Algorithm Implementation/Mathematics/Fibonacci Number Program

Fibonacci number Computational complexity of mathematical operations Fibonacci is similar to a "hello world" for many functional programming languages, since it can involve paradigms like pattern matching, memoization, and bog-standard tail recursion (which is equivalent to iteration). However, iteration or tail-recursion in linear time is only the first step: more clever exponentiation runs in logarithmic time. Although the Binet/Lucas formula is technically also exponentiation, its use of floating-point numbers makes it less attractive than the matrix-based solution. In addition, the above discussion of complexity and indeed most of the code here assumes that both addition and multiplication are done in a single step, which is not the case for big, exponentially-growing numbers easily created by fibonacci calculation. C An explanation of many of the following above can be found in nayuki's post. Recursive version unsigned int fib(unsigned int n) { // This is exactly the same as a ternary expression if (n < 2) return n; else return fib(n - 1) + fib(n - 2); } Tail recursive version // C++ default arguments used. If implemented in C, call with fib(n, 0, 1) instead. unsigned int fib(unsigned int n, unsigned int acc = 0, unsigned int prev = 1) { if (n < 1) return acc; else return fib(n - 1, prev + acc, acc); } Using Binet's formula #include <math.h> const static double phi = (1 + sqrt(5)) / 2; long fib(unsigned int n) { return (pow(phi, n) - pow(1 - phi, n)) / sqrt(5); } Iterative version unsigned int fib(unsigned int n) { unsigned int i = 0, j = 1, t; if (n == 0) return 0; for (k = 1; k < n; ++k) { t = i + j; i = j; j = t; } return j; } Alternate iterative version // This version is more "parallel" in the sense of a more unrolled loop. unsigned int fib(int n) { unsigned int i = 0, j = 1, k = 1; for (; n >= 3; n -= 3) { i = j + k; j = i + k; k = i + j; } return n == 0 ? i : n == 1 ? j : k; } Matrix exponentiation by squaring // A 2x2 matrix: m00, m01; m10, m11. typedef struct mat22_s { unsigned int m00, m01, m10, m11; } mat22_t; // Matrix multiplication. inline static mat22_t mat22_mul(mat22_t a, mat22_t b) { return (mat22_t){ a.m00 * b.m00 + a.m01 * b.m10, a.m00 * b.m01 + a.m01 * b.m11, a.m10 * b.m00 + a.m11 * b.m10, a.m10 * b.m01 + a.m11 * b.m11}; } // This is a less concise version written for clarity. The compiler should be able to optimize the boilerplate out. unsigned int fib(unsigned int n) { // Matrix holds F(N-1), F(N); F(N), F(N+1). // This is an identity matrix, also the 0th power of matrix. mat22_t result = {1, 0, 0, 1}; mat22_t matrix = {1, 1, 1, 0}; for (; n > 0; n /= 2) { if (n % 2 == 1) { result = mat22_mul(matrix, result); } matrix = mat22_mul(matrix, matrix); } return result.m10; } Matrix exponentiation, compact // A moderately compact version of the matrix calculation. unsigned int fib(unsigned int n) { // [a b] is "result"; [c d] is "matrix". unsigned int a = 1, b = 0, c = 0, d = 1, t; if (n == 0) return 0; n = n - 1; while (n > 0) { if (n % 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; n = n / 2; } return a + b; } A similar alternative is based on Lucas numbers. Matrix-derived fast doubling // Nayuki's fast doubling code. Runs from high to low bits. #include <limits.h> static inline unsigned int log2i(unsigned int n) { #if defined(__has_builtin) && __has_builtin(__builtin_clz) return sizeof (unsigned int) * CHAR_BIT - __builtin_clz(n) - 1; #else return sizeof (unsigned int) * CHAR_BIT - 1; // pessimistic guess #endif } unsigned int fib(unsigned int n) { // Two numbers for iteration. At the end, a = F(N) and b = F(N+1). unsigned int a = 0, b = 1; // log2i is not reliable for n = 0. We know better. unsigned int mask = n == 0 ? 0 : 1 << log2i(n); for (; mask > 0; mask /= 2) { // F(2k) = F(k) * (2 * F(k+1) + F(k)) unsigned int new_a = a * (b * 2 - a); // F(2k+1) = F(k+1)**2 + F(k)**2 unsigned int new_b = a * a + b * b; a = new_a; b = new_b; if (n & mask) { new_b = a + b; a = b; b = new_b; } } return a; } C# C# code is essentially the same as C with some static method specifiers. 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) - Math.Pow(-phi, -n)) / sqrt5); } Using long numbers static Num FibonacciNumber(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(); } } D Basic Recursive Version ulong fib(uint n){ return (n < 2) ? n : fib(n - 1) + fib(n - 2); } Iterative Version ulong fib(uint n) { ulong fib0 = 0; ulong fib1 = 1; for (auto i = 2; i <= n; i++) { auto tmp = fib0; fib0 = fib1; fib1 = tmp + fib1; } return (n > 0 ? fib1 : 0); } Memoized Recursive Version ulong fib(uint n){ static ulong[] memo; if (n < 0) return n; if (n < memo.length) return memo[n]; auto result = (n < 2) ? n : fib(n - 1) + fib(n - 2); memo.length = n + 1; memo[n] = result; return result; } 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 < 2I then x else fib(x - 1I) + fib(x - 2I) This version uses F# System.Numerics.BigInteger type Memoized recursive open System.Collections.Generic let rec fib n = let memo = Dictionary<_, _>() let rec fibInner = function | n when n = 0I -> 0I | n when n = 1I -> 1I | n -> fib(n - 1I) + fib(n - 2I) if memo.ContainsKey(n) then memo.