81
\$\begingroup\$

Create the shortest program or function that finds the factorial of a non-negative integer.

The factorial, represented with ! is defined as such

$$n!:=\begin{cases}1 & n=0\\n\cdot(n-1)!&n>0\end{cases}$$

In plain English the factorial of 0 is 1 and the factorial of n, where n is larger than 0 is n times the factorial of one less than n.

Your code should perform input and output using a standard methods.

Requirements:

  • Does not use any built-in libraries that can calculate the factorial (this includes any form of eval)
  • Can calculate factorials for numbers up to 125
  • Can calculate the factorial for the number 0 (equal to 1)
  • Completes in under a minute for numbers up to 125

The shortest submission wins, in the case of a tie the answer with the most votes at the time wins.

\$\endgroup\$
13
  • 13
    \$\begingroup\$ How many of the given answers can actually compute up to 125! without integer overflow? Wasn't that one of the requirements? Are results as exponential approximations acceptable (ie 125 ! = 1.88267718 × 10^209)? \$\endgroup\$
    – Ami
    Feb 6 '11 at 22:43
  • 7
    \$\begingroup\$ @SHiNKiROU, even golfscript can manage 125! less than 1/10th of a second and it's and interpreted interpreted language! \$\endgroup\$
    – gnibbler
    Feb 8 '11 at 3:21
  • 6
    \$\begingroup\$ @ugoren the two-character solution to the other question uses a built-in factorial function. That's not allowed in this version of the challenge. \$\endgroup\$ Jan 7 '14 at 3:18
  • 5
    \$\begingroup\$ Completes in under a minute seems a very hardware-dependent requirement. Completes in under a minute on what hardware? \$\endgroup\$
    – sergiol
    Aug 24 '17 at 18:05
  • 4
    \$\begingroup\$ @sergiol Incredibly that hasn't been an issue in the last 2 years, I suspect most languages can get it done in under a minute. \$\endgroup\$ Aug 24 '17 at 21:20

191 Answers 191

1
2 3 4 5
7
71
\$\begingroup\$

Golfscript -- 12 chars

{,1\{)*}/}:f

Getting started with Golfscript -- Factorial in step by step

Here's something for the people who are trying to learn golfscript. The prerequisite is a basic understanding of golfscript, and the ability to read golfscript documentation.

So we want to try out our new tool golfscript. It's always good to start with something simple, so we're beginning with factorial. Here's an initial attempt, based on a simple imperative pseudocode:

# pseudocode: f(n){c=1;while(n>1){c*=n;n--};return c}
{:n;1:c;{n 1>}{n c*:c;n 1-:n;}while c}:f

Whitespace is very rarely used in golfscript. The easiest trick to get rid of whitespace is to use different variable names. Every token can be used as a variable (see the syntax page). Useful tokens to use as variables are special characters like |, &, ? -- generally anything not used elsewhere in the code. These are always parsed as single character tokens. In contrast, variables like n will require a space to push a number to the stack after. Numbers are essentially preinitialized variables.

As always, there are going to be statements which we can change, without affecting the end result. In golfscript, everything evaluates to true except 0, [], "", and {} (see this). Here, we can change the loop exit condition to simply {n} (we loop an additional time, and terminate when n=0).

As with golfing any language, it helps to know the available functions. Luckily the list is very short for golfscript. We can change 1- to ( to save another character. At present the code looks like this: (we could be using 1 instead of | here if we wanted, which would drop the initialization.)

{:n;1:|;{n}{n|*:|;n(:n;}while|}:f

It is important to use the stack well to get the shortest solutions (practice practice practice). Generally, if values are only used in a small segment of code, it may not be necessary to store them into variables. By removing the running product variable and simply using the stack, we can save quite a lot of characters.

{:n;1{n}{n*n(:n;}while}:f

Here's something else to think about. We're removing the variable n from the stack at the end of the loop body, but then pushing it immediately after. In fact, before the loop begins we also remove it from the stack. We should instead leave it on the stack, and we can keep the loop condition blank.

{1\:n{}{n*n(:n}while}:f

Maybe we can even eliminate the variable completely. To do this, we will need to keep the variable on the stack at all times. This means that we need two copies of the variable on the stack at the end of the condition check so we don't lose it after the check. Which means that we'll have a redundant 0 on the stack after the loop ends, but that is easy to fix.

