Algorithm Implementation/Mathematics/Fibonacci Number Program
C
Recursive version
unsigned int fib(unsigned int n){ if (n < 2) return n; else return fib(n - 1) + fib(n - 2); }
Recursive version 2
unsigned int fib(unsigned int n){ return (n < 2) ? n : fib(n - 1) + fib(n - 2); }
Lucas form
float fib(unsigned int n){ float fi = (1 + sqrt(5))/2; return (pow(fi,(float)n) - pow(-fi,-(float)n))/sqrt(5); }
Iterative version
unsigned int fib(unsigned int n) { unsigned int i = 1, j = 0, k, t; for (k = 1; k <= n; k++) { t = i + j; i = j; j = t; } return j; }
Exponentiation by squaring
unsigned int fib(unsigned int n){ unsigned int i = n - 1, a = 1, b = 0, c = 0, d = 1, t; if (n <= 0) return 0; while (i > 0){ if (i % 2 == 1){ t = d*(b + a) + c*b; a = d*b + c*a; b = t; } t = d*(2*c + d); c = c*c + d*d; d = t; i = i / 2; } return a + b; }
Alternate exponentiation by squaring
unsigned int fib(unsigned int n){ unsigned int i = n - 1, a = 1, b = 0, c = 0, d = 1, t; if (n <= 0) return 0; while (i > 0){ while (i % 2 == 0){ t = d*(2*c + d); c = c*c + d*d; d = t; i = i / 2; } t = d*(b + a) + c*b; a = d*b + c*a; b = t; i--; } return a + b; }
C#
Iterative version
static int fib(int n) { int fib0 = 0, fib1 = 1; for (int i = 2; i <= n; i++) { int tmp = fib0; fib0 = fib1; fib1 = tmp + fib1; } return (n > 0 ? fib1 : 0); }
Binet's formula
static int fibBINET(int n) { double sqrt5 = Math.Sqrt(5.0); double phi = (1 + sqrt5 ) / 2; return (int)((Math.Pow(phi, n+1) - Math.Pow(1-phi, n+1)) / sqrt5); }
Using long numbers
static Num FibbonaciNumber(int n)
{
Num n1 = new Num(0);
Num n2 = new Num(1);
Num n3 = new Num(1);
for (int i = 2; i <= n; i++)
{
n3 = n2 + n1;
n1 = n2;
n2 = n3;
}
return n3;
}
struct Num
{
const int digit_base = 0x40000000; // 2^30
List<int> digits;
public int Length { get { return digits.Count; } }
public int this[int index] { get { return digits[index]; } private set { digits[index] = value; } }
public Num(int i)
{
digits = new List<int>();
while (i > 0)
{
digits.Add(i % digit_base);
i /= digit_base;
}
}
public static Num operator +(Num a, Num b)
{
Num n = new Num();
n.digits = new List<int>();
int l = Math.Min(a.Length,b.Length);
int remainder = 0;
for (int i = 0; i < l; i++)
{
n.digits.Add((a[i] + b[i] + remainder) % digit_base);
remainder = (a[i] + b[i] + remainder) / digit_base;
}
Num longer = a.Length > b.Length ? a : b;
for (; l < longer.Length; l++)
{
n.digits.Add((longer[l] + remainder) % digit_base);
remainder = (longer[l] + remainder) / digit_base;
}
if (remainder > 0) n.digits.Add(remainder);
return n;
}
public override string ToString()
{
StringBuilder sb = new StringBuilder();
for (int i = Length - 1; i >= 0; i--)
{
sb.AppendFormat("{0:D" + (digit_base.ToString().Length-1) + "}", this[i]);
}
return sb.ToString();
}
}
Erlang
fib(0) -> 0; fib(1) -> 1; fib(N) -> fib(N-1) + fib(N-2).
Arithmetic version
fib(N) -> S = math:sqrt(5), round(math:pow(((1 + S) / 2), N) / S).
algorithm taken from the Pascal "more efficient" version, below
F#
Simple recursive
let rec fib x = if x < 2 then x else fib(x - 1) + fib(x - 2)
This version overflows pretty quickly, to compute larger numbers Int64 or bigint can be used.
