_Line, Crypto.Types.Big_Numbers;
procedure Fibonacci is
X: Positive := Positive'Value(Ada.Command_Line.Argument(1));
Bit_Length: Positive := 1 + (696 * X) / 1000; -- that number of bits is sufficient to store the full result.
package LN is new Crypto.Types.Big_Numbers (Bit_Length + (32 - Bit_Length mod 32)); -- the actual number of bits has to be a multiple of 32 use LN;
function Fib(P: Positive) return Big_Unsigned is Previous: Big_Unsigned := Big_Unsigned_Zero; Result: Big_Unsigned := Big_Unsigned_One; Tmp: Big_Unsigned; begin -- Result = 1 = Fibonacci(1) for I in 1 .. P-1 loop Tmp := Result; Result := Previous + Result; Previous := Tmp; -- Result = Fibonacci(I+1)) end loop; return Result; end Fib;
begin
Ada.Text_IO.Put("Fibonacci(" & Integer'Image(X) & " ) = "); Ada.Text_IO.Put_Line(LN.Utils.To_String(Fib(X)));
end Fibonacci;</lang>
- Output:
> ./fibonacci 777 Fibonacci( 777 ) = 1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082
Fast method using fast matrix exponentiation
<lang Ada> with ada.text_io; use ada.text_io;
procedure fast_fibo is -- We work with biggest natural integers in a 64 bits machine type Big_Int is mod 2**64;
-- We provide an index type for accessing the fibonacci sequence terms type Index is new Big_Int;
-- fibo is a generic function that needs a modulus type since it will return -- the n'th term of the fibonacci sequence modulus this type (use Big_Int to get the -- expected behaviour in this particular task) generic type ring_element is mod <>; with function "*" (a, b : ring_element) return ring_element is <>; function fibo (n : Index) return ring_element; function fibo (n : Index) return ring_element is
type matrix is array (1 .. 2, 1 .. 2) of ring_element;
-- f is the matrix you apply to a column containing (F_n, F_{n+1}) to get -- the next one containing (F_{n+1},F_{n+2}) -- could be a more general matrix (given as a generic parameter) to deal with -- other linear sequences of order 2 f : constant matrix := (1 => (0, 1), 2 => (1, 1));
function "*" (a, b : matrix) return matrix is (1 => (a(1,1)*b(1,1)+a(1,2)*b(2,1), a(1,1)*b(1,2)+a(1,2)*b(2,2)), 2 => (a(2,1)*b(1,1)+a(2,2)*b(2,1), a(2,1)*b(1,2)+a(2,2)*b(2,2)));
function square (m : matrix) return matrix is (m * m);
-- Fast_Pow could be non recursive but it doesn't really matter since -- the number of calls is bounded up by the size (in bits) of Big_Int (e.g 64) function fast_pow (m : matrix; n : Index) return matrix is (if n = 0 then (1 => (1, 0), 2 => (0, 1)) -- = identity matrix elsif n mod 2 = 0 then square (fast_pow (m, n / 2)) else m * square (fast_pow (m, n / 2)));
begin return fast_pow (f, n)(2, 1); end fibo;
function Big_Int_Fibo is new fibo (Big_Int); begin -- calculate instantly F_n with n=10^15 (modulus 2^64 ) put_line (Big_Int_Fibo (10**15)'img); end fast_fibo; </lang>
AdvPL
Recursive
<lang AdvPL>
- include "totvs.ch"
User Function fibb(a,b,n) return(if(--n>0,fibb(b,a+b,n),a)) </lang>
Iterative
<lang AdvPL>
- include "totvs.ch"
User Function fibb(n) local fnow:=0, fnext:=1, tempf while (--n>0) tempf:=fnow+fnext fnow:=fnext fnext:=tempf end while return(fnext) </lang>
Aime
<lang aime>integer fibs(integer n) {
integer w;
if (n == 0) { w = 0; } elif (n == 1) { w = 1; } else { integer a, b, i;
i = 1; a = 0; b = 1; while (i < n) { w = a + b; a = b; b = w; i += 1; } }
return w;
} </lang>
ALGOL 68
Analytic
<lang algol68>PROC analytic fibonacci = (LONG INT n)LONG INT:(
LONG REAL sqrt 5 = long sqrt(5); LONG REAL p = (1 + sqrt 5) / 2; LONG REAL q = 1/p; ROUND( (p**n + q**n) / sqrt 5 )
);
FOR i FROM 1 TO 30 WHILE
print(whole(analytic fibonacci(i),0));
- WHILE # i /= 30 DO
print(", ")
OD; print(new line)</lang>
- Output:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040
Iterative
<lang algol68>PROC iterative fibonacci = (INT n)INT:
CASE n+1 IN 0, 1, 1, 2, 3, 5 OUT INT even:=3, odd:=5; FOR i FROM odd+1 TO n DO (ODD i|odd|even) := odd + even OD; (ODD n|odd|even) ESAC;
FOR i FROM 0 TO 30 WHILE
print(whole(iterative fibonacci(i),0));
- WHILE # i /= 30 DO
print(", ")
OD; print(new line)</lang>
- Output:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040
Recursive
<lang algol68>PROC recursive fibonacci = (INT n)INT:
( n < 2 | n | fib(n-1) + fib(n-2));</lang>
Generative
<lang algol68>MODE YIELDINT = PROC(INT)VOID;
PROC gen fibonacci = (INT n, YIELDINT yield)VOID: (
INT even:=0, odd:=1; yield(even); yield(odd); FOR i FROM odd+1 TO n DO yield( (ODD i|odd|even) := odd + even ) OD
);
main:(
# FOR INT n IN # gen fibonacci(30, # ) DO ( # ## (INT n)VOID:( print((" ",whole(n,0))) # OD # )); print(new line)
)</lang>
- Output:
1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
Array (Table) Lookup
This uses a pre-generated list, requiring much less run-time processor usage, but assumes that INT is only 31 bits wide. <lang algol68>[]INT const fibonacci = []INT( -1836311903, 1134903170,
-701408733, 433494437, -267914296, 165580141, -102334155, 63245986, -39088169, 24157817, -14930352, 9227465, -5702887, 3524578, -2178309, 1346269, -832040, 514229, -317811, 196418, -121393, 75025, -46368, 28657, -17711, 10946, -6765, 4181, -2584, 1597, -987, 610, -377, 233, -144, 89, -55, 34, -21, 13, -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903
)[@-46];
PROC VOID value error := stop;
PROC lookup fibonacci = (INT i)INT: (
IF LWB const fibonacci <= i AND i<= UPB const fibonacci THEN const fibonacci[i] ELSE value error; SKIP FI
);
FOR i FROM 0 TO 30 WHILE
print(whole(lookup fibonacci(i),0));
- WHILE # i /= 30 DO
print(", ")
OD; print(new line)</lang>
- Output:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040
ALGOL W
<lang algolw>begin
% return the nth Fibonacci number % integer procedure Fibonacci( integer value n ) ; begin integer fn, fn1, fn2; fn2 := 1; fn1 := 0; fn := 0; for i := 1 until n do begin fn := fn1 + fn2; fn2 := fn1; fn1 := fn end ; fn end Fibonacci ;
for i := 0 until 10 do writeon( i_w := 3, s_w := 0, Fibonacci( i ) )
end.</lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55
ALGOL-M
Note that the 21st Fibonacci number (= 10946) is the largest that can be calculated without overflowing ALGOL-M's integer data type.
Iterative
<lang algol>INTEGER FUNCTION FIBONACCI( X ); INTEGER X; BEGIN
INTEGER M, N, A, I; M := 0; N := 1; FOR I := 2 STEP 1 UNTIL X DO BEGIN A := N; N := M + N; M := A; END; FIBONACCI := N;
END;</lang>
Naively recursive
<lang algol>INTEGER FUNCTION FIBONACCI( X ); INTEGER X; BEGIN
IF X < 3 THEN FIBONACCI := 1 ELSE FIBONACCI := FIBONACCI( X - 2 ) + FIBONACCI( X - 1 );
END;</lang>
Alore
<lang Alore>def fib(n as Int) as Int
if n < 2 return 1 end return fib(n-1) + fib(n-2)
end</lang>
AntLang
<lang AntLang>/Sequence fib:{<0;1> {x,<x[-1]+x[-2]>}/ range[x]} /nth fibn:{fib[x][x]}</lang>
Apex
<lang Apex> /*
author: snugsfbay date: March 3, 2016 description: Create a list of x numbers in the Fibonacci sequence. - user may specify the length of the list - enforces a minimum of 2 numbers in the sequence because any fewer is not a sequence - enforces a maximum of 47 because further values are too large for integer data type - Fibonacci sequence always starts with 0 and 1 by definition
- /
public class FibNumbers{
final static Integer MIN = 2; //minimum length of sequence final static Integer MAX = 47; //maximum length of sequence
/*
description: method to create a list of numbers in the Fibonacci sequence param: user specified integer representing length of sequence should be 2-47, inclusive. - Sequence starts with 0 and 1 by definition so the minimum length could be as low as 2. - For 48th number in sequence or greater, code would require a Long data type rather than an Integer. return: list of integers in sequence.
- /
public static List<Integer> makeSeq(Integer len){
List<Integer> fib = new List<Integer>{0,1}; // initialize list with first two values Integer i; if(len<MIN || len==null || len>MAX) { if (len>MAX){ len=MAX; //set length to maximum if user entered too high a value }else{ len=MIN; //set length to minimum if user entered too low a value or none } } //This could be refactored using teneray operator, but we want code coverage to be reflected for each condition //start with initial list size to find previous two values in the sequence, continue incrementing until list reaches user defined length for(i=fib.size(); i<len; i++){ fib.add(fib[i-1]+fib[i-2]); //create new number based on previous numbers and add that to the list }
return fib; }
} </lang>
APL
Naive Recursive
<lang APL> fib←{⍵≤1:⍵ ⋄ (∇ ⍵-1)+∇ ⍵-2} </lang>
Read this as: In the variable "fib", store the function that says, if the argument is less than or equal to 1, return the argument. Else, calculate the value you get when you recursively call the current function with the argument of the current argument minus one and add that to the value you get when you recursively call the current function with the argument of the current function minus two.
This naive solution requires Dyalog APL because GNU APL does not support this syntax for conditional guards.
Array
Since APL is an array language we'll use the following identity:
In APL: <lang APL> ↑+.×/N/⊂2 2⍴1 1 1 0 </lang> Plugging in 4 for N gives the following result:
Here's what happens: We replicate the 2-by-2 matrix N times and then apply inner product-replication. The First removes the shell from the Enclose. At this point we're basically done, but we need to pick out only in order to complete the task. Here's one way: <lang APL> ↑0 1↓↑+.×/N/⊂2 2⍴1 1 1 0 </lang>
Analytic
An alternative approach, using Binet's formula (which was apparently known long before Binet): <lang apl>⌊.5+(((1+PHI)÷2)*⍳N)÷PHI←5*.5</lang>
AppleScript
Imperative
<lang applescript>set fibs to {} set x to (text returned of (display dialog "What fibbonaci number do you want?" default answer "3")) set x to x as integer repeat with y from 1 to x if (y = 1 or y = 2) then copy 1 to the end of fibs else copy ((item (y - 1) of fibs) + (item (y - 2) of fibs)) to the end of fibs end if end repeat return item x of fibs</lang>
Functional
The simple recursive version is famously slow:
<lang AppleScript>on fib(n)
if n < 1 then 0 else if n < 3 then 1 else fib(n - 2) + fib(n - 1) end if
end fib</lang>
but we can combine enumFromTo(m, n) with the accumulator of a higher-order fold/reduce function to memoize the series:
(ES6 memoized fold example)
(Memoized fold example)
<lang AppleScript>-------------------- FIBONACCI SEQUENCE --------------------
-- fib :: Int -> Int on fib(n)
-- lastTwo : (Int, Int) -> (Int, Int) script lastTwo on |λ|([a, b]) [b, a + b] end |λ| end script item 1 of foldl(lastTwo, {0, 1}, enumFromTo(1, n))
end fib
TEST ---------------------------
on run
fib(32) --> 2178309
end run
GENERIC FUNCTIONS ---------------------
-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)
if m ≤ n then set lst to {} repeat with i from m to n set end of lst to i end repeat lst else {} end if
end enumFromTo
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs repeat with i from 1 to lng set v to |λ|(v, item i of xs, i, xs) end repeat return v end tell
end foldl
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn</lang>
- Output:
2178309
Arendelle
( fibonacci , 1; 1 ) [ 98 , // 100 numbers of fibonacci ( fibonacci[ @fibonacci? ] , @fibonacci[ @fibonacci - 1 ] + @fibonacci[ @fibonacci - 2 ] ) "Index: | @fibonacci? | => | @fibonacci[ @fibonacci? - 1 ] |" ]
ARM Assembly
Expects to be called with in R0, and will return in the same register. <lang armasm>fibonacci:
push {r1-r3} mov r1, #0 mov r2, #1
fibloop:
mov r3, r2 add r2, r1, r2 mov r1, r3 sub r0, r0, #1 cmp r0, #1 bne fibloop mov r0, r2 pop {r1-r3} mov pc, lr</lang>
ArnoldC
<lang ArnoldC>IT'S SHOWTIME
HEY CHRISTMAS TREE f1 YOU SET US UP @I LIED TALK TO THE HAND f1
HEY CHRISTMAS TREE f2 YOU SET US UP @NO PROBLEMO
HEY CHRISTMAS TREE f3 YOU SET US UP @I LIED
STICK AROUND @NO PROBLEMO
GET TO THE CHOPPER f3 HERE IS MY INVITATION f1 GET UP f2 ENOUGH TALK TALK TO THE HAND f3
GET TO THE CHOPPER f1 HERE IS MY INVITATION f2 ENOUGH TALK
GET TO THE CHOPPER f2 HERE IS MY INVITATION f3 ENOUGH TALK
CHILL
YOU HAVE BEEN TERMINATED</lang>
Arturo
Recursive
<lang arturo>fib: @(x){ if x<2 { 1 }{ [fib x-1] + [fib x-2] } }</lang>
Recursive with Memoization
<lang arturo>Fib $(memoize [x]{ if x<2 { 1 }{ $(Fib x-1) + $(Fib x-2) } })</lang>
AsciiDots
<lang AsciiDots>
/--#$--\ | | >-*>{+}/ | \+-/ 1 |
- 1
| # | | . .
</lang>
ATS
Recursive
<lang ATS> fun fib_rec(n: int): int =
if n >= 2 then fib_rec(n-1) + fib_rec(n-2) else n
</lang>
Iterative
<lang ATS> (*
- This one is also referred to as being tail-recursive
- )
fun fib_trec(n: int): int = if n > 0 then (fix loop (i:int, r0:int, r1:int): int => if i > 1 then loop (i-1, r1, r0+r1) else r1)(n, 0, 1) else 0 </lang>
Iterative and Verified
<lang ATS> (*
- This implementation is verified!
- )
dataprop FIB (int, int) =
| FIB0 (0, 0) | FIB1 (1, 1) | {n:nat} {r0,r1:int} FIB2 (n+2, r0+r1) of (FIB (n, r0), FIB (n+1, r1))
// end of [FIB] // end of [dataprop]
fun fibats{n:nat}
(n: int (n))
- [r:int] (FIB (n, r) | int r) = let
fun loop {i:nat | i <= n}{r0,r1:int} ( pf0: FIB (i, r0), pf1: FIB (i+1, r1) | ni: int (n-i), r0: int r0, r1: int r1 ) : [r:int] (FIB (n, r) | int r) = if (ni > 0) then loop{i+1}(pf1, FIB2 (pf0, pf1) | ni - 1, r1, r0 + r1) else (pf0 | r0) // end of [if] // end of [loop]
in
loop {0} (FIB0 (), FIB1 () | n, 0, 1)
end // end of [fibats] </lang>
Matrix-based
<lang ATS> (* ****** ****** *) // // How to compile: // patscc -o fib fib.dats // (* ****** ****** *) //
- include
"share/atspre_staload.hats" // (* ****** ****** *) // abst@ype int3_t0ype =
(int, int, int)
// typedef int3 = int3_t0ype // (* ****** ****** *)
extern fun int3 : (int, int, int) -<> int3 extern fun int3_1 : int3 -<> int extern fun mul_int3_int3: (int3, int3) -<> int3
(* ****** ****** *)
local
assume int3_t0ype = (int, int, int)
in (* in-of-local *) // implement int3 (x, y, z) = @(x, y, z) // implement int3_1 (xyz) = xyz.1 // implement mul_int3_int3 (
@(a,b,c), @(d,e,f)
) =
(a*d + b*e, a*e + b*f, b*e + c*f)
// end // end of [local]
(* ****** ****** *) // implement gnumber_int<int3> (n) = int3(n, 0, n) // implement gmul_val<int3> = mul_int3_int3 // (* ****** ****** *) // fun fib (n: intGte(0)): int =
int3_1(gpow_int_val<int3> (n, int3(1, 1, 0)))
// (* ****** ****** *)
implement main0 () = { // val N = 10 val () = println! ("fib(", N, ") = ", fib(N)) val N = 20 val () = println! ("fib(", N, ") = ", fib(N)) val N = 30 val () = println! ("fib(", N, ") = ", fib(N)) val N = 40 val () = println! ("fib(", N, ") = ", fib(N)) // } (* end of [main0] *) </lang>
AutoHotkey
Search autohotkey.com: sequence
Iterative
<lang AutoHotkey>Loop, 5
MsgBox % fib(A_Index)
Return
fib(n) {
If (n < 2) Return n i := last := this := 1 While (i <= n) { new := last + this last := this this := new i++ } Return this
}</lang>
Recursive and iterative
Source: AutoHotkey forum by Laszlo <lang AutoHotkey>/* Important note: the recursive version would be very slow without a global or static array. The iterative version handles also negative arguments properly.
- /
FibR(n) { ; n-th Fibonacci number (n>=0, recursive with static array Fibo)
Static Return n<2 ? n : Fibo%n% ? Fibo%n% : Fibo%n% := FibR(n-1)+FibR(n-2)
}
Fib(n) { ; n-th Fibonacci number (n < 0 OK, iterative)
a := 0, b := 1 Loop % abs(n)-1 c := b, b += a, a := c Return n=0 ? 0 : n>0 || n&1 ? b : -b
}</lang>
AutoIt
Iterative
<lang AutoIt>#AutoIt Version: 3.2.10.0 $n0 = 0 $n1 = 1 $n = 10 MsgBox (0,"Iterative Fibonacci ", it_febo($n0,$n1,$n))
Func it_febo($n_0,$n_1,$N)
$first = $n_0 $second = $n_1 $next = $first + $second $febo = 0 For $i = 1 To $N-3 $first = $second $second = $next $next = $first + $second Next if $n==0 Then $febo = 0 ElseIf $n==1 Then $febo = $n_0 ElseIf $n==2 Then $febo = $n_1 Else $febo = $next EndIf Return $febo
EndFunc </lang>
Recursive
<lang AutoIt>#AutoIt Version: 3.2.10.0 $n0 = 0 $n1 = 1 $n = 10 MsgBox (0,"Recursive Fibonacci ", rec_febo($n0,$n1,$n)) Func rec_febo($r_0,$r_1,$R)
if $R<3 Then if $R==2 Then
Return $r_1
ElseIf $R==1 Then
Return $r_0
ElseIf $R==0 Then
Return 0
EndIf Return $R Else Return rec_febo($r_0,$r_1,$R-1) + rec_febo($r_0,$r_1,$R-2) EndIf
EndFunc </lang>
AWK
As in many examples, this one-liner contains the function as well as testing with input from stdin, output to stdout. <lang awk>$ awk 'func fib(n){return(n<2?n:fib(n-1)+fib(n-2))}{print "fib("$1")="fib($1)}' 10 fib(10)=55</lang>
Axe
A recursive solution is not practical in Axe because there is no concept of variable scope in Axe.
Iterative solution: <lang axe>Lbl FIB r₁→N 0→I 1→J For(K,1,N)
I+J→T J→I T→J
End J Return</lang>
Babel
In Babel, we can define fib using a stack-based approach that is not recursive:
<lang babel>fib { <- 0 1 { dup <- + -> swap } -> times zap } <</lang>
foo x < puts x in foo. In this case, x is the code list between the curly-braces. This is how you define callable code in Babel. The definition works by initializing the stack with 0, 1. On each iteration of the times loop, the function duplicates the top element. In the first iteration, this gives 0, 1, 1. Then it moves down the stack with the <- operator, giving 0, 1 again. It adds, giving 1. then it moves back up the stack, giving 1, 1. Then it swaps. On the next iteration this gives:
1, 1, 1 (dup) 1, 1, (<-) 2 (+) 2, 1 (->) 1, 2 (swap)
And so on. To test fib:
<lang babel>{19 iter - fib !} 20 times collect ! lsnum !</lang>
- Output:
( 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 )
bash
Iterative
<lang bash> $ fib=1;j=1;while((fib<100));do echo $fib;((k=fib+j,fib=j,j=k));done </lang>
1 1 2 3 5 8 13 21 34 55 89
Recursive
<lang bash>fib() {
if [ $1 -le 0 ] then echo 0 return 0 fi if [ $1 -le 2 ] then echo 1 else a=$(fib $[$1-1]) b=$(fib $[$1-2]) echo $(($a+$b)) fi
} </lang>
BASIC
Applesoft BASIC
Same code as Commodore BASIC
Entering a value of N > 183, produces an error message:
?OVERFLOW ERROR IN 40
BASIC256
<lang BASIC256>
- Basic-256 ver 1.1.4
- iterative Fibonacci sequence
- Matches sequence A000045 in the OEIS, https://oeis.org/A000045/list
- Return the Nth Fibonacci number
input "N = ",f limit = 500 # set upper limit - can be changed, removed f = int(f) if f > limit then f = limit a = 0 : b = 1 : c = 0 : n = 0 # initial values
while n < f
print n + chr(9) + c # chr(9) = tab a = b b = c c = a + b n += 1
end while
print " " print n + chr(9) + c </lang>
Commodore BASIC
<lang basic>10 INPUT "ENTER VALUE OF N"; N 20 N1 = 0 : N2 = 1 30 FOR K=1 TO N 40 SUM = N1+N2 50 N1 = N2 60 N2 = SUM 70 NEXT K 80 PRINT N1</lang>
Integer BASIC
Only works with quite small values of . <lang basic> 10 INPUT N
20 A=0 30 B=1 40 FOR I=2 TO N 50 C=B 60 B=A+B 70 A=C 80 NEXT I 90 PRINT B
100 END</lang>
IS-BASIC
<lang IS-BASIC>100 PROGRAM "Fibonac.bas" 110 FOR I=0 TO 20 120 PRINT "F";I,FIB(I) 130 NEXT 140 DEF FIB(N) 150 NUMERIC I 160 LET A=0:LET B=1 170 FOR I=1 TO N 180 LET T=A+B:LET A=B:LET B=T 190 NEXT 200 LET FIB=A 210 END DEF</lang>
QBasic
Iterative
<lang qbasic>FUNCTION itFib (n)
n1 = 0 n2 = 1 FOR k = 1 TO ABS(n) sum = n1 + n2 n1 = n2 n2 = sum NEXT k IF n < 0 THEN itFib = n1 * ((-1) ^ ((-n) + 1)) ELSE itFib = n1 END IF
END FUNCTION</lang>
Next version calculates each value once, as needed, and stores the results in an array for later retreival (due to the use of REDIM PRESERVE
, it requires QuickBASIC 4.5 or newer):
<lang qbasic>DECLARE FUNCTION fibonacci& (n AS INTEGER)
REDIM SHARED fibNum(1) AS LONG
fibNum(1) = 1
'*****sample inputs***** PRINT fibonacci(0) 'no calculation needed PRINT fibonacci(13) 'figure F(2)..F(13) PRINT fibonacci(-42) 'figure F(14)..F(42) PRINT fibonacci(47) 'error: too big '*****sample inputs*****
FUNCTION fibonacci& (n AS INTEGER)
DIM a AS INTEGER a = ABS(n) SELECT CASE a CASE 0 TO 46 SHARED fibNum() AS LONG DIM u AS INTEGER, L0 AS INTEGER u = UBOUND(fibNum) IF a > u THEN REDIM PRESERVE fibNum(a) AS LONG FOR L0 = u + 1 TO a fibNum(L0) = fibNum(L0 - 1) + fibNum(L0 - 2) NEXT END IF IF n < 0 THEN fibonacci = fibNum(a) * ((-1) ^ (a + 1)) ELSE fibonacci = fibNum(n) END IF CASE ELSE 'limited to signed 32-bit int (LONG) 'F(47)=&hB11924E1 ERROR 6 'overflow END SELECT
END FUNCTION</lang>
- Output:
(unhandled error in final input prevents output)
0 233 -267914296
Recursive
This example can't handle n < 0.
<lang qbasic>FUNCTION recFib (n)
IF (n < 2) THEN
recFib = n
ELSE
recFib = recFib(n - 1) + recFib(n - 2)
END IF
END FUNCTION</lang>
Array (Table) Lookup
This uses a pre-generated list, requiring much less run-time processor usage. (Since the sequence never changes, this is probably the best way to do this in "the real world". The same applies to other sequences like prime numbers, and numbers like pi and e.)
<lang qbasic>DATA -1836311903,1134903170,-701408733,433494437,-267914296,165580141,-102334155 DATA 63245986,-39088169,24157817,-14930352,9227465,-5702887,3524578,-2178309 DATA 1346269,-832040,514229,-317811,196418,-121393,75025,-46368,28657,-17711 DATA 10946,-6765,4181,-2584,1597,-987,610,-377,233,-144,89,-55,34,-21,13,-8,5,-3 DATA 2,-1,1,0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765 DATA 10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269 DATA 2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986 DATA 102334155,165580141,267914296,433494437,701408733,1134903170,1836311903
DIM fibNum(-46 TO 46) AS LONG
FOR n = -46 TO 46
READ fibNum(n)
NEXT
'*****sample inputs***** FOR n = -46 TO 46
PRINT fibNum(n),
NEXT PRINT '*****sample inputs*****</lang>
Fibonacci from Russia
Fibonacci from Russia
<lang qbasic>'FibRus.bas DIM F(80) AS DOUBLE F(1) = 0: F(2) = 1 'OPEN "FibRus.txt" FOR OUTPUT AS #1 FOR i = 3 TO 80
F(i) = F(i-1)+F(i-2)
NEXT i
FOR i = 1 TO 80
f$ = STR$(F(i)): LF = 22 - LEN(f$) n$ = "" FOR j = 1 TO LF: n$ = " " + n$: NEXT f$ = n$ + f$ PRINT i, f$: ' PRINT #1, i, f$
NEXT i
1 0 2 1 3 1 4 2 5 3 6 5 7 8 8 13 9 21 ... 24 28657 25 46368 26 75025 ... 36 9227465 37 14930352 38 24157817 ... 48 2971215073 49 4807526976 50 7778742049 ... 60 956722026041 61 1548008755920 62 2504730781961 ... 76 2111485077978050 77 3416454622906707 78 5527939700884757 79 8944394323791464 80 1.447233402467622D+16</lang>
Sinclair ZX81 BASIC
Analytic
<lang basic> 10 INPUT N
20 PRINT INT (0.5+(((SQR 5+1)/2)**N)/SQR 5)</lang>
Iterative
<lang basic> 10 INPUT N
20 LET A=0 30 LET B=1 40 FOR I=2 TO N 50 LET C=B 60 LET B=A+B 70 LET A=C 80 NEXT I 90 PRINT B</lang>
Tail recursive
<lang basic> 10 INPUT N
20 LET A=0 30 LET B=1 40 GOSUB 70 50 PRINT B 60 STOP 70 IF N=1 THEN RETURN 80 LET C=B 90 LET B=A+B
100 LET A=C 110 LET N=N-1 120 GOSUB 70 130 RETURN</lang>
Batch File
Recursive version <lang dos>::fibo.cmd @echo off if "%1" equ "" goto :eof call :fib %1 echo %errorlevel% goto :eof
- fib
setlocal enabledelayedexpansion if %1 geq 2 goto :ge2 exit /b %1
- ge2
set /a r1 = %1 - 1 set /a r2 = %1 - 2 call :fib !r1! set r1=%errorlevel% call :fib !r2! set r2=%errorlevel% set /a r0 = r1 + r2 exit /b !r0!</lang>
- Output:
>for /L %i in (1,5,20) do fibo.cmd %i >fibo.cmd 1 1 >fibo.cmd 6 8 >fibo.cmd 11 89 >fibo.cmd 16 987
Battlestar
<lang c> // Fibonacci sequence, recursive version fun fibb
loop a = funparam[0] break (a < 2)
a--
// Save "a" while calling fibb a -> stack
// Set the parameter and call fibb funparam[0] = a call fibb
// Handle the return value and restore "a" b = funparam[0] stack -> a
// Save "b" while calling fibb again b -> stack
a--
// Set the parameter and call fibb funparam[0] = a call fibb
// Handle the return value and restore "b" c = funparam[0] stack -> b
// Sum the results b += c a = b
funparam[0] = a
break end
end
// vim: set syntax=c ts=4 sw=4 et: </lang>
BBC BASIC
<lang bbcbasic> PRINT FNfibonacci_r(1), FNfibonacci_i(1)
PRINT FNfibonacci_r(13), FNfibonacci_i(13) PRINT FNfibonacci_r(26), FNfibonacci_i(26) END DEF FNfibonacci_r(N) IF N < 2 THEN = N = FNfibonacci_r(N-1) + FNfibonacci_r(N-2) DEF FNfibonacci_i(N) LOCAL F, I, P, T IF N < 2 THEN = N P = 1 FOR I = 1 TO N T = F F += P P = T NEXT = F
</lang>
- Output:
1 1 233 233 121393 121393
bc
iterative
<lang bc>#! /usr/bin/bc -q
define fib(x) {
if (x <= 0) return 0; if (x == 1) return 1;
a = 0; b = 1; for (i = 1; i < x; i++) { c = a+b; a = b; b = c; } return c;
} fib(1000) quit</lang>
beeswax
<lang beeswax> #>'#{; _`Enter n: `TN`Fib(`{`)=`X~P~K#{;
#>~P~L#MM@>+@'q@{; b~@M<</lang>
Example output:
Notice the UInt64 wrap-around at Fib(94)
!
<lang julia>julia> beeswax("n-th Fibonacci number.bswx") Enter n: i0
Fib(0)=0 Program finished!
julia> beeswax("n-th Fibonacci number.bswx") Enter n: i10
Fib(10)=55 Program finished!
julia> beeswax("n-th Fibonacci number.bswx") Enter n: i92
Fib(92)=7540113804746346429 Program finished!
julia> beeswax("n-th Fibonacci number.bswx") Enter n: i93
Fib(93)=12200160415121876738 Program finished!
julia> beeswax("n-th Fibonacci number.bswx") Enter n: i94
Fib(94)=1293530146158671551 Program finished!</lang>
Befunge
<lang befunge>00:.1:.>:"@"8**++\1+:67+`#@_v
^ .:\/*8"@"\%*8"@":\ <</lang>
Bracmat
Recursive
<lang bracmat>fib=.!arg:<2|fib$(!arg+-2)+fib$(!arg+-1)</lang>
fib$30 832040
Iterative
<lang bracmat>(fib=
last i this new
. !arg:<2
| 0:?last:?i & 1:?this & whl ' ( !i+1:<!arg:?i & !last+!this:?new & !this:?last & !new:?this ) & !this
)</lang>
fib$777 1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082
Brainf***
The first cell contains n (10), the second cell will contain fib(n) (55), and the third cell will contain fib(n-1) (34). <lang bf>++++++++++ >>+<<[->[->+>+<<]>[-<+>]>[-<+>]<<<]</lang>
The following generates n fibonacci numbers and prints them, though not in ascii. It does have a limit due to the cells usually being 1 byte in size. <lang bf>+++++ +++++ #0 set to n >> + Init #2 to 1 << [ - #Decrement counter in #0 >>. Notice: This doesn't print it in ascii To look at results you can pipe into a file and look with a hex editor
Copying sequence to save #2 in #4 using #5 as restore space >>[-] Move to #4 and clear >[-] Clear #5 <<< #2 [ Move loop - >> + > + <<< Subtract #2 and add #4 and #5 ] >>> [ Restore loop - <<< + >>> Subtract from #5 and add to #2 ]
<<<< Back to #1 Non destructive add sequence using #3 as restore value [ Loop to add - > + > + << Subtract #1 and add to value #2 and restore space #3 ] >> [ Loop to restore #1 from #3 - << + >> Subtract from restore space #3 and add in #1 ]
<< [-] Clear #1 >>> [ Loop to move #4 to #1 - <<< + >>> Subtract from #4 and add to #1 ] <<<< Back to #0 ]</lang>
Brat
Recursive
<lang brat>fibonacci = { x |
true? x < 2, x, { fibonacci(x - 1) + fibonacci(x - 2) }
}</lang>
Tail Recursive
<lang brat>fib_aux = { x, next, result |
true? x == 0, result, { fib_aux x - 1, next + result, next }
}
fibonacci = { x |
fib_aux x, 1, 0
}</lang>
Memoization
<lang brat>cache = hash.new
fibonacci = { x |
true? cache.key?(x) { cache[x] } {true? x < 2, x, { cache[x] = fibonacci(x - 1) + fibonacci(x - 2) }}
}</lang>
Burlesque
<lang burlesque> {0 1}{^^++[+[-^^-]\/}30.*\[e!vv </lang>
<lang burlesque> 0 1{{.+}c!}{1000.<}w! </lang>
C
Recursive
<lang c>long long fibb(long long a, long long b, int n) {
return (--n>0)?(fibb(b, a+b, n)):(a);
}</lang>
Iterative
<lang c>long long int fibb(int n) { int fnow = 0, fnext = 1, tempf; while(--n>0){ tempf = fnow + fnext; fnow = fnext; fnext = tempf; } return fnext; }</lang>
Analytic
<lang c>#include <tgmath.h>
- define PHI ((1 + sqrt(5))/2)
long long unsigned fib(unsigned n) {
return floor( (pow(PHI, n) - pow(1 - PHI, n))/sqrt(5) );
}</lang>
Generative
<lang c>#include <stdio.h> typedef enum{false=0, true=!0} bool; typedef void iterator;
- include <setjmp.h>
/* declare label otherwise it is not visible in sub-scope */
- define LABEL(label) jmp_buf label; if(setjmp(label))goto label;
- define GOTO(label) longjmp(label, true)
/* the following line is the only time I have ever required "auto" */
- define FOR(i, iterator) { auto bool lambda(i); yield_init = (void *)λ iterator; bool lambda(i)
- define DO {
- define YIELD(x) if(!yield(x))return
- define BREAK return false
- define CONTINUE return true
- define OD CONTINUE; } }
static volatile void *yield_init; /* not thread safe */
- define YIELDS(type) bool (*yield)(type) = yield_init
iterator fibonacci(int stop){
YIELDS(int); int f[] = {0, 1}; int i; for(i=0; i<stop; i++){ YIELD(f[i%2]); f[i%2]=f[0]+f[1]; }
}
main(){
printf("fibonacci: "); FOR(int i, fibonacci(16)) DO printf("%d, ",i); OD; printf("...\n");
}</lang>
- Output:
fibonacci: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, ...
