125
\$\begingroup\$

The Fibonacci sequence is a sequence of numbers, where every number in the sequence is the sum of the two numbers preceding it. The first two numbers in the sequence are both 1.

Here are the first few terms

1 1 2 3 5 8 13 21 34 55 89 ...

Write the shortest code that either:

  • Generates the Fibonacci sequence without end.

  • Given n calculates the nth term of the sequence. (Either 1 or zero indexed)

You may use standard forms of input and output.

(I gave both options in case one is easier to do in your chosen language than the other.)


For the function that takes an n, a reasonably large return value (the largest Fibonacci number that fits your computer's normal word size, at a minimum) has to be supported.


Leaderboard

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\$\endgroup\$
1
  • 1
    \$\begingroup\$ I am sort of waiting for a response like "f", 1 byte, in my math based golf language. \$\endgroup\$ – Zsolt Szilagy Aug 11 '20 at 11:57

260 Answers 260

1 2
3
4 5
9
3
\$\begingroup\$

Java, 90 characters and just two variables

There was one before with 55 characters, but it used a variable without declaring it and had no output. This one has both and (the actual logic) is shorter. And as a little bonus it looks absolutely horrific code-style-wise and depends on compiler quirks, yay!

interface A{static void main(String[]x){for(int a,b=a=1;;System.out.println(b=a+(a=b)));}}

The special features I used are:

  • Using an interface instead of a class. The program can still be run as normal, but I don't need to write "public" twice. This saves 10 characters
  • Declaring multiple variables at once: int a,b;
  • Initializing multiple variables at once and in the declaration, needs a second a: b=a=1;
  • Everything is done in the for head, the body is empty: for(...);
    The first and third block of for are intended for variable initialization and variable incrementation, but they can hold any commands.
  • The whole logic is inside the output: System.out.println(b=a+(a=b))
  • Just two variables without recursion! This is done by using the way the compiler works: The assignment to b first reads the value of a, then it evaluates the right side of the +, where it reads the value of b and writes it into a, but the left side of the + still has the old value of a that gets added to the value of b after assigning the value of b to a. Then that sum gets written to b while a already holds the old value of b.
    I was lucky that the compiler works this way, because it could also have first evaluated the expression in the brackets, like for example C does, then it just lists all powers of 2 instead of the Fibonacci numbers.

In a dream programming language this would just be: b=a+(a=b

\$\endgroup\$
1
  • \$\begingroup\$ Dammit, referring to this answer just helped me in actually useful code, but I can't upvote it! :D \$\endgroup\$ – Fabian Röling May 27 '19 at 21:56
3
\$\begingroup\$

Flurry, 54 bytes

<>{<>(){}(<>())}{{}[<>{<>(){}{}[<><<>()>]}{({})}]{{}}}

This is an anonymous function that takes a number \$n\$ as a Church numeral and pushes \$F_n\$ to the stack as a Church numeral.

It can be tested by invoking the interpreter as follows:

$ ./Flurry -inn -c '<>{<>(){}(<>())}{{}[<>{<>(){}{}[<><<>()>]}{({})}]{{}}}{}' 10
55

Note that this is very slow for large inputs (>30 or so) because Church numeral addition is \$O(n)\$.

Explanation

The basic idea is to loop \$n\$ times and keep track of the smaller number using the stack and the larger number using function parameters, so the function to iterate looks like this:

next = λb. pop() + push(b)

Flurry is mostly combinator-based, and so doesn't really have lambdas. The closest approximation is to write a function that pushes its argument onto the stack (using the {...} monad) and then pops from the stack to read the argument back (using the {} nilad). So the function λx. f x gets represented as:

{ f {} }

However, this translation doesn't work if f depends on the stack, which is true in this case, so we need to use a combinator-based approach instead. Rather than popping using {} directly, we can write an anonymous function that ignores its argument and returns a value popped from the stack:

  λa. pop()
{ <> () {} {} }

Here, <> and () represent the S and K combinators respectively.

