Fibonacci sequence
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
%<:0D:>~$<:01:~%>=<:a94fad42221f2702:>~>
}:_s:{x{={~$x+%{=>~>x~-x<:0D:~>~>~^:_s:?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()- 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
* 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- 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
* 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- 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.
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 SUBROUTINE68000 Assembly
Input is in D0, and the output is also in D0. There is no protection from 32-bit overflow, so use at your own risk. (I used this C compiler to create this in ARM Assembly and translated it manually into 68000 Assembly. It wasn't that difficult since both CPUs have similar architectures.)
fib:
MOVEM.L D4-D5,-(SP)
MOVE.L D0,D4
MOVEQ #0,D5
CMP.L #2,D0
BCS .bar
MOVEQ #0,D5
.foo:
MOVE.L D4,D0
SUBQ.L #1,D0
JSR fib
SUBQ.L #2,D4
ADD.L D0,D5
CMP.L #1,D4
BHI .foo
.bar:
MOVE.L D5,D0
ADD.L D4,D0
MOVEM.L (SP)+,D4-D5
RTS
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 .
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 subroutineThe range may be easily be extended from the 13th (=233) to the 24th (=46368) Fibonacci number with 16-bit addition using DAD. The code here assumes the CP/M operating system.
;-------------------------------------------------------
; useful equates
;-------------------------------------------------------
bdos equ 5 ; BDOS entry
cr equ 13 ; ASCII carriage return
lf equ 10 ; ASCII line feed
space equ 32 ; ASCII space char
conout equ 2 ; BDOS console output function
putstr equ 9 ; BDOS print string function
nterms equ 20 ; number of terms (max=24) to display
;-------------------------------------------------------
; main code begins here
;-------------------------------------------------------
org 100h
lxi h,0 ; save CP/M's stack
dad sp
shld oldstk
lxi sp,stack ; set our own stack
lxi d,signon ; print signon message
mvi c,putstr
call bdos
mvi a,0 ; start with Fib(0)
mloop: push a ; save our count
call fib
call putdec
mvi a,space ; separate the numbers
call putchr
pop a ; get our count back
inr a ; increment it
cpi nterms+1 ; have we reached our limit?
jnz mloop ; no, keep going
lhld oldstk ; all done; get CP/M's stack back
sphl ; restore it
ret ; back to command processor
;-------------------------------------------------------
; calculate nth Fibonacci number (max n = 24)
; entry: A = n
; exit: HL = Fib(n)
;-------------------------------------------------------
fib: mov c,a ; C holds n
lxi d,0 ; DE holds Fib(n-2)
lxi h,1 ; HL holds Fib(n-1)
ana a ; Fib(0) is a special case
jnz fib2 ; n > 0 so calculate normally
xchg ; otherwise return with HL=0
ret
fib2: dcr c
jz fib3 ; we're done
push h ; save Fib(n-1)
dad d ; HL = Fib(n), soon to be Fib(n-1)
pop d ; DE = old F(n-1), now Fib(n-2)
jmp fib2 ; ready for next pass
fib3: ret
;-------------------------------------------------------
; console output of char in A register
;-------------------------------------------------------
putchr: push h
push d
push b
mov e,a
mvi c,conout
call bdos
pop b
pop d
pop h
ret
;---------------------------------------------------------
; Output decimal number to console
; HL holds 16-bit unsigned binary number to print
;---------------------------------------------------------
putdec: push b
push d
push h
lxi b,-10
lxi d,-1
putdec2:
dad b
inx d
jc putdec2
lxi b,10
dad b
xchg
mov a,h
ora l
cnz putdec ; recursive call
mov a,e
adi '0'
call putchr
pop h
pop d
pop b
ret
;-------------------------------------------------------
; data area
;-------------------------------------------------------
signon: db 'Fibonacci number sequence:',cr,lf,'$'
oldstk: dw 0
stack equ $+128 ; 64-level stack
;
end- Output:
Fibonacci number sequence: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
8086 Assembly
Calculating the values at runtime
Is it cheating to write it in C for 64-bit x86 then port it to 16-bit x86?
The max input is 24 before you start having 16-bit overflow.
fib:
; WRITTEN IN C WITH X86-64 CLANG 3.3 AND DOWNGRADED TO 16-BIT X86
; INPUT: DI = THE NUMBER YOU WISH TO CALC THE FIBONACCI NUMBER FOR.
; OUTPUTS TO AX
push BP
push BX
push AX
mov BX, DI ;COPY INPUT TO BX
xor AX, AX ;MOV AX,0
test BX, BX ;SET FLAGS ACCORDING TO BX
je LBB0_4 ;IF BX == 0 RETURN 0
cmp BX, 1 ;IF BX == 1 RETURN 1
jne LBB0_3
mov AX, 1 ;ELSE, SET AX = 1 AND RETURN
jmp LBB0_4
LBB0_3:
lea DI, WORD PTR [BX - 1] ;DI = BX - 1
call fib ;RETURN FIB(BX-1)
mov BP, AX ;STORE THIS IN BP
add BX, -2
mov DI, BX
call fib ;GET FIB(DI - 2)
add AX, BP ;RETURN FIB(DI - 1) + FIB(DI - 2)
LBB0_4:
add sp,2
pop BX
pop BP
ret
Using A Lookup Table
With old computers it was common to use lookup tables to fetch pre-calculated values that would otherwise take some time to compute. The elements of the table are ordered by index, so you can simply create a function that takes an offset as the parameter and returns the element of the array at that offset.
Although lookup tables are very fast, there are some drawbacks to using them. For one, you end up taking up a lot of space. We're wasting a lot of bytes to store very low numbers at the beginning (each takes up 4 bytes regardless of how many digits you see). Unfortunately, when using lookup tables you have very little choice, since trying to conditionally change the scaling of the index would more than likely take more code than encoding all data as the maximum size regardless of the contents, as was done here. This keeps it simple for the CPU, which isn't aware of the intended size of each entry of the table.
For the purpose of this example, assume that both this code and the table are in the .CODE segment.
getfib:
;input: BX = the desired fibonacci number (in other words, the "n" in "F(n)")
; DS must point to the segment where the fibonacci table is stored
;outputs to DX:AX (DX = high word, AX = low word)
push ds
cmp bx,41 ;bounds check
ja IndexOutOfBounds
shl bx,1
shl bx,1 ;multiply by 4, since this is a table of dwords
mov ax,@code
mov ds,ax
mov si,offset fib
mov ax,[ds:si] ;fetch the low word into AX
mov dx,2[ds:si] ;fetch the high word into DX
pop ds
ret
IndexOutOfBounds:
stc ;set carry to indicate an error
mov ax,0FFFFh ;return FFFF as the error code
pop ds
ret
;table of the first 41 fibonacci numbers
fib dword 0, 1, 1, 2, 3, 5, 8, 13
dword 21, 34, 55, 89, 144, 233, 377, 610
dword 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657
dword 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269
dword 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986
dword 102334155
8th
An iterative solution:
: 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 ;
ABAP
Iterative
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.
Impure Functional
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 ) ).
ACL2
Fast, tail recursive solution:
(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))
- Output:
>(first-fibs 20) (1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765)
Action!
Action! language does not support recursion. Therefore an iterative approach has been proposed.
INT FUNC Fibonacci(INT n)
INT curr,prev,tmp
IF n>=-1 AND n<=1 THEN
RETURN (n)
FI
prev=0
IF n>0 THEN
curr=1
DO
tmp=prev
prev=curr
curr==+tmp
n==-1
UNTIL n=1
OD
ELSE
curr=-1
DO
tmp=prev
prev=curr
curr==+tmp
n==+1
UNTIL n=-1
OD
FI
RETURN (curr)
PROC Main()
BYTE n
INT f
Put(125) ;clear screen
FOR n=0 TO 22
DO
f=Fibonacci(n)
Position(2,n+1)
PrintF("Fib(%I)=%I",n,f)
IF n>0 THEN
f=Fibonacci(-n)
Position(21,n+1)
PrintF("Fib(%I)=%I",-n,f)
FI
OD
RETURN- Output:
Screenshot from Atari 8-bit computer
Fib(0)=0 Fib(1)=1 Fib(-1)=-1 Fib(2)=1 Fib(-2)=-1 Fib(3)=2 Fib(-3)=-2 Fib(4)=3 Fib(-4)=-3 Fib(5)=5 Fib(-5)=-5 Fib(6)=8 Fib(-6)=-8 Fib(7)=13 Fib(-7)=-13 Fib(8)=21 Fib(-8)=-21 Fib(9)=34 Fib(-9)=-34 Fib(10)=55 Fib(-10)=-55 Fib(11)=89 Fib(-11)=-89 Fib(12)=144 Fib(-12)=-144 Fib(13)=233 Fib(-13)=-233 Fib(14)=377 Fib(-14)=-377 Fib(15)=610 Fib(-15)=-610 Fib(16)=987 Fib(-16)=-987 Fib(17)=1597 Fib(-17)=-1597 Fib(18)=2584 Fib(-18)=-2584 Fib(19)=4181 Fib(-19)=-4181 Fib(20)=6765 Fib(-20)=-6765 Fib(21)=10946 Fib(-21)=-10946 Fib(22)=17711 Fib(-22)=-17711
ActionScript
public function fib(n:uint):uint
{
if (n < 2)
return n;
return fib(n - 1) + fib(n - 2);
}
Ada
Recursive
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;
Iterative, build-in integers
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;
- 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].
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;
- Output:
> ./fibonacci 777 Fibonacci( 777 ) = 1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082
Fast method using fast matrix exponentiation
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;
AdvPL
Recursive
#include "totvs.ch"
User Function fibb(a,b,n)
return(if(--n>0,fibb(b,a+b,n),a))Iterative
#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)Agda
module FibonacciSequence where
open import Data.Nat using (ℕ ; zero ; suc ; _+_)
rec_fib : (m : ℕ) -> (a : ℕ) -> (b : ℕ) -> ℕ
rec_fib zero a b = a
rec_fib (suc k) a b = rec_fib k b (a + b)
fib : (n : ℕ) -> ℕ
fib n = rec_fib n zero (suc zero)
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;
}ALGOL 60
begin
comment Fibonacci sequence;
integer procedure fibonacci(n); value n; integer n;
begin
integer i, fn, fn1, fn2;
fn2 := 1;
fn1 := 0;
fn := 0;
for i := 1 step 1 until n do begin
fn := fn1 + fn2;
fn2 := fn1;
fn1 := fn
end;
fibonacci := fn
end fibonacci;
integer i;
for i := 0 step 1 until 20 do outinteger(1,fibonacci(i))
end- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
ALGOL 68
Analytic
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)- 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
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)- 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
PROC recursive fibonacci = (INT n)INT:
( n < 2 | n | fib(n-1) + fib(n-2));Generative
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)
)- 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.
[]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)- 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
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.- 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
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;Naively recursive
INTEGER FUNCTION FIBONACCI( X ); INTEGER X;
BEGIN
IF X < 3 THEN
FIBONACCI := 1
ELSE
FIBONACCI := FIBONACCI( X - 2 ) + FIBONACCI( X - 1 );
END;Alore
def fib(n as Int) as Int
if n < 2
return 1
end
return fib(n-1) + fib(n-2)
endAmazing Hopper
Analitic, Recursive and Iterative mode.
#include <hbasic.h>
#define TERM1 1.61803398874989
#define TERM2 -0.61803398874989
#context get Fibonacci number with analitic mode
GetArgs(n)
get Inv of (M_SQRT5), Mul by( Pow (TERM 1, n), Minus( Pow(TERM 2, n) ) );
then Return\\
#proto fibonacci_recursive(__X__)
#synon _fibonacci_recursive getFibonaccinumberwithrecursivemodeof
#proto fibonacci_iterative(__X__)
#synon _fibonacci_iterative getFibonaccinumberwithiterativemodeof
Begin
Option Stack 1024
get Arg Number(2, n), and Take( n );
then, get Fibonacci number with analitic mode, and Print It with a Newl.
secondly, get Fibonacci number with recursive mode of(n), and Print It with a Newl.
finally, get Fibonacci number with iterative mode of (n), and Print It with a Newl.
End
Subrutines
fibonacci_recursive(n)
Iif ( var(n) Is Le? (2), 1 , \
get Fibonacci number with recursive mode of( var(n) Minus (1));\
get Fibonacci number with recursive mode of( var(n) Minus (2)); and Add It )
Return
fibonacci_iterative(n)
A=0
B=1
For Up( I:=2, n, 1 )
C=B
Let ( B: = var(A) Plus (B) )
A=C
Next
Return(B)- Output:
$ hopper src/fibo1.bas 25 75025 75025 75025
AntLang
/Sequence
fib:{<0;1> {x,<x[-1]+x[-2]>}/ range[x]}
/nth
fibn:{fib[x][x]}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;
}
}APL
Naive Recursive
fib←{⍵≤1:⍵ ⋄ (∇ ⍵-1)+∇ ⍵-2}
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:
↑+.×/N/⊂2 2⍴1 1 1 0
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:
↑0 1↓↑+.×/N/⊂2 2⍴1 1 1 0
Analytic
An alternative approach, using Binet's formula (which was apparently known long before Binet):
⌊.5+(((1+PHI)÷2)*⍳N)÷PHI←5*.5
AppleScript
Imperative
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
Functional
The simple recursive version is famously slow:
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
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)
-------------------- 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
- 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.