[n] else let res = fibInner n memo.[n] <- res res Tail recursive let fib n = let rec fibInner (n, a, b) = if (n = 0I) then a else fibInner ((n - 1I), b, (a + b)) fibInner (n, 0I, 1I) Iterative let fib n = if n < 2I then n else let mutable fib1 = 0I let mutable fib2 = 1I let mutable i = 2I let mutable tmp = 0I while (i <= n) do i <- i + 1I tmp <- fib1 fib1 <- fib2 fib2 <- tmp + fib2 fib2 Infinite Sequence Generator let fibSeq = Seq.unfold (fun (a, b) -> Some(a, (b, a + b))) (0I, 1I) let fib n = fibSeq |> (Seq.skip n) |> Seq.head Forth : fib ( n -- fib ) 0 1 rot 0 ?do over + swap loop drop ; Go Recursive Solution func fib(n int) int { if n < 2 { return n; } return fib(n-1) + fib(n-2); } Iterative Solution func fib(n int) int { if n == 0 { return 0 } a, b := 0, 1 for i := 1; i < n; i++ { a, b = b, a+b } return b } 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)] Or : fibs = 0 : 1 : zipWith (+) fibs (tail fibs) Or : fibs = map fst $ iterate (\(a,b)->(b,a+b)) (0,1) 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 nayuki's fast doubling again, but recursive this time Dyalog APL Basic Tail-Recursive Version fibonacci?{ ??0 1 ?=0:??? (1??,+/?)? ?-1 } Array-Oriented Version fibonacci?{+/{?!??}(??)-?IO} Other Versions See Fibonacci at the Dynamic Functions Database 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); } Variations on the recursive version /* Recursive versions. Horribly inefficient. Use iterative/memoized versions instead */ public long fib(long n) { if (n < 2) { return 1; } return fib(n - 1) + fib(n - 2); } public long fib2(long n) { return (n < 2) ? n : getValue(n - 1) + getValue(n - 2); } 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; } Simpler Iterative Version /** * returns the Nth number in the Fibonacci sequence */ public int fibonacci(int N) { int lo = 0; int hi = 1; for (int i = 0; i < N; i++) { hi = lo + hi; lo = hi - lo; } return lo; } 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]; } Iterative Memoized version public int fib(int n) { if (n < 2) { return n; } int[] f = new int[n + 1]; f[0] = 0; f[1] = 1; for (int i = 2; i <= n; i++) { f[i] = f[i - 1] + f[i - 2]; } return f[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; $a; } PHP Iterative version function generate_fibonacci_sequence($length) { for($l = [0, 1], $i = 2, $x = 0; $i < $length; $i++) { $l[] = $l[$x++] + $l[$x]; } return $l; } Recursive version function fib($n) { if ($n < 2) { return $n; } return fib($n - 1) + fib($n - 2); } Ternary version function fib($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 return fib(n - 1) + fib(n - 2) Or : def fib( n ): return n if n < 2 else fib( n - 1 ) + fib( n - 2 ) Recursive with memoization m = {0: 1, 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): phi = (1 + sqrt(5))/2 return int((phi**n - (-phi)**-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 Iterative version using Generator def fib(n): a,b = 1,0 for i in range(n): yield b a, b = b, a+b 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 >> 1 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 Lucas sequence identities, recursion As with the iterative version, this solution is also O(log n) with arbitrary precision. def fib(n): def fib_inner(n): if n == 0: return 0, 2 m = n >> 1 # q = 2*(-1)**m q = -2 if (m & 1) == 1 else 2 u, v = fib_inner(m) u, v = u * v, v * v - q if n & 1 == 1: # u, v of 2m+1 u1 = (u + v) >> 1 return u1, 2*u + u1 else: # u, v of 2m return u, v if n <= 0: return 0 # the outermost loop is unrolled # to avoid calculating an unnecessary v m = n >> 1 u, v = fib_inner(m) if n & 1 == 1: # u of m+1 u1 = (u + v) >> 1 # u of 2m+1 return u*u + u1*u1 else: # u of 2m return u * v 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 Recursive def fib n return n if n < 2 fib(n - 1) + fib(n - 2) end 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]=1 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 = 50 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 Private Function fibonacci(ByVal i as integer) As Integer If i < 1 Then Return -1 ElseIf i < 2 Then Return i Else Return fibonacci(i-1) + fibonacci(i-2) End If End Function 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); } Binet's 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;