Fast method for a single large value
<lang c>#include <stdlib.h>
- include <stdio.h>
- include <gmp.h>
typedef struct node node; struct node { int n; mpz_t v; node *next; };
- define CSIZE 37
node *cache[CSIZE];
// very primitive linked hash table node * find_cache(int n) { int idx = n % CSIZE; node *p;
for (p = cache[idx]; p && p->n != n; p = p->next); if (p) return p;
p = malloc(sizeof(node)); p->next = cache[idx]; cache[idx] = p;
if (n < 2) { p->n = n; mpz_init_set_ui(p->v, 1); } else { p->n = -1; // -1: value not computed yet mpz_init(p->v); } return p; }
mpz_t tmp1, tmp2; mpz_t *fib(int n) { int x; node *p = find_cache(n);
if (p->n < 0) { p->n = n; x = n / 2;
mpz_mul(tmp1, *fib(x-1), *fib(n - x - 1)); mpz_mul(tmp2, *fib(x), *fib(n - x)); mpz_add(p->v, tmp1, tmp2); } return &p->v; }
int main(int argc, char **argv) { int i, n; if (argc < 2) return 1;
mpz_init(tmp1); mpz_init(tmp2);
for (i = 1; i < argc; i++) { n = atoi(argv[i]); if (n < 0) { printf("bad input: %s\n", argv[i]); continue; }
// about 75% of time is spent in printing gmp_printf("%Zd\n", *fib(n)); } return 0; }</lang>
- Output:
% ./a.out 0 1 2 3 4 5 1 1 2 3 5 8 % ./a.out 10000000 | wc -c # count length of output, including the newline 1919488
C#
Recursive
<lang csharp> public static ulong Fib(uint n) {
return (n < 2)? n : Fib(n - 1) + Fib(n - 2);
} </lang>
Tail-Recursive
<lang csharp> public static ulong Fib(uint n) {
return Fib(0, 1, n);
}
private static ulong Fib(ulong a, ulong b, uint n) {
return (n < 1)? a :(n == 1)? b : Fib(b, a + b, n - 1);
} </lang>
Iterative
<lang csharp> public static ulong Fib(uint x) {
if (x == 0) return 0;
ulong prev = 0; ulong next = 1; for (int i = 1; i < x; i++) { ulong sum = prev + next; prev = next; next = sum; } return next;
} </lang>
Eager-Generative
<lang csharp> public static IEnumerable<long> Fibs(uint x) {
IList<ulong> fibs = new List<ulong>();
ulong prev = -1; ulong next = 1; for (int i = 0; i < x; i++) { long sum = prev + next; prev = next; next = sum; fibs.Add(sum); } return fibs;
} </lang>
Lazy-Generative
<lang csharp> public static IEnumerable<ulong> Fibs(uint x) {
ulong prev = -1; ulong next = 1; for (uint i = 0; i < x; i++) { ulong sum = prev + next; prev = next; next = sum; yield return sum; }
} </lang>
Analytic
Only works to the 92th fibonacci number. <lang csharp> private static double Phi = ((1d + Math.Sqrt(5d))/2d); private static double D = 1d/Math.Sqrt(5d);
ulong Fib(uint n) {
if(n > 92) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 93."); return (ulong)((Phi^n) - (1d - Phi)^n))*D);
} </lang>
Matrix
Algorithm is based on
- .
Needs System.Windows.Media.Matrix
or similar Matrix class.
Calculates in .
<lang csharp>
public static ulong Fib(uint n) {
var M = new Matrix(1,0,0,1); var N = new Matrix(1,1,1,0); for (uint i = 1; i < n; i++) M *= N; return (ulong)M[0][0];
}
</lang>
Needs System.Windows.Media.Matrix
or similar Matrix class.
Calculates in .
<lang csharp>
private static Matrix M;
private static readonly Matrix N = new Matrix(1,1,1,0);
public static ulong Fib(uint n) {
M = new Matrix(1,0,0,1); MatrixPow(n-1); return (ulong)M[0][0];
}
private static void MatrixPow(double n){
if (n > 1) { MatrixPow(n/2); M *= M; } if (n % 2 == 0) M *= N;
} </lang>
Array (Table) Lookup
<lang csharp> private static int[] fibs = new int[]{ -1836311903, 1134903170,
-701408733, 433494437, -267914296, 165580141, -102334155, 63245986, -39088169, 24157817, -14930352, 9227465, -5702887, 3524578, -2178309, 1346269, -832040, 514229, -317811, 196418, -121393, 75025, -46368, 28657, -17711, 10946, -6765, 4181, -2584, 1597, -987, 610, -377, 233, -144, 89, -55, 34, -21, 13, -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903};
public static int Fib(int n) {
if(n < -46 || n > 46) throw new ArgumentOutOfRangeException("n", n, "Has to be between -46 and 47.") return fibs[n+46];
} </lang>
Arbitrary Precision
This large step recurrence routine can calculate the two millionth Fibonacci number in under 1 / 5 second at tio.run. This routine can generate the fifty millionth Fibonacci number in under 30 seconds at tio.run. The unused conventional iterative method times out at two million on tio.run, you can only go to around 1,290,000 or so to keep the calculation time (plus string conversion time) under the 60 second timeout limit there. When using this large step recurrence method, it takes around 5 seconds to convert the two millionth Fibonacci number (417975 digits) into a string (so that one may count those digits).
<lang csharp>using System; using System.Collections.Generic; using System.Numerics;
static class QuikFib {
// A sparse array of values calculated along the way private static SortedList<int, BigInteger> sl = new SortedList<int, BigInteger>();
// Square a BigInteger public static BigInteger sqr(BigInteger n) { return n * n; }
// Helper routine for Fsl(). It adds an entry to the sorted list when necessary public static void IfNec(int n) { if (!sl.ContainsKey(n)) sl.Add(n, Fsl(n)); }
// This routine is semi-recursive, but doesn't need to evaluate every number up to n. // Algorithm from here: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html#section3 public static BigInteger Fsl(int n) { if (n < 2) return n; int n2 = n >> 1, pm = n2 + ((n & 1) << 1) - 1; IfNec(n2); IfNec(pm); return n2 > pm ? (2 * sl[pm] + sl[n2]) * sl[n2] : sqr(sl[n2]) + sqr(sl[pm]); }
// Conventional iteration method (not used here) public static BigInteger Fm(BigInteger n) { if (n < 2) return n; BigInteger cur = 0, pre = 1; for (int i = 0; i <= n - 1; i++) { BigInteger sum = cur + pre; pre = cur; cur = sum; } return cur; }
public static void Main() { int num = 2_000_000; DateTime st = DateTime.Now; BigInteger v = Fsl(num); Console.WriteLine("{0:n3} ms to calculate the {1:n0}th Fibonacci number,", (DateTime.Now - st).TotalMilliseconds, num); st = DateTime.Now; string vs = v.ToString(); Console.WriteLine("{0:n3} seconds to convert to a string.", (DateTime.Now - st).TotalSeconds); Console.WriteLine("number of digits is {0}", vs.Length); if (vs.Length < 10000) { st = DateTime.Now; Console.WriteLine(vs); Console.WriteLine("{0:n3} ms to write it to the console.", (DateTime.Now - st).TotalMilliseconds); } else Console.WriteLine("partial: {0}...{1}", vs.Substring(1, 35), vs.Substring(vs.Length - 35)); }
}</lang>
- Output:
179.978 ms to calculate the 2,000,000th Fibonacci number, 4.728 seconds to convert to a string. number of digits is 417975 partial: 53129491750764154305166065450382516...91799493108960825129188777803453125
C++
Using unsigned int, this version only works up to 48 before fib overflows. <lang cpp>#include <iostream>
int main() {
unsigned int a = 1, b = 1; unsigned int target = 48; for(unsigned int n = 3; n <= target; ++n) { unsigned int fib = a + b; std::cout << "F("<< n << ") = " << fib << std::endl; a = b; b = fib; }
return 0;
}</lang>
This version does not have an upper bound.
<lang cpp>#include <iostream>
- include <gmpxx.h>
int main() {
mpz_class a = mpz_class(1), b = mpz_class(1); mpz_class target = mpz_class(100); for(mpz_class n = mpz_class(3); n <= target; ++n) { mpz_class fib = b + a; if ( fib < b ) { std::cout << "Overflow at " << n << std::endl; break; } std::cout << "F("<< n << ") = " << fib << std::endl; a = b; b = fib; } return 0;
}</lang>
Version using transform: <lang cpp>#include <algorithm>
- include <vector>
- include <functional>
- include <iostream>
unsigned int fibonacci(unsigned int n) {
if (n == 0) return 0; std::vector<int> v(n+1); v[1] = 1; transform(v.begin(), v.end()-2, v.begin()+1, v.begin()+2, std::plus<int>()); // "v" now contains the Fibonacci sequence from 0 up return v[n];
}</lang>
Far-fetched version using adjacent_difference: <lang cpp>#include <numeric>
- include <vector>
- include <functional>
- include <iostream>
unsigned int fibonacci(unsigned int n) {
if (n == 0) return 0; std::vector<int> v(n, 1); adjacent_difference(v.begin(), v.end()-1, v.begin()+1, std::plus<int>()); // "array" now contains the Fibonacci sequence from 1 up return v[n-1];
} </lang>
Version which computes at compile time with metaprogramming: <lang cpp>#include <iostream>
template <int n> struct fibo {
enum {value=fibo<n-1>::value+fibo<n-2>::value};
};
template <> struct fibo<0> {
enum {value=0};
};
template <> struct fibo<1> {
enum {value=1};
};
int main(int argc, char const *argv[])
{
std::cout<<fibo<12>::value<<std::endl; std::cout<<fibo<46>::value<<std::endl; return 0;
}</lang>
The following version is based on fast exponentiation: <lang cpp>#include <iostream>
inline void fibmul(int* f, int* g) {
int tmp = f[0]*g[0] + f[1]*g[1]; f[1] = f[0]*g[1] + f[1]*(g[0] + g[1]); f[0] = tmp;
}
int fibonacci(int n) {
int f[] = { 1, 0 }; int g[] = { 0, 1 }; while (n > 0) { if (n & 1) // n odd { fibmul(f, g); --n; } else { fibmul(g, g); n >>= 1; } } return f[1];
}
int main() {
for (int i = 0; i < 20; ++i) std::cout << fibonacci(i) << " "; std::cout << std::endl;
}</lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181
Using Zeckendorf Numbers
The nth fibonacci is represented as Zeckendorf 1 followed by n-1 zeroes. Here I define a class N which defines the operations increment ++() and comparison <=(other N) for Zeckendorf Numbers. <lang cpp> // Use Zeckendorf numbers to display Fibonacci sequence. // Nigel Galloway October 23rd., 2012 int main(void) {
char NG[22] = {'1',0}; int x = -1; N G; for (int fibs = 1; fibs <= 20; fibs++) { for (;G <= N(NG); ++G) x++; NG[fibs] = '0'; NG[fibs+1] = 0; std::cout << x << " "; } std::cout << std::endl; return 0;
} </lang>
- Output:
1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946
Using Standard Template Library
Possibly less "Far-fetched version". <lang cpp> // Use Standard Template Library to display Fibonacci sequence. // Nigel Galloway March 30th., 2013
- include <algorithm>
- include <iostream>
- include <iterator>
int main() {
int x = 1, y = 1; generate_n(std::ostream_iterator<int>(std::cout, " "), 21, [&]{int n=x; x=y; y+=n; return n;}); return 0;
} </lang>
- Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946
Cat
<lang cat>define fib {
dup 1 <= [] [dup 1 - fib swap 2 - fib +] if
}</lang>
Chapel
<lang chapel>iter fib() {
var a = 0, b = 1;
while true { yield a; (a, b) = (b, b + a); }
}</lang>
Chef
<lang chef>Stir-Fried Fibonacci Sequence.
An unobfuscated iterative implementation. It prints the first N + 1 Fibonacci numbers, where N is taken from standard input.
Ingredients. 0 g last 1 g this 0 g new 0 g input
Method. Take input from refrigerator. Put this into 4th mixing bowl. Loop the input. Clean the 3rd mixing bowl. Put last into 3rd mixing bowl. Add this into 3rd mixing bowl. Fold new into 3rd mixing bowl. Clean the 1st mixing bowl. Put this into 1st mixing bowl. Fold last into 1st mixing bowl. Clean the 2nd mixing bowl. Put new into 2nd mixing bowl. Fold this into 2nd mixing bowl. Put new into 4th mixing bowl. Endloop input until looped. Pour contents of the 4th mixing bowl into baking dish.
Serves 1.</lang>
Clio
Clio is pure and functions are lazy and memoized by default
<lang clio>fn fib n:
if n < 2: n else: (n - 1 -> fib) + (n - 2 -> fib)
[0:100] -> * fib -> * print</lang>
Clojure
Lazy Sequence
This is implemented idiomatically as an infinitely long, lazy sequence of all Fibonacci numbers: <lang Clojure>(defn fibs []
(map first (iterate (fn a b [b (+ a b)]) [0 1])))</lang>
Thus to get the nth one: <lang Clojure>(nth (fibs) 5)</lang> So long as one does not hold onto the head of the sequence, this is unconstrained by length.
The one-line implementation may look confusing at first, but on pulling it apart it actually solves the problem more "directly" than a more explicit looping construct. <lang Clojure>(defn fibs []
(map first ;; throw away the "metadata" (see below) to view just the fib numbers (iterate ;; create an infinite sequence of [prev, curr] pairs (fn a b ;; to produce the next pair, call this function on the current pair [b (+ a b)]) ;; new prev is old curr, new curr is sum of both previous numbers [0 1]))) ;; recursive base case: prev 0, curr 1</lang>
A more elegant solution is inspired by the Haskell implementation of an infinite list of Fibonacci numbers: <lang Clojure>(def fib (lazy-cat [0 1] (map + fib (rest fib))))</lang> Then, to see the first ten, <lang Clojure>user> (take 10 fib) (0 1 1 2 3 5 8 13 21 34)</lang>
Iterative
Here's a simple interative process (using a recursive function) that carries state along with it (as args) until it reaches a solution: <lang Clojure>;; max is which fib number you'd like computed (0th, 1st, 2nd, etc.)
- n is which fib number you're on for this call (0th, 1st, 2nd, etc.)
- j is the nth fib number (ex. when n = 5, j = 5)
- i is the nth - 1 fib number
(defn- fib-iter
[max n i j] (if (= n max) j (recur max (inc n) j (+ i j))))
(defn fib
[max] (if (< max 2) max (fib-iter max 1 0N 1N)))</lang>
"defn-" means that the function is private (for use only inside this library). The "N" suffixes on integers tell Clojure to use arbitrary precision ints for those.
Doubling Algorithm (Fast)
Based upon the doubling algorithm which computes in O(log (n)) time as described here https://www.nayuki.io/page/fast-fibonacci-algorithms Implementation credit: https://stackoverflow.com/questions/27466311/how-to-implement-this-fast-doubling-fibonacci-algorithm-in-clojure/27466408#27466408 <lang clojure> (defn fib [n]
(letfn [(fib* [n] (if (zero? n) [0 1] (let [[a b] (fib* (quot n 2)) c (*' a (-' (*' 2 b) a)) d (+' (*' b b) (*' a a))] (if (even? n) [c d] [d (+' c d)]))))] (first (fib* n))))
</lang>
Recursive
A naive slow recursive solution:
<lang Clojure>(defn fib [n]
(case n 0 0 1 1 (+ (fib (- n 1)) (fib (- n 2)))))</lang>
This can be improved to an O(n) solution, like the iterative solution, by memoizing the function so that numbers that have been computed are cached. Like a lazy sequence, this also has the advantage that subsequent calls to the function use previously cached results rather than recalculating.
<lang Clojure>(def fib
(memoize (fn [n] (case n 0 0 1 1 (+ (fib (- n 1)) (fib (- n 2)))))))</lang>
Using core.async
<lang Clojure>(ns fib.core) (require '[clojure.core.async
:refer [<! >! >!! <!! timeout chan alt! go]])
(defn fib [c]
(loop [a 0 b 1] (>!! c a) (recur b (+ a b))))
(defn -main []
(let [c (chan)] (go (fib c)) (dorun (for [i (range 10)] (println (<!! c))))))
</lang>
CMake
Iteration uses a while() loop. Memoization uses global properties.
<lang cmake>set_property(GLOBAL PROPERTY fibonacci_0 0) set_property(GLOBAL PROPERTY fibonacci_1 1) set_property(GLOBAL PROPERTY fibonacci_next 2)
- var = nth number in Fibonacci sequence.
function(fibonacci var n)
# If the sequence is too short, compute more Fibonacci numbers. get_property(next GLOBAL PROPERTY fibonacci_next) if(NOT next GREATER ${n}) # a, b = last 2 Fibonacci numbers math(EXPR i "${next} - 2") get_property(a GLOBAL PROPERTY fibonacci_${i}) math(EXPR i "${next} - 1") get_property(b GLOBAL PROPERTY fibonacci_${i})
while(NOT next GREATER ${n}) math(EXPR i "${a} + ${b}") # i = next Fibonacci number set_property(GLOBAL PROPERTY fibonacci_${next} ${i}) set(a ${b}) set(b ${i}) math(EXPR next "${next} + 1") endwhile() set_property(GLOBAL PROPERTY fibonacci_next ${next}) endif()
get_property(answer GLOBAL PROPERTY fibonacci_${n}) set(${var} ${answer} PARENT_SCOPE)
endfunction(fibonacci)</lang>
<lang cmake># Test program: print 0th to 9th and 25th to 30th Fibonacci numbers. set(s "") foreach(i RANGE 0 9)
fibonacci(f ${i}) set(s "${s} ${f}")
endforeach(i) set(s "${s} ... ") foreach(i RANGE 25 30)
fibonacci(f ${i}) set(s "${s} ${f}")
endforeach(i) message(${s})</lang>
0 1 1 2 3 5 8 13 21 34 ... 75025 121393 196418 317811 514229 832040
COBOL
Iterative
<lang cobol>Program-ID. Fibonacci-Sequence. Data Division. Working-Storage Section.
01 FIBONACCI-PROCESSING. 05 FIBONACCI-NUMBER PIC 9(36) VALUE 0. 05 FIB-ONE PIC 9(36) VALUE 0. 05 FIB-TWO PIC 9(36) VALUE 1. 01 DESIRED-COUNT PIC 9(4). 01 FORMATTING. 05 INTERM-RESULT PIC Z(35)9. 05 FORMATTED-RESULT PIC X(36). 05 FORMATTED-SPACE PIC x(35).
Procedure Division.
000-START-PROGRAM. Display "What place of the Fibonacci Sequence would you like (<173)? " with no advancing. Accept DESIRED-COUNT. If DESIRED-COUNT is less than 1 Stop run. If DESIRED-COUNT is less than 2 Move FIBONACCI-NUMBER to INTERM-RESULT Move INTERM-RESULT to FORMATTED-RESULT Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT Display FORMATTED-RESULT Stop run. Subtract 1 from DESIRED-COUNT. Move FIBONACCI-NUMBER to INTERM-RESULT. Move INTERM-RESULT to FORMATTED-RESULT. Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT. Display FORMATTED-RESULT. Perform 100-COMPUTE-FIBONACCI until DESIRED-COUNT = zero. Stop run. 100-COMPUTE-FIBONACCI. Compute FIBONACCI-NUMBER = FIB-ONE + FIB-TWO. Move FIB-TWO to FIB-ONE. Move FIBONACCI-NUMBER to FIB-TWO. Subtract 1 from DESIRED-COUNT. Move FIBONACCI-NUMBER to INTERM-RESULT. Move INTERM-RESULT to FORMATTED-RESULT. Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT. Display FORMATTED-RESULT.</lang>
Recursive
<lang cobol> >>SOURCE FREE IDENTIFICATION DIVISION. PROGRAM-ID. fibonacci-main.
DATA DIVISION. WORKING-STORAGE SECTION. 01 num PIC 9(6) COMP. 01 fib-num PIC 9(6) COMP.
PROCEDURE DIVISION.
ACCEPT num CALL "fibonacci" USING CONTENT num RETURNING fib-num DISPLAY fib-num .
END PROGRAM fibonacci-main.
IDENTIFICATION DIVISION. PROGRAM-ID. fibonacci RECURSIVE.
DATA DIVISION. LOCAL-STORAGE SECTION. 01 1-before PIC 9(6) COMP. 01 2-before PIC 9(6) COMP.
LINKAGE SECTION. 01 num PIC 9(6) COMP.
01 fib-num PIC 9(6) COMP BASED.
PROCEDURE DIVISION USING num RETURNING fib-num.
ALLOCATE fib-num EVALUATE num WHEN 0 MOVE 0 TO fib-num WHEN 1 MOVE 1 TO fib-num WHEN OTHER SUBTRACT 1 FROM num CALL "fibonacci" USING CONTENT num RETURNING 1-before SUBTRACT 1 FROM num CALL "fibonacci" USING CONTENT num RETURNING 2-before ADD 1-before TO 2-before GIVING fib-num END-EVALUATE .
END PROGRAM fibonacci.</lang>
CoffeeScript
Analytic
<lang coffeescript>fib_ana = (n) ->
sqrt = Math.sqrt phi = ((1 + sqrt(5))/2) Math.round((Math.pow(phi, n)/sqrt(5)))</lang>
Iterative
<lang coffeescript>fib_iter = (n) ->
return n if n < 2 [prev, curr] = [0, 1] [prev, curr] = [curr, curr + prev] for i in [1..n] curr</lang>
Recursive
<lang coffeescript>fib_rec = (n) ->
if n < 2 then n else fib_rec(n-1) + fib_rec(n-2)</lang>
Comefrom0x10
Recursion is is not possible in Comefrom0x10.
Iterative
<lang cf0x10>stop = 6 a = 1 i = 1 # start a # print result
fib
comefrom if i is 1 # start b = 1 comefrom fib # start of loop i = i + 1 next_b = a + b a = b b = next_b
comefrom fib if i > stop</lang>
Common Lisp
Note that Common Lisp uses bignums, so this will never overflow.
Iterative
<lang lisp>(defun fibonacci-iterative (n &aux (f0 0) (f1 1))
(case n (0 f0) (1 f1) (t (loop for n from 2 to n for a = f0 then b and b = f1 then result for result = (+ a b) finally (return result)))))</lang>
Simpler one: <lang lisp>(defun fibonacci (n)
(let ((a 0) (b 1) (c n)) (loop for i from 2 to n do
(setq c (+ a b) a b b c))
c))</lang>
Not a function, just printing out the entire (for some definition of "entire") sequence with a for var =
loop:<lang lisp>(loop for x = 0 then y and y = 1 then (+ x y) do (print x))</lang>
Recursive
<lang lisp>(defun fibonacci-recursive (n)
(if (< n 2) n (+ (fibonacci-recursive (- n 2)) (fibonacci-recursive (- n 1)))))</lang>
<lang lisp>(defun fibonacci-tail-recursive ( n &optional (a 0) (b 1))
(if (= n 0) a (fibonacci-tail-recursive (- n 1) b (+ a b))))</lang>
Tail recursive and squaring: <lang lisp>(defun fib (n &optional (a 1) (b 0) (p 0) (q 1))
(if (= n 1) (+ (* b p) (* a q)) (fib (ash n -1) (if (evenp n) a (+ (* b q) (* a (+ p q)))) (if (evenp n) b (+ (* b p) (* a q))) (+ (* p p) (* q q)) (+ (* q q) (* 2 p q))))) ;p is Fib(2^n-1), q is Fib(2^n).
(print (fib 100000))</lang>
Alternate solution
I use Allegro CL 10.1
<lang lisp>
- Project
- Fibonacci sequence
(defun fibonacci (nr)
(cond ((= nr 0) 1) ((= nr 1) 1) (t (+ (fibonacci (- nr 1)) (fibonacci (- nr 2))))))
(format t "~a" "First 10 Fibonacci numbers") (dotimes (n 10) (if (< n 1) (terpri)) (if (< n 9) (format t "~a" " ")) (write(+ n 1)) (format t "~a" ": ") (write (fibonacci n)) (terpri)) </lang> Output:
First 10 Fibonacci numbers 1: 1 2: 1 3: 2 4: 3 5: 5 6: 8 7: 13 8: 21 9: 34 10: 55
Solution with methods and eql specializers
<lang lisp> (defmethod fib (n)
(declare ((integer 0 *) n)) (+ (fib (- n 1)) (fib (- n 2))))
(defmethod fib ((n (eql 0))) 0)
(defmethod fib ((n (eql 1))) 1) </lang>
List-based iterative
This solution uses a list to keep track of the Fibonacci sequence for 0 or a positive integer. <lang lisp>(defun fibo (n)
(cond ((< n 0) nil) ((< n 2) n) (t (let ((leo '(1 0))) (loop for i from 2 upto n do (setf leo (cons (+ (first leo) (second leo)) leo)) finally (return (first leo)))))))</lang>
- Output:
> (fibo 0) 0 > (fibo 1) 1 > (fibo 10) 55 > (fibo 100) 354224848179261915075 > (fibo 1000) 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875 > (fibo -10) NIL
List-based recursive
This solution computes Fibonacci numbers as either:
- a list starting from the first element;
- a single number;
- an interval from i-th to j-th element.
Options #2 and #3 can take negative parameters,
but i (lowest index in range) must be greater
than j (highest index in range).
Values are represented internally by a reversed list that grows from the head (and that's why we reverse it back when we return it).
<lang lisp>(defparameter *fibo-start* '(1 1)) ; elements 1 and 2
- Helper functions
(defun grow-fibo (fibo)
(cons (+ (first fibo) (second fibo)) fibo))
(defun generate-fibo (fibo n) ; n must be > 1
(if (equal (list-length fibo) n) fibo (generate-fibo (grow-fibo fibo) n)))
- User functions
(defun fibo (n)
(cond ((= n 0) 0) ((= (abs n) 1) 1) (t (let ((result (first (generate-fibo *fibo-start* (abs n))))) (if (and (< n -1) (evenp n)) (- result) result)))))
(defun fibo-list (n)
(cond ((< n 1) nil) ((= n 1) '(1)) (t (reverse (generate-fibo *fibo-start* n)))))
(defun fibo-range (lower upper)
(if (<= upper lower) nil (reverse (generate-fibo (list (fibo (1+ lower)) (fibo lower)) (1+ (- upper lower))))))</lang>
- Output:
> (fibo 100) 354224848179261915075 > (fibo -150) -9969216677189303386214405760200 > (fibo-list 20) (1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765) > (fibo-range -10 15) (-55 34 -21 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610) > (fibo-range 0 20) (0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765)
Computer/zero Assembly
To find the th Fibonacci number, set the initial value of count equal to –2 and run the program. The machine will halt with the answer stored in the accumulator. Since Computer/zero's word length is only eight bits, the program will not work with values of greater than 13. <lang czasm>loop: LDA y ; higher No.
STA temp ADD x ; lower No. STA y LDA temp STA x
LDA count SUB one BRZ done
STA count JMP loop
done: LDA y
STP
one: 1 count: 8 ; n = 10 x: 1 y: 1 temp: 0</lang>
Corescript
<lang corescript>print Fibonacci Sequence: var previous = 1 var number = 0 var temp = (blank)
- fib
if number > 50000000000:kill print (number) set temp = (add number previous) set previous = (number) set number = (temp) goto fib
- kill
stop</lang>
D
Here are four versions of Fibonacci Number calculating functions. FibD has an argument limit of magnitude 84 due to floating point precision, the others have a limit of 92 due to overflow (long).The traditional recursive version is inefficient. It is optimized by supplying a static storage to store intermediate results. A Fibonacci Number generating function is added. All functions have support for negative arguments. <lang d>import std.stdio, std.conv, std.algorithm, std.math;
long sgn(alias unsignedFib)(int n) { // break sign manipulation apart
immutable uint m = (n >= 0) ? n : -n; if (n < 0 && (n % 2 == 0)) return -unsignedFib(m); else return unsignedFib(m);
}
long fibD(uint m) { // Direct Calculation, correct for abs(m) <= 84
enum sqrt5r = 1.0L / sqrt(5.0L); // 1 / sqrt(5) enum golden = (1.0L + sqrt(5.0L)) / 2.0L; // (1 + sqrt(5)) / 2 return roundTo!long(pow(golden, m) * sqrt5r);
}
long fibI(in uint m) pure nothrow { // Iterative
long thisFib = 0; long nextFib = 1; foreach (i; 0 .. m) { long tmp = nextFib; nextFib += thisFib; thisFib = tmp; } return thisFib;
}
long fibR(uint m) { // Recursive
return (m < 2) ? m : fibR(m - 1) + fibR(m - 2);
}
long fibM(uint m) { // memoized Recursive
static long[] fib = [0, 1]; while (m >= fib.length ) fib ~= fibM(m - 2) + fibM(m - 1); return fib[m];
}
alias sgn!fibD sfibD; alias sgn!fibI sfibI; alias sgn!fibR sfibR; alias sgn!fibM sfibM;
auto fibG(in int m) { // generator(?)
immutable int sign = (m < 0) ? -1 : 1; long yield; return new class { final int opApply(int delegate(ref int, ref long) dg) { int idx = -sign; // prepare for pre-increment foreach (f; this) if (dg(idx += sign, f)) break; return 0; } final int opApply(int delegate(ref long) dg) { long f0, f1 = 1; foreach (p; 0 .. m * sign + 1) { if (sign == -1 && (p % 2 == 0)) yield = -f0; else yield = f0; if (dg(yield)) break; auto temp = f1; f1 = f0 + f1; f0 = temp; } return 0; } };
}
void main(in string[] args) {
int k = args.length > 1 ? to!int(args[1]) : 10; writefln("Fib(%3d) = ", k); writefln("D : %20d <- %20d + %20d", sfibD(k), sfibD(k - 1), sfibD(k - 2)); writefln("I : %20d <- %20d + %20d", sfibI(k), sfibI(k - 1), sfibI(k - 2)); if (abs(k) < 36 || args.length > 2) // set a limit for recursive version writefln("R : %20d <- %20d + %20d", sfibR(k), sfibM(k - 1), sfibM(k - 2)); writefln("O : %20d <- %20d + %20d", sfibM(k), sfibM(k - 1), sfibM(k - 2)); foreach (i, f; fibG(-9)) writef("%d:%d | ", i, f);
}</lang>
- Output:
for n = 85
Fib( 85) = D : 259695496911122586 <- 160500643816367088 + 99194853094755497 I : 259695496911122585 <- 160500643816367088 + 99194853094755497 O : 259695496911122585 <- 160500643816367088 + 99194853094755497 0:0 | -1:1 | -2:-1 | -3:2 | -4:-3 | -5:5 | -6:-8 | -7:13 | -8:-21 | -9:34 |
Matrix Exponentiation Version
<lang d>import std.bigint;
T fibonacciMatrix(T=BigInt)(size_t n) {
int[size_t.sizeof * 8] binDigits; size_t nBinDigits; while (n > 0) { binDigits[nBinDigits] = n % 2; n /= 2; nBinDigits++; }
T x=1, y, z=1; foreach_reverse (b; binDigits[0 .. nBinDigits]) { if (b) { x = (x + z) * y; y = y ^^ 2 + z ^^ 2; } else { auto x_old = x; x = x ^^ 2 + y ^^ 2; y = (x_old + z) * y; } z = x + y; }
return y;
}
void main() {
10_000_000.fibonacciMatrix;
}</lang>
Faster Version
For N = 10_000_000 this is about twice faster (run-time about 2.20 seconds) than the matrix exponentiation version. <lang d>import std.bigint, std.math;
// Algorithm from: Takahashi, Daisuke, // "A fast algorithm for computing large Fibonacci numbers". // Information Processing Letters 75.6 (30 November 2000): 243-246. // Implementation from: // pythonista.wordpress.com/2008/07/03/pure-python-fibonacci-numbers BigInt fibonacci(in ulong n) in {
assert(n > 0, "fibonacci(n): n must be > 0.");
} body {
if (n <= 2) return 1.BigInt; BigInt F = 1; BigInt L = 1; int sign = -1; immutable uint n2 = cast(uint)n.log2.floor; auto mask = 2.BigInt ^^ (n2 - 1); foreach (immutable i; 1 .. n2) { auto temp = F ^^ 2; F = (F + L) / 2; F = 2 * F ^^ 2 - 3 * temp - 2 * sign; L = 5 * temp + 2 * sign; sign = 1; if (n & mask) { temp = F; F = (F + L) / 2; L = F + 2 * temp; sign = -1; } mask /= 2; } if ((n & mask) == 0) { F *= L; } else { F = (F + L) / 2; F = F * L - sign; } return F;
}
void main() {
10_000_000.fibonacci;
}</lang>
Dart
<lang dart>int fib(int n) {
if (n==0 || n==1) { return n; } var prev=1; var current=1; for (var i=2; i<n; i++) { var next = prev + current; prev = current; current = next; } return current;
}
int fibRec(int n) => n==0 || n==1 ? n : fibRec(n-1) + fibRec(n-2);
main() {
print(fib(11)); print(fibRec(11));
}</lang>
Dc
This needs a modern Dc with r
(swap) and #
(comment).
It easily can be adapted to an older Dc, but it will impact readability a lot.
<lang dc>[ # todo: n(<2) -- 1 and break 2 levels
d - # 0 1 + # 1 q
] s1
[ # todo: n(>-1) -- F(n)
d 0=1 # n(!=0) d 1=1 # n(!in {0,1}) 2 - d 1 + # (n-2) (n-1) lF x # (n-2) F(n-1) r # F(n-1) (n-2) lF x # F(n-1)+F(n-2) +
] sF
33 lF x f</lang>
- Output:
5702887
Delphi
Iterative
<lang Delphi> function FibonacciI(N: Word): UInt64; var
Last, New: UInt64; I: Word;
begin
if N < 2 then Result := N else begin Last := 0; Result := 1; for I := 2 to N do begin New := Last + Result; Last := Result; Result := New; end; end;
end; </lang>
Recursive
<lang Delphi> function Fibonacci(N: Word): UInt64; begin
if N < 2 then Result := N else Result := Fibonacci(N - 1) + Fibonacci(N - 2);
end; </lang>
Matrix
Algorithm is based on
- .
<lang Delphi> function fib(n: Int64): Int64;
type TFibMat = array[0..1] of array[0..1] of Int64;
function FibMatMul(a,b: TFibMat): TFibMat; var i,j,k: integer; tmp: TFibMat; begin for i := 0 to 1 do for j := 0 to 1 do begin
tmp[i,j] := 0; for k := 0 to 1 do tmp[i,j] := tmp[i,j] + a[i,k] * b[k,j];
end; FibMatMul := tmp; end;
function FibMatExp(a: TFibMat; n: Int64): TFibmat; begin if n <= 1 then fibmatexp := a else if (n mod 2 = 0) then FibMatExp := FibMatExp(FibMatMul(a,a), n div 2) else if (n mod 2 = 1) then FibMatExp := FibMatMul(a, FibMatExp(FibMatMul(a,a), n div 2)); end;
var
matrix: TFibMat;
begin
matrix[0,0] := 1; matrix[0,1] := 1; matrix[1,0] := 1; matrix[1,1] := 0; if n > 1 then matrix := fibmatexp(matrix,n-1); fib := matrix[0,0];
end; </lang>
DIBOL-11
<lang DIBOL-11>
START ;First 10 Fibonacci NUmbers
RECORD FIB1, D10, 0 FIB2, D10, 1 FIBNEW, D10 LOOPCNT, D2, 1
RECORD HEADER
, A32, "First 10 Fibonacci Numbers."
RECORD OUTPUT LOOPOUT, A2
, A3, " : "
FIBOUT, A10
PROC
OPEN(8,O,'TT:')
WRITES(8,HEADER)
LOOP, FIBNEW = FIB1 + FIB2 LOOPOUT = LOOPCNT, 'ZX' FIBOUT = FIBNEW, 'ZZZZZZZZZX'
WRITES(8,OUTPUT)
FIB1 = FIB2 FIB2 = FIBNEW LOOPCNT = LOOPCNT + 1 IF LOOPCNT .LE. 10 GOTO LOOP
CLOSE 8 END
</lang>
DWScript
<lang Delphi>function fib(N : Integer) : Integer; begin
if N < 2 then Result := 1 else Result := fib(N-2) + fib(N-1);
End;</lang>
Dyalect
<lang Dyalect>func fib(n) {
if n < 2 { return n } else { return fib(n - 1) + fib(n - 2) }
}
print(fib(30))</lang>
E
<lang e>def fib(n) {
var s := [0, 1] for _ in 0..!n { def [a, b] := s s := [b, a+b] } return s[0]
}</lang>
(This version defines fib(0) = 0 because OEIS A000045 does.)