Haskell
List version
fib n = fibs 0 1 !! n where fibs a b = a : fibs b (a + b)
Tail-recursive version
fib n | n < 0 = undefined | otherwise = fib' n 0 1 where fib' 0 a _ = a fib' n a b = fib' (n - 1) b (a + b)
Simple recursive version
fib 0 = 0 fib 1 = 1 fib n = fib (n-1) + fib (n-2)
Awesome recursive version
fibs = 0 : 1 : [ a + b | (a, b) <- zip fibs (tail fibs)]
Closed-form version
Defines arithmetic operations on a custom data type, and then uses it to run the explicit formula without going via floating point - no rounding or truncation. Calculates the ten millionth fibonacci number in a few seconds (it has roughly two million digits).
module Fib where -- A type for representing a + b * sqrt n -- The n is encoded in the type. data PlusRoot n a = a :+/ a deriving (Eq, Read, Show) infix 6 :+/ -- Fetch the n in the root. class WithRoot n where getRoot :: Num b => PlusRoot n a -> b instance (WithRoot n, Num a) => Num (PlusRoot n a) where (a :+/ b) + (c :+/ d) = (a + c) :+/ (b + d) x@(a :+/ b) * (c :+/ d) = (a * c + getRoot x * b * d) :+/ (a * d + b * c) negate (a :+/ b) = negate a :+/ negate b fromInteger = (:+/ 0) . fromInteger -- I could implement these with (Ord a) but then we can't use the type -- with e.g. complex numbers. abs _ = error "PlusRoot.abs: unimplemented" signum _ = error "PlusRoot.signum: unimplemented" instance (WithRoot n, Fractional a) => Fractional (PlusRoot n a) where fromRational = (:+/ 0) . fromRational recip x@(a :+/ b) = (a / r) :+/ (negate b / r) where r = a*a - getRoot x * b*b -- Type parameter to PlusRoot. It would be easy to declare similar -- types for Two or whatever, and get all the above arithmetic for free. newtype Five = Five Five instance WithRoot Five where getRoot _ = 5 -- The formula is phi^n - xi^n / sqrt 5 -- but it's always an integer, i.e. phi^n - xi^n is always a multiple -- of sqrt 5, so the division isn't strictly necessary - just grab the -- relevant coefficient. fib :: Integer -> Integer fib n = case phi^n - xi^n of -- The 'round' here is to make the types match; as discussed previously -- n must be an integer so no actual rounding is done. _ :+/ n -> round n where phi :: PlusRoot Five Rational phi = (1 :+/ 1) / 2 xi = (1 :+/ negate 1) / 2
For other versions, see Haskell/Overview.
Dyalog APL
Basic Tail-Recursive Version
fibonacci?{
??0 1
?=0:???
(1??,+/?)? ?-1
}
Array-Oriented Version
fibonacci?{+/{?!??}(??)-?IO}
Other Versions
Io
Generic method version
fib := method(n, if(n < 4, (n + n % 2) / 2, (n % 2 * 2 - 1) * fib((n + n % 3) / 2 - 1) ** 2 + fib((n - n % 3) / 2 + 1) ** 3 ) )
Polymorphic method version
Number fibonacci := method((self - 1) fibonacci + (self -2) fibonacci) 1 fibonacci = 1 0 fibonacci = 0
Java
Recursive version
public void run(int n)
{
if (n <= 0)
{
return;
}
run(n,1,0);
}
private void run(int n, int eax, int ebx)
{
n--;
if (n == 0)
{
System.out.println(eax+ebx);
return;
}
run(n,ebx,eax+ebx);
}
2nd recursive version
public int fib(int n)
{
if (n>1) return fib(n-1) + fib(n-2);
return n;
}
Iterative version
/**
* Source based on
* http://20bits.com/2007/05/08/introduction-to-dynamic-programming/
* as at 9-May-2007
*/
private long fibonacci(int n)
{
long n2 = 0;
long n1 = 1;
long tmp;
for (int i=n ; i>2 ; i--) {
tmp = n2;
n2 = n1;
n1 = n1 + tmp;
}
return n2 + n1;
}
Memoized version
private int[] fibs; // array for memoized fibonacci numbers
public int fib(int n) {
if (n < 2) {
return n;
}
if (fibs == null) { // initialise array to first size asked for
fibs = new int[n + 1];
} else if (fibs.length < n) { // expand array
int[] newfibs = new int[n + 1]; // inefficient if looping through values of n
System.arraycopy(fibs, 0, newfibs, 0, fibs.length);
fibs = newfibs;
}
if (fibs[n] == 0) {
fibs[n] = fib(n - 1) + fib(n - 2);
}
return fibs[n];
}
Linotte
Fibonacci:
Principal :
Rôles :
n :: nombre
Actions :
"Entrez un nombre :" !
n ?
fibo(n) !