Of course, the next thing we do with the popped value is apply it to the successor function <><<>()> to compute addition:

  λa. pop()(suc)
{ <> () {} {} [<><<>()>] }

Now all we need is to write a function that pushes its argument to the stack and returns it unchanged:

  λa. push(a)
{ ({}) }

We can then put these functions together using the S combinator:

  λb. pop() + push(b)
  λb. S (λb. pop()(suc)) (λb. push(b)) b
<> {<>(){}{}[<><<>()>]} {({})}

The main function is responsible for setting up the initial values and calling next the appropriate number of times. In other words, it looks like this:

main = λn. push(0); n (next) (1)

Since we don't care about the return value, we can just use function application to sequence the two actions:

main = λn. push(0) (n (next) (1))

Again, we can't keep lambda variables on the stack because we're already doing stack manipulation, so we'll have to create intermediate functions. First, one that ignores its argument and pushes 0 to the stack:

  λa. push(0)
{ <> () {} (0) }

Second, one that applies its argument to next and 1 (which we can finally use stack lambdas for):

  λa. a next 1
{ {} next 1 }

Again, we can put these together with S:

<> {<>(){}(0)} {{}[next]1}

Substituting the definitions of 0, next, and 1 gives us the final result:

<>{<>(){}(<>())}{{}[<>{<>(){}{}[<><<>()>]}{({})}]{{}}}
\$\endgroup\$
4
  • 1
    \$\begingroup\$ Is it some kind of functional brain-flak? :D \$\endgroup\$ – Bubbler Aug 10 '20 at 23:13
  • \$\begingroup\$ @Bubbler Pretty much. \$\endgroup\$ – Esolanging Fruit Aug 10 '20 at 23:46
  • \$\begingroup\$ Given the git description is "Brain-flak, but functional.", I'd say yes \$\endgroup\$ – Jo King Aug 11 '20 at 0:52
  • \$\begingroup\$ I just opened a chat room for Flurry. \$\endgroup\$ – Bubbler Aug 12 '20 at 23:49
3
\$\begingroup\$

convey, 8 bytes

Generates the sequence.

v+"}
1"1

Try it online!

enter image description here

The values (initially 1 and 1) follow the conveyor belts indicated by the arrow heads. " duplicates the input into both outputs, + adds them, and } writes them to the output.

\$\endgroup\$
2
\$\begingroup\$

Python, 36

f=lambda x:x>1and f(x-1)+f(x-2)or x
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2
\$\begingroup\$

Python

a,b,n=0,1,10
while n:a,b,n=b,a+b,n-1;print b
\$\endgroup\$
3
  • 1
    \$\begingroup\$ a=b=1<newline>while 1:print a;a,b=b,a+b 30 Characters \$\endgroup\$ – st0le Feb 1 '11 at 6:19
  • \$\begingroup\$ @st0le that's actually 31 characters. I've spent like 5 minutes recounting my solution, which is identical to yours, until I came to the conclusion you are wrong :) \$\endgroup\$ – Mikle Nov 24 '11 at 18:02
  • \$\begingroup\$ The fibonacci sequence starts with 0. So you can't do a=b=1. It should be something like a,b=0,1\nwhile 1:print a;a,b=b,a+b which is 33 characters. \$\endgroup\$ – Bakuriu Sep 1 '12 at 9:42
2
\$\begingroup\$

Python, 34 chars first variant, 31 chars for second variant,

a,b=1,1
while 1:print a;a,b=b,a+b

Second variant:

f=lambda x:x<2 or f(x-2)+f(x-1)
\$\endgroup\$
1
  • \$\begingroup\$ You can remove the space between 2 and or \$\endgroup\$ – Cyoce Oct 14 '16 at 0:52
2
\$\begingroup\$

Python O(1) Nth number, 91 char

48 characters for the import, a newline, 42 for the rest. I know it's longer than most here and that the question is a bit old, but I looked through the answers and I didn't see any that use the constant-time floating-point calculation.

from math import trunc as t,pow as p,sqrt as s
r=s(5);i=(1+r)/2;f=lambda n:t(p(i,n)/r+.5)

From there you call f(n) for the nth number in the sequence. Eventually it loses precision, and is only accurate up through f(70) (190,392,490,709,135). i is the constant Phi.