EasyLang
<lang>intvars n = number input a = 0 b = 1 while n > 1
h = a + b a = b b = h n -= 1
. print b</lang>
EchoLisp
Use memoization with the recursive version. <lang scheme> (define (fib n)
(if (< n 2) n (+ (fib (- n 2)) (fib (1- n)))))
(remember 'fib #(0 1))
(for ((i 12)) (write (fib i))) 0 1 1 2 3 5 8 13 21 34 55 89 </lang>
ECL
Analytic
<lang ECL>//Calculates Fibonacci sequence up to n steps using Binet's closed form solution
FibFunction(UNSIGNED2 n) := FUNCTION
REAL Sqrt5 := Sqrt(5);
REAL Phi := (1+Sqrt(5))/2;
REAL Phi_Inv := 1/Phi;
UNSIGNED FibValue := ROUND( ( POWER(Phi,n)-POWER(Phi_Inv,n) ) /Sqrt5);
RETURN FibValue;
END;
FibSeries(UNSIGNED2 n) := FUNCTION Fib_Layout := RECORD UNSIGNED5 FibNum; UNSIGNED5 FibValue; END; FibSeq := DATASET(n+1, TRANSFORM ( Fib_Layout , SELF.FibNum := COUNTER-1 , SELF.FibValue := IF(SELF.FibNum<2,SELF.FibNum, FibFunction(SELF.FibNum) ) ) ); RETURN FibSeq; END; }</lang>
EDSAC order code
This program calculates the nth—by default the tenth—number in the Fibonacci sequence and displays it (in binary) in the first word of storage tank 3. <lang edsac>[ Fibonacci sequence
================== A program for the EDSAC Calculates the nth Fibonacci number and displays it at the top of storage tank 3 The default value of n is 10 To calculate other Fibonacci numbers, set the starting value of the count to n-2 Works with Initial Orders 2 ]
T56K [ set load point ] GK [ set theta ]
[ Orders ]
[ 0 ] T20@ [ a = 0 ]
A17@ [ a += y ] U18@ [ temp = a ] A16@ [ a += x ] T17@ [ y = a; a = 0 ] A18@ [ a += temp ] T16@ [ x = a; a = 0 ] A19@ [ a = count ] S15@ [ a -= 1 ] U19@ [ count = a ] E@ [ if a>=0 go to θ ] T20@ [ a = 0 ] A17@ [ a += y ] T96F [ C(96) = a; a = 0] ZF [ halt ]
[ Data ]
[ 15 ] P0D [ const: 1 ] [ 16 ] P0F [ var: x = 0 ] [ 17 ] P0D [ var: y = 1 ] [ 18 ] P0F [ var: temp = 0 ] [ 19 ] P4F [ var: count = 8 ] [ 20 ] P0F [ used to clear a ]
EZPF [ begin execution ]</lang>
- Output:
00000000000110111
Eiffel
<lang eiffel> class APPLICATION
create make
feature
fibonacci (n: INTEGER): INTEGER require non_negative: n >= 0 local i, n2, n1, tmp: INTEGER do n2 := 0 n1 := 1 from i := 1 until i >= n loop tmp := n1 n1 := n2 + n1 n2 := tmp i := i + 1 end Result := n1 if n = 0 then Result := 0 end end
feature {NONE} -- Initialization
make -- Run application. do print (fibonacci (0)) print (" ") print (fibonacci (1)) print (" ") print (fibonacci (2)) print (" ") print (fibonacci (3)) print (" ") print (fibonacci (4)) print ("%N") end
end </lang>
Ela
Tail-recursive function: <lang Ela>fib = fib' 0 1
where fib' a b 0 = a fib' a b n = fib' b (a + b) (n - 1)</lang>
Infinite (lazy) list: <lang Ela>fib = fib' 1 1
where fib' x y = & x :: fib' y (x + y)</lang>
Elena
ELENA 5.0 : <lang elena>import extensions;
fibu(n) {
int[] ac := new int[]{ 0,1 }; if (n < 2) { ^ ac[n] } else { for(int i := 2, i <= n, i+=1) { int t := ac[1]; ac[1] := ac[0] + ac[1]; ac[0] := t }; ^ ac[1] }
}
public program() {
for(int i := 0, i <= 10, i+=1) { console.printLine(fibu(i)) }
}</lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55
Alternative version using yieldable method
<lang elena>import extensions;
public FibonacciGenerator {
yieldable next() { long n_2 := 1l; long n_1 := 1l;
yield:n_2; yield:n_1;
while(true) { long n := n_2 + n_1;
yield:n;
n_2 := n_1; n_1 := n } }
}
public program() {
auto e := new FibonacciGenerator(); for(int i := 0, i < 10, i += 1) { console.printLine(e.next()) }; console.readChar()
}</lang>
Elixir
<lang Elixir>defmodule Fibonacci do
def fib(0), do: 0 def fib(1), do: 1 def fib(n), do: fib(0, 1, n-2) def fib(_, prv, -1), do: prv def fib(prvprv, prv, n) do next = prv + prvprv fib(prv, next, n-1) end
end
IO.inspect Enum.map(0..10, fn i-> Fibonacci.fib(i) end)</lang>
Using Stream: <lang Elixir> Stream.unfold({0,1}, fn {a,b} -> {a,{b,a+b}} end) |> Enum.take(10) </lang>
- Output:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
Elm
Naïve recursive implementation. <lang haskell>fibonacci : Int -> Int fibonacci n = if n < 2 then
n else fibonacci(n - 2) + fibonacci(n - 1)</lang>
Emacs Lisp
version 1
<lang Emacs Lisp> (defun fib (n a b c)
(if (< c n) (fib n b (+ a b) (+ 1 c) ) (if (= c n) b a) ))
(defun fibonacci (n) (if (< n 2) n (fib n 0 1 1) )) </lang>
version 2
<lang Emacs Lisp> (defun fibonacci (n)
(let ( (vec) (i) (j) (k) ) (if (< n 2) n (progn
(setq vec (make-vector (+ n 1) 0) i 0 j 1 k 2) (setf (aref vec 1) 1) (while (<= k n) (setf (aref vec k) (+ (elt vec i) (elt vec j) )) (setq i (+ 1 i) j (+ 1 j) k (+ 1 k) )) (elt vec n) )))) </lang> Eval: <lang Emacs Lisp> (insert
(mapconcat '(lambda (n) (format "%d" (fibonacci n) ))
(number-sequence 0 15) " ") ) </lang> Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610
Erlang
Recursive
<lang Erlang> -module(fib). -export([fib/1]).
fib(0) -> 0; fib(1) -> 1; fib(N) -> fib(N-1) + fib(N-2). </lang>
Iterative
<lang Erlang> -module(fiblin). -export([fib/1])
fib(0) -> 0; fib(1) -> 1; fib(2) -> 1; fib(3) -> 2; fib(4) -> 3; fib(5) -> 5;
fib(N) when is_integer(N) -> fib(N - 6, 5, 8). fib(N, A, B) -> if N < 1 -> B; true -> fib(N-1, B, A+B) end.
</lang>
Evaluate: <lang Erlang> io:write([fiblin:fib(X) || X <- lists:seq(1,10) ]). </lang>
Output:
[1,1,2,3,5,8,13,21,34,55]ok
Iterative 2
<lang Erlang> fib(N) -> fib(N, 0, 1).
fib(0, Result, _Next) -> Result; fib(Iter, Result, Next) -> fib(Iter-1, Next, Result+Next).
</lang>
ERRE
<lang ERRE>!------------------------------------------- ! derived from my book "PROGRAMMARE IN ERRE" ! iterative solution !-------------------------------------------
PROGRAM FIBONACCI
!$DOUBLE
!VAR F1#,F2#,TEMP#,COUNT%,N%
BEGIN !main
INPUT("Number",N%) F1=0 F2=1 REPEAT TEMP=F2 F2=F1+F2 F1=TEMP COUNT%=COUNT%+1 UNTIL COUNT%=N% PRINT("FIB(";N%;")=";F2)
! Obviously a FOR loop or a WHILE loop can ! be used to solve this problem
END PROGRAM </lang>
- Output:
Number? 20 FIB( 20 )= 6765
Euphoria
'Recursive' version
<lang Euphoria> function fibor(integer n)
if n<2 then return n end if return fibor(n-1)+fibor(n-2)
end function </lang>
'Iterative' version
<lang Euphoria> function fiboi(integer n) integer f0=0, f1=1, f
if n<2 then return n end if for i=2 to n do f=f0+f1 f0=f1 f1=f end for return f
end function </lang>
'Tail recursive' version
<lang Euphoria> function fibot(integer n, integer u = 1, integer s = 0)
if n < 1 then return s else return fibot(n-1,u+s,u) end if
end function
-- example: ? fibot(10) -- says 55 </lang>
'Paper tape' version
<lang Euphoria> include std/mathcons.e -- for PINF constant
enum ADD, MOVE, GOTO, OUT, TEST, TRUETO
global sequence tape = { 0, 1, { ADD, 2, 1 }, { TEST, 1, PINF }, { TRUETO, 0 }, { OUT, 1, "%.0f\n" }, { MOVE, 2, 1 }, { MOVE, 0, 2 }, { GOTO, 3 } }
global integer ip global integer test global atom accum
procedure eval( sequence cmd ) atom i = 1 while i <= length( cmd ) do switch cmd[ i ] do case ADD then accum = tape[ cmd[ i + 1 ] ] + tape[ cmd[ i + 2 ] ] i += 2
case OUT then printf( 1, cmd[ i + 2], tape[ cmd[ i + 1 ] ] ) i += 2
case MOVE then if cmd[ i + 1 ] = 0 then tape[ cmd[ i + 2 ] ] = accum else tape[ cmd[ i + 2 ] ] = tape[ cmd[ i + 1 ] ] end if i += 2
case GOTO then ip = cmd[ i + 1 ] - 1 -- due to ip += 1 in main loop i += 1
case TEST then if tape[ cmd[ i + 1 ] ] = cmd[ i + 2 ] then test = 1 else test = 0 end if i += 2
case TRUETO then if test then if cmd[ i + 1 ] = 0 then abort(0) else ip = cmd[ i + 1 ] - 1 end if end if
end switch i += 1 end while end procedure
test = 0 accum = 0 ip = 1
while 1 do
-- embedded sequences (assumed to be code) are evaluated -- atoms (assumed to be data) are ignored
if sequence( tape[ ip ] ) then eval( tape[ ip ] ) end if ip += 1 end while
</lang>
F#
This is a fast [tail-recursive] approach using the F# big integer support: <lang fsharp> let fibonacci n : bigint =
let rec f a b n = match n with | 0 -> a | 1 -> b | n -> (f b (a + b) (n - 1)) f (bigint 0) (bigint 1) n
> fibonacci 100;; val it : bigint = 354224848179261915075I</lang> Lazy evaluated using sequence workflow: <lang fsharp>let rec fib = seq { yield! [0;1];
for (a,b) in Seq.zip fib (Seq.skip 1 fib) -> a+b}</lang>
The above is extremely slow due to the nested recursions on sequences, which aren't very efficient at the best of times. The above takes seconds just to compute the 30th Fibonacci number!
Lazy evaluation using the sequence unfold anamorphism is much much better as to efficiency: <lang fsharp>let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I) fibonacci |> Seq.nth 10000 </lang>
Approach similar to the Matrix algorithm in C#, with some shortcuts involved. Since it uses exponentiation by squaring, calculations of fib(n) where n is a power of 2 are particularly quick. Eg. fib(2^20) was calculated in a little over 4 seconds on this poster's laptop. <lang fsharp> open System open System.Diagnostics open System.Numerics
/// Finds the highest power of two which is less than or equal to a given input. let inline prevPowTwo (x : int) =
let mutable n = x n <- n - 1 n <- n ||| (n >>> 1) n <- n ||| (n >>> 2) n <- n ||| (n >>> 4) n <- n ||| (n >>> 8) n <- n ||| (n >>> 16) n <- n + 1 match x with | x when x = n -> x | _ -> n/2
/// Evaluates the nth Fibonacci number using matrix arithmetic and /// exponentiation by squaring. let crazyFib (n : int) =
let powTwo = prevPowTwo n
/// Applies 2n rule repeatedly until another application of the rule would /// go over the target value (or the target value has been reached). let rec iter1 i q r s = match i with | i when i < powTwo -> iter1 (i*2) (q*q + r*r) (r * (q+s)) (r*r + s*s) | _ -> i, q, r, s
/// Applies n+1 rule until the target value is reached. let rec iter2 (i, q, r, s) = match i with | i when i < n -> iter2 ((i+1), (q+r), q, r) | _ -> q
match n with | 0 -> 1I | _ -> iter1 1 1I 1I 0I |> iter2
</lang>
Factor
Iterative
<lang factor>: fib ( n -- m )
dup 2 < [ [ 0 1 ] dip [ swap [ + ] keep ] times drop ] unless ;</lang>
Recursive
<lang factor>: fib ( n -- m )
dup 2 < [ [ 1 - fib ] [ 2 - fib ] bi + ] unless ;</lang>
Tail-Recursive
<lang factor>: fib2 ( x y n -- a )
dup 1 < [ 2drop ] [ [ swap [ + ] keep ] dip 1 - fib2 ] if ;
- fib ( n -- m ) [ 0 1 ] dip fib2 ;</lang>
Matrix
<lang factor>USE: math.matrices
- fib ( n -- m )
dup 2 < [ [ { { 0 1 } { 1 1 } } ] dip 1 - m^n second second ] unless ;</lang>
Falcon
Iterative
<lang falcon>function fib_i(n)
if n < 2: return n
fibPrev = 1 fib = 1 for i in [2:n] tmp = fib fib += fibPrev fibPrev = tmp end return fib
end</lang>
Recursive
<lang falcon>function fib_r(n)
if n < 2 : return n return fib_r(n-1) + fib_r(n-2)
end</lang>
Tail Recursive
<lang falcon>function fib_tr(n)
return fib_aux(n,0,1)
end function fib_aux(n,a,b)
switch n case 0 : return a default: return fib_aux(n-1,a+b,a) end
end</lang>
FALSE
<lang false>[[$0=~][1-@@\$@@+\$44,.@]#]f: 20n: {First 20 numbers} 0 1 n;f;!%%44,. {Output: "0,1,1,2,3,5..."}</lang>
Fancy
<lang fancy>class Fixnum {
def fib { match self -> { case 0 -> 0 case 1 -> 1 case _ -> self - 1 fib + (self - 2 fib) } }
}
15 times: |x| {
x fib println
} </lang>
Fantom
Ints have a limit of 64-bits, so overflow errors occur after computing Fib(92) = 7540113804746346429.
<lang fantom> class Main {
static Int fib (Int n) { if (n < 2) return n fibNums := [1, 0] while (fibNums.size <= n) { fibNums.insert (0, fibNums[0] + fibNums[1]) } return fibNums.first }
public static Void main () { 20.times |n| { echo ("Fib($n) is ${fib(n)}") } }
} </lang>
Fexl
<lang Fexl>
- (fib n) = the nth Fibonacci number
\fib=
( \loop== (\x\y\n le n 0 x; \z=(+ x y) \n=(- n 1) loop y z n ) loop 0 1 )
- Now test it:
for 0 20 (\n say (fib n)) </lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
FOCAL
<lang focal>01.10 TYPE "FIBONACCI NUMBERS" ! 01.20 ASK "N =", N 01.30 SET A=0 01.40 SET B=1 01.50 FOR I=2,N; DO 2.0 01.60 TYPE "F(N) ", %8, B, ! 01.70 QUIT
02.10 SET T=B 02.20 SET B=A+B 02.30 SET A=T</lang>
- Output:
FIBONACCI NUMBERS N =:20 F(N) = 6765
Forth
<lang forth>: fib ( n -- fib )
0 1 rot 0 ?do over + swap loop drop ;</lang>
Since there are only a fixed and small amount of Fibonacci numbers that fit in a machine word, this FORTH version creates a table of Fibonacci numbers at compile time. It stops compiling numbers when there is arithmetic overflow (the number turns negative, indicating overflow.)
<lang forth>: F-start, here 1 0 dup , ;
- F-next, over + swap
dup 0> IF dup , true ELSE false THEN ;
- computed-table ( compile: 'start 'next / run: i -- x )
create >r execute BEGIN r@ execute not UNTIL rdrop does> swap cells + @ ;
' F-start, ' F-next, computed-table fibonacci 2drop here swap - cell/ Constant #F/64 \ # of fibonacci numbers generated
16 fibonacci . 987 ok
- F/64 . 93 ok
92 fibonacci . 7540113804746346429 ok \ largest number generated.</lang>
Fortran
FORTRAN IV
<lang fortran>C FIBONACCI SEQUENCE - FORTRAN IV
NN=46 DO 1 I=0,NN 1 WRITE(*,300) I,IFIBO(I) 300 FORMAT(1X,I2,1X,I10) END
C
FUNCTION IFIBO(N) IF(N) 9,1,2 1 IFN=0 GOTO 9 2 IF(N-1) 9,3,4 3 IFN=1 GOTO 9 4 IFNM1=0 IFN=1 DO 5 I=2,N IFNM2=IFNM1 IFNM1=IFN 5 IFN=IFNM1+IFNM2 9 IFIBO=IFN END</lang>
- Output:
0 0 1 1 2 1 3 2 4 3 5 5 6 8 7 13 8 21 9 34 10 55 ... 45 1134903170 46 1836311903
FORTRAN 77
<lang fortran>
FUNCTION IFIB(N) IF (N.EQ.0) THEN ITEMP0=0 ELSE IF (N.EQ.1) THEN ITEMP0=1 ELSE IF (N.GT.1) THEN ITEMP1=0 ITEMP0=1 DO 1 I=2,N ITEMP2=ITEMP1 ITEMP1=ITEMP0 ITEMP0=ITEMP1+ITEMP2 1 CONTINUE ELSE ITEMP1=1 ITEMP0=0 DO 2 I=-1,N,-1 ITEMP2=ITEMP1 ITEMP1=ITEMP0 ITEMP0=ITEMP2-ITEMP1 2 CONTINUE END IF IFIB=ITEMP0 END
</lang> Test program <lang fortran>
EXTERNAL IFIB CHARACTER*10 LINE PARAMETER ( LINE = '----------' ) WRITE(*,900) 'N', 'F[N]', 'F[-N]' WRITE(*,900) LINE, LINE, LINE DO 1 N = 0, 10 WRITE(*,901) N, IFIB(N), IFIB(-N) 1 CONTINUE 900 FORMAT(3(X,A10)) 901 FORMAT(3(X,I10)) END
</lang>
- Output:
N F[N] F[-N] ---------- ---------- ---------- 0 0 0 1 1 1 2 1 -1 3 2 2 4 3 -3 5 5 5 6 8 -8 7 13 13 8 21 -21 9 34 34 10 55 -55
Recursive
In ISO Fortran 90 or later, use a RECURSIVE function: <lang fortran>module fibonacci contains
recursive function fibR(n) result(fib) integer, intent(in) :: n integer :: fib select case (n) case (:0); fib = 0 case (1); fib = 1 case default; fib = fibR(n-1) + fibR(n-2) end select end function fibR</lang>
Iterative
In ISO Fortran 90 or later: <lang fortran> function fibI(n)
integer, intent(in) :: n integer, parameter :: fib0 = 0, fib1 = 1 integer :: fibI, back1, back2, i select case (n) case (:0); fibI = fib0 case (1); fibI = fib1 case default fibI = fib1 back1 = fib0 do i = 2, n back2 = back1 back1 = fibI fibI = back1 + back2 end do end select end function fibI
end module fibonacci</lang>
Test program <lang fortran>program fibTest
use fibonacci do i = 0, 10 print *, fibr(i), fibi(i) end do
end program fibTest</lang>
- Output:
0 0 1 1 1 1 2 2 3 3 5 5 8 8 13 13 21 21 34 34 55 55
Free Pascal
See also: Pascal <lang Pascal>type /// domain for Fibonacci function /// where result is within nativeUInt // You can not name it fibonacciDomain, // since the Fibonacci function itself // is defined for all whole numbers // but the result beyond F(n) exceeds high(nativeUInt). fibonacciLeftInverseRange = {$ifdef CPU64} 0..93 {$else} 0..47 {$endif};
{** implements Fibonacci sequence iteratively
\param n the index of the Fibonacci number to calculate \returns the Fibonacci value at n } function fibonacci(const n: fibonacciLeftInverseRange): nativeUInt; type /// more meaningful identifiers than simple integers relativePosition = (previous, current, next); var /// temporary iterator variable i: longword; /// holds preceding fibonacci values f: array[relativePosition] of nativeUInt; begin f[previous] := 0; f[current] := 1;
// note, in Pascal for-loop-limits are inclusive for i := 1 to n do begin f[next] := f[previous] + f[current]; f[previous] := f[current]; f[current] := f[next]; end;
// assign to previous, bc f[current] = f[next] for next iteration fibonacci := f[previous]; end;</lang>
FreeBASIC
Extended sequence coded big integer. <lang FreeBASIC>'Fibonacci extended 'Freebasic version 24 Windows Dim Shared ADDQmod(0 To 19) As Ubyte Dim Shared ADDbool(0 To 19) As Ubyte
For z As Integer=0 To 19
ADDQmod(z)=(z Mod 10+48) ADDbool(z)=(-(10<=z))
Next z
Function plusINT(NUM1 As String,NUM2 As String) As String
Dim As Byte flag #macro finish() three=Ltrim(three,"0") If three="" Then Return "0" If flag=1 Then Swap NUM2,NUM1 Return three Exit Function #endmacro var lenf=Len(NUM1) var lens=Len(NUM2) If lens>lenf Then Swap NUM2,NUM1 Swap lens,lenf flag=1 End If var diff=lenf-lens-Sgn(lenf-lens) var three="0"+NUM1 var two=String(lenf-lens,"0")+NUM2 Dim As Integer n2 Dim As Ubyte addup,addcarry addcarry=0 For n2=lenf-1 To diff Step -1 addup=two[n2]+NUM1[n2]-96 three[n2+1]=addQmod(addup+addcarry) addcarry=addbool(addup+addcarry) Next n2 If addcarry=0 Then finish() End If If n2=-1 Then three[0]=addcarry+48 finish() End If For n2=n2 To 0 Step -1 addup=two[n2]+NUM1[n2]-96 three[n2+1]=addQmod(addup+addcarry) addcarry=addbool(addup+addcarry) Next n2 three[0]=addcarry+48 finish()
End Function
Function fibonacci(n As Integer) As String
Dim As String sl,l,term sl="0": l="1" If n=1 Then Return "0" If n=2 Then Return "1" n=n-2 For x As Integer= 1 To n term=plusINT(l,sl) sl=l l=term Next x Function =term
End Function
'============== EXAMPLE =============== print "THE SEQUENCE TO 10:" print For n As Integer=1 To 10
Print "term";n;": "; fibonacci(n)
Next n print print "Selected Fibonacci number" print "Fibonacci 500" print print fibonacci(500) Sleep</lang>
- Output:
THE SEQUENCE TO 10: term 1: 0 term 2: 1 term 3: 1 term 4: 2 term 5: 3 term 6: 5 term 7: 8 term 8: 13 term 9: 21 term 10: 34 Selected Fibonacci number Fibonacci 500 86168291600238450732788312165664788095941068326060883324529903470149056115823592 713458328176574447204501
Frink
All of Frink's integers can be arbitrarily large. <lang frink> fibonacciN[n] := {
a = 0 b = 1 count = 0 while count < n { [a,b] = [b, a + b] count = count + 1 } return a
} </lang>
FRISC Assembly
To find the nth Fibonacci number, call this subroutine with n in register R0: the answer will be returned in R0 too. Contents of other registers are preserved. <lang friscasm>FIBONACCI PUSH R1
PUSH R2 PUSH R3
MOVE 0, R1 MOVE 1, R2
FIB_LOOP SUB R0, 1, R0
JP_Z FIB_DONE
MOVE R2, R3 ADD R1, R2, R2 MOVE R3, R1
JP FIB_LOOP
FIB_DONE MOVE R2, R0
POP R3 POP R2 POP R1
RET</lang>
FunL
Recursive
<lang funl>def
fib( 0 ) = 0 fib( 1 ) = 1 fib( n ) = fib( n - 1 ) + fib( n - 2 )</lang>
Tail Recursive
<lang funl>def fib( n ) =
def _fib( 0, prev, _ ) = prev _fib( 1, _, next ) = next _fib( n, prev, next ) = _fib( n - 1, next, next + prev )
_fib( n, 0, 1 )</lang>
Lazy List
<lang funl>val fib =
def _fib( a, b ) = a # _fib( b, a + b )
_fib( 0, 1 )
println( fib(10000) )</lang>
- Output:
33644764876431783266621612005107543310302148460680063906564769974680081442166662368155595513633734025582065332680836159373734790483865268263040892463056431887354544369559827491606602099884183933864652731300088830269235673613135117579297437854413752130520504347701602264758318906527890855154366159582987279682987510631200575428783453215515103870818298969791613127856265033195487140214287532698187962046936097879900350962302291026368131493195275630227837628441540360584402572114334961180023091208287046088923962328835461505776583271252546093591128203925285393434620904245248929403901706233888991085841065183173360437470737908552631764325733993712871937587746897479926305837065742830161637408969178426378624212835258112820516370298089332099905707920064367426202389783111470054074998459250360633560933883831923386783056136435351892133279732908133732642652633989763922723407882928177953580570993691049175470808931841056146322338217465637321248226383092103297701648054726243842374862411453093812206564914032751086643394517512161526545361333111314042436854805106765843493523836959653428071768775328348234345557366719731392746273629108210679280784718035329131176778924659089938635459327894523777674406192240337638674004021330343297496902028328145933418826817683893072003634795623117103101291953169794607632737589253530772552375943788434504067715555779056450443016640119462580972216729758615026968443146952034614932291105970676243268515992834709891284706740862008587135016260312071903172086094081298321581077282076353186624611278245537208532365305775956430072517744315051539600905168603220349163222640885248852433158051534849622434848299380905070483482449327453732624567755879089187190803662058009594743150052402532709746995318770724376825907419939632265984147498193609285223945039707165443156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875
Iterative
<lang funl>def fib( n ) =
a, b = 0, 1
for i <- 1..n a, b = b, a+b
a</lang>
Binet's Formula
<lang funl>import math.sqrt
def fib( n ) =
phi = (1 + sqrt( 5 ))/2 int( (phi^n - (-phi)^-n)/sqrt(5) + .5 )</lang>
Matrix Exponentiation
<lang funl>def mul( a, b ) =
res = array( a.length(), b(0).length() )
for i <- 0:a.length(), j <- 0:b(0).length() res( i, j ) = sum( a(i, k)*b(k, j) | k <- 0:b.length() )
vector( res )
def
pow( _, 0 ) = ((1, 0), (0, 1)) pow( x, 1 ) = x pow( x, n ) | 2|n = pow( mul(x, x), n\2 ) | otherwise = mul(x, pow( mul(x, x), (n - 1)\2 ) )
def fib( n ) = pow( ((0, 1), (1, 1)), n )(0, 1)
for i <- 0..10
println( fib(i) )</lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55
Futhark
Iterative
<lang Futhark> fun main(n: int): int =
loop((a,b) = (0,1)) = for _i < n do (b, a + b) in a
</lang>
FutureBasic
Iterative
<lang futurebasic> include "Tlbx Timer.incl" include "ConsoleWindow"
local fn Fibonacci( n as long ) as Long begin globals dim as long s1, s2// static end globals
dim as long temp
if ( n < 2 )
s1 = n exit fn
else
temp = s1 + s2 s2 = s1 s1 = temp exit fn
end if end fn = s1
dim as long i dim as UnsignedWide start, finish
Microseconds( @start ) for i = 0 to 40 print i; ". "; fn Fibonacci(i) next i Microseconds( @finish ) print "Compute time:"; (finish.lo - start.lo ) / 1000; " ms" </lang> Output:
0. 0 1. 1 2. 1 3. 2 4. 3 5. 5 6. 8 7. 13 8. 21 9. 34 10. 55 11. 89 12. 144 13. 233 14. 377 15. 610 16. 987 17. 1597 18. 2584 19. 4181 20. 6765 21. 10946 22. 17711 23. 28657 24. 46368 25. 75025 26. 121393 27. 196418 28. 317811 29. 514229 30. 832040 31. 1346269 32. 2178309 33. 3524578 34. 5702887 35. 9227465 36. 14930352 37. 24157817 38. 39088169 39. 63245986 40. 102334155 41. 165580141 42. 267914296 43. 433494437 44. 701408733 45. 1134903170 46. 1836311903 47. 2971215073 48. 4.80752698e+9 49. 7.77874205e+9 50. 1.2586269e+10 51. 2.03650111e+10 52. 3.29512801e+10 53. 5.33162912e+10 54. 8.62675713e+10 55. 1.39583862e+11 56. 2.25851434e+11 57. 3.65435296e+11 58. 5.9128673e+11 59. 9.56722026e+11 60. 1.54800876e+12 61. 2.50473078e+12 62. 4.05273954e+12 63. 6.55747032e+12 64. 1.06102099e+13 65. 1.71676802e+13 66. 2.777789e+13 67. 4.49455702e+13 68. 7.27234602e+13 69. 1.1766903e+14 70. 1.90392491e+14 71. 3.08061521e+14 72. 4.98454012e+14 73. 8.06515533e+14 74. 1.30496954e+15 75. 2.11148508e+15 76. 3.41645462e+15 77. 5.5279397e+15 78. 8.94439432e+15 79. 1.4472334e+16 80. 2.34167283e+16 81. 3.78890624e+16 82. 6.13057907e+16 83. 9.91948531e+16 84. 1.60500644e+17 85. 2.59695497e+17 86. 4.20196141e+17 87. 6.79891638e+17 88. 1.10008778e+18 89. 1.77997942e+18 90. 2.88006719e+18 91. 4.66004661e+18 92. 7.5401138e+18 93. 1.22001604e+19 94. 1.97402742e+19 95. 3.19404346e+19 96. 5.16807089e+19 97. 8.36211435e+19 98. 1.35301852e+20 99. 2.18922996e+20 100. 3.54224848e+20 Compute time: 15 ms
GAP
<lang gap>fib := function(n)
local a; a := [[0, 1], [1, 1]]^n; return a[1][2];
end;</lang> GAP has also a buit-in function for that. <lang gap>Fibonacci(n);</lang>
Gecho
<lang gecho>0 1 dup wover + dup wover + dup wover + dup wover +</lang> Prints the first several fibonacci numbers...
GFA Basic
<lang> ' ' Compute nth Fibonacci number ' ' open a window for display OPENW 1 CLEARW 1 ' Display some fibonacci numbers ' Fib(46) is the largest number GFA Basic can reach ' (long integers are 4 bytes) FOR i%=0 TO 46
PRINT "fib(";i%;")=";@fib(i%)
NEXT i% ' wait for a key press and tidy up ~INP(2) CLOSEW 1 ' ' Function to compute nth fibonacci number ' n must be in range 0 to 46, inclusive ' FUNCTION fib(n%)
LOCAL n0%,n1%,nn%,i% n0%=0 n1%=1 SELECT n% CASE 0 RETURN n0% CASE 1 RETURN n1% DEFAULT FOR i%=2 TO n% nn%=n0%+n1% n0%=n1% n1%=nn% NEXT i% RETURN nn% ENDSELECT
ENDFUNC </lang>
GML
<lang gml>///fibonacci(n) //Returns the nth fibonacci number
var n, numb; n = argument0;
if (n == 0)
{ numb = 0; }
else
{ var fm2, fm1; fm2 = 0; fm1 = 1; numb = 1; repeat(n-1) { numb = fm2+fm1; fm2 = fm1; fm1 = numb; } }
return numb;</lang>
Go
Recursive
<lang go>func fib(a int) int {
if a < 2 { return a } return fib(a - 1) + fib(a - 2)
}</lang>
Iterative
<lang go>import ( "math/big" )
func fib(n uint64) *big.Int { if n < 2 { return big.NewInt(int64(n)) } a, b := big.NewInt(0), big.NewInt(1) for n--; n > 0; n-- { a.Add(a, b) a, b = b, a } return b }</lang>
Iterative using a closure
<lang go>func fibNumber() func() int { fib1, fib2 := 0, 1 return func() int { fib1, fib2 = fib2, fib1 + fib2 return fib1 } }
func fibSequence(n int) int { f := fibNumber() fib := 0 for i := 0; i < n; i++ { fib = f() } return fib }</lang>
Using a goroutine and channel
<lang go>func fib(c chan int) { a, b := 0, 1 for { c <- a a, b = b, a+b } }
func main() { c := make(chan int) go fib(c) for i := 0; i < 10; i++ { fmt.Println(<-c) } } </lang>
Groovy
Full "extra credit" solutions.
Recursive
A recursive closure must be pre-declared. <lang groovy>def rFib rFib = {
it == 0 ? 0 : it == 1 ? 1 : it > 1 ? rFib(it-1) + rFib(it-2) /*it < 0*/: rFib(it+2) - rFib(it+1)
}</lang>
Iterative
<lang groovy>def iFib = {
it == 0 ? 0 : it == 1 ? 1 : it > 1 ? (2..it).inject([0,1]){i, j -> [i[1], i[0]+i[1]]}[1] /*it < 0*/: (-1..it).inject([0,1]){i, j -> [i[1]-i[0], i[0]]}[0]
}</lang>
Analytic
<lang groovy>final φ = (1 + 5**(1/2))/2 def aFib = { (φ**it - (-φ)**(-it))/(5**(1/2)) as BigInteger }</lang>
Test program: <lang groovy>def time = { Closure c ->
def start = System.currentTimeMillis() def result = c() def elapsedMS = (System.currentTimeMillis() - start)/1000 printf '(%6.4fs elapsed)', elapsedMS result
}
print " F(n) elapsed time "; (-10..10).each { printf ' %3d', it }; println() print "--------- -----------------"; (-10..10).each { print ' ---' }; println() [recursive:rFib, iterative:iFib, analytic:aFib].each { name, fib ->
printf "%9s ", name def fibList = time { (-10..10).collect {fib(it)} } fibList.each { printf ' %3d', it } println()
}</lang>
- Output:
F(n) elapsed time -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 --------- ----------------- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- recursive (0.0080s elapsed) -55 34 -21 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13 21 34 55 iterative (0.0040s elapsed) -55 34 -21 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13 21 34 55 analytic (0.0030s elapsed) -55 34 -21 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13 21 34 55
Harbour
Recursive
<lang Harbour>
- include "harbour.ch"
Function fibb(a,b,n) return(if(--n>0,fibb(b,a+b,n),a)) </lang>
Iterative
<lang Harbour>
- include "harbour.ch"
Function fibb(n) local fnow:=0, fnext:=1, tempf while (--n>0) tempf:=fnow+fnext fnow:=fnext fnext:=tempf end while return(fnext) </lang>
Haskell
Analytic
<lang haskell>main :: IO () main =
print [ floor (0.01 + (1 / p ** n + p ** n) / sqrt 5) | let p = (1 + sqrt 5) / 2 , n <- [0 .. 42] ]</lang>
Recursive
Simple definition, very inefficient.