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, lrArnoldC
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 TERMINATEDArturo
Recursive
fib: $[x][
if? x<2 [1]
else [(fib x-1) + (fib x-2)]
]
loop 1..25 [x][
print ["Fibonacci of" x "=" fib x]
]
AsciiDots
/--#$--\
| |
>-*>{+}/
| \+-/
1 |
# 1
| #
| |
. .ATS
Recursive
fun fib_rec(n: int): int =
if n >= 2 then fib_rec(n-1) + fib_rec(n-2) else nIterative
(*
** 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 0Iterative and Verified
(*
** 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]Matrix-based
(* ****** ****** *)
//
// 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] *)AutoHotkey
Search autohotkey.com: sequence
Iterative
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
}
Recursive and iterative
Source: AutoHotkey forum by Laszlo
/*
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
}
AutoIt
Iterative
#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
Recursive
#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
AWK
As in many examples, this one-liner contains the function as well as testing with input from stdin, output to stdout.
$ awk 'func fib(n){return(n<2?n:fib(n-1)+fib(n-2))}{print "fib("$1")="fib($1)}'
10
fib(10)=55
Axe
A recursive solution is not practical in Axe because there is no concept of variable scope in Axe.
Iterative solution:
Lbl FIB
r₁→N
0→I
1→J
For(K,1,N)
I+J→T
J→I
T→J
End
J
ReturnBabel
In Babel, we can define fib using a stack-based approach that is not recursive:
fib { <- 0 1 { dup <- + -> swap } -> times zap } <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:
{19 iter - fib !} 20 times collect ! lsnum !- Output:
( 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 )
bash
Iterative
$ fib=1;j=1;while((fib<100));do echo $fib;((k=fib+j,fib=j,j=k));done
1 1 2 3 5 8 13 21 34 55 89
Recursive
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
}
BASIC
Applesoft BASIC
Same code as Commodore BASIC
Entering a value of N > 183, produces an error message:
?OVERFLOW ERROR IN 220
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) + cBBC BASIC
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
- Output:
1 1
233 233
121393 121393
bootBASIC
Variables in bootBASIC are 2 byte unsigned integers that roll over if there is an overflow. Entering a number greater than 24 will result in an incorrect outcome.
10 print "Enter a ";
20 print "number ";
30 print "greater ";
40 print "than 1";
50 print " and less";
60 print " than 25";
70 input z
80 b=1
90 a=0
100 n=2
110 f=a+b
120 a=b
130 b=f
140 n=n+1
150 if n-z-1 goto 110
160 print "The ";
170 print z ;
180 print "th ";
190 print "Fibonacci ";
200 print "Number is ";
210 print f
Chipmunk Basic
100 cls
110 for i = 0 to 20 : print fibor(i); : next i
120 print
130 for i = 0 to 20 : print fiboi(i); : next i
140 print
150 for i = 0 to 20 : print fiboa(i); : next i
160 end
170 sub fibor(n) : 'Recursive
180 if n < 2 then
190 fibor = n
200 else
210 fibor = fibor(n-1)+fibor(n-2)
220 endif
230 end sub
240 sub fiboi(n) : 'Iterative
250 n1 = 0
260 n2 = 1
270 for k = 1 to abs(n)
280 sum = n1+n2
290 n1 = n2
300 n2 = sum
310 next k
320 if n < 0 then
330 fiboi = n1*((-1)^((-n)+1))
340 else
350 fiboi = n1
360 endif
370 end sub
380 sub fiboa(n) : 'Analytic
390 fiboa = int(0.5+(((sqr 5+1)/2)^n)/sqr 5)
400 end sub
- 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 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
Commodore BASIC
100 PRINT CHR$(147); CHR$(18); "**** FIBONACCI GENERATOR ****"
110 INPUT "MIN, MAX"; N1, N2
120 IF N1 > N2 THEN T = N1: N1 = N2: N2 = T
130 A = 0: B = 1: S = SGN(N1)
140 FOR I = S TO N1 STEP S
150 : IF S > 0 THEN T = A + B: A = B: B = T
160 : IF S < 0 THEN T = B - A: B = A: A = T
170 NEXT I
180 PRINT
190 PRINT STR$(A); : REM STR$() PREVENTS TRAILING SPACE
200 IF N2 = N1 THEN 250
210 FOR I = N1 + 1 TO N2
220 : T = A + B: A = B: B = T
230 : PRINT ","; STR$(A);
240 NEXT I
250 PRINT
- Output:
**** FIBONACCI GENERATOR **** MIN, MAX? -6,6 -8, 5,-3, 2,-1, 1, 0, 1, 1, 2, 3, 5, 8 READY.
Craft Basic
let a = 1
let b = 1
print "Fibonacci Sequence"
for i = 0 to 20
let s = a + b
let a = b
let b = s
print s
next i
- Output:
Fibonacci Sequence 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657
FreeBASIC
Extended sequence coded big integer.
'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
- 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
FTCBASIC
define a = 1, b = 1, s = 0, i = 0
cls
print "Fibonacci Sequence"
do
let s = a + b
let a = b
let b = s
+1 i
print s
loop i < 20
pause
end
GFA Basic
'
' 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
GW-BASIC
Iterative
10 ' SAVE"FIBONA", A
20 ' Secuencia de Fibonacci
30 ' Var
40 DEFDBL D
50 IMAXFIBO% = 76
60 DNUM1 = 1: DNUM2 = DNUM1
70 CLS
80 PRINT "Este programa calcula la serie de Fibonacci."
90 PRINT DNUM1; DNUM2;
100 FOR I% = 1 TO IMAXFIBO%
110 DNUM3 = DNUM1 + DNUM2
120 PRINT DNUM3;
130 DNUM1 = DNUM2: DNUM2 = DNUM3
140 NEXT I%
150 PRINT
160 PRINT "Fin de la ejecución del programa."
170 END
- Output:
Este programa calcula la serie de Fibonacci. 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 Fin de la ejecución del programa. Ok
Binet formula
10 ' SAVE"FIBINF", A
20 ' Secuencia de Fibonacci mediante la fórmula de Binet
30 ' Var
40 DEFDBL D
50 IMAXFIBO% = 77
60 DSQR5 = SQR(5)
70 DPIV1 = (1 + DSQR5) / 2
80 DPIV2 = (1 - DSQR5) / 2
90 DNUM1 = DPIV1: DNUM2 = DPIV2
100 CLS
110 PRINT "Este programa calcula la serie de Fibonacci."
120 FOR I% = 1 TO IMAXFIBO%
130 DNUM1 = DNUM1 * DPIV1
140 DNUM2 = DNUM2 * DPIV2
150 PRINT FIX(((DNUM1 - DNUM2) / DSQR5)+.5);
160 NEXT I%
170 PRINT
180 PRINT "Fin de la ejecución del programa."
190 END
- Output:
Este programa calcula la serie de Fibonacci. 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 2178310 3524579 5702889 9227468 14930357 24157826 39088183 63246010 102334195 165580207 267914406 433494620 701409036 1134903671 1836312733 2971216446 4807529247 7778745802 12586275225 20365021312 32951296999 53316319058 86267617266 139583938281 225851558714 365435502118 591287069122 956722584654 1548009675479 2504732295250 4052742027550 6557474414738 10610216591046 17167691246479 27777908226979 44945600103607 72723509350188 117669111103547 190392623123089 308061738545741 498454368657289 806516118510594 1304970505463907 2111486653578091 3416457206941612 5527943938022908 8944401270367342 Fin de la ejecución del programa. Ok
Integer BASIC
Only works with quite small values of .
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
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 DEFLiberty BASIC
Iterative/Recursive
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- 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
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- 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.
Microsoft Small Basic
Iterative
' 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- Output:
fibo(139)=50095301248058391139327916261
Binet's Formula
' 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- 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
Minimal BASIC
110 REM THE ARRAY F HOLDS THE FIBONACCI NUMBERS
120 DIM F(22)
130 LET F(0) = 0
140 LET F(1) = 1
150 LET N = 1
160 REM COMPUTE THE NEXT FIBBONACCI NUMBER
170 LET F(N+1) = F(N)+F(N-1)
180 LET N = N+1
190 PRINT F(N-2);
200 REM STOP AFTER PRINTING 20 NUMBERS
210 IF N < 22 THEN 170
220 END
MSX Basic
100 CLS
110 FOR N = 0 TO 15: GOSUB 130: PRINT FIBOI; : NEXT N
120 END
130 REM Iterative Fibonacci sequence
140 N1 = 0
150 N2 = 1
160 FOR K = 1 TO ABS(N)
170 SUM = N1 + N2
180 N1 = N2
190 N2 = SUM
200 NEXT K
210 IF N < 0 THEN FIBOI = N1 * ((-1) ^ ((-N) + 1)) ELSE FIBOI = N1
220 RETURN
Palo Alto Tiny BASIC
10 REM FIBONACCI SEQUENCE
20 INPUT "ENTER N FOR FIB(N)"N
30 LET A=0,B=1
40 FOR I=2 TO N
50 LET T=B,B=A+B,A=T
60 NEXT I
70 PRINT B
80 STOP
- Output:
2 runs.
ENTER N FOR FIB(N):9
34
ENTER N FOR FIB(N):13
233
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
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 FUNCTIONPureBasic
Macro based calculation
Macro Fibonacci (n)
Int((Pow(((1+Sqr(5))/2),n)-Pow(((1-Sqr(5))/2),n))/Sqr(5))
EndMacro
Recursive
Procedure FibonacciReq(n)
If n<2
ProcedureReturn n
Else
ProcedureReturn FibonacciReq(n-1)+FibonacciReq(n-2)
EndIf
EndProcedure
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
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
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
QB64
CBTJD: 2020/03/13
_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
Fibonacci from Russia
DIM F(80) AS DOUBLE 'FibRus.bas DANILIN
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
- Output:
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
QBasic
Iterative
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
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):
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
- Output:
(unhandled error in final input prevents output)
0 233 -267914296
Recursive
This example can't handle n < 0.
FUNCTION recFib (n)
IF (n < 2) THEN
recFib = n
ELSE
recFib = recFib(n - 1) + recFib(n - 2)
END IF
END FUNCTION
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.)
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*****
Quite BASIC
100 CLS
110 rem The array F holds the Fibonacci numbers
120 ARRAY f : rem DIM f(22) para Quite BASIC and MSX-BASIC
130 LET f(0) = 0
140 LET f(1) = 1
150 LET n = 1
160 rem Compute the NEXT Fibbonacci number
170 LET f(n+1) = f(n)+f(n-1)
180 LET n = n+1
190 PRINT f(n-2);" ";
200 rem STOP after printing 20 numbers
210 IF n < 22 THEN GOTO 170
Run BASIC
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 functionS-BASIC
Note that the 23rd Fibonacci number (=28657) is the largest that can be generated without overflowing S-BASIC's integer data type.
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
- Output:
0 1 1 2 3 5 8 13 21 34 55
Sinclair ZX81 BASIC
Analytic
10 INPUT N
20 PRINT INT (0.5+(((SQR 5+1)/2)**N)/SQR 5)
Iterative
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
Tail recursive
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
smart BASIC
The Iterative method is slow (relatively) and the Recursive method doubly so since it references the recursion 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
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
Softbridge BASIC
Iterative
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
TI-83 BASIC
Sequence table
[Y=]
nMin=0
u(n)=u(n-1)+u(n-2)
u(nMin)={1,0}
[TABLE]
n u(n)
------- -------
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 144Iterative
{0,1
While 1
Disp Ans(1
{Ans(2),sum(Ans
EndBinet's formula
Prompt N
.5(1+√(5 //golden ratio
(Ans^N–(-Ans)^-N)/√(5TI-89 BASIC
Recursive
Optimized implementation (too slow to be usable for n higher than about 12).
fib(n)
when(n<2, n, fib(n-1) + fib(n-2))Iterative
Unoptimized implementation (I think the for loop can be eliminated, but I'm not sure).
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
EndFuncTiny BASIC
10 LET A = 0
20 LET B = 1
30 PRINT "Which F_n do you want?"
40 INPUT N
50 IF N = 0 THEN GOTO 140
60 IF N = 1 THEN GOTO 120
70 LET C = B + A
80 LET A = B
90 LET B = C
100 LET N = N - 1
110 GOTO 60
120 PRINT B
130 END
140 PRINT 0
150 END
True BASIC
FUNCTION fibonacci (n)
LET n1 = 0
LET n2 = 1
FOR k = 1 TO ABS(n)
LET sum = n1 + n2
LET n1 = n2
LET n2 = sum
NEXT k
IF n < 0 THEN
LET fibonacci = n1 * ((-1) ^ ((-n) + 1))
ELSE
LET fibonacci = n1
END IF
END FUNCTION
PRINT fibonacci(0) ! 0
PRINT fibonacci(13) ! 233
PRINT fibonacci(-42) !-267914296
PRINT fibonacci(47) ! 2971215073
END
VBA
Like Visual Basic .NET, but with keyword "Public" and type Variant (subtype Currency) instead of Decimal:
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
With Currency type, maximum value is fibo(73).
The (slow) recursive version:
Public Function RFib(Term As Integer) As Long
If Term < 2 Then RFib = Term Else RFib = RFib(Term - 1) + RFib(Term - 2)
End FunctionWith 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:
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
Invocation:
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
- 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 190392490709135 308061521170129 498454011879264 806515533049393
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.
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
- 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).
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
Recursive
Function Seq(ByVal Term As Integer)
If Term < 2 Then Return Term
Return Seq(Term - 1) + Seq(Term - 2)
End Function
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.
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
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.
