Fibonacci sequence: Difference between revisions
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=={{header|F_Sharp|F#}}== |
=={{header|F_Sharp|F#}}== |
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This is a fast [tail-recursive] approach using the F# big integer support |
This is a fast [tail-recursive] approach using the F# big integer support: |
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<lang fsharp> |
<lang fsharp> |
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let fibonacci n : bigint = |
let fibonacci n : bigint = |
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> fibonacci 100;; |
> fibonacci 100;; |
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val it : bigint = 354224848179261915075I</lang> |
val it : bigint = 354224848179261915075I</lang> |
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Lazy evaluated using sequence workflow |
Lazy evaluated using sequence workflow: |
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<lang fsharp>let rec fib = seq { yield! [0;1]; |
<lang fsharp>let rec fib = seq { yield! [0;1]; |
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for (a,b) in Seq.zip fib (Seq.skip 1 fib) -> a+b}</lang> |
for (a,b) in Seq.zip fib (Seq.skip 1 fib) -> a+b}</lang> |
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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! |
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⚫ | |||
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<lang fsharp>let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I) |
<lang fsharp>let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I) |
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fibonacci |> Seq.nth 10000 |
fibonacci |> Seq.nth 10000 |
Revision as of 01:57, 10 May 2014
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
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.
- Cf.
- References
- Wikipedia, Fibonacci number
- Wikipedia, Lucas number
- MathWorld, Fibonacci Number
- Some identities for r-Fibonacci numbers
- OEIS Fibonacci numbers
- OEIS Lucas numbers
0815
<lang 0815> %<:0D:>~$<:01:~%>=<:a94fad42221f2702:>~> }:_s:{x{={~$x+%{=>~>x~-x<:0D:~>~>~^:_s:? </lang>
ACL2
Fast, tail recursive solution: <lang Lisp>(defun fast-fib-r (n a b)
(if (or (zp n) (zp (1- n))) b (fast-fib-r (1- n) b (+ a b))))
(defun fast-fib (n)
(fast-fib-r n 1 1))
(defun first-fibs-r (n i)
(declare (xargs :measure (nfix (- n i)))) (if (zp (- n i)) nil (cons (fast-fib i) (first-fibs-r n (1+ i)))))
(defun first-fibs (n)
(first-fibs-r n 0))</lang>
Output:
>(first-fibs 20) (1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765)
ActionScript
<lang actionscript>public function fib(n:uint):uint {
if (n < 2) return n; return fib(n - 1) + fib(n - 2);
}</lang>
AppleScript
<lang applescript>set fibs to {} set x to (text returned of (display dialog "What fibbonaci number do you want?" default answer "3")) set x to x as integer repeat with y from 1 to x if (y = 1 or y = 2) then copy 1 to the end of fibs else copy ((item (y - 1) of fibs) + (item (y - 2) of fibs)) to the end of fibs end if end repeat return item x of fibs</lang>
Ada
Recursive
<lang Ada>with Ada.Text_IO, Ada.Command_Line;
procedure Fib is
X: Positive := Positive'Value(Ada.Command_Line.Argument(1));
function Fib(P: Positive) return Positive is begin if P <= 2 then return 1; else return Fib(P-1) + Fib(P-2); end if; end Fib;
begin
Ada.Text_IO.Put("Fibonacci(" & Integer'Image(X) & " ) = "); Ada.Text_IO.Put_Line(Integer'Image(Fib(X)));
end Fib;</lang>
Iterative, build-in integers
<lang ada>with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Fibonacci is
function Fibonacci (N : Natural) return Natural is This : Natural := 0; That : Natural := 1; Sum : Natural; begin for I in 1..N loop Sum := This + That; That := This; This := Sum; end loop; return This; end Fibonacci;
begin
for N in 0..10 loop Put_Line (Positive'Image (Fibonacci (N))); end loop;
end Test_Fibonacci;</lang> Sample output:
0 1 1 2 3 5 8 13 21 34 55
Iterative, long integers
Using the big integer implementation from a cryptographic library [1].
<lang Ada>with Ada.Text_IO, Ada.Command_Line, Crypto.Types.Big_Numbers;
procedure Fibonacci is
X: Positive := Positive'Value(Ada.Command_Line.Argument(1));
Bit_Length: Positive := 1 + (696 * X) / 1000; -- that number of bits is sufficient to store the full result.
package LN is new Crypto.Types.Big_Numbers (Bit_Length + (32 - Bit_Length mod 32)); -- the actual number of bits has to be a multiple of 32 use LN;
function Fib(P: Positive) return Big_Unsigned is Previous: Big_Unsigned := Big_Unsigned_Zero; Result: Big_Unsigned := Big_Unsigned_One; Tmp: Big_Unsigned; begin -- Result = 1 = Fibonacci(1) for I in 1 .. P-1 loop Tmp := Result; Result := Previous + Result; Previous := Tmp; -- Result = Fibonacci(I+1)) end loop; return Result; end Fib;
begin
Ada.Text_IO.Put("Fibonacci(" & Integer'Image(X) & " ) = "); Ada.Text_IO.Put_Line(LN.Utils.To_String(Fib(X)));
end Fibonacci;</lang>
Output:
> ./fibonacci 777 Fibonacci( 777 ) = 1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082
Aime
<lang aime>integer fibs(integer n) {
integer w;
if (n == 0) { w = 0; } elif (n == 1) { w = 1; } else { integer a, b, i;
i = 1; a = 0; b = 1; while (i < n) { w = a + b; a = b; b = w; i += 1; } }
return w;
} </lang>
ALGOL 68
Analytic
<lang algol68>PROC analytic fibonacci = (LONG INT n)LONG INT:(
LONG REAL sqrt 5 = long sqrt(5); LONG REAL p = (1 + sqrt 5) / 2; LONG REAL q = 1/p; ROUND( (p**n + q**n) / sqrt 5 )
);
FOR i FROM 1 TO 30 WHILE
print(whole(analytic fibonacci(i),0));
- WHILE # i /= 30 DO
print(", ")
OD; print(new line)</lang> Output:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040
Iterative
<lang algol68>PROC iterative fibonacci = (INT n)INT:
CASE n+1 IN 0, 1, 1, 2, 3, 5 OUT INT even:=3, odd:=5; FOR i FROM odd+1 TO n DO (ODD i|odd|even) := odd + even OD; (ODD n|odd|even) ESAC;
FOR i FROM 0 TO 30 WHILE
print(whole(iterative fibonacci(i),0));
- WHILE # i /= 30 DO
print(", ")
OD; print(new line)</lang> Output:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040
Recursive
<lang algol68>PROC recursive fibonacci = (INT n)INT:
( n < 2 | n | fib(n-1) + fib(n-2));</lang>
Generative
- note: This specimen retains the original Python coding style.
<lang algol68>MODE YIELDINT = PROC(INT)VOID;
PROC gen fibonacci = (INT n, YIELDINT yield)VOID: (
INT even:=0, odd:=1; yield(even); yield(odd); FOR i FROM odd+1 TO n DO yield( (ODD i|odd|even) := odd + even ) OD
);
main:(
# FOR INT n IN # gen fibonacci(30, # ) DO ( # ## (INT n)VOID:( print((" ",whole(n,0))) # OD # )); print(new line)
)</lang> Output:
1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
Array (Table) Lookup
This uses a pre-generated list, requiring much less run-time processor usage, but assumes that INT is only 31 bits wide. <lang algol68>[]INT const fibonacci = []INT( -1836311903, 1134903170,
-701408733, 433494437, -267914296, 165580141, -102334155, 63245986, -39088169, 24157817, -14930352, 9227465, -5702887, 3524578, -2178309, 1346269, -832040, 514229, -317811, 196418, -121393, 75025, -46368, 28657, -17711, 10946, -6765, 4181, -2584, 1597, -987, 610, -377, 233, -144, 89, -55, 34, -21, 13, -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903
)[@-46];
PROC VOID value error := stop;
PROC lookup fibonacci = (INT i)INT: (
IF LWB const fibonacci <= i AND i<= UPB const fibonacci THEN const fibonacci[i] ELSE value error; SKIP FI
);
FOR i FROM 0 TO 30 WHILE
print(whole(lookup fibonacci(i),0));
- WHILE # i /= 30 DO
print(", ")
OD; print(new line)</lang> Output:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040
Alore
<lang Alore>def fib(n as Int) as Int
if n < 2 return 1 end return fib(n-1) + fib(n-2)
end</lang>
AutoHotkey
Search autohotkey.com: sequence
Iterative
<lang AutoHotkey>Loop, 5
MsgBox % fib(A_Index)
Return
fib(n) {
If (n < 2) Return n i := last := this := 1 While (i <= n) { new := last + this last := this this := new i++ } Return this
}</lang>
Recursive and iterative
Source: AutoHotkey forum by Laszlo <lang AutoHotkey>/* Important note: the recursive version would be very slow without a global or static array. The iterative version handles also negative arguments properly.
- /
FibR(n) { ; n-th Fibonacci number (n>=0, recursive with static array Fibo)
Static Return n<2 ? n : Fibo%n% ? Fibo%n% : Fibo%n% := FibR(n-1)+FibR(n-2)
}
Fib(n) { ; n-th Fibonacci number (n < 0 OK, iterative)
a := 0, b := 1 Loop % abs(n)-1 c := b, b += a, a := c Return n=0 ? 0 : n>0 || n&1 ? b : -b
}</lang>
AutoIt
Iterative
<lang AutoIt>#AutoIt Version: 3.2.10.0 $n0 = 0 $n1 = 1 $n = 10 MsgBox (0,"Iterative Fibonacci ", it_febo($n0,$n1,$n))
Func it_febo($n_0,$n_1,$N)
$first = $n_0 $second = $n_1 $next = $first + $second $febo = 0 For $i = 1 To $N-3 $first = $second $second = $next $next = $first + $second Next if $n==0 Then $febo = 0 ElseIf $n==1 Then $febo = $n_0 ElseIf $n==2 Then $febo = $n_1 Else $febo = $next EndIf Return $febo
EndFunc </lang>
Recursive
<lang AutoIt>#AutoIt Version: 3.2.10.0 $n0 = 0 $n1 = 1 $n = 10 MsgBox (0,"Recursive Fibonacci ", rec_febo($n0,$n1,$n)) Func rec_febo($r_0,$r_1,$R)
if $R<3 Then if $R==2 Then
Return $r_1
ElseIf $R==1 Then
Return $r_0
ElseIf $R==0 Then
Return 0
EndIf Return $R Else Return rec_febo($r_0,$r_1,$R-1) + rec_febo($r_0,$r_1,$R-2) EndIf
EndFunc </lang>
AWK
As in many examples, this one-liner contains the function as well as testing with input from stdin, output to stdout. <lang awk>$ awk 'func fib(n){return(n<2?n:fib(n-1)+fib(n-2))}{print "fib("$1")="fib($1)}' 10 fib(10)=55</lang>
Babel
<lang babel>((main
{{iter fib !} 20 times collect ! rev
{%d " " . <<} each})
(collect { -1 take })
(fib
{{dup 2 <} {fnord} {dup <- 2 - fib ! -> 1 - fib ! + } ifte}))</lang>
This code produces the following output when run:
<lang babel>1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765</lang>
BASIC
Iterative
<lang qbasic>FUNCTION itFib (n)
n1 = 0 n2 = 1 FOR k = 1 TO ABS(n) sum = n1 + n2 n1 = n2 n2 = sum NEXT k IF n < 0 THEN itFib = n1 * ((-1) ^ ((-n) + 1)) ELSE itFib = n1 END IF
END FUNCTION</lang>
Next version calculates each value once, as needed, and stores the results in an array for later retreival (due to the use of REDIM PRESERVE
, it requires QuickBASIC 4.5 or newer):
<lang qbasic>DECLARE FUNCTION fibonacci& (n AS INTEGER)
REDIM SHARED fibNum(1) AS LONG
fibNum(1) = 1
'*****sample inputs***** PRINT fibonacci(0) 'no calculation needed PRINT fibonacci(13) 'figure F(2)..F(13) PRINT fibonacci(-42) 'figure F(14)..F(42) PRINT fibonacci(47) 'error: too big '*****sample inputs*****
FUNCTION fibonacci& (n AS INTEGER)
DIM a AS INTEGER a = ABS(n) SELECT CASE a CASE 0 TO 46 SHARED fibNum() AS LONG DIM u AS INTEGER, L0 AS INTEGER u = UBOUND(fibNum) IF a > u THEN REDIM PRESERVE fibNum(a) AS LONG FOR L0 = u + 1 TO a fibNum(L0) = fibNum(L0 - 1) + fibNum(L0 - 2) NEXT END IF IF n < 0 THEN fibonacci = fibNum(a) * ((-1) ^ (a + 1)) ELSE fibonacci = fibNum(n) END IF CASE ELSE 'limited to signed 32-bit int (LONG) 'F(47)=&hB11924E1 ERROR 6 'overflow END SELECT
END FUNCTION</lang>
Outputs (unhandled error in final input prevents output):
0 233 -267914296
Recursive
This example can't handle n < 0.
<lang qbasic>FUNCTION recFib (n)
IF (n < 2) THEN
recFib = n
ELSE
recFib = recFib(n - 1) + recFib(n - 2)
END IF
END FUNCTION</lang>
Array (Table) Lookup
This uses a pre-generated list, requiring much less run-time processor usage. (Since the sequence never changes, this is probably the best way to do this in "the real world". The same applies to other sequences like prime numbers, and numbers like pi and e.)
<lang qbasic>DATA -1836311903,1134903170,-701408733,433494437,-267914296,165580141,-102334155 DATA 63245986,-39088169,24157817,-14930352,9227465,-5702887,3524578,-2178309 DATA 1346269,-832040,514229,-317811,196418,-121393,75025,-46368,28657,-17711 DATA 10946,-6765,4181,-2584,1597,-987,610,-377,233,-144,89,-55,34,-21,13,-8,5,-3 DATA 2,-1,1,0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765 DATA 10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269 DATA 2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986 DATA 102334155,165580141,267914296,433494437,701408733,1134903170,1836311903
DIM fibNum(-46 TO 46) AS LONG
FOR n = -46 TO 46
READ fibNum(n)
NEXT
'*****sample inputs***** FOR n = -46 TO 46
PRINT fibNum(n),
NEXT PRINT '*****sample inputs*****</lang>
Batch File
Recursive version <lang dos>::fibo.cmd @echo off if "%1" equ "" goto :eof call :fib %1 echo %errorlevel% goto :eof
- fib
setlocal enabledelayedexpansion if %1 geq 2 goto :ge2 exit /b %1
- ge2
set /a r1 = %1 - 1 set /a r2 = %1 - 2 call :fib !r1! set r1=%errorlevel% call :fib !r2! set r2=%errorlevel% set /a r0 = r1 + r2 exit /b !r0!</lang>
Output:
>for /L %i in (1,5,20) do fibo.cmd %i >fibo.cmd 1 1 >fibo.cmd 6 8 >fibo.cmd 11 89 >fibo.cmd 16 987
BBC BASIC
<lang bbcbasic> PRINT FNfibonacci_r(1), FNfibonacci_i(1)
PRINT FNfibonacci_r(13), FNfibonacci_i(13) PRINT FNfibonacci_r(26), FNfibonacci_i(26) END DEF FNfibonacci_r(N) IF N < 2 THEN = N = FNfibonacci_r(N-1) + FNfibonacci_r(N-2) DEF FNfibonacci_i(N) LOCAL F, I, P, T IF N < 2 THEN = N P = 1 FOR I = 1 TO N T = F F += P P = T NEXT = F
</lang> Output:
1 1 233 233 121393 121393
bc
iterative
<lang bc>#! /usr/bin/bc -q
define fib(x) {
if (x <= 0) return 0; if (x == 1) return 1;
a = 0; b = 1; for (i = 1; i < x; i++) { c = a+b; a = b; b = c; } return c;
} fib(1000) quit</lang>
Befunge
<lang befunge>00:.1:.>:"@"8**++\1+:67+`#@_v
^ .:\/*8"@"\%*8"@":\ <</lang>
Brainf***
The first cell contains n (10), the second cell will contain fib(n) (55), and the third cell will contain fib(n-1) (34). <lang bf>++++++++++ >>+<<[->[->+>+<<]>[-<+>]>[-<+>]<<<]</lang>
The following generates n fibonacci numbers and prints them, though not in ascii. It does have a limit due to the cells usually being 1 byte in size. <lang bf>+++++ +++++ #0 set to n >> + Init #2 to 1 << [ - #Decrement counter in #0 >>. Notice: This doesn't print it in ascii To look at results you can pipe into a file and look with a hex editor
Copying sequence to save #2 in #4 using #5 as restore space >>[-] Move to #4 and clear >[-] Clear #5 <<< #2 [ Move loop - >> + > + <<< Subtract #2 and add #4 and #5 ] >>> [ Restore loop - <<< + >>> Subtract from #5 and add to #2 ]
<<<< Back to #1 Non destructive add sequence using #3 as restore value [ Loop to add - > + > + << Subtract #1 and add to value #2 and restore space #3 ] >> [ Loop to restore #1 from #3 - << + >> Subtract from restore space #3 and add in #1 ]
<< [-] Clear #1 >>> [ Loop to move #4 to #1 - <<< + >>> Subtract from #4 and add to #1 ] <<<< Back to #0 ]</lang>
Bracmat
Recursive
<lang bracmat>fib=.!arg:<2|fib$(!arg+-2)+fib$(!arg+-1)</lang>
fib$30 832040
Iterative
<lang bracmat>(fib=
last i this new
. !arg:<2
| 0:?last:?i & 1:?this & whl ' ( !i+1:<!arg:?i & !last+!this:?new & !this:?last & !new:?this ) & !this
)</lang>
fib$777 1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082
Brat
Recursive
<lang brat>fibonacci = { x |
true? x < 2, x, { fibonacci(x - 1) + fibonacci(x - 2) }
}</lang>
Tail Recursive
<lang brat>fib_aux = { x, next, result |
true? x == 0, result, { fib_aux x - 1, next + result, next }
}
fibonacci = { x |
fib_aux x, 1, 0
}</lang>
Memoization
<lang brat>cache = hash.new
fibonacci = { x |
true? cache.key?(x) { cache[x] } {true? x < 2, x, { cache[x] = fibonacci(x - 1) + fibonacci(x - 2) }}
}</lang>
Burlesque
<lang burlesque> {0 1}{^^++[+[-^^-]\/}30.*\[e!vv </lang>
<lang burlesque> 0 1{{.+}c!}{1000.<}w! </lang>
C
Recursive
<lang c>long long int fibb(long long int a, long long int b, int n) { return (--n>0)?(fibb(b, a+b, n)):(a); }</lang>
Iterative
<lang c>long long int fibb(int n) { int fnow = 0, fnext = 1, tempf; while(--n>0){ tempf = fnow + fnext; fnow = fnext; fnext = tempf; } return fnext; }</lang>
Analytic
<lang c>#include <tgmath.h>
- define PHI ((1 + sqrt(5))/2)
long long unsigned fib(unsigned n) {
return floor( (pow(PHI, n) - pow(1 - PHI, n))/sqrt(5) );
}</lang>
Generative
<lang c>#include <stdio.h> typedef enum{false=0, true=!0} bool; typedef void iterator;
- include <setjmp.h>
/* declare label otherwise it is not visible in sub-scope */
- define LABEL(label) jmp_buf label; if(setjmp(label))goto label;
- define GOTO(label) longjmp(label, true)
/* the following line is the only time I have ever required "auto" */
- define FOR(i, iterator) { auto bool lambda(i); yield_init = (void *)λ iterator; bool lambda(i)
- define DO {
- define YIELD(x) if(!yield(x))return
- define BREAK return false
- define CONTINUE return true
- define OD CONTINUE; } }
static volatile void *yield_init; /* not thread safe */
- define YIELDS(type) bool (*yield)(type) = yield_init
iterator fibonacci(int stop){
YIELDS(int); int f[] = {0, 1}; int i; for(i=0; i<stop; i++){ YIELD(f[i%2]); f[i%2]=f[0]+f[1]; }
}
main(){
printf("fibonacci: "); FOR(int i, fibonacci(16)) DO printf("%d, ",i); OD; printf("...\n");
}</lang> Output:
fibonacci: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, ...