<lang haskell>fib x =
if x < 1 then 0 else if x < 2 then 1 else fib (x - 1) + fib (x - 2)</lang>
Recursive with Memoization
Very fast. <lang haskell>fib x =
if x < 1 then 0 else if x == 1 then 1 else fibs !! (x - 1) + fibs !! (x - 2) where fibs = map fib [0 ..]</lang>
Recursive with Memoization using memoized library
Even faster and simpler is to use a defined memoizer (e.g. from MemoTrie package): <lang haskell>import Data.MemoTrie fib :: Integer -> Integer fib = memo f where
f 0 = 0 f 1 = 1 f n = fib (n-1) + fib (n-2)</lang>
You can rewrite this without introducing f explicitly <lang haskell>import Data.MemoTrie fib :: Integer -> Integer fib = memo $ \x -> case x of
0 -> 0 1 -> 1 n -> fib (n-1) + fib (n-2)</lang>
Or using LambdaCase extension you can write it even shorter: <lang haskell>{-# Language LambdaCase #-} import Data.MemoTrie fib :: Integer -> Integer fib = memo $ \case
0 -> 0 1 -> 1 n -> fib (n-1) + fib (n-2)</lang>
The version that supports negative numbers: <lang haskell>{-# Language LambdaCase #-} import Data.MemoTrie fib :: Integer -> Integer fib = memo $ \case
0 -> 0 1 -> 1 n | n>0 -> fib (n-1) + fib (n-2) | otherwise -> fib (n+2) - fib (n+1)</lang>
Iterative
<lang haskell>fib n = go n 0 1
where go n a b | n == 0 = a | otherwise = go (n - 1) b (a + b)</lang>
With lazy lists
This is a standard example how to use lazy lists. Here's the (infinite) list of all Fibonacci numbers:
<lang haskell>fib = 0 : 1 : zipWith (+) fib (tail fib)</lang> Or alternatively: <lang haskell>fib = 0 : 1 : (zipWith (+) <*> tail) fib </lang>
The nth Fibonacci number is then just fib !! n
. The above is equivalent to
<lang haskell>fib = 0 : 1 : next fib where next (a: t@(b:_)) = (a+b) : next t</lang>
Also
<lang haskell>fib = 0 : scanl (+) 1 fib</lang>
As a fold
Accumulator holds last two members of the series:
<lang haskell>import Data.List (foldl') --'
fib :: Integer -> Integer fib n =
fst $ foldl' --' (\(a, b) _ -> (b, a + b)) (0, 1) [1 .. n]</lang>
With matrix exponentiation
Adapting the (rather slow) code from Matrix exponentiation operator, we can simply write:
<lang haskell>import Data.List (transpose)
fib
:: (Integral b, Num a) => b -> a
fib 0 = 0 -- this line is necessary because "something ^ 0" returns "fromInteger 1", which unfortunately -- in our case is not our multiplicative identity (the identity matrix) but just a 1x1 matrix of 1 fib n = (last . head . unMat) (Mat [[1, 1], [1, 0]] ^ n)
-- Code adapted from Matrix exponentiation operator task --------------------- (<+>)
:: Num c => [c] -> [c] -> [c]
(<+>) = zipWith (+)
(<*>)
:: Num a => [a] -> [a] -> a
(<*>) = (sum .) . zipWith (*)
newtype Mat a = Mat
{ unMat :: a } deriving (Eq)
instance Show a =>
Show (Mat a) where show xm = "Mat " ++ show (unMat xm)
instance Num a =>
Num (Mat a) where negate xm = Mat $ map (map negate) $ unMat xm xm + ym = Mat $ zipWith (<+>) (unMat xm) (unMat ym) xm * ym = Mat [ [ xs Main.<*> ys -- to distinguish from standard applicative operator | ys <- transpose $ unMat ym ] | xs <- unMat xm ] fromInteger n = Mat fromInteger n abs = undefined signum = undefined
-- TEST ---------------------------------------------------------------------- main :: IO () main = (print . take 10 . show . fib) (10 ^ 5)</lang>
So, for example, the hundred-thousandth Fibonacci number starts with the digits:
- Output:
"2597406934"
With recurrence relations
Using Fib[m=3n+r]
recurrence identities:
<lang haskell>import Control.Arrow ((&&&))
fibstep :: (Integer, Integer) -> (Integer, Integer) fibstep (a, b) = (b, a + b)
fibnums :: [Integer] fibnums = map fst $ iterate fibstep (0, 1)
fibN2 :: Integer -> (Integer, Integer) fibN2 m
| m < 10 = iterate fibstep (0, 1) !! fromIntegral m
fibN2 m = fibN2_next (n, r) (fibN2 n)
where (n, r) = quotRem m 3
fibN2_next (n, r) (f, g)
| r == 0 = (a, b) -- 3n ,3n+1 | r == 1 = (b, c) -- 3n+1,3n+2 | r == 2 = (c, d) -- 3n+2,3n+3 (*) where a = 5 * f ^ 3 + if even n then 3 * f else (-3 * f) -- 3n b = g ^ 3 + 3 * g * f ^ 2 - f ^ 3 -- 3n+1 c = g ^ 3 + 3 * g ^ 2 * f + f ^ 3 -- 3n+2 d = 5 * g ^ 3 + if even n then (-3 * g) else 3 * g -- 3(n+1) (*)
main :: IO () main = print $ (length &&& take 20) . show . fst $ fibN2 (10 ^ 2)</lang>
- Output:
(21,"35422484817926191507")
(fibN2 n)
directly calculates a pair (f,g)
of two consecutive Fibonacci numbers, (Fib[n], Fib[n+1])
, from recursively calculated such pair at about n/3
:
<lang haskell> *Main> (length &&& take 20) . show . fst $ fibN2 (10^6)
(208988,"19532821287077577316")</lang>
The above should take less than 0.1s to calculate on a modern box.
Other identities that could also be used are here. In particular, for (n-1,n) ---> (2n-1,2n) transition which is equivalent to the matrix exponentiation scheme, we have
<lang haskell>f (n,(a,b)) = (2*n,(a*a+b*b,2*a*b+b*b)) -- iterate f (1,(0,1)) ; b is nth</lang>
and for (n,n+1) ---> (2n,2n+1) (derived from d'Ocagne's identity, for example),
<lang haskell>g (n,(a,b)) = (2*n,(2*a*b-a*a,a*a+b*b)) -- iterate g (1,(1,1)) ; a is nth</lang>
Haxe
Iterative
<lang haxe>static function fib(steps:Int, handler:Int->Void) { var current = 0; var next = 1;
for (i in 1...steps) { handler(current);
var temp = current + next; current = next; next = temp; } handler(current); }</lang>
As Iterator
<lang haxe>class FibIter { private var current = 0; private var nextItem = 1; private var limit:Int;
public function new(limit) this.limit = limit;
public function hasNext() return limit > 0;
public function next() { limit--; var ret = current; var temp = current + nextItem; current = nextItem; nextItem = temp; return ret; } }</lang>
Used like: <lang haxe>for (i in new FibIter(10)) Sys.println(i);</lang>
HicEst
<lang hicest>REAL :: Fibonacci(10)
Fibonacci = ($==2) + Fibonacci($-1) + Fibonacci($-2) WRITE(ClipBoard) Fibonacci ! 0 1 1 2 3 5 8 13 21 34</lang>
Hoon
<lang hoon>|= n=@ud =/ a=@ud 0 =/ b=@ud 1 |- ?: =(n 0) a $(a b, b (add a b), n (dec n))</lang>
Hope
Recursive
<lang hope>dec f : num -> num; --- f 0 <= 0; --- f 1 <= 1; --- f(n+2) <= f n + f(n+1);</lang>
Tail-recursive
<lang hope>dec fib : num -> num; --- fib n <= l (1, 0, n)
whererec l == \(a,b,succ c) => if c<1 then a else l((a+b),a,c) |(a,b,0) => 0;</lang>
With lazy lists
This language, being one of Haskell's ancestors, also has lazy lists. Here's the (infinite) list of all Fibonacci numbers:
<lang hope>dec fibs : list num;
--- fibs <= fs whererec fs == 0::1::map (+) (tail fs||fs);</lang>
The nth Fibonacci number is then just fibs @ n
.
Hy
Recursive implementation. <lang clojure>(defn fib [n]
(if (< n 2) n (+ (fib (- n 2)) (fib (- n 1)))))</lang>
Icon and Unicon
Icon has built-in support for big numbers. First, a simple recursive solution augmented by caching for non-negative input. This examples computes fib(1000) if there is no integer argument.
<lang Icon>procedure main(args)
write(fib(integer(!args) | 1000)
end
procedure fib(n)
static fCache initial { fCache := table() fCache[0] := 0 fCache[1] := 1 } /fCache[n] := fib(n-1) + fib(n-2) return fCache[n]
end</lang>
The above solution is similar to the one provided fib in memrfncs
Now, an O(logN) solution. For large N, it takes far longer to convert the result to a string for output than to do the actual computation. This example computes fib(1000000) if there is no integer argument.
<lang Icon>procedure main(args)
write(fib(integer(!args) | 1000000))
end
procedure fib(n)
return fibMat(n)[1]
end
procedure fibMat(n)
if n <= 0 then return [0,0] if n = 1 then return [1,0] fp := fibMat(n/2) c := fp[1]*fp[1] + fp[2]*fp[2] d := fp[1]*(fp[1]+2*fp[2]) if n%2 = 1 then return [c+d, d] else return [d, c]
end</lang>
IDL
Recursive
<lang idl>function fib,n
if n lt 3 then return,1L else return, fib(n-1)+fib(n-2)
end</lang>
Execution time O(2^n) until memory is exhausted and your machine starts swapping. Around fib(35) on a 2GB Core2Duo.
Iterative
<lang idl>function fib,n
psum = (csum = 1uL) if n lt 3 then return,csum for i = 3,n do begin nsum = psum + csum psum = csum csum = nsum endfor return,nsum
end</lang>
Execution time O(n). Limited by size of uLong to fib(49)
Analytic
<lang idl>function fib,n
q=1/( p=(1+sqrt(5))/2 ) return,round((p^n+q^n)/sqrt(5))
end</lang>
Execution time O(1), only limited by the range of LongInts to fib(48).
Idris
Analytic
<lang idris>fibAnalytic : Nat -> Double fibAnalytic n =
floor $ ((pow goldenRatio n) - (pow (-1.0/goldenRatio) n)) / sqrt(5) where goldenRatio : Double goldenRatio = (1.0 + sqrt(5)) / 2.0</lang>
Recursive
<lang idris>fibRecursive : Nat -> Nat fibRecursive Z = Z fibRecursive (S Z) = (S Z) fibRecursive (S (S n)) = fibRecursive (S n) + fibRecursive n </lang>
Iterative
<lang idris>fibIterative : Nat -> Nat fibIterative n = fibIterative' n Z (S Z)
where fibIterative' : Nat -> Nat -> Nat -> Nat fibIterative' Z a _ = a fibIterative' (S n) a b = fibIterative' n b (a + b) </lang>
Lazy
<lang idris>fibLazy : Lazy (List Nat) fibLazy = 0 :: 1 :: zipWith (+) fibLazy (
case fibLazy of (x::xs) => xs [] => []) </lang>
J
The Fibonacci Sequence essay on the J Wiki presents a number of different ways of obtaining the nth Fibonacci number. Here is one: <lang j> fibN=: (-&2 +&$: -&1)^:(1&<) M."0</lang> Examples: <lang j> fibN 12 144
fibN i.31
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040</lang>
(This implementation is doubly recursive except that results are cached across function calls.)
Java
Iterative
<lang java>public static long itFibN(int n) {
if (n < 2) return n; long ans = 0; long n1 = 0; long n2 = 1; for(n--; n > 0; n--) { ans = n1 + n2; n1 = n2; n2 = ans; } return ans;
}</lang>
<lang java> /**
* O(log(n)) */
public static long fib(long n) {
if (n <= 0)
return 0;
long i = (int) (n - 1); long a = 1, b = 0, c = 0, d = 1, tmp1,tmp2;
while (i > 0) {
if (i % 2 != 0) {
tmp1 = d * b + c * a;
tmp2 = d * (b + a) + c * b; a = tmp1; b = tmp2; }
tmp1 = (long) (Math.pow(c, 2) + Math.pow(d, 2)); tmp2 = d * (2 * c + d);
c = tmp1; d = tmp2;
i = i / 2; } return a + b;
} </lang>
Recursive
<lang java>public static long recFibN(final int n) {
return (n < 2) ? n : recFibN(n - 1) + recFibN(n - 2);
}</lang>
Analytic
This method works up to the 92nd Fibonacci number. After that, it goes out of range. <lang java>public static long anFibN(final long n) {
double p = (1 + Math.sqrt(5)) / 2; double q = 1 / p; return (long) ((Math.pow(p, n) + Math.pow(q, n)) / Math.sqrt(5));
}</lang>
Tail-recursive
<lang java>public static long fibTailRec(final int n) {
return fibInner(0, 1, n);
}
private static long fibInner(final long a, final long b, final int n) {
return n < 1 ? a : n == 1 ? b : fibInner(b, a + b, n - 1);
}</lang>
Streams
<lang java5> import java.util.function.LongUnaryOperator; import java.util.stream.LongStream;
public class FibUtil {
public static LongStream fibStream() { return LongStream.iterate( 1l, new LongUnaryOperator() { private long lastFib = 0; @Override public long applyAsLong( long operand ) { long ret = operand + lastFib; lastFib = operand; return ret; } }); } public static long fib(long n) { return fibStream().limit( n ).reduce((prev, last) -> last).getAsLong(); }
} </lang>
JavaScript
ES5
Recursive
Basic recursive function: <lang javascript>function fib(n) {
return n<2?n:fib(n-1)+fib(n-2);
}</lang> Can be rewritten as: <lang javascript>function fib(n) {
if (n<2) { return n; } else { return fib(n-1)+fib(n-2); }
}</lang>
One possibility familiar to Scheme programmers is to define an internal function for iteration through anonymous tail recursion: <lang javascript>function fib(n) {
return function(n,a,b) { return n>0 ? arguments.callee(n-1,b,a+b) : a; }(n,0,1);
}</lang>
Iterative
<lang javascript>function fib(n) {
var a = 0, b = 1, t; while (n-- > 0) { t = a; a = b; b += t; console.log(a); } return a;
}</lang>
Memoization
With the keys of a dictionary,
<lang javascript>var fib = (function(cache){
return cache = cache || {}, function(n){ if (cache[n]) return cache[n]; else return cache[n] = n == 0 ? 0 : n < 0 ? -fib(-n) : n <= 2 ? 1 : fib(n-2) + fib(n-1); };
})(); </lang>
with the indices of an array,
<lang javascript>(function () {
'use strict';
function fib(n) { return Array.apply(null, Array(n + 1)) .map(function (_, i, lst) { return lst[i] = ( i ? i < 2 ? 1 : lst[i - 2] + lst[i - 1] : 0 ); })[n]; }
return fib(32);
})();</lang>
- Output:
2178309
Y-Combinator
<lang javascript>function Y(dn) {
return (function(fn) { return fn(fn); }(function(fn) { return dn(function() { return fn(fn).apply(null, arguments); }); }));
} var fib = Y(function(fn) {
return function(n) { if (n === 0 || n === 1) { return n; } return fn(n - 1) + fn(n - 2); };
});</lang>
Generators
<lang javascript>function* fibonacciGenerator() {
var prev = 0; var curr = 1; while (true) { yield curr; curr = curr + prev; prev = curr - prev; }
} var fib = fibonacciGenerator();</lang>
ES6
Memoized
If we want access to the whole preceding series, as well as a memoized route to a particular member, we can use an accumulating fold.
<lang JavaScript>(() => {
'use strict';
// Nth member of fibonacci series
// fib :: Int -> Int function fib(n) { return mapAccumL(([a, b]) => [ [b, a + b], b ], [0, 1], range(1, n))[0][0]; };
// GENERIC FUNCTIONS
// mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y]) let mapAccumL = (f, acc, xs) => { return xs.reduce((a, x) => { let pair = f(a[0], x);
return [pair[0], a[1].concat(pair[1])]; }, [acc, []]); }
// range :: Int -> Int -> Maybe Int -> [Int] let range = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i);
// TEST return fib(32);
// --> 2178309
})();</lang>
Otherwise, a simple fold will suffice.
(Memoized fold example)
<lang JavaScript>(() => {
'use strict';
// fib :: Int -> Int let fib = n => range(1, n) .reduce(([a, b]) => [b, a + b], [0, 1])[0];
// GENERIC [m..n]
// range :: Int -> Int -> [Int] let range = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i);
// TEST return fib(32);
// --> 2178309
})();</lang>
- Output:
2178309
Joy
Recursive
<lang joy>DEFINE fib == [small] [] [pred dup pred] [+] binrec.</lang>
Iterative
<lang joy>DEFINE fib == [1 0] dip [swap [+] unary] times popd.</lang>
jq
jq does not (yet) have infinite-precision integer arithmetic, and currently the following algorithms only give exact answers up to fib(78). At a certain point, integers are converted to floats, but floating point precision for fib(n) fails after n = 1476: in jq, fib(1476) evaluates to 1.3069892237633987e+308
Recursive
<lang jq>def nth_fib_naive(n):
if (n < 2) then n else nth_fib_naive(n - 1) + nth_fib_naive(n - 2) end;</lang>
Tail Recursive
Recent versions of jq (after July 1, 2014) include basic optimizations for tail recursion, and nth_fib is defined here to take advantage of TCO. For example, nth_fib(10000000) completes with only 380KB (that's K) of memory. However nth_fib can also be used with earlier versions of jq. <lang jq>def nth_fib(n):
# input: [f(i-2), f(i-1), countdown] def fib: (.[0] + .[1]) as $sum | .[2] as $n | if ($n <= 0) then $sum else [ .[1], $sum, $n - 1 ] | fib end; [-1, 1, n] | fib;
</lang>
Example:<lang jq> (range(0;5), 50) | [., nth_fib(.)] </lang> yields: <lang jq>[0,0] [1,1] [2,1] [3,2] [4,3] [50,12586269025]</lang>
Binet's Formula
<lang jq>def fib_binet(n):
(5|sqrt) as $rt | ((1 + $rt)/2) as $phi | (($phi | log) * n | exp) as $phin | (if 0 == (n % 2) then 1 else -1 end) as $sign | ( ($phin - ($sign / $phin) ) / $rt ) + .5 | floor;</lang>
Generator
The following is a jq generator which produces the first n terms of the Fibonacci sequence efficiently, one by one. Notice that it is simply a variant of the above tail-recursive function. The function is in effect turned into a generator by changing "( _ | fib )" to "$sum, (_ | fib)".<lang jq># Generator def fibonacci(n):
# input: [f(i-2), f(i-1), countdown] def fib: (.[0] + .[1]) as $sum | if .[2] == 0 then $sum else $sum, ([ .[1], $sum, .[2] - 1 ] | fib) end; [-1, 1, n] | fib;</lang>
Julia
Recursive
<lang Julia>fib(n) = n < 2 ? n : fib(n-1) + fib(n-2)</lang>
Iterative
<lang Julia>function fib(n)
x,y = (0,1) for i = 1:n x,y = (y, x+y) end x
end</lang>
Matrix form
<lang Julia>fib(n) = ([1 1 ; 1 0]^n)[1,2]</lang>
K
Recursive
<lang K>{:[x<3;1;_f[x-1]+_f[x-2]]}</lang>
Recursive with memoization
Using a (global) dictionary c.
<lang K>{c::.();{v:c[a:`$$x];:[x<3;1;:[_n~v;c[a]:_f[x-1]+_f[x-2];v]]}x}</lang>
Analytic
<lang K>phi:(1+_sqrt(5))%2 {_((phi^x)-((1-phi)^x))%_sqrt[5]}</lang>
Sequence to n
<lang K>{(x(|+\)\1 1)[;1]}</lang> <lang K>{x{x,+/-2#x}/!2}</lang>
Kabap
Sequence to n
<lang Kabap> // Calculate the $n'th Fibonacci number
// Set this to how many in the sequence to generate $n = 10;
// These are what hold the current calculation $a = 0; $b = 1;
// This holds the complete sequence that is generated $sequence = "";
// Prepare a loop $i = 0;
- calcnextnumber;
$i = $i++;
// Do the calculation for this loop iteration $b = $a + $b; $a = $b - $a;
// Add the result to the sequence $sequence = $sequence << $a;
// Make the loop run a fixed number of times if $i < $n; { $sequence = $sequence << ", "; goto calcnextnumber; }
// Use the loop counter as the placeholder $i--;
// Return the sequence return = "Fibonacci number " << $i << " is " << $a << " (" << $sequence << ")"; </lang>
Kotlin
<lang scala>enum class Fibonacci {
ITERATIVE { override fun invoke(n: Long) = if (n < 2) { n } else { var n1 = 0L var n2 = 1L var i = n do { val sum = n1 + n2 n1 = n2 n2 = sum } while (i-- > 1) n1 } }, RECURSIVE { override fun invoke(n: Long): Long = if (n < 2) n else this(n - 1) + this(n - 2) };
abstract operator fun invoke(n: Long): Long
}
fun main(a: Array<String>) {
val r = 0..30L Fibonacci.values().forEach { print("${it.name}: ") r.forEach { i -> print(" " + it(i)) } println() }
}</lang>
- Output:
ITERATIVE: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 RECURSIVE: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
L++
<lang lisp>(defn int fib (int n) (return (? (< n 2) n (+ (fib (- n 1)) (fib (- n 2)))))) (main (prn (fib 30)))</lang>
LabVIEW
This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.
lambdatalk
<lang scheme> 1) basic version {def fib1
{lambda {:n} {if {< :n 3} then 1 else {+ {fib1 {- :n 1}} {fib1 {- :n 2}}} }}}
{fib1 16} -> 987 (CPU ~ 16ms) {fib1 30} = 832040 (CPU > 12000ms)
2) tail-recursive version {def fib2
{def fib2.r {lambda {:a :b :i} {if {< :i 1} then :a else {fib2.r :b {+ :a :b} {- :i 1}} }}} {lambda {:n} {fib2.r 0 1 :n}}}
{fib2 16} -> 987 (CPU ~ 1ms) {fib2 30} -> 832040 (CPU ~2ms) {fib2 1000} -> 4.346655768693743e+208 (CPU ~ 22ms)
3) Dijkstra Algorithm {def fib3
{def fib3.r {lambda {:a :b :p :q :count} {if {= :count 0} then :b else {if {= {% :count 2} 0} then {fib3.r :a :b {+ {* :p :p} {* :q :q}} {+ {* :q :q} {* 2 :p :q}} {/ :count 2}} else {fib3.r {+ {* :b :q} {* :a :q} {* :a :p}} {+ {* :b :p} {* :a :q}} :p :q {- :count 1}} }}}} {lambda {:n} {fib3.r 1 0 0 1 :n} }}
{fib3 16} -> 987 (CPU ~ 2ms) {fib3 30} -> 832040 (CPU ~ 2ms) {fib3 1000} -> 4.346655768693743e+208 (CPU ~ 3ms)
4) memoization {def fib4
{def fib4.m {array.new}} // init an empty array {def fib4.r {lambda {:n} {if {< :n 2} then {array.get {array.set! {fib4.m} :n 1} :n} // init with 1,1 else {if {equal? {array.get {fib4.m} :n} undefined} // if not exists then {array.get {array.set! {fib4.m} :n {+ {fib4.r {- :n 1}} {fib4.r {- :n 2}}}} :n} // compute it else {array.get {fib4.m} :n} }}}} // else get it {lambda {:n} {fib4.r :n} {fib4.m} }} // display the number AND all its predecessors
-> fib4 {fib4 90} -> 4660046610375530000 [1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418, 317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155, 165580141,267914296,433494437,701408733,1134903170,1836311903,2971215073,4807526976,7778742049,12586269025, 20365011074,32951280099,53316291173,86267571272,139583862445,225851433717,365435296162,591286729879,956722026041, 1548008755920,2504730781961,4052739537881,6557470319842,10610209857723,17167680177565,27777890035288,44945570212853, 72723460248141,117669030460994,190392490709135,308061521170129,498454011879264,806515533049393,1304969544928657, 2111485077978050,3416454622906707,5527939700884757,8944394323791464,14472334024676220,23416728348467684, 37889062373143900,61305790721611580,99194853094755490,160500643816367070,259695496911122560,420196140727489660, 679891637638612200,1100087778366101900,1779979416004714000,2880067194370816000,4660046610375530000]
5) Binet's formula (non recursive) {def fib5
{lambda {:n} {let { {:n :n} {:sqrt5 {sqrt 5}} } {round {/ {- {pow {/ {+ 1 :sqrt5} 2} :n} {pow {/ {- 1 :sqrt5} 2} :n}} :sqrt5}}} }}
{fib5 16} -> 987 (CPU ~ 1ms) {fib5 30} -> 832040 (CPU ~ 1ms) {fib5 1000} -> 4.346655768693743e+208 (CPU ~ 1ms)
</lang>
Lang5
<lang lang5>[] '__A set : dip swap __A swap 2 compress collapse '__A set execute
__A -1 extract nip ; : nip swap drop ; : tuck swap over ;
- -rot rot rot ; : 0= 0 == ; : 1+ 1 + ; : 1- 1 - ; : sum '+ reduce ;
- bi 'keep dip execute ; : keep over 'execute dip ;
- fib dup 1 > if dup 1- fib swap 2 - fib + then ;
- fib dup 1 > if "1- fib" "2 - fib" bi + then ;</lang>
langur
<lang langur>val .fibonacci = f if(.x < 2: .x ; self(.x - 1) + self(.x - 2))
writeln map .fibonacci, series 2..20</lang>
- Output:
[1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]
Lasso
<lang Lasso> define fibonacci(n::integer) => {
#n < 1 ? return false
local( swap = 0, n1 = 0, n2 = 1 )
loop(#n) => {
#swap = #n1 + #n2; #n2 = #n1; #n1 = #swap;
} return #n1
}
fibonacci(0) //->output false fibonacci(1) //->output 1 fibonacci(2) //->output 1 fibonacci(3) //->output 2 </lang>
Latitude
Recursive
<lang latitude>fibo := {
takes '[n]. if { n <= 1. } then { n. } else { fibo (n - 1) + fibo (n - 2). }.
}.</lang>
Memoization
<lang latitude>fibo := {
takes '[n]. cache := here cache. { cache slot? (n ordinal). } ifFalse { cache slot (n ordinal) = if { n <= 1. } then { n. } else { fibo (n - 1) + fibo (n - 2). }. }. cache slot (n ordinal).
} tap {
;; Attach the cache to the method object itself. #'self cache := Object clone.
}.</lang>
Lean
It runs on Lean 3.4.2 or later:
<lang lean> -- Our first implementation is the usual recursive definition: def fib1 : ℕ → ℕ | 0 := 0 | 1 := 1 | (n + 2) := fib1 n + fib1 (n + 1)
-- We can give a second more efficient implementation using an auxiliary function:
def fib_aux : ℕ → ℕ → ℕ → ℕ
| 0 a b := b
| (n + 1) a b := fib_aux n (a + b) a
def fib2 : ℕ → ℕ | n := fib_aux n 1 0
-- Use #eval to check computations:
- eval fib1 20
- eval fib2 20</lang>
LFE
Recursive
<lang lisp> (defun fib
((0) 0) ((1) 1) ((n) (+ (fib (- n 1)) (fib (- n 2)))))
</lang>
Iterative
<lang lisp> (defun fib
((n) (when (>= n 0)) (fib n 0 1)))
(defun fib
((0 result _) result) ((n result next) (fib (- n 1) next (+ result next))))
</lang>
Liberty BASIC
Iterative/Recursive
<lang lb> for i = 0 to 15
print fiboR(i),fiboI(i)
next i
function fiboR(n)
if n <= 1 then fiboR = n else fiboR = fiboR(n-1) + fiboR(n-2) end if
end function
function fiboI(n)
a = 0 b = 1 for i = 1 to n temp = a + b a = b b = temp next i fiboI = a
end function </lang>
- Output:
0 0 1 1 1 1 2 2 3 3 5 5 8 8 13 13 21 21 34 34 55 55 89 89 144 144 233 233 377 377 610 610
Iterative/Negative
<lang lb> print "Rosetta Code - Fibonacci sequence": print print " n Fn" for x=-12 to 12 '68 max
print using("### ", x); using("##############", FibonacciTerm(x))
next x print [start] input "Enter a term#: "; n$ n$=lower$(trim$(n$)) if n$="" then print "Program complete.": end print FibonacciTerm(val(n$)) goto [start]
function FibonacciTerm(n)
n=int(n) FTa=0: FTb=1: FTc=-1 select case case n=0 : FibonacciTerm=0 : exit function case n=1 : FibonacciTerm=1 : exit function case n=-1 : FibonacciTerm=-1 : exit function case n>1 for x=2 to n FibonacciTerm=FTa+FTb FTa=FTb: FTb=FibonacciTerm next x exit function case n<-1 for x=-2 to n step -1 FibonacciTerm=FTa+FTc FTa=FTc: FTc=FibonacciTerm next x exit function end select
end function </lang>
- Output:
Rosetta Code - Fibonacci sequence n Fn -12 -144 -11 -89 -10 -55 -9 -34 -8 -21 -7 -13 -6 -8 -5 -5 -4 -3 -3 -2 -2 -1 -1 -1 0 0 1 1 2 1 3 2 4 3 5 5 6 8 7 13 8 21 9 34 10 55 11 89 12 144 Enter a term#: 12 144 Enter a term#: Program complete.
Lingo
Recursive
<lang lingo>on fib (n)
if n<2 then return n return fib(n-1)+fib(n-2)
end</lang>
Iterative
<lang lingo>on fib (n)
if n<2 then return n fibPrev = 0 fib = 1 repeat with i = 2 to n tmp = fib fib = fib + fibPrev fibPrev = tmp end repeat return fib
end</lang>
Analytic
<lang lingo>on fib (n)
sqrt5 = sqrt(5.0) p = (1+sqrt5)/2 q = 1 - p return integer((power(p,n)-power(q,n))/sqrt5)
end</lang>
Lisaac
<lang Lisaac>- fib(n : UINTEGER_32) : UINTEGER_64 <- (
+ result : UINTEGER_64; (n < 2).if { result := n; } else { result := fib(n - 1) + fib(n - 2); }; result
);</lang>
LiveCode
<lang LiveCode>-- Iterative, translation of the basic version. function fibi n
put 0 into aa put 1 into b repeat with i = 1 to n put aa + b into temp put b into aa put temp into b end repeat return aa
end fibi
-- Recursive function fibr n
if n <= 1 then return n else return fibr(n-1) + fibr(n-2) end if
end fibr</lang>
LLVM
<lang llvm>; This is not strictly LLVM, as it uses the C library function "printf".
- LLVM does not provide a way to print values, so the alternative would be
- to just load the string into memory, and that would be boring.
- Additional comments have been inserted, as well as changes made from the output produced by clang such as putting more meaningful labels for the jumps
$"PRINT_LONG" = comdat any @"PRINT_LONG" = linkonce_odr unnamed_addr constant [5 x i8] c"%ld\0A\00", comdat, align 1
- --- The declaration for the external C printf function.
declare i32 @printf(i8*, ...)
- --------------------------------------------------------------------
- -- Function for calculating the nth fibonacci numbers
- --------------------------------------------------------------------
define i32 @fibonacci(i32) {
%2 = alloca i32, align 4 ;-- allocate local copy of n %3 = alloca i32, align 4 ;-- allocate a %4 = alloca i32, align 4 ;-- allocate b store i32 %0, i32* %2, align 4 ;-- store copy of n store i32 0, i32* %3, align 4 ;-- a := 0 store i32 1, i32* %4, align 4 ;-- b := 1 br label %loop
loop:
%5 = load i32, i32* %2, align 4 ;-- load n %6 = icmp sgt i32 %5, 0 ;-- n > 0 br i1 %6, label %loop_body, label %exit
loop_body:
%7 = load i32, i32* %3, align 4 ;-- load a %8 = load i32, i32* %4, align 4 ;-- load b %9 = add nsw i32 %7, %8 ;-- t = a + b store i32 %8, i32* %3, align 4 ;-- store a = b store i32 %9, i32* %4, align 4 ;-- store b = t %10 = load i32, i32* %2, align 4 ;-- load n %11 = add nsw i32 %10, -1 ;-- decrement n store i32 %11, i32* %2, align 4 ;-- store n br label %loop
exit:
%12 = load i32, i32* %3, align 4 ;-- load a ret i32 %12 ;-- return a
}
- --------------------------------------------------------------------
- -- Main function for printing successive fibonacci numbers
- --------------------------------------------------------------------
define i32 @main() {
%1 = alloca i32, align 4 ;-- allocate index store i32 0, i32* %1, align 4 ;-- index := 0 br label %loop
loop:
%2 = load i32, i32* %1, align 4 ;-- load index %3 = icmp sle i32 %2, 12 ;-- index <= 12 br i1 %3, label %loop_body, label %exit
loop_body:
%4 = load i32, i32* %1, align 4 ;-- load index %5 = call i32 @fibonacci(i32 %4) %6 = call i32 (i8*, ...) @printf(i8* getelementptr inbounds ([5 x i8], [5 x i8]* @"PRINT_LONG", i32 0, i32 0), i32 %5) %7 = load i32, i32* %1, align 4 ;-- load index %8 = add nsw i32 %7, 1 ;-- increment index store i32 %8, i32* %1, align 4 ;-- store index br label %loop
exit:
ret i32 0 ;-- return EXIT_SUCCESS
}</lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144
Logo
<lang logo>to fib :n [:a 0] [:b 1]
if :n < 1 [output :a] output (fib :n-1 :b :a+:b)
end</lang>
LOLCODE
<lang LOLCODE> HAI 1.2 HOW DUZ I fibonacci YR N
EITHER OF BOTH SAEM N AN 1 AN BOTH SAEM N AN 0 O RLY? YA RLY, FOUND YR 1 NO WAI I HAS A N1 I HAS A N2 N1 R DIFF OF N AN 1 N2 R DIFF OF N AN 2 N1 R fibonacci N1 N2 R fibonacci N2 FOUND YR SUM OF N1 AN N2 OIC
IF U SAY SO KTHXBYE </lang>
LSL
Rez a box on the ground, and add the following as a New Script. <lang LSL>integer Fibonacci(integer n) { if(n<2) { return n; } else { return Fibonacci(n-1)+Fibonacci(n-2); } } default { state_entry() { integer x = 0; for(x=0 ; x<35 ; x++) { llOwnerSay("Fibonacci("+(string)x+")="+(string)Fibonacci(x)); } } }</lang> Output:
Fibonacci(0)=0 Fibonacci(1)=1 Fibonacci(2)=1 Fibonacci(3)=2 Fibonacci(4)=3 Fibonacci(5)=5 Fibonacci(6)=8 Fibonacci(7)=13 Fibonacci(8)=21 Fibonacci(9)=34 Fibonacci(10)=55 Fibonacci(11)=89 Fibonacci(12)=144 Fibonacci(13)=233 Fibonacci(14)=377 Fibonacci(15)=610 Fibonacci(16)=987 Fibonacci(17)=1597 Fibonacci(18)=2584 Fibonacci(19)=4181 Fibonacci(20)=6765 Fibonacci(21)=10946 Fibonacci(22)=17711 Fibonacci(23)=28657 Fibonacci(24)=46368 Fibonacci(25)=75025 Fibonacci(26)=121393 Fibonacci(27)=196418 Fibonacci(28)=317811 Fibonacci(29)=514229 Fibonacci(30)=832040 Fibonacci(31)=1346269 Fibonacci(32)=2178309 Fibonacci(33)=3524578 Fibonacci(34)=5702887
Lua
Recursive
<lang lua> --calculates the nth fibonacci number. Breaks for negative or non-integer n. function fibs(n)
return n < 2 and n or fibs(n - 1) + fibs(n - 2)
end </lang>
Pedantic Recursive
<lang lua> --more pedantic version, returns 0 for non-integer n function pfibs(n)
if n ~= math.floor(n) then return 0 elseif n < 0 then return pfibs(n + 2) - pfibs(n + 1) elseif n < 2 then return n else return pfibs(n - 1) + pfibs(n - 2) end
end </lang>
Tail Recursive
<lang lua> function a(n,u,s) if n<2 then return u+s end return a(n-1,u+s,u) end function trfib(i) return a(i-1,1,0) end </lang>
Table Recursive
<lang lua> fib_n = setmetatable({1, 1}, {__index = function(z,n) return n<=0 and 0 or z[n-1] + z[n-2] end}) </lang>
Table Recursive 2
<lang lua> -- table recursive done properly (values are actually saved into table; -- also the first element of Fibonacci sequence is 0, so the initial table should be {0, 1}). fib_n = setmetatable({0, 1}, {
__index = function(t,n) if n <= 0 then return 0 end t[n] = t[n-1] + t[n-2] return t[n] end
}) </lang>
Iterative
<lang lua> function ifibs(n)
local p0,p1=0,1 for _=1,n do p0,p1 = p1,p0+p1 end return p0
end </lang>
Luck
<lang luck>function fib(x: int): int = (
let cache = {} in let fibc x = if x<=1 then x else ( if x not in cache then cache[x] = fibc(x-1) + fibc(x-2); cache[x] ) in fibc(x)
);; for x in range(10) do print(fib(x))</lang>
Lush
<lang lush>(de fib-rec (n)
(if (< n 2) n (+ (fib-rec (- n 2)) (fib-rec (- n 1)))))</lang>
M2000 Interpreter
Return decimal type and use an Inventory (as closure) to store known return values. All closures are in scope in every recursive call (we use here lambda(), but we can use fib(), If we make Fib1=fib then we have to use lambda() for recursion. <lang M2000 Interpreter> Inventory K=0:=0,1:=1 fib=Lambda K (x as decimal)-> {
If Exist(K, x) Then =Eval(K) :Exit Def Ret as Decimal Ret=If(x>1->Lambda(x-1)+Lambda(x-2), x) Append K, x:=Ret =Ret
} \\ maximum 139 For i=1 to 139 {
Print Fib(i)
} </lang>
Here an example where we use a BigNum class to make a Group which hold a stack of values, and take 14 digits per item in stack. We can use inventory to hold groups, so we use the fast fib() function from code above, where we remove the type definition of Ret variable, and set two first items in inventory as groups.