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2
  • 4
    \$\begingroup\$ actually it's O(log n) since pow has that complexity... \$\endgroup\$ – JBernardo Jul 10 '11 at 2:03
  • 3
    \$\begingroup\$ @JBernado and even bigger since pow for bigint is more complicated story. \$\endgroup\$ – shabunc Aug 19 '11 at 8:49
2
\$\begingroup\$

Perl, 51 (Loopless)

The following code uses Binet's formula to give the Nth Fibonacci number without using any loops.

print((($p=5**.5/2+.5)**($n=<>)-(-1/$p)**$n)/5**.5)
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1
  • \$\begingroup\$ The second term starts small and only becomers smaller later on, so it can always be replaced by integer rounding. Golfing that a bit gives 30 byes: say.5+(.5+5**.5/2)**<>/5**.5|0 \$\endgroup\$ – Ton Hospel Mar 6 '16 at 13:35
2
\$\begingroup\$

PHP - 109 97 88 49 characters

<?for($a=$b++;;$b+=$a=$b-$a){$s+=$b%2*$b;echo$a;}
\$\endgroup\$
2
  • \$\begingroup\$ I have confirmed this works without using the optional parameters, but what exactly are they for? \$\endgroup\$ – Kevin Brown Mar 17 '11 at 1:42
  • \$\begingroup\$ @Bass5098: So that the function works when called in a unary context, I presume. If PHP uses JS-style argument passing where you can supply fewer arguments than the function declares, and you can perform meaningful computations involving undefined (or the PHP equivalent thereof), then cool! \$\endgroup\$ – Chris Jester-Young Mar 17 '11 at 16:35
2
\$\begingroup\$

Perl - 39 chars

($a,$b)=($b,$a+$b||1),print"$b
"while$=
\$\endgroup\$
2
\$\begingroup\$

C#

Generated as a stream (65 chars):

IEnumerable<int>F(){for(int c=1,s=1;;){s+=c=s-c;yield return c;}}

Could be reduced to 61 characters using non-generic IEnumerable. Of course, if you include the required System.Collections.Generic, then it's a few more characters.

\$\endgroup\$
2
\$\begingroup\$

Mathematica,26 chars

If[#>1,#0[#-1]+#0[#-2],#]&
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3
  • \$\begingroup\$ Damn, and I just thought I had come up with a new shortest Fibonacci implementation in Mathematica. +1 :) \$\endgroup\$ – Martin Ender Jan 30 '15 at 21:08
  • \$\begingroup\$ What does the trailing & do? \$\endgroup\$ – Cyoce Oct 14 '16 at 0:55
  • \$\begingroup\$ @Cyoce making it a function, instead of an expression \$\endgroup\$ – Keyu Gan Feb 1 '18 at 0:40
2
\$\begingroup\$

F# - 42 chars

Seq.unfold(fun(a,b)->Some(a,(b,a+b)))(0,1)
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1
  • \$\begingroup\$ Nice, I didn't know about Seq.unfold! =) \$\endgroup\$ – Roujo Feb 8 '16 at 20:32
2
\$\begingroup\$

JAGL V1.0 - 13 / 11

1d{cdc+dcPd}u

Infinite Fibonacci sequence. Or, if not required to print:

11 bytes

1d{cdc+cd}u
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2
\$\begingroup\$

Octave, 26 chars

f=@(n)([1 0]*[1 1;1 0]^n)(2)

Basically, a copy of my solution from Calculating (3 + sqrt(5))^n exactly.

[a b] x [1 1 ;1 0] equals [a+b a]

, so

[f(1) f(0)] x [1 1 ;1 0]^n equals [f(n+1) f(n)]

It's a disaster to do unnecessary* loops in Octave/Matlab. It's neither elegant, nor fast, let alone golfy.


*All loops that can be vectorized are unnecessary :).