Fast method for a single large value
<lang c>#include <stdlib.h>
- include <stdio.h>
- include <gmp.h>
typedef struct node node; struct node { int n; mpz_t v; node *next; };
- define CSIZE 37
node *cache[CSIZE];
// very primitive linked hash table node * find_cache(int n) { int idx = n % CSIZE; node *p;
for (p = cache[idx]; p && p->n != n; p = p->next); if (p) return p;
p = malloc(sizeof(node)); p->next = cache[idx]; cache[idx] = p;
if (n < 2) { p->n = n; mpz_init_set_ui(p->v, 1); } else { p->n = -1; // -1: value not computed yet mpz_init(p->v); } return p; }
mpz_t tmp1, tmp2; mpz_t *fib(int n) { int x; node *p = find_cache(n);
if (p->n < 0) { p->n = n; x = n / 2;
mpz_mul(tmp1, *fib(x-1), *fib(n - x - 1)); mpz_mul(tmp2, *fib(x), *fib(n - x)); mpz_add(p->v, tmp1, tmp2); } return &p->v; }
int main(int argc, char **argv) { int i, n; if (argc < 2) return 1;
mpz_init(tmp1); mpz_init(tmp2);
for (i = 1; i < argc; i++) { n = atoi(argv[i]); if (n < 0) { printf("bad input: %s\n", argv[i]); continue; }
// about 75% of time is spent in printing gmp_printf("%Zd\n", *fib(n)); } return 0; }</lang>
- Output:
% ./a.out 0 1 2 3 4 5 1 1 2 3 5 8 % ./a.out 10000000 | wc -c # count length of output, including the newline 1919488
C++
Using unsigned int, this version only works up to 48 before fib overflows. <lang cpp>#include <iostream>
int main() {
unsigned int a = 1, b = 1; unsigned int target = 48; for(unsigned int n = 3; n <= target; ++n) { unsigned int fib = a + b; std::cout << "F("<< n << ") = " << fib << std::endl; a = b; b = fib; }
return 0;
}</lang>
This version does not have an upper bound.
<lang cpp>#include <iostream>
- include <gmpxx.h>
int main() {
mpz_class a = mpz_class(1), b = mpz_class(1); mpz_class target = mpz_class(100); for(mpz_class n = mpz_class(3); n <= target; ++n) { mpz_class fib = b + a; if ( fib < b ) { std::cout << "Overflow at " << n << std::endl; break; } std::cout << "F("<< n << ") = " << fib << std::endl; a = b; b = fib; } return 0;
}</lang>
Version using transform: <lang cpp>#include <algorithm>
- include <vector>
- include <functional>
- include <iostream>
unsigned int fibonacci(unsigned int n) {
if (n == 0) return 0; std::vector<int> v(n+1); v[1] = 1; transform(v.begin(), v.end()-2, v.begin()+1, v.begin()+2, std::plus<int>()); // "v" now contains the Fibonacci sequence from 0 up return v[n];
}</lang>
Far-fetched version using adjacent_difference: <lang cpp>#include <numeric>
- include <vector>
- include <functional>
- include <iostream>
unsigned int fibonacci(unsigned int n) {
if (n == 0) return 0; std::vector<int> v(n, 1); adjacent_difference(v.begin(), v.end()-1, v.begin()+1, std::plus<int>()); // "array" now contains the Fibonacci sequence from 1 up return v[n-1];
} </lang>
Version which computes at compile time with metaprogramming: <lang cpp>#include <iostream>
template <int n> struct fibo {
enum {value=fibo<n-1>::value+fibo<n-2>::value};
};
template <> struct fibo<0> {
enum {value=0};
};
template <> struct fibo<1> {
enum {value=1};
};
int main(int argc, char const *argv[])
{
std::cout<<fibo<12>::value<<std::endl; std::cout<<fibo<46>::value<<std::endl; return 0;
}</lang>
The following version is based on fast exponentiation: <lang cpp>#include <iostream>
inline void fibmul(int* f, int* g) {
int tmp = f[0]*g[0] + f[1]*g[1]; f[1] = f[0]*g[1] + f[1]*(g[0] + g[1]); f[0] = tmp;
}
int fibonacci(int n) {
int f[] = { 1, 0 }; int g[] = { 0, 1 }; while (n > 0) { if (n & 1) // n odd { fibmul(f, g); --n; } else { fibmul(g, g); n >>= 1; } } return f[1];
}
int main() {
for (int i = 0; i < 20; ++i) std::cout << fibonacci(i) << " "; std::cout << std::endl;
}</lang> Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181
Using Zeckendorf Numbers
The nth fibonacci is represented as Zeckendorf 1 followed by n-1 zeroes. Here I define a class N which defines the operations increment ++() and comparison <=(other N) for Zeckendorf Numbers. <lang cpp> // Use Zeckendorf numbers to display Fibonacci sequence. // Nigel Galloway October 23rd., 2012 int main(void) {
char NG[22] = {'1',0}; int x = -1; N G; for (int fibs = 1; fibs <= 20; fibs++) { for (;G <= N(NG); ++G) x++; NG[fibs] = '0'; NG[fibs+1] = 0; std::cout << x << " "; } std::cout << std::endl; return 0;
} </lang>
- Output:
1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946
Using Standard Template Library
Possibly less "Far-fetched version". <lang cpp> // Use Standard Template Library to display Fibonacci sequence. // Nigel Galloway March 30th., 2013
- include <algorithm>
- include <iostream>
- include <iterator>
int main() {
int x = 1, y = 1; generate_n(std::ostream_iterator<int>(std::cout, " "), 21, [&]{int n=x; x=y; y+=n; return n;}); return 0;
} </lang>
- Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946
C#
Recursive
<lang csharp> public static ulong Fib(uint n) {
return (n < 2)? n : Fib(n - 1) + Fib(n - 2);
} </lang>
Tail-Recursive
<lang csharp> public static ulong Fib(uint n) {
return Fib(0, 1, n);
}
private static ulong Fib(ulong a, ulong b, uint n) {
return (n < 1)? a :(n == 1)? b : Fib(b, a + b, n - 1);
} </lang>
Iterative
<lang csharp> public static ulong Fib(uint x) {
int prev = -1; int next = 1; for (int i = 0; i < x; i++) { int sum = prev + next; prev = next; next = sum; fibs.Add(sum); } return fibs;
} </lang>
Eager-Generative
<lang csharp> public static IEnumerable<long> Fibs(uint x) {
IList<ulong> fibs = new List<ulong>();
ulong prev = -1; ulong next = 1; for (int i = 0; i < x; i++) { long sum = prev + next; prev = next; next = sum; fibs.Add(sum); } return fibs;
} </lang>
Lazy-Generative
<lang csharp> public static IEnumerable<ulong> Fibs(uint x) {
ulong prev = -1; ulong next = 1; for (uint i = 0; i < x; i++) { ulong sum = prev + next; prev = next; next = sum; yield return sum; }
} </lang>
Analytic
Only works to the 92th fibonacci number. <lang csharp> private static double Phi = ((1d + Math.Sqrt(5d))/2d); private static double D = 1d/Math.Sqrt(5d);
ulong Fib(uint n) {
if(n > 92) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 93."); return (ulong)((Phi^n) - (1d - Phi)^n))*D);
} </lang>
Matrix
Algorithm is based on
- .
Needs System.Windows.Media.Matrix
or similar Matrix class.
Calculates in .
<lang csharp>
public static ulong Fib(uint n) {
var M = new Matrix(1,0,0,1); var N = new Matrix(1,1,1,0); for (uint i = 1; i < n; i++) M *= N; return (ulong)M[0][0];
}
</lang>
Needs System.Windows.Media.Matrix
or similar Matrix class.
Calculates in .
<lang csharp>
private static Matrix M;
private static readonly Matrix N = new Matrix(1,1,1,0);
public static ulong Fib(uint n) {
M = new Matrix(1,0,0,1); MatrixPow(n-1); return (ulong)M[0][0];
}
private static void MatrixPow(double n){
if (n > 1) { MatrixPow(n/2); M *= M; } if (n % 2 == 0) M *= N;
} </lang>
Array (Table) Lookup
<lang csharp> private static int[] fibs = new int[]{ -1836311903, 1134903170,
-701408733, 433494437, -267914296, 165580141, -102334155, 63245986, -39088169, 24157817, -14930352, 9227465, -5702887, 3524578, -2178309, 1346269, -832040, 514229, -317811, 196418, -121393, 75025, -46368, 28657, -17711, 10946, -6765, 4181, -2584, 1597, -987, 610, -377, 233, -144, 89, -55, 34, -21, 13, -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903};
public static int Fib(int n) {
if(n < -46 || n > 46) throw new ArgumentOutOfRangeException("n", n, "Has to be between -46 and 47.") return fibs[n+46];
} </lang>
Cat
<lang cat>define fib {
dup 1 <= [] [dup 1 - fib swap 2 - fib +] if
}</lang>
Chapel
<lang chapel>iter fib() {
var a = 0, b = 1;
while true { yield a; (a, b) = (b, b + a); }
}</lang>
Chef
<lang chef>Stir-Fried Fibonacci Sequence.
An unobfuscated iterative implementation. It prints the first N + 1 Fibonacci numbers, where N is taken from standard input.
Ingredients. 0 g last 1 g this 0 g new 0 g input
Method. Take input from refrigerator. Put this into 4th mixing bowl. Loop the input. Clean the 3rd mixing bowl. Put last into 3rd mixing bowl. Add this into 3rd mixing bowl. Fold new into 3rd mixing bowl. Clean the 1st mixing bowl. Put this into 1st mixing bowl. Fold last into 1st mixing bowl. Clean the 2nd mixing bowl. Put new into 2nd mixing bowl. Fold this into 2nd mixing bowl. Put new into 4th mixing bowl. Endloop input until looped. Pour contents of the 4th mixing bowl into baking dish.
Serves 1.</lang>
CMake
Iteration uses a while() loop. Memoization uses global properties.
<lang cmake>set_property(GLOBAL PROPERTY fibonacci_0 0) set_property(GLOBAL PROPERTY fibonacci_1 1) set_property(GLOBAL PROPERTY fibonacci_next 2)
- var = nth number in Fibonacci sequence.
function(fibonacci var n)
# If the sequence is too short, compute more Fibonacci numbers. get_property(next GLOBAL PROPERTY fibonacci_next) if(NOT next GREATER ${n}) # a, b = last 2 Fibonacci numbers math(EXPR i "${next} - 2") get_property(a GLOBAL PROPERTY fibonacci_${i}) math(EXPR i "${next} - 1") get_property(b GLOBAL PROPERTY fibonacci_${i})
while(NOT next GREATER ${n}) math(EXPR i "${a} + ${b}") # i = next Fibonacci number set_property(GLOBAL PROPERTY fibonacci_${next} ${i}) set(a ${b}) set(b ${i}) math(EXPR next "${next} + 1") endwhile() set_property(GLOBAL PROPERTY fibonacci_next ${next}) endif()
get_property(answer GLOBAL PROPERTY fibonacci_${n}) set(${var} ${answer} PARENT_SCOPE)
endfunction(fibonacci)</lang>
<lang cmake># Test program: print 0th to 9th and 25th to 30th Fibonacci numbers. set(s "") foreach(i RANGE 0 9)
fibonacci(f ${i}) set(s "${s} ${f}")
endforeach(i) set(s "${s} ... ") foreach(i RANGE 25 30)
fibonacci(f ${i}) set(s "${s} ${f}")
endforeach(i) message(${s})</lang>
0 1 1 2 3 5 8 13 21 34 ... 75025 121393 196418 317811 514229 832040
Clojure
This is implemented idiomatically as an infinitely long, lazy sequence of all Fibonacci numbers: <lang Clojure>(defn fibs []
(map first (iterate (fn a b [b (+ a b)]) [0 1])))</lang>
Thus to get the nth one: <lang Clojure>(nth (fibs) 5)</lang> So long as one does not hold onto the head of the sequence, this is unconstrained by length.
The one-line implementation may look confusing at first, but on pulling it apart it actually solves the problem more "directly" than a more explicit looping construct. <lang Clojure>(defn fibs []
(map first ;; throw away the "metadata" (see below) to view just the fib numbers (iterate ;; create an infinite sequence of [prev, curr] pairs (fn a b ;; to produce the next pair, call this function on the current pair [b (+ a b)]) ;; new prev is old curr, new curr is sum of both previous numbers [0 1]))) ;; recursive base case: prev 0, curr 1</lang>
A more elegant solution is inspired by the Haskell implementation of an infinite list of Fibonacci numbers: <lang Clojure>(def fib (lazy-cat [0 1] (map + fib (rest fib))))</lang> Then, to see the first ten, <lang Clojure>user> (take 10 fib) (0 1 1 2 3 5 8 13 21 34)</lang>
Here's a simple interative process (using a recursive function) that carries state along with it (as args) until it reaches a solution: <lang Clojure>;; max is which fib number you'd like computed (0th, 1st, 2nd, etc.)
- n is which fib number you're on for this call (0th, 1st, 2nd, etc.)
- j is the nth fib number (ex. when n = 5, j = 5)
- i is the nth - 1 fib number
(defn- fib-iter
[max n i j] (if (= n max) j (recur max (inc n) j (+ i j))))
(defn fib
[max] (if (< max 2) max (fib-iter max 1 0N 1N)))</lang>
"defn-" means that the function is private (for use only inside this library). The "N" suffixes on integers tell Clojure to use arbitrary precision ints for those.
COBOL
Iterative
<lang cobol>Program-ID. Fibonacci-Sequence. Data Division. Working-Storage Section.
01 FIBONACCI-PROCESSING. 05 FIBONACCI-NUMBER PIC 9(36) VALUE 0. 05 FIB-ONE PIC 9(36) VALUE 0. 05 FIB-TWO PIC 9(36) VALUE 1. 01 DESIRED-COUNT PIC 9(4). 01 FORMATTING. 05 INTERM-RESULT PIC Z(35)9. 05 FORMATTED-RESULT PIC X(36). 05 FORMATTED-SPACE PIC x(35).
Procedure Division.
000-START-PROGRAM. Display "What place of the Fibonacci Sequence would you like (<173)? " with no advancing. Accept DESIRED-COUNT. If DESIRED-COUNT is less than 1 Stop run. If DESIRED-COUNT is less than 2 Move FIBONACCI-NUMBER to INTERM-RESULT Move INTERM-RESULT to FORMATTED-RESULT Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT Display FORMATTED-RESULT Stop run. Subtract 1 from DESIRED-COUNT. Move FIBONACCI-NUMBER to INTERM-RESULT. Move INTERM-RESULT to FORMATTED-RESULT. Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT. Display FORMATTED-RESULT. Perform 100-COMPUTE-FIBONACCI until DESIRED-COUNT = zero. Stop run. 100-COMPUTE-FIBONACCI. Compute FIBONACCI-NUMBER = FIB-ONE + FIB-TWO. Move FIB-TWO to FIB-ONE. Move FIBONACCI-NUMBER to FIB-TWO. Subtract 1 from DESIRED-COUNT. Move FIBONACCI-NUMBER to INTERM-RESULT. Move INTERM-RESULT to FORMATTED-RESULT. Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT. Display FORMATTED-RESULT.</lang>
Recursive
<lang cobol> >>SOURCE FREE IDENTIFICATION DIVISION. PROGRAM-ID. fibonacci-main.
DATA DIVISION. WORKING-STORAGE SECTION. 01 num PIC 9(6) COMP. 01 fib-num PIC 9(6) COMP.
PROCEDURE DIVISION.
ACCEPT num CALL "fibonacci" USING CONTENT num RETURNING fib-num DISPLAY fib-num .
END PROGRAM fibonacci-main.
IDENTIFICATION DIVISION. PROGRAM-ID. fibonacci RECURSIVE.
DATA DIVISION. LOCAL-STORAGE SECTION. 01 1-before PIC 9(6) COMP. 01 2-before PIC 9(6) COMP.
LINKAGE SECTION. 01 num PIC 9(6) COMP.
01 fib-num PIC 9(6) COMP BASED.
PROCEDURE DIVISION USING num RETURNING fib-num.
ALLOCATE fib-num EVALUATE num WHEN 0 MOVE 0 TO fib-num WHEN 1 MOVE 1 TO fib-num WHEN OTHER SUBTRACT 1 FROM num CALL "fibonacci" USING CONTENT num RETURNING 1-before SUBTRACT 1 FROM num CALL "fibonacci" USING CONTENT num RETURNING 2-before ADD 1-before TO 2-before GIVING fib-num END-EVALUATE .
END PROGRAM fibonacci.</lang>
CoffeeScript
Analytic
<lang coffeescript>fib_ana = (n) ->
sqrt = Math.sqrt phi = ((1 + sqrt(5))/2) return Math.round((Math.pow(phi, n)/sqrt(5)))</lang>
Iterative
<lang coffeescript>fib_iter = (n) ->
if n < 2 return n [prev, curr] = 0, 1 for i in [1..n] [prev, curr] = [curr, curr + prev] return curr</lang>
Recursive
<lang coffeescript>fib_rec = (n) ->
if n < 2 return n else return fib_rec(n-1) + fib_rec(n-2)</lang>
Common Lisp
Note that Common Lisp uses bignums, so this will never overflow.
Iterative
<lang lisp>(defun fibonacci-iterative (n &aux (f0 0) (f1 1))
(case n (0 f0) (1 f1) (t (loop for n from 2 to n for a = f0 then b and b = f1 then result for result = (+ a b) finally (return result)))))</lang>
Simpler one: <lang lisp>(defun fibonacci (n)
(let ((a 0) (b 1) (c n)) (loop for i from 2 to n do
(setq c (+ a b) a b b c))
c))</lang>
Not a function, just printing out the entire (for some definition of "entire") sequence with a for var =
loop:<lang lisp>(loop for x = 0 then y and y = 1 then (+ x y) do (print x))</lang>
Recursive
<lang lisp>(defun fibonacci-recursive (n)
(if (< n 2) n (+ (fibonacci-recursive (- n 2)) (fibonacci-recursive (- n 1)))))</lang>
<lang lisp>(defun fibonacci-tail-recursive ( n &optional (a 0) (b 1))
(if (= n 0) a (fibonacci-tail-recursive (- n 1) b (+ a b))))</lang>
Tail recursive and squaring: <lang lisp>(defun fib (n &optional (a 1) (b 0) (p 0) (q 1))
(if (= n 1) (+ (* b p) (* a q)) (fib (ash n -1) (if (evenp n) a (+ (* b q) (* a (+ p q)))) (if (evenp n) b (+ (* b p) (* a q))) (+ (* p p) (* q q)) (+ (* q q) (* 2 p q))))) ;p is Fib(2^n-1), q is Fib(2^n).