Imports System
Imports System.Collections.Generic
Imports BI = System.Numerics.BigInteger
Module Module1
' A sparse array of values calculated along the way
Dim sl As SortedList(Of Integer, BI) = New SortedList(Of Integer, BI)()
' Square a BigInteger
Function sqr(ByVal n As BI) As BI
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 BI
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 BI) As BI
If n < 2 Then Return n
Dim cur As BI = 0, pre As BI = 1
For i As Integer = 0 To n - 1
Dim sum As BI = cur + pre : pre = cur : cur = sum : Next : Return cur
End Function
Sub Main()
Dim vlen As Integer, num As Integer = 2_000_000, digs As Integer = 35
Dim sw As System.Diagnostics.Stopwatch = System.Diagnostics.Stopwatch.StartNew()
Dim v As BI = Fsl(num) : sw.[Stop]()
Console.Write("{0:n3} ms to calculate the {1:n0}th Fibonacci number, ", sw.Elapsed.TotalMilliseconds, num)
vlen = CInt(Math.Ceiling(BI.Log10(v))) : Console.WriteLine("number of digits is {0}", vlen)
If vlen < 10000 Then
sw.Restart() : Console.WriteLine(v) : sw.[Stop]()
Console.WriteLine("{0:n3} ms to write it to the console.", sw.Elapsed.TotalMilliseconds)
Else
Console.Write("partial: {0}...{1}", v / BI.Pow(10, vlen - digs), v Mod BI.Pow(10, digs))
End If
End Sub
End Module
- Output:
120.374 ms to calculate the 2,000,000th Fibonacci number, number of digits is 417975 partial: 85312949175076415430516606545038251...91799493108960825129188777803453125
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.
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
Yabasic
Iterative
sub fibonacciI (n)
local n1, n2, k, sum
n1 = 0
n2 = 1
for k = 1 to abs(n)
sum = n1 + n2
n1 = n2
n2 = sum
next k
if n < 1 then
return n1 * ((-1) ^ ((-n) + 1))
else
return n1
end if
end sub
Recursive
Only positive numbers
sub fibonacciR(n)
if n <= 1 then
return n
else
return fibonacciR(n-1) + fibonacciR(n-2)
end if
end sub
Analytic
Only positive numbers
sub fibonacciA (n)
return int(0.5 + (((sqrt(5) + 1) / 2) ^ n) / sqrt(5))
end sub
Binet's formula
Fibonacci sequence using the Binet formula
sub fibonacciB(n)
local sq5, phi1, phi2, dn1, dn2, k
sq5 = sqrt(5)
phi1 = (1 + sq5) / 2
phi2 = (1 - sq5) / 2
dn1 = phi1: dn2 = phi2
for k = 0 to n
dn1 = dn1 * phi1
dn2 = dn2 * phi2
print int(((dn1 - dn2) / sq5) + .5);
next k
end sub
ZX Spectrum Basic
Iterative
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
Analytic
10 DEF FN f(x)=INT (0.5+(((SQR 5+1)/2)^x)/SQR 5)
Batch File
Recursive version
::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!
- 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
// 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:
bc
iterative
#! /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
BCPL
get "libhdr"
let fib(n) = n<=1 -> n, valof
$( let a=0 and b=1
for i=2 to n
$( let c=a
a := b
b := a+c
$)
resultis b
$)
let start() be
for i=0 to 10 do
writef("F_%N*T= %N*N", i, fib(i))- Output:
F_0 = 0 F_1 = 1 F_2 = 1 F_3 = 2 F_4 = 3 F_5 = 5 F_6 = 8 F_7 = 13 F_8 = 21 F_9 = 34 F_10 = 55
beeswax
#>'#{;
_`Enter n: `TN`Fib(`{`)=`X~P~K#{;
#>~P~L#MM@>+@'q@{;
b~@M<Example output:
Notice the UInt64 wrap-around at Fib(94)!
julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i0
Fib(0)=0
Program finished!
julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i10
Fib(10)=55
Program finished!
julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i92
Fib(92)=7540113804746346429
Program finished!
julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i93
Fib(93)=12200160415121876738
Program finished!
julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i94
Fib(94)=1293530146158671551
Program finished!
Befunge
00:.1:.>:"@"8**++\1+:67+`#@_v
^ .:\/*8"@"\%*8"@":\ <
BlitzMax
local a:int = 0, b:int = 1, c:int = 1, n:int
n = int(input( "Enter n: "))
if n = 0 then
print 0
end
else if n = 1
print 1
end
end if
while n>2
a = b
b = c
c = a + b
n = n - 1
wend
print c
Blue
: fib ( nth:ecx -- result:edi ) 1 0
: compute ( times:ecx accum:eax scratch:edi -- result:edi ) xadd latest loop ;
: example ( -- ) 11 fib drop ;BQN
All given functions return the nth element in the sequence.
Recursive
A primitive recursive can be done with predicates:
Fib ← {𝕩>1 ? (𝕊 𝕩-1) + 𝕊 𝕩-2; 𝕩}
Or, it can be done with the Choose(◶) modifier:
Fib2 ← {(𝕩-1) (𝕩>1)◶⟨𝕩, +○𝕊⟩ 𝕩-2}
Iterative
An iterative solution can be made with the Repeat(⍟) modifier:
{⊑(+`⌽)⍟𝕩 0‿1}
Bracmat
Recursive
fib=.!arg:<2|fib$(!arg+-2)+fib$(!arg+-1)fib$30 832040
Iterative
(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
)fib$777 1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082
Brainf***
The first cell contains n (10), the second cell will contain fib(n) (55), and the third cell will contain fib(n-1) (34).
++++++++++
>>+<<[->[->+>+<<]>[-<+>]>[-<+>]<<<]
The following generates n fibonacci numbers and prints them, though not in ascii. It does have a limit due to the cells usually being 1 byte in size.
+++++ +++++ #0 set to n
>> + Init #2 to 1
<<
[
- #Decrement counter in #0
>>. Notice: This doesn't print it in ascii
To look at results you can pipe into a file and look with a hex editor
Copying sequence to save #2 in #4 using #5 as restore space
>>[-] Move to #4 and clear
>[-] Clear #5
<<< #2
[ Move loop
- >> + > + <<< Subtract #2 and add #4 and #5
]
>>>
[ Restore loop
- <<< + >>> Subtract from #5 and add to #2
]
<<<< Back to #1
Non destructive add sequence using #3 as restore value
[ Loop to add
- > + > + << Subtract #1 and add to value #2 and restore space #3
]
>>
[ Loop to restore #1 from #3
- << + >> Subtract from restore space #3 and add in #1
]
<< [-] Clear #1
>>>
[ Loop to move #4 to #1
- <<< + >>> Subtract from #4 and add to #1
]
<<<< Back to #0
]
Brat
Recursive
fibonacci = { x |
true? x < 2, x, { fibonacci(x - 1) + fibonacci(x - 2) }
}Tail Recursive
fib_aux = { x, next, result |
true? x == 0,
result,
{ fib_aux x - 1, next + result, next }
}
fibonacci = { x |
fib_aux x, 1, 0
}Memoization
cache = hash.new
fibonacci = { x |
true? cache.key?(x)
{ cache[x] }
{true? x < 2, x, { cache[x] = fibonacci(x - 1) + fibonacci(x - 2) }}
}Bruijn
:import std/Combinator .
:import std/Math .
:import std/List .
# unary/Church fibonacci (moderately fast but very high space complexity)
fib-unary [0 [[[2 0 [2 (1 0)]]]] k i]
:test (fib-unary (+6u)) ((+8u))
# ternary fibonacci using infinite list iteration (very fast)
fib-list \index fibs
fibs head <$> (iterate &[[0 : (1 + 0)]] ((+0) : (+1)))
:test (fib-list (+6)) ((+8))
# recursive fib (very slow)
fib-rec y [[0 <? (+1) (+0) (0 <? (+2) (+1) rec)]]
rec (1 --0) + (1 --(--0))
:test (fib-rec (+6)) ((+8))Performance using HigherOrder reduction without optimizations:
> :time fib-list (+1000)
0.9 seconds
> :time fib-unary (+50u)
1.7 seconds
> :time fib-rec (+25)
5.1 seconds
> :time fib-list (+50)
0.0006 secondsBurlesque
{0 1}{^^++[+[-^^-]\/}30.*\[e!vv0 1{{.+}c!}{1000.<}w!C
Recursive
long long fibb(long long a, long long b, int n) {
return (--n>0)?(fibb(b, a+b, n)):(a);
}
Iterative
long long int fibb(int n) {
int fnow = 0, fnext = 1, tempf;
while(--n>0){
tempf = fnow + fnext;
fnow = fnext;
fnext = tempf;
}
return fnext;
}
Analytic
#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) );
}
Generative
#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");
}
- 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
#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;
}
- 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
public static ulong Fib(uint n) {
return (n < 2)? n : Fib(n - 1) + Fib(n - 2);
}
Tail-Recursive
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);
}
Iterative
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;
}
Iterative
using System; using System.Text; // FIBRUS.cs Russia
namespace Fibrus { class Program { static void Main()
{ long fi1=1; long fi2=1; long fi3=1; int da; int i; int d;
for (da=1; da<=78; da++) // rextester.com/MNGUV70257
{ d = 20-Convert.ToInt32((Convert.ToString(fi3)).Length);
for (i=1; i<d; i++) Console.Write(".");
Console.Write(fi3); Console.Write(" "); Console.WriteLine(da);
fi3 = fi2 + fi1;
fi1 = fi2;
fi2 = fi3;
}}}}
- Output:
..................1 1 ..................2 2 ..................3 3 ... ...5527939700884757 76 ...8944394323791464 77 ..14472334024676221 78
Eager-Generative
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;
}
Lazy-Generative
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;
}
}
Analytic
This returns digits up to the 93rd Fibonacci number, but the digits become inaccurate past the 71st. There is custom rounding applied to the result that allows the function to be accurate at the 71st number instead of topping out at the 70th.
static double r5 = Math.Sqrt(5.0), Phi = (r5 + 1.0) / 2.0;
static ulong fib(uint n) {
if (n > 71) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 72.");
double r = Math.Pow(Phi, n) / r5;
return (ulong)(n < 64 ? Math.Round(r) : Math.Floor(r)); }
To get to the 93rd Fibonacci number, one must use the decimal type, rather than the double type, like this:
static decimal Sqrt_dec(decimal x, decimal g) { decimal t, lg;
do { t = x / g; lg = g; g = (t + g) / 2M; } while (lg != g);
return g; }
static decimal Pow_dec (decimal bas, uint exp) {
if (exp == 0) return 1M;
decimal tmp = Pow_dec(bas, exp >> 1); tmp *= tmp;
if ((exp & 1) == 1) tmp *= bas; return tmp; }
static decimal r5 = Sqrt_dec(5.0M, (decimal)Math.Sqrt(5.0)),
Phi = (r5 + 1.0M) / 2.0M;
static ulong fib(uint n) {
if (n > 93) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 94.");
decimal r = Pow_dec(Phi, n) / r5;
return (ulong)(n < 64 ? Math.Round(r) : Math.Floor(r)); }
Note that the Math.Pow() function and the Math.Sqrt() function must be replaced with ones returning the decimal type.
If one allows the fib() function to return the decimal type, one can reach the 138th Fibonacci number. However, the accuracy is lost after the 128th.
static decimal Sqrt_dec(decimal x, decimal g) { decimal t, lg;
do { t = x / g; lg = g; g = (t + g) / 2M; } while (lg != g);
return g; }
static decimal Pow_dec (decimal bas, uint exp) {
if (exp == 0) return 1M;
decimal tmp = Pow_dec(bas, exp >> 1); tmp *= tmp;
if ((exp & 1) == 1) tmp *= bas; return tmp; }
static decimal r5 = Sqrt_dec(5.0M, (decimal)Math.Sqrt(5.0)),
Phi = (r5 + 1.0M) / 2.0M;
static decimal fib(uint n) {
if (n > 128) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 129.");
decimal r = Pow_dec(Phi, n) / r5;
return n < 64 ? Math.Round(r) : Math.Floor(r); }Matrix
Algorithm is based on
- .
Needs System.Windows.Media.Matrix or similar Matrix class.
Calculates in .
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];
}Needs System.Windows.Media.Matrix or similar Matrix class.
Calculates in .
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;
}Array (Table) Lookup
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];
}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).
using System;
using System.Collections.Generic;
using BI = System.Numerics.BigInteger;
class Program
{
// A sparse array of values calculated along the way
static SortedList<int, BI> sl = new SortedList<int, BI>();
// 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
static BI 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]);
// Helper routine for Fsl(). It adds an entry to the sorted list when necessary
void IfNec(int x) { if (!sl.ContainsKey(x)) sl.Add(x, Fsl(x)); }
// Helper function to square a BigInteger
BI sqr(BI x) { return x * x; }
}
// Conventional iteration method (not used here)
public static BI Fm(BI n)
{
if (n < 2) return n; BI cur = 0, pre = 1;
for (int i = 0; i <= n - 1; i++) { BI sum = cur + pre; pre = cur; cur = sum; }
return cur;
}
public static void Main()
{
int num = 2_000_000, digs = 35, vlen;
var sw = System.Diagnostics.Stopwatch.StartNew(); var v = Fsl(num); sw.Stop();
Console.Write("{0:n3} ms to calculate the {1:n0}th Fibonacci number, ",
sw.Elapsed.TotalMilliseconds, num);
Console.WriteLine("number of digits is {0}", vlen = (int)Math.Ceiling(BI.Log10(v)));
if (vlen < 10000) {
sw.Restart(); Console.WriteLine(v); sw.Stop();
Console.WriteLine("{0:n3} ms to write it to the console.", sw.Elapsed.TotalMilliseconds);
} else
Console.Write("partial: {0}...{1}", v / BI.Pow(10, vlen - digs), v % BI.Pow(10, digs));
}
}- Output:
137.209 ms to calculate the 2,000,000th Fibonacci number, number of digits is 417975 partial: 85312949175076415430516606545038251...91799493108960825129188777803453125
Shift PowerMod
Illustrated here is an algorithm to compute a Fibonacci number directly, without needing to calculate any of the Fibonacci numbers less than the desired result. It uses shifting and the power mod function (BigInteger.ModPow() in C#). It calculates more quickly than the large step recurrence routine (illustrated above) for smaller Fibonacci numbers (less than 2800 digits or so, around Fibonacci(13000)), but gets slower for larger ones, as the intermediate BigIntegers created are very large, much larger than the Fibonacci result.