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2
  • 2
    \$\begingroup\$ I don't think you need the [1 0]. Picking the second item out of the matrix will give you the right number anyway. \$\endgroup\$ – Andrew Apr 16 '15 at 21:37
  • 2
    \$\begingroup\$ Thank you for notice, f=@(n)([1 1;1 0]^n)(3) is six characters shorter indeed (Octave enumerates items in matrix top-down and then left-right when indexing with a single number, so the value at first row, second index is at index 3). \$\endgroup\$ – pawel.boczarski Apr 16 '15 at 22:22
2
\$\begingroup\$

ArnoldC, 451 bytes

IT'S SHOWTIME
HEY CHRISTMAS TREE a
YOU SET US UP 1
HEY CHRISTMAS TREE b
YOU SET US UP 1
HEY CHRISTMAS TREE c
YOU SET US UP 1
STICK AROUND c
TALK TO THE HAND a
GET TO THE CHOPPER a
HERE IS MY INVITATION a
GET UP b
ENOUGH TALK
TALK TO THE HAND b
GET TO THE CHOPPER b
HERE IS MY INVITATION b
GET UP a
ENOUGH TALK
GET TO THE CHOPPER c
HERE IS MY INVITATION 1e300
LET OFF SOME STEAM BENNET a
ENOUGH TALK
CHILL
YOU HAVE BEEN TERMINATED

This is actually my first ArnoldC program. Horrible for golfing, but great for lolz!

Produces an stream of Fibonacci numbers up to 1.1253474885494065e+274.

Explanation

IT'S SHOWTIME               #start program

HEY CHRISTMAS TREE a        #declare a...
YOU SET US UP 1             #and set it to 1
HEY CHRISTMAS TREE b        #declare b...
YOU SET US UP 1             #and set it to 1
HEY CHRISTMAS TREE c        #declare c...
YOU SET US UP 1             #and set it to 1

STICK AROUND c              #while c is truthy
TALK TO THE HAND a          #output a
GET TO THE CHOPPER a        #assign a to...
HERE IS MY INVITATION a     #a...
GET UP b                    #plus b
ENOUGH TALK                 #end assignment
TALK TO THE HAND b          #output b
GET TO THE CHOPPER b        #assign b to...
HERE IS MY INVITATION b     #b...
GET UP a                    #plus a
ENOUGH TALK                 #end assignment
GET TO THE CHOPPER c        #assign c to...
HERE IS MY INVITATION 1e300 #whether 1e300...
LET OFF SOME STEAM BENNET a #is greater than a (returns 0 or 1)
ENOUGH TALK                 #end assignment
CHILL                       #end while

YOU HAVE BEEN TERMINATED    #end program
\$\endgroup\$
2
\$\begingroup\$

Ruby, 28 bytes

->f{loop{f<<p(f[-1]+f[-2])}}

Usage:

->f{loop{f<<p(f[-1]+f[-2])}}[[-1,1]]
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2
\$\begingroup\$

𝔼𝕊𝕄𝕚𝕟, 3 chars / 6 bytes (noncompetitive)

Мȫï

Try it here (Firefox only).

More builtins!

math.js + numbers.js = hella functions

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2
\$\begingroup\$

PARI/GP, 9 bytes

fibonacci

Alternate solution (21 bytes), for those disliking the built-in:

n->([1,1;1,0]^n)[1,2]

Alternate alternate solution (21 bytes):

n->imag(quadgen(5)^n)

I also posted all three solutions (in ungolfed form) to Rosetta Code's Fibonacci page.

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2
\$\begingroup\$

Reng v.2.1, 18 bytes

(Noncompeting, postdates question)

11{:nAo}#xxx:)+x5h

11 initializes the stack with 2 1s. {:nAo}#x sets the command x to mean "duplicate and output as number" (:n) then "output a newline" (Ao, A = 10). Then, xx prints the initial 2 1s. : duplicates the TOS and ) rotates the stack, so it becomes b a b. + adds the two figures, making it b (a+b). x prints and leaves this new result on the stack. 5h jumps back 5 spaces, and the loop continues.

Try it out here! Or check out the github!

\$\endgroup\$
0
2
\$\begingroup\$

Fuzzy Octo Guacamole, 11 bytes

01(!aZrZo;)

This takes the infinite route.

Explanation:

01 pushes 0 and then 1 to the stack.

( starts a infinite loop.

! sets the register, saving the value on the top of the stack and storing it. It doesn't pop though.

a adds the 2 values.

ZrZ reverses the stack, pushes the register contents, and reverses again. This pushes the stored number to the bottom of the stack.

o; peeks and prints.

) ends the infinite loop.

Then the whole things starts again from the (.