(print (fib 100000))</lang>
D
Here are four versions of Fibonacci Number calculating functions. FibD has an argument limit of magnitude 84 due to floating point precision, the others have a limit of 92 due to overflow (long).The traditional recursive version is inefficient. It is optimized by supplying a static storage to store intermediate results. A Fibonacci Number generating function is added. All functions have support for negative arguments. <lang d>import std.stdio, std.conv, std.algorithm, std.math;
long sgn(alias unsignedFib)(int n) { // break sign manipulation apart
immutable uint m = (n >= 0) ? n : -n; if (n < 0 && (n % 2 == 0)) return -unsignedFib(m); else return unsignedFib(m);
}
long fibD(uint m) { // Direct Calculation, correct for abs(m) <= 84
enum sqrt5r = 1.0L / sqrt(5.0L); // 1 / sqrt(5) enum golden = (1.0L + sqrt(5.0L)) / 2.0L; // (1 + sqrt(5)) / 2 return roundTo!long(pow(golden, m) * sqrt5r);
}
long fibI(in uint m) pure nothrow { // Iterative
long thisFib = 0; long nextFib = 1; foreach (i; 0 .. m) { long tmp = nextFib; nextFib += thisFib; thisFib = tmp; } return thisFib;
}
long fibR(uint m) { // Recursive
return (m < 2) ? m : fibR(m - 1) + fibR(m - 2);
}
long fibM(uint m) { // memoized Recursive
static long[] fib = [0, 1]; while (m >= fib.length ) fib ~= fibM(m - 2) + fibM(m - 1); return fib[m];
}
alias sgn!fibD sfibD; alias sgn!fibI sfibI; alias sgn!fibR sfibR; alias sgn!fibM sfibM;
auto fibG(in int m) { // generator(?)
immutable int sign = (m < 0) ? -1 : 1; long yield; return new class { final int opApply(int delegate(ref int, ref long) dg) { int idx = -sign; // prepare for pre-increment foreach (f; this) if (dg(idx += sign, f)) break; return 0; } final int opApply(int delegate(ref long) dg) { long f0, f1 = 1; foreach (p; 0 .. m * sign + 1) { if (sign == -1 && (p % 2 == 0)) yield = -f0; else yield = f0; if (dg(yield)) break; auto temp = f1; f1 = f0 + f1; f0 = temp; } return 0; } };
}
void main(in string[] args) {
int k = args.length > 1 ? to!int(args[1]) : 10; writefln("Fib(%3d) = ", k); writefln("D : %20d <- %20d + %20d", sfibD(k), sfibD(k - 1), sfibD(k - 2)); writefln("I : %20d <- %20d + %20d", sfibI(k), sfibI(k - 1), sfibI(k - 2)); if (abs(k) < 36 || args.length > 2) // set a limit for recursive version writefln("R : %20d <- %20d + %20d", sfibR(k), sfibM(k - 1), sfibM(k - 2)); writefln("O : %20d <- %20d + %20d", sfibM(k), sfibM(k - 1), sfibM(k - 2)); foreach (i, f; fibG(-9)) writef("%d:%d | ", i, f);
}</lang> Output for n = 85:
Fib( 85) = D : 259695496911122586 <- 160500643816367088 + 99194853094755497 I : 259695496911122585 <- 160500643816367088 + 99194853094755497 O : 259695496911122585 <- 160500643816367088 + 99194853094755497 0:0 | -1:1 | -2:-1 | -3:2 | -4:-3 | -5:5 | -6:-8 | -7:13 | -8:-21 | -9:34 |
Matrix Exponentiation Version
<lang d>import std.bigint;
T fibonacciMatrix(T=BigInt)(size_t n) {
int[size_t.sizeof * 8] binDigits; size_t nBinDigits; while (n > 0) { binDigits[nBinDigits] = n % 2; n /= 2; nBinDigits++; }
T x=1, y, z=1; foreach_reverse (b; binDigits[0 .. nBinDigits]) { if (b) { x = (x + z) * y; y = y ^^ 2 + z ^^ 2; } else { auto x_old = x; x = x ^^ 2 + y ^^ 2; y = (x_old + z) * y; } z = x + y; }
return y;
}
void main() {
10_000_000.fibonacciMatrix;
}</lang>
Faster Version
For N = 10_000_000 this is about twice faster (run-time about 2.20 seconds) than the matrix exponentiation version. <lang d>import std.bigint, std.math;
// Algorithm from: Takahashi, Daisuke, // "A fast algorithm for computing large Fibonacci numbers". // Information Processing Letters 75.6 (30 November 2000): 243-246. // Implementation from: // pythonista.wordpress.com/2008/07/03/pure-python-fibonacci-numbers BigInt fibonacci(in ulong n) in {
assert(n > 0, "fibonacci(n): n must be > 0.");
} body {
if (n <= 2) return 1.BigInt; BigInt F = 1; BigInt L = 1; int sign = -1; immutable uint n2 = cast(uint)n.log2.floor; auto mask = 2.BigInt ^^ (n2 - 1); foreach (immutable i; 1 .. n2) { auto temp = F ^^ 2; F = (F + L) / 2; F = 2 * F ^^ 2 - 3 * temp - 2 * sign; L = 5 * temp + 2 * sign; sign = 1; if (n & mask) { temp = F; F = (F + L) / 2; L = F + 2 * temp; sign = -1; } mask /= 2; } if ((n & mask) == 0) { F *= L; } else { F = (F + L) / 2; F = F * L - sign; } return F;
}
void main() {
10_000_000.fibonacci;
}</lang>
Dart
<lang dart>int fib(int n) {
if(n==0 || n==1) { return n; } int prev=1; int current=1; for(int i=2;i<n;i++) { int next=prev+current; prev=current; current=next; } return current;
}
int fibRec(int n) => n==0||n==1 ? n : fibRec(n-1)+fibRec(n-2);
main() {
print(fib(11)); print(fibRec(11));
}</lang>
Delphi
Iterative
<lang Delphi> function FibonacciI(N: Word): UInt64; var
Last, New: UInt64; I: Word;
begin
if N < 2 then Result := N else begin Last := 0; Result := 1; for I := 2 to N do begin New := Last + Result; Last := Result; Result := New; end; end;
end; </lang>
Recursive
<lang Delphi> function Fibonacci(N: Word): UInt64; begin
if N < 2 then Result := N else Result := Fibonacci(N - 1) + Fibonacci(N - 2);
end; </lang>
Matrix
Algorithm is based on
- .
<lang Delphi> function fib(n:Int64):int64;
type TFibMat = array[0..1] of array[0..1] of int64;
function FibMatMul(a,b:TFibMat):TFibMat; var i,j,k:integer; tmp:TFibMat; begin for i:=0 to 1 do for j:=0 to 1 do begin tmp[i,j]:=0; for k:=0 to 1 do tmp[i,j]:=tmp[i,j]+a[i,k]*b[k,j]; end; FibMatMul:=tmp; end;
function FibMatExp(a:TFibMat;n:int64):TFibmat; begin if n<=1 then fibmatexp:=a else if (n mod 2 = 0) then FibMatExp:=FibMatExp(FibMatMul(a,a), n div 2) else if (n mod 2 = 1) then FibMatExp:=FibMatMul(a,FibMatExp(FibMatMul(a,a),(n) div 2)); end;
var
matrix:TFibMat;
begin matrix[0,0]:=1; matrix[0,1]:=1; matrix[1,0]:=1; matrix[1,1]:=0; if n>1 then matrix:=fibmatexp(matrix,n-1); fib:=matrix[0,0]; end; </lang>
DWScript
<lang Delphi>function fib(N : Integer) : Integer; begin
if N < 2 then Result := 1 else Result := fib(N-2) + fib(N-1);
End;</lang>
E
<lang e>def fib(n) {
var s := [0, 1] for _ in 0..!n { def [a, b] := s s := [b, a+b] } return s[0]
}</lang>
(This version defines fib(0) = 0 because OEIS A000045 does.)
ECL
Analytic
<lang ECL>//Calculates Fibonacci sequence up to n steps using Binet's closed form solution
FibFunction(UNSIGNED2 n) := FUNCTION
REAL Sqrt5 := Sqrt(5);
REAL Phi := (1+Sqrt(5))/2;
REAL Phi_Inv := 1/Phi;
UNSIGNED FibValue := ROUND( ( POWER(Phi,n)-POWER(Phi_Inv,n) ) /Sqrt5);
RETURN FibValue;
END;
FibSeries(UNSIGNED2 n) := FUNCTION Fib_Layout := RECORD UNSIGNED5 FibNum; UNSIGNED5 FibValue; END; FibSeq := DATASET(n+1, TRANSFORM ( Fib_Layout , SELF.FibNum := COUNTER-1 , SELF.FibValue := IF(SELF.FibNum<2,SELF.FibNum, FibFunction(SELF.FibNum) ) ) ); RETURN FibSeq; END; }</lang>
Eiffel
<lang eiffel> class APPLICATION
create make
feature
fibonacci (n: INTEGER): INTEGER require non_negative: n >= 0 local i, n2, n1, tmp: INTEGER do n2 := 0 n1 := 1 from i := 1 until i >= n loop tmp := n1 n1 := n2 + n1 n2 := tmp i := i + 1 end Result := n1 if n = 0 then Result := 0 end end
feature {NONE} -- Initialization
make -- Run application. do print (fibonacci (0)) print (" ") print (fibonacci (1)) print (" ") print (fibonacci (2)) print (" ") print (fibonacci (3)) print (" ") print (fibonacci (4)) print ("%N") end
end </lang>
Ela
Tail-recursive function: <lang Ela>fib = fib' 0 1
where fib' a b 0 = a fib' a b n = fib' b (a + b) (n - 1)</lang>
Infinite (lazy) list: <lang Ela>fib = fib' 1 1
where fib' x y = & x :: fib' y (x + y)</lang>
Erlang
Recursive: <lang erlang>fib(0) -> 0; fib(1) -> 1; fib(N) when N > 1 -> fib(N-1) + fib(N-2).</lang>
Tail-recursive (iterative): <lang erlang>fib(N) -> fib(N,0,1). fib(0,Res,_) -> Res; fib(N,Res,Next) when N > 0 -> fib(N-1, Next, Res+Next).</lang>
Euphoria
'Recursive' version
<lang Euphoria> function fibor(integer n)
if n<2 then return n end if return fibor(n-1)+fibor(n-2)
end function </lang>
'Iterative' version
<lang Euphoria> function fiboi(integer n) integer f0=0, f1=1, f
if n<2 then return n end if for i=2 to n do f=f0+f1 f0=f1 f1=f end for return f
end function </lang>
'Tail recursive' version
<lang Euphoria> function fibot(integer n, integer u = 1, integer s = 0)
if n < 1 then return s else return fibot(n-1,u+s,u) end if
end function
-- example: ? fibot(10) -- says 55 </lang>
'Paper tape' version
<lang Euphoria> include std/mathcons.e -- for PINF constant
enum ADD, MOVE, GOTO, OUT, TEST, TRUETO
global sequence tape = { 0, 1, { ADD, 2, 1 }, { TEST, 1, PINF }, { TRUETO, 0 }, { OUT, 1, "%.0f\n" }, { MOVE, 2, 1 }, { MOVE, 0, 2 }, { GOTO, 3 } }
global integer ip global integer test global atom accum
procedure eval( sequence cmd ) atom i = 1 while i <= length( cmd ) do switch cmd[ i ] do case ADD then accum = tape[ cmd[ i + 1 ] ] + tape[ cmd[ i + 2 ] ] i += 2
case OUT then printf( 1, cmd[ i + 2], tape[ cmd[ i + 1 ] ] ) i += 2
case MOVE then if cmd[ i + 1 ] = 0 then tape[ cmd[ i + 2 ] ] = accum else tape[ cmd[ i + 2 ] ] = tape[ cmd[ i + 1 ] ] end if i += 2
case GOTO then ip = cmd[ i + 1 ] - 1 -- due to ip += 1 in main loop i += 1
case TEST then if tape[ cmd[ i + 1 ] ] = cmd[ i + 2 ] then test = 1 else test = 0 end if i += 2
case TRUETO then if test then if cmd[ i + 1 ] = 0 then abort(0) else ip = cmd[ i + 1 ] - 1 end if end if
end switch i += 1 end while end procedure
test = 0 accum = 0 ip = 1
while 1 do
-- embedded sequences (assumed to be code) are evaluated -- atoms (assumed to be data) are ignored
if sequence( tape[ ip ] ) then eval( tape[ ip ] ) end if ip += 1 end while
</lang>
FALSE
<lang false>[[$0=~][1-@@\$@@+\$44,.@]#]f: 20n: {First 20 numbers} 0 1 n;f;!%%44,. {Output: "0,1,1,2,3,5..."}</lang>
Factor
Iterative
<lang factor>: fib ( n -- m )
dup 2 < [ [ 0 1 ] dip [ swap [ + ] keep ] times drop ] unless ;</lang>
Recursive
<lang factor>: fib ( n -- m )
dup 2 < [ [ 1 - fib ] [ 2 - fib ] bi + ] unless ;</lang>
Tail-Recursive
<lang factor>: fib2 ( x y n -- a )
dup 1 < [ 2drop ] [ [ swap [ + ] keep ] dip 1 - fib2 ] if ;
- fib ( n -- m ) [ 0 1 ] dip fib2 ;</lang>
Matrix
<lang factor>USE: math.matrices
- fib ( n -- m )
dup 2 < [ [ { { 0 1 } { 1 1 } } ] dip 1 - m^n second second ] unless ;</lang>
Fancy
<lang fancy>class Fixnum {
def fib { match self -> { case 0 -> 0 case 1 -> 1 case _ -> self - 1 fib + (self - 2 fib) } }
}
15 times: |x| {
x fib println
} </lang>
Falcon
Iterative
<lang falcon>function fib_i(n)
if n < 2: return n
fibPrev = 1 fib = 1 for i in [2:n] tmp = fib fib += fibPrev fibPrev = tmp end return fib
end</lang>
Recursive
<lang falcon>function fib_r(n)
if n < 2 : return n return fib_r(n-1) + fib_r(n-2)
end</lang>
Tail Recursive
<lang falcon>function fib_tr(n)
return fib_aux(n,0,1)
end function fib_aux(n,a,b)
switch n case 0 : return a default: return fib_aux(n-1,a+b,a) end
end</lang>
Fantom
Ints have a limit of 64-bits, so overflow errors occur after computing Fib(92) = 7540113804746346429.
<lang fantom> class Main {
static Int fib (Int n) { if (n < 2) return n fibNums := [1, 0] while (fibNums.size <= n) { fibNums.insert (0, fibNums[0] + fibNums[1]) } return fibNums.first }
public static Void main () { 20.times |n| { echo ("Fib($n) is ${fib(n)}") } }
} </lang>
Fexl
<lang Fexl>
- fibonacci is the infinite list of all Fibonacci numbers.
- Note that this program uses the symbols "1" and "+". You can specify
- the definitions of those symbols however you like, allowing you to use
- any system of arithmetic you need. For example, you can use either the
- built-in long arithmetic, or infinite precision arithmetic, by simply
- defining "1" and "+" appropriately.
\fibonacci =
( \fibonacci == (\x\y item x; \z = (+ x y) fibonacci y z ) fibonacci 1 1 )
- OK, so that's the list of *all* Fibonacci numbers. If you want the nth number,
- you can extract it with the fib function as follows. By the way, this *does*
- have the effect of caching, so once a particular point in the sequence is
- calculated, it doesn't have to be calculated again.
- (fib n) is the nth fibonacci number, starting with n==0, or 1 if n is negative.
\fib = (\n list_at fibonacci n 1) </lang>
Forth
<lang forth>: fib ( n -- fib )
0 1 rot 0 ?do over + swap loop drop ;</lang>
Since there are only a fixed and small amount of Fibonacci numbers that fit in a machine word, this FORTH version creates a table of Fibonacci numbers at compile time. It stops compiling numbers when there is arithmetic overflow (the number turns negative, indicating overflow.)
<lang forth>: F-start, here 1 0 dup , ;
- F-next, over + swap
dup 0> IF dup , true ELSE false THEN ;
- computed-table ( compile: 'start 'next / run: i -- x )
create >r execute BEGIN r@ execute not UNTIL rdrop does> swap cells + @ ;
' F-start, ' F-next, computed-table fibonacci 2drop here swap - cell/ Constant #F/64 \ # of fibonacci numbers generated
16 fibonacci . 987 ok
- F/64 . 93 ok
92 fibonacci . 7540113804746346429 ok \ largest number generated.</lang>
Fortran
Recursive
In ISO Fortran 90 or later, use a RECURSIVE function: <lang fortran>module fibonacci contains
recursive function fibR(n) result(fib) integer, intent(in) :: n integer :: fib select case (n) case (:0); fib = 0 case (1); fib = 1 case default; fib = fibR(n-1) + fibR(n-2) end select end function fibR</lang>
Iterative
In ISO Fortran 90 or later: <lang fortran> function fibI(n)
integer, intent(in) :: n integer, parameter :: fib0 = 0, fib1 = 1 integer :: fibI, back1, back2, i select case (n) case (:0); fibI = fib0 case (1); fibI = fib1 case default fibI = fib1 back1 = fib0 do i = 2, n back2 = back1 back1 = fibI fibI = back1 + back2 end do end select end function fibI
end module fibonacci</lang>
Test program <lang fortran>program fibTest
use fibonacci do i = 0, 10 print *, fibr(i), fibi(i) end do
end program fibTest</lang>
Output
0 0 1 1 1 1 2 2 3 3 5 5 8 8 13 13 21 21 34 34 55 55
freebasic
Extended sequence coded big integer. <lang freebasic>
'Fibonacci extended 'Freebasic version 24 Windows Dim Shared ADDQmod(0 To 19) As Ubyte Dim Shared ADDbool(0 To 19) As Ubyte For z As Integer=0 To 19
ADDQmod(z)=(z Mod 10+48) ADDbool(z)=(-(10<=z))
Next z Function plusINT(NUM1 As String,NUM2 As String) As String
Dim As Byte flag #macro finish() three=Ltrim(three,"0") If three="" Then Return "0" If flag=1 Then Swap NUM2,NUM1 Return three Exit Function #endmacro var lenf=Len(NUM1) var lens=Len(NUM2) If lens>lenf Then Swap NUM2,NUM1 Swap lens,lenf flag=1 End If var diff=lenf-lens-Sgn(lenf-lens) var three="0"+NUM1 var two=String(lenf-lens,"0")+NUM2 Dim As Integer n2 Dim As Ubyte addup,addcarry addcarry=0 For n2=lenf-1 To diff Step -1 addup=two[n2]+NUM1[n2]-96 three[n2+1]=addQmod(addup+addcarry) addcarry=addbool(addup+addcarry) Next n2 If addcarry=0 Then finish() End If If n2=-1 Then three[0]=addcarry+48 finish() End If For n2=n2 To 0 Step -1 addup=two[n2]+NUM1[n2]-96 three[n2+1]=addQmod(addup+addcarry) addcarry=addbool(addup+addcarry) Next n2 three[0]=addcarry+48 finish()
End Function
Function fibonacci(n As Integer) As String
Dim As String sl,l,term sl="0": l="1" If n=1 Then Return "0" If n=2 Then Return "1" n=n-2 For x As Integer= 1 To n term=plusINT(l,sl) sl=l l=term Next x Function =term
End Function
'============== EXAMPLE =============== print "THE SEQUENCE TO 10:" print For n As Integer=1 To 10
Print "term";n;": "; fibonacci(n)
Next n print print "Selected Fibonacci number" print "Fibonacci 500" print print fibonacci(500) Sleep </lang> Output
THE SEQUENCE TO 10: term 1: 0 term 2: 1 term 3: 1 term 4: 2 term 5: 3 term 6: 5 term 7: 8 term 8: 13 term 9: 21 term 10: 34 Selected Fibonacci number Fibonacci 500 86168291600238450732788312165664788095941068326060883324529903470149056115823592 713458328176574447204501
Frink
All of Frink's integers can be arbitrarily large. <lang frink> fibonacciN[n] := {
a = 0 b = 1 count = 0 while count < n { [a,b] = [b, a + b] count = count + 1 } return a
} </lang>
F#
This is a fast [tail-recursive] approach using the F# big integer support: <lang fsharp> let fibonacci n : bigint =
let rec f a b n = match n with | 0 -> a | 1 -> b | n -> (f b (a + b) (n - 1)) f (bigint 0) (bigint 1) n
> fibonacci 100;; val it : bigint = 354224848179261915075I</lang> Lazy evaluated using sequence workflow: <lang fsharp>let rec fib = seq { yield! [0;1];
for (a,b) in Seq.zip fib (Seq.skip 1 fib) -> a+b}</lang>
The above is extremely slow due to the nested recursions on sequences, which aren't very efficient at the best of times. The above takes seconds just to compute the 30th Fibonacci number!
Lazy evaluation using the sequence unfold anamorphism is much much better as to efficiency: <lang fsharp>let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I) fibonacci |> Seq.nth 10000 </lang>
Approach similar to the Matrix algorithm in C#, with some shortcuts involved. Since it uses exponentiation by squaring, calculations of fib(n) where n is a power of 2 are particularly quick. Eg. fib(2^20) was calculated in a little over 4 seconds on this poster's laptop. <lang fsharp> open System open System.Diagnostics open System.Numerics
/// Finds the highest power of two which is less than or equal to a given input. let inline prevPowTwo (x : int) =
let mutable n = x n <- n - 1 n <- n ||| (n >>> 1) n <- n ||| (n >>> 2) n <- n ||| (n >>> 4) n <- n ||| (n >>> 8) n <- n ||| (n >>> 16) n <- n + 1 match x with | x when x = n -> x | _ -> n/2
/// Evaluates the nth Fibonacci number using matrix arithmetic and /// exponentiation by squaring. let crazyFib (n : int) =
let powTwo = prevPowTwo n
/// Applies 2n rule repeatedly until another application of the rule would /// go over the target value (or the target value has been reached). let rec iter1 i q r s = match i with | i when i < powTwo -> iter1 (i*2) (q*q + r*r) (r * (q+s)) (r*r + s*s) | _ -> i, q, r, s
/// Applies n+1 rule until the target value is reached. let rec iter2 (i, q, r, s) = match i with | i when i < n -> iter2 ((i+1), (q+r), q, r) | _ -> q
match n with | 0 -> 1I | _ -> iter1 1 1I 1I 0I |> iter2
</lang>
GAP
<lang gap>fib := function(n)
local a; a := [[0, 1], [1, 1]]^n; return a[1][2];
end;</lang> GAP has also a buit-in function for that. <lang gap>Fibonacci(n);</lang>
Gecho
<lang gecho>0 1 dup wover + dup wover + dup wover + dup wover +</lang> Prints the first several fibonacci numbers...