Also included is a routine that returns an array of Fibonacci numbers (fibTab()). It reuses the intermediate large shifted BigIntegers on suceeding iterations, therfore it is a little more efficient than calling the oneshot (oneFib()) routine repeatedly from a loop.
using System;
using BI = System.Numerics.BigInteger;
class Program {
// returns the nth Fibonacci number without calculating 0..n-1
static BI oneFib(int n) {
BI z = (BI)1 << ++n;
return BI.ModPow(z, n, (z << n) - z - 1) % z;
}
// returns an array of Fibonacci numbers from the 0th to the nth
static BI[] fibTab(int n) {
var res = new BI[++n];
BI z = (BI)1 << 1, zz = z << 1;
for (int i = 0; i < n; ) {
res[i] = BI.ModPow(z, ++i, zz - z - 1) % z;
z <<= 1; zz <<= 2;
}
return res;
}
static void Main(string[] args) {
int n = 20;
Console.WriteLine("Fibonacci numbers 0..{0}: {1}", n, string.Join(" ",fibTab(n)));
n = 1000;
Console.WriteLine("Fibonacci({0}): {1}", n, oneFib(n));
}
}- Output:
Fibonacci numbers 0..20: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 Fibonacci(1000): 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
C++
Using unsigned int, this version only works up to 48 before fib overflows.
#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;
}
This version does not have an upper bound.
#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;
}Version using transform:
#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];
}Far-fetched version using adjacent_difference:
#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];
}Version which computes at compile time with metaprogramming:
#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;
}The following version is based on fast exponentiation:
#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;
}- 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.
// 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;
}- 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".
// 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;
}- Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946
Cat
define fib {
dup 1 <=
[]
[dup 1 - fib swap 2 - fib +]
if
}Chapel
iter fib() {
var a = 0, b = 1;
while true {
yield a;
(a, b) = (b, b + a);
}
}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.Chez Scheme
(define fib (lambda (n) (cond ((> n 1) (+ (fib (- n 1)) (fib (- n 2))))
((= n 1) 1)
((= n 0) 0))))Clio
Clio is pure and functions are lazy and memoized by default
fn fib n:
if n < 2: n
else: (n - 1 -> fib) + (n - 2 -> fib)
[0:100] -> * fib -> * printClojure
Lazy Sequence
This is implemented idiomatically as an infinitely long, lazy sequence of all Fibonacci numbers:
(defn fibs []
(map first (iterate (fn [[a b]] [b (+ a b)]) [0 1])))Thus to get the nth one:
(nth (fibs) 5)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.
(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 1A more elegant solution is inspired by the Haskell implementation of an infinite list of Fibonacci numbers:
(def fib (lazy-cat [0 1] (map + fib (rest fib))))Then, to see the first ten,
user> (take 10 fib)
(0 1 1 2 3 5 8 13 21 34)Iterative
Here's a simple interative process (using a recursive function) that carries state along with it (as args) until it reaches a solution:
;; 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)))"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
(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))))Recursive
A naive slow recursive solution:
(defn fib [n]
(case n
0 0
1 1
(+ (fib (- n 1))
(fib (- n 2)))))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.
(def fib
(memoize
(fn [n]
(case n
0 0
1 1
(+ (fib (- n 1))
(fib (- n 2)))))))Using core.async
(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))))))CLU
% Generate Fibonacci numbers
fib = iter () yields (int)
a: int := 0
b: int := 1
while true do
yield (a)
a, b := b, a+b
end
end fib
% Grab the n'th value from an iterator
nth = proc [T: type] (g: itertype () yields (T), n: int) returns (T)
for v: T in g() do
if n<=0 then return (v) end
n := n-1
end
end nth
% Print a few values
start_up = proc ()
po: stream := stream$primary_output()
% print values coming out of the fibonacci iterator
% (which are generated one after the other without delay)
count: int := 0
for f: int in fib() do
stream$putl(po, "F(" || int$unparse(count) || ") = " || int$unparse(f))
count := count + 1
if count = 15 then break end
end
% print a few random fibonacci numbers
% (to do this it has to restart at the beginning for each
% number, making it O(N))
fibs: sequence[int] := sequence[int]$[20,30,50]
for n: int in sequence[int]$elements(fibs) do
stream$putl(po, "F(" || int$unparse(n) || ") = "
|| int$unparse(nth[int](fib, n)))
end
end start_up- Output:
F(0) = 0 F(1) = 1 F(2) = 1 F(3) = 2 F(4) = 3 F(5) = 5 F(6) = 8 F(7) = 13 F(8) = 21 F(9) = 34 F(10) = 55 F(11) = 89 F(12) = 144 F(13) = 233 F(14) = 377 F(20) = 6765 F(30) = 832040 F(50) = 12586269025
CMake
Iteration uses a while() loop. Memoization uses global properties.
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)# 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})0 1 1 2 3 5 8 13 21 34 ... 75025 121393 196418 317811 514229 832040
COBOL
Iterative
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.Recursive
>>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.CoffeeScript
Analytic
fib_ana = (n) ->
sqrt = Math.sqrt
phi = ((1 + sqrt(5))/2)
Math.round((Math.pow(phi, n)/sqrt(5)))Iterative
fib_iter = (n) ->
return n if n < 2
[prev, curr] = [0, 1]
[prev, curr] = [curr, curr + prev] for i in [1..n]
currRecursive
fib_rec = (n) ->
if n < 2 then n else fib_rec(n-1) + fib_rec(n-2)Comefrom0x10
Recursion is is not possible in Comefrom0x10.
Iterative
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 > stopCommon Lisp
Note that Common Lisp uses bignums, so this will never overflow.
Iterative
(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)))))Simpler one:
(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))
Not a function, just printing out the entire (for some definition of "entire") sequence with a for var = loop:
(loop for x = 0 then y and y = 1 then (+ x y) do (print x))Recursive
(defun fibonacci-recursive (n)
(if (< n 2)
n
(+ (fibonacci-recursive (- n 2)) (fibonacci-recursive (- n 1)))))
(defun fibonacci-tail-recursive ( n &optional (a 0) (b 1))
(if (= n 0)
a
(fibonacci-tail-recursive (- n 1) b (+ a b))))Tail recursive and squaring:
(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))Alternate solution
I use Allegro CL 10.1
;; 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))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
(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)List-based iterative
This solution uses a list to keep track of the Fibonacci sequence for 0 or a positive integer.
(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)))))))- 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).
(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))))))- 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.
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: 0Coq
Fixpoint rec_fib (m : nat) (a : nat) (b : nat) : nat :=
match m with
| 0 => a
| S k => rec_fib k b (a + b)
end.
Definition fib (n : nat) : nat :=
rec_fib n 0 1 .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
stopCowgol
include "cowgol.coh";
sub fibonacci(n: uint32): (a: uint32) is
a := 0;
var b: uint32 := 1;
while n > 0 loop
var c := a + b;
a := b;
b := c;
n := n - 1;
end loop;
end sub;
# test
var i: uint32 := 0;
while i < 20 loop
print_i32(fibonacci(i));
print_char(' ');
i := i + 1;
end loop;
print_nl();- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181
Crystal
Recursive
def fib(n)
n < 2 ? n : fib(n - 1) + fib(n - 2)
endIterative
def fibIterative(n, prevFib = 0, fib = 1)
return n if n < 2
n.times do
prevFib, fib = fib, prevFib + fib
end
prevFib
endTail Recursive
def fibTailRecursive(n, prevFib = 0, fib = 1)
n == 0 ? prevFib : fibTailRecursive(n - 1, fib, prevFib + fib)
endAnalytic
def fibBinet(n)
(((5 ** 0.5 + 1) / 2) ** n / 5 ** 0.5).round.to_i
endD
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.
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);
}- 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
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;
}Faster Version
For N = 10_000_000 this is about twice faster (run-time about 2.20 seconds) than the matrix exponentiation version.
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;
}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));
}Datalog
Simple recurive implementation for Souffle.
.decl Fib(i:number, x:number)
Fib(0, 0).
Fib(1, 1).
Fib(i+2,x+y) :- Fib(i+1, x), Fib(i, y), i+2<=40, i+2>=2.
Fib(i-2,y-x) :- Fib(i-1, x), Fib(i, y), i-2>=-40, i-2<0.DBL
;
; Fibonacci sequence for DBL version 4 by Dario B.
;
RECORD
FIB1, D10
FIB2, D10
FIBN, D10
J, D5
A2, A2
A5, A5
PROC
;----------------------------------------------------------------
XCALL FLAGS (0007000000,1) ;Suppress STOP message
OPEN (1,O,'TT:')
DISPLAY (1,'First 10 Fibonacci Numbers:',10)
FIB2=1
FOR J=1 UNTIL 10
DO BEGIN
FIBN=FIB1+FIB2
A2=J,'ZX'
A5=FIBN,'ZZZZX'
DISPLAY (1,A2,' : ',A5,10)
FIB1=FIB2
FIB2=FIBN
END
CLOSE 1
ENDDc
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.
[ # 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- Output:
5702887
Delphi
Iterative
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;Recursive
function Fibonacci(N: Word): UInt64;
begin
if N < 2 then
Result := N
else
Result := Fibonacci(N - 1) + Fibonacci(N - 2);
end;Matrix
Algorithm is based on
- .
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;DIBOL-11
; Redone to include the first two values that
; are noot computed.
START ;First 15 Fibonacci NUmbers
RECORD
FIB1, D10, 0
FIB2, D10, 1
FIBNEW, D10
LOOPCNT, D2, 3
RECORD HEADER
, A32, "First 15 Fibonacci Numbers."
RECORD OUTPUT
LOOPOUT, A2
, A3, " : "
FIBOUT, A10
PROC
OPEN(8,O,'TT:')
WRITES(8,HEADER)
; The First Two are given.
FIBOUT = 0
LOOPOUT = 1
WRITES(8,OUTPUT)
FIBOUT = 1
LOOPOUT = 2
WRITES(8,OUTPUT)
; The Rest are Computed.
LOOP,
FIBNEW = FIB1 + FIB2
LOOPOUT = LOOPCNT, 'ZX'
FIBOUT = FIBNEW, 'ZZZZZZZZZX'
WRITES(8,OUTPUT)
FIB1 = FIB2
FIB2 = FIBNEW
LOOPCNT = LOOPCNT + 1
IF LOOPCNT .LE. 15 GOTO LOOP
CLOSE 8
ENDDWScript
function fib(N : Integer) : Integer;
begin
if N < 2 then Result := 1
else Result := fib(N-2) + fib(N-1);
End;Dyalect
func fib(n) {
if n < 2 {
return n
} else {
return fib(n - 1) + fib(n - 2)
}
}
print(fib(30))E
def fib(n) {
var s := [0, 1]
for _ in 0..!n {
def [a, b] := s
s := [b, a+b]
}
return s[0]
}(This version defines fib(0) = 0 because OEIS A000045 does.)
EasyLang
func fib n .
if n < 2
return n
.
prev = 0
val = 1
for i = 2 to n
h = prev + val
prev = val
val = h
.
return val
.
print fib 36Recursive (inefficient):
func fib n .
if n < 2
return n
.
return fib (n - 2) + fib (n - 1)
.
print fib 36EchoLisp
Use memoization with the recursive version.
(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 89ECL
Analytic
//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; }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.
[ 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 ]- Output:
00000000000110111
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
endEla
Tail-recursive function:
fib = fib' 0 1
where fib' a b 0 = a
fib' a b n = fib' b (a + b) (n - 1)Infinite (lazy) list:
fib = fib' 1 1
where fib' x y = & x :: fib' y (x + y)Elena
ELENA 6.x :
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))
}
}- Output:
0 1 1 2 3 5 8 13 21 34 55
Alternative version using yieldable method
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()
}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)Using Stream:
Stream.unfold({0,1}, fn {a,b} -> {a,{b,a+b}} end) |> Enum.take(10)- Output:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
Elm
Naïve recursive implementation.
fibonacci : Int -> Int
fibonacci n = if n < 2 then
n
else
fibonacci(n - 2) + fibonacci(n - 1)
version 2
fib : Int -> number
fib n =
case n of
0 -> 0
1 -> 1
_ -> fib (n-1) + fib (n-2)- Output:
elm repl > fib 40 102334155 : number > List.map (\elem -> fib elem) (List.range 1 40) [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] : List number
Emacs Lisp
version 1
(defun fib (n a b c)
(cond
((< c n) (fib n b (+ a b) (+ 1 c)))
((= c n) b)
(t a)))
(defun fibonacci (n)
(if (< n 2)
n
(fib n 0 1 1)))version 2
(defun fibonacci (n)
(let (vec i j k)
(if (< n 2)
n
(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))))Eval:
(insert
(mapconcat (lambda (n) (format "%d" (fibonacci n)))
(number-sequence 0 15) " "))- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610
Erlang
Recursive
-module(fib).