<lang M2000 Interpreter>
Class BigNum {
a=stack Function Digits { =len(.a)*14-(14-len(str$(stackitem(.a,len(.a)) ,""))) } Operator "+" (n) { \\ we get a copy, but .a is pointer \\ we make a copy, and get a new pointer .a<=stack(.a) acc=0 carry=0 const d=100000000000000@ k=min.data(Len(.a), len(n.a)) i=each(.a, 1,k ) j=each(n.a, 1,k) while i, j { acc=stackitem(i)+stackitem(j)+carry carry= acc div d return .a, i^+1:=acc mod d } if len(.a)<len(n.a) Then { i=each(n.a, k+1, -1) while i { acc=stackitem(i)+carry carry= acc div d stack .a {data acc mod d} } } ELse.if len(.a)>len(n.a) Then { i=each(.a, k+1, -1) while i { acc=stackitem(i)+carry carry= acc div d Return .a, i^+1:=acc mod d if carry else exit } } if carry then stack .a { data carry} } Function tostring$ { if len(.a)=0 then ="0" : Exit if len(.a)=1 then =str$(Stackitem(.a),"") : Exit document buf$=str$(Stackitem(.a, len(.a)),"") for i=len(.a)-1 to 1 { Stack .a { buf$=str$(StackItem(i), "00000000000000") } } =buf$ } class: Module BigNum (s$) { s$=filter$(s$,"+-.,") if s$<>"" Then { repeat { If len(s$)<14 then Stack .a { Data val(s$) }: Exit Stack .a { Data val(Right$(s$, 14)) } S$=Left$(S$, len(S$)-14) } Until S$="" } }
}
Inventory K=0:=BigNum("0"),1:=BigNum("1") fib=Lambda K (x as decimal)-> {
If Exist(K, x) Then =Eval(K) :Exit Ret=If(x>1->Lambda(x-1)+Lambda(x-2), bignum(str$(x,""))) Append K, x:=Ret =Ret
} \\ Using this to handle form refresh by code Set Fast! For i=1 to 4000 {
N=Fib(i) Print i Print N.tostring$() Refresh
}
</lang>
M4
<lang m4>define(`fibo',`ifelse(0,$1,0,`ifelse(1,$1,1, `eval(fibo(decr($1)) + fibo(decr(decr($1))))')')')dnl define(`loop',`ifelse($1,$2,,`$3($1) loop(incr($1),$2,`$3')')')dnl loop(0,15,`fibo')</lang>
Maple
<lang maple> > f := n -> ifelse(n<3,1,f(n-1)+f(n-2)); > f(2);
1
> f(3);
2
</lang>
Mathematica / Wolfram Language
The Wolfram Language already has a built-in function Fibonacci, but a simple recursive implementation would be
<lang mathematica>fib[0] = 0 fib[1] = 1 fib[n_Integer] := fib[n - 1] + fib[n - 2]</lang>
An optimization is to cache the values already calculated:
<lang mathematica>fib[0] = 0 fib[1] = 1 fib[n_Integer] := fib[n] = fib[n - 1] + fib[n - 2]</lang>
The above implementations may be too simplistic, as the first is incredibly slow for any reasonable range due to nested recursions and while the second is faster it uses an increasing amount of memory. The following uses recursion much more effectively while not using memory:
<lang mathematica>fibi[prvprv_Integer, prv_Integer, rm_Integer] :=
If[rm < 1, prvprv, fibi[prv, prvprv + prv, rm - 1]]
fib[n_Integer] := fibi[0, 1, n]</lang>
However, the recursive approaches in Mathematica are limited by the limit set for recursion depth (default 1024 or 4096 for the above cases), limiting the range for 'n' to about 1000 or 2000. The following using an iterative approach has an extremely high limit (greater than a million):
<lang mathematica>fib[n_Integer] := Block[{tmp, prvprv = 0, prv = 1},
For[i = 0, i < n, i++, tmp = prv; prv += prvprv; prvprv = tmp]; Return[prvprv]]</lang>
If one wanted a list of Fibonacci numbers, the following is quite efficient:
<lang mathematica>fibi[{prvprv_Integer, prv_Integer}] := {prv, prvprv + prv} fibList[n_Integer] := Map[Take[#, 1] &, NestList[fibi, {0, 1}, n]] // Flatten</lang>
Output from the last with "fibList[100]":
<lang mathematica>{0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, \ 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, \ 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, \ 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, \ 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, \ 2971215073, 4807526976, 7778742049, 12586269025, 20365011074, \ 32951280099, 53316291173, 86267571272, 139583862445, 225851433717, \ 365435296162, 591286729879, 956722026041, 1548008755920, \ 2504730781961, 4052739537881, 6557470319842, 10610209857723, \ 17167680177565, 27777890035288, 44945570212853, 72723460248141, \ 117669030460994, 190392490709135, 308061521170129, 498454011879264, \ 806515533049393, 1304969544928657, 2111485077978050, \ 3416454622906707, 5527939700884757, 8944394323791464, \ 14472334024676221, 23416728348467685, 37889062373143906, \ 61305790721611591, 99194853094755497, 160500643816367088, \ 259695496911122585, 420196140727489673, 679891637638612258, \ 1100087778366101931, 1779979416004714189, 2880067194370816120, \ 4660046610375530309, 7540113804746346429, 12200160415121876738, \ 19740274219868223167, 31940434634990099905, 51680708854858323072, \ 83621143489848422977, 135301852344706746049, 218922995834555169026, \ 354224848179261915075}</lang>
The Wolfram Language can also solve recurrence equations using the built-in function RSolve
<lang mathematica>fib[n] /. RSolve[{fib[n] == fib[n - 1] + fib[n - 2], fib[0] == 0,
fib[1] == 1}, fib[n], n]1</lang>
which evaluates to the built-in function Fibonacci[n]
This function can also be expressed as
<lang mathematica>Fibonacci[n] // FunctionExpand // FullSimplify</lang>
which evaluates to
<lang mathematica>(2^-n ((1 + Sqrt[5])^n - (-1 + Sqrt[5])^n Cos[n π]))/Sqrt[5]</lang>
and is defined for all real or complex values of n.
MATLAB
Matrix
<lang MATLAB>function f = fib(n)
f = [1 1 ; 1 0]^(n-1); f = f(1,1);
end</lang>
Iterative
<lang MATLAB>function F = fibonacci(n)
Fn = [1 0]; %Fn(1) is F_{n-2}, Fn(2) is F_{n-1} F = 0; %F is F_{n} for i = (1:abs(n)) Fn(2) = F; F = sum(Fn); Fn(1) = Fn(2); end if n < 0 F = F*((-1)^(n+1)); end
end</lang>
Dramadah Matrix Method
The MATLAB help file suggests an interesting method of generating the Fibonacci numbers. Apparently the determinate of the Dramadah Matrix of type 3 (MATLAB designation) and size n-by-n is the nth Fibonacci number. This method is implimented below.
<lang MATLAB>function number = fibonacci2(n)
if n == 1 number = 1; elseif n == 0 number = 0; elseif n < 0 number = ((-1)^(n+1))*fibonacci2(-n);; else number = det(gallery('dramadah',n,3)); end
end</lang>
Tartaglia/Pascal Triangle Method
<lang Matlab> function number = fibonacci(n) %construct the Tartaglia/Pascal Triangle
pt=tril(ones(n)); for r = 3 : n % Every element is the addition of the two elements % on top of it. That means the previous row. for c = 2 : r-1 pt(r, c) = pt(r-1, c-1) + pt(r-1, c); end end number=trace(rot90(pt));
end </lang>
Maxima
<lang maxima>/* fib(n) is built-in; here is an implementation */ fib2(n) := (matrix([0, 1], [1, 1])^^n)[1, 2]$
fib2(100)-fib(100); 0
fib2(-10); -55</lang>
MAXScript
Iterative
<lang maxscript>fn fibIter n = (
if n < 2 then ( n ) else ( fib = 1 fibPrev = 1 for num in 3 to n do ( temp = fib fib += fibPrev fibPrev = temp ) fib )
)</lang>
Recursive
<lang maxscript>fn fibRec n = (
if n < 2 then ( n ) else ( fibRec (n - 1) + fibRec (n - 2) )
)</lang>
Mercury
Mercury is both a logic language and a functional language. As such there are two possible interfaces for calculating a Fibonacci number. This code shows both styles. Note that much of the code here is ceremony put in place to have this be something which can actually compile. The actual Fibonacci number generation is contained in the predicate fib/2
and in the function fib/1
. The predicate main/2
illustrates first the unification semantics of the predicate form and the function call semantics of the function form.
The provided code uses a very naive form of generating a Fibonacci number. A more realistic implementation would use memoization to cache previous results, exchanging time for space. Also, in the case of supplying both a function implementation and a predicate implementation, one of the two would be implemented in terms of the other. Examples of this are given as comments below.
fib.m
<lang mercury> % The following code is derived from the Mercury Tutorial by Ralph Becket. % http://www.mercury.csse.unimelb.edu.au/information/papers/book.pdf
- - module fib.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - implementation.
- - import_module int.
- - pred fib(int::in, int::out) is det.
fib(N, X) :-
( if N =< 2 then X = 1 else fib(N - 1, A), fib(N - 2, B), X = A + B ).
- - func fib(int) = int is det.
fib(N) = X :- fib(N, X).
main(!IO) :-
fib(40, X), write_string("fib(40, ", !IO), write_int(X, !IO), write_string(")\n", !IO), write_string("fib(40) = ", !IO), write_int(fib(40), !IO), write_string("\n", !IO).
</lang>
Iterative algorithm
The much faster iterative algorithm can be written as:
<lang mercury>
- - pred fib_acc(int::in, int::in, int::in, int::in, int::out) is det.
fib_acc(N, Limit, Prev2, Prev1, Res) :-
( N < Limit -> % limit not reached, continue computation. ( N =< 2 -> Res0 = 1 ; Res0 = Prev2 + Prev1 ), fib_acc(N+1, Limit, Prev1, Res0, Res) ; % Limit reached, return the sum of the two previous results. Res = Prev2 + Prev1 ).
</lang>
This predicate can be called as <lang mercury>fib_acc(1, 40, 1, 1, Result)</lang> It has several inputs which form the loop, the first is the current number, the second is a limit, ie when to stop counting. And the next two are accumulators for the last and next-to-last results.
Memoization
But what if you want the speed of the fib_acc with the recursive (more declarative) definition of fib? Then use memoization, because Mercury is a pure language fib(N, F) will always give the same F for the same N, guaranteed. Therefore memoization asks the compiler to use a table to remember the value for F for any N, and it's a one line change:
<lang mercury>
- - pragma memo(fib/2).
- - pred fib(int::in, int::out) is det.
fib(N, X) :-
( if N =< 2 then X = 1 else fib(N - 1, A), fib(N - 2, B), X = A + B ).
</lang>
We've shown the definition of fib/2 again, but the only change here is the memoization pragma (see the reference manual). This is not part of the language specification and different Mercury implementations are allowed to ignore it, however there is only one implementation so in practice memoization is fully supported.
Memoization trades speed for space, a table of results is constructed and kept in memory. So this version of fib consumes more memory than than fib_acc. It is also slightly slower than fib_acc since it must manage its table of results but it is much much faster than without memoization. Memoization works very well for the Fibonacci sequence because in the naive version the same results are calculated over and over again.
Metafont
<lang metafont>vardef fibo(expr n) = if n=0: 0 elseif n=1: 1 else:
fibo(n-1) + fibo(n-2)
fi enddef;
for i=0 upto 10: show fibo(i); endfor end</lang>
Microsoft Small Basic
Iterative
<lang smallbasic>' Fibonacci sequence - 31/07/2018
n = 139 f1 = 0 f2 = 1 TextWindow.WriteLine("fibo(0)="+f1) TextWindow.WriteLine("fibo(1)="+f2) For i = 2 To n f3 = f1 + f2 TextWindow.WriteLine("fibo("+i+")="+f3) f1 = f2 f2 = f3 EndFor</lang>
- Output:
fibo(139)=50095301248058391139327916261
Binet's Formula
<lang smallbasic>' Fibonacci sequence - Binet's Formula - 31/07/2018
n = 69 sq5=Math.SquareRoot(5) phi1=(1+sq5)/2 phi2=(1-sq5)/2 phi1n=phi1 phi2n=phi2 For i = 2 To n phi1n=phi1n*phi1 phi2n=phi2n*phi2 TextWindow.Write(Math.Floor((phi1n-phi2n)/sq5)+" ") EndFor</lang>
- Output:
1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 102334155 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025 20365011074 32951280099 53316291173 86267571272 139583862445 225851433717 365435296162 591286729879 956722026041 1548008755920 2504730781961 4052739537881 6557470319842 10610209857723 17167680177565 27777890035288 44945570212853 72723460248141 117669030460994
min
<lang min>(
(2 <) ((0 1 (dup rollup +)) dip pred times nip) unless
) :fib</lang>
MiniScript
<lang MiniScript>fibonacci = function(n)
if n < 2 then return n ans = 0 n1 = 0 n2 = 1 for i in range(n-1, 1) ans = n1 + n2 n1 = n2 n2 = ans end for return ans
end function
print fibonacci(6)</lang>
MIPS Assembly
This is the iterative approach to the Fibonacci sequence. <lang MIPS> .text main: li $v0, 5 # read integer from input. The read integer will be stroed in $v0 syscall
beq $v0, 0, is1 beq $v0, 1, is1
li $s4, 1 # the counter which has to equal to $v0
li $s0, 1 li $s1, 1
loop: add $s2, $s0, $s1 addi $s4, $s4, 1 beq $v0, $s4, iss2
add $s0, $s1, $s2 addi $s4, $s4, 1 beq $v0, $s4, iss0
add $s1, $s2, $s0 addi $s4, $s4, 1 beq $v0, $s4, iss1
b loop
iss0: move $a0, $s0 b print
iss1: move $a0, $s1 b print
iss2: move $a0, $s2 b print
is1: li $a0, 1
b print
print: li $v0, 1 syscall li $v0, 10 syscall </lang>
Mirah
<lang mirah>def fibonacci(n:int)
return n if n < 2 fibPrev = 1 fib = 1 3.upto(Math.abs(n)) do oldFib = fib fib = fib + fibPrev fibPrev = oldFib end fib * (n<0 ? int(Math.pow(n+1, -1)) : 1)
end
puts fibonacci 1 puts fibonacci 2 puts fibonacci 3 puts fibonacci 4 puts fibonacci 5 puts fibonacci 6 puts fibonacci 7 </lang>
МК-61/52
<lang>П0 1 lg Вx <-> + L0 03 С/П БП 03</lang>
Instruction: n В/О С/П, where n is serial number of the number of Fibonacci sequence; С/П for the following numbers.
ML
Standard ML
Recursion
This version is tail recursive. <lang sml>fun fib n =
let
fun fib' (0,a,b) = a | fib' (n,a,b) = fib' (n-1,a+b,a)
in
fib' (n,0,1)
end</lang>
MLite
Recursion
Tail recursive. <lang ocaml>fun fib
(0, x1, x2) = x2 | (n, x1, x2) = fib (n-1, x2, x1+x2) | n = fib (n, 0, 1)</lang>
ML/I
<lang ML/I>MCSKIP "WITH" NL "" Fibonacci - recursive MCSKIP MT,<> MCINS %. MCDEF FIB WITHS () AS <MCSET T1=%A1. MCGO L1 UNLESS 2 GR T1 %T1.<>MCGO L0 %L1.%FIB(%T1.-1)+FIB(%T1.-2).> fib(0) is FIB(0) fib(1) is FIB(1) fib(2) is FIB(2) fib(3) is FIB(3) fib(4) is FIB(4) fib(5) is FIB(5)</lang>
Modula-2
<lang modula2>MODULE Fibonacci; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE Fibonacci(n : LONGINT) : LONGINT; VAR
a,b,c : LONGINT;
BEGIN
IF n<0 THEN RETURN 0 END;
a:=1; b:=1;
WHILE n>0 DO c := a + b; a := b; b := c; DEC(n) END;
RETURN a
END Fibonacci;
VAR
buf : ARRAY[0..63] OF CHAR; i : INTEGER; r : LONGINT;
BEGIN
FOR i:=0 TO 10 DO r := Fibonacci(i);
FormatString("%l\n", buf, r); WriteString(buf); END;
ReadChar
END Fibonacci.</lang>
Modula-3
Recursive
<lang modula3>PROCEDURE Fib(n: INTEGER): INTEGER =
BEGIN IF n < 2 THEN RETURN n; ELSE RETURN Fib(n-1) + Fib(n-2); END; END Fib;</lang>
Iterative (with negatives)
<lang modula3>PROCEDURE IterFib(n: INTEGER): INTEGER =
VAR
limit := ABS(n); prev := 0; curr, next: INTEGER;
BEGIN
(* trivial case *) IF n = 0 THEN RETURN 0; END;
IF n > 0 THEN (* positive case *)
curr := 1; FOR i := 2 TO limit DO next := prev + curr; prev := curr; curr := next; END;
ELSE (* negative case *)
curr := -1; FOR i := 2 TO limit DO next := prev - curr; prev := curr; curr := next; END;
END;
RETURN curr;
END IterFib;</lang>
Monicelli
Recursive version. It includes a main that reads a number N from standard input and prints the Nth Fibonacci number. <lang monicelli>
- Main
Lei ha clacsonato voglio un nonnulla, Necchi mi porga un nonnulla il nonnulla come se fosse brematurata la supercazzola bonaccia con il nonnulla o scherziamo? un nonnulla a posterdati
- Fibonacci function 'bonaccia'
blinda la supercazzola Necchi bonaccia con antani Necchi o scherziamo? che cos'è l'antani? minore di 3: vaffanzum 1! o tarapia tapioco: voglio unchiamo, Necchi come se fosse brematurata la supercazzola bonaccia con antani meno 1 o scherziamo? voglio duechiamo, Necchi come se fosse brematurata la supercazzola bonaccia con antani meno 2 o scherziamo? vaffanzum unchiamo più duechiamo! e velocità di esecuzione </lang>
MontiLang
Reads number from standard input and prints to that number in the fibonacci sequence <lang MontiLang>0 VAR a . 1 VAR b . INPUT TOINT FOR :
a b + VAR c . a PRINT . b VAR a . c VAR b .
ENDFOR</lang>
Forth-style solution
<lang MontiLang>def over
swap dup rot swap
enddef
|Enter a number to obtain Fibonacci sequence: | input nip var count . 0 1 FOR count
over out |, | out . + swap
ENDFOR . print input clear</lang>
Simpler
<lang MontiLang>|Enter a number to obtain Fibonacci sequence: | input nip 1 - var count . 0 1 FOR count
out |, | out . dup rot +
ENDFOR print input /# wait until press ENTER #/ clear /# empties the stack #/</lang>
MUMPS
Iterative
<lang MUMPS>FIBOITER(N)
;Iterative version to get the Nth Fibonacci number ;N must be a positive integer ;F is the tree containing the values ;I is a loop variable. QUIT:(N\1'=N)!(N<0) "Error: "_N_" is not a positive integer." NEW F,I SET F(0)=0,F(1)=1 QUIT:N<2 F(N) FOR I=2:1:N SET F(I)=F(I-1)+F(I-2) QUIT F(N)</lang>
USER>W $$FIBOITER^ROSETTA(30) 832040
Nanoquery
Iterative
<lang nanoquery>def fibIter(n)
if (n < 2) return n end if $fib = 1 $fibPrev = 1
for num in range(2, n - 1)
fib += fibPrev fibPrev = fib - fibPrev end for return fib
end</lang>
Nemerle
Recursive
<lang Nemerle>using System; using System.Console;
module Fibonacci {
Fibonacci(x : long) : long { |x when x < 2 => x |_ => Fibonacci(x - 1) + Fibonacci(x - 2) } Main() : void { def num = Int64.Parse(ReadLine()); foreach (n in $[0 .. num]) WriteLine("{0}: {1}", n, Fibonacci(n)); }
}</lang>
Tail Recursive
<lang Nemerle>Fibonacci(x : long, current : long, next : long) : long {
match(x) { |0 => current |_ => Fibonacci(x - 1, next, current + next) }
}
Fibonacci(x : long) : long {
Fibonacci(x, 0, 1)
}</lang>
NESL
Recursive
<lang nesl>function fib(n) = if n < 2 then n else fib(n - 2) + fib(n - 1);</lang>
NetRexx
<lang NetRexx>/* NetRexx */
options replace format comments java crossref savelog symbols
numeric digits 210000 /*prepare for some big 'uns. */ parse arg x y . /*allow a single number or range.*/ if x == then do /*no input? Then assume -30-->+30*/
x = -30 y = -x end
if y == then y = x /*if only one number, show fib(n)*/ loop k = x to y /*process each Fibonacci request.*/
q = fib(k) w = q.length /*if wider than 25 bytes, tell it*/ say 'Fibonacci' k"="q if w > 25 then say 'Fibonacci' k "has a length of" w end k
exit
/*-------------------------------------FIB subroutine (non-recursive)---*/ method fib(arg) private static
parse arg n na = n.abs
if na < 2 then return na /*handle special cases. */ a = 0 b = 1
loop j = 2 to na s = a + b a = b b = s end j
if n > 0 | na // 2 == 1 then return s /*if positive or odd negative... */ else return -s /*return a negative Fib number. */
</lang>
NewLISP
Iterative
<lang newLISP>(define (fibonacci n)
(let (L '(0 1)) (dotimes (i n) (setq L (list (L 1) (apply + L)))) (L 1)) )
</lang>
Recursive
<lang newLISP>(define (fibonacci n) (if (< n 2) 1
(+ (fibonacci (- n 1)) (fibonacci (- n 2)))))
</lang>
Matrix multiplication
<lang newLISP>(define (fibonacci n)
(letn (f '((0 1) (1 1)) fib f) (dotimes (i n) (set 'fib (multiply fib f))) (fib 0 1)) )
(print(fibonacci 10)) ;;89</lang>
NGS
Iterative
<lang NGS>F fib(n:Int) { n < 2 returns n local a = 1, b = 1 # i is automatically local because of for() for(i=2; i<n; i=i+1) { local next = a + b a = b b = next } b }</lang>
Nial
Iterative
On my machine, about 1.7s for 100,000 iterations, n=92. Maybe a few percent faster than iterative Python. Note that n>92 produces overflow; Python keeps going - single iteration with n=1,000,000 takes it about 15s.
<lang nial>fibi is op n {
if n<2 then n else x1:=0; x2:=1; for i with tell (n - 1) do x:=x1+x2; x1:=x2; x2:=x; endfor; x2 endif};</lang>
Iterative using fold. Slightly faster, <1.6s:
<lang nial>fibf is op n {1 pick ((n- 1) fold [1 pick, +] 0 1)};</lang>
Tacit verion of above. Slightly faster still, <1.4s:
<lang nial>fibf2 is 1 pick fold [1 pick, +] reverse (0 1 hitch) (-1+);</lang>
Recursive
Really slow (over 8s for single iteration, n=33). (Similar to time for recursive python version with n=37.)
<lang nial>fibr is op n {fork [2>, +, + [fibr (-1 +), fibr (-2 +)]] n};</lang>
...or tacit version. More than twice as fast (?) but still slow:
<lang nial>fibr2 is fork [2>, +, + [fibr2 (-1 +), fibr2 (-2 +)]];</lang>
Matrix
Matrix inner product (ip). This appears to be the fastest, about 1.0s for 100,000 iterations, n=92: Note that n>92 produces negative result.
<lang nial>fibm is op n {floor (0 1 pick (reduce ip (n reshape [2 2 reshape 1 1 1 0])))};</lang>
Could it look a little more like J? (Maybe 5% slower than above.)
<lang nial>$ is reshape; ~ is tr f op a b {b f a}; % Goes before verb, rather than after like in J; _ is floor; % Not really J, but J-ish? (Cannot redefine "<.".);
fibm2 is _(0 1 pick reduce ip([2 2$1 1 1 0](~$)));</lang>
Alternate, not involving replicating matrix n times, but maybe 50% slower than the fastest matrix version above - similar speed to iterative:
<lang nial>fibm3 is op n {a:=2 2$1 1 1 0; _(0 1 pick ((n- 1) fold (a ip) a))};</lang>
Nim
Analytic
<lang nim>proc Fibonacci(n: int): int64 =
var fn = float64(n) var p: float64 = (1.0 + sqrt(5.0)) / 2.0 var q: float64 = 1.0 / p return int64((pow(p, fn) + pow(q, fn)) / sqrt(5.0))</lang>
Iterative
<lang nim>proc Fibonacci(n: int): int =
var first = 0 second = 1
for i in 0 .. <n: swap first, second second += first
result = first</lang>
Recursive
<lang nim>proc Fibonacci(n: int): int64 =
if n <= 2: result = 1 else: result = Fibonacci(n - 1) + Fibonacci(n - 2)</lang>
Tail-recursive
<lang nim>proc Fibonacci(n: int, current: int64, next: int64): int64 =
if n == 0: result = current else: result = Fibonacci(n - 1, next, current + next)
proc Fibonacci(n: int): int64 =
result = Fibonacci(n, 0, 1)</lang>
Continuations
<lang nim>iterator fib: int {.closure.} =
var a = 0 var b = 1 while true: yield a swap a, b b = a + b
var f = fib for i in 0.. <10:
echo f()</lang>
Oberon-2
<lang oberon2> MODULE Fibonacci; IMPORT
Out := NPCT:Console;
PROCEDURE Fibs(VAR r: ARRAY OF LONGREAL); VAR
i: LONGINT;
BEGIN
r[0] := 1.0; r[1] := 1.0; FOR i := 2 TO LEN(r) - 1 DO r[i] := r[i - 2] + r[i - 1]; END
END Fibs;
PROCEDURE FibsR(n: LONGREAL): LONGREAL; BEGIN
IF n < 2. THEN RETURN n ELSE RETURN FibsR(n - 1) + FibsR(n - 2) END
END FibsR;
PROCEDURE Show(r: ARRAY OF LONGREAL); VAR
i: LONGINT;
BEGIN
Out.String("First ");Out.Int(LEN(r),0);Out.String(" Fibonacci numbers");Out.Ln; FOR i := 0 TO LEN(r) - 1 DO Out.LongRealFix(r[i],8,0) END; Out.Ln
END Show;
PROCEDURE Gen(s: LONGINT); VAR
x: POINTER TO ARRAY OF LONGREAL;
BEGIN
NEW(x,s); Fibs(x^); Show(x^)
END Gen;
PROCEDURE GenR(s: LONGINT); VAR
i: LONGINT;
BEGIN
Out.String("First ");Out.Int(s,0);Out.String(" Fibonacci numbers (Recursive)");Out.Ln; FOR i := 1 TO s DO Out.LongRealFix(FibsR(i),8,0) END; Out.Ln
END GenR;
BEGIN
Gen(10); Gen(20); GenR(10); GenR(20);
END Fibonacci. </lang>
- Output:
First 10 Fibonacci numbers 1. 1. 2. 3. 5. 8. 13. 21. 34. 55. First 20 Fibonacci numbers 1. 1. 2. 3. 5. 8. 13. 21. 34. 55. 89. 144. 233. 377. 610. 987. 1597. 2584. 4181. 6765. First 10 Fibonacci numbers (Recursive) 1. 1. 2. 3. 5. 8. 13. 21. 34. 55. First 20 Fibonacci numbers (Recursive) 1. 1. 2. 3. 5. 8. 13. 21. 34. 55. 89. 144. 233. 377. 610. 987. 1597. 2584. 4181. 6765.
Objeck
Recursive
<lang objeck>bundle Default {
class Fib { function : Main(args : String[]), Nil { for(i := 0; i <= 10; i += 1;) { Fib(i)->PrintLine(); }; } function : native : Fib(n : Int), Int { if(n < 2) { return n; }; return Fib(n-1) + Fib(n-2); } }
}</lang>
Objective-C
Recursive
<lang objc>-(long)fibonacci:(int)position {
long result = 0; if (position < 2) { result = position; } else { result = [self fibonacci:(position -1)] + [self fibonacci:(position -2)]; } return result;
}</lang>
Iterative
<lang objc>+(long)fibonacci:(int)index {
long beforeLast = 0, last = 1; while (index > 0) { last += beforeLast; beforeLast = last - beforeLast; --index; } return last;
}</lang>
OCaml
Iterative
<lang ocaml>let fib_iter n =
if n < 2 then n else let fib_prev = ref 1 and fib = ref 1 in for num = 2 to n - 1 do let temp = !fib in fib := !fib + !fib_prev; fib_prev := temp done; !fib</lang>
Recursive
<lang ocaml>let rec fib_rec n =
if n < 2 then n else fib_rec (n - 1) + fib_rec (n - 2)
let rec fib = function
0 -> 0 | 1 -> 1 | n -> if n > 0 then fib (n-1) + fib (n-2) else fib (n+2) - fib (n+1)
</lang>
Arbitrary Precision
Using OCaml's Num module.
<lang ocaml>open Num
let fib =
let rec fib_aux f0 f1 = function | 0 -> f0 | 1 -> f1 | n -> fib_aux f1 (f1 +/ f0) (n - 1) in fib_aux (num_of_int 0) (num_of_int 1)
(* support for negatives *) let fib n =
if n < 0 && n mod 2 = 0 then minus_num (fib (abs n)) else fib (abs n)
(* It can be called from the command line with an argument *) (* Result is send to standart output *) let n = int_of_string Sys.argv.(1) in print_endline (string_of_num (fib n)) </lang>
compile with:
ocamlopt nums.cmxa -o fib fib.ml
Output:
$ ./fib 0 0 $ ./fib 10 55 $ ./fib 399 108788617463475645289761992289049744844995705477812699099751202749393926359816304226 $ ./fib -6 -8
O(log(n)) with arbitrary precision
This performs log2(N) matrix multiplys. Each multiplication is not constant-time but increases sub-linearly, about O(log(N)). <lang ocaml>open Num
let mul (a,b,c) (d,e,f) = let bxe = b*/e in
(a*/d +/ bxe, a*/e +/ b*/f, bxe +/ c*/f)
let id = (Int 1, Int 0, Int 1) let rec pow a n =
if n=0 then id else let b = pow a (n/2) in if (n mod 2) = 0 then mul b b else mul a (mul b b)
let fib n =
let (_,y,_) = (pow (Int 1, Int 1, Int 0) n) in string_of_num y
Printf.printf "fib %d = %s\n" 300 (fib 300)</lang>
Output:
fib 300 = 222232244629420445529739893461909967206666939096499764990979600
Octave
Recursive <lang octave>% recursive function fibo = recfibo(n)
if ( n < 2 ) fibo = n; else fibo = recfibo(n-1) + recfibo(n-2); endif
endfunction</lang>
Iterative <lang octave>% iterative function fibo = iterfibo(n)
if ( n < 2 ) fibo = n; else f = zeros(2,1); f(1) = 0; f(2) = 1; for i = 2 : n t = f(2); f(2) = f(1) + f(2); f(1) = t; endfor fibo = f(2); endif
endfunction</lang>
Testing <lang octave>% testing for i = 0 : 20
printf("%d %d\n", iterfibo(i), recfibo(i));
endfor</lang>
Oforth
<lang Oforth>: fib 0 1 rot #[ tuck + ] times drop ;</lang>
OPL
<lang opl>FIBON: REM Fibonacci sequence is generated to the Organiser II floating point variable limit. REM CLEAR/ON key quits. REM Mikesan - http://forum.psion2.org/ LOCAL A,B,C A=1 :B=1 :C=1 PRINT A, DO
C=A+B A=B B=C PRINT A,
UNTIL GET=1</lang>
Order
Recursive
<lang c>#include <order/interpreter.h>
- define ORDER_PP_DEF_8fib_rec \
ORDER_PP_FN(8fn(8N, \
8if(8less(8N, 2), \ 8N, \ 8add(8fib_rec(8sub(8N, 1)), \ 8fib_rec(8sub(8N, 2))))))
ORDER_PP(8fib_rec(10))</lang>
Tail recursive version (example supplied with language): <lang c>#include <order/interpreter.h>
- define ORDER_PP_DEF_8fib \
ORDER_PP_FN(8fn(8N, \
8fib_iter(8N, 0, 1)))
- define ORDER_PP_DEF_8fib_iter \
ORDER_PP_FN(8fn(8N, 8I, 8J, \
8if(8is_0(8N), \ 8I, \ 8fib_iter(8dec(8N), 8J, 8add(8I, 8J)))))
ORDER_PP(8to_lit(8fib(8nat(5,0,0))))</lang>
Memoization
<lang c>#include <order/interpreter.h>
- define ORDER_PP_DEF_8fib_memo \
ORDER_PP_FN(8fn(8N, \
8tuple_at(0, 8fib_memo_inner(8N, 8seq))))
- define ORDER_PP_DEF_8fib_memo_inner \
ORDER_PP_FN(8fn(8N, 8M, \
8cond((8less(8N, 8seq_size(8M)), 8pair(8seq_at(8N, 8M), 8M)) \ (8equal(8N, 0), 8pair(0, 8seq(0))) \ (8equal(8N, 1), 8pair(1, 8seq(0, 1))) \ (8else, \ 8lets((8S, 8fib_memo_inner(8sub(8N, 2), 8M)) \ (8T, 8fib_memo_inner(8dec(8N), 8tuple_at(1, 8S))) \ (8U, 8add(8tuple_at(0, 8S), 8tuple_at(0, 8T))), \ 8pair(8U, \ 8seq_append(8tuple_at(1, 8T), 8seq(8U))))))))
ORDER_PP( 8for_each_in_range(8fn(8N,
8print(8to_lit(8fib_memo(8N)) (,) 8space)), 1, 21)
)</lang>
Oz
Iterative
Using mutable references (cells). <lang oz>fun{FibI N}
Temp = {NewCell 0} A = {NewCell 0} B = {NewCell 1}
in
for I in 1..N do Temp := @A + @B A := @B B := @Temp end @A
end</lang>
Recursive
Inefficient (blows up the stack). <lang oz>fun{FibR N}
if N < 2 then N else {FibR N-1} + {FibR N-2} end
end</lang>
Tail-recursive
Using accumulators. <lang oz>fun{Fib N}
fun{Loop N A B} if N == 0 then
B
else
{Loop N-1 A+B A}
end end
in
{Loop N 1 0}
end</lang>
Lazy-recursive
<lang oz>declare
fun lazy {FiboSeq} {LazyMap {Iterate fun {$ [A B]} [B A+B] end [0 1]} Head} end
fun {Head A|_} A end
fun lazy {Iterate F I} I|{Iterate F {F I}} end
fun lazy {LazyMap Xs F} case Xs of X|Xr then {F X}|{LazyMap Xr F} [] nil then nil end end
in
{Show {List.take {FiboSeq} 8}}</lang>
PARI/GP
Built-in
<lang parigp>fibonocci(n)</lang>
Matrix
<lang parigp>fib(n)=([1,1;1,0]^n)[1,2]</lang>
Analytic
This uses the Binet form. <lang parigp>fib(n)=my(phi=(1+sqrt(5))/2);round((phi^n-phi^-n)/sqrt(5))</lang> The second term can be dropped since the error is always small enough to be subsumed by the rounding. <lang parigp>fib(n)=round(((1+sqrt(5))/2)^n/sqrt(5))</lang>
Algebraic
This is an exact version of the above formula. quadgen(5)
represents and the number is stored in the form . imag
takes the coefficient of . This uses the relation
and hence real(quadgen(5)^n)
would give the (n-1)-th Fibonacci number.
<lang parigp>fib(n)=imag(quadgen(5)^n)</lang>
A more direct translation (note that ) would be <lang parigp>fib(n)=my(phi=quadgen(5));(phi^n-(-1/phi)^n)/(2*phi-1)</lang>
Combinatorial
This uses the generating function. It can be trivially adapted to give the first n Fibonacci numbers. <lang parigp>fib(n)=polcoeff(x/(1-x-x^2)+O(x^(n+1)),n)</lang>
Binary powering
<lang parigp>fib(n)={
if(n<=0, if(n,(-1)^(n+1)*fib(n),0) , my(v=lucas(n-1)); (2*v[1]+v[2])/5 )
}; lucas(n)={
if (!n, return([2,1])); my(v=lucas(n >> 1), z=v[1], t=v[2], pr=v[1]*v[2]); n=n%4; if(n%2, if(n==3,[v[1]*v[2]+1,v[2]^2-2],[v[1]*v[2]-1,v[2]^2+2]) , if(n,[v[1]^2+2,v[1]*v[2]+1],[v[1]^2-2,v[1]*v[2]-1]) )
};</lang>
Recursive
<lang parigp>fib(n)={
if(n<2, n , fib(n-1)+fib(n) )
};</lang>
Anonymous recursion
This uses self()
which gives a self-reference.
<lang parigp>fib(n)={
if(n<2, n , my(s=self()); s(n-2)+s(n-1) )
};</lang>
It can be used without being named: <lang parigp>apply(n->if(n<2,n,my(s=self());s(n-2)+s(n-1)), [1..10])</lang> gives
- Output:
%1 = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
Memoization
<lang parigp>F=[]; fib(n)={
if(n>#F, F=concat(F, vector(n-#F)); F[n]=fib(n-1)+fib(n-2) , if(n<2, n , if(F[n],F[n],F[n]=fib(n-1)+fib(n-2)) ) );
}</lang>
Iterative
<lang parigp>fib(n)={
if(n<0,return((-1)^(n+1)*fib(n))); my(a=0,b=1,t); while(n, t=a+b; a=b; b=t; n-- ); a
};</lang>
Chebyshev
This solution uses Chebyshev polynomials of the second kind (Chyebyshev U-polynomials). <lang parigp>fib(n)=n--;polchebyshev(n,2,I/2)*I^n;</lang> or <lang parigp>fib(n)=abs(polchebyshev(n-1,2,I/2));</lang>
Anti-Hadamard matrix
All n×n (0,1) lower Hessenberg matrices have determinant at most F(n). The n×n anti-Hadamard matrix[1] matches this upper bound, and hence can be used as an inefficient method for computing Fibonacci numbers of positive index. These matrices are the same as Matlab's type-3 "Dramadah" matrices, following a naming suggestion of C. L. Mallows according to Graham & Sloane.