GML
<lang gml>///fibonacci(n) //Returns the nth fibonacci number
var n, numb; n = argument0;
if (n == 0)
{ numb = 0; }
else
{ var fm2, fm1; fm2 = 0; fm1 = 1; numb = 1; repeat(n-1) { numb = fm2+fm1; fm2 = fm1; fm1 = numb; } }
return numb;</lang>
Go
Recursive
<lang go>func fib(a int) int {
if a < 2 { return a } return fib(a - 1) + fib(a - 2)
}</lang>
Iterative
<lang go>import ( "math/big" )
func fib(n uint64) *big.Int { if n < 2 { return big.NewInt(int64(n)) } a, b := big.NewInt(0), big.NewInt(1) for n--; n > 0; n-- { a.Add(a, b) a, b = b, a } return b }</lang>
Groovy
Recursive
A recursive closure must be pre-declared. <lang groovy>def rFib rFib = { it < 1 ? 0 : it == 1 ? 1 : rFib(it-1) + rFib(it-2) }</lang>
Iterative
<lang groovy>def iFib = { it < 1 ? 0 : it == 1 ? 1 : (2..it).inject([0,1]){i, j -> [i[1], i[0]+i[1]]}[1] }</lang>
Test program: <lang groovy>(0..20).each { println "${it}: ${rFib(it)} ${iFib(it)}" }</lang>
Output:
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 11: 89 89 12: 144 144 13: 233 233 14: 377 377 15: 610 610 16: 987 987 17: 1597 1597 18: 2584 2584 19: 4181 4181 20: 6765 6765
Haxe
Iterative
<lang haxe>static function fib(steps:Int, handler:Int->Void) { var current = 0; var next = 1;
for (i in 1...steps) { handler(current);
var temp = current + next; current = next; next = temp; } handler(current); }</lang>
As Iterator
<lang haxe>class FibIter { public var current:Int; private var nextItem:Int; private var limit:Int;
public function new(limit) { current = 0; nextItem = 1; this.limit = limit; }
public function hasNext() return limit > 0
public function next() { limit--; var ret = current; var temp = current + nextItem; current = nextItem; nextItem = temp; return ret; } }</lang>
Used like:
<lang haxe>for (i in new FibIter(10)) Sys.println(i);</lang>
Haskell
With lazy lists
This is a standard example how to use lazy lists. Here's the (infinite) list of all Fibonacci numbers:
<lang haskell>fib = 0 : 1 : zipWith (+) fib (tail fib)</lang>
The nth Fibonacci number is then just fib !! n
. The above is equivalent to
<lang haskell>fib = 0 : 1 : next fib where next (a: t@(b:_)) = (a+b) : next t</lang>
Also
<lang haskell>fib = 0 : scanl (+) 1 fib</lang>
With matrix exponentiation
With the (rather slow) code from Matrix exponentiation operator
<lang haskell>import Data.List
xs <+> ys = zipWith (+) xs ys xs <*> ys = sum $ zipWith (*) xs ys
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 <*> ys | ys <- transpose $ unMat ym] | xs <- unMat xm] fromInteger n = Mat fromInteger n</lang>
we can simply write
<lang haskell>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</lang>
So, for example, the hundred-thousandth Fibonacci number starts with the digits
*Main> take 10 $ show $ fib (10^5) "2597406934"
With recurrence relations
Using Fib[m=3n+r]
recurrence identities:
<lang haskell>fibsteps (a,b) n
| n <= 0 = (a,b) | True = fibsteps (b, a+b) (n-1)
fibnums :: [Integer] fibnums = map fst $ iterate (`fibsteps` 1) (0,1)
fibN2 :: Integer -> (Integer, Integer) fibN2 m | m < 10 = fibsteps (0,1) 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 d = ( 5*g^3 + if even n then (- 3*g) else 3*g ) -- 3(n+1) (*) b = ( g^3 + 3 * g * f^2 - f^3 ) -- 3n+1 c = ( g^3 + 3 * g^2 * f + f^3 ) -- 3n+2 </lang>
(fibN2 n)
directly calculates a pair (f,g)
of two consecutive Fibonacci numbers, (Fib[n], Fib[n+1])
, from recursively calculated such pair at about n/3
:
<lang haskell> *Main> take 10 $ show $ fst $ fibN2 (10^6)
"1953282128"</lang>
The above should take less than 0.1s on modern PC to calculate. Other identities that could also be used are here.
ghci; functional; recursive; one-line
Simple the definition, not efficient.
<lang haskell>let fib x = if x < 1 then 0 else (if x < 3 then 1 else (fib(x - 1) + fib(x - 2)))</lang>
Hope
Recursive
<lang hope>dec f : num -> num; --- f 0 <= 0; --- f 1 <= 1; --- f(n+2) <= f n + f(n+1);</lang>
Tail-recursive
<lang hope>dec fib : num -> num; --- fib n <= l (1, 0, n)
whererec l == \(a,b,succ c) => if c<1 then a else l((a+b),a,c) |(a,b,0) => 0;</lang>
With lazy lists
This language, being one of Haskell's ancestors, also has lazy lists. Here's the (infinite) list of all Fibonacci numbers:
<lang hope>dec fibs : list num;
--- fibs <= fs whererec fs == 0::1::map (+) (tail fs||fs);</lang>
The nth Fibonacci number is then just fibs @ n
.
HicEst
<lang hicest>REAL :: Fibonacci(10)
Fibonacci = ($==2) + Fibonacci($-1) + Fibonacci($-2) WRITE(ClipBoard) Fibonacci ! 0 1 1 2 3 5 8 13 21 34</lang>
Icon and Unicon
Icon has built-in support for big numbers. First, a simple recursive solution augmented by caching for non-negative input. This examples computes fib(1000) if there is no integer argument.
<lang Icon>procedure main(args)
write(fib(integer(!args) | 1000)
end
procedure fib(n)
static fCache initial { fCache := table() fCache[0] := 0 fCache[1] := 1 } /fCache[n] := fib(n-1) + fib(n-2) return fCache[n]
end</lang>
The above solution is similar to the one provided fib in memrfncs
Now, an O(logN) solution. For large N, it takes far longer to convert the result to a string for output than to do the actual computation. This example computes fib(1000000) if there is no integer argument.
<lang Icon>procedure main(args)
write(fib(integer(!args) | 1000000))
end
procedure fib(n)
return fibMat(n)[1]
end
procedure fibMat(n)
if n <= 0 then return [0,0] if n = 1 then return [1,0] fp := fibMat(n/2) c := fp[1]*fp[1] + fp[2]*fp[2] d := fp[1]*(fp[1]+2*fp[2]) if n%2 = 1 then return [c+d, d] else return [d, c]
end</lang>
IDL
Recursive
<lang idl>function fib,n
if n lt 3 then return,1L else return, fib(n-1)+fib(n-2)
end</lang>
Execution time O(2^n) until memory is exhausted and your machine starts swapping. Around fib(35) on a 2GB Core2Duo.
Iterative
<lang idl>function fib,n
psum = (csum = 1uL) if n lt 3 then return,csum for i = 3,n do begin nsum = psum + csum psum = csum csum = nsum endfor return,nsum
end</lang>
Execution time O(n). Limited by size of uLong to fib(49)
Analytic
<lang idl>function fib,n
q=1/( p=(1+sqrt(5))/2 ) return,round((p^n+q^n)/sqrt(5))
end</lang>
Execution time O(1), only limited by the range of LongInts to fib(48).
J
The Fibonacci Sequence essay on the J Wiki presents a number of different ways of obtaining the nth Fibonacci number. Here is one: <lang j> fibN=: (-&2 +&$: -&1)^:(1&<) M."0</lang> Examples: <lang j> fibN 12 144
fibN i.31
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040</lang>
(This implementation is doubly recursive except that results are cached across function calls.)
Java
Iterative
<lang java>public static long itFibN(int n) {
if (n < 2) return n; long ans = 0; long n1 = 0; long n2 = 1; for(n--; n > 0; n--) { ans = n1 + n2; n1 = n2; n2 = ans; } return ans;
}</lang>
Recursive
<lang java>public static long recFibN(final int n) {
return (n < 2) ? n : recFibN(n - 1) + recFibN(n - 2);
}</lang>
Analytic
This method works up to the 92nd Fibonacci number. After that, it goes out of range. <lang java>public static long anFibN(final long n) {
double p = (1 + Math.sqrt(5)) / 2; double q = 1 / p; return (long) ((Math.pow(p, n) + Math.pow(q, n)) / Math.sqrt(5));
}</lang>
Tail-recursive
<lang java>public static long fibTailRec(final int n) {
return fibInner(0, 1, n);
}
private static long fibInner(final long a, final long b, final int n) {
return n < 1 ? a : n == 1 ? b : fibInner(b, a + b, n - 1);
}</lang>
JavaScript
Recursive
One possibility familiar to Scheme programmers is to define an internal function for iteration through anonymous tail recursion: <lang javascript>function fib(n) {
return function(n,a,b) { return n>0 ? arguments.callee(n-1,b,a+b) : a; }(n,0,1);
}</lang>
Iterative
<lang javascript>function fib(n) {
var a = 0, b = 1, t; while (n-- > 0) { t = a; a = b; b += t; } return a;
}
var i; for (i = 0; i < 10; ++i)
alert(fib(i));</lang>
Memoization
<lang javascript>var fib = (function(cache){
return cache = cache || {}, function(n){ if (cache[n]) return cache[n]; else return cache[n] = n == 0 ? 0 : n < 0 ? -fib(-n) : n <= 2 ? 1 : fib(n-2) + fib(n-1); };
})(); </lang>
Y-Combinator
<lang javascript>function Y(dn) {
return (function(fn) { return fn(fn); }(function(fn) { return dn(function() { return fn(fn).apply(null, arguments); }); }));
} var fib = Y(function(fn) {
return function(n) { if (n === 0 || n === 1) { return n; } return fn(n - 1) + fn(n - 2); };
});</lang>
Joy
Recursive
<lang joy>DEFINE fib == [small] [] [pred dup pred] [+] binrec.</lang>
Iterative
<lang joy>DEFINE fib == [1 0] dip [swap [+] unary] times popd.</lang>
Julia
Recursive
<lang Julia>fib(n) = n < 2 ? n : fib(n-1) + fib(n-2)</lang>
Iterative
<lang Julia>function fib(n)
x,y = (0,1) for i = 1:n x,y = (y, x+y) end x
end</lang>
Matrix form
<lang Julia>fib(n) = ([1 1 ; 1 0]^n)[1,2]</lang>
K
Recursive
<lang K>{:[x<3;1;_f[x-1]+_f[x-2]]}</lang>
Recursive with memoization
Using a (global) dictionary c.
<lang K>{c::.();{v:c[a:`$$x];:[x<3;1;:[_n~v;c[a]:_f[x-1]+_f[x-2];v]]}x}</lang>
Analytic
<lang K>phi:(1+_sqrt(5))%2 {_((phi^x)-((1-phi)^x))%_sqrt[5]}</lang>
Sequence to n
<lang K>{(x(|+\)\1 1)[;1]}</lang> <lang K>{x{x,+/-2#x}/!2}</lang>
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.
Lang5
<lang lang5>[] '__A set : dip swap __A swap 2 compress collapse '__A set execute
__A -1 extract nip ; : nip swap drop ; : tuck swap over ;
- -rot rot rot ; : 0= 0 == ; : 1+ 1 + ; : 1- 1 - ; : sum '+ reduce ;
- bi 'keep dip execute ; : keep over 'execute dip ;
- fib dup 1 > if dup 1- fib swap 2 - fib + then ;
- fib dup 1 > if "1- fib" "2 - fib" bi + then ;</lang>
Liberty BASIC
<lang lb>for i = 0 to 15
print fiboR(i),fiboI(i)
next i
function fiboR(n)
if n <= 1 then fiboR = n else fiboR = fiboR(n-1) + fiboR(n-2) end if
end function
function fiboI(n)
a = 0 b = 1 for i = 1 to n temp = a + b a = b b = temp next i fiboI = a
end function
</lang>
Lisaac
<lang Lisaac>- fib(n : UINTEGER_32) : UINTEGER_64 <- (
+ result : UINTEGER_64; (n < 2).if { result := n; } else { result := fib(n - 1) + fib(n - 2); }; result
);</lang>
Lasso
<lang Lasso> define_tag("fibonacci", -required="n", -type="integer");
if(#n < 1); return false; /if; local("swap") = 0; local("n1") = 0; local("n2") = 1; loop(#n); #swap = #n1 + #n2; #n2 = #n1; #n1 = #swap; /loop; return(#n1);
/define_tag;
fibonacci(0); //->output false fibonacci(1); //->output 1 fibonacci(2); //->output 1 fibonacci(3); //->output 2 </lang>
Logo
<lang logo>to fib :n [:a 0] [:b 1]
if :n < 1 [output :a] output (fib :n-1 :b :a+:b)
end</lang>
Lua
<lang lua>--calculates the nth fibonacci number. Breaks for negative or non-integer n. function fibs(n)
return n < 2 and n or fibs(n - 1) + fibs(n - 2)
end
--more pedantic version, returns 0 for non-integer n function pfibs(n)
if n ~= math.floor(n) then return 0 elseif n < 0 then return pfibs(n + 2) - pfibs(n + 1) elseif n < 2 then return n else return pfibs(n - 1) + pfibs(n - 2) end
end
--tail-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,1,0) end
--table-recursive fib_n = setmetatable({1, 1}, {__index = function(z,n) return z[n-1] + z[n-2] end})</lang>
LSL
Rez a box on the ground, and add the following as a New Script. <lang LSL>integer Fibonacci(integer n) { if(n<2) { return n; } else { return Fibonacci(n-1)+Fibonacci(n-2); } } default { state_entry() { integer x = 0; for(x=0 ; x<35 ; x++) { llOwnerSay("Fibonacci("+(string)x+")="+(string)Fibonacci(x)); } } }</lang> Output:
Fibonacci(0)=0 Fibonacci(1)=1 Fibonacci(2)=1 Fibonacci(3)=2 Fibonacci(4)=3 Fibonacci(5)=5 Fibonacci(6)=8 Fibonacci(7)=13 Fibonacci(8)=21 Fibonacci(9)=34 Fibonacci(10)=55 Fibonacci(11)=89 Fibonacci(12)=144 Fibonacci(13)=233 Fibonacci(14)=377 Fibonacci(15)=610 Fibonacci(16)=987 Fibonacci(17)=1597 Fibonacci(18)=2584 Fibonacci(19)=4181 Fibonacci(20)=6765 Fibonacci(21)=10946 Fibonacci(22)=17711 Fibonacci(23)=28657 Fibonacci(24)=46368 Fibonacci(25)=75025 Fibonacci(26)=121393 Fibonacci(27)=196418 Fibonacci(28)=317811 Fibonacci(29)=514229 Fibonacci(30)=832040 Fibonacci(31)=1346269 Fibonacci(32)=2178309 Fibonacci(33)=3524578 Fibonacci(34)=5702887
M4
<lang m4>define(`fibo',`ifelse(0,$1,0,`ifelse(1,$1,1, `eval(fibo(decr($1)) + fibo(decr(decr($1))))')')')dnl define(`loop',`ifelse($1,$2,,`$3($1) loop(incr($1),$2,`$3')')')dnl loop(0,15,`fibo')</lang>
Mathematica / Wolfram Language
The Wolfram Language already has a built-in function Fibonacci, but a simple recursive implementation would be
<lang mathematica>fib[0] = 0 fib[1] = 1 fib[n_Integer] := fib[n - 1] + fib[n - 2]</lang>
An optimization is to cache the values already calculated:
<lang mathematica>fib[0] = 0 fib[1] = 1 fib[n_Integer] := fib[n] = fib[n - 1] + fib[n - 2]</lang>
The above implementations may be too simplistic, as the first is incredibly slow for any reasonable range due to nested recursions and while the second is faster it uses an increasing amount of memory. The following uses recursion much more effectively while not using memory:
<lang mathematica>fibi[prvprv_Integer, prv_Integer, rm_Integer] :=
If[rm < 1, prvprv, fibi[prv, prvprv + prv, rm - 1]]
fib[n_Integer] := fibi[0, 1, n]</lang>
However, the recursive approaches in Mathematica are limited by the limit set for recursion depth (default 1024 or 4096 for the above cases), limiting the range for 'n' to about 1000 or 2000. The following using an iterative approach has an extremely high limit (greater than a million):
<lang mathematica>fib[n_Integer] := Block[{tmp, prvprv = 0, prv = 1},
For[i = 0, i < n, i++, tmp = prv; prv += prvprv; prvprv = tmp]; Return[prvprv]]</lang>
If one wanted a list of Fibonacci numbers, the following is quite efficient:
<lang mathematica>fibi[{prvprv_Integer, prv_Integer}] := {prv, prvprv + prv} fibList[n_Integer] := Map[Take[#, 1] &, NestList[fibi, {0, 1}, n]] // Flatten</lang>
Output from the last with "fibList[100]":
<lang mathematica>{0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, \ 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, \ 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, \ 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, \ 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, \ 2971215073, 4807526976, 7778742049, 12586269025, 20365011074, \ 32951280099, 53316291173, 86267571272, 139583862445, 225851433717, \ 365435296162, 591286729879, 956722026041, 1548008755920, \ 2504730781961, 4052739537881, 6557470319842, 10610209857723, \ 17167680177565, 27777890035288, 44945570212853, 72723460248141, \ 117669030460994, 190392490709135, 308061521170129, 498454011879264, \ 806515533049393, 1304969544928657, 2111485077978050, \ 3416454622906707, 5527939700884757, 8944394323791464, \ 14472334024676221, 23416728348467685, 37889062373143906, \ 61305790721611591, 99194853094755497, 160500643816367088, \ 259695496911122585, 420196140727489673, 679891637638612258, \ 1100087778366101931, 1779979416004714189, 2880067194370816120, \ 4660046610375530309, 7540113804746346429, 12200160415121876738, \ 19740274219868223167, 31940434634990099905, 51680708854858323072, \ 83621143489848422977, 135301852344706746049, 218922995834555169026, \ 354224848179261915075}</lang>
MATLAB
Iterative
<lang MATLAB>function F = fibonacci(n)
Fn = [1 0]; %Fn(1) is F_{n-2}, Fn(2) is F_{n-1} F = 0; %F is F_{n} for i = (1:abs(n)) Fn(2) = F; F = sum(Fn); Fn(1) = Fn(2); end if n < 0 F = F*((-1)^(n+1)); end
end</lang>
Dramadah Matrix Method
The MATLAB help file suggests an interesting method of generating the Fibonacci numbers. Apparently the determinate of the Dramadah Matrix of type 3 (MATLAB designation) and size n-by-n is the nth Fibonacci number. This method is implimented below.
<lang MATLAB>function number = fibonacci2(n)
if n == 1 number = 1; elseif n == 0 number = 0; elseif n < 0 number = ((-1)^(n+1))*fibonacci2(-n);; else number = det(gallery('dramadah',n,3)); end
end</lang>
Maxima
<lang maxima>/* fib(n) is built-in; here is an implementation */ fib2(n) := (matrix([0, 1], [1, 1])^^n)[1, 2]$
fib2(100)-fib(100); 0
fib2(-10); -55</lang>
MAXScript
Iterative
<lang maxscript>fn fibIter n = (
if n < 2 then ( n ) else ( fib = 1 fibPrev = 1 for num in 3 to n do ( temp = fib fib += fibPrev fibPrev = temp ) fib )
)</lang>
Recursive
<lang maxscript>fn fibRec n = (
if n < 2 then ( n ) else ( fibRec (n - 1) + fibRec (n - 2) )
)</lang>
Mercury
Mercury is both a logic language and a functional language. As such there are two possible interfaces for calculating a Fibonacci number. This code shows both styles. Note that much of the code here is ceremony put in place to have this be something which can actually compile. The actual Fibonacci number generation is contained in the predicate fib/2
and in the function fib/1
. The predicate main/2
illustrates first the unification semantics of the predicate form and the function call semantics of the function form.