Fibo :
Rôles :
* n :: nombre
Actions :
si n < 2 alors retourne n
retourne fibo(n-1) + fibo(n-2)
Lexico (in spanish)
clase Fib
publicos:
mensajes:
Fib nop
Fibonacci(deme n es una cantidad) es_funcion cantidad
{
los objetos uno, dos, tres, i, respuesta son cantidades
copie 0 en uno
copie 1 en dos
variando i desde 1 hasta n haga:
{
copie uno en respuesta
copie uno + dos en tres
copie dos en uno
copie tres en dos
}
retornar uno
}
/**********************************/
tarea
{
el objeto f es un Fib
muestre "el 5: ", f.Fibonacci(doy 5)
}
Lua
function fib(n) local a, b = 0, 1 while n > 0 do a, b = b, a + b n = n - 1 end return a end
Recursive version
function fib(n) if n > 1 then n = fib(n - 1) + fib(n - 2) end return n end
Matlab
Recursive snippet
function F = fibonacci_recursive(n) if n < 2 F = n; else F = fibonacci_recursive(n-1) + fibonacci_recursive(n-2); end
Iterative snippet
function F = fibonacci_iterative(n) first = 0; second = 1; third = 0; for q = 1:n, third = first + second; first = second; second = third; end F = first;
Maxima
Recursive version
fib(n):=
if n < 2 then
n
else
fib(n - 1) + fib(n - 2)
$
Lucas form
fib(n):=(%phi^n-(-%phi)^-n)/sqrt(5);
Iterative version
fib(n) := block(
[i,j,k],
i : 1,
j : 0,
for k from 1 thru n do
[i,j] : [j,i + j],
return(j)
)$
Exponentiation by squaring
fib(n) := block(
[i,F,A],
if n <= 0 then
return(0),
i : n - 1,
F : matrix([1,0],[0,1]),
A : matrix([0,1],[1,1]),
while i > 0 do block(
if oddp(i) then
F : F.A,
A : A^^2,
i : quotient(i,2)
),
return(F[2,2])
)$
O'Caml
let fib n = let rec fibonacci n = match n with | 0 -> (0, 0) | 1 -> (0, 1) | m -> let (a, b) = fibonacci (m-1) in (b, a+b) in let (_, k) = fibonacci n in k;;
Pascal
function F(n: integer): integer; begin case n of 1,2: Result:=1 else Result:=F(n-1)+F(n-2) end; end;
A bit more efficient
function F(n: integer): integer; begin Result:=Round(Power((1+sqrt(5))/2, n)/sqrt(5)); end;
Note that Power is usually defined in Math, which is not included by default.
For most compilers it's possible to improve performance by using the Math.IntPower instead of the Math.Power.
Iterative version also for negative arguments
function fib(n:integer):extended; var i:integer; fib0,fib1:extended; begin fib0:=0; fib1:=1; for i:=1 to abs(n) do begin fib0:=fib0+fib1; fib1:=fib0-fib1; end; if (n<0)and(not odd(n)) then fib0:=-fib0; fib:=fib0; end:
Perl
sub fib { my ($n, $a, $b) = (shift, 0, 1); ($a, $b) = ($b, $a + $b) while $n-- > 0; $a; }
Recursive versions
sub fib { my $n = shift; return $n if $n < 2; return fib($n - 1) + fib($n - 2); } # returns F_n in a scalar context # returns all elements in the sequence up to F_n in a list context # only one recursive call sub fib { my ($n) = @_; return (0) if ($n == 0); return (0, 1) if ($n == 1); my @fib = fib($n - 1); return (@fib, $fib[-1] + $fib[-2]); }
Binary recursion, snippet
sub fibo; sub fibo {$_ [0] < 2 ? $_ [0] : fibo ($_ [0] - 1) + fibo ($_ [0] - 2)}
Runs in Θ(F(n)) time, which is Ω(1.6n).