This leads us to our optimal while loop solution!

{1\{.}{.@*\(}while;}:f

Now we still want to make this shorter. The obvious target should be the word while. Looking at the documentation, there are two viable alternatives -- unfold and do. When you have a choice of different routes to take, try and weigh the benefits of both. Unfold is 'pretty much a while loop', so as an estimate we'll cut down the 5 character while by 4 into /. As for do, we cut while by 3 characters, and get to merge the two blocks, which might save another character or two.

There's actually a big drawback to using a do loop. Since the condition check is done after the body is executed once, the value of 0 will be wrong, so we may need an if statement. I'll tell you now that unfold is shorter (some solutions with do are provided at the end). Go ahead and try it, the code we already have requires minimal changes.

{1\{}{.@*\(}/;}:f

Great! Our solution is now super-short and we're done here, right? Nope. This is 17 characters, and J has 12 characters. Never admit defeat!


Now you're thinking with... recursion

Using recursion means we must use a branching structure. Unfortunate, but as factorial can be expressed so succinctly recursively, this seems like a viable alternative to iteration.

# pseudocode: f(n){return n==0?n*f(n-1):1}
{:n{n.(f*}1if}:f # taking advantage of the tokeniser

Well that was easy -- had we tried recursion earlier we may not have even looked at using a while loop! Still, we're only at 16 characters.


Arrays

Arrays are generally created in two ways -- using the [ and ] characters, or with the , function. If executed with an integer at the top of the stack, , returns an array of that length with arr[i]=i.

For iterating over arrays, we have three options:

  1. {block}/: push, block, push, block, ...
  2. {block}%: [ push, block, push, block, ... ] (this has some nuances, e.g. intermediate values are removed from the stack before each push)
  3. {block}*: push, push, block, push, block, ...

The golfscript documentation has an example of using {+}* to sum the contents of an array. This suggests we can use {*}* to get the product of an array.

{,{*}*}:f

Unfortunately, it isn't quite that simple. All the elements are off by one ([0 1 2] instead of [1 2 3]). We can use {)}% to rectify this issue.

{,{)}%{*}*}:f

Well not quite. This doesn't handle zero correctly. We can calculate (n+1)!/(n+1) to rectify this, although this costs far too much.

{).,{)}%{*}*\/}:f

We can also try to handle n=0 in the same bucket as n=1. This is actual extremely short to do, try and work out the shortest you can.

Not so good is sorting, at 7 characters: [1\]$1=. Note that this sorting technique does has useful purposes, such as imposing boundaries on a number (e.g. `[0\100]$1=)
Here's the winner, with only 3 characters: .!+

If we want to have the increment and multiplication in the same block, we should iterate over every element in the array. Since we aren't building an array, this means we should be using {)*}/, which brings us to the shortest golfscript implementation of factorial! At 12 characters long, this is tied with J!

{,1\{)*}/}:f


Bonus solutions

Starting with a straightforward if solution for a do loop:

{.{1\{.@*\(.}do;}{)}if}:f

We can squeeze a couple extra out of this. A little complicated, so you'll have to convince yourself these ones work. Make sure you understand all of these.

{1\.!!{{.@*\(.}do}*+}:f
{.!{1\{.@*\(.}do}or+}:f
{.{1\{.@*\(.}do}1if+}:f

A better alternative is to calculate (n+1)!/(n+1), which eliminates the need for an if structure.

{).1\{.@*\(.}do;\/}:f

But the shortest do solution here takes a few characters to map 0 to 1, and everything else to itself -- so we don't need any branching. This sort of optimization is extremely easy to miss.

{.!+1\{.@*\(.}do;}:f

For anyone interested, a few alternative recursive solutions with the same length as above are provided here:

{.!{.)f*0}or+}:f
{.{.)f*0}1if+}:f
{.{.(f*}{)}if}:f

*note: I haven't actually tested many of the pieces of code in this post, so feel free to inform if there are errors.