-export([fib/1]).
fib(0) -> 0;
fib(1) -> 1;
fib(N) -> fib(N-1) + fib(N-2).Iterative
-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.Evaluate:
io:write([fiblin:fib(X) || X <- lists:seq(1,10) ]).Output:
[1,1,2,3,5,8,13,21,34,55]ok
Iterative 2
fib(N) -> fib(N, 0, 1).
fib(0, Result, _Next) -> Result;
fib(Iter, Result, Next) -> fib(Iter-1, Next, Result+Next).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- Output:
Number? 20 FIB( 20 )= 6765
Euphoria
'Recursive' version
function fibor(integer n)
if n<2 then return n end if
return fibor(n-1)+fibor(n-2)
end function'Iterative' version
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'Tail recursive' version
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'Paper tape' version
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 whileExcel
LAMBDA
Binding the name FIBONACCI to the following lambda in the Excel worksheet Name Manager:
(See The LAMBDA worksheet function)
FIBONACCI
=LAMBDA(n,
APPLYN(n - 2)(
LAMBDA(xs,
APPENDROWS(xs)(
SUM(
LASTNROWS(2)(xs)
)
)
)
)({1;1})
)And assuming that the following names are also bound to reusable generic lambdas in the Name manager:
APPENDROWS
=LAMBDA(xs,
LAMBDA(ys,
LET(
nx, ROWS(xs),
rowIndexes, SEQUENCE(nx + ROWS(ys)),
colIndexes, SEQUENCE(
1,
MAX(COLUMNS(xs), COLUMNS(ys))
),
IFERROR(
IF(rowIndexes <= nx,
INDEX(xs, rowIndexes, colIndexes),
INDEX(ys, rowIndexes - nx, colIndexes)
),
NA()
)
)
)
)
APPLYN
=LAMBDA(n,
LAMBDA(f,
LAMBDA(x,
IF(0 < n,
APPLYN(n - 1)(f)(
f(x)
),
x
)
)
)
)
LASTNROWS
=LAMBDA(n,
LAMBDA(xs,
LET(
nRows, COUNTA(xs),
x, MIN(nRows, n),
IF(0 < n,
INDEX(
xs,
SEQUENCE(
x, 1,
1 + nRows - x, 1
)
),
NA()
)
)
)
)- Output:
The FIBONACCI(n) lambda defines a column of integers.
Here we obtain a row, by composing FIBONACCI with the built-in TRANSPOSE function:
| fx | =TRANSPOSE(FIBONACCI(15)) | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | ||
| 1 | |||||||||||||||||
| 2 | 15 Fibonacci terms: | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 | 34 | 55 | 89 | 144 | 233 | 377 | 610 | |
Or as a fold, obtaining just the Nth term of the Fibonacci series:
FIBONACCI2
=LAMBDA(n,
INDEX(
FOLDL(
LAMBDA(ab,
LAMBDA(_,
APPEND(INDEX(ab, 2))(SUM(ab))
)
)
)({0;1})(
ENUMFROMTO(1)(n)
),
1
)
)Assuming the following generic bindings in the Excel worksheet Name manager:
APPEND
=LAMBDA(xs,
LAMBDA(ys,
LET(
nx, ROWS(xs),
rowIndexes, SEQUENCE(nx + ROWS(ys)),
colIndexes, SEQUENCE(
1,
MAX(COLUMNS(xs), COLUMNS(ys))
),
IF(rowIndexes <= nx,
INDEX(xs, rowIndexes, colIndexes),
INDEX(ys, rowIndexes - nx, colIndexes)
)
)
)
)
ENUMFROMTO
=LAMBDA(a,
LAMBDA(z,
SEQUENCE(1 + z - a, 1, a, 1)
)
)
FOLDL
=LAMBDA(op,
LAMBDA(a,
LAMBDA(xs,
IF(
2 > ROWS(xs),
op(a)(xs),
FOLDL(op)(
op(a)(
HEAD(xs)
)
)(
TAIL(xs)
)
)
)
)
)
HEAD
=LAMBDA(xs,
INDEX(xs, 1, SEQUENCE(1, COLUMNS(xs)))
)
TAIL
=LAMBDA(xs,
INDEX(
xs,
SEQUENCE(ROWS(xs) - 1, 1, 2, 1),
SEQUENCE(1, COLUMNS(xs))
)
)- Output:
| fx | =FIBONACCI2(A2) | ||
|---|---|---|---|
| A | B | ||
| 1 | N | Fibonacci | |
| 2 | 32 | 2178309 | |
| 3 | 64 | 10610209857723 | |
F#
This is a fast [tail-recursive] approach using the F# big integer support:
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 = 354224848179261915075ILazy evaluated using sequence workflow:
let rec fib = seq { yield! [0;1];
for (a,b) in Seq.zip fib (Seq.skip 1 fib) -> a+b}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:
let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I)
fibonacci |> Seq.nth 10000Approach 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.
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
|> iter2Factor
Iterative
: fib ( n -- m )
dup 2 < [
[ 0 1 ] dip [ swap [ + ] keep ] times
drop
] unless ;Recursive
: fib ( n -- m )
dup 2 < [
[ 1 - fib ] [ 2 - fib ] bi +
] unless ;Tail-Recursive
: fib2 ( x y n -- a )
dup 1 <
[ 2drop ]
[ [ swap [ + ] keep ] dip 1 - fib2 ]
if ;
: fib ( n -- m ) [ 0 1 ] dip fib2 ;Matrix
USE: math.matrices
: fib ( n -- m )
dup 2 < [
[ { { 0 1 } { 1 1 } } ] dip 1 - m^n
second second
] unless ;Falcon
Iterative
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
endRecursive
function fib_r(n)
if n < 2 : return n
return fib_r(n-1) + fib_r(n-2)
endTail Recursive
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
endFALSE
[[$0=~][1-@@\$@@+\$44,.@]#]f:
20n: {First 20 numbers}
0 1 n;f;!%%44,. {Output: "0,1,1,2,3,5..."}Fancy
class Fixnum {
def fib {
match self -> {
case 0 -> 0
case 1 -> 1
case _ -> self - 1 fib + (self - 2 fib)
}
}
}
15 times: |x| {
x fib println
}Fantom
Ints have a limit of 64-bits, so overflow errors occur after computing Fib(92) = 7540113804746346429.
class Main
{
static Int fib (Int n)
{
if (n < 2) return n
fibNums := [1, 0]
while (fibNums.size <= n)
{
fibNums.insert (0, fibNums[0] + fibNums[1])
}
return fibNums.first
}
public static Void main ()
{
20.times |n|
{
echo ("Fib($n) is ${fib(n)}")
}
}
}Fe
Recursive:
(= fib (fn (n)
(if (< n 2) n
(+ (fib (- n 1)) (fib (- n 2))))))Iterative:
(= fib (fn (n)
(let p0 0)
(let p1 1)
(while (< 0 n)
(= n (- n 1))
(let tmp (+ p0 p1))
(= p0 p1)
(= p1 tmp))
p0))Fermat
Func Fibonacci(n) = if n<0 then -(-1)^n*Fibonacci(-n) else if n<2 then n else
Array fib[n+1];
fib[1] := 0;
fib[2] := 1;
for i = 2, n do
fib[i+1]:=fib[i]+fib[i-1]
od;
Return(fib[n+1]);
fi;
fi;
.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))- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
Fish
Outputs Fibonacci numbers until stopped.
10::n' 'o&+&$10.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- Output:
FIBONACCI NUMBERS N =:20 F(N) = 6765
Forth
: fib ( n -- fib )
0 1 rot 0 ?do over + swap loop drop ;Or, for negative-index support:
: fib ( n -- Fn ) 0 1 begin
rot dup 0 = if drop drop exit then
dup 0 > if 1 - rot rot dup rot +
else 1 + rot rot over - swap then
again ;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.)
: 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.Fortran
FORTRAN IV
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- 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
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
ENDTest program
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- 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:
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 fibRIterative
In ISO Fortran 90 or later:
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 fibonacciTest program
program fibTest
use fibonacci
do i = 0, 10
print *, fibr(i), fibi(i)
end do
end program fibTest- 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
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;Frink
All of Frink's integers can be arbitrarily large.
fibonacciN[n] :=
{
a = 0
b = 1
count = 0
while count < n
{
[a,b] = [b, a + b]
count = count + 1
}
return a
}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.
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
RETFunL
Recursive
def
fib( 0 ) = 0
fib( 1 ) = 1
fib( n ) = fib( n - 1 ) + fib( n - 2 )Tail Recursive
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 )Lazy List
val fib =
def _fib( a, b ) = a # _fib( b, a + b )
_fib( 0, 1 )
println( fib(10000) )- Output:
33644764876431783266621612005107543310302148460680063906564769974680081442166662368155595513633734025582065332680836159373734790483865268263040892463056431887354544369559827491606602099884183933864652731300088830269235673613135117579297437854413752130520504347701602264758318906527890855154366159582987279682987510631200575428783453215515103870818298969791613127856265033195487140214287532698187962046936097879900350962302291026368131493195275630227837628441540360584402572114334961180023091208287046088923962328835461505776583271252546093591128203925285393434620904245248929403901706233888991085841065183173360437470737908552631764325733993712871937587746897479926305837065742830161637408969178426378624212835258112820516370298089332099905707920064367426202389783111470054074998459250360633560933883831923386783056136435351892133279732908133732642652633989763922723407882928177953580570993691049175470808931841056146322338217465637321248226383092103297701648054726243842374862411453093812206564914032751086643394517512161526545361333111314042436854805106765843493523836959653428071768775328348234345557366719731392746273629108210679280784718035329131176778924659089938635459327894523777674406192240337638674004021330343297496902028328145933418826817683893072003634795623117103101291953169794607632737589253530772552375943788434504067715555779056450443016640119462580972216729758615026968443146952034614932291105970676243268515992834709891284706740862008587135016260312071903172086094081298321581077282076353186624611278245537208532365305775956430072517744315051539600905168603220349163222640885248852433158051534849622434848299380905070483482449327453732624567755879089187190803662058009594743150052402532709746995318770724376825907419939632265984147498193609285223945039707165443156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875
Iterative
def fib( n ) =
a, b = 0, 1
for i <- 1..n
a, b = b, a+b
aBinet's Formula
import math.sqrt
def fib( n ) =
phi = (1 + sqrt( 5 ))/2
int( (phi^n - (-phi)^-n)/sqrt(5) + .5 )Matrix Exponentiation
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) )- Output:
0 1 1 2 3 5 8 13 21 34 55
Futhark
Iterative
fun main(n: int): int =
loop((a,b) = (0,1)) = for _i < n do
(b, a + b)
in aFutureBasic
Iterative
window 1, @"Fibonacci Sequence", (0,0,480,620)
local fn Fibonacci( n as long ) as long
static long s1
static long s2
long temp
if ( n < 2 )
s1 = n
exit fn
else
temp = s1 + s2
s2 = s1
s1 = temp
exit fn
end if
end fn = s1
long i
CFTimeInterval t
t = fn CACurrentMediaTime
for i = 0 to 40
print i;@".\t";fn Fibonacci(i)
next i
print : printf @"Compute time: %.3f ms",(fn CACurrentMediaTime-t)*1000
HandleEventsOutput:
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 Compute time: 2.143 ms
Recursive
Cost is a time penalty
local fn Fibonacci( n as NSInteger ) as NSInteger
NSInteger result
if n < 2 then result = n : exit fn
result = fn Fibonacci( n-1 ) + fn Fibonacci( n-2 )
end fn = result
window 1
NSInteger i
CFTimeInterval t
t = fn CACurrentMediaTime
for i = 0 to 40
print i;@".\t";fn Fibonacci(i)
next
print : printf @"Compute time: %.3f ms",(fn CACurrentMediaTime-t)*1000
HandleEvents- 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 Compute time: 2844.217 ms
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website.
In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.
Recursive
Recursive version is slow, it is O(2n), or of exponential order.
It is because it makes a lot of recursive calls.
To illustrate this, the following is a functions that makes a tree or the recursive calls:
Iterative (3 variables)
It is O(n), or of linear order.
Iterative (2 variables)
It is O(n), or of linear order.
Iterative, using a list
It is O(n), or of linear order.
Using matrix multiplication
Divide and conquer
It is an optimized version of the matrix multiplication algorithm, with an order of O(lg(n))
Closed-form
It has an order of O(lg(n))
GAP
fib := function(n)
local a;
a := [[0, 1], [1, 1]]^n;
return a[1][2];
end;GAP has also a buit-in function for that.
Fibonacci(n);Gecho
0 1 dup wover + dup wover + dup wover + dup wover +Prints the first several fibonacci numbers...
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;Go
Recursive
func fib(a int) int {
if a < 2 {
return a
}
return fib(a - 1) + fib(a - 2)
}Iterative
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
}Iterative using a closure
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
}Using a goroutine and channel
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)
}
}Grain
Recursive
import String from "string"
import File from "sys/file"
let rec fib = n => if (n < 2) {
n
} else {
fib(n - 1) + fib(n - 2)
}
for (let mut i = 0; i <= 20; i += 1) {
File.fdWrite(File.stdout, Pervasives.toString(fib(i)))
ignore(File.fdWrite(File.stdout, " "))
}Iterative
import File from "sys/file"
let fib = j => {
let mut fnow = 0, fnext = 1
for (let mut n = 0; n <= j; n += 1) {
if (n == 0 || n == 1) {
let output1 = " " ++ toString(n)
ignore(File.fdWrite(File.stdout, output1))
} else {
let tempf = fnow + fnext
fnow = fnext
fnext = tempf
let output2 = " " ++ toString(fnext)
ignore(File.fdWrite(File.stdout, output2))
}
}
}
fib(20)- Output:
Iterative with Buffer
import Buffer from "buffer"
import String from "string"
let fib = j => {
// set-up minimal, growable buffer
let buf = Buffer.make(j * 2)
let mut fnow = 0, fnext = 1
for (let mut n = 0; n <= j; n += 1) {
if (n == 0 || n == 1) {
Buffer.addChar(' ', buf)
Buffer.addString(toString(n), buf)
} else {
let tempf = fnow + fnext
fnow = fnext
fnext = tempf
Buffer.addChar(' ', buf)
Buffer.addString(toString(fnext), buf)
}
}
// stringify buffer and return
Buffer.toString(buf)
}
let output = fib(20)
print(output)- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
Groovy
Full "extra credit" solutions.