Binary recursion with special Perl "caching", snippet
use Memoize; memoize 'fibo'; sub fibo; sub fibo {$_ [0] < 2 ? $_ [0] : fibo ($_ [0] - 1) + fibo ($_ [0] - 2)}
Iterative, snippet
sub fibo { my ($n, $a, $b) = (shift, 0, 1); ($a, $b) = ($b, $a + $b) while $n-- > 0; $a; }
PHP
function generate_fibonacci_sequence( $length ) { for( $l = array(0,1), $i = 2, $x = 0; $i < $length; $i++ ) $l[] = $l[$x++] + $l[$x]; return $l; }
Recursive version
function fib( $n ){ return ( $n < 2 ) ? $n : fib( $n-1 )+fib( $n-2 );}
OOP version
class fibonacci { public $Begin = 0; public $Next; public $Amount; public $i; public function __construct( $Begin, $Amount ) { $this->Begin = 0; $this->Next = 1; $this->Amount = $Amount; } public function _do() { for( $this->i = 0; $this->i < $this->Amount; $this->i++ ) { $Value = ( $this->Begin + $this->Next ); echo $this->Begin . ' + ' . $this->Next . ' = ' . $Value . '<br />'; $this->Begin = $this->Next; $this->Next = $Value; } } } $Fib = new fibonacci( 0, 6 ); echo $Fib->_do();
Alternate version
function fib($n) { return round(pow(1.6180339887498948482, $n) / 2.2360679774998); }
Python
Recursive version
def fib(n): if n < 2: return n else: return fib(n - 1) + fib(n - 2)
Recursive with memoization
m = {0: 0, 1: 1} def fib(n): #assert n >= 0 if n not in m: m[n] = fib(n-1) + fib(n-2) return m[n]
Lucas form
def fib(n): fi = (1 + sqrt(5))/2 return (fi**n - (-fi)**-n)/sqrt(5)
Iterative version
def fib(n): i,j = 1,0 for k in range(1,n + 1): i,j = j, i + j return j
Exponentiation by squaring
def fib(n): if n <= 0: return 0 i = n - 1 a,b = 1,0 c,d = 0,1 while i > 0: if i % 2 == 1: a,b = d*b + c*a, d*(b + a) + c*b c,d = c**2 + d**2, d*(2*c + d) i = i / 2 return a + b
Lucas sequence identities
def fib(n): if n <= 0: return 0 # n = 2**r*s where s is odd s, r = n, 0 while s & 1 == 0: r, s = r+1, s/2 # calculate the bit reversal t of (odd) s # e.g. 19 (10011) <=> 25 (11001) t = 0 while s > 0: if s & 1 == 1: t, s = t+1, s-1 else: t, s = t*2, s/2 # use the same bit reversal process # to calculate the sth Fibonacci number # using Lucas sequence identities u, v, q = 0, 2, 2 while t > 0: if t & 1 == 1: # u, v of x+1 u, v = (u + v) / 2, (5*u + v) / 2 q, t = -q, t-1 else: # u, v of 2*x u, v = u * v, v * v - q q, t = 2, t/2 # double s until we have # the 2**r*sth Fibonacci number while r > 0: u, v = u * v, v * v - q q, r = 2, r-1 return u
REBOL
Recursive version
fib: func [n [integer!]] [ either n < 2 [n] [(fib n - 1) + (fib n - 2)] ]
Ruby
class Integer def fib @n = self.abs if @n < 2 return @n else return (@n-1).fib + (@n-2).fib end end end
Alternate:
class Integer def fib @n = self.abs (@n<2)?(return @n):(return (@n-1).fib+(@n-2).fib) end end # you run it like this puts 10.fib # output: 55 puts 15.fib # output: 610
Generator
class FibGenerator def initialize(n) @n = n end def each a, b = 1, 1 @n.times do yield a a, b = b, a+b end end include Enumerable end def fibs(n) FibGenerator.new(n) end #use like this fibs(6).each do |x| puts x end
Arithmetic version
def f(n) ((((1+Math.sqrt(5))/2)**n)/Math.sqrt(5)+0.5).floor end
Memoized Version
fibmemo=Hash.new{|h,k| h[k-1]+h[k-2]} fibmemo[0]=0 fibmemo[1]=1 def fib n fibmemo[n] end
Scheme
Tree-recursive version
(define (fib n) (if (<= n 1) n (+ (fib (- n 1)) (fib (- n 2)))))
Iterative (tail-recursive) version
(define (fib n) (define (iter a b count) (if (<= count 0) a (iter b (+ a b) (- count 1)))) (iter 0 1 n))