As a a side note, this is quite fast to hit the max long size possible in Python. The last number it prints is 12200160415121876738, and it repeats that forever.

\$\endgroup\$
2
\$\begingroup\$

Python 2, 43 bytes

def f(n):k=9**n;return k**-~-~n/~-(k*~-k)%k
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2
\$\begingroup\$

Python 33 characters

x,y=0,1
while 1:print x;x,y=y,x+y

This will be an infinite loop!


Python 31 characters

def f(a=[1,0]):a[:]=a[1],sum(a)

demonstration

for _ in range(10):
    f(); print f.func_defaults[0][0]

0
1
1
2
3
5
8
13
21
34
\$\endgroup\$
2
  • 2
    \$\begingroup\$ The loop for your "31 char" program would need to be included in the score. \$\endgroup\$ – mbomb007 Nov 4 '16 at 15:00
  • \$\begingroup\$ ^^ also, we usually give the count in bytes, not characters. \$\endgroup\$ – FlipTack Dec 28 '16 at 12:25
2
\$\begingroup\$

Cylon (Non-Competing), 12 bytes

The language is in development, Im just putting this up here.

1:øÌ[:ì+Á])r

An explanation:

1    ;pushes a 1 to the stack
:    ;duplicates the top of the stack
ø    ;reads a number from stdin, pushing it to the stack
Ì    ;non-pushing loop, doesn't push counter to the stack, but deletes it
[    ;start of function, to be pushed to the stack
  :  ;duplicate top of stack
  ì  ;rotate the stack, moving the copy to the back
  +  ;replaces top two objects on the stack with their sum
  Á  ;push the result to the shadowing stack (non-consuming)
]    ;end of function
)    ;switch to shadowed stack
r    ;standard library call, reverses a stack
     ;stack implicitly printed
\$\endgroup\$
2
\$\begingroup\$

bc, 21

for(b=1;b=a+(a=b);)a

The trailing newline is significant.

Outputs the entire sequence. bc has arbitrary precision arithmetic, so this continues forever.

\$\endgroup\$
2
\$\begingroup\$

OIL, 46 bytes

This program writes an infinite unstoppable stream of fibonacci numbers. It is mostly copied from the standard library but fit to the requirements and golfed.

14
add
17
17
14
swap
17
17
4
17




11
6
0
0
1
\$\endgroup\$
1
  • 2
    \$\begingroup\$ Welcome to the site! This is a cool answer, I've never heard of the language before. :) \$\endgroup\$ – James May 4 '17 at 17:29
2
\$\begingroup\$

Python 2, 30 bytes

f=lambda n:n<3or f(n-2)+f(n-1)

Try it online!

One catch: this outputs True instead of 1. This is allowed by this meta consensus.

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Gaia, 6 bytes

0₁@+₌ₓ

I might make a built-in for this in the future, but built-ins are boring anyway.

Explanation

0₁      Push 0 and 1
  @     Push an input
   +₌ₓ  Add the top two stack elements, without popping them, (input) times
        Implicitly print the top stack element.
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Husk, 7 2 bytes (non-competing)

İf

Try it online!

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Python 2, 49 40 chars

a,b=0,1
exec"a,b=b,b+a;"*input()
print b

Function form, 44 chars

def f(n):a,b=0,1;exec"a,b=b,b+a;"*n;return b

My take on this challenge. Didn't find this kind of an answer yet. I hope it's a valid one.

Print's n:th Fibonacci number. Functions by multiplying the string inside exec n times and then executing it as Python.

Edit: input() instead of int(raw_input())

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    \$\begingroup\$ If I'm not mistaken, you don't need to the () around exec in Python 2. \$\endgroup\$ – Stephen Jul 21 '17 at 14:07
  • \$\begingroup\$ @StepHen That seems to be true, thanks, down by 2 chars \$\endgroup\$ – SydB Jul 21 '17 at 14:11
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    \$\begingroup\$ Oh, I almost forgot: this would be considered a snippet because it preassumes the value of n. Generally you must write either a full program that gets input, or a function. So, you would have to replace n with input(). See this meta post for more information. \$\endgroup\$ – Stephen Jul 21 '17 at 14:15
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