Recursive
A recursive closure must be pre-declared.
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)
}Iterative
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]
}Analytic
final φ = (1 + 5**(1/2))/2
def aFib = { (φ**it - (-φ)**(-it))/(5**(1/2)) as BigInteger }Test program:
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()
}- 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
#include "harbour.ch"
Function fibb(a,b,n)
return(if(--n>0,fibb(b,a+b,n),a))Iterative
#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)Haskell
Analytic
Using Binet's formula and exact real arithmetic library we can calculate arbitrary Fibonacci number exactly.
import Data.CReal
phi = (1 + sqrt 5) / 2
fib :: (Integral b) => b -> CReal 0
fib n = (phi^^n - (-phi)^^(-n))/sqrt 5Let's try it for large numbers:
λ> fib 10 :: CReal 0
55
(0.01 secs, 137,576 bytes)
λ> fib 100 :: CReal 0
354224848179261915075
(0.01 secs, 253,152 bytes)
λ> fib 10000 :: CReal 0
33644764876431783266621612005107543310302148460680063906564769974680081442166662368155595513633734025582065332680836159373734790483865268263040892463056431887354544369559827491606602099884183933864652731300088830269235673613135117579297437854413752130520504347701602264758318906527890855154366159582987279682987510631200575428783453215515103870818298969791613127856265033195487140214287532698187962046936097879900350962302291026368131493195275630227837628441540360584402572114334961180023091208287046088923962328835461505776583271252546093591128203925285393434620904245248929403901706233888991085841065183173360437470737908552631764325733993712871937587746897479926305837065742830161637408969178426378624212835258112820516370298089332099905707920064367426202389783111470054074998459250360633560933883831923386783056136435351892133279732908133732642652633989763922723407882928177953580570993691049175470808931841056146322338217465637321248226383092103297701648054726243842374862411453093812206564914032751086643394517512161526545361333111314042436854805106765843493523836959653428071768775328348234345557366719731392746273629108210679280784718035329131176778924659089938635459327894523777674406192240337638674004021330343297496902028328145933418826817683893072003634795623117103101291953169794607632737589253530772552375943788434504067715555779056450443016640119462580972216729758615026968443146952034614932291105970676243268515992834709891284706740862008587135016260312071903172086094081298321581077282076353186624611278245537208532365305775956430072517744315051539600905168603220349163222640885248852433158051534849622434848299380905070483482449327453732624567755879089187190803662058009594743150052402532709746995318770724376825907419939632265984147498193609285223945039707165443156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875
(0.02 secs, 4,847,128 bytes)
λ> fib (-10) :: CReal 0
-55
(0.01 secs, 138,408 bytes)Recursive
Simple definition, very inefficient.
fib x =
if x < 1
then 0
else if x < 2
then 1
else fib (x - 1) + fib (x - 2)Recursive with Memoization
Very fast.
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 ..]Recursive with Memoization using memoized library
Even faster and simpler is to use a defined memoizer (e.g. from MemoTrie package):
import Data.MemoTrie
fib :: Integer -> Integer
fib = memo f where
f 0 = 0
f 1 = 1
f n = fib (n-1) + fib (n-2)You can rewrite this without introducing f explicitly
import Data.MemoTrie
fib :: Integer -> Integer
fib = memo $ \x -> case x of
0 -> 0
1 -> 1
n -> fib (n-1) + fib (n-2)Or using LambdaCase extension you can write it even shorter:
{-# Language LambdaCase #-}
import Data.MemoTrie
fib :: Integer -> Integer
fib = memo $ \case
0 -> 0
1 -> 1
n -> fib (n-1) + fib (n-2)The version that supports negative numbers:
{-# 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)Iterative
fib n = go n 0 1
where
go n a b
| n == 0 = a
| otherwise = go (n - 1) b (a + b)With lazy lists
This is a standard example how to use lazy lists. Here's the (infinite) list of all Fibonacci numbers:
fib = 0 : 1 : zipWith (+) fib (tail fib)Or alternatively:
fib = 0 : 1 : (zipWith (+) <*> tail) fibThe nth Fibonacci number is then just fib !! n. The above is equivalent to
fib = 0 : 1 : next fib where next (a: t@(b:_)) = (a+b) : next tAlso
fib = 0 : scanl (+) 1 fibAs a fold
Accumulator holds last two members of the series:
import Data.List (foldl') --'
fib :: Integer -> Integer
fib n =
fst $
foldl' --'
(\(a, b) _ -> (b, a + b))
(0, 1)
[1 .. n]As an unfold
Create an infinite list of integers using an unfold. The nth fibonacci number is the nth number in the list.
import Data.List (unfoldr)
fibs :: [Integer]
fibs = unfoldr (\(x, y) -> Just (x, (y, x + y))) (0, 1)
fib n :: Integer -> Integer
fib n = fibs !! nWith matrix exponentiation
Adapting the (rather slow) code from Matrix exponentiation operator, we can simply write:
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)So, for example, the hundred-thousandth Fibonacci number starts with the digits:
- Output:
"2597406934"
With recurrence relations
Using Fib[m=3n+r] recurrence identities:
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)- 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:
*Main> (length &&& take 20) . show . fst $ fibN2 (10^6)
(208988,"19532821287077577316")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
f (n,(a,b)) = (2*n,(a*a+b*b,2*a*b+b*b)) -- iterate f (1,(0,1)) ; b is nthand for (n,n+1) ---> (2n,2n+1) (derived from d'Ocagne's identity, for example),
g (n,(a,b)) = (2*n,(2*a*b-a*a,a*a+b*b)) -- iterate g (1,(1,1)) ; a is nthHaxe
Iterative
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);
}As Iterator
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;
}
}Used like:
for (i in new FibIter(10))
Sys.println(i);HicEst
REAL :: Fibonacci(10)
Fibonacci = ($==2) + Fibonacci($-1) + Fibonacci($-2)
WRITE(ClipBoard) Fibonacci ! 0 1 1 2 3 5 8 13 21 34Hoon
|= n=@ud
=/ a=@ud 0
=/ b=@ud 1
|-
?: =(n 0) a
$(a b, b (add a b), n (dec n))Hope
Recursive
dec f : num -> num;
--- f 0 <= 0;
--- f 1 <= 1;
--- f(n+2) <= f n + f(n+1);Tail-recursive
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;With lazy lists
This language, being one of Haskell's ancestors, also has lazy lists. Here's the (infinite) list of all Fibonacci numbers:
dec fibs : list num;
--- fibs <= fs whererec fs == 0::1::map (+) (tail fs||fs);The nth Fibonacci number is then just fibs @ n.
Hy
Recursive implementation.
(defn fib [n]
(if (< n 2)
n
(+ (fib (- n 2)) (fib (- n 1)))))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.
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.
IDL
Recursive
function fib,n
if n lt 3 then return,1L else return, fib(n-1)+fib(n-2)
endExecution time O(2^n) until memory is exhausted and your machine starts swapping. Around fib(35) on a 2GB Core2Duo.
Iterative
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
endExecution time O(n). Limited by size of uLong to fib(49)
Analytic
function fib,n
q=1/( p=(1+sqrt(5))/2 )
return,round((p^n+q^n)/sqrt(5))
endExecution time O(1), only limited by the range of LongInts to fib(48).
Idris
Analytic
fibAnalytic : Nat -> Double
fibAnalytic n =
floor $ ((pow goldenRatio n) - (pow (-1.0/goldenRatio) n)) / sqrt(5)
where goldenRatio : Double
goldenRatio = (1.0 + sqrt(5)) / 2.0Recursive
fibRecursive : Nat -> Nat
fibRecursive Z = Z
fibRecursive (S Z) = (S Z)
fibRecursive (S (S n)) = fibRecursive (S n) + fibRecursive nIterative
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)Lazy
fibLazy : Lazy (List Nat)
fibLazy = 0 :: 1 :: zipWith (+) fibLazy (
case fibLazy of
(x::xs) => xs
[] => [])J
The Fibonacci Sequence essay on the J Wiki presents a number of different ways of obtaining the nth Fibonacci number.
This implementation is doubly recursive except that results are cached across function calls:
fibN=: (-&2 +&$: <:)^:(1&<) M."0Iteration:
fibN=: [: {."1 +/\@|.@]^:[&0 1Examples:
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 832040Java
Iterative
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;
}/**
* 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;
}Recursive
public static long recFibN(final int n)
{
return (n < 2) ? n : recFibN(n - 1) + recFibN(n - 2);
}Caching-recursive
A variant on recursive, that caches previous results, reducing complexity from O(n2) to simply O(n). Leveraging Java’s Map.computeIfAbsent makes this thread-safe, and the implementation pretty trivial.
public class Fibonacci {
static final Map<Integer, Long> cache = new HashMap<>();
static {
cache.put(1, 1L);
cache.put(2, 1L);
}
public static long get(int n)
{
return (n < 2) ? n : impl(n);
}
private static long impl(int n)
{
return cache.computeIfAbsent(n, k -> impl(k-1) + impl(k-2));
}
}Analytic
This method works up to the 92nd Fibonacci number. After that, it goes out of range.
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));
}Tail-recursive
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);
}Streams
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();
}
}JavaScript
ES5
Recursive
Basic recursive function:
function fib(n) {
return n<2?n:fib(n-1)+fib(n-2);
}Can be rewritten as:
function fib(n) {
if (n<2) { return n; } else { return fib(n-1)+fib(n-2); }
}One possibility familiar to Scheme programmers is to define an internal function for iteration through anonymous tail recursion:
function fib(n) {
return function(n,a,b) {
return n>0 ? arguments.callee(n-1,b,a+b) : a;
}(n,0,1);
}Iterative
function fib(n) {
var a = 0, b = 1, t;
while (n-- > 0) {
t = a;
a = b;
b += t;
console.log(a);
}
return a;
}Memoization
With the keys of a dictionary,
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);
};
})();with the indices of an array,
(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);
})();
- Output:
2178309
Y-Combinator
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);
};
});Generators
function* fibonacciGenerator() {
var prev = 0;
var curr = 1;
while (true) {
yield curr;
curr = curr + prev;
prev = curr - prev;
}
}
var fib = fibonacciGenerator();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.
(() => {
'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
})();Otherwise, a simple fold will suffice.
(Memoized fold example)
(() => {
'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
})();- Output:
2178309
Joy
Recursive
DEFINE fib == [small] [] [pred dup pred] [+] binrec.Iterative
DEFINE fib == [1 0] dip [swap [+] unary] times popd.jq
Works with gojq, the Go implementation of jq
The C implementation of jq does not (yet) have infinite-precision integer arithmetic, and so using it, the following algorithms only give exact answers up to fib(78). By contrast, using the Go implementation of jq and the definition of nth_fib given below:
nth_fib(pow(2;20)) | tostring | [length, .[:10], .[-10:]]yields
[219140,"1186800606","0691163707"]
in about 20 seconds on a 3GHz machine.
Using either the C or Go implementations, 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
def nth_fib_naive(n):
if (n < 2) then n
else nth_fib_naive(n - 1) + nth_fib_naive(n - 2)
end;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.
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;Example:
(range(0;5), 50) | [., nth_fib(.)]yields:
[0,0]
[1,1]
[2,1]
[3,2]
[4,3]
[50,12586269025]Binet's Formula
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) ) / $rt ) + .5
| floor;Generator
The following is a jq generator which produces the first n terms of the Fibonacci sequence efficiently, one by one. Notice that it is simply a variant of the above tail-recursive function. The function is in effect turned into a generator by changing "( _ | fib )" to "$sum, (_ | fib)".
# 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;Julia
Recursive
fib(n) = n < 2 ? n : fib(n-1) + fib(n-2)Iterative
function fib(n)
x,y = (0,1)
for i = 1:n x,y = (y, x+y) end
x
endMatrix form
fib(n) = ([1 1 ; 1 0]^n)[1,2]K
Recursive
{:[x<3;1;_f[x-1]+_f[x-2]]}Recursive with memoization
Using a (global) dictionary c.
{c::.();{v:c[a:`$$x];:[x<3;1;:[_n~v;c[a]:_f[x-1]+_f[x-2];v]]}x}Analytic
phi:(1+_sqrt(5))%2
{_((phi^x)-((1-phi)^x))%_sqrt[5]}Sequence to n
{(x(|+\)\1 1)[;1]}{x{x,+/-2#x}/!2}Kabap
Sequence to n
// 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 << " (" << $sequence << ")";Klingphix
:Fibonacci
dup 0 less
( ["Invalid argument"]
[1 1 rot 2 sub [drop over over add] for]
) if
;
30 Fibonacci pstack print nl
msec print nl
"bertlham " inputKotlin
enum class Fibonacci {
ITERATIVE {
override fun get(n: Int): Long = if (n < 2) {
n.toLong()
} else {
var n1 = 0L
var n2 = 1L
repeat(n) {
val sum = n1 + n2
n1 = n2
n2 = sum
}
n1
}
},
RECURSIVE {
override fun get(n: Int): Long = if (n < 2) n.toLong() else this[n - 1] + this[n - 2]
},
CACHING {
val cache: MutableMap<Int, Long> = mutableMapOf(0 to 0L, 1 to 1L)
override fun get(n: Int): Long = if (n < 2) n.toLong() else impl(n)
private fun impl(n: Int): Long = cache.computeIfAbsent(n) { impl(it-1) + impl(it-2) }
},
;
abstract operator fun get(n: Int): Long
}
fun main() {
val r = 0..30
for (fib in Fibonacci.values()) {
print("${fib.name.padEnd(10)}:")
for (i in r) { print(" " + fib[i]) }
println()
}
}- 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 CACHING : 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++
(defn int fib (int n) (return (? (< n 2) n (+ (fib (- n 1)) (fib (- n 2))))))
(main (prn (fib 30)))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
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
Iterative
fp.fib = ($n) -> {
if($n < 2) {
return $n
}
$prev = 1
$cur = 1
$i = 2
while($i < $n) {
$tmp = $cur
$cur += $prev
$prev = $tmp
$i += 1
}
return $cur
}Recursive
fp.fib = ($n) -> {
if($n < 2) {
return $n
}
return parser.op(fp.fib($n - 1) + fp.fib($n - 2))
}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 ;langur
val fibonacci = fn x:if(x < 2: x ; self(x - 1) + self(x - 2))
writeln map(fibonacci, series(2..20))- Output:
[1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]
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 2Latitude
Recursive
fibo := {
takes '[n].
if { n <= 1. } then {
n.