Named-let, Iterative version
(define (fib n) (let loop ((a 0) (b 1) (count n)) (if (<= count 0) a (loop b (+ a b) (- count 1))))))
Lucas form
(define fib (let* ((sqrt5 (inexact->exact (sqrt 5))) (fi (/ (+ sqrt5 1) 2))) (lambda (n) (round (/ (- (expt fi n) (expt (- fi 1) n)) sqrt5)))))
Logarithmic-time Version
This version squares the Fibonacci transformation, allowing calculations in log2(n) time:
(define (fib-log n) "Fibonacci, in logarithmic time." (define (fib-iter a b p q count) (cond ((= count 0) b) ((even? count) (fib-iter a b (+ (* p p) (* q q)) (+ (* 2 p q) (* q q)) (/ count 2))) (else (fib-iter (+ (* b q) (* a q) (* a p)) (+ (* b p) (* a q)) p q (- count 1))))) (fib-iter 1 0 0 1 n))
UCBLogo
to fib :n output (cascade :n [?1+?2] 1 [?1] 0) end
Recursive version
to fib :n if :n<2 [output 1] output (fib :n-1)+(fib :n-2) end
VB.NET
Array oriented version
Dim i As Integer = 2 Dim sequencelength As Integer = 20 Dim fibonacci(sequencelength) As Integer fibonacci(0) = 0 fibonacci(1) = 1 While i <> sequencelength fibonacci(i) = fibonacci(i - 1) + fibonacci(i - 2) i += 1 End While
Recursive Version
Public Function fibonacci(ByVal i as integer) As Integer
If i < 2 Then
Return i
Else
Return fibonacci(i-1) + fibonacci(i-2)
End If
JavaScript
Recursive version
function fib(n) { return n < 2 ? n : fib(n - 1) + fib(n - 2); }
Alternative recursive version
function fib(n, prev, cur) { if (prev == null) prev = 0; if (cur == null) cur = 1; if (n < 2) return cur; return fib(--n, cur, cur + prev); }
Prev and cur is optional arguments.
Iterative version
function fibonacci(n) { var i = 1, j = 0, k, t; for (k = 1; k <= Math.abs(n); k++) { t = i + j; i = j; j = t; } if (n < 0 && n % 2 === 0) j = -j; return j; }
This example supports negative arguments.
Lucas form
function fibonacci(n) { var sqrt5 = Math.sqrt(5); var fi = (1 + sqrt5) / 2; return Math.round((Math.pow(fi, n) - Math.pow(-fi, -n)) / sqrt5); }
Binets formula
function fibonacci(n) { var sqrt5 = Math.sqrt(5); var fi = (1 + sqrt5) / 2; return Math.round((Math.pow(fi, n + 1) - Math.pow(1 - fi, n + 1)) / sqrt5); }
Algorithm from the Pascal "more efficient" version
function fibonacci(n) { var sqrt5 = Math.sqrt(5); return Math.round(Math.pow(((1 + sqrt5) / 2), n) / sqrt5); }
Common Lisp
Lucas form
(defun fib (n) (cond ((= n 0) 0) ((or (= n 1) (= n 2)) 1) ((= 0 (mod n 2)) (- (expt (fib (+ (truncate n 2) 1)) 2) (expt (fib (- (truncate n 2) 1)) 2))) (t (+ (expt (fib (truncate n 2)) 2) (expt (fib (+ (truncate n 2) 1)) 2))))) (fib (parse-integer (second *posix-argv*))) ;
Recursive version
(defun fib (x) (if (or (zerop x) (= x 1)) 1 (+ (fib (- x 1)) (fib (- x 2))))) (print (fib 10))
PostScript
Iterative
20 % how many Fibonacci numbers to print
1 dup
3 -1 roll
{
dup
3 -1 roll
dup
4 1 roll
add
3 -1 roll
=
}
repeat
Stack recursion
This example uses recursion on the stack.
% the procedure
/fib
{
dup dup 1 eq exch 0 eq or not
{
dup 1 sub fib
exch 2 sub fib
add
} if
} def
% prints the first twenty fib numbers
/ntimes 20 def
/i 0 def
ntimes {
i fib =
/i i 1 add def
} repeat
PL/SQL
Iterative snippet
CREATE OR REPLACE PROCEDURE fibonacci(lim NUMBER) AS fibupper NUMBER(38); fiblower NUMBER(38); fibnum NUMBER(38); i NUMBER(38); BEGIN fiblower := 0; fibupper := 1; fibnum := 1; FOR i IN 1 .. lim LOOP fibnum := fiblower + fibupper; fiblower := fibupper; fibupper := fibnum; DBMS_OUTPUT.PUT_LINE(fibnum); END LOOP; END;