\$\endgroup\$
4
  • 8
    \$\begingroup\$ Interesting, the seems to be a bug in the spoiler markdown when you use code in a spoiler... Anyone cares to mention this on Meta? \$\endgroup\$
    – Ivo Flipse
    Feb 7 '11 at 11:55
  • 8
    \$\begingroup\$ I find it interesting how golfscript - a golfing language - allows multi-letter variable names and "punishes" you for using 1 letter with necessary whitespace \$\endgroup\$
    – Cyoce
    Feb 4 '16 at 15:45
  • \$\begingroup\$ I see a trailing 1 when I try {:n;1:c;{n 1>}{n c*:c;n 1-:n;}while c}:f \$\endgroup\$
    – Jay Lee
    Mar 23 '20 at 19:19
  • \$\begingroup\$ All this just to tie J... take my +1... \$\endgroup\$ May 7 at 19:37
49
\$\begingroup\$

Python - 27

Just simply:

f=lambda x:0**x or x*f(x-1)
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6
  • 27
    \$\begingroup\$ Good trick: 0**x. \$\endgroup\$
    – Alexandru
    Jul 11 '11 at 20:25
  • \$\begingroup\$ What about math.factorial? It isn't a built-in, is it? \$\endgroup\$
    – user63571
    Jan 19 '17 at 20:21
  • 1
    \$\begingroup\$ @JackBates that counts as a builtin, as you didn't write the code to compute the factorial. \$\endgroup\$
    – FlipTack
    Jan 21 '17 at 17:51
  • 2
    \$\begingroup\$ Can anyone tell me what's the trick behind 0**x? \$\endgroup\$
    – pavi2410
    Aug 4 '19 at 5:36
  • 1
    \$\begingroup\$ @Pavitra: 00=1, and it's the first thing that evaluates so it gets returned. For any other n, 0n=0, thus the first operand of or is falsey, such that the second operand gets evaluated. \$\endgroup\$
    – univalence
    Aug 5 '19 at 18:13
48
\$\begingroup\$

Haskell, 17

f n=product[1..n]
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11
  • 2
    \$\begingroup\$ I don't know Haskell... But Will this calculate factorial for 0 \$\endgroup\$
    – The King
    Feb 7 '11 at 12:11
  • 11
    \$\begingroup\$ @The King: yes it will. [1..0] ==> [] and product [] ==> 1 \$\endgroup\$
    – J B
    Feb 7 '11 at 12:12
  • 5
    \$\begingroup\$ I would argue this uses the "built-in library" that the problem prohibits. Still, the other method f 0=1;f n=n*f$n-1 is 17 characters as well. \$\endgroup\$ Jul 26 '11 at 1:20
  • 7
    \$\begingroup\$ @eternalmatt: that part of the restrictions is underspecified to me. Both product and, say, (*) or (-) "can calculate the factorial", and they're all defined through the Prelude. Why would one be cool and not the other? \$\endgroup\$
    – J B
    Jul 27 '11 at 9:28
  • 2
    \$\begingroup\$ @YoYoYonnY: I count 17 characters as well, for less (subjective) readability. IMHO it's fine in the comments. \$\endgroup\$
    – J B
    Feb 8 '16 at 13:32
32
\$\begingroup\$

APL (4)

×/∘⍳

Works as an anonymous function:

    ×/∘⍳ 5
120

If you want to give it a name, 6 characters:

f←×/∘⍳
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4
  • \$\begingroup\$ I don't speak APL, what is going on here? \$\endgroup\$ Jan 6 '14 at 21:11
  • \$\begingroup\$ @MichaelStern: makes an index vector, i.e. ⍳5 is 1 2 3 4 5. × is (obviously) multiply, / is reduce, and is function composition. So, ×/∘⍳ is a function that takes an argument x and gives the product of the numbers [1..x]. \$\endgroup\$
    – marinus
    Jan 6 '14 at 23:09
  • \$\begingroup\$ Ah, the same approach as in @Yves Klett's Mathematica solution. Very nice. \$\endgroup\$ Jan 7 '14 at 3:14
  • \$\begingroup\$ @NBZ: That didn't exist yet in 2011 when this question was written, nor in 2012 when I wrote this answer. Trains were only added in Dyalog 14.0 which came out in 2014. \$\endgroup\$
    – marinus
    Jan 25 '16 at 18:38
19
\$\begingroup\$

J (12)

A standard definition in J:

f=:*/@:>:@i.

Less than 1sec for 125!