} else {
fibo (n - 1) + fibo (n - 2).
}.
}.Memoization
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.
}.
Lean
It runs on Lean 3.4.2:
-- 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
It runs on Lean 4:
-- Naive version
def fib1 (n : Nat) : Nat :=
match n with
| 0 => 0
| 1 => 1
| (k + 2) => fib1 k + fib1 (k + 1)
-- More efficient version
def fib_aux (n : Nat) (a : Nat) (b : Nat) : Nat :=
match n with
| 0 => b
| (k + 1) => fib_aux k (a + b) a
def fib2 (n : Nat) : Nat :=
fib_aux n 1 0
-- Examples
#eval fib1 20
#eval fib2 20LFE
Recursive
(defun fib
((0) 0)
((1) 1)
((n)
(+ (fib (- n 1))
(fib (- n 2)))))Iterative
(defun fib
((n) (when (>= n 0))
(fib n 0 1)))
(defun fib
((0 result _)
result)
((n result next)
(fib (- n 1) next (+ result next))))
Lingo
Recursive
on fib (n)
if n<2 then return n
return fib(n-1)+fib(n-2)
endIterative
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
endAnalytic
on fib (n)
sqrt5 = sqrt(5.0)
p = (1+sqrt5)/2
q = 1 - p
return integer((power(p,n)-power(q,n))/sqrt5)
endLisaac
- fib(n : UINTEGER_32) : UINTEGER_64 <- (
+ result : UINTEGER_64;
(n < 2).if {
result := n;
} else {
result := fib(n - 1) + fib(n - 2);
};
result
);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 fibrLLVM
; 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
}- Output:
0 1 1 2 3 5 8 13 21 34 55 89 144
Logo
to fib "loop
make "fib1 0
make "fib2 1
type [You requested\ ]
type :loop
print [\ Fibonacci Numbers]
type :fib1
type [\ ]
type :fib2
type [\ ]
make "loop :loop - 2
repeat :loop [ make "fibnew :fib1 + :fib2 type :fibnew type [\ ] make "fib1 :fib2 make "fib2 :fibnew ]
print [\ ]
endLOLCODE
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
KTHXBYELSL
Rez a box on the ground, and add the following as a New Script.
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));
}
}
}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
--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)
endPedantic Recursive
--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
endTail Recursive
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) endTable Recursive
fib_n = setmetatable({1, 1}, {__index = function(z,n) return n<=0 and 0 or z[n-1] + z[n-2] end})Table Recursive 2
-- 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
})Iterative
function ifibs(n)
local p0,p1=0,1
for _=1,n do p0,p1 = p1,p0+p1 end
return p0
endLuck
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))Lush
(de fib-rec (n)
(if (< n 2)
n
(+ (fib-rec (- n 2)) (fib-rec (- n 1)))))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.
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)
}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.
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
}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')MAD
NORMAL MODE IS INTEGER
INTERNAL FUNCTION(N)
ENTRY TO FIB.
A = 0
B = 1
THROUGH LOOP, FOR N=N, -1, N.E.0
C = A + B
A = B
LOOP B = C
FUNCTION RETURN A
END OF FUNCTION
THROUGH SHOW, FOR I=0, 1, I.GE.20
SHOW PRINT FORMAT FNUM, I, FIB.(I)
VECTOR VALUES FNUM = $4HFIB(,I2,4H) = ,I4*$
END OF PROGRAM- Output:
FIB( 0) = 0 FIB( 1) = 1 FIB( 2) = 1 FIB( 3) = 2 FIB( 4) = 3 FIB( 5) = 5 FIB( 6) = 8 FIB( 7) = 13 FIB( 8) = 21 FIB( 9) = 34 FIB(10) = 55 FIB(11) = 89 FIB(12) = 144 FIB(13) = 233 FIB(14) = 377 FIB(15) = 610 FIB(16) = 987 FIB(17) = 1597 FIB(18) = 2584 FIB(19) = 4181
Maple
> f := n -> ifelse(n<3,1,f(n-1)+f(n-2));
> f(2);
1
> f(3);
2Mathematica / Wolfram Language
The Wolfram Language already has a built-in function Fibonacci, but a simple recursive implementation would be
fib[0] = 0
fib[1] = 1
fib[n_Integer] := fib[n - 1] + fib[n - 2]An optimization is to cache the values already calculated:
fib[0] = 0
fib[1] = 1
fib[n_Integer] := fib[n] = fib[n - 1] + fib[n - 2]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:
fibi[prvprv_Integer, prv_Integer, rm_Integer] :=
If[rm < 1, prvprv, fibi[prv, prvprv + prv, rm - 1]]
fib[n_Integer] := fibi[0, 1, n]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):
fib[n_Integer] := Block[{tmp, prvprv = 0, prv = 1},
For[i = 0, i < n, i++, tmp = prv; prv += prvprv; prvprv = tmp];
Return[prvprv]]If one wanted a list of Fibonacci numbers, the following is quite efficient:
fibi[{prvprv_Integer, prv_Integer}] := {prv, prvprv + prv}
fibList[n_Integer] := Map[Take[#, 1] &, NestList[fibi, {0, 1}, n]] // FlattenOutput from the last with "fibList[100]":
{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}The Wolfram Language can also solve recurrence equations using the built-in function RSolve
fib[n] /. RSolve[{fib[n] == fib[n - 1] + fib[n - 2], fib[0] == 0,
fib[1] == 1}, fib[n], n][[1]]which evaluates to the built-in function Fibonacci[n]
This function can also be expressed as
Fibonacci[n] // FunctionExpand // FullSimplifywhich evaluates to
(2^-n ((1 + Sqrt[5])^n - (-1 + Sqrt[5])^n Cos[n π]))/Sqrt[5]and is defined for all real or complex values of n.
MATLAB
Matrix
function f = fib(n)
f = [1 1 ; 1 0]^(n-1);
f = f(1,1);
endIterative
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
endDramadah 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.
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
endTartaglia/Pascal Triangle Method
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));
endMaxima
/* 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);
-55MAXScript
Iterative
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
)
)Recursive
fn fibRec n =
(
if n < 2 then
(
n
)
else
(
fibRec (n - 1) + fibRec (n - 2)
)
)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
% 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).Iterative algorithm
The much faster iterative algorithm can be written as:
:- 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
).This predicate can be called as
fib_acc(1, 40, 1, 1, Result)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:
:- 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 ).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
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
min
(
(2 <)
(pred (1 0 (over + swap)) dip times pop)
unless
) ^fibMiniScript
An efficient solution (for n >= 0):
fibonacci = function(n)
if n < 2 then return n
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)And for comparison, a recursive solution (also for n >= 0):
rfib = function(n)
if n < 1 then return 0
if n == 1 then return 1
return rfib(n-1) + rfib(n-2)
end function
print rfib(6)MiniZinc
function var int: fibonacci(int: n) =
let {
array[0..n] of var int: fibonacci;
constraint forall(a in 0..n)(
fibonacci[a] = if (a == 0 \/ a == 1) then
a
else
fibonacci[a-1]+fibonacci[a-2]
endif
)
} in fibonacci[n];
var int: fib = fibonacci(6);
solve satisfy;
output [show(fib),"\n"];MIPS Assembly
This is the iterative approach to the Fibonacci sequence.
.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
syscallMirah
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 7МК-61/52
П0 1 lg Вx <-> + L0 03 С/П БП
03Instruction: 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
Tail Recursion
This version is tail recursive.
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)
endRecursion
fun fib n = if n < 2 then n else fib (n - 1) + fib (n - 2)MLite
Recursion
Tail recursive.
fun fib
(0, x1, x2) = x2
| (n, x1, x2) = fib (n-1, x2, x1+x2)
| n = fib (n, 0, 1)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)Modula-2
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.Modula-3
Recursive
PROCEDURE Fib(n: INTEGER): INTEGER =
BEGIN
IF n < 2 THEN
RETURN n;
ELSE
RETURN Fib(n-1) + Fib(n-2);
END;
END Fib;Iterative (with negatives)
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;Monicelli
Recursive version. It includes a main that reads a number N from standard input and prints the Nth Fibonacci number.
# 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 esecuzioneMontiLang
Reads number from standard input and prints to that number in the fibonacci sequence
0 VAR a .
1 VAR b .
INPUT TOINT
FOR :
a b + VAR c .
a PRINT .
b VAR a .
c VAR b .
ENDFORForth-style solution
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
clearSimpler
|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 #/MUMPS
Iterative
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)USER>W $$FIBOITER^ROSETTA(30) 832040
Nanoquery
Iterative
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
endNemerle
Recursive
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));
}
}Tail Recursive
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)
}NESL
Recursive
function fib(n) = if n < 2 then n else fib(n - 2) + fib(n - 1);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. */NewLISP
Iterative
(define (fibonacci n)
(let (L '(0 1))
(dotimes (i n)
(setq L (list (L 1) (apply + L))))
(L 1)) )Recursive
(define (fibonacci n)
(if (< n 2) 1
(+ (fibonacci (- n 1))
(fibonacci (- n 2)))))Matrix multiplication
(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)) ;;89With a stack
;;; Global variable (bigints); can be isolated in a namespace if need be
(setq stack '(0L 1L))
;
;;; If the stack is too short, complete it; then read from it
;;; Adding at the end of a list is optimized in NewLisp
(define (fib n)
(while (<= (length stack) n)
(push (+ (stack -1) (stack -2)) stack -1))
(stack n))
;
;;; Test (~ 7+ s on my mediocre laptop)
;(println (time (fib 50000)))
;;; or
(println (length (fib 50000)))
;;; outputs 10450 (digits)NGS
Iterative
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
}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.
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};Iterative using fold. Slightly faster, <1.6s:
fibf is op n {1 pick ((n- 1) fold [1 pick, +] 0 1)};Tacit verion of above. Slightly faster still, <1.4s:
fibf2 is 1 pick fold [1 pick, +] reverse (0 1 hitch) (-1+);Recursive
Really slow (over 8s for single iteration, n=33). (Similar to time for recursive python version with n=37.)
fibr is op n {fork [2>, +, + [fibr (-1 +), fibr (-2 +)]] n};...or tacit version. More than twice as fast (?) but still slow:
fibr2 is fork [2>, +, + [fibr2 (-1 +), fibr2 (-2 +)]];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.
fibm is op n {floor (0 1 pick (reduce ip (n reshape [2 2 reshape 1 1 1 0])))};Could it look a little more like J? (Maybe 5% slower than above.)
$ 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$1 1 1 0](~$)));Alternate, not involving replicating matrix n times, but maybe 50% slower than the fastest matrix version above - similar speed to iterative:
fibm3 is op n {a:=2 2$1 1 1 0; _(0 1 pick ((n- 1) fold (a ip) a))};Nim
Analytic
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))Iterative
proc Fibonacci(n: int): int =
var
first = 0
second = 1
for i in 0 .. <n:
swap first, second
second += first
result = firstRecursive
proc Fibonacci(n: int): int64 =
if n <= 2:
result = 1
else:
result = Fibonacci(n - 1) + Fibonacci(n - 2)Tail-recursive
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)Continuations
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()Nix
fibonacci = n:
if n <= 1 then n else (fibonacci (n - 1) + fibonacci (n - 2));Oberon-2
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.- 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
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);
}
}
}Objective-C
Recursive
-(long)fibonacci:(int)position
{
long result = 0;
if (position < 2) {
result = position;
} else {
result = [self fibonacci:(position -1)] + [self fibonacci:(position -2)];
}
return result;
}Iterative
+(long)fibonacci:(int)index {
long beforeLast = 0, last = 1;
while (index > 0) {
last += beforeLast;
beforeLast = last - beforeLast;
--index;
}
return last;
}OCaml
Tail-Recursive (fast)
let fib n =
let rec aux i a b =
if i = 0 then a else aux (pred i) b (a + b)
in
aux n 0 1Iterative
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;
!fibRecursive
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)Arbitrary Precision
Using OCaml's Num module.
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))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)).