The provided code uses a very naive form of generating a Fibonacci number. A more realistic implementation would use memoization to cache previous results, exchanging time for space. Also, in the case of supplying both a function implementation and a predicate implementation, one of the two would be implemented in terms of the other. Examples of this are given as comments below.
fib.m
<lang mercury> % The following code is derived from the Mercury Tutorial by Ralph Becket. % http://www.mercury.csse.unimelb.edu.au/information/papers/book.pdf
- - module fib.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - implementation.
- - import_module int.
- - pred fib(int::in, int::out) is det.
fib(N, X) :-
( if N =< 2 then X = 1 else fib(N - 1, A), fib(N - 2, B), X = A + B ).
- - func fib(int) = int is det.
fib(N) = X :- fib(N, X).
main(!IO) :-
fib(40, X), write_string("fib(40, ", !IO), write_int(X, !IO), write_string(")\n", !IO), write_string("fib(40) = ", !IO), write_int(fib(40), !IO), write_string("\n", !IO).
</lang>
Iterative algorithm
The much faster iterative algorithm can be written as:
<lang mercury>
- - pred fib_acc(int::in, int::in, int::in, int::in, int::out) is det.
fib_acc(N, Limit, Prev2, Prev1, Res) :-
( N < Limit -> % limit not reached, continue computation. ( N =< 2 -> Res0 = 1 ; Res0 = Prev2 + Prev1 ), fib_acc(N+1, Limit, Prev1, Res0, Res) ; % Limit reached, return the sum of the two previous results. Res = Prev2 + Prev1 ).
</lang>
This predicate can be called as <lang mercury>fib_acc(1, 40, 1, 1, Result)</lang> It has several inputs which form the loop, the first is the current number, the second is a limit, ie when to stop counting. And the next two are accumulators for the last and next-to-last results.
Memoization
But what if you want the speed of the fib_acc with the recursive (more declarative) definition of fib? Then use memoization, because Mercury is a pure language fib(N, F) will always give the same F for the same N, guaranteed. Therefore memoization asks the compiler to use a table to remember the value for F for any N, and it's a one line change:
<lang mercury>
- - pragma memo(fib/2).
- - pred fib(int::in, int::out) is det.
fib(N, X) :-
( if N =< 2 then X = 1 else fib(N - 1, A), fib(N - 2, B), X = A + B ).
</lang>
We've shown the definition of fib/2 again, but the only change here is the memoization pragma (see the reference manual). This is not part of the language specification and different Mercury implementations are allowed to ignore it, however there is only one implementation so in practice memoization is fully supported.
Memoization trades speed for space, a table of results is constructed and kept in memory. So this version of fib consumes more memory than than fib_acc. It is also slightly slower than fib_acc since it must manage its table of results but it is much much faster than without memoization. Memoization works very well for the Fibonacci sequence because in the naive version the same results are calculated over and over again.
Metafont
<lang metafont>vardef fibo(expr n) = if n=0: 0 elseif n=1: 1 else:
fibo(n-1) + fibo(n-2)
fi enddef;
for i=0 upto 10: show fibo(i); endfor end</lang>
Mirah
<lang mirah>def fibonacci(n:int)
return n if n < 2 fibPrev = 1 fib = 1 3.upto(Math.abs(n)) do oldFib = fib fib = fib + fibPrev fibPrev = oldFib end fib * (n<0 ? int(Math.pow(n+1, -1)) : 1)
end
puts fibonacci 1 puts fibonacci 2 puts fibonacci 3 puts fibonacci 4 puts fibonacci 5 puts fibonacci 6 puts fibonacci 7 </lang>
МК-61/52
<lang>П0 1 lg Вx <-> + L0 03 С/П БП 03</lang>
Instruction: n В/О С/П, where n is serial number of the number of Fibonacci sequence; С/П for the following numbers.
ML/I
<lang ML/I>MCSKIP "WITH" NL "" Fibonacci - recursive MCSKIP MT,<> MCINS %. MCDEF FIB WITHS () AS <MCSET T1=%A1. MCGO L1 UNLESS 2 GR T1 %T1.<>MCGO L0 %L1.%FIB(%T1.-1)+FIB(%T1.-2).> fib(0) is FIB(0) fib(1) is FIB(1) fib(2) is FIB(2) fib(3) is FIB(3) fib(4) is FIB(4) fib(5) is FIB(5)</lang>
Modula-3
Recursive
<lang modula3>PROCEDURE Fib(n: INTEGER): INTEGER =
BEGIN IF n < 2 THEN RETURN n; ELSE RETURN Fib(n-1) + Fib(n-2); END; END Fib;</lang>
MUMPS
Iterative
<lang MUMPS>FIBOITER(N)
;Iterative version to get the Nth Fibonacci number ;N must be a positive integer ;F is the tree containing the values ;I is a loop variable. QUIT:(N\1'=N)!(N<0) "Error: "_N_" is not a positive integer." NEW F,I SET F(0)=0,F(1)=1 QUIT:N<2 F(N) FOR I=2:1:N SET F(I)=F(I-1)+F(I-2) QUIT F(N)</lang>
USER>W $$FIBOITER^ROSETTA(30) 832040
Nemerle
Recursive
<lang Nemerle>using System; using System.Console;
module Fibonacci {
Fibonacci(x : long) : long { |x when x < 2 => x |_ => Fibonacci(x - 1) + Fibonacci(x - 2) } Main() : void { def num = Int64.Parse(ReadLine()); foreach (n in $[0 .. num]) WriteLine("{0}: {1}", n, Fibonacci(n)); }
}</lang>
Tail Recursive
<lang Nemerle>Fibonacci(x : long, current : long, next : long) : long {
match(x) { |0 => current |_ => Fibonacci(x - 1, next, current + next) }
}
Fibonacci(x : long) : long {
Fibonacci(x, 0, 1)
}</lang>
NetRexx
<lang NetRexx>/* NetRexx */
options replace format comments java crossref savelog symbols
numeric digits 210000 /*prepare for some big 'uns. */ parse arg x y . /*allow a single number or range.*/ if x == then do /*no input? Then assume -30-->+30*/
x = -30 y = -x end
if y == then y = x /*if only one number, show fib(n)*/ loop k = x to y /*process each Fibonacci request.*/
q = fib(k) w = q.length /*if wider than 25 bytes, tell it*/ say 'Fibonacci' k"="q if w > 25 then say 'Fibonacci' k "has a length of" w end k
exit
/*-------------------------------------FIB subroutine (non-recursive)---*/ method fib(arg) private static
parse arg n na = n.abs
if na < 2 then return na /*handle special cases. */ a = 0 b = 1
loop j = 2 to na s = a + b a = b b = s end j
if n > 0 | na // 2 == 1 then return s /*if positive or odd negative... */ else return -s /*return a negative Fib number. */
</lang>
NewLISP
Iterative
<lang newLISP>(define (fibonacci n)
(let (a 0 b 1 c n i 2) (while (<= i n) (setq c (+ a b) a b b c) (++ i)) c))</lang>
Recursive
<lang newLisp>(define (fibonacci n) (if (< n 2) 1 (+ (fibonacci (- n 1)) (fibonacci (- n 2))))) (print(fibonacci 10)) ;;89</lang>
Nimrod
Analytic
<lang nimrod>proc Fibonacci(n: int): int64 =
var fn = float64(n) var p: float64 = (1.0 + sqrt(5.0)) / 2.0 var q: float64 = 1.0 / p return int64((pow(p, fn) + pow(q, fn)) / sqrt(5.0))</lang>
Iterative
<lang nimrod>proc Fibonacci(n: int): int =
var first = 0 second = 1
for i in 0 .. <n: swap first, second second += first
result = first</lang>
Recursive
<lang nimrod>proc Fibonacci(n: int): int64 =
if n <= 2: result = 1 else: result = Fibonacci(n - 1) + Fibonacci(n - 2)</lang>
Tail-recursive
<lang nimrod>proc Fibonacci(n: int, current: int64, next: int64): int64 =
if n == 0: result = current else: result = Fibonacci(n - 1, next, current + next)
proc Fibonacci(n: int): int64 =
result = Fibonacci(n, 0, 1)</lang>
Continuations
<lang nimrod>iterator fib: int {.closure.} =
var a = 0 var b = 1 while true: yield a swap a, b b = a + b
var f = fib for i in 0.. <10:
echo f()</lang>
Objeck
Recursive
<lang objeck>bundle Default {
class Fib { function : Main(args : String[]), Nil { for(i := 0; i <= 10; i += 1;) { Fib(i)->PrintLine(); }; } function : native : Fib(n : Int), Int { if(n < 2) { return n; }; return Fib(n-1) + Fib(n-2); } }
}</lang>
Objective-C
Recursive
<lang objc>-(long)fibonacci:(int)position {
long result = 0; if (position < 2) { result = position; } else { result = [self fibonacci:(position -1)] + [self fibonacci:(position -2)]; } return result;
}</lang>
Iterative
<lang objc>+(long)fibonacci:(int)index {
long beforeLast = 0, last = 1; while (index > 0) { last += beforeLast; beforeLast = last - beforeLast; --index; } return last;
}</lang>
OCaml
Iterative
<lang ocaml>let fib_iter n =
if n < 2 then n else let fib_prev = ref 1 and fib = ref 1 in for num = 2 to n - 1 do let temp = !fib in fib := !fib + !fib_prev; fib_prev := temp done; !fib</lang>
Recursive
<lang ocaml>let rec fib_rec n =
if n < 2 then n else fib_rec (n - 1) + fib_rec (n - 2)</lang>
The previous way is the naive form, because for most n the fib_rec is called twice, and it is not tail recursive because it adds the result of two function calls. The next version resolves these problems through accumulator argument technique. It is computationally equivalent to the iterative version above (tail recursion is effectively iteration):
<lang ocaml>let fib n =
let rec fib_aux n a b = match n with | 0 -> a | _ -> fib_aux (n-1) b (a+b) in fib_aux n 0 1</lang>
Arbitrary Precision
Using OCaml's Num module.
<lang ocaml>open Num
let fib =
let rec fib_aux f0 f1 = function | 0 -> f0 | 1 -> f1 | n -> fib_aux f1 (f1 +/ f0) (n - 1) in fib_aux (num_of_int 0) (num_of_int 1)</lang>
compile with:
ocamlopt nums.cmxa -o fib fib.ml
O(log(n)) with arbitrary precision
<lang ocaml>open Num
let mul (a,b,c) (d,e,f) =
(a*/d +/ b*/e, a*/e +/ b*/f, b*/e +/ c*/f)
let rec pow a n =
if n=1 then a else let b = pow a (n/2) in if (n mod 2) = 0 then mul b b else mul a (mul b b)
let fib n =
let (_,y,_) = (pow (Int 1, Int 1, Int 0) n) in string_of_num y
Printf.printf "fib %d = %s\n" 300 (fib 300)</lang>
Output:
fib 300 = 22223224462942044552973989346190996720666693909649976499097960
Octave
Recursive <lang octave>% recursive function fibo = recfibo(n)
if ( n < 2 ) fibo = n; else fibo = recfibo(n-1) + recfibo(n-2); endif
endfunction</lang>
Iterative <lang octave>% iterative function fibo = iterfibo(n)
if ( n < 2 ) fibo = n; else f = zeros(2,1); f(1) = 0; f(2) = 1; for i = 2 : n t = f(2); f(2) = f(1) + f(2); f(1) = t; endfor fibo = f(2); endif
endfunction</lang>
Testing <lang octave>% testing for i = 0 : 20
printf("%d %d\n", iterfibo(i), recfibo(i));
endfor</lang>
OPL
<lang opl>FIBON: REM Fibonacci sequence is generated to the Organiser II floating point variable limit. REM This method was derived from (not copied...) the original OPL manual that came with the CM and XP in the mid 1980s. REM CLEAR/ON key quits. REM Mikesan - http://forum.psion2.org/ LOCAL A,B,C A=1 :B=1 :C=1 PRINT A, DO
C=A+B A=B B=C PRINT A,
UNTIL GET=1</lang>
Order
Recursive
<lang c>#include <order/interpreter.h>
- define ORDER_PP_DEF_8fib_rec \
ORDER_PP_FN(8fn(8N, \
8if(8less(8N, 2), \ 8N, \ 8add(8fib_rec(8sub(8N, 1)), \ 8fib_rec(8sub(8N, 2))))))
ORDER_PP(8fib_rec(10))</lang>
Tail recursive version (example supplied with language): <lang c>#include <order/interpreter.h>
- define ORDER_PP_DEF_8fib \
ORDER_PP_FN(8fn(8N, \
8fib_iter(8N, 0, 1)))
- define ORDER_PP_DEF_8fib_iter \
ORDER_PP_FN(8fn(8N, 8I, 8J, \
8if(8is_0(8N), \ 8I, \ 8fib_iter(8dec(8N), 8J, 8add(8I, 8J)))))
ORDER_PP(8to_lit(8fib(8nat(5,0,0))))</lang>
Memoization
<lang c>#include <order/interpreter.h>
- define ORDER_PP_DEF_8fib_memo \
ORDER_PP_FN(8fn(8N, \
8tuple_at(0, 8fib_memo_inner(8N, 8seq))))
- define ORDER_PP_DEF_8fib_memo_inner \
ORDER_PP_FN(8fn(8N, 8M, \
8cond((8less(8N, 8seq_size(8M)), 8pair(8seq_at(8N, 8M), 8M)) \ (8equal(8N, 0), 8pair(0, 8seq(0))) \ (8equal(8N, 1), 8pair(1, 8seq(0, 1))) \ (8else, \ 8lets((8S, 8fib_memo_inner(8sub(8N, 2), 8M)) \ (8T, 8fib_memo_inner(8dec(8N), 8tuple_at(1, 8S))) \ (8U, 8add(8tuple_at(0, 8S), 8tuple_at(0, 8T))), \ 8pair(8U, \ 8seq_append(8tuple_at(1, 8T), 8seq(8U))))))))
ORDER_PP( 8for_each_in_range(8fn(8N,
8print(8to_lit(8fib_memo(8N)) (,) 8space)), 1, 21)
)</lang>
Oz
Iterative
Using mutable references (cells). <lang oz>fun{FibI N}
Temp = {NewCell 0} A = {NewCell 0} B = {NewCell 1}
in
for I in 1..N do Temp := @A + @B A := @B B := @Temp end @A
end</lang>
Recursive
Inefficient (blows up the stack). <lang oz>fun{FibR N}
if N < 2 then N else {FibR N-1} + {FibR N-2} end
end</lang>
Tail-recursive
Using accumulators. <lang oz>fun{Fib N}
fun{Loop N A B} if N == 0 then
B
else
{Loop N-1 A+B A}
end end
in
{Loop N 1 0}
end</lang>
Lazy-recursive
<lang oz>declare
fun lazy {FiboSeq} {LazyMap {Iterate fun {$ [A B]} [B A+B] end [0 1]} Head} end
fun {Head A|_} A end
fun lazy {Iterate F I} I|{Iterate F {F I}} end
fun lazy {LazyMap Xs F} case Xs of X|Xr then {F X}|{LazyMap Xr F} [] nil then nil end end
in
{Show {List.take {FiboSeq} 8}}</lang>
PARI/GP
Built-in
<lang parigp>fibonocci(n)</lang>
Matrix
<lang parigp>([1,1;1,0]^n)[1,2]</lang>
Analytic
This uses the Binet form. <lang parigp>fib(n)=my(phi=(1+sqrt(5))/2);round((phi^n-phi^-n)/sqrt(5))</lang> The second term can be dropped since the error is always small enough to be subsumed by the rounding. <lang parigp>fib(n)=round(((1+sqrt(5))/2)^n/sqrt(5))</lang>
Algebraic
This is an exact version of the above formula. quadgen(5)
represents and the number is stored in the form . imag
takes the coefficient of . This uses the relation
and hence real(quadgen(5)^n)
would give the (n-1)-th Fibonacci number.
<lang parigp>fib(n)=imag(quadgen(5)^n)</lang>
A more direct translation (note that ) would be <lang parigp>fib(n)=my(phi=quadgen(5));(phi^n-(-1/phi)^n)/(2*phi-1)</lang>
Combinatorial
This uses the generating function. It can be trivially adapted to give the first n Fibonacci numbers. <lang parigp>fib(n)=polcoeff(x/(1-x-x^2)+O(x^(n+1)),n)</lang>
Binary powering
<lang parigp>fib(n)={
if(n<=0, if(n,(-1)^(n+1)*fib(n),0) , my(v=lucas(n-1)); (2*v[1]+v[2])/5 )
}; lucas(n)={
if (!n, return([2,1])); my(v=lucas(n >> 1), z=v[1], t=v[2], pr=v[1]*v[2]); n=n%4; if(n%2, if(n==3,[v[1]*v[2]+1,v[2]^2-2],[v[1]*v[2]-1,v[2]^2+2]) , if(n,[v[1]^2+2,v[1]*v[2]+1],[v[1]^2-2,v[1]*v[2]-1]) )
};</lang>
Recursive
<lang parigp>fib(n)={
if(n<2, if(n<0, (-1)^(n+1)*fib(n) , n ) , fib(n-1)+fib(n) )
};</lang>
Iterative
<lang parigp>fib(n)={
if(n<0,return((-1)^(n+1)*fib(n))); my(a=0,b=1,t); while(n, t=a+b; a=b; b=t; n-- ); a
};</lang>
One-by-one
This code is purely for amusement and requires n > 1. It tests numbers in order to see if they are Fibonacci numbers, and waits until it has seen n of them. <lang parigp>fib(n)=my(k=0);while(n--,k++;while(!issquare(5*k^2+4)&&!issquare(5*k^2-4),k++));k</lang>
Pascal
Recursive
<lang pascal>function fib(n: integer): integer;
begin if (n = 0) or (n = 1) then fib := n else fib := fib(n-1) + fib(n-2) end;</lang>
Iterative
<lang pascal>function fib(n: integer): integer;
var f0, f1, f2, k: integer; begin f0 := 0; f1 := 1; for k := 2 to n do begin f2:= f0 + f1; f0 := f1; f1 := f2; end; fib := f2; end;</lang>
Perl
Iterative
<lang perl>
- Iterative Fibonacci with bignum support.
- Multi-licensed under your choice of:
- 1. The GNU Free Documentation License (GFDL).
- 2. The MIT/X11 license.
- 3. The GNU General Publice License (GPL).
- 4. The Public Domain as understood by the CC-Zero public domain dedication.
use strict; use warnings;
use Math::BigInt try => 'GMP';
sub fib_iter {
my ($n) = @_;
my $this_fib = Math::BigInt->new(0); my $next_fib = Math::BigInt->new(1);
my $pos = 0;
while ($pos < $n) { ($this_fib, $next_fib) = ($next_fib, $this_fib+$next_fib); } continue { $pos++; }
return $this_fib;
} </lang>
Recursive
<lang perl>sub fibRec {
my $n = shift; $n < 2 ? $n : fibRec($n - 1) + fibRec($n - 2);
}</lang>
Perl 6
Iterative
<lang perl6>sub fib (Int $n --> Int) {
$n > 1 or return $n; my ($prev, $this) = 0, 1; ($prev, $this) = $this, $this + $prev for 1 ..^ $n; return $this;
}</lang>
Recursive
<lang perl6>proto fib (Int $n --> Int) {*} multi fib (0) { 0 } multi fib (1) { 1 } multi fib ($n) { fib($n - 1) + fib($n - 2) }</lang> (Unfortunately, rakudo does not yet implement the is cached trait, so this remains an inefficient solution.)