Eg:

 f 0
 1
 f 5
 120
  f 125x
 1882677176888926099743767702491600857595403
 6487149242588759823150835315633161359886688
 2932889495923133646405445930057740630161919
 3413805978188834575585470555243263755650071
 31770880000000000000000000000000000000
\$\endgroup\$
6
  • \$\begingroup\$ why not just */>:i. ? \$\endgroup\$
    – Andbdrew
    Aug 21 '11 at 11:26
  • 3
    \$\begingroup\$ There's no reason it can't be an anonymous function right? Like ([:*/1+i.) for 10 points, or even 8 as the parentheses are only needed for calling the function, not for the definition. \$\endgroup\$
    – jpjacobs
    Nov 4 '14 at 22:05
  • 1
    \$\begingroup\$ in the last one, f 125x what does the x do? Is it a special kind of number? \$\endgroup\$
    – Cyoce
    Jan 21 '16 at 7:11
  • 1
    \$\begingroup\$ @Cyoce, yes, it's extended precision integer. \$\endgroup\$
    – Eelvex
    Jan 21 '16 at 19:23
  • 1
    \$\begingroup\$ What about f=:[:*/1+i. which saves one byte? \$\endgroup\$
    – Leaky Nun
    Jun 24 '16 at 8:52
18
\$\begingroup\$

Golfscript - 13 chars (SYM)

defines the function !

{),()\{*}/}:!             # happy robot version \{*}/ 

alternate 13 char version

{),()+{*}*}:! 

whole program version is 10 chars

~),()+{*}*

testcases take less than 1/10 second:

input:

0!

output

1

input

125!

output

188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000
\$\endgroup\$
2
  • 1
    \$\begingroup\$ +1 for symbolic golf entry! I wish I could upvote more than once. :-D \$\endgroup\$ Feb 7 '11 at 0:56
  • \$\begingroup\$ @ChrisJester-Young I'll do it for you. \$\endgroup\$
    – Cyoce
    Jan 21 '16 at 7:30
15
\$\begingroup\$

Mornington Crescent, 1827 1698 chars

I felt like learning a new language today, and this is what I landed on... (Why do I do this to myself?) This entry won't be winning any prizes, but it beats all 0 other answers so far using the same language!

Take Northern Line to Bank
Take Central Line to Holborn
Take Piccadilly Line to Heathrow Terminals 1, 2, 3
Take Piccadilly Line to Acton Town
Take District Line to Acton Town
Take District Line to Parsons Green
Take District Line to Bank
Take District Line to Parsons Green
Take District Line to Acton Town
Take District Line to Hammersmith
Take Circle Line to Aldgate
Take Circle Line to Aldgate
Take Metropolitan Line to Chalfont & Latimer
Take Metropolitan Line to Aldgate
Take Circle Line to Hammersmith
Take District Line to Acton Town
Take Piccadilly Line to Bounds Green
Take Piccadilly Line to Acton Town
Take Piccadilly Line to Bounds Green
Take Piccadilly Line to Acton Town
Take District Line to Acton Town
Take District Line to Bank
Take Circle Line to Hammersmith
Take District Line to Upminster
Take District Line to Parsons Green
Take District Line to Notting Hill Gate
Take Circle Line to Notting Hill Gate
Take Circle Line to Bank
Take Circle Line to Temple
Take Circle Line to Aldgate
Take Circle Line to Aldgate
Take Metropolitan Line to Chalfont & Latimer
Take Metropolitan Line to Chalfont & Latimer
Take Metropolitan Line to Aldgate
Take Circle Line to Hammersmith
Take District Line to Upminster
Take District Line to Bank
Take District Line to Upney
Take District Line to Upminster
Take District Line to Hammersmith
Take District Line to Upminster
Take District Line to Upney
Take District Line to Bank
Take Circle Line to Embankment
Take Circle Line to Embankment
Take Northern Line to Angel
Take Northern Line to Moorgate
Take Metropolitan Line to Chalfont & Latimer
Take Metropolitan Line to Moorgate
Take Circle Line to Moorgate
Take Northern Line to Mornington Crescent

Try it online!

Anyone who's journeyed around London will understand that instantly of course, so I'm sure I don't need to give a full explanation.

Most of the work at the start is in handling the 0 case. After initialising the product at 1, I can use that to calculate max(input, 1) to get the new input, taking advantage of the fact that 0! = 1! Then the main loop can begin.

(EDIT: A whole bunch of trips have been saved by stripping the 1 from "Heathrow Terminals 1, 2, 3" instead of generating it by dividing 7 (Sisters) by itself. I also use a cheaper method to generate the -1 in the next step.)