<lang parigp>matantihadamard(n)={
matrix(n,n,i,j, my(t=j-i+1); if(t<1,t%2,t<3) );
} fib(n)=matdet(matantihadamard(n))</lang>
Testing adjacent bits
The Fibonacci numbers can be characterized (for n > 0) as the number of n-bit strings starting and ending with 1 without adjacent 0s. This inefficient, exponential-time algorithm demonstrates: <lang parigp>fib(n)= {
my(g=2^(n+1)-1); sum(i=2^(n-1),2^n-1, bitor(i,i<<1)==g );
}</lang>
One-by-one
This code is purely for amusement and requires n > 1. It tests numbers in order to see if they are Fibonacci numbers, and waits until it has seen n of them. <lang parigp>fib(n)=my(k=0);while(n--,k++;while(!issquare(5*k^2+4)&&!issquare(5*k^2-4),k++));k</lang>
Pascal
Analytic
<lang pascal>function fib(n: integer):longInt; const
Sqrt5 = sqrt(5.0); C1 = ln((Sqrt5+1.0)*0.5);//ln( 1.618..)
//C2 = ln((1.0-Sqrt5)*0.5);//ln(-0.618 )) tsetsetse
C2 = ln((Sqrt5-1.0)*0.5);//ln(+0.618 ))
begin
IF n>0 then begin IF odd(n) then fib := round((exp(C1*n) + exp(C2*n) )/Sqrt5) else fib := round((exp(C1*n) - exp(C2*n) )/Sqrt5) end else Fibdirekt := 0
end;</lang>
Recursive
<lang pascal>function fib(n: integer): integer;
begin if (n = 0) or (n = 1) then fib := n else fib := fib(n-1) + fib(n-2) end;</lang>
Iterative
<lang pascal>function fib(n: integer): integer; var
f0, f1, tmpf0, k: integer;
begin
f1 := n; IF f1 >1 then begin k := f1-1; f0 := 0; f1 := 1; repeat tmpf0 := f0; f0 := f1; f1 := f1+tmpf0; dec(k); until k = 0; end else IF f1 < 0 then f1 := 0; fib := f1;
end;</lang>
Analytic2
<lang pascal>function FiboMax(n: integer):Extended; //maXbox begin
result:= (pow((1+SQRT5)/2,n)-pow((1-SQRT5)/2,n))/SQRT5
end;</lang>
<lang pascal>function Fibo_BigInt(n: integer): string; //maXbox
var tbig1, tbig2, tbig3: TInteger; begin result:= '0' tbig1:= TInteger.create(1); //temp tbig2:= TInteger.create(0); //result (a) tbig3:= Tinteger.create(1); //b for it:= 1 to n do begin tbig1.assign(tbig2)
tbig2.assign(tbig3); tbig1.add(tbig3); tbig3.assign(tbig1); end;
result:= tbig2.toString(false) tbig3.free; tbig2.free; tbig1.free; end;</lang>
writeln(floattoStr(FiboMax(555))) >>>4.3516638122555E115
writeln(Fibo_BigInt(555)) >>>43516638122555047989641805373140394725407202037260729735885664398655775748034950972577909265605502785297675867877570
Perl
Iterative
<lang perl>sub fib_iter {
my $n = shift; use bigint try => "GMP,Pari"; my ($v2,$v1) = (-1,1); ($v2,$v1) = ($v1,$v2+$v1) for 0..$n; $v1;
}</lang>
Recursive
<lang perl>sub fibRec {
my $n = shift; $n < 2 ? $n : fibRec($n - 1) + fibRec($n - 2);
}</lang>
Modules
Quite a few modules have ways to do this. Performance is not typically an issue with any of these until 100k or so. With GMP available, the first three are much faster at large values. <lang perl># Uses GMP method so very fast use Math::AnyNum qw/fibonacci/; say fibonacci(10000);
- Uses GMP method, so also very fast
use Math::GMP; say Math::GMP::fibonacci(10000);
- Binary ladder, GMP if available, Pure Perl otherwise
use ntheory qw/lucasu/; say lucasu(1, -1, 10000);
- All Perl
use Math::NumSeq::Fibonacci; my $seq = Math::NumSeq::Fibonacci->new; say $seq->ith(10000);
- All Perl
use Math::Big qw/fibonacci/; say 0+fibonacci(10000); # Force scalar context
- Perl, gives floating point *approximation*
use Math::Fibonacci qw/term/; say term(10000);</lang>
Phix
<lang Phix>function fibonacci(integer n) -- iterative, works for -ve numbers atom a=0, b=1
if n=0 then return 0 end if if abs(n)>=79 then ?9/0 end if -- inaccuracies creep in above 78 for i=1 to abs(n)-1 do {a,b} = {b,a+b} end for if n<0 and remainder(n,2)=0 then return -b end if return b
end function
for i=0 to 28 do
if i then puts(1,", ") end if printf(1,"%d", fibonacci(i))
end for puts(1,"\n")</lang>
- Output:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811
Using native integers/atoms, errors creep in above 78, so the same program converted to use mpfr:
<lang Phix>-- demo\rosetta\fibonacci.exw include mpfr.e
mpz res = NULL, prev, next integer lastn atom t0 = time()
function fibonampz(integer n) -- resumable, works for -ve numbers, yields mpz integer absn = abs(n)
if res=NULL or absn!=abs(lastn)+1 then if res=NULL then prev = mpz_init(0) res = mpz_init(1) next = mpz_init() else if n==lastn then return res end if end if mpz_fib2_ui(res,prev,absn) else if lastn<0 and remainder(lastn,2)=0 then mpz_mul_si(res,res,-1) end if mpz_add(next,res,prev) {prev,res,next} = {res,next,prev} end if if n<0 and remainder(n,2)=0 then mpz_mul_si(res,res,-1) end if lastn = n return res
end function
for i=0 to 28 do
if i then puts(1,", ") end if printf(1,"%s", {mpz_get_str(fibonampz(i))})
end for puts(1,"\n") printf(1,"%s\n", {mpz_get_str(fibonampz(705))})
string s = mpz_get_str(fibonampz(4784969)) integer l = length(s) s[40..-40] = "..." ?{l,s} ?elapsed(time()-t0) </lang>
- Output:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811 970066202977562212558683426760773016559904631977220423547980211057068777324159443678590358026859129109599109446646966713225742014317926940054191330 {1000000,"107273956418004772293648135962250043219...407167474856539211500699706378405156269"} "2.1s"
Phixmonti
<lang Phixmonti>def Fibonacci
dup 0 < if "Invalid argument: " print else 1 1 rot 2 - for drop over over + endfor endif
enddef
10 Fibonacci pstack print nl -10 Fibonacci print</lang>
PHP
Iterative
<lang php>function fibIter($n) {
if ($n < 2) { return $n; } $fibPrev = 0; $fib = 1; foreach (range(1, $n-1) as $i) { list($fibPrev, $fib) = array($fib, $fib + $fibPrev); } return $fib;
}</lang>
Recursive
<lang php>function fibRec($n) {
return $n < 2 ? $n : fibRec($n-1) + fibRec($n-2);
}</lang>
PicoLisp
Recursive
<lang PicoLisp>(de fibo (N)
(if (>= 2 N) 1 (+ (fibo (dec N)) (fibo (- N 2))) ) )</lang>
Recursive with Cache
Using a recursive version doesn't need to be slow, as the following shows: <lang PicoLisp>(de fibo (N)
(cache '(NIL) N # Use a cache to accelerate (if (>= 2 N) N (+ (fibo (dec N)) (fibo (- N 2))) ) ) )
(bench (fibo 1000))</lang> Output: <lang PicoLisp>0.012 sec -> 43466557686937456435688527675040625802564660517371780402481729089536555417949 05189040387984007925516929592259308032263477520968962323987332247116164299644090 6533187938298969649928516003704476137795166849228875</lang>
Iterative
<lang PicoLisp>(de fib (N)
(let (A 0 B 1) (do N (prog1 B (setq B (+ A B) A @)) ) ) )</lang>
Coroutines
<lang PicoLisp>(co 'fibo
(let (A 0 B 1) (yield 'ready) (while (yield (swap 'B (+ (swap 'A B) B)) ) ) ) )
(do 15
(printsp (yield 'next 'fibo)) )
(prinl) (yield NIL 'fibo)</lang>
- Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610
Pike
Iterative
<lang pike>int fibIter(int n) {
int fibPrev, fib, i; if (n < 2) { return 1; } fibPrev = 0; fib = 1; for (i = 1; i < n; i++) { int oldFib = fib; fib += fibPrev; fibPrev = oldFib; } return fib;
}</lang>
Recursive
<lang pike>int fibRec(int n) {
if (n < 2) { return(1); } return( fib(n-2) + fib(n-1) );
}</lang>
PIR
Recursive:
<lang pir>.sub fib
.param int n .local int nt .local int ft if n < 2 goto RETURNN nt = n - 1 ft = fib( nt ) dec nt nt = fib(nt) ft = ft + nt .return( ft )
RETURNN:
.return( n ) end
.end
.sub main :main
.local int counter .local int f counter=0
LOOP:
if counter > 20 goto DONE f = fib(counter) print f print "\n" inc counter goto LOOP
DONE:
end
.end</lang>
Iterative (stack-based):
<lang pir>.sub fib
.param int n .local int counter .local int f .local pmc fibs .local int nmo .local int nmt
fibs = new 'ResizableIntegerArray' if n == 0 goto RETURN0 if n == 1 goto RETURN1 push fibs, 0 push fibs, 1 counter = 2
FIBLOOP:
if counter > n goto DONE nmo = pop fibs nmt = pop fibs f = nmo + nmt push fibs, nmt push fibs, nmo push fibs, f inc counter goto FIBLOOP
RETURN0:
.return( 0 ) end
RETURN1:
.return( 1 ) end
DONE:
f = pop fibs .return( f ) end
.end
.sub main :main
.local int counter .local int f counter=0
LOOP:
if counter > 20 goto DONE f = fib(counter) print f print "\n" inc counter goto LOOP
DONE:
end
.end</lang>
PL/I
<lang pli>/* Form the n-th Fibonacci number, n > 1. */ get list(n); f1 = 0; f2 = 1; do i = 2 to n;
f3 = f1 + f2; put skip edit('fibo(',i,')=',f3)(a,f(5),a,f(5)); f1 = f2; f2 = f3;
end;</lang>
PL/pgSQL
Recursive
<lang SQL>CREATE OR REPLACE FUNCTION fib(n INTEGER) RETURNS INTEGER AS $$ BEGIN
IF (n < 2) THEN RETURN n; END IF; RETURN fib(n - 1) + fib(n - 2);
END; $$ LANGUAGE plpgsql;</lang>
Calculated
<lang SQL>CREATE OR REPLACE FUNCTION fibFormula(n INTEGER) RETURNS INTEGER AS $$ BEGIN
RETURN round(pow((pow(5, .5) + 1) / 2, n) / pow(5, .5));
END; $$ LANGUAGE plpgsql;</lang>
Linear
<lang SQL>CREATE OR REPLACE FUNCTION fibLinear(n INTEGER) RETURNS INTEGER AS $$ DECLARE
prevFib INTEGER := 0; fib INTEGER := 1;
BEGIN
IF (n < 2) THEN RETURN n; END IF;
WHILE n > 1 LOOP SELECT fib, prevFib + fib INTO prevFib, fib; n := n - 1; END LOOP;
RETURN fib;
END; $$ LANGUAGE plpgsql;</lang>
Tail recursive
<lang SQL>CREATE OR REPLACE FUNCTION fibTailRecursive(n INTEGER, prevFib INTEGER DEFAULT 0, fib INTEGER DEFAULT 1) RETURNS INTEGER AS $$ BEGIN
IF (n = 0) THEN RETURN prevFib; END IF; RETURN fibTailRecursive(n - 1, fib, prevFib + fib);
END; $$ LANGUAGE plpgsql;</lang>
PL/SQL
<lang PL/SQL>Create or replace Function fnu_fibonnaci(p_iNumber integer) return integer is
nuFib integer; nuP integer; nuQ integer;
Begin
if p_iNumber is not null then if p_iNumber=0 then nuFib:=0; Elsif p_iNumber=1 then nuFib:=1; Else nuP:=0; nuQ:=1; For nuI in 2..p_iNumber loop nuFib:=nuP+nuQ; nuP:=nuQ; nuQ:=nuFib; End loop; End if; End if; return(nuFib);
End fnu_fibonnaci;</lang>
Pop11
<lang pop11>define fib(x); lvars a , b;
1 -> a; 1 -> b; repeat x - 1 times (a + b, b) -> (b, a); endrepeat; a;
enddefine;</lang>
PostScript
Enter the desired number for "n" and run through your favorite postscript previewer or send to your postscript printer:
<lang postscript>%!PS
% We want the 'n'th fibonacci number /n 13 def
% Prepare output canvas: /Helvetica findfont 20 scalefont setfont 100 100 moveto
%define the function recursively: /fib { dup
3 lt { pop 1 } { dup 1 sub fib exch 2 sub fib add } ifelse } def (Fib\() show n (....) cvs show (\)=) show n fib (.....) cvs show
showpage</lang>
Potion
Recursive
Starts with int and upgrades on-the-fly to doubles. <lang potion>recursive = (n):
if (n <= 1): 1. else: recursive (n - 1) + recursive (n - 2)..
n = 40 ("fib(", n, ")= ", recursive (n), "\n") join print</lang>
recursive(40)= 165580141 real 0m2.851s
Iterative
<lang potion>iterative = (n) :
curr = 0 prev = 1 tmp = 0 n times: tmp = curr curr = curr + prev prev = tmp . curr
.</lang>
Matrix based
<lang potion>sqr = (x): x * x.
- Based on the fact that
- F2n = Fn(2Fn+1 - Fn)
- F2n+1 = Fn ^2 + Fn+1 ^2
matrix = (n) :
algorithm = (n) : "computes (Fn, Fn+1)" if (n < 2): return ((0, 1), (1, 1)) at(n). # n = e + {0, 1} q = algorithm(n / 2) # q = (Fe/2, Fe/2+1) q = (q(0) * (2 * q(1) - q(0)), sqr(q(0)) + sqr(q(1))) # q => (Fe, Fe+1) if (n % 2 == 1) : # q => (Fe+{0, 1}, Fe+1+{0,1}) = (Fn, Fn+1) q = (q(1), q(1) + q(0)) . q . algorithm(n)(0)
.</lang>
Handling negative values
<lang>fibonacci = (n) :
myFavorite = matrix if (n >= 0) : myFavorite(n) . else : n = n * -1 if (n % 2 == 1) : myFavorite(n) . else : myFavorite(n) * -1 . .
.</lang>
PowerBASIC
There seems to be a limitation (dare I say, bug?) in PowerBASIC regarding how large numbers are stored. 10E17 and larger get rounded to the nearest 10. For F(n), where ABS(n) > 87, is affected like this:
actual: displayed: F(88) 1100087778366101931 1100087778366101930 F(89) 1779979416004714189 1779979416004714190 F(90) 2880067194370816120 2880067194370816120 F(91) 4660046610375530309 4660046610375530310 F(92) 7540113804746346429 7540113804746346430
<lang powerbasic>FUNCTION fibonacci (n AS LONG) AS QUAD
DIM u AS LONG, a AS LONG, L0 AS LONG, outP AS QUAD STATIC fibNum() AS QUAD
u = UBOUND(fibNum) a = ABS(n)
IF u < 1 THEN REDIM fibNum(1) fibNum(1) = 1 u = 1 END IF
SELECT CASE a CASE 0 TO 92 IF a > u THEN REDIM PRESERVE fibNum(a) FOR L0 = u + 1 TO a fibNum(L0) = fibNum(L0 - 1) + fibNum(L0 - 2) IF 88 = L0 THEN fibNum(88) = fibNum(88) + 1 NEXT END IF IF n < 0 THEN fibonacci = fibNum(a) * ((-1)^(a+1)) ELSE fibonacci = fibNum(a) END IF CASE ELSE 'Even without the above-mentioned bug, we're still limited to 'F(+/-92), due to data type limits. (F(93) = &hA94F AD42 221F 2702) ERROR 6 END SELECT
END FUNCTION
FUNCTION PBMAIN () AS LONG
DIM n AS LONG #IF NOT %DEF(%PB_CC32) OPEN "out.txt" FOR OUTPUT AS 1 #ENDIF FOR n = -92 TO 92 #IF %DEF(%PB_CC32) PRINT STR$(n); ": "; FORMAT$(fibonacci(n), "#") #ELSE PRINT #1, STR$(n) & ": " & FORMAT$(fibonacci(n), "#") #ENDIF NEXT CLOSE
END FUNCTION</lang>
PowerShell
Iterative
<lang powershell> function FibonacciNumber ( $count ) {
$answer = @(0,1) while ($answer.Length -le $count) { $answer += $answer[-1] + $answer[-2] } return $answer
} </lang>
An even shorter version that eschews function calls altogether:
<lang powershell> $count = 8 $answer = @(0,1) 0..($count - $answer.Length) | Foreach { $answer += $answer[-1] + $answer[-2] } $answer </lang>
Recursive
<lang powershell>function fib($n) {
switch ($n) { 0 { return 0 } 1 { return 1 } { $_ -lt 0 } { return [Math]::Pow(-1, -$n + 1) * (fib (-$n)) } default { return (fib ($n - 1)) + (fib ($n - 2)) } }
}</lang>
Processing
<lang processing>void setup() {
size(400, 400); fill(255, 64); frameRate(2);
} void draw() {
int num = fibonacciNum(frameCount); println(frameCount, num); rect(0,0,num, num); if(frameCount==14) frameCount = -1; // restart
} int fibonacciNum(int n) {
return (n < 2) ? n : fibonacciNum(n - 1) + fibonacciNum(n - 2);
}</lang>
On the nth frame, the nth Fibonacci number is printed to the console and a square of that size is drawn on the sketch surface. The sketch restarts to keep drawing within the window size.
- Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377
Prolog
<lang prolog> fib(1, 1) :- !. fib(0, 0) :- !. fib(N, Value) :-
A is N - 1, fib(A, A1), B is N - 2, fib(B, B1), Value is A1 + B1.
</lang>
This naive implementation works, but is very slow for larger values of N. Here are some simple measurements (in SWI-Prolog): <lang prolog>?- time(fib(0,F)). % 2 inferences, 0.000 CPU in 0.000 seconds (88% CPU, 161943 Lips) F = 0.
?- time(fib(10,F)). % 265 inferences, 0.000 CPU in 0.000 seconds (98% CPU, 1458135 Lips) F = 55.
?- time(fib(20,F)). % 32,836 inferences, 0.016 CPU in 0.016 seconds (99% CPU, 2086352 Lips) F = 6765.
?- time(fib(30,F)). % 4,038,805 inferences, 1.122 CPU in 1.139 seconds (98% CPU, 3599899 Lips) F = 832040.
?- time(fib(40,F)). % 496,740,421 inferences, 138.705 CPU in 140.206 seconds (99% CPU, 3581264 Lips) F = 102334155.</lang>
As you can see, the calculation time goes up exponentially as N goes higher.
Poor man's memoization
The performance problem can be readily fixed by the addition of two lines of code (the first and last in this version): <lang prolog>%:- dynamic fib/2. % This is ISO, but GNU doesn't like it.
- - dynamic(fib/2). % Not ISO, but works in SWI, YAP and GNU unlike the ISO declaration.
fib(1, 1) :- !. fib(0, 0) :- !. fib(N, Value) :-
A is N - 1, fib(A, A1), B is N - 2, fib(B, B1), Value is A1 + B1, asserta((fib(N, Value) :- !)).</lang>
Let's take a look at the execution costs now:
<lang prolog>?- time(fib(0,F)). % 2 inferences, 0.000 CPU in 0.000 seconds (90% CPU, 160591 Lips) F = 0.
?- time(fib(10,F)). % 37 inferences, 0.000 CPU in 0.000 seconds (96% CPU, 552610 Lips) F = 55.
?- time(fib(20,F)). % 41 inferences, 0.000 CPU in 0.000 seconds (96% CPU, 541233 Lips) F = 6765.
?- time(fib(30,F)). % 41 inferences, 0.000 CPU in 0.000 seconds (95% CPU, 722722 Lips) F = 832040.
?- time(fib(40,F)). % 41 inferences, 0.000 CPU in 0.000 seconds (96% CPU, 543572 Lips) F = 102334155.</lang>
In this case by asserting the new N,Value pairing as a rule in the database we're making the classic time/space tradeoff. Since the space costs are (roughly) linear by N and the time costs are exponential by N, the trade-off is desirable. You can see the poor man's memoizing easily:
<lang prolog>?- listing(fib).
- - dynamic fib/2.
fib(40, 102334155) :- !. fib(39, 63245986) :- !. fib(38, 39088169) :- !. fib(37, 24157817) :- !. fib(36, 14930352) :- !. fib(35, 9227465) :- !. fib(34, 5702887) :- !. fib(33, 3524578) :- !. fib(32, 2178309) :- !. fib(31, 1346269) :- !. fib(30, 832040) :- !. fib(29, 514229) :- !. fib(28, 317811) :- !. fib(27, 196418) :- !. fib(26, 121393) :- !. fib(25, 75025) :- !. fib(24, 46368) :- !. fib(23, 28657) :- !. fib(22, 17711) :- !. fib(21, 10946) :- !. fib(20, 6765) :- !. fib(19, 4181) :- !. fib(18, 2584) :- !. fib(17, 1597) :- !. fib(16, 987) :- !. fib(15, 610) :- !. fib(14, 377) :- !. fib(13, 233) :- !. fib(12, 144) :- !. fib(11, 89) :- !. fib(10, 55) :- !. fib(9, 34) :- !. fib(8, 21) :- !. fib(7, 13) :- !. fib(6, 8) :- !. fib(5, 5) :- !. fib(4, 3) :- !. fib(3, 2) :- !. fib(2, 1) :- !. fib(1, 1) :- !. fib(0, 0) :- !. fib(A, D) :- B is A+ -1, fib(B, E), C is A+ -2, fib(C, F), D is E+F, asserta((fib(A, D):-!)).</lang>
All of the interim N/Value pairs have been asserted as facts for quicker future use, speeding up the generation of the higher Fibonacci numbers.
Continuation passing style
Works with SWI-Prolog and module lambda, written by Ulrich Neumerkel found there http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl <lang Prolog>:- use_module(lambda). fib(N, FN) :- cont_fib(N, _, FN, \_^Y^_^U^(U = Y)).
cont_fib(N, FN1, FN, Pred) :- ( N < 2 -> call(Pred, 0, 1, FN1, FN) ; N1 is N - 1, P = \X^Y^Y^U^(U is X + Y), cont_fib(N1, FNA, FNB, P), call(Pred, FNA, FNB, FN1, FN) ). </lang>
With lazy lists
Works with SWI-Prolog and others that support freeze/2
.
<lang Prolog>fib([0,1|X]) :-
ffib(0,1,X).
ffib(A,B,X) :-
freeze(X, (C is A+B, X=[C|Y], ffib(B,C,Y)) ).</lang>
The predicate fib(Xs)
unifies Xs
with an infinite list whose values are the Fibonacci sequence. The list can be used like this:
<lang Prolog>?- fib(X), length(A,15), append(A,_,X), writeln(A). [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]</lang>
Generators idiom
<lang Prolog>take( 0, Next, Z-Z, Next). take( N, Next, [A|B]-Z, NZ):- N>0, !, next( Next, A, Next1),
N1 is N-1, take( N1, Next1, B-Z, NZ).
next( fib(A,B), A, fib(B,C)):- C is A+B.
%% usage: ?- take(15, fib(0,1), _X-[], G), writeln(_X). %% [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377] %% G = fib(610, 987)</lang>
Yet another implementation
One of my favorites; loosely similar to the first example, but without the performance penalty, and needs nothing special to implement. Not even a dynamic database predicate. Attributed to M.E. for the moment, but simply because I didn't bother to search for the many people who probably did it like this long before I did. If someone knows who came up with it first, please let us know.
<lang Prolog>% Fibonacci sequence generator fib(C, [P,S], C, N) :- N is P + S. fib(C, [P,S], Cv, V) :- succ(C, Cn), N is P + S, !, fib(Cn, [S,N], Cv, V).
fib(0, 0). fib(1, 1). fib(C, N) :- fib(2, [0,1], C, N). % Generate from 3rd sequence on</lang> Looking at performance:
?- time(fib(30,X)). % 86 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 832040 ?- time(fib(40,X)). % 116 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 102334155 ?- time(fib(100,X)). % 296 inferences, 0.000 CPU in 0.001 seconds (0% CPU, Infinite Lips) X = 354224848179261915075
What I really like about this one, is it is also a generator- i.e. capable of generating all the numbers in sequence needing no bound input variables or special Prolog predicate support (such as freeze/3 in the previous example):
?- time(fib(X,Fib)). % 0 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = Fib, Fib = 0 ; % 1 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = Fib, Fib = 1 ; % 3 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 2, Fib = 1 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 3, Fib = 2 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 4, Fib = 3 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = Fib, Fib = 5 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 6, Fib = 8 ...etc.
It stays at 5 inferences per iteration after X=3. Also, quite useful:
?- time(fib(100,354224848179261915075)). % 296 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) true . ?- time(fib(X,354224848179261915075)). % 394 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 100 .
Pure
Tail Recursive
<lang pure>fib n = loop 0 1 n with
loop a b n = if n==0 then a else loop b (a+b) (n-1);
end;</lang>
PureBasic
Macro based calculation
<lang PureBasic>Macro Fibonacci (n) Int((Pow(((1+Sqr(5))/2),n)-Pow(((1-Sqr(5))/2),n))/Sqr(5)) EndMacro</lang>
Recursive
<lang PureBasic>Procedure FibonacciReq(n)
If n<2 ProcedureReturn n Else ProcedureReturn FibonacciReq(n-1)+FibonacciReq(n-2) EndIf
EndProcedure</lang>
Recursive & optimized with a static hash table
This will be much faster on larger n's, this as it uses a table to store known parts instead of recalculating them. On my machine the speedup compares to above code is
Fib(n) Speedup 20 2 25 23 30 217 40 25847 46 1156741
<lang PureBasic>Procedure Fibonacci(n)
Static NewMap Fib.i() Protected FirstRecursion If MapSize(Fib())= 0 ; Init the hash table the first run Fib("0")=0: Fib("1")=1 FirstRecursion = #True EndIf If n >= 2 Protected.s s=Str(n) If Not FindMapElement(Fib(),s) ; Calculate only needed parts Fib(s)= Fibonacci(n-1)+Fibonacci(n-2) EndIf n = Fib(s) EndIf If FirstRecursion ; Free the memory when finalizing the first call ClearMap(Fib()) EndIf ProcedureReturn n
EndProcedure</lang>
Example
Fibonacci(0)= 0 Fibonacci(1)= 1 Fibonacci(2)= 1 Fibonacci(3)= 2 Fibonacci(4)= 3 Fibonacci(5)= 5 FibonacciReq(0)= 0 FibonacciReq(1)= 1 FibonacciReq(2)= 1 FibonacciReq(3)= 2 FibonacciReq(4)= 3 FibonacciReq(5)= 5
Purity
The following takes a natural number and generates an initial segment of the Fibonacci sequence of that length:
<lang Purity> data Fib1 = FoldNat
< const (Cons One (Cons One Empty)), (uncurry Cons) . ((uncurry Add) . (Head, Head . Tail), id) >
</lang>
This following calculates the Fibonacci sequence as an infinite stream of natural numbers:
<lang Purity> type (Stream A?,,,Unfold) = gfix X. A? . X? data Fib2 = Unfold ((outl, (uncurry Add, outl))) ((curry id) One One) </lang>
As a histomorphism:
<lang Purity> import Histo
data Fib3 = Histo . Memoize
< const One, (p1 => < const One, (p2 => Add (outl $p1) (outl $p2)). UnmakeCofree > (outr $p1)) . UnmakeCofree >
</lang>
Python
Iterative positive and negative
<lang python>def fib(n,x=[0,1]):
for i in range(abs(n)-1): x=[x[1],sum(x)] return x[1]*pow(-1,abs(n)-1) if n<0 else x[1] if n else 0
for i in range(-30,31): print fib(i),</lang> Output:
-832040 514229 -317811 196418 -121393 75025 -46368 28657 -17711 10946 -6765 4181 -2584 1597 -987 610 -377 233 -144 89 -55 34 -21 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
Analytic
Binet's formula: <lang python>from math import *
def analytic_fibonacci(n):
sqrt_5 = sqrt(5); p = (1 + sqrt_5) / 2; q = 1/p; return int( (p**n + q**n) / sqrt_5 + 0.5 )
for i in range(1,31):
print analytic_fibonacci(i),</lang>
Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
Iterative
<lang python>def fibIter(n):
if n < 2: return n fibPrev = 1 fib = 1 for num in range(2, n): fibPrev, fib = fib, fib + fibPrev return fib</lang>
Recursive
<lang python>def fibRec(n):
if n < 2: return n else: return fibRec(n-1) + fibRec(n-2)</lang>
Recursive with Memoization
<lang python>def fibMemo():
pad = {0:0, 1:1} def func(n): if n not in pad: pad[n] = func(n-1) + func(n-2) return pad[n] return func
fm = fibMemo() for i in range(1,31):
print fm(i),</lang>
Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
Better Recursive doesn't need Memoization
The recursive code as written two sections above is incredibly slow and inefficient due to the nested recursion calls. Although the memoization above makes the code run faster, it is at the cost of extra memory use. The below code is syntactically recursive but actually encodes the efficient iterative process, and thus doesn't require memoization:
<lang python>def fibFastRec(n):
def fib(prvprv, prv, c): if c < 1: return prvprv else: return fib(prv, prvprv + prv, c - 1) return fib(0, 1, n)</lang>
However, although much faster and not requiring memory, the above code can only work to a limited 'n' due to the limit on stack recursion depth by Python; it is better to use the iterative code above or the generative one below.
Generative
<lang python>def fibGen(n):
a, b = 0, 1 while n>0: yield a a, b, n = b, a+b, n-1</lang>
Example use
<lang python> >>> [i for i in fibGen(11)]
[0,1,1,2,3,5,8,13,21,34,55] </lang>
Matrix-Based
Translation of the matrix-based approach used in F#. <lang python> def prevPowTwo(n):
'Gets the power of two that is less than or equal to the given input' if ((n & -n) == n): return n else: n -= 1 n |= n >> 1 n |= n >> 2 n |= n >> 4 n |= n >> 8 n |= n >> 16 n += 1 return (n/2)
def crazyFib(n):
'Crazy fast fibonacci number calculation' powTwo = prevPowTwo(n) q = r = i = 1 s = 0 while(i < powTwo): i *= 2 q, r, s = q*q + r*r, r * (q + s), (r*r + s*s) while(i < n): i += 1 q, r, s = q+r, q, r return q
</lang>
Large step recurrence
This is much faster for a single, large value of n: <lang python>def fib(n, c={0:1, 1:1}):
if n not in c: x = n // 2 c[n] = fib(x-1) * fib(n-x-1) + fib(x) * fib(n - x) return c[n]
fib(10000000) # calculating it takes a few seconds, printing it takes eons</lang>
Same as above but slightly faster
Putting the dictionary outside the function makes this about 2 seconds faster, could just make a wrapper: <lang python>F = {0: 0, 1: 1, 2: 1} def fib(n):
if n in F: return F[n] f1 = fib(n // 2 + 1) f2 = fib((n - 1) // 2) F[n] = (f1 * f1 + f2 * f2 if n & 1 else f1 * f1 - f2 * f2) return F[n]</lang>
Generative with Recursion
This can get very slow and uses a lot of memory. Can be sped up by caching the generator results. <lang python>def fib():
"""Yield fib[n+1] + fib[n]""" yield 1 # have to start somewhere lhs, rhs = fib(), fib() yield next(lhs) # move lhs one iteration ahead while True: yield next(lhs)+next(rhs)
f=fib() print [next(f) for _ in range(9)]</lang>
Output:
[1, 1, 2, 3, 5, 8, 13, 21, 34]
Another version of recursive generators solution, starting from 0 <lang Python>from itertools import islice
def fib():
yield 0 yield 1 a, b = fib(), fib() next(b) while True: yield next(a)+next(b)
print(tuple(islice(fib(), 10)))</lang>
As a scan or a fold
itertools.accumulate
The Fibonacci series can be defined quite simply and efficiently as a scan or accumulation, in which the accumulator is a pair of the two last numbers.
<lang python>Fibonacci accumulation
from itertools import accumulate, chain from operator import add
- fibs :: Integer :: [Integer]
def fibs(n):
An accumulation of the first n integers in the Fibonacci series. The accumulator is a pair of the two preceding numbers. def go(ab, _): return ab[1], add(*ab)
return [xy[1] for xy in accumulate( chain( [(0, 1)], range(1, n) ), go )]
- MAIN ---
if __name__ == '__main__':
print( 'First twenty: ' + repr( fibs(20) ) )</lang>
- Output:
First twenty: [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]
functools.reduce
A fold can be understood as an amnesic scan, and functools.reduce can provide a useful and efficient re-write of the scanning version above, if we only need the Nth term in the series:
<lang python>Nth Fibonacci term (by folding)
from functools import reduce from operator import add
- nthFib :: Integer -> Integer
def nthFib(n):
Nth integer in the Fibonacci series. def go(ab, _): return ab[1], add(*ab) return reduce(go, range(1, n), (0, 1))[1]
- MAIN ---
if __name__ == '__main__':
print( '1000th term: ' + repr( nthFib(1000) ) )</lang>
- Output:
1000th term: 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
QB64
CBTJD: 2020/03/13 <lang qbasic>_DEFINE F AS _UNSIGNED _INTEGER64 CLS PRINT PRINT "Enter 40 to more easily see the difference in calculation speeds." PRINT INPUT "Enter n for Fibonacci(n): ", n PRINT PRINT " Analytic Method (Fastest): F("; LTRIM$(STR$(n)); ") ="; fA(n) PRINT "Iterative Method (Fast): F("; LTRIM$(STR$(n)); ") ="; fI(n) PRINT "Recursive Method (Slow): F("; LTRIM$(STR$(n)); ") ="; fR(n) END
' === Analytic Fibonacci Function (Fastest) FUNCTION fA (n)
fA = INT(0.5 + (((SQR(5) + 1) / 2) ^ n) / SQR(5))
END FUNCTION
' === Iterative Fibonacci Function (Fast) FUNCTION fI (n)
FOR i = 1 TO n IF i < 3 THEN a = 1: b = 1 t = fI + b: fI = b: b = t NEXT
END FUNCTION
' === Recursive Fibonacci function (Slow) FUNCTION fR (n)
IF n <= 1 THEN fR = n ELSE fR = fR(n - 1) + fR(n - 2) END IF
END FUNCTION </lang>
Qi
Recursive
<lang qi> (define fib
0 -> 0 1 -> 1 N -> (+ (fib-r (- N 1)) (fib-r (- N 2))))
</lang>
Iterative
<lang qi> (define fib-0
V2 V1 0 -> V2 V2 V1 N -> (fib-0 V1 (+ V2 V1) (1- N)))
(define fib
N -> (fib-0 0 1 N))
</lang>
R
Iterative positive and negative
<lang python>fib=function(n,x=c(0,1)) {
if (abs(n)>1) for (i in seq(abs(n)-1)) x=c(x[2],sum(x)) if (n<0) return(x[2]*(-1)^(abs(n)-1)) else if (n) return(x[2]) else return(0)
}
sapply(seq(-31,31),fib)</lang> Output:
[1] 1346269 -832040 514229 -317811 196418 -121393 75025 -46368 28657 [10] -17711 10946 -6765 4181 -2584 1597 -987 610 -377 [19] 233 -144 89 -55 34 -21 13 -8 5 [28] -3 2 -1 1 0 1 1 2 3 [37] 5 8 13 21 34 55 89 144 233 [46] 377 610 987 1597 2584 4181 6765 10946 17711 [55] 28657 46368 75025 121393 196418 317811 514229 832040 1346269
Other methods
<lang R># recursive recfibo <- function(n) {
if ( n < 2 ) n else Recall(n-1) + Recall(n-2)
}
- print the first 21 elements
print.table(lapply(0:20, recfibo))
- iterative
iterfibo <- function(n) {
if ( n < 2 ) n else { f <- c(0, 1) for (i in 2:n) { t <- f[2] f[2] <- sum(f) f[1] <- t } f[2] }
}
print.table(lapply(0:20, iterfibo))
- iterative but looping replaced by map-reduce'ing
funcfibo <- function(n) {
if (n < 2) n else { generator <- function(f, ...) { c(f[2], sum(f)) } Reduce(generator, 2:n, c(0,1))[2] }
}
print.table(lapply(0:20, funcfibo))</lang>
Note that an idiomatic way to implement such low level, basic arithmethic operations in R is to implement them C and then call the compiled code.