Analytic
<lang perl6>sub fib (Int $n --> Int) {
constant φ1 = 1 / constant φ = (1 + sqrt 5)/2; constant invsqrt5 = 1 / sqrt 5;
floor invsqrt5 * (φ**$n + φ1**$n);
}</lang>
List Generator (built in)
This constructs the fibonacci sequence as a lazy infinite array. <lang perl6>my @fib := 0, 1, *+* ... *;</lang>
If you really need a function for it: <lang perl6>sub fib ($n) { @fib[$n] }</lang>
To support negative indices: <lang perl6>my @neg_fib := 0, 1, *-* ... *; sub fib ($n) { $n >= 0 and @fib[$n] or @neg_fib[-$n]; }</lang>
PHP
Iterative
<lang php>function fibIter($n) {
if ($n < 2) { return $n; } $fibPrev = 0; $fib = 1; foreach (range(1, $n-1) as $i) { list($fibPrev, $fib) = array($fib, $fib + $fibPrev); } return $fib;
}</lang>
Recursive
<lang php>function fibRec($n) {
return $n < 2 ? $n : fibRec($n-1) + fibRec($n-2);
}</lang>
PicoLisp
Recursive
<lang PicoLisp>(de fibo (N)
(if (>= 2 N) 1 (+ (fibo (dec N)) (fibo (- N 2))) ) )</lang>
Recursive with Cache
Using a recursive version doesn't need to be slow, as the following shows: <lang PicoLisp>(de fibo (N)
(cache '(NIL) N # Use a cache to accelerate (if (>= 2 N) N (+ (fibo (dec N)) (fibo (- N 2))) ) ) )
(bench (fibo 1000))</lang> Output: <lang PicoLisp>0.012 sec -> 43466557686937456435688527675040625802564660517371780402481729089536555417949 05189040387984007925516929592259308032263477520968962323987332247116164299644090 6533187938298969649928516003704476137795166849228875</lang>
Iterative
Recursive can only go so far until a stack overflow brings the whole thing crashing down. <lang PicoLisp>(de fibo (N)
(let (I 1 J 0) (do N (let (Tmp J) (inc 'J I) (setq I Tmp) ) ) J) )</lang>
PIR
Recursive:
<lang pir>.sub fib
.param int n .local int nt .local int ft if n < 2 goto RETURNN nt = n - 1 ft = fib( nt ) dec nt nt = fib(nt) ft = ft + nt .return( ft )
RETURNN:
.return( n ) end
.end
.sub main :main
.local int counter .local int f counter=0
LOOP:
if counter > 20 goto DONE f = fib(counter) print f print "\n" inc counter goto LOOP
DONE:
end
.end</lang>
Iterative (stack-based):
<lang pir>.sub fib
.param int n .local int counter .local int f .local pmc fibs .local int nmo .local int nmt
fibs = new 'ResizableIntegerArray' if n == 0 goto RETURN0 if n == 1 goto RETURN1 push fibs, 0 push fibs, 1 counter = 2
FIBLOOP:
if counter > n goto DONE nmo = pop fibs nmt = pop fibs f = nmo + nmt push fibs, nmt push fibs, nmo push fibs, f inc counter goto FIBLOOP
RETURN0:
.return( 0 ) end
RETURN1:
.return( 1 ) end
DONE:
f = pop fibs .return( f ) end
.end
.sub main :main
.local int counter .local int f counter=0
LOOP:
if counter > 20 goto DONE f = fib(counter) print f print "\n" inc counter goto LOOP
DONE:
end
.end</lang>
Pike
Iterative
<lang pike>int fibIter(int n) {
int fibPrev, fib, i; if (n < 2) { return 1; } fibPrev = 0; fib = 1; for (i = 1; i < n; i++) { int oldFib = fib; fib += fibPrev; fibPrev = oldFib; } return fib;
}</lang>
Recursive
<lang pike>int fibRec(int n) {
if (n < 2) { return(1); } return( fib(n-2) + fib(n-1) );
}</lang>
PL/I
<lang PL/I>/* Form the n-th Fibonacci number, n > 1. */ get list (n); f1 = 0; f2, f3 = 1; do i = 1 to n-2;
f3 = f1 + f2; f1 = f2; f2 = f3;
end; put skip list (f3);</lang>
PL/SQL
<lang PL/SQL>Create or replace Function fnu_fibonnaci(p_iNumber integer) return integer is
nuFib integer; nuP integer; nuQ integer;
Begin
if p_iNumber is not null then if p_iNumber=0 then nuFib:=0; Elsif p_iNumber=1 then nuFib:=1; Else nuP:=0; nuQ:=1; For nuI in 2..p_iNumber loop nuFib:=nuP+nuQ; nuP:=nuQ; nuQ:=nuFib; End loop; End if; End if; return(nuFib);
End fnu_fibonnaci;</lang>
Pop11
<lang pop11>define fib(x); lvars a , b;
1 -> a; 1 -> b; repeat x - 1 times (a + b, b) -> (b, a); endrepeat; a;
enddefine;</lang>
PostScript
Enter the desired number for "n" and run through your favorite postscript previewer or send to your postscript printer:
<lang postscript>%!PS
% We want the 'n'th fibonacci number /n 13 def
% Prepare output canvas: /Helvetica findfont 20 scalefont setfont 100 100 moveto
%define the function recursively: /fib { dup
3 lt { pop 1 } { dup 1 sub fib exch 2 sub fib add } ifelse } def (Fib\() show n (....) cvs show (\)=) show n fib (.....) cvs show
showpage</lang>
Potion
Starts with int and upgrades on-the-fly to doubles. <lang potion>fib = (n):
if (n <= 1): 1. else: fib (n - 1) + fib (n - 2)..
n = 40 ("fib(", n, ")= ", fib (n), "\n") join print</lang>
fib(40)= 165580141 real 0m2.851s
PowerBASIC
There seems to be a limitation (dare I say, bug?) in PowerBASIC regarding how large numbers are stored. 10E17 and larger get rounded to the nearest 10. For F(n), where ABS(n) > 87, is affected like this:
actual: displayed: F(88) 1100087778366101931 1100087778366101930 F(89) 1779979416004714189 1779979416004714190 F(90) 2880067194370816120 2880067194370816120 F(91) 4660046610375530309 4660046610375530310 F(92) 7540113804746346429 7540113804746346430
<lang powerbasic>FUNCTION fibonacci (n AS LONG) AS QUAD
DIM u AS LONG, a AS LONG, L0 AS LONG, outP AS QUAD STATIC fibNum() AS QUAD
u = UBOUND(fibNum) a = ABS(n)
IF u < 1 THEN REDIM fibNum(1) fibNum(1) = 1 u = 1 END IF
SELECT CASE a CASE 0 TO 92 IF a > u THEN REDIM PRESERVE fibNum(a) FOR L0 = u + 1 TO a fibNum(L0) = fibNum(L0 - 1) + fibNum(L0 - 2) IF 88 = L0 THEN fibNum(88) = fibNum(88) + 1 NEXT END IF IF n < 0 THEN fibonacci = fibNum(a) * ((-1)^(a+1)) ELSE fibonacci = fibNum(a) END IF CASE ELSE 'Even without the above-mentioned bug, we're still limited to 'F(+/-92), due to data type limits. (F(93) = &hA94F AD42 221F 2702) ERROR 6 END SELECT
END FUNCTION
FUNCTION PBMAIN () AS LONG
DIM n AS LONG #IF NOT %DEF(%PB_CC32) OPEN "out.txt" FOR OUTPUT AS 1 #ENDIF FOR n = -92 TO 92 #IF %DEF(%PB_CC32) PRINT STR$(n); ": "; FORMAT$(fibonacci(n), "#") #ELSE PRINT #1, STR$(n) & ": " & FORMAT$(fibonacci(n), "#") #ENDIF NEXT CLOSE
END FUNCTION</lang>
PowerShell
Iterative
<lang powershell>function fib ($n) {
if ($n -eq 0) { return 0 } if ($n -eq 1) { return 1 }
$m = 1 if ($n -lt 0) { if ($n % 2 -eq -1) { $m = 1 } else { $m = -1 }
$n = -$n }
$a = 0 $b = 1
for ($i = 1; $i -lt $n; $i++) { $c = $a + $b $a = $b $b = $c } return $m * $b
}</lang>
Recursive
<lang powershell>function fib($n) {
switch ($n) { 0 { return 0 } 1 { return 1 } { $_ -lt 0 } { return [Math]::Pow(-1, -$n + 1) * (fib (-$n)) } default { return (fib ($n - 1)) + (fib ($n - 2)) } }
}</lang>
Prolog
<lang prolog> fib(1, 1) :- !. fib(0, 0) :- !. fib(N, Value) :-
A is N - 1, fib(A, A1), B is N - 2, fib(B, B1), Value is A1 + B1.
</lang>
This naive implementation works, but is very slow for larger values of N. Here are some simple measurements (in SWI-Prolog): <lang prolog>?- time(fib(0,F)). % 2 inferences, 0.000 CPU in 0.000 seconds (88% CPU, 161943 Lips) F = 0.
?- time(fib(10,F)). % 265 inferences, 0.000 CPU in 0.000 seconds (98% CPU, 1458135 Lips) F = 55.
?- time(fib(20,F)). % 32,836 inferences, 0.016 CPU in 0.016 seconds (99% CPU, 2086352 Lips) F = 6765.
?- time(fib(30,F)). % 4,038,805 inferences, 1.122 CPU in 1.139 seconds (98% CPU, 3599899 Lips) F = 832040.
?- time(fib(40,F)). % 496,740,421 inferences, 138.705 CPU in 140.206 seconds (99% CPU, 3581264 Lips) F = 102334155.</lang>
As you can see, the calculation time goes up exponentially as N goes higher.
Poor man's memoization
The performance problem can be readily fixed by the addition of two lines of code (the first and last in this version): <lang prolog>%:- dynamic fib/2. % This is ISO, but GNU doesn't like it.
- - dynamic(fib/2). % Not ISO, but works in SWI, YAP and GNU unlike the ISO declaration.
fib(1, 1) :- !. fib(0, 0) :- !. fib(N, Value) :-
A is N - 1, fib(A, A1), B is N - 2, fib(B, B1), Value is A1 + B1, asserta((fib(N, Value) :- !)).</lang>
Let's take a look at the execution costs now:
<lang prolog>?- time(fib(0,F)). % 2 inferences, 0.000 CPU in 0.000 seconds (90% CPU, 160591 Lips) F = 0.
?- time(fib(10,F)). % 37 inferences, 0.000 CPU in 0.000 seconds (96% CPU, 552610 Lips) F = 55.
?- time(fib(20,F)). % 41 inferences, 0.000 CPU in 0.000 seconds (96% CPU, 541233 Lips) F = 6765.
?- time(fib(30,F)). % 41 inferences, 0.000 CPU in 0.000 seconds (95% CPU, 722722 Lips) F = 832040.
?- time(fib(40,F)). % 41 inferences, 0.000 CPU in 0.000 seconds (96% CPU, 543572 Lips) F = 102334155.</lang>
In this case by asserting the new N,Value pairing as a rule in the database we're making the classic time/space tradeoff. Since the space costs are (roughly) linear by N and the time costs are exponential by N, the trade-off is desirable. You can see the poor man's memoizing easily:
<lang prolog>?- listing(fib).
- - dynamic fib/2.
fib(40, 102334155) :- !. fib(39, 63245986) :- !. fib(38, 39088169) :- !. fib(37, 24157817) :- !. fib(36, 14930352) :- !. fib(35, 9227465) :- !. fib(34, 5702887) :- !. fib(33, 3524578) :- !. fib(32, 2178309) :- !. fib(31, 1346269) :- !. fib(30, 832040) :- !. fib(29, 514229) :- !. fib(28, 317811) :- !. fib(27, 196418) :- !. fib(26, 121393) :- !. fib(25, 75025) :- !. fib(24, 46368) :- !. fib(23, 28657) :- !. fib(22, 17711) :- !. fib(21, 10946) :- !. fib(20, 6765) :- !. fib(19, 4181) :- !. fib(18, 2584) :- !. fib(17, 1597) :- !. fib(16, 987) :- !. fib(15, 610) :- !. fib(14, 377) :- !. fib(13, 233) :- !. fib(12, 144) :- !. fib(11, 89) :- !. fib(10, 55) :- !. fib(9, 34) :- !. fib(8, 21) :- !. fib(7, 13) :- !. fib(6, 8) :- !. fib(5, 5) :- !. fib(4, 3) :- !. fib(3, 2) :- !. fib(2, 1) :- !. fib(1, 1) :- !. fib(0, 0) :- !. fib(A, D) :- B is A+ -1, fib(B, E), C is A+ -2, fib(C, F), D is E+F, asserta((fib(A, D):-!)).</lang>
All of the interim N/Value pairs have been asserted as facts for quicker future use, speeding up the generation of the higher Fibonacci numbers.
Continuation passing style
Works with SWI-Prolog and module lambda, written by Ulrich Neumerkel found there http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl <lang Prolog>:- use_module(lambda). fib(N, FN) :- cont_fib(N, _, FN, \_^Y^_^U^(U = Y)).
cont_fib(N, FN1, FN, Pred) :- ( N < 2 -> call(Pred, 0, 1, FN1, FN) ; N1 is N - 1, P = \X^Y^Y^U^(U is X + Y), cont_fib(N1, FNA, FNB, P), call(Pred, FNA, FNB, FN1, FN) ). </lang>
With lazy lists
Works with SWI-Prolog and others that support freeze/2
.
<lang Prolog>fib([0,1|X]) :-
ffib(0,1,X).
ffib(A,B,X) :-
freeze(X, (C is A+B, X=[C|Y], ffib(B,C,Y)) ).</lang>
The predicate fib(Xs)
unifies Xs
with an infinite list whose values are the Fibonacci sequence. The list can be used like this:
<lang Prolog>?- fib(X), length(A,15), append(A,_,X), writeln(A). [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]</lang>
Generators idiom
<lang Prolog>take( 0, Next, Z-Z, Next). take( N, Next, [A|B]-Z, NZ):- N>0, !, next( Next, A, Next1),
N1 is N-1, take( N1, Next1, B-Z, NZ).
next( fib(A,B), A, fib(B,C)):- C is A+B.
%% usage: ?- take(15, fib(0,1), _X-[], G), writeln(_X). %% [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377] %% G = fib(610, 987)</lang>
Yet another implementation
One of my favorites; loosely similar to the first example, but without the performance penalty, and needs nothing special to implement. Not even a dynamic database predicate. Attributed to M.E. for the moment, but simply because I didn't bother to search for the many people who probably did it like this long before I did. If someone knows who came up with it first, please let us know.
<lang Prolog>% Fibonacci sequence generator fib(C, [P,S], C, N) :- N is P + S. fib(C, [P,S], Cv, V) :- succ(C, Cn), N is P + S, !, fib(Cn, [S,N], Cv, V).
fib(0, 0). fib(1, 1). fib(C, N) :- fib(2, [0,1], C, N). % Generate from 3rd sequence on</lang> Looking at performance:
?- time(fib(30,X)). % 86 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 832040 ?- time(fib(40,X)). % 116 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 102334155 ?- time(fib(100,X)). % 296 inferences, 0.000 CPU in 0.001 seconds (0% CPU, Infinite Lips) X = 354224848179261915075
What I really like about this one, is it is also a generator- i.e. capable of generating all the numbers in sequence needing no bound input variables or special Prolog predicate support (such as freeze/3 in the previous example):
?- time(fib(X,Fib)). % 0 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = Fib, Fib = 0 ; % 1 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = Fib, Fib = 1 ; % 3 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 2, Fib = 1 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 3, Fib = 2 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 4, Fib = 3 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = Fib, Fib = 5 ; % 5 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 6, Fib = 8 ...etc.
It stays at 5 inferences per iteration after X=3. Also, quite useful:
?- time(fib(100,354224848179261915075)). % 296 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) true . ?- time(fib(X,354224848179261915075)). % 394 inferences, 0.000 CPU in 0.000 seconds (?% CPU, Infinite Lips) X = 100 .
Pure
Tail Recursive
<lang pure>fib n = loop 0 1 n with
loop a b n = if n==0 then a else loop b (a+b) (n-1);
end;</lang>
PureBasic
Macro based calculation
<lang PureBasic>Macro Fibonacci (n) Int((Pow(((1+Sqr(5))/2),n)-Pow(((1-Sqr(5))/2),n))/Sqr(5)) EndMacro</lang>
Recursive
<lang PureBasic>Procedure FibonacciReq(n)
If n<2 ProcedureReturn n Else ProcedureReturn FibonacciReq(n-1)+FibonacciReq(n-2) EndIf
EndProcedure</lang>
Recursive & optimized with a static hash table
This will be much faster on larger n's, this as it uses a table to store known parts instead of recalculating them. On my machine the speedup compares to above code is
Fib(n) Speedup 20 2 25 23 30 217 40 25847 46 1156741
<lang PureBasic>Procedure Fibonacci(n)
Static NewMap Fib.i() Protected FirstRecursion If MapSize(Fib())= 0 ; Init the hash table the first run Fib("0")=0: Fib("1")=1 FirstRecursion = #True EndIf If n >= 2 Protected.s s=Str(n) If Not FindMapElement(Fib(),s) ; Calculate only needed parts Fib(s)= Fibonacci(n-1)+Fibonacci(n-2) EndIf n = Fib(s) EndIf If FirstRecursion ; Free the memory when finalizing the first call ClearMap(Fib()) EndIf ProcedureReturn n
EndProcedure</lang>
Example
Fibonacci(0)= 0 Fibonacci(1)= 1 Fibonacci(2)= 1 Fibonacci(3)= 2 Fibonacci(4)= 3 Fibonacci(5)= 5 FibonacciReq(0)= 0 FibonacciReq(1)= 1 FibonacciReq(2)= 1 FibonacciReq(3)= 2 FibonacciReq(4)= 3 FibonacciReq(5)= 5
Purity
The following takes a natural number and generates an initial segment of the Fibonacci sequence of that length:
<lang Purity> data Fib1 = FoldNat
< const (Cons One (Cons One Empty)), (uncurry Cons) . ((uncurry Add) . (Head, Head . Tail), id) >
</lang>
This following calculates the Fibonacci sequence as an infinite stream of natural numbers:
<lang Purity> type (Stream A?,,,Unfold) = gfix X. A? . X? data Fib2 = Unfold ((outl, (uncurry Add, outl))) ((curry id) One One) </lang>
As a histomorphism:
<lang Purity> import Histo
data Fib3 = Histo . Memoize
< const One, (p1 => < const One, (p2 => Add (outl $p1) (outl $p2)). UnmakeCofree > (outr $p1)) . UnmakeCofree >
</lang>
Python
Analytic
<lang python>from math import *
def analytic_fibonacci(n):
sqrt_5 = sqrt(5); p = (1 + sqrt_5) / 2; q = 1/p; return int( (p**n + q**n) / sqrt_5 + 0.5 )
for i in range(1,31):
print analytic_fibonacci(i),</lang>
Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
Iterative
<lang python>def fibIter(n):
if n < 2: return n fibPrev = 1 fib = 1 for num in xrange(2, n): fibPrev, fib = fib, fib + fibPrev return fib</lang>
Recursive
<lang python>def fibRec(n):
if n < 2: return n else: return fibRec(n-1) + fibRec(n-2)</lang>
Recursive with Memoization
<lang python>def fibMemo():
pad = {0:0, 1:1} def func(n): if n not in pad: pad[n] = func(n-1) + func(n-2) return pad[n] return func
fm = fibMemo() for i in range(1,31):
print fm(i),</lang>
Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040
Better Recursive doesn't need Memoization
The recursive code as written two sections above is incredibly slow and inefficient due to the nested recursion calls. Although the memoization above makes the code run faster, it is at the cost of extra memory use. The below code uses much more efficient recursion that doesn't require memoization:
<lang python>def fibFastRec(n):
def fib(prvprv, prv, c): if c < 1: return prvprv else: return fib(prv, prvprv + prv, c - 1) return fib(0, 1, n)</lang>
However, although much faster and not requiring memory, the above code can only process to a limited 'n' due to the limit on stack recursion depth by Python; it is better to use the iterative approach above or the generative one below.
Generative
<lang python>def fibGen():
f0, f1 = 0, 1 while True: yield f0 f0, f1 = f1, f0+f1</lang>
Example use
<lang python>>>> fg = fibGen() >>> for x in range(9):
print fg.next()
0 1 1 2 3 5 8 13 21 >>></lang>
Matrix-Based
Translation of the matrix-based approach used in F#. <lang python> def prevPowTwo(n):
'Gets the power of two that is less than or equal to the given input' if ((n & -n) == n): return n else: n -= 1 n |= n >> 1 n |= n >> 2 n |= n >> 4 n |= n >> 8 n |= n >> 16 n += 1 return (n/2)
def crazyFib(n):
'Crazy fast fibonacci number calculation' powTwo = prevPowTwo(n) q = r = i = 1 s = 0 while(i < powTwo): i *= 2 q, r, s = q*q + r*r, r * (q + s), (r*r + s*s) while(i < n): i += 1 q, r, s = q+r, q, r return q
</lang>
Large step recurrence
This is much faster for a single, large value of n: <lang python>def fib(n, c={0:1, 1:1}):
if n not in c: x = n // 2 c[n] = fib(x-1) * fib(n-x-1) + fib(x) * fib(n - x) return c[n]
fib(10000000) # calculating it takes a few seconds, printing it takes eons</lang>
Generative with Recursion
This can get very slow and uses a lot of memory. Can be sped up by caching the generator results. <lang python>def fib():
"""Yield fib[n+1] + fib[n]""" yield 1 # have to start somewhere lhs, rhs = fib(), fib() yield next(lhs) # move lhs one iteration ahead while True: yield next(lhs)+next(rhs)
f=fib() print [next(f) for _ in range(9)]</lang>
Output:
[1, 1, 2, 3, 5, 8, 13, 21, 34]
Qi
Recursive
<lang qi> (define fib
0 -> 0 1 -> 1 N -> (+ (fib-r (- N 1)) (fib-r (- N 2))))
</lang>
Iterative
<lang qi> (define fib-0
V2 V1 0 -> V2 V2 V1 N -> (fib-0 V1 (+ V2 V1) (1- N)))
(define fib
N -> (fib-0 0 1 N))
</lang>
R
<lang R># recursive recfibo <- function(n) {
if ( n < 2 ) n else Recall(n-1) + Recall(n-2)
}
- print the first 21 elements
print.table(lapply(0:20, recfibo))
- iterative
iterfibo <- function(n) {
if ( n < 2 ) n else { f <- c(0, 1) for (i in 2:n) { t <- f[2] f[2] <- sum(f) f[1] <- t } f[2] }
}
print.table(lapply(0:20, iterfibo))
- iterative but looping replaced by map-reduce'ing
funcfibo <- function(n) {
if (n < 2) n else { generator <- function(f, ...) { c(f[2], sum(f)) } Reduce(generator, 2:n, c(0,1))[2] }
}
print.table(lapply(0:20, funcfibo))</lang>
Note that an idiomatic way to implement such low level, basic arithmethic operations in R is to implement them C and then call the compiled code.