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)Output:
fib 300 = 222232244629420445529739893461909967206666939096499764990979600
Matrix Exponentiation
open List
;;
let rec bin n =
if n < 2 then [n mod 2 = 1]
else bin (n/2) @ [n mod 2 = 1]
;;
let cut = function
| [] -> []
| _ :: x -> x
;;
let multiply a b =
let ((p, q), (r, s)) = a in
let ((t, u), (v, w)) = b in
((p*t+q*v, p*u+q*w), (r*t+s*v, r*u+s*w))
;;
let fib n =
let rec f p q r =
if length r = 1 then
if nth r 0 then (multiply p q, q)
else (p, q)
else
let (pp, qq) = f p q (cut r) in
let qqq = multiply qq qq in
if nth r 0 then (multiply pp qqq, qqq)
else (pp, qqq) in
f ((1L, 0L), (0L, 1L)) ((1L, 1L), (1L, 0L)) (bin n) |> fst |> fst |> snd
;;Octave
Recursive
% recursive
function fibo = recfibo(n)
if ( n < 2 )
fibo = n;
else
fibo = recfibo(n-1) + recfibo(n-2);
endif
endfunctionTesting
% testing
for i = 0 : 20
printf("%d %d\n", i, recfibo(i));
endforIterative
% 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
endfunctionTesting
% testing
for i = 0 : 20
printf("%d %d\n", i, iterfibo(i));
endforAnalytic
function retval = fibanalytic(n)
retval = round(((5 .^ 0.5 + 1) / 2) .^ n / 5 .^ 0.5);
endfunctionTail Recursive
function retval = fibtailrecursive(n, prevfib = 0, fib = 1)
if (n == 0)
retval = prevfib;
else
retval = fibtailrecursive(n - 1, fib, prevfib + fib);
endif
endfunctionOdin
Fibinacci Sequence - Iterative Solution
Odin Build: dev-2023-07-nightly:3072479c
package fib
import "core:fmt"
main :: proc() {
fmt.println("\nFibonacci Seq - starting n and n+1 from 0 and 1:")
fmt.println("------------------------------------------------")
for j: u128 = 0; j <= 20; j += 1 {
fmt.println("n:", j, "\tFib:", fibi(j))
}
}
fibi :: proc(n: u128) -> u128 {
if n < 2 {
return n
}
a, b: u128 = 0, 1
for _ in 2..=n {
a += b
a, b = b, a
}
return b
}- Output:
First 20 Fibonacci Numbers Starting n and n+1 from 0 and 1 ------------------------------- n: 0 Fib: 0 n: 1 Fib: 1 n: 2 Fib: 1 n: 3 Fib: 2 n: 4 Fib: 3 n: 5 Fib: 5 n: 6 Fib: 8 n: 7 Fib: 13 n: 8 Fib: 21 n: 9 Fib: 34 n: 10 Fib: 55 n: 11 Fib: 89 n: 12 Fib: 144 n: 13 Fib: 233 n: 14 Fib: 377 n: 15 Fib: 610 n: 16 Fib: 987 n: 17 Fib: 1597 n: 18 Fib: 2584 n: 19 Fib: 4181 n: 20 Fib: 6765 -------------------------------
Oforth
: fib 0 1 rot #[ tuck + ] times drop ;Ol
Same as Scheme example(s).
Onyx (wasm)
use core.iter
use core { printf }
// Procedural Simple For-Loop Style
fib_for_loop :: (n: i32) -> i32 {
a: i32 = 0;
b: i32 = 1;
for 0 .. n {
tmp := a;
a = b;
b = tmp + b;
}
return a;
}
FibState :: struct { a, b: u64 }
// Functional Folding Style
fib_by_fold :: (n: i32) => {
end_state :=
iter.counter()
|> iter.take(n)
|> iter.fold(
FibState.{ a = 0, b = 1 },
(_, state) => FibState.{
a = state.b,
b = state.a + state.b
}
);
return end_state.a;
}
// Custom Iterator Style
fib_iterator :: (n: i32) =>
iter.generator(
&.{ a = cast(u64) 0, b = cast(u64) 1, counter = n },
(state: & $Ctx) -> (u64, bool) {
if state.counter <= 0 {
return 0, false;
}
tmp := state.a;
state.a = state.b;
state.b = state.b + tmp;
state.counter -= 1;
return tmp, true;
}
);
main :: () {
printf("\nBy For Loop:\n");
for i in 0 .. 21 {
printf("{} ", fib_for_loop(i));
}
printf("\n\nBy Iterator:\n");
for i in 0 .. 21 {
printf("{} ", fib_by_fold(i));
}
printf("\n\nBy Fold:\n");
for value, index in fib_iterator(21) {
printf("{} ", value);
}
}- Output:
For-Loop: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 Functional Fold: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 Custom Iterator: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
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=1Order
Recursive
#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))Tail recursive version (example supplied with language):
#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))))Memoization
#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)
)Oz
Iterative
Using mutable references (cells).
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
endRecursive
Inefficient (blows up the stack).
fun{FibR N}
if N < 2 then N
else {FibR N-1} + {FibR N-2}
end
endTail-recursive
Using accumulators.
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}
endLazy-recursive
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}}PARI/GP
Built-in
fibonacci(n)Matrix
fib(n)=([1,1;1,0]^n)[1,2]Analytic
This uses the Binet form.
fib(n)=my(phi=(1+sqrt(5))/2);round((phi^n-phi^-n)/sqrt(5))The second term can be dropped since the error is always small enough to be subsumed by the rounding.
fib(n)=round(((1+sqrt(5))/2)^n/sqrt(5))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.
fib(n)=imag(quadgen(5)^n)A more direct translation (note that ) would be
fib(n)=my(phi=quadgen(5));(phi^n-(-1/phi)^n)/(2*phi-1)Combinatorial
This uses the generating function. It can be trivially adapted to give the first n Fibonacci numbers.
fib(n)=polcoeff(x/(1-x-x^2)+O(x^(n+1)),n)Binary powering
This is an efficient method (similar to the one used internally by fibonacci()), although running it without compilation won't give competitive speed.
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])
)
};Recursive
fib(n)={
if(n<2,
n
,
fib(n-1)+fib(n)
)
};Anonymous recursion
This uses self() which gives a self-reference.
fib(n)={
if(n<2,
n
,
my(s=self());
s(n-2)+s(n-1)
)
};It can be used without being named:
apply(n->if(n<2,n,my(s=self());s(n-2)+s(n-1)), [1..10])gives
- Output:
%1 = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
Memoization
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))
)
);
}Iterative
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
};Trigonometric
This solution uses the complex hyperbolic sine.
fib(n)=real(2/sqrt(5)/I^n*sinh(n*log(I*(1+sqrt(5))/2)))\/1;Chebyshev
This solution uses Chebyshev polynomials of the second kind (Chyebyshev U-polynomials).
fib(n)=n--;polchebyshev(n,2,I/2)*I^n;or
fib(n)=abs(polchebyshev(n-1,2,I/2));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.
matantihadamard(n)={
matrix(n,n,i,j,
my(t=j-i+1);
if(t<1,t%2,t<3)
);
}
fib(n)=matdet(matantihadamard(n))
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:
fib(n)=
{
my(g=2^(n+1)-1);
sum(i=2^(n-1),2^n-1,
bitor(i,i<<1)==g
);
}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.
fib(n)=my(k=0);while(n--,k++;while(!issquare(5*k^2+4)&&!issquare(5*k^2-4),k++));kParaCL
fibbonachi = func(x) : fibb
{
res = 1;
if (x > 1)
res = fibb(x - 1) + fibb(x - 2);
res;
}Pascal
Analytic
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;Recursive
function fib(n: integer): integer;
begin
if (n = 0) or (n = 1)
then
fib := n
else
fib := fib(n-1) + fib(n-2)
end;Iterative
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;Analytic2
function FiboMax(n: integer):Extended; //maXbox
begin
result:= (pow((1+SQRT5)/2,n)-pow((1-SQRT5)/2,n))/SQRT5
end;
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;writeln(floattoStr(FiboMax(555))) >>>4.3516638122555E115
writeln(Fibo_BigInt(555)) >>>43516638122555047989641805373140394725407202037260729735885664398655775748034950972577909265605502785297675867877570
Doubling (iterative)
The Clojure solution gives a recursive version of the Fibonacci doubling algorithm. The code below is an iterative version, written in Free Pascal. The unsigned 64-bit integer type can hold Fibonacci numbers up to F[93].
program Fibonacci_console;
{$mode objfpc}{$H+}
uses SysUtils;
function Fibonacci( n : word) : uint64;
{
Starts with the pair F[0],F[1]. At each iteration, uses the doubling formulae
to pass from F[k],F[k+1] to F[2k],F[2k+1]. If the current bit of n (starting
from the high end) is 1, there is a further step to F[2k+1],F[2k+2].
}
var
marker, half_n : word;
f, g : uint64; // pair of consecutive Fibonacci numbers
t, u : uint64; // -----"-----
begin
// The values of F[0], F[1], F[2] are assumed to be known
case n of
0 : result := 0;
1, 2 : result := 1;
else begin
half_n := n shr 1;
marker := 1;
while marker <= half_n do marker := marker shl 1;
// First time: current bit is 1 by construction,
// so go straight from F[0],F[1] to F[1],F[2].
f := 1; // = F[1]
g := 1; // = F[2]
marker := marker shr 1;
while marker > 1 do begin
t := f*(2*g - f);
u := f*f + g*g;
if (n and marker = 0) then begin
f := t;
g := u;
end
else begin
f := u;
g := t + u;
end;
marker := marker shr 1;
end;
// Last time: we need only one of the pair.
if (n and marker = 0) then
result := f*(2*g - f)
else
result := f*f + g*g;
end; // end else (i.e. n > 2)
end; // end case
end;
// Main program
var
n : word;
begin
for n := 0 to 93 do
WriteLn( SysUtils.Format( 'F[%2u] = %20u', [n, Fibonacci(n)]));
end.- Output:
F[ 0] = 0 F[ 1] = 1 F[ 2] = 1 F[ 3] = 2 F[ 4] = 3 F[ 5] = 5 F[ 6] = 8 F[ 7] = 13 [...] F[90] = 2880067194370816120 F[91] = 4660046610375530309 F[92] = 7540113804746346429 F[93] = 12200160415121876738
PascalABC.NET
Uses functionality from Fibonacci n-step number sequences#PascalABC.NET
// Fibonacci sequence. Nigel Galloway: August 30th., 2022
begin
unfold(nFib,0bi,1bi).ElementAt(1000).Println;
end.43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
Perl
Iterative
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;
}Recursive
sub fibRec {
my $n = shift;
$n < 2 ? $n : fibRec($n - 1) + fibRec($n - 2);
}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.
# 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);Array accumulation
This solution accumulates all Fibonacci numbers up to n into an array of n+1 elements (to account for the zeroth Fibonacci number). When the loop reaches n, the function returns the last element of the array, i.e. the n-th Fibonacci number. This function only works for positive integers, but it can be easily extended into negatives.
Note that, without the use of big integer libraries, pure Perl switches to floats in scientific notation above n=93 and treats any number as infinite above n=1476 (see output). This behaviour could vary across Perl implementations.
sub fibonacci {
my $n = shift;
return 0 if $n < 1;
return 1 if $n == 1;
my @numbers = (0, 1);
push @numbers, $numbers[-1] + $numbers[-2] foreach 2 .. $n;
return $numbers[-1];
}
print "Fibonacci($_) -> ", (fibonacci $_), "\n"
foreach (0 .. 20, 50, 93, 94, 100, 200, 1000, 1476, 1477);- 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(50) -> 12586269025 Fibonacci(93) -> 12200160415121876738 Fibonacci(94) -> 1.97402742198682e+19 Fibonacci(100) -> 3.54224848179262e+20 Fibonacci(200) -> 2.8057117299251e+41 Fibonacci(1000) -> 4.34665576869374e+208 Fibonacci(1476) -> 1.3069892237634e+308 Fibonacci(1477) -> Inf
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")
- 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:
-- demo\rosetta\fibonacci.exw with javascript_semantics 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))}) -- not surprisingly JavaScript BigInt takes a bit -- longer for 0.1M digits that gmp does for 1.0M: integer big = iff(platform()==JS?478495:4784969) string s = mpz_get_str(fibonampz(big)) ?shorten(s) ?elapsed(time()-t0)
- 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
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 printPHP
Iterative
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 $fib;
}Recursive
function fibRec($n) {
return $n < 2 ? $n : fibRec($n-1) + fibRec($n-2);
}Picat
Function
tabling for speed.
go =>
println([fib_fun(I) : I in 1..10]),
F1=fib_fun(2**10),
println(f1=F1),
nl.
table
fib_fun(0) = 0.
fib_fun(1) = 1.
fib_fun(N) = fib_fun(N-1) + fib_fun(N-2).- Output:
[1,1,2,3,5,8,13,21,34,55] f1 = 4506699633677819813104383235728886049367860596218604830803023149600030645708721396248792609141030396244873266580345011219530209367425581019871067646094200262285202346655868899711089246778413354004103631553925405243
Array
fib_array(0,[0]).
fib_array(1,[0,1]).
fib_array(N,A) :-
N > 1,
A = new_array(N),
A[1] = 1,
A[2] = 1,
foreach(I in 3..N)
A[I] = A[I-1] + A[I-2]
end.Loop
fib_loop(N) = Curr =>
Curr = 0,
Prev = 1,
foreach(_I in 1..N)
Tmp = Curr,
Curr := Curr + Prev,
Prev := Tmp
end.Formula
Works for n <= 70.
fib_formula(N) = round((0.5 + 0.5*sqrt(5))**N / sqrt(5))."Lazy lists"
go =>fib_lazy(X),
A = new_list(15),
append(A,_,X),
println(A),
fib_lazy([0,1|X]) :-
ffib(0,1,X).
ffib(A,B,X) :-
freeze(X, (C is A+B, X=[C|Y], ffib(B,C,Y)) ).- Output:
[0,1,1,2,3,5,8,13,21,34,55,89,144,233,377]
Generators idiom
go =>
take(15, $fib_gen(0,1), $T-[], _G),
println(T).
take(