- Output:
All three solutions print
[1] 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 [16] 610 987 1597 2584 4181 6765
Ra
<lang Ra> class FibonacciSequence **Prints the nth fibonacci number**
on start
args := program arguments
if args empty print .fibonacci(8)
else
try print .fibonacci(integer.parse(args[0]))
catch FormatException print to Console.error made !, "Input must be an integer" exit program with error code
catch OverflowException print to Console.error made !, "Number too large" exit program with error code
define fibonacci(n as integer) as integer is shared **Returns the nth fibonacci number**
test assert fibonacci(0) = 0 assert fibonacci(1) = 1 assert fibonacci(2) = 1 assert fibonacci(3) = 2 assert fibonacci(4) = 3 assert fibonacci(5) = 5 assert fibonacci(6) = 8 assert fibonacci(7) = 13 assert fibonacci(8) = 21
body
a, b := 0, 1
for n a, b := b, a + b
return a </lang>
Racket
Tail Recursive
<lang Racket> (define (fib n)
(let loop ((cnt 0) (a 0) (b 1)) (if (= n cnt) a (loop (+ cnt 1) b (+ a b)))))
</lang>
<lang Racket> (define (fib n (a 0) (b 1))
(if (< n 2) 1 (+ a (fib (- n 1) b (+ a b)))))
</lang>
Matrix Form
<lang Racket>
- lang racket
(require math/matrix)
(define (fibmat n) (matrix-ref
(matrix-expt (matrix ([1 1] [1 0])) n) 1 0))
(fibmat 1000) </lang>
Raku
(formerly Perl 6)
List Generator
This constructs the fibonacci sequence as a lazy infinite list. <lang perl6>constant @fib = 0, 1, *+* ... *;</lang>
If you really need a function for it: <lang perl6>sub fib ($n) { @fib[$n] }</lang>
To support negative indices: <lang perl6>constant @neg-fib = 0, 1, *-* ... *; sub fib ($n) { $n >= 0 ?? @fib[$n] !! @neg-fib[-$n] }</lang>
Iterative
<lang perl6>sub fib (Int $n --> Int) {
$n > 1 or return $n; my ($prev, $this) = 0, 1; ($prev, $this) = $this, $this + $prev for 1 ..^ $n; return $this;
}</lang>
Recursive
<lang perl6>proto fib (Int $n --> Int) {*} multi fib (0) { 0 } multi fib (1) { 1 } multi fib ($n) { fib($n - 1) + fib($n - 2) }</lang>
Analytic
<lang perl6>sub fib (Int $n --> Int) {
constant φ1 = 1 / constant φ = (1 + sqrt 5)/2; constant invsqrt5 = 1 / sqrt 5;
floor invsqrt5 * (φ**$n + φ1**$n);
}</lang>
Retro
Recursive
<lang Retro>: fib ( n-m ) dup [ 0 = ] [ 1 = ] bi or if; [ 1- fib ] sip [ 2 - fib ] do + ;</lang>
Iterative
<lang Retro>: fib ( n-N )
[ 0 1 ] dip [ over + swap ] times drop ;</lang>
REXX
With 210,000 numeric decimal digits, this REXX program can handle Fibonacci numbers past one million.
[Generally speaking, some REXX interpreters can handle up to around eight million decimal digits.]
This version of the REXX program can also handle negative Fibonacci numbers. <lang rexx>/*REXX program calculates the Nth Fibonacci number, N can be zero or negative. */ numeric digits 210000 /*be able to handle ginormous numbers. */ parse arg x y . /*allow a single number or a range. */ if x== | x=="," then do; x=-40; y=+40; end /*No input? Then use range -40 ──► +40*/ if y== | y=="," then y=x /*if only one number, display fib(X).*/ w= max(length(x), length(y) ) /*W: used for making formatted output.*/ fw= 10 /*Minimum maximum width. Sounds ka─razy*/
do j=x to y; q= fib(j) /*process all of the Fibonacci requests*/ L= length(q) /*obtain the length (decimal digs) of Q*/ fw= max(fw, L) /*fib number length, or the max so far.*/ say 'Fibonacci('right(j,w)") = " right(q,fw) /*right justify Q.*/ if L>10 then say 'Fibonacci('right(j, w)") has a length of" L end /*j*/ /* [↑] list a Fib. sequence of x──►y */
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ fib: procedure; parse arg n; an= abs(n) /*use │n│ (the absolute value of N).*/
a= 0; b= 1; if an<2 then return an /*handle two special cases: zero & one.*/ /* [↓] this method is non─recursive. */ do k=2 to an; $= a+b; a= b; b= $ /*sum the numbers up to │n│ */ end /*k*/ /* [↑] (only positive Fibs nums used).*/ /* [↓] an//2 [same as] (an//2==1).*/ if n>0 | an//2 then return $ /*Positive or even? Then return sum. */ return -$ /*Negative and odd? Return negative sum*/</lang>
- output when using the default inputs:
Fibonacci(-40) = -102334155 Fibonacci(-39) = 63245986 Fibonacci(-38) = -39088169 Fibonacci(-37) = 24157817 Fibonacci(-36) = -14930352 Fibonacci(-35) = 9227465 Fibonacci(-34) = -5702887 Fibonacci(-33) = 3524578 Fibonacci(-32) = -2178309 Fibonacci(-31) = 1346269 Fibonacci(-30) = -832040 Fibonacci(-29) = 514229 Fibonacci(-28) = -317811 Fibonacci(-27) = 196418 Fibonacci(-26) = -121393 Fibonacci(-25) = 75025 Fibonacci(-24) = -46368 Fibonacci(-23) = 28657 Fibonacci(-22) = -17711 Fibonacci(-21) = 10946 Fibonacci(-20) = -6765 Fibonacci(-19) = 4181 Fibonacci(-18) = -2584 Fibonacci(-17) = 1597 Fibonacci(-16) = -987 Fibonacci(-15) = 610 Fibonacci(-14) = -377 Fibonacci(-13) = 233 Fibonacci(-12) = -144 Fibonacci(-11) = 89 Fibonacci(-10) = -55 Fibonacci( -9) = 34 Fibonacci( -8) = -21 Fibonacci( -7) = 13 Fibonacci( -6) = -8 Fibonacci( -5) = 5 Fibonacci( -4) = -3 Fibonacci( -3) = 2 Fibonacci( -2) = -1 Fibonacci( -1) = 1 Fibonacci( 0) = 0 Fibonacci( 1) = 1 Fibonacci( 2) = 1 Fibonacci( 3) = 2 Fibonacci( 4) = 3 Fibonacci( 5) = 5 Fibonacci( 6) = 8 Fibonacci( 7) = 13 Fibonacci( 8) = 21 Fibonacci( 9) = 34 Fibonacci( 10) = 55 Fibonacci( 11) = 89 Fibonacci( 12) = 144 Fibonacci( 13) = 233 Fibonacci( 14) = 377 Fibonacci( 15) = 610 Fibonacci( 16) = 987 Fibonacci( 17) = 1597 Fibonacci( 18) = 2584 Fibonacci( 19) = 4181 Fibonacci( 20) = 6765 Fibonacci( 21) = 10946 Fibonacci( 22) = 17711 Fibonacci( 23) = 28657 Fibonacci( 24) = 46368 Fibonacci( 25) = 75025 Fibonacci( 26) = 121393 Fibonacci( 27) = 196418 Fibonacci( 28) = 317811 Fibonacci( 29) = 514229 Fibonacci( 30) = 832040 Fibonacci( 31) = 1346269 Fibonacci( 32) = 2178309 Fibonacci( 33) = 3524578 Fibonacci( 34) = 5702887 Fibonacci( 35) = 9227465 Fibonacci( 36) = 14930352 Fibonacci( 37) = 24157817 Fibonacci( 38) = 39088169 Fibonacci( 39) = 63245986 Fibonacci( 40) = 102334155
output when the following was used as input: 10000
Fibonacci(10000) = 3364476487643178326662161200510754331030214846068006390656476997468008144216666236815559551363373402558206533268083615937373479048386526826304089246305643188735454436955982749160660209988418393386465273130008883026923567361313511757929743785441375213052050434770160226475831890652789085515436615958298727968298751063120057542878345321551510387081829896979161312785626503319548714021428753269818796204693609787990035096230229102636813149319527563022783762844154036058440257211433496118002309120828704608892396232883546150577658327125254609359112820392528539343462090424524892940390 170623388899108584106518317336043747073790855263176432573399371287193758774689747992630583706574283016163740896917842637862421283525811282051637029808933209990570792006436742620238978311147005407499845925036063356093388383192338678305613643535189213327973290813373264265263398976392272340788292817795358057099369104917547080893184105614632233821746563732124822638309210329770164805472624384237486241145309381220656491403275108664339451751216152654536133311131404243685480510676584349352383695965342807176877532834823434555736671973139274627362910821067928078471803532913117677892465908993863545932789 452377767440619224033763867400402133034329749690202832814593341882681768389307200363479562311710310129195316979460763273758925353077255237594378843450406771555577905645044301664011946258097221672975861502696844314695203461493229110597067624326851599283470989128470674086200858713501626031207190317208609408129832158107728207635318662461127824553720853236530577595643007251774431505153960090516860322034916322264088524885243315805153484962243484829938090507048348244932745373262456775587908918719080366205800959474315005240253270974699531877072437682590741993963226598414749819360928522394503970716544 3156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875 Fibonacci(10000) has a length of 2090 decimal digits
Ring
<lang ring> give n x = fib(n) see n + " Fibonacci is : " + x
func fib nr if nr = 0 return 0 ok
if nr = 1 return 1 ok if nr > 1 return fib(nr-1) + fib(nr-2) ok
</lang>
Rockstar
Iterative (minimized)
<lang Rockstar> Fibonacci takes Number
FNow is 0 FNext is 1 While FNow is less than Number Say FNow Put FNow into Temp Put FNow into FNext Put FNext plus Temp into FNext
Say Fibonacci taking 1000 (prints out highest number in Fibonacci sequence less than 1000) </lang>
Iterative (idiomatic)
<lang Rockstar> Love takes Time My love was addictions Put my love into your heart Build it up Until my love is as strong as Time Whisper my love Put my love into a river Put your heart into my love Put it with a river into your heart
Shout; Love taking 1000 (years, years)
</lang>
The semicolon and the comment (years, years)
in this version are there only for poetic effect
Recursive
<lang Rockstar> The Italian takes a lover, a kiss, a promise love is population hate is information If a lover is love Give back a kiss
If a lover is hate Give back a promise
Knock a lover down Put a promise with a kiss into my heart Give back The Italian taking a lover, a promise, my heart
Listen to your heart your mind is everything your soul is opportunity Whisper The Italian taking your heart, your mind, your soul </lang>
Ruby
Iterative
<lang ruby>def fib(n)
if n < 2 n else prev, fib = 0, 1 (n-1).times do prev, fib = fib, fib + prev end fib end
end
p (0..10).map { |i| fib(i) }</lang> Output:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
Recursive
<lang ruby>def fib(n, sequence=[1])
return sequence.last if n == 0
current_number, last_number = sequence.last(2) sequence << current_number + (last_number or 0)
fib(n-1, sequence)
end </lang>
Recursive with Memoization
<lang ruby># Use the Hash#default_proc feature to
- lazily calculate the Fibonacci numbers.
fib = Hash.new do |f, n|
f[n] = if n <= -2 (-1)**(n + 1) * f[n.abs] elsif n <= 1 n.abs else f[n - 1] + f[n - 2] end
end
- examples: fib[10] => 55, fib[-10] => (-55/1)</lang>
Matrix
<lang ruby>require 'matrix'
- To understand why this matrix is useful for Fibonacci numbers, remember
- that the definition of Matrix.**2 for any Matrix[[a, b], [c, d]] is
- is [[a*a + b*c, a*b + b*d], [c*a + d*b, c*b + d*d]]. In other words, the
- lower right element is computing F(k - 2) + F(k - 1) every time M is multiplied
- by itself (it is perhaps easier to understand this by computing M**2, 3, etc, and
- watching the result march up the sequence of Fibonacci numbers).
M = Matrix[[0, 1], [1,1]]
- Matrix exponentiation algorithm to compute Fibonacci numbers.
- Let M be Matrix [[0, 1], [1, 1]]. Then, the lower right element of M**k is
- F(k + 1). In other words, the lower right element of M is F(2) which is 1, and the
- lower right element of M**2 is F(3) which is 2, and the lower right element
- of M**3 is F(4) which is 3, etc.
- This is a good way to compute F(n) because the Ruby implementation of Matrix.**(n)
- uses O(log n) rather than O(n) matrix multiplications. It works by squaring squares
- ((m**2)**2)... as far as possible
- and then multiplying that by by M**(the remaining number of times). E.g., to compute
- M**19, compute partial = ((M**2)**2) = M**16, and then compute partial*(M**3) = M**19.
- That's only 5 matrix multiplications of M to compute M*19.
def self.fib_matrix(n)
return 0 if n <= 0 # F(0) return 1 if n == 1 # F(1) # To get F(n >= 2), compute M**(n - 1) and extract the lower right element. return CS::lower_right(M**(n - 1))
end
- Matrix utility to return
- the lower, right-hand element of a given matrix.
def self.lower_right matrix
return nil if matrix.row_size == 0 return matrix[matrix.row_size - 1, matrix.column_size - 1]
end</lang>
Generative
<lang ruby>fib = Enumerator.new do |y|
f0, f1 = 0, 1 loop do y << f0 f0, f1 = f1, f0 + f1 end
end</lang>
Usage:
p fib.lazy.drop(8).next # => 21
"Fibers are primitives for implementing light weight cooperative concurrency in Ruby. Basically they are a means of creating code blocks that can be paused and resumed, much like threads. The main difference is that they are never preempted and that the scheduling must be done by the programmer and not the VM." [2]
<lang ruby>fib = Fiber.new do
a,b = 0,1 loop do Fiber.yield a a,b = b,a+b end
end 9.times {puts fib.resume}</lang>
using a lambda <lang ruby>def fib_gen
a, b = 1, 1 lambda {ret, a, b = a, b, a+b; ret}
end</lang>
irb(main):034:0> fg = fib_gen => #<Proc:0xb7cdf750@(irb):22> irb(main):035:0> 9.times { puts fg.call} 1 1 2 3 5 8 13 21 34 => 9
Binet's Formula
<lang ruby>def fib
phi = (1 + Math.sqrt(5)) / 2 ((phi**self - (-1 / phi)**self) / Math.sqrt(5)).to_i
end</lang>
1.9.3p125 :001 > def fib 1.9.3p125 :002?> phi = (1 + Math.sqrt(5)) / 2 1.9.3p125 :003?> ((phi**self - (-1 / phi)**self) / Math.sqrt(5)).to_i 1.9.3p125 :004?> end => nil 1.9.3p125 :005 > (0..10).map(&:fib) => [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
Run BASIC
<lang runbasic>for i = 0 to 10
print i;" ";fibR(i);" ";fibI(i)
next i end
function fibR(n) if n < 2 then fibR = n else fibR = fibR(n-1) + fibR(n-2) end function function fibI(n) b = 1 for i = 1 to n t = a + b a = b b = t next i
fibI = a end function</lang>
Rust
Iterative
<lang rust>use std::mem; fn main() {
let mut prev = 0; // Rust needs this type hint for the checked_add method let mut curr = 1usize;
while let Some(n) = curr.checked_add(prev) { prev = curr; curr = n; println!("{}", n); }
}</lang>
Recursive
<lang rust>use std::mem; fn main() {
fibonacci(0,1);
}
fn fibonacci(mut prev: usize, mut curr: usize) {
mem::swap(&mut prev, &mut curr); if let Some(n) = curr.checked_add(prev) { println!("{}", n); fibonacci(prev, n); }
}</lang>
Recursive (with pattern matching)
<lang rust>fn fib(n: u32) -> u32 {
match n { 0 => 0, 1 => 1, n => fib(n - 1) + fib(n - 2), }
}</lang>
Tail recursive (with pattern matching)
<lang rust>fn fib_tail_recursive(nth: usize) -> usize {
fn fib_tail_iter(n: usize, prev_fib: usize, fib: usize) -> usize { match n { 0 => prev_fib, n => fib_tail_iter(n - 1, fib, prev_fib + fib), } } fib_tail_iter(nth, 0, 1)
}</lang>
Analytic
This uses a feature from nightly Rust which makes it possible to (cleanly) return an iterator without the additional overhead of putting it on the heap. In stable Rust, we'd need to return a Box<Iterator<Item=u64>>
which has the cost of an additional allocation and the overhead of dynamic dispatch. The version below does not require the use of the heap and is done entirely through static dispatch.
<lang rust>#![feature(conservative_impl_trait)]
fn main() {
for num in fibonacci_gen(10) { println!("{}", num); }
}
fn fibonacci_gen(terms: i32) -> impl Iterator<Item=u64> {
let sqrt_5 = 5.0f64.sqrt(); let p = (1.0 + sqrt_5) / 2.0; let q = 1.0/p; (1..terms).map(move |n| ((p.powi(n) + q.powi(n)) / sqrt_5 + 0.5) as u64)
}</lang>
Using an Iterator
Iterators are very idiomatic in rust, though they may be overkill for such a simple problem. <lang rust>use std::mem;
struct Fib {
prev: usize, curr: usize,
}
impl Fib {
fn new() -> Self { Fib {prev: 0, curr: 1} }
}
impl Iterator for Fib {
type Item = usize; fn next(&mut self) -> Option<Self::Item>{ mem::swap(&mut self.curr, &mut self.prev); self.curr.checked_add(self.prev).map(|n| { self.curr = n; n }) }
}
fn main() {
for num in Fib::new() { println!("{}", num); }
}</lang>
S-BASIC
Note that the 23rd Fibonacci number (=28657) is the largest that can be generated without overflowing S-BASIC's integer data type. <lang basic> rem - iterative function to calculate nth fibonacci number function fibonacci(n = integer) = integer var f, i, p1, p2 = integer p1 = 0 p2 = 1 if n = 0 then
f = 0
else
for i = 1 to n f = p1 + p2 p2 = p1 p1 = f next i
end = f
rem - exercise the function var i = integer for i = 0 to 10
print fibonacci(i);
next i
end </lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55
SAS
Iterative
This code builds a table fib
holding the first few values of the Fibonacci sequence.
<lang sas>data fib;
a=0; b=1; do n=0 to 20; f=a; output; a=b; b=f+a; end; keep n f;
run;</lang>
Naive recursive
This code provides a simple example of defining a function and using it recursively. One of the members of the sequence is written to the log.
<lang sas>options cmplib=work.f;
proc fcmp outlib=work.f.p;
function fib(n); if n = 0 or n = 1 then return(1); else return(fib(n - 2) + fib(n - 1)); endsub;
run;
data _null_;
x = fib(5); put 'fib(5) = ' x;
run;</lang>
Sather
The implementations use the arbitrary precision class INTI. <lang sather>class MAIN is
-- RECURSIVE -- fibo(n :INTI):INTI pre n >= 0 is if n < 2.inti then return n; end; return fibo(n - 2.inti) + fibo(n - 1.inti); end;
-- ITERATIVE -- fibo_iter(n :INTI):INTI pre n >= 0 is n3w :INTI;
if n < 2.inti then return n; end; last ::= 0.inti; this ::= 1.inti; loop (n - 1.inti).times!; n3w := last + this; last := this; this := n3w; end; return this; end;
main is loop i ::= 0.upto!(16); #OUT + fibo(i.inti) + " "; #OUT + fibo_iter(i.inti) + "\n"; end; end;
end;</lang>
Scala
Recursive
<lang scala>def fib(i: Int): Int = i match {
case 0 => 0 case 1 => 1 case _ => fib(i - 1) + fib(i - 2)
}</lang>
Lazy sequence
<lang scala>lazy val fib: LazyList[Int] = 0 #:: 1 #:: fib.zip(fib.tail).map { case (a, b) => a + b }</lang>
Tail recursive
<lang scala>import scala.annotation.tailrec @tailrec final def fib(x: Int, prev: BigInt = 0, next: BigInt = 1): BigInt = x match {
case 0 => prev case _ => fib(x - 1, next, next + prev)
}</lang>
foldLeft
<lang scala>// Fibonacci using BigInt with LazyList.foldLeft optimized for GC (Scala v2.13 and above) // Does not run out of memory for very large Fibonacci numbers def fib(n: Int): BigInt = {
def series(i: BigInt, j: BigInt): LazyList[BigInt] = i #:: series(j, i + j)
series(1, 0).take(n).foldLeft(BigInt("0"))(_ + _)
}
// Small test (0 to 13) foreach {n => print(fib(n).toString + " ")}
// result: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 </lang>
Iterator
<lang scala>val it: Iterator[Int] = Iterator.iterate((0, 1)) { case (a, b) => (b, a + b) }.map(_._1)
def fib(n: Int): Int = it.drop(n).next()
// example: println(it.take(13).mkString(",")) // prints: 0,1,1,2,3,5,8,13,21,34,55,89,144</lang>
Scheme
Iterative
<lang scheme>(define (fib-iter n)
(do ((num 2 (+ num 1)) (fib-prev 1 fib) (fib 1 (+ fib fib-prev))) ((>= num n) fib)))</lang>
Recursive
<lang scheme>(define (fib-rec n)
(if (< n 2) n (+ (fib-rec (- n 1)) (fib-rec (- n 2)))))</lang>
This version is tail recursive: <lang scheme>(define (fib n)
(let loop ((a 0) (b 1) (n n)) (if (= n 0) a (loop b (+ a b) (- n 1)))))
</lang>
Recursive Sequence Generator
Although the tail recursive version above is quite efficient, it only generates the final nth Fibonacci number and not the sequence up to that number without wasteful repeated calls to the procedure/function.
The following procedure generates the sequence of Fibonacci numbers using a simplified version of a lazy list/stream - since no memoization is requried, it just implements future values by using a zero parameter lambda "thunk" with a closure containing the last and the pre-calculated next value of the sequence; in this way it uses almost no memory during the sequence generation other than as required for the last and the next values of the sequence (note that the test procedure does not generate a linked list to contain the elements of the sequence to show, but rather displays each one by one in sequence):
<lang scheme> (define (fib)
(define (nxt lv nv) (cons nv (lambda () (nxt nv (+ lv nv))))) (cons 0 (lambda () (nxt 0 1))))
- test...
(define (show-stream-take n strm)
(define (shw-nxt n strm) (begin (display (car strm)) (if (> n 1) (begin (display " ") (shw-nxt (- n 1) ((cdr strm)))) (display ")")))) (begin (display "(") (shw-nxt n strm)))
(show-stream-take 30 (fib))</lang>
- Output:
(0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229)
Dijkstra Algorithm
<lang scheme>;;; Fibonacci numbers using Edsger Dijkstra's algorithm
(define (fib n)
(define (fib-aux a b p q count) (cond ((= count 0) b) ((even? count) (fib-aux a b (+ (* p p) (* q q)) (+ (* q q) (* 2 p q)) (/ count 2))) (else (fib-aux (+ (* b q) (* a q) (* a p)) (+ (* b p) (* a q)) p q (- count 1))))) (fib-aux 1 0 0 1 n))</lang>
Scilab
<lang> clear
n=46 f1=0; f2=1 printf("fibo(%d)=%d\n",0,f1) printf("fibo(%d)=%d\n",1,f2) for i=2:n f3=f1+f2 printf("fibo(%d)=%d\n",i,f3) f1=f2 f2=f3 end</lang>
- Output:
... fibo(43)=433494437 fibo(44)=701408733 fibo(45)=1134903170 fibo(46)=1836311903
sed
<lang sed>#!/bin/sed -f
- First we need to convert each number into the right number of ticks
- Start by marking digits
s/[0-9]/<&/g
- We have to do the digits manually.
s/0//g; s/1/|/g; s/2/||/g; s/3/|||/g; s/4/||||/g; s/5/|||||/g s/6/||||||/g; s/7/|||||||/g; s/8/||||||||/g; s/9/|||||||||/g
- Multiply by ten for each digit from the front.
- tens
s/|</<||||||||||/g t tens
- Done with digit markers
s/<//g
- Now the actual work.
- split
- Convert each stretch of n >= 2 ticks into two of n-1, with a mark between
s/|\(|\+\)/\1-\1/g
- Convert the previous mark and the first tick after it to a different mark
- giving us n-1+n-2 marks.
s/-|/+/g
- Jump back unless we're done.
t split
- Get rid of the pluses, we're done with them.
s/+//g
- Convert back to digits
- back
s/||||||||||/</g s/<\([0-9]*\)$/<0\1/g s/|||||||||/9/g; s/|||||||||/9/g; s/||||||||/8/g; s/|||||||/7/g; s/||||||/6/g; s/|||||/5/g; s/||||/4/g; s/|||/3/g; s/||/2/g; s/|/1/g; s/</|/g t back s/^$/0/</lang>
Seed7
Recursive
<lang seed7>const func integer: fib (in integer: number) is func
result var integer: result is 1; begin if number > 2 then result := fib(pred(number)) + fib(number - 2); elsif number = 0 then result := 0; end if; end func;</lang>
Original source: [3]
Iterative
This funtion uses a bigInteger result:
<lang seed7>const func bigInteger: fib (in integer: number) is func
result var bigInteger: result is 1_; local var integer: i is 0; var bigInteger: a is 0_; var bigInteger: c is 0_; begin for i range 1 to pred(number) do c := a; a := result; result +:= c; end for; end func;</lang>
Original source: [4]
SequenceL
Recursive
<lang sequencel>fibonacci(n) := n when n < 2 else fibonacci(n - 1) + fibonacci(n - 2);</lang> Based on: [5]
Tail Recursive
<lang sequencel>fibonacci(n) := fibonacciHelper(0, 1, n);
fibonacciHelper(prev, next, n) := prev when n < 1 else next when n = 1 else fibonacciHelper(next, next + prev, n - 1);</lang>
Matrix
<lang sequencel>fibonacci(n) := fibonacciHelper([[1,0],[0,1]], n);
fibonacciHelper(M(2), n) := let N := [[1,1],[1,0]]; in M[1,1] when n <= 1 else fibonacciHelper(matmul(M, N), n - 1);
matmul(A(2), B(2)) [i,j] := sum( A[i,all] * B[all,j] );</lang> Based on the C# version: [6]
Using the SequenceL Matrix Multiply solution: [7]
SETL
<lang setl>$ Print out the first ten Fibonacci numbers $ This uses Set Builder Notation, it roughly means $ 'collect fib(n) forall n in {0,1,2,3,4,5,6,7,8,9,10}' print({fib(n) : n in {0..10}});
$ Iterative Fibonacci function proc fib(n);
A := 0; B := 1; C := n; for i in {0..n} loop C := A + B; A := B; B := C; end loop; return C;
end proc;</lang>
Shen
<lang Shen>(define fib
0 -> 0 1 -> 1 N -> (+ (fib (+ N 1)) (fib (+ N 2))) where (< N 0) N -> (+ (fib (- N 1)) (fib (- N 2))))</lang>
Sidef
Iterative
<lang ruby>func fib_iter(n) {
var (a, b) = (0, 1) { (a, b) = (b, a+b) } * n return a
}</lang>
Recursive
<lang ruby>func fib_rec(n) {
n < 2 ? n : (__FUNC__(n-1) + __FUNC__(n-2))
}</lang>
Recursive with memoization
<lang ruby>func fib_mem (n) is cached {
n < 2 ? n : (__FUNC__(n-1) + __FUNC__(n-2))
}</lang>
Closed-form
<lang ruby>func fib_closed(n) {
define S = (1.25.sqrt + 0.5) define T = (-S + 1) (S**n - T**n) / (-T + S) -> round
}</lang>
Built-in
<lang ruby>say fib(12) #=> 144</lang>
Simula
Straightforward iterative implementation. <lang simula>INTEGER PROCEDURE fibonacci(n); INTEGER n; BEGIN
INTEGER lo, hi, temp, i; lo := 0; hi := 1; FOR i := 1 STEP 1 UNTIL n - 1 DO BEGIN temp := hi; hi := hi + lo; lo := temp END; fibonacci := hi
END;</lang>
SkookumScript
Built-in
SkookumScript's Integer
class has a fast built-in fibonnaci()
method.
<lang javascript>42.fibonacci</lang>
SkookumScript is designed to work in tandem with C++ and its strength is at the high-level stage-direction of things. So when confronted with benchmarking scripting systems it is genrally better to make a built-in call. Though in most practical cases this isn't necessary.
Recursive
Simple recursive method in same 42.fibonacci
form as built-in form above.
<lang javascript>// Assuming code is in Integer.fibonacci() method () Integer
[ if this < 2 [this] else [[this - 1].fibonacci + [this - 2].fibonacci] ]</lang>
Recursive procedure in fibonacci(42)
form.
<lang javascript>// Assuming in fibonacci(n) procedure (Integer n) Integer
[ if n < 2 [n] else [fibonacci(n - 1) + fibonacci(n - 2)] ]</lang>
Iterative
Iterative method in 42.fibonacci
form.
<lang javascript>// Assuming code is in Integer.fibonacci() method () Integer
[ if this < 2 [this] else [ !prev: 1 !next: 1 2.to_pre this [ !sum : prev + next prev := next next := sum ] next ] ]</lang>
Optimized iterative method in 42.fibonacci
form.
Though the best optimiation is to write it in C++ as with the built-in form that comes with SkookumScript.
<lang javascript>// Bind : is faster than assignment := // loop is faster than to_pre (which uses a closure) () Integer
[ if this < 2 [this] else [ !prev: 1 !next: 1 !sum !count: this - 2 loop [ if count = 0 [exit] count-- sum : prev + next prev : next next : sum ] next ] ]</lang>
Slate
<lang slate>n@(Integer traits) fib [
n <= 0 ifTrue: [^ 0]. n = 1 ifTrue: [^ 1]. (n - 1) fib + (n - 2) fib
].
slate[15]> 10 fib = 55. True</lang>
Smalltalk
<lang smalltalk>|fibo| fibo := [ :i |
|ac t| ac := Array new: 2. ac at: 1 put: 0 ; at: 2 put: 1. ( i < 2 ) ifTrue: [ ac at: (i+1) ] ifFalse: [ 2 to: i do: [ :l | t := (ac at: 2). ac at: 2 put: ( (ac at: 1) + (ac at: 2) ). ac at: 1 put: t ]. ac at: 2. ]
].
0 to: 10 do: [ :i |
(fibo value: i) displayNl
]</lang>
smart BASIC
The Iterative method is slow (relatively) and the Recursive method doubly so since it references the Iterative function twice.
The N-th Term (fibN) function is much faster as it utilizes Binet's Formula.
- fibR: Fibonacci Recursive
- fibI: Fibonacci Iterative
- fibN: Fibonacci N-th Term
<lang qbasic>FOR i = 0 TO 15
PRINT fibR(i),fibI(i),fibN(i)
NEXT i
/* Recursive Method */ DEF fibR(n)
IF n <= 1 THEN fibR = n ELSE fibR = fibR(n-1) + fibR(n-2) ENDIF
END DEF
/* Iterative Method */ DEF fibI(n)
a = 0 b = 1 FOR i = 1 TO n temp = a + b a = b b = temp NEXT i fibI = a
END DEF
/* N-th Term Method */ DEF fibN(n)
uphi = .5 + SQR(5)/2 lphi = .5 - SQR(5)/2 fibN = (uphi^n-lphi^n)/SQR(5)
END DEF</lang>
SNOBOL4
Recursive
<lang snobol> define("fib(a)") :(fib_end) fib fib = lt(a,2) a :s(return) fib = fib(a - 1) + fib(a - 2) :(return) fib_end
while a = trim(input) :f(end) output = a " " fib(a) :(while) end</lang>
Tail-recursive
<lang SNOBOL4> define('trfib(n,a,b)') :(trfib_end) trfib trfib = eq(n,0) a :s(return)
trfib = trfib(n - 1, a + b, a) :(return)
trfib_end</lang>
Iterative
<lang SNOBOL4> define('ifib(n)f1,f2') :(ifib_end) ifib ifib = le(n,2) 1 :s(return)
f1 = 1; f2 = 1
if1 ifib = gt(n,2) f1 + f2 :f(return)
f1 = f2; f2 = ifib; n = n - 1 :(if1)
ifib_end</lang>
Analytic
Note: Snobol4+ lacks built-in sqrt( ) function.
<lang SNOBOL4> define('afib(n)s5') :(afib_end) afib s5 = sqrt(5)
afib = (((1 + s5) / 2) ^ n - ((1 - s5) / 2) ^ n) / s5 afib = convert(afib,'integer') :(return)
afib_end</lang>
Test and display all, Fib 1 .. 10
<lang SNOBOL4>loop i = lt(i,10) i + 1 :f(show)
s1 = s1 fib(i) ' ' ; s2 = s2 trfib(i,0,1) ' ' s3 = s3 ifib(i) ' '; s4 = s4 afib(i) ' ' :(loop)
show output = s1; output = s2; output = s3; output = s4 end</lang>
Output:
1 1 2 3 5 8 13 21 34 55 1 1 2 3 5 8 13 21 34 55 1 1 2 3 5 8 13 21 34 55 1 1 2 3 5 8 13 21 34 55
SNUSP
This is modular SNUSP (which introduces @ and # for threading).
Iterative
<lang snusp> @!\+++++++++# /<<+>+>-\ fib\==>>+<<?!/>!\ ?/\
#<</?\!>/@>\?-<<</@>/@>/>+<-\ \-/ \ !\ !\ !\ ?/#</lang>
Recursive
<lang snusp> /========\ />>+<<-\ />+<-\ fib==!/?!\-?!\->+>+<<?/>>-@\=====?/<@\===?/<#
| #+==/ fib(n-2)|+fib(n-1)| \=====recursion======/!========/</lang>
Softbridge BASIC
Iterative
<lang basic> Function Fibonacci(n)
x = 0 y = 1 i = 0 n = ABS(n) If n < 2 Then Fibonacci = n Else Do Until (i = n) sum = x+y x=y y=sum i=i+1 Loop Fibonacci = x End If
End Function </lang>
Spin
Iterative
<lang spin>con
_clkmode = xtal1 + pll16x _clkfreq = 80_000_000
obj
ser : "FullDuplexSerial.spin"
pub main | i
ser.start(31, 30, 0, 115200)
repeat i from 0 to 10 ser.dec(fib(i)) ser.tx(32)
waitcnt(_clkfreq + cnt) ser.stop cogstop(0)
pub fib(i) : b | a
b := a := 1 repeat i a := b + (b := a)</lang>
- Output:
1 1 2 3 5 8 13 21 34 55 89
SPL
Analytic
<lang spl>fibo(n)=
s5 = #.sqrt(5) <= (((1+s5)/2)^n-((1-s5)/2)^n)/s5
.</lang>
Iterative
<lang spl>fibo(n)=
? n<2, <= n f2 = 0 f1 = 1 > i, 2..n f = f1+f2 f2 = f1 f1 = f < <= f
.</lang>
Recursive
<lang spl>fibo(n)=
? n<2, <= n <= fibo(n-1)+fibo(n-2)
.</lang>
SQL
Analytic
As a running sum: <lang SQL> select round ( exp ( sum (ln ( ( 1 + sqrt( 5 ) ) / 2)
) over ( order by level ) ) / sqrt( 5 ) ) fibo
from dual connect by level <= 10; </lang>
FIB ---------- 1 1 2 3 5 8 13 21 34 55 10 rows selected.
As a power: <lang SQL> select round ( power( ( 1 + sqrt( 5 ) ) / 2, level ) / sqrt( 5 ) ) fib from dual connect by level <= 10; </lang>
FIB ---------- 1 1 2 3 5 8 13 21 34 55 10 rows selected.