- Output:
All three solutions print
[1] 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 [16] 610 987 1597 2584 4181 6765
Racket
Tail Recursive
<lang Racket> (define (fibo number)
(define (fibo-rec number n i) (if (<= number 0) i (fibo-rec (- number 1) (+ n i) n))) (fibo-rec number 1 0))
</lang>
REALbasic
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 92nd iteration. <lang vb>Function fibo(n as integer) As UInt64
dim noOne as UInt64 = 1 dim noTwo as UInt64 = 1 dim sum As UInt64
for i as integer = 1 to n sum = noOne + noTwo noTwo = noOne noOne = sum Next
Return noOne
End Function</lang>
Retro
Recursive
<lang Retro>: fib ( n-m ) dup [ 0 = ] [ 1 = ] bi or if; [ 1- fib ] sip [ 2 - fib ] do + ;</lang>
Iterative
<lang Retro>: fib ( n-N )
[ 0 1 ] dip [ over + swap ] times drop ;</lang>
REXX
With 210,000 numeric digits, this REXX program can handle Fibonacci numbers past one million.
[Generally speaking, REXX can handle up to around 8 million digits.]
This version of the REXX program can also handle negative Fibonacci numbers.
<lang rexx>/*REXX program calculates the Nth Fibonacci number, N can be zero or neg*/
numeric digits 210000 /*prepare for some big 'uns. */
parse arg x y . /*allow a single number or range.*/
if x== then do; x=-40; y=abs(x); end /*No input? Use range -40 ──► +40*/
if y== then y=x /*if only one number, show fib(n)*/
w=max(length(x),length(y)) /*used for making output pretty. */
fw=10 /*minmum maximum width. Ka-razy*/
do j=x to y; q=fib(j) /*process each Fibonacci request.*/ fw=max(fw,length(q)) /*fib# length or the max so far. */ say 'Fibonacci('right(j,w)") = " right(q,fw) /*right justify Q.*/ if length(q)>10 then say 'Fibonacci('right(j,w)") has a length of" l end /*j*/
exit /*stick a fork in it, we're done.*/ /*─────────────────────────────────────FIB subroutine (non-recursive)───*/ fib: procedure; parse arg n; na=abs(n); a=0; b=1 if na<2 then return na /*handle couple of special cases.*/
do k=2 to na; s=a+b /*sum the numbers up to │n│ */ parse value b s with a b /*faster version of: a=b; s=b */ end /*k*/
if n>0 | na//2==1 then return s /*if positive or odd negative ...*/
return -s /* return a negative Fib number.*/</lang>
output using the default input:
Fibonacci(-40) = -102334155 Fibonacci(-39) = 63245986 Fibonacci(-38) = -39088169 Fibonacci(-37) = 24157817 Fibonacci(-36) = -14930352 Fibonacci(-35) = 9227465 Fibonacci(-34) = -5702887 Fibonacci(-33) = 3524578 Fibonacci(-32) = -2178309 Fibonacci(-31) = 1346269 Fibonacci(-30) = -832040 Fibonacci(-29) = 514229 Fibonacci(-28) = -317811 Fibonacci(-27) = 196418 Fibonacci(-26) = -121393 Fibonacci(-25) = 75025 Fibonacci(-24) = -46368 Fibonacci(-23) = 28657 Fibonacci(-22) = -17711 Fibonacci(-21) = 10946 Fibonacci(-20) = -6765 Fibonacci(-19) = 4181 Fibonacci(-18) = -2584 Fibonacci(-17) = 1597 Fibonacci(-16) = -987 Fibonacci(-15) = 610 Fibonacci(-14) = -377 Fibonacci(-13) = 233 Fibonacci(-12) = -144 Fibonacci(-11) = 89 Fibonacci(-10) = -55 Fibonacci( -9) = 34 Fibonacci( -8) = -21 Fibonacci( -7) = 13 Fibonacci( -6) = -8 Fibonacci( -5) = 5 Fibonacci( -4) = -3 Fibonacci( -3) = 2 Fibonacci( -2) = -1 Fibonacci( -1) = 1 Fibonacci( 0) = 0 Fibonacci( 1) = 1 Fibonacci( 2) = 1 Fibonacci( 3) = 2 Fibonacci( 4) = 3 Fibonacci( 5) = 5 Fibonacci( 6) = 8 Fibonacci( 7) = 13 Fibonacci( 8) = 21 Fibonacci( 9) = 34 Fibonacci( 10) = 55 Fibonacci( 11) = 89 Fibonacci( 12) = 144 Fibonacci( 13) = 233 Fibonacci( 14) = 377 Fibonacci( 15) = 610 Fibonacci( 16) = 987 Fibonacci( 17) = 1597 Fibonacci( 18) = 2584 Fibonacci( 19) = 4181 Fibonacci( 20) = 6765 Fibonacci( 21) = 10946 Fibonacci( 22) = 17711 Fibonacci( 23) = 28657 Fibonacci( 24) = 46368 Fibonacci( 25) = 75025 Fibonacci( 26) = 121393 Fibonacci( 27) = 196418 Fibonacci( 28) = 317811 Fibonacci( 29) = 514229 Fibonacci( 30) = 832040 Fibonacci( 31) = 1346269 Fibonacci( 32) = 2178309 Fibonacci( 33) = 3524578 Fibonacci( 34) = 5702887 Fibonacci( 35) = 9227465 Fibonacci( 36) = 14930352 Fibonacci( 37) = 24157817 Fibonacci( 38) = 39088169 Fibonacci( 39) = 63245986 Fibonacci( 40) = 102334155
output when the following was used as input: 10000
Fibonacci(10000) = 3364476487643178326662161200510754331030214846068006390656476997468008144216666236815559551363373402558206533268083615937373479048386526826304089246305643188735454436955982749160660209988418393386465273130008883026923567361313511757929743785441375213052050434770160226475831890652789085515436615958298727968298751063120057542878345321551510387081829896979161312785626503319548714021428753269818796204693609787990035096230229102636813149319527563022783762844154036058440257211433496118002309120828704608892396232883546150577658327125254609359112820392528539343462090424524892940390 170623388899108584106518317336043747073790855263176432573399371287193758774689747992630583706574283016163740896917842637862421283525811282051637029808933209990570792006436742620238978311147005407499845925036063356093388383192338678305613643535189213327973290813373264265263398976392272340788292817795358057099369104917547080893184105614632233821746563732124822638309210329770164805472624384237486241145309381220656491403275108664339451751216152654536133311131404243685480510676584349352383695965342807176877532834823434555736671973139274627362910821067928078471803532913117677892465908993863545932789 452377767440619224033763867400402133034329749690202832814593341882681768389307200363479562311710310129195316979460763273758925353077255237594378843450406771555577905645044301664011946258097221672975861502696844314695203461493229110597067624326851599283470989128470674086200858713501626031207190317208609408129832158107728207635318662461127824553720853236530577595643007251774431505153960090516860322034916322264088524885243315805153484962243484829938090507048348244932745373262456775587908918719080366205800959474315005240253270974699531877072437682590741993963226598414749819360928522394503970716544 3156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875 Fibonacci(10000) has a length of 2090
Ruby
Iterative
<lang ruby>def fibIter(n)
return 0 if n == 0 fibPrev, fib = 1, 1 (n.abs - 2).times { fibPrev, fib = fib, fib + fibPrev } fib * (n<0 ? (-1)**(n+1) : 1)
end</lang>
Recursive
<lang ruby>def fibRec(n)
if n <= -2 (-1)**(n+1) * fibRec(n.abs) elsif n <= 1 n.abs else fibRec(n-1) + fibRec(n-2) end
end</lang>
Recursive with Memoization
<lang ruby># Use the Hash#default_proc feature to
- lazily calculate the Fibonacci numbers.
fib = Hash.new do |f, n|
f[n] = if n <= -2 (-1)**(n+1) * f[n.abs] elsif n <= 1 n.abs else f[n-1] + f[n-2] end
end
- examples: fib[10] => 55, fib[-10] => (-55/1)</lang>
Matrix
<lang ruby>require 'matrix'
- To understand why this matrix is useful for Fibonacci numbers, remember
- that the definition of Matrix.**2 for any Matrix[[a, b], [c, d]] is
- is [[a*a + b*c, a*b + b*d], [c*a + d*b, c*b + d*d]]. In other words, the
- lower right element is computing F(k - 2) + F(k - 1) every time M is multiplied
- by itself (it is perhaps easier to understand this by computing M**2, 3, etc, and
- watching the result march up the sequence of Fibonacci numbers).
M = Matrix[[0, 1], [1,1]]
- Matrix exponentiation algorithm to compute Fibonacci numbers.
- Let M be Matrix [[0, 1], [1, 1]]. Then, the lower right element of M**k is
- F(k + 1). In other words, the lower right element of M is F(2) which is 1, and the
- lower right element of M**2 is F(3) which is 2, and the lower right element
- of M**3 is F(4) which is 3, etc.
- This is a good way to compute F(n) because the Ruby implementation of Matrix.**(n)
- uses O(log n) rather than O(n) matrix multiplications. It works by squaring squares
- ((m**2)**2)... as far as possible
- and then multiplying that by by M**(the remaining number of times). E.g., to compute
- M**19, compute partial = ((M**2)**2) = M**16, and then compute partial*(M**3) = M**19.
- That's only 5 matrix multiplications of M to compute M*19.
def self.fibMatrix(n)
return 0 if n <= 0 # F(0) return 1 if n == 1 # F(1) # To get F(n >= 2), compute M**(n - 1) and extract the lower right element. return CS::lower_right(M**(n - 1))
end
- Matrix utility to return
- the lower, right-hand element of a given matrix.
def self.lower_right matrix
return nil if matrix.row_size == 0 return matrix[matrix.row_size - 1, matrix.column_size - 1]
end</lang>
Generative
<lang ruby>require 'generator'
def fibGen
Generator.new do |g| f0, f1 = 0, 1 loop do g.yield f0 f0, f1 = f1, f0+f1 end end
end</lang>
Usage:
irb(main):012:0> fg = fibGen => #<Generator:0xb7d3ead4 @cont_next=nil, @queue=[0], @cont_endp=nil, @index=0, @block=#<Proc:0xb7d41680@(irb):4>, @cont_yield=#<Continuation:0xb7d3e8a4>> irb(main):013:0> 9.times { puts fg.next } 0 1 1 2 3 5 8 13 21 => 9
"Fibers are primitives for implementing light weight cooperative concurrency in Ruby. Basically they are a means of creating code blocks that can be paused and resumed, much like threads. The main difference is that they are never preempted and that the scheduling must be done by the programmer and not the VM." [2]
<lang ruby>fib = Fiber.new do
a,b = 0,1 loop do Fiber.yield a a,b = b,a+b end
end 9.times {puts fib.resume}</lang>
using a lambda <lang ruby>def fib_gen
a, b = 1, 1 lambda {ret, a, b = a, b, a+b; ret}
end</lang>
irb(main):034:0> fg = fib_gen => #<Proc:0xb7cdf750@(irb):22> irb(main):035:0> 9.times { puts fg.call} 1 1 2 3 5 8 13 21 34 => 9
Binet's Formula
<lang ruby>def fib
phi = (1 + Math.sqrt(5)) / 2 ((phi**self - (-1 / phi)**self) / Math.sqrt(5)).to_i
end</lang>
1.9.3p125 :001 > def fib 1.9.3p125 :002?> phi = (1 + Math.sqrt(5)) / 2 1.9.3p125 :003?> ((phi**self - (-1 / phi)**self) / Math.sqrt(5)).to_i 1.9.3p125 :004?> end => nil 1.9.3p125 :005 > (0..10).map(&:fib) => [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
Run BASIC
<lang runbasic>for i = 0 to 10
print i;" ";fibR(i);" ";fibI(i)
next i end
function fibR(n) if n < 2 then fibR = n else fibR = fibR(n-1) + fibR(n-2) end function function fibI(n) b = 1 for i = 1 to n t = a + b a = b b = t next i
fibI = a end function</lang>
Rust
Iterative
<lang cpp> fn fib(n: int, f: fn (num: i64) -> bool) -> (i64, int) {
if n < 0 { // Let these variables be mutated, otherwise too slow let mut n1:i64 = 0, n2:i64 = -1, i:int = 0, tmp:i64; while i > n { f(n1); tmp = n1-n2; if (tmp > 0 && n2 > 0) { //Detect overflow io::println("\nReached the limit of i64, halting"); return (n1, i); } n1 = n2; n2 = tmp; i -= 1; } (n1+n2, n) } else if n > 0 { // And these variables let mut n1:i64 = 0, n2:i64 = 1, i:int = 0, tmp:i64; while i < n { f(n1); tmp = n1+n2; if (tmp < 0) { //Detect overflow io::println("\nReached the limit of i64, halting"); return (n1, i); } n1 = n2; n2 = tmp; i += 1; } (n2-n1, n) } else { f(0); (0,1) }
}
fn main() {
let args = os::args(); let n = if args.len() == 1 { 10 } else if args.len() > 1 { // Convert from a string match (int::from_str(args[1])) { Some(num) => num, None => 10 //Fall back to default } } else { /* Required to use the if as an expression. * We know that args.len() is always >= 1, the compiler * does not. fail lets it know that we can't get past here. */ fail ~"No arguments given, somehow..."; };
/* Use the loop protocol to be able to do things * with the sequence given, in this case, print them out. * The loop itself returns a tuple with where it got to and * what the number is. */ let (result, n) = for fib(n) |num| { //print out the sequence io::print(fmt!("%? ", num)); };
io::println(fmt!("\nThe %dth fibonacci number is: %?", n, result));
} </lang>
Recursive
Minimalist tail-recursive version, no overflow checking:
// v0.8 <lang rust>fn main() {
fn fib(n: int) -> int {
fn _fib(n: int, a: int, b: int) -> int { match (n, a, b) {
(0, _, _) => a, _ => _fib(n-1, a+b, a)
} }
_fib(n, 0, 1)
}
for n in range(0,20) { println(fmt!("%d", fib(n))) }
}</lang>
SAS
<lang sas>/* building a table with fibonacci sequence */ data fib; a=0; b=1; do n=0 to 20;
f=a; output; a=b; b=f+a;
end; keep n f; run; </lang>
Sather
The implementations use the arbitrary precision class INTI. <lang sather>class MAIN is
-- RECURSIVE -- fibo(n :INTI):INTI pre n >= 0 is if n < 2.inti then return n; end; return fibo(n - 2.inti) + fibo(n - 1.inti); end;
-- ITERATIVE -- fibo_iter(n :INTI):INTI pre n >= 0 is n3w :INTI;
if n < 2.inti then return n; end; last ::= 0.inti; this ::= 1.inti; loop (n - 1.inti).times!; n3w := last + this; last := this; this := n3w; end; return this; end;
main is loop i ::= 0.upto!(16); #OUT + fibo(i.inti) + " "; #OUT + fibo_iter(i.inti) + "\n"; end; end;
end;</lang>
Scala
Recursive
<lang scala>def fib(i:Int):Int = i match{
case 0 => 0 case 1 => 1 case _ => fib(i-1) + fib(i-2)
}</lang>
Lazy sequence
<lang scala>lazy val fib: Stream[Int] = 0 #:: 1 #:: fib.zip(fib.tail).map{case (a,b) => a + b}</lang>
Tail recursive
<lang scala>def fib(i:Int, a:Int=1, b:Int=0):Int = i match{
case 1 => b case _ => fib(i-1, b, a+b)
}</lang>
foldLeft
<lang scala>// Fibonacci using BigInt with Stream.foldLeft optimized for GC (Scala v2.9 and above) // Does not run out of memory for very large Fibonacci numbers def fib(n:Int) = {
def series(i:BigInt,j:BigInt):Stream[BigInt] = i #:: series(j, i+j)
series(1,0).take(n).foldLeft(BigInt("0"))(_+_)
}
// Small test (0 to 13) foreach {n => print(fib(n).toString + " ")}
// result: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 </lang>
Scheme
Iterative
<lang scheme>(define (fib-iter n)
(do ((num 2 (+ num 1)) (fib-prev 1 fib) (fib 1 (+ fib fib-prev))) ((>= num n) fib)))</lang>
Recursive
<lang scheme>(define (fib-rec n)
(if (< n 2) n (+ (fib-rec (- n 1)) (fib-rec (- n 2)))))</lang>
This version is tail recursive: <lang scheme>(define (fib n)
(let loop ((a 0) (b 1) (n n)) (if (= n 0) a (loop b (+ a b) (- n 1)))))
</lang>
Dijkstra Algorithm
<lang scheme>;;; Fibonacci numbers using Edsger Dijkstra's algorithm
(define (fib n)
(define (fib-aux a b p q count) (cond ((= count 0) b) ((even? count) (fib-aux a b (+ (* p p) (* q q)) (+ (* q q) (* 2 p q)) (/ count 2))) (else (fib-aux (+ (* b q) (* a q) (* a p)) (+ (* b p) (* a q)) p q (- count 1))))) (fib-aux 1 0 0 1 n))</lang>
sed
<lang sed>#!/bin/sed -f
- First we need to convert each number into the right number of ticks
- Start by marking digits
s/[0-9]/<&/g
- We have to do the digits manually.
s/0//g; s/1/|/g; s/2/||/g; s/3/|||/g; s/4/||||/g; s/5/|||||/g s/6/||||||/g; s/7/|||||||/g; s/8/||||||||/g; s/9/|||||||||/g
- Multiply by ten for each digit from the front.
- tens
s/|</<||||||||||/g t tens
- Done with digit markers
s/<//g
- Now the actual work.
- split
- Convert each stretch of n >= 2 ticks into two of n-1, with a mark between
s/|\(|\+\)/\1-\1/g
- Convert the previous mark and the first tick after it to a different mark
- giving us n-1+n-2 marks.
s/-|/+/g
- Jump back unless we're done.
t split
- Get rid of the pluses, we're done with them.
s/+//g
- Convert back to digits
- back
s/||||||||||/</g s/<\([0-9]*\)$/<0\1/g s/|||||||||/9/g; s/|||||||||/9/g; s/||||||||/8/g; s/|||||||/7/g; s/||||||/6/g; s/|||||/5/g; s/||||/4/g; s/|||/3/g; s/||/2/g; s/|/1/g; s/</|/g t back s/^$/0/</lang>
Seed7
Recursive
<lang seed7>const func integer: fib (in integer: number) is func
result var integer: result is 1; begin if number > 2 then result := fib(pred(number)) + fib(number - 2); elsif number = 0 then result := 0; end if; end func;</lang>
Original source: [3]
Iterative
This funtion uses a bigInteger result:
<lang seed7>const func bigInteger: fib (in integer: number) is func
result var bigInteger: result is 1_; local var integer: i is 0; var bigInteger: a is 0_; var bigInteger: c is 0_; begin for i range 1 to pred(number) do c := a; a := result; result +:= c; end for; end func;</lang>
Original source: [4]
Shen
<lang Shen>(define fib
0 -> 0 1 -> 1 N -> (+ (fib (+ N 1)) (fib (+ N 2))) where (< N 0) N -> (+ (fib (- N 1)) (fib (- N 2))))</lang>
Sidef
Iterative
<lang ruby>func fib_iter(n) {
var fib = [1, 1]; { fib = [fib[-1], fib[-2] + fib[-1]] } * (n - fib.len); return fib[-1];
}</lang>
Recursive
<lang ruby>func fib_rec(n) {
n < 2 ? n : (__FUNC__(n-1) + __FUNC__(n-2));
}</lang>
Recursive with memoization
<lang ruby>func fib_mem (n) {
static c = []; n < 2 && return n; c[n] := (__FUNC__(n-1) + __FUNC__(n-2));
}</lang>
Closed-form solution
<lang ruby>func fib_closed(n) {
define S (1.25.sqrt + 0.5); define T (-S + 1); (S**n - T**n) / (-T + S) -> roundf(0);
}</lang>
Slate
<lang slate>n@(Integer traits) fib [
n <= 0 ifTrue: [^ 0]. n = 1 ifTrue: [^ 1]. (n - 1) fib + (n - 2) fib
].
slate[15]> 10 fib = 55. True</lang>
Smalltalk
<lang smalltalk>|fibo| fibo := [ :i |
|ac t| ac := Array new: 2. ac at: 1 put: 0 ; at: 2 put: 1. ( i < 2 ) ifTrue: [ ac at: (i+1) ] ifFalse: [ 2 to: i do: [ :l | t := (ac at: 2). ac at: 2 put: ( (ac at: 1) + (ac at: 2) ). ac at: 1 put: t ]. ac at: 2. ]
].
0 to: 10 do: [ :i |
(fibo value: i) displayNl
]</lang>
SNOBOL4
Recursive
<lang snobol> define("fib(a)") :(fib_end) fib fib = lt(a,2) a :s(return) fib = fib(a - 1) + fib(a - 2) :(return) fib_end
while a = trim(input) :f(end) output = a " " fib(a) :(while) end</lang>
Tail-recursive
<lang SNOBOL4> define('trfib(n,a,b)') :(trfib_end) trfib trfib = eq(n,0) a :s(return)
trfib = trfib(n - 1, a + b, a) :(return)
trfib_end</lang>
Iterative
<lang SNOBOL4> define('ifib(n)f1,f2') :(ifib_end) ifib ifib = le(n,2) 1 :s(return)
f1 = 1; f2 = 1
if1 ifib = gt(n,2) f1 + f2 :f(return)
f1 = f2; f2 = ifib; n = n - 1 :(if1)
ifib_end</lang>
Analytic
Note: Snobol4+ lacks built-in sqrt( ) function.
<lang SNOBOL4> define('afib(n)s5') :(afib_end) afib s5 = sqrt(5)
afib = (((1 + s5) / 2) ^ n - ((1 - s5) / 2) ^ n) / s5 afib = convert(afib,'integer') :(return)
afib_end</lang>
Test and display all, Fib 1 .. 10
<lang SNOBOL4>loop i = lt(i,10) i + 1 :f(show)
s1 = s1 fib(i) ' ' ; s2 = s2 trfib(i,0,1) ' ' s3 = s3 ifib(i) ' '; s4 = s4 afib(i) ' ' :(loop)
show output = s1; output = s2; output = s3; output = s4 end</lang>
Output:
1 1 2 3 5 8 13 21 34 55 1 1 2 3 5 8 13 21 34 55 1 1 2 3 5 8 13 21 34 55 1 1 2 3 5 8 13 21 34 55
SNUSP
This is modular SNUSP (which introduces @ and # for threading).
Iterative
<lang snusp> @!\+++++++++# /<<+>+>-\ fib\==>>+<<?!/>!\ ?/\
#<</?\!>/@>\?-<<</@>/@>/>+<-\ \-/ \ !\ !\ !\ ?/#</lang>
Recursive
<lang snusp> /========\ />>+<<-\ />+<-\ fib==!/?!\-?!\->+>+<<?/>>-@\=====?/<@\===?/<#
| #+==/ fib(n-2)|+fib(n-1)| \=====recursion======/!========/</lang>
Softbridge BASIC
Iterative
<lang basic> Function Fibonacci(n)
x = 0 y = 1 i = 0 n = ABS(n) If n < 2 Then Fibonacci = n Else Do Until (i = n) sum = x+y x=y y=sum i=i+1 Loop Fibonacci = x End If
End Function </lang>
Standard ML
Recursion
This version is tail recursive. <lang sml>fun fib n =
let
fun fib' (0,a,b) = a | fib' (n,a,b) = fib' (n-1,a+b,a)
in
fib' (n,0,1)
end</lang>
StreamIt
<lang streamit>void->int feedbackloop Fib {
join roundrobin(0,1); body in->int filter { work pop 1 push 1 peek 2 { push(peek(0) + peek(1)); pop(); } }; loop Identity<int>; split duplicate; enqueue(0); enqueue(1);
}</lang>
Tcl
Simple Version
These simple versions do not handle negative numbers -- they will return N for N < 2
Iterative
<lang tcl>proc fibiter n {
if {$n < 2} {return $n} set prev 1 set fib 1 for {set i 2} {$i < $n} {incr i} { lassign [list $fib [incr fib $prev]] prev fib } return $fib
}</lang>
Recursive
<lang tcl>proc fib {n} {
if {$n < 2} then {expr {$n}} else {expr {[fib [expr {$n-1}]]+[fib [expr {$n-2}]]} }
}</lang>
The following
: defining a procedure in the ::tcl::mathfunc
namespace allows that proc to be used as a function in expr
expressions.
<lang tcl>proc tcl::mathfunc::fib {n} {
if { $n < 2 } { return $n } else { return [expr {fib($n-1) + fib($n-2)}] }
}
- or, more tersely
proc tcl::mathfunc::fib {n} {expr {$n<2 ? $n : fib($n-1) + fib($n-2)}}</lang>
E.g.:
<lang tcl>expr {fib(7)} ;# ==> 13
namespace path tcl::mathfunc #; or, interp alias {} fib {} tcl::mathfunc::fib fib 7 ;# ==> 13</lang>
Tail-Recursive
In Tcl 8.6 a tailcall function is available to permit writing tail-recursive functions in Tcl. This makes deeply recursive functions practical. The availability of large integers also means no truncation of larger numbers. <lang tcl>proc fib-tailrec {n} {
proc fib:inner {a b n} { if {$n < 1} { return $a } elseif {$n == 1} { return $b } else { tailcall fib:inner $b [expr {$a + $b}] [expr {$n - 1}] } } return [fib:inner 0 1 $n]
}</lang>
% fib-tailrec 100 354224848179261915075
Handling Negative Numbers
Iterative
<lang tcl>proc fibiter n {
if {$n < 0} { set n [expr {abs($n)}] set sign [expr {-1**($n+1)}] } else { set sign 1 } if {$n < 2} {return $n} set prev 1 set fib 1 for {set i 2} {$i < $n} {incr i} { lassign [list $fib [incr fib $prev]] prev fib } return [expr {$sign * $fib}]
} fibiter -5 ;# ==> 5 fibiter -6 ;# ==> -8</lang>
Recursive
<lang tcl>proc tcl::mathfunc::fib {n} {expr {$n<-1 ? -1**($n+1) * fib(abs($n)) : $n<2 ? $n : fib($n-1) + fib($n-2)}} expr {fib(-5)} ;# ==> 5 expr {fib(-6)} ;# ==> -8</lang>
For the Mathematically Inclined
This works up to , after which the limited precision of IEEE double precision floating point arithmetic starts to show.
<lang tcl>proc fib n {expr {round((.5 + .5*sqrt(5)) ** $n / sqrt(5))}}</lang>
TI-83 BASIC
Unoptimized fibonacci program
<lang ti83b> :Disp "0" //Dirty, I know, however this does not interfere with the code
:Disp "1" :Disp "1" :1→A :1→B :0→C :Goto 1 :Lbl 1 :A+B→C :Disp C :B→A :C→B :Goto 1 </lang>
Optimized fibonacci program, compute fibonacci for N <lang ti83b>:Prompt N
- 0→A
- 1→B
- For(I,1,N)
:A→C :B→A :C+B→B
- End
- A </lang>
Binet's formula <lang ti83b>:Prompt N
- .5(1+√(5))→P
- (P^N–(-1/P)^N)/√(5) </lang>
TI-89 BASIC
Recursive
Optimized implementation (too slow to be usable for n higher than about 12).
<lang ti89b>fib(n) when(n<2, n, fib(n-1) + fib(n-2))</lang>
Iterative
Unoptimized implementation (I think the for loop can be eliminated, but I'm not sure).
<lang ti89b>fib(n) Func Local a,b,c,i 0→a 1→b For i,1,n
a→c b→a c+b→b
EndFor a EndFunc</lang>
TSE SAL
<lang TSE SAL>
// library: math: get: series: fibonacci <description></description> <version control></version control> <version>1.0.0.0.3</version> <version control></version control> (filenamemacro=getmasfi.s) [<Program>] [<Research>] [kn, ri, su, 20-01-2013 22:04:02] INTEGER PROC FNMathGetSeriesFibonacciI( INTEGER nI )
// // Method: // // 1. Take the sum of the last 2 terms // // 2. Let the sum be the last term // and goto step 1 // INTEGER I = 0 INTEGER minI = 1 INTEGER maxI = nI INTEGER term1I = 0 INTEGER term2I = 1 INTEGER term3I = 0 // FOR I = minI TO maxI // // make value 3 equal to sum of two previous values 1 and 2 // term3I = term1I + term2I // // make value 1 equal to next value 2 // term1I = term2I // // make value 2 equal to next value 3 // term2I = term3I // ENDFOR // RETURN( term3I ) //
END
PROC Main()
STRING s1[255] = "3" REPEAT IF ( NOT ( Ask( " = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF Warn( FNMathGetSeriesFibonacciI( Val( s1 ) ) ) // gives e.g. 3 UNTIL FALSE
END
</lang>
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT ASK "What fibionacci number do you want?": searchfib="" IF (searchfib!='digits') STOP Loop n=0,{searchfib}
IF (n==0) THEN fib=fiba=n ELSEIF (n==1) THEN fib=fibb=n ELSE fib=fiba+fibb, fiba=fibb, fibb=fib ENDIF IF (n!=searchfib) CYCLE PRINT "fibionacci number ",n,"=",fib
ENDLOOP </lang> Output:
What fibionacci number do you want? >12 fibionacci number 12=144
Output:
What fibionacci number do you want? >31 fibionacci number 31=1346269
Output:
What fibionacci number do you want? >46 fibionacci 46=1836311903
UnixPipes
<lang bash>echo 1 |tee last fib ; tail -f fib | while read x do
cat last | tee -a fib | xargs -n 1 expr $x + |tee last
done</lang>
UNIX Shell
<lang bash>#!/bin/bash
a=0 b=1 max=$1
for (( n=1; "$n" <= "$max"; $((n++)) )) do
a=$(($a + $b)) echo "F($n): $a" b=$(($a - $b))
done</lang>
Recursive:
<lang bash>fib() {
local n=$1 [ $n -lt 2 ] && echo -n $n || echo -n $(( $( fib $(( n - 1 )) ) + $( fib $(( n - 2 )) ) ))
}</lang>
Ursala
All three methods are shown here, and all have unlimited precision. <lang Ursala>#import std
- import nat
iterative_fib = ~&/(0,1); ~&r->ll ^|\predecessor ^/~&r sum
recursive_fib = {0,1}^?<a/~&a sum^|W/~& predecessor^~/~& predecessor
analytical_fib =
%np+ -+
mp..round; ..mp2str; sep`+; ^CNC/~&hh take^\~&htt %np@t, (mp..div^|\~& mp..sub+ ~~ @rlX mp..pow_ui)^lrlPGrrPX/~& -+ ^\~& ^(~&,mp..sub/1.E0)+ mp..div\2.E0+ mp..add/1.E0, mp..sqrt+ ..grow/5.E0+-+-</lang>
The analytical method uses arbitrary precision floating point arithmetic from the mpfr library and then converts the result to a natural number. Sufficient precision for an exact result is always chosen based on the argument. This test program computes the first twenty Fibonacci numbers by all three methods. <lang Ursala>#cast %nLL
examples = <.iterative_fib,recursive_fib,analytical_fib>* iota20</lang> output:
< <0,0,0>, <1,1,1>, <1,1,1>, <2,2,2>, <3,3,3>, <5,5,5>, <8,8,8>, <13,13,13>, <21,21,21>, <34,34,34>, <55,55,55>, <89,89,89>, <144,144,144>, <233,233,233>, <377,377,377>, <610,610,610>, <987,987,987>, <1597,1597,1597>, <2584,2584,2584>, <4181,4181,4181>>
V
Generate n'th fib by using binary recursion
<lang v>[fib
[small?] [] [pred dup pred] [+] binrec].</lang>
Vala
Recursive
Using int, but could easily replace with double, long, ulong, etc. <lang vala> int fibRec(int n){ if (n < 2) return n; else return fibRec(n - 1) + fibRec(n - 2); } </lang>
Iterative
Using int, but could easily replace with double, long, ulong, etc. <lang vala> int fibIter(int n){ if (n < 2) return n;
int last = 0; int cur = 1; int next;
for (int i = 1; i < n; ++i){ next = last + cur; last = cur; cur = next; }
return cur; } </lang>
VBA
Like Visual Basic .NET, but with keyword "Public" and type Long instead of Decimal:
<lang VBA> Public Function Fib(n As Integer) As Long
Dim fib0, fib1, sum As Long Dim i As Integer fib0 = 0 fib1 = 1 For i = 1 To n sum = fib0 + fib1 fib0 = fib1 fib1 = sum Next Fib = fib0
End Function </lang>
The (slow) recursive version:
<lang VBA> Public Function RFib(Term As Integer) As Long
If Term < 2 Then RFib = Term Else RFib = RFib(Term - 1) + RFib(Term - 2)
End Function </lang>
VBScript
Non-recursive, object oriented, generator
Defines a generator class, with a default Get property. Uses Currency for larger-than-Long values. Tests for overflow and switches to Double. Overflow information also available from class.
Class Definition:
<lang vb>class generator dim t1 dim t2 dim tn dim cur_overflow
Private Sub Class_Initialize cur_overflow = false t1 = ccur(0) t2 = ccur(1) tn = ccur(t1 + t2) end sub
public default property get generated on error resume next
generated = ccur(tn) if err.number <> 0 then generated = cdbl(tn) cur_overflow = true end if t1 = ccur(t2) if err.number <> 0 then t1 = cdbl(t2) cur_overflow = true end if t2 = ccur(tn) if err.number <> 0 then t2 = cdbl(tn) cur_overflow = true end if tn = ccur(t1+ t2) if err.number <> 0 then tn = cdbl(t1) + cdbl(t2) cur_overflow = true end if on error goto 0 end property
public property get overflow overflow = cur_overflow end property
end class</lang>
Invocation:
<lang vb>dim fib set fib = new generator dim i for i = 1 to 100 wscript.stdout.write " " & fib if fib.overflow then wscript.echo exit for end if next</lang>
Output:
<lang vbscript> 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 102334155 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025 20365011074 32951280099 53316291173 86267571272 139583862445 225851433717 365435296162 591286729879 956722026041 1548008755920 2504730781961 4052739537881 6557470319842 10610209857723 17167680177565 27777890035288 44945570212853 72723460248141 117669030460994 190392490709135 308061521170129 498454011879264 806515533049393</lang>
Vedit macro language
Iterative
Calculate fibonacci(#1). Negative values return 0. <lang vedit>:FIBONACCI:
- 11 = 0
- 12 = 1
Repeat(#1) {
#10 = #11 + #12 #11 = #12 #12 = #10
} Return(#11)</lang>
Visual Basic .NET
Platform: .NET
<lang vbnet>Function Fib(ByVal n As Integer) As Decimal
Dim fib0, fib1, sum As Decimal Dim i As Integer fib0 = 0 fib1 = 1 For i = 1 To n sum = fib0 + fib1 fib0 = fib1 fib1 = sum Next Fib = fib0
End Function</lang>
Recursive
<lang vbnet>Function Seq(ByVal Term As Integer)
If Term < 2 Then Return Term Return Seq(Term - 1) + Seq(Term - 2)
End Function</lang>
Wart
Recursive, all at once
<lang python>def (fib n)
if (n < 2) n (+ (fib n-1) (fib n-2))</lang>
Recursive, using cases
<lang python>def (fib n)
(+ (fib n-1) (fib n-2))
def (fib n) :case (n < 2)
n</lang>
Recursive, using memoization
<lang python>def (fib n saved)
# all args in wart are optional, and we expect callers to not provide saved default saved :to (table 0 0 1 1) # pre-populate base cases default saved.n :to (+ (fib n-1 saved) (fib n-2 saved)) saved.n</lang>
Whitespace
Iterative
This program generates Fibonacci numbers until it is forced to terminate. <lang Whitespace>
</lang>
It was generated from the following pseudo-Assembly. <lang asm>push 0 push 1
0:
swap dup onum push 10 ochr copy 1 add jump 0</lang>
- Output:
$ wspace fib.ws | head -n 6 0 1 1 2 3 5
Recursive
This program takes a number n on standard input and outputs the nth member of the Fibonacci sequence. <lang Whitespace>
</lang> <lang asm>; Read n. push 0 dup inum load
- Call fib(n), ouput the result and a newline, then exit.
call 0 onum push 10 ochr exit
0:
dup push 2 sub jn 1 ; Return if n < 2. dup push 1 sub call 0 ; Call fib(n - 1). swap ; Get n back into place. push 2 sub call 0 ; Call fib(n - 2). add ; Leave the sum on the stack.
1:
ret</lang>
- Output:
$ echo 10 | wspace fibrec.ws 55
Wrapl
Generator
<lang wrapl>DEF fib() (
VAR seq <- [0, 1]; EVERY SUSP seq:values; REP SUSP seq:put(seq:pop + seq[1])[-1];
);</lang> To get the 17th number: <lang wrapl>16 SKIP fib();</lang> To get the list of all 17 numbers: <lang wrapl>ALL 17 OF fib();</lang>
Iterator
Using type match signature to ensure integer argument: <lang wrapl>TO fib(n @ Integer.T) (
VAR seq <- [0, 1]; EVERY 3:to(n) DO seq:put(seq:pop + seq[1]); RET seq[-1];
);</lang>
x86 Assembly
<lang asm>TITLE i hate visual studio 4 (Fibs.asm)
- __ __/--------\
- >__ \ / | |\
- \ \___/ @ \ / \__________________
- \____ \ / \\\
- \____ Coded with love by
- |||
- \ Alexander Alvonellos |||
- | 9/29/2011 / ||
- | | MM
- | |--------------| |
- |< | |< |
- | | | |
- |mmmmmm| |mmmmm|
- Epic Win.
INCLUDE Irvine32.inc
.data BEERCOUNT = 48; Fibs dd 0, 1, BEERCOUNT DUP(0);
.code main PROC
- I am not responsible for this code.
- They made me write it, against my will.
;Here be dragons mov esi, offset Fibs; offset array; ;;were to start (start) mov ecx, BEERCOUNT; ;;count of items (how many) mov ebx, 4; ;;size (in number of bytes) call DumpMem;
mov ecx, BEERCOUNT; ;//http://www.wolframalpha.com/input/?i=F ib%5B47%5D+%3E+4294967295 mov esi, offset Fibs NextPlease:; mov eax, [esi]; ;//Get me the data from location at ESI add eax, [esi+4]; ;//add into the eax the data at esi + another double (next mem loc) mov [esi+8], eax; ;//Move that data into the memory location after the second number add esi, 4; ;//Update the pointer loop NextPlease; ;//Thank you sir, may I have another?
;Here be dragons
mov esi, offset Fibs; offset array; ;;were to start (start)
mov ecx, BEERCOUNT; ;;count of items (how many)
mov ebx, 4; ;;size (in number of bytes)
call DumpMem;
exit ; exit to operating system main ENDP
END main</lang>
XQuery
<lang xquery>declare function local:fib($n as xs:integer) as xs:integer {
if($n < 2) then $n else local:fib($n - 1) + local:fib($n - 2)
};</lang>
zkl
A slight tweak to the task; creates a function that continuously generates fib numbers <lang zkl>var fibShift=fcn(ab){ab.append(ab.sum()).pop(0)}.fp(L(0,1));</lang>
zkl: do(15){ fibShift().print(",") } 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377, zkl: do(5){ fibShift().print(",") } 610,987,1597,2584,4181,
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