Recursive
Oracle 12c required <lang sql> SQL> with fib(e,f) as (select 1, 1 from dual union all select e+f,e from fib where e <= 55) select f from fib;
F
1 1 2 3 5 8 13 21 34 55
10 rows selected. </lang>
<lang postgresql>CREATE FUNCTION fib(n int) RETURNS numeric AS $$
-- This recursive with generates endless list of Fibonacci numbers. WITH RECURSIVE fibonacci(current, previous) AS ( -- Initialize the current with 0, so the first value will be 0. -- The previous value is set to 1, because its only goal is not -- special casing the zero case, and providing 1 as the second -- number in the sequence. -- -- The numbers end with dots to make them numeric type in -- Postgres. Numeric type has almost arbitrary precision -- (technically just 131,072 digits, but that's good enough for -- most purposes, including calculating huge Fibonacci numbers) SELECT 0., 1. UNION ALL -- To generate Fibonacci number, we need to add together two -- previous Fibonacci numbers. Current number is saved in order -- to be accessed in the next iteration of recursive function. SELECT previous + current, current FROM fibonacci ) -- The user is only interested in current number, not previous. SELECT current FROM fibonacci -- We only need one number, so limit to 1 LIMIT 1 -- Offset the query by the requested argument to get the correct -- position in the list. OFFSET n
$$ LANGUAGE SQL RETURNS NULL ON NULL INPUT IMMUTABLE;</lang>
SSEM
Calculates the tenth Fibonacci number. To calculate the nth, change the initial value of the counter to n-1 (subject to the restriction that the answer must be small enough to fit in a signed 32-bit integer, the SSEM's only data type). The algorithm is basically straightforward, but the absence of an Add instruction makes the implementation a little more complicated than it would otherwise be. <lang ssem>10101000000000100000000000000000 0. -21 to c acc = -n 01101000000001100000000000000000 1. c to 22 temp = acc 00101000000001010000000000000000 2. Sub. 20 acc -= m 10101000000001100000000000000000 3. c to 21 n = acc 10101000000000100000000000000000 4. -21 to c acc = -n 10101000000001100000000000000000 5. c to 21 n = acc 01101000000000100000000000000000 6. -22 to c acc = -temp 00101000000001100000000000000000 7. c to 20 m = acc 11101000000000100000000000000000 8. -23 to c acc = -count 00011000000001010000000000000000 9. Sub. 24 acc -= -1 00000000000000110000000000000000 10. Test skip next if acc<0 10011000000000000000000000000000 11. 25 to CI goto (15 + 1) 11101000000001100000000000000000 12. c to 23 count = acc 11101000000000100000000000000000 13. -23 to c acc = -count 11101000000001100000000000000000 14. c to 23 count = acc 00011000000000000000000000000000 15. 24 to CI goto (-1 + 1) 10101000000000100000000000000000 16. -21 to c acc = -n 10101000000001100000000000000000 17. c to 21 n = acc 10101000000000100000000000000000 18. -21 to c acc = -n 00000000000001110000000000000000 19. Stop 00000000000000000000000000000000 20. 0 var m = 0 10000000000000000000000000000000 21. 1 var n = 1 00000000000000000000000000000000 22. 0 var temp = 0 10010000000000000000000000000000 23. 9 var count = 9 11111111111111111111111111111111 24. -1 const -1 11110000000000000000000000000000 25. 15 const 15</lang>
Stata
<lang stata>program fib args n clear qui set obs `n' qui gen a=1 qui replace a=a[_n-1]+a[_n-2] in 3/l end</lang>
An implementation using dyngen.
<lang stata>program fib args n clear qui set obs `n' qui gen a=. dyngen { update a=a[_n-1]+a[_n-2], missval(1) } end
fib 10 list</lang>
Output
+----+ | a | |----| 1. | 1 | 2. | 1 | 3. | 2 | 4. | 3 | 5. | 5 | |----| 6. | 8 | 7. | 13 | 8. | 21 | 9. | 34 | 10. | 55 | +----+
Mata
<lang stata>. mata
- function fib(n) {
return((((1+sqrt(5))/2):^n-((1-sqrt(5))/2):^n)/sqrt(5))
}
- fib(0..10)
1 2 3 4 5 6 7 8 9 10 11 +--------------------------------------------------------+ 1 | 0 1 1 2 3 5 8 13 21 34 55 | +--------------------------------------------------------+
- end</lang>
StreamIt
<lang streamit>void->int feedbackloop Fib {
join roundrobin(0,1); body in->int filter { work pop 1 push 1 peek 2 { push(peek(0) + peek(1)); pop(); } }; loop Identity<int>; split duplicate; enqueue(0); enqueue(1);
}</lang>
SuperCollider
Recursive
nth fibonacci term for positive n <lang SuperCollider> f = { |n| if(n < 2) { n } { f.(n-1) + f.(n-2) } }; (0..20).collect(f) </lang>
nth fibonacci term for positive and negative n. <lang SuperCollider> f = { |n| var u = neg(sign(n)); if(abs(n) < 2) { n } { f.(2 * u + n) + f.(u + n) } }; (-20..20).collect(f) </lang>
Analytic
<lang SuperCollider>( f = { |n| var sqrt5 = sqrt(5); var p = (1 + sqrt5) / 2; var q = reciprocal(p); ((p ** n) + (q ** n) / sqrt5 + 0.5).trunc }; (0..20).collect(f) )</lang>
Iterative
<lang SuperCollider> f = { |n| var a = [1, 1]; n.do { a = a.addFirst(a[0] + a[1]) }; a.reverse }; f.(18) </lang>
Swift
Analytic
<lang Swift>import Cocoa
func fibonacci(n: Int) -> Int {
let square_root_of_5 = sqrt(5.0) let p = (1 + square_root_of_5) / 2 let q = 1 / p return Int((pow(p,CDouble(n)) + pow(q,CDouble(n))) / square_root_of_5 + 0.5)
}
for i in 1...30 {
println(fibonacci(i))
}</lang>
Iterative
<lang Swift>func fibonacci(n: Int) -> Int {
if n < 2 { return n } var fibPrev = 1 var fib = 1 for num in 2...n { (fibPrev, fib) = (fib, fib + fibPrev) } return fib
}</lang> Sequence: <lang swift>func fibonacci() -> SequenceOf<UInt> {
return SequenceOf {() -> GeneratorOf<UInt> in var window: (UInt, UInt, UInt) = (0, 0, 1) return GeneratorOf { window = (window.1, window.2, window.1 + window.2) return window.0 } }
}</lang>
Recursive
<lang Swift>func fibonacci(n: Int) -> Int {
if n < 2 { return n } else { return fibonacci(n-1) + fibonacci(n-2) }
}
println(fibonacci(30))</lang>
Tailspin
Recursive simple
The simplest exponential-time recursive algorithm only handling positive N. Note that the "#" is the tailspin internal recursion which sends the value to the matchers. In this case where there is no initial block and no templates state, we could equivalently write the templates name "nthFibonacci" in place of the "#" to do a normal recursion. <lang tailspin> templates nthFibonacci
when <=0|=1> do $ ! otherwise ($ - 1 -> #) + ($ - 2 -> #) !
end nthFibonacci </lang>
Iterative, mutable state
We could use the templates internal mutable state, still only positive N. <lang tailspin> templates nthFibonacci
@: {n0: 0, n1: 1}; 1..$ -> @: {n0: $@.n1, n1: $@.n0 + $@.n1}; $@.n0!
end nthFibonacci </lang> To handle negatives, we can keep track of the sign and send it to the matchers. <lang tailspin> templates nthFibonacci
@: {n0: 0, n1: 1}; def sign: $ -> \(<0..> 1! <> -1!\); 1..$*$sign -> $sign -> # $@.n0! <=1> @: {n0: $@.n1, n1: $@.n0 + $@.n1}; <=-1> @: {n0: $@.n1 - $@.n0, n1: $@.n0};
end nthFibonacci </lang>
State machine
Instead of mutating state, we could just recurse internally on a state structure. <lang tailspin> templates nthFibonacci
{ N: $, n0: 0, n1: 1 } -> # <{ N: <=0> }> $.n0 ! <{ N: <1..>}> { N: $.N - 1, n0: $.n1, n1: $.n0 + $.n1} -> # <> { N: $.N + 1, n1: $.n0, n0: $.n1 - $.n0} -> #
end nthFibonacci
8 -> nthFibonacci -> '$; ' -> !OUT::write -5 -> nthFibonacci -> '$; ' -> !OUT::write -6 -> nthFibonacci -> '$; ' -> !OUT::write </lang>
- Output:
21 5 -8
Tcl
Simple Version
These simple versions do not handle negative numbers -- they will return N for N < 2
Iterative
<lang tcl>proc fibiter n {
if {$n < 2} {return $n} set prev 1 set fib 1 for {set i 2} {$i < $n} {incr i} { lassign [list $fib [incr fib $prev]] prev fib } return $fib
}</lang>
Recursive
<lang tcl>proc fib {n} {
if {$n < 2} then {expr {$n}} else {expr {[fib [expr {$n-1}]]+[fib [expr {$n-2}]]} }
}</lang>
The following
: defining a procedure in the ::tcl::mathfunc
namespace allows that proc to be used as a function in expr
expressions.
<lang tcl>proc tcl::mathfunc::fib {n} {
if { $n < 2 } { return $n } else { return [expr {fib($n-1) + fib($n-2)}] }
}
- or, more tersely
proc tcl::mathfunc::fib {n} {expr {$n<2 ? $n : fib($n-1) + fib($n-2)}}</lang>
E.g.:
<lang tcl>expr {fib(7)} ;# ==> 13
namespace path tcl::mathfunc #; or, interp alias {} fib {} tcl::mathfunc::fib fib 7 ;# ==> 13</lang>
Tail-Recursive
In Tcl 8.6 a tailcall function is available to permit writing tail-recursive functions in Tcl. This makes deeply recursive functions practical. The availability of large integers also means no truncation of larger numbers. <lang tcl>proc fib-tailrec {n} {
proc fib:inner {a b n} { if {$n < 1} { return $a } elseif {$n == 1} { return $b } else { tailcall fib:inner $b [expr {$a + $b}] [expr {$n - 1}] } } return [fib:inner 0 1 $n]
}</lang>
% fib-tailrec 100 354224848179261915075
Handling Negative Numbers
Iterative
<lang tcl>proc fibiter n {
if {$n < 0} { set n [expr {abs($n)}] set sign [expr {-1**($n+1)}] } else { set sign 1 } if {$n < 2} {return $n} set prev 1 set fib 1 for {set i 2} {$i < $n} {incr i} { lassign [list $fib [incr fib $prev]] prev fib } return [expr {$sign * $fib}]
} fibiter -5 ;# ==> 5 fibiter -6 ;# ==> -8</lang>
Recursive
<lang tcl>proc tcl::mathfunc::fib {n} {expr {$n<-1 ? -1**($n+1) * fib(abs($n)) : $n<2 ? $n : fib($n-1) + fib($n-2)}} expr {fib(-5)} ;# ==> 5 expr {fib(-6)} ;# ==> -8</lang>
For the Mathematically Inclined
This works up to , after which the limited precision of IEEE double precision floating point arithmetic starts to show.
<lang tcl>proc fib n {expr {round((.5 + .5*sqrt(5)) ** $n / sqrt(5))}}</lang>
Tern
Recursive
<lang tern>func fib(n) {
if (n < 2) { return 1; } return fib(n - 1) + fib(n - 2);
}</lang>
Coroutine
<lang tern>func fib(n) {
let a = 1; let b = 2; until(n-- <= 0) { yield a; (a, b) = (b, a + b); }
}</lang>
TI-83 BASIC
Iterative: <lang ti83b>{0,1 While 1 Disp Ans(1 {Ans(2),sum(Ans End</lang>
Binet's formula: <lang ti83b>Prompt N .5(1+√(5 //golden ratio (Ans^N–(-Ans)^-N)/√(5</lang>
TI-89 BASIC
Recursive
Optimized implementation (too slow to be usable for n higher than about 12).
<lang ti89b>fib(n) when(n<2, n, fib(n-1) + fib(n-2))</lang>
Iterative
Unoptimized implementation (I think the for loop can be eliminated, but I'm not sure).
<lang ti89b>fib(n) Func Local a,b,c,i 0→a 1→b For i,1,n
a→c b→a c+b→b
EndFor a EndFunc</lang>
TSE SAL
<lang TSE SAL>
// library: math: get: series: fibonacci <description></description> <version control></version control> <version>1.0.0.0.3</version> <version control></version control> (filenamemacro=getmasfi.s) [<Program>] [<Research>] [kn, ri, su, 20-01-2013 22:04:02] INTEGER PROC FNMathGetSeriesFibonacciI( INTEGER nI )
// // Method: // // 1. Take the sum of the last 2 terms // // 2. Let the sum be the last term // and goto step 1 // INTEGER I = 0 INTEGER minI = 1 INTEGER maxI = nI INTEGER term1I = 0 INTEGER term2I = 1 INTEGER term3I = 0 // FOR I = minI TO maxI // // make value 3 equal to sum of two previous values 1 and 2 // term3I = term1I + term2I // // make value 1 equal to next value 2 // term1I = term2I // // make value 2 equal to next value 3 // term2I = term3I // ENDFOR // RETURN( term3I ) //
END
PROC Main()
STRING s1[255] = "3" REPEAT IF ( NOT ( Ask( " = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF Warn( FNMathGetSeriesFibonacciI( Val( s1 ) ) ) // gives e.g. 3 UNTIL FALSE
END
</lang>
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT ASK "What fibionacci number do you want?": searchfib="" IF (searchfib!='digits') STOP Loop n=0,{searchfib}
IF (n==0) THEN fib=fiba=n ELSEIF (n==1) THEN fib=fibb=n ELSE fib=fiba+fibb, fiba=fibb, fibb=fib ENDIF IF (n!=searchfib) CYCLE PRINT "fibionacci number ",n,"=",fib
ENDLOOP </lang> Output:
What fibionacci number do you want? >12 fibionacci number 12=144
Output:
What fibionacci number do you want? >31 fibionacci number 31=1346269
Output:
What fibionacci number do you want? >46 fibionacci 46=1836311903
UNIX Shell
<lang bash>#!/bin/bash
a=0 b=1 max=$1
for (( n=1; "$n" <= "$max"; $((n++)) )) do
a=$(($a + $b)) echo "F($n): $a" b=$(($a - $b))
done</lang>
Recursive:
<lang bash>fib() {
local n=$1 [ $n -lt 2 ] && echo -n $n || echo -n $(( $( fib $(( n - 1 )) ) + $( fib $(( n - 2 )) ) ))
}</lang>
UnixPipes
<lang bash>echo 1 |tee last fib ; tail -f fib | while read x do
cat last | tee -a fib | xargs -n 1 expr $x + |tee last
done</lang>
Ursa
Iterative
<lang ursa>def fibIter (int n) if (< n 2) return n end if decl int fib fibPrev num set fib (set fibPrev 1) for (set num 2) (< num n) (inc num) set fib (+ fib fibPrev) set fibPrev (- fib fibPrev) end for return fib end</lang>
Ursala
All three methods are shown here, and all have unlimited precision. <lang Ursala>#import std
- import nat
iterative_fib = ~&/(0,1); ~&r->ll ^|\predecessor ^/~&r sum
recursive_fib = {0,1}^?<a/~&a sum^|W/~& predecessor^~/~& predecessor
analytical_fib =
%np+ -+
mp..round; ..mp2str; sep`+; ^CNC/~&hh take^\~&htt %np@t, (mp..div^|\~& mp..sub+ ~~ @rlX mp..pow_ui)^lrlPGrrPX/~& -+ ^\~& ^(~&,mp..sub/1.E0)+ mp..div\2.E0+ mp..add/1.E0, mp..sqrt+ ..grow/5.E0+-+-</lang>
The analytical method uses arbitrary precision floating point arithmetic from the mpfr library and then converts the result to a natural number. Sufficient precision for an exact result is always chosen based on the argument. This test program computes the first twenty Fibonacci numbers by all three methods. <lang Ursala>#cast %nLL
examples = <.iterative_fib,recursive_fib,analytical_fib>* iota20</lang> output:
< <0,0,0>, <1,1,1>, <1,1,1>, <2,2,2>, <3,3,3>, <5,5,5>, <8,8,8>, <13,13,13>, <21,21,21>, <34,34,34>, <55,55,55>, <89,89,89>, <144,144,144>, <233,233,233>, <377,377,377>, <610,610,610>, <987,987,987>, <1597,1597,1597>, <2584,2584,2584>, <4181,4181,4181>>
V
Generate n'th fib by using binary recursion
<lang v>[fib
[small?] [] [pred dup pred] [+] binrec].</lang>
Vala
Recursive
Using int, but could easily replace with double, long, ulong, etc. <lang vala> int fibRec(int n){ if (n < 2) return n; else return fibRec(n - 1) + fibRec(n - 2); } </lang>
Iterative
Using int, but could easily replace with double, long, ulong, etc. <lang vala> int fibIter(int n){ if (n < 2) return n;
int last = 0; int cur = 1; int next;
for (int i = 1; i < n; ++i){ next = last + cur; last = cur; cur = next; }
return cur; } </lang>
VAX Assembly
<lang VAX Assembly> 0000 0000 1 .entry main,0
7E 7CFD 0002 2 clro -(sp) ;result buffer 5E DD 0005 3 pushl sp ;pointer to buffer 10 DD 0007 4 pushl #16 ;descriptor: len of buffer 5B 5E D0 0009 5 movl sp, r11 ;-> descriptor 000C 6 7E 01 7D 000C 7 movq #1, -(sp) ;init 0,1 000F 8 loop: 7E 6E 04 AE C1 000F 9 addl3 4(sp), (sp), -(sp) ;next element on stack 17 1D 0014 10 bvs ret ;vs - overflow set, exit 0016 11 5B DD 0016 12 pushl r11 ;-> descriptor by ref 04 AE DF 0018 13 pushal 4(sp) ;-> fib on stack by ref 00000000'GF 02 FB 001B 14 calls #2, g^ots$cvt_l_ti ;convert integer to string 5B DD 0022 15 pushl r11 ; 00000000'GF 01 FB 0024 16 calls #1, g^lib$put_output ;show result E2 11 002B 17 brb loop 002D 18 ret: 04 002D 19 ret 002E 20 .end main
$ run fib ...
14930352 24157817 39088169 63245986 102334155 165580141 267914296 433494437 701408733 1134903170 1836311903
$ </lang>
VBA
Like Visual Basic .NET, but with keyword "Public" and type Variant (subtype Currency) instead of Decimal: <lang vb>Public Function Fib(ByVal n As Integer) As Variant
Dim fib0 As Variant, fib1 As Variant, sum As Variant Dim i As Integer fib0 = 0 fib1 = 1 For i = 1 To n sum = fib0 + fib1 fib0 = fib1 fib1 = sum Next i Fib = fib0
End Function </lang> With Currency type, maximum value is fibo(73).
The (slow) recursive version:
<lang VBA> Public Function RFib(Term As Integer) As Long
If Term < 2 Then RFib = Term Else RFib = RFib(Term - 1) + RFib(Term - 2)
End Function </lang>
With Long type, maximum value is fibo(46).
VBScript
Non-recursive, object oriented, generator
Defines a generator class, with a default Get property. Uses Currency for larger-than-Long values. Tests for overflow and switches to Double. Overflow information also available from class.
Class Definition:
<lang vb>class generator dim t1 dim t2 dim tn dim cur_overflow
Private Sub Class_Initialize cur_overflow = false t1 = ccur(0) t2 = ccur(1) tn = ccur(t1 + t2) end sub
public default property get generated on error resume next
generated = ccur(tn) if err.number <> 0 then generated = cdbl(tn) cur_overflow = true end if t1 = ccur(t2) if err.number <> 0 then t1 = cdbl(t2) cur_overflow = true end if t2 = ccur(tn) if err.number <> 0 then t2 = cdbl(tn) cur_overflow = true end if tn = ccur(t1+ t2) if err.number <> 0 then tn = cdbl(t1) + cdbl(t2) cur_overflow = true end if on error goto 0 end property
public property get overflow overflow = cur_overflow end property
end class</lang>
Invocation:
<lang vb>dim fib set fib = new generator dim i for i = 1 to 100 wscript.stdout.write " " & fib if fib.overflow then wscript.echo exit for end if next</lang>
Output:
<lang vbscript> 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 102334155 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025 20365011074 32951280099 53316291173 86267571272 139583862445 225851433717 365435296162 591286729879 956722026041 1548008755920 2504730781961 4052739537881 6557470319842 10610209857723 17167680177565 27777890035288 44945570212853 72723460248141 117669030460994 190392490709135 308061521170129 498454011879264 806515533049393</lang>
Vedit macro language
Iterative
Calculate fibonacci(#1). Negative values return 0. <lang vedit>:FIBONACCI:
- 11 = 0
- 12 = 1
Repeat(#1) {
#10 = #11 + #12 #11 = #12 #12 = #10
} Return(#11)</lang>
Unlimited precision
<lang vedit>// Fibonacci, unlimited precision. // input: #1 = n // return: fibonacci(n) in text register 10 //
- fibo_unlimited:
if (#1 < 2) {
Num_Str(#1, 10) return
} else {
Buf_Switch(Buf_Free) IC('0') IN IC('1') IN #10 = #1 While (#10 > 1) { #12 = 0 // carry out
#15 = 1 // column (ones, tens, hundreds...) Repeat (ALL) { // Sum all columns Line(-1) // n-1 Goto_col(#15) if (At_EOL) { // all digits added break } #11 = Cur_Char - '0' + #12 // digit of (n-1) + carry Line(-1) // n-2 Goto_Col(#15) if (!At_EOL) { // may contain fewer digits than n-1 #11 += Cur_Char - '0' } Goto_Line(3) // sum EOL #12 = #11 / 10 // carry out Ins_Char((#11 % 10) + '0') #15++ // next column } if (#12) { Goto_Line(3) EOL Ins_Char(#12 + '0') // any extra digit from carry
}
BOF Del_Line(1) // Next n Line(1) EOL Ins_Newline #10--
} Goto_Line(2) // Results on line 2
} // Copy the results to text register 10 in reverse order Reg_Empty(10) While(!At_EOL) {
Reg_Copy_Block(10, CP, CP+1, INSERT) Char()
} Buf_Quit(OK) return</lang>
Test: <lang vedit>#1 = Get_Num("n: ", STATLINE) Ins_Text("fibonacci(") Num_Ins(#1, LEFT+NOCR) Ins_Text(") = ") Call("fibo_unlimited") Reg_Ins(10) IN return</lang>
- Output:
fibonacci(1000) = 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
Visual Basic
Maximum integer value (7*10^28) can be obtained by using decimal type, but decimal type is only a sub type of the variant type. <lang vb>Sub fibonacci()
Const n = 139 Dim i As Integer Dim f1 As Variant, f2 As Variant, f3 As Variant 'for Decimal f1 = CDec(0): f2 = CDec(1) 'for Decimal setting Debug.Print "fibo("; 0; ")="; f1 Debug.Print "fibo("; 1; ")="; f2 For i = 2 To n f3 = f1 + f2 Debug.Print "fibo("; i; ")="; f3 f1 = f2 f2 = f3 Next i
End Sub 'fibonacci</lang>
- Output:
fibo( 0 )= 0 fibo( 1 )= 1 fibo( 2 )= 1 ... fibo( 137 )= 19134702400093278081449423917 fibo( 138 )= 30960598847965113057878492344 fibo( 139 )= 50095301248058391139327916261
Visual Basic .NET
Platform: .NET
Iterative
With Decimal type, maximum value is fibo(139). <lang vbnet>Function Fib(ByVal n As Integer) As Decimal
Dim fib0, fib1, sum As Decimal Dim i As Integer fib0 = 0 fib1 = 1 For i = 1 To n sum = fib0 + fib1 fib0 = fib1 fib1 = sum Next Fib = fib0
End Function</lang>
Recursive
<lang vbnet>Function Seq(ByVal Term As Integer)
If Term < 2 Then Return Term Return Seq(Term - 1) + Seq(Term - 2)
End Function</lang>
BigInteger
There is no real maximum value of BigInterger class, except the memory to store the number. Within a minute, fibo(2000000) is a number with 417975 digits. <lang vbnet> Function FiboBig(ByVal n As Integer) As BigInteger
' Fibonacci sequence with BigInteger Dim fibn2, fibn1, fibn As BigInteger Dim i As Integer fibn = 0 fibn2 = 0 fibn1 = 1 If n = 0 Then Return fibn2 ElseIf n = 1 Then Return fibn1 ElseIf n >= 2 Then For i = 2 To n fibn = fibn2 + fibn1 fibn2 = fibn1 fibn1 = fibn Next i Return fibn End If Return 0 End Function 'FiboBig
Sub fibotest() Dim i As Integer, s As String i = 2000000 ' 2 millions s = FiboBig(i).ToString Console.WriteLine("fibo(" & i & ")=" & s & " - length=" & Len(s)) End Sub 'fibotest</lang>
BigInteger, speedier method
This method doesn't need to iterate the entire list, and is much faster. The 2000000 (two millionth) Fibonacci number can be found in a fraction of a second.
Algorithm from here, see section 3, Finding Fibonacci Numbers Fully.
<lang vbnet>Imports System
Imports System.Collections.Generic
Imports System.Numerics
Module Module1
' A sparse array of values calculated along the way Dim sl As SortedList(Of Integer, BigInteger) = New SortedList(Of Integer, BigInteger)()
' Square a BigInteger Function sqr(ByVal n As BigInteger) As BigInteger Return n * n End Function
' Helper routine for Fsl(). It adds an entry to the sorted list when necessary Sub IfNec(n as integer) If Not sl.ContainsKey(n) Then sl.Add(n, Fsl(n)) End Sub
' This routine is semi-recursive, but doesn't need to evaluate every number up to n. ' Algorithm from here: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html#section3 Function Fsl(ByVal n As Integer) As BigInteger If n < 2 Then Return n Dim n2 As Integer = n >> 1, pm As Integer = n2 + ((n And 1) << 1) - 1 : IfNec(n2) : IfNec(pm) Return If(n2 > pm, (2 * sl(pm) + sl(n2)) * sl(n2), sqr(sl(n2)) + sqr(sl(pm))) End Function
' Conventional iteration method (not used here) Function Fm(ByVal n As BigInteger) As BigInteger If n < 2 Then Return n Dim cur As BigInteger = 0, pre As BigInteger = 1 For i As Integer = 0 To n - 1 Dim sum As BigInteger = cur + pre pre = cur : cur = sum Next : Return cur End Function
Sub Main() Dim num As Integer = 2_000_000 Dim st As DateTime = DateTime.Now Dim v As BigInteger = Fsl(num) Console.WriteLine("{0:n3} ms to calculate the {1:n0}th Fibonacci number,", (DateTime.Now - st).TotalMilliseconds, num) st = DateTime.Now Dim vs As String = v.ToString() Console.WriteLine("{0:n3} seconds to convert to a string.", (DateTime.Now - st).TotalSeconds) Console.WriteLine("number of digits is {0}", vs.Length) If vs.Length < 10000 Then st = DateTime.Now Console.WriteLine(vs) Console.WriteLine("{0:n3} ms to write it to the console.", (DateTime.Now - st).TotalMilliseconds) Else Console.WriteLine("partial: {0}...{1}", vs.Substring(1, 35), vs.Substring(vs.Length - 35)) End If End Sub
End Module</lang>
- Output:
177.831 ms to calculate the 2,000,000th Fibonacci number, 4.649 seconds to convert to a string. number of digits is 417975 partial: 53129491750764154305166065450382516...91799493108960825129188777803453125
Vlang
<lang vlang>fn fibonacci(n int) int {
if n == 0 || n == 1 { return n } if n > 1 { return fibonacci(n - 2) + fibonacci(n - 1) } else { return fibonacci(n + 2) - fibonacci(n + 1) }
}
// does not support negative numbers fn fib_iter(i int) int {
if i < 2 { return i } mut a := 0 mut b := 1 mut c := 0 for n := i - 1; n > 0; n-- { a += b c = a a = b b = c } return b
}
fn main() {
arr := [10, -5, 4, 3, -8] for val in arr { println('fibonacci(${val : 3d}) = ${fibonacci(val) : 3d}') }
}</lang>
- Output:
fibonacci( 10) = 55 fibonacci( -5) = 5 fibonacci( 4) = 3 fibonacci( 3) = 2 fibonacci( -8) = -21
Wart
Recursive, all at once
<lang python>def (fib n)
if (n < 2) n (+ (fib n-1) (fib n-2))</lang>
Recursive, using cases
<lang python>def (fib n)
(+ (fib n-1) (fib n-2))
def (fib n) :case (n < 2)
n</lang>
Recursive, using memoization
<lang python>def (fib n saved)
# all args in Wart are optional, and we expect callers to not provide `saved` default saved :to (table 0 0 1 1) # pre-populate base cases default saved.n :to (+ (fib n-1 saved) (fib n-2 saved)) saved.n</lang>
WDTE
Memoized Recursive
<lang WDTE>let memo fib n => n { > 1 => + (fib (- n 1)) (fib (- n 2)) };</lang>
Iterative
<lang WDTE>let s => import 'stream'; let a => import 'arrays';
let fib n => ( let reducer p n => [a.at p 1; + (a.at p 0) (a.at p 1)]; s.range 1 n -> s.reduce [0; 1] reducer -> a.at 1 ; );</lang>
Whitespace
Iterative
This program generates Fibonacci numbers until it is forced to terminate. <lang Whitespace>
</lang>
It was generated from the following pseudo-Assembly. <lang asm>push 0 push 1
0:
swap dup onum push 10 ochr copy 1 add jump 0</lang>
- Output:
$ wspace fib.ws | head -n 6 0 1 1 2 3 5
Recursive
This program takes a number n on standard input and outputs the nth member of the Fibonacci sequence. <lang Whitespace>
</lang> <lang asm>; Read n. push 0 dup inum load
- Call fib(n), ouput the result and a newline, then exit.
call 0 onum push 10 ochr exit
0:
dup push 2 sub jn 1 ; Return if n < 2. dup push 1 sub call 0 ; Call fib(n - 1). swap ; Get n back into place. push 2 sub call 0 ; Call fib(n - 2). add ; Leave the sum on the stack.
1:
ret</lang>
- Output:
$ echo 10 | wspace fibrec.ws 55
Wrapl
Generator
<lang wrapl>DEF fib() (
VAR seq <- [0, 1]; EVERY SUSP seq:values; REP SUSP seq:put(seq:pop + seq[1])[-1];
);</lang> To get the 17th number: <lang wrapl>16 SKIP fib();</lang> To get the list of all 17 numbers: <lang wrapl>ALL 17 OF fib();</lang>
Iterator
Using type match signature to ensure integer argument: <lang wrapl>TO fib(n @ Integer.T) (
VAR seq <- [0, 1]; EVERY 3:to(n) DO seq:put(seq:pop + seq[1]); RET seq[-1];
);</lang>
Wren
<lang ecmascript>// iterative (quick) var fibItr = Fn.new { |n|
if (n < 2) return n var a = 0 var b = 1 for (i in 2..n) { var c = a + b a = b b = c } return b
}
// recursive (slow) var fibRec fibRec = Fn.new { |n|
if (n < 2) return n return fibRec.call(n-1) + fibRec.call(n-2)
}
System.print("Iterative: %(fibItr.call(36))") System.print("Recursive: %(fibRec.call(36))")</lang>
- Output:
Iterative: 14930352 Recursive: 14930352
x86 Assembly
<lang asm>TITLE i hate visual studio 4 (Fibs.asm)
- __ __/--------\
- >__ \ / | |\
- \ \___/ @ \ / \__________________
- \____ \ / \\\
- \____ Coded with love by
- |||
- \ Alexander Alvonellos |||
- | 9/29/2011 / ||
- | | MM
- | |--------------| |
- |< | |< |
- | | | |
- |mmmmmm| |mmmmm|
- Epic Win.
INCLUDE Irvine32.inc
.data BEERCOUNT = 48; Fibs dd 0, 1, BEERCOUNT DUP(0);
.code main PROC
- I am not responsible for this code.
- They made me write it, against my will.
;Here be dragons mov esi, offset Fibs; offset array; ;;were to start (start) mov ecx, BEERCOUNT; ;;count of items (how many) mov ebx, 4; ;;size (in number of bytes) call DumpMem;
mov ecx, BEERCOUNT; ;//http://www.wolframalpha.com/input/?i=F ib%5B47%5D+%3E+4294967295 mov esi, offset Fibs NextPlease:; mov eax, [esi]; ;//Get me the data from location at ESI add eax, [esi+4]; ;//add into the eax the data at esi + another double (next mem loc) mov [esi+8], eax; ;//Move that data into the memory location after the second number add esi, 4; ;//Update the pointer loop NextPlease; ;//Thank you sir, may I have another?
;Here be dragons
mov esi, offset Fibs; offset array; ;;were to start (start)
mov ecx, BEERCOUNT; ;;count of items (how many)
mov ebx, 4; ;;size (in number of bytes)
call DumpMem;
exit ; exit to operating system main ENDP
END main</lang>
xEec
This will display the first 93 numbers of the sequence. <lang xEec> h#1 h#1 h#1 o# h#10 o$ p >f
o# h#10 o$ p ma h? jnext p t
jnf </lang>
XLISP
Analytic
Uses Binet's method, based on the golden ratio, which almost feels like cheating—but the task specification doesn't require any particular algorithm, and this one is straightforward and fast. <lang lisp>(DEFUN FIBONACCI (N)
(FLOOR (+ (/ (EXPT (/ (+ (SQRT 5) 1) 2) N) (SQRT 5)) 0.5)))</lang>
To test it, we'll define a RANGE function and ask for the first 50 numbers in the sequence: <lang lisp>(DEFUN RANGE (X Y)
(IF (<= X Y) (CONS X (RANGE (+ X 1) Y))))
(PRINT (MAPCAR FIBONACCI (RANGE 1 50)))</lang>
- Output:
(1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 102334155 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025)
Tail recursive
Alternatively, this approach is reasonably efficient: <lang lisp>(defun fibonacci (x)
(defun fib (a b n) (if (= n 2) b (fib b (+ a b) (- n 1)) ) ) (if (< x 2) x (fib 1 1 x) ) )</lang>
Xojo
Pass n to this function where n is the desired number of iterations. This example uses the UInt64 datatype which is as unsigned 64 bit integer. As such, it overflows after the 93rd iteration. <lang vb>Function fibo(n As Integer) As UInt64
Dim noOne As UInt64 = 1 Dim noTwo As UInt64 = 1 Dim sum As UInt64
For i As Integer = 3 To n sum = noOne + noTwo noTwo = noOne noOne = sum Next
Return noOne
End Function</lang>
XPL0
<lang XPL0>func Fib1(N); \Return Nth Fibonacci number using iteration int N; int Fn, F0, F1; [F0:= 0; F1:= 1; Fn:= N; while N > 1 do
[Fn:= F0 + F1; F0:= F1; F1:= Fn; N:= N-1; ];
return Fn; ];
func Fib2(N); \Return Nth Fibonacci number using recursion int N; return if N < 2 then N else Fib2(N-1) + Fib2(N-2);
int N; [for N:= 0 to 20 do [IntOut(0, Fib1(N)); ChOut(0, ^ )];
CrLf(0); for N:= 0 to 20 do [IntOut(0, Fib2(N)); ChOut(0, ^ )]; CrLf(0);
]</lang>
- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
XQuery
<lang xquery>declare function local:fib($n as xs:integer) as xs:integer {
if($n < 2) then $n else local:fib($n - 1) + local:fib($n - 2)
};</lang>
zkl
A slight tweak to the task; creates a function that continuously generates fib numbers <lang zkl>var fibShift=fcn(ab){ab.append(ab.sum()).pop(0)}.fp(L(0,1));</lang>
zkl: do(15){ fibShift().print(",") } 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377, zkl: do(5){ fibShift().print(",") } 610,987,1597,2584,4181,
ZX Spectrum Basic
Iterative
<lang zxbasic>10 REM Only positive numbers 20 LET n=10 30 LET n1=0: LET n2=1 40 FOR k=1 TO n 50 LET sum=n1+n2 60 LET n1=n2 70 LET n2=sum 80 NEXT k 90 PRINT n1</lang>
Analytic
<lang zxbasic>10 DEF FN f(x)=INT (0.5+(((SQR 5+1)/2)^x)/SQR 5)</lang>
- ↑ R. L. Graham and N. J. A. Sloane, Anti-Hadamard matrices, Linear Algebra Appl. 62 (1984), 113–137.
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