Decrementing is expensive in Mornington Crescent (although less expensive than the Tube itself). To make things more efficient I generate a -1 by taking the NOT of a parsed 0 and store that in Hammersmith for much of the loop.


I put some significant work into this, but since this is my first attempt at golfing in Mornington Crescent (in fact my first attempt in any language), I expect I missed a few optimisations here and there. If you're interested in programming in this language yourself (and why wouldn't you be?), Esoteric IDE - with its debug mode and watch window - is a must!

\$\endgroup\$
1
  • \$\begingroup\$ Why did this get so less upvotes? \$\endgroup\$
    – null
    Jul 25 '20 at 12:10
14
\$\begingroup\$

MATL, 2 bytes

:p

Explained:

:    % generate list 1,2,3,...,i, where i is an implicit input
p    % calculate the product of of all the list entries (works on an empty list too)

Try it online!

\$\endgroup\$
3
  • 13
    \$\begingroup\$ ​​​​​​​​​:O​​​​ \$\endgroup\$ Jan 25 '16 at 12:14
  • \$\begingroup\$ I was going to post exactly this :-) You may want to modify the link to include the code and an example input \$\endgroup\$
    – Luis Mendo
    Jan 31 '16 at 4:56
  • 5
    \$\begingroup\$ @AndrasDeak, No, that would output all numbers from 1 to i... \$\endgroup\$
    – yyny
    Feb 6 '16 at 16:09
13
\$\begingroup\$

Perl 6: 13 chars

$f={[*]1..$_}

[*] is same as Haskell product, and 1..$_ is a count-up from 1 to $_, the argument.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ It's not allowed to not use a space after [*] anymore ("Two terms in a row" error message). \$\endgroup\$ Dec 28 '13 at 19:24
  • \$\begingroup\$ You don't need to set a variable, a bare code block is an acceptable answer as it implicitly forms a function. Also does this still work for 0? \$\endgroup\$
    – Phil H
    Apr 28 '18 at 7:24
11
\$\begingroup\$

Python, 28 bytes

f=lambda x:x/~x+1or x*f(x-1)

(based off Alexandru's solution)

\$\endgroup\$
9
\$\begingroup\$

F#: 26 chars

There's no inbuilt product function in F#, but you can make one with a fold

let f n=Seq.fold(*)1{1..n}
\$\endgroup\$
0
9
\$\begingroup\$

Matlab, 15

f=@(x)prod(1:x)

Test Cases

>> f(0)
ans =
     1
>> f(4)
ans =
    24
>> tic,f(125),toc
ans =
  1.8827e+209
Elapsed time is 0.000380 seconds.
\$\endgroup\$
9
\$\begingroup\$

Java, 85 Chars

BigInteger f(int n){return n<2?BigInteger.ONE:new BigInteger(""+n).multiply(f(n-1));}
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2
  • 1
    \$\begingroup\$ This misses the imports: import java.math.*; (so, +19 bytes). \$\endgroup\$ Dec 24 '17 at 10:48
  • \$\begingroup\$ Fair point. ............ \$\endgroup\$
    – st0le
    Dec 29 '17 at 19:36
8
\$\begingroup\$

Ruby - 21 chars

f=->n{n>1?n*f[n-1]:1}

Test

irb(main):009:0> f=->n{n>1?n*f[n-1]:1}
=> #<Proc:0x25a6d48@(irb):9 (lambda)>
irb(main):010:0> f[125]
=> 18826771768889260997437677024916008575954036487149242588759823150835315633161
35988668829328894959231336464054459300577406301619193413805978188834575585470555
24326375565007131770880000000000000000000000000000000
\$\endgroup\$
0
8
\$\begingroup\$

JavaScript, 25

function f(n)!n||n*f(n-1)

CoffeeScript, 19

f=(n)->!n||n*f(n-1)

Returns true in the case of n=0, but JavaScript will type-coerce that to 1 anyway.

\$\endgroup\$
5
  • \$\begingroup\$ Don't you need a return statement in the JavaScript function? \$\endgroup\$ Jul 12 '11 at 19:19
  • \$\begingroup\$ Update: Holy smoke, you don't need a return! But why not? \$\endgroup\$ Jul 12 '11 at 19:43
  • \$\begingroup\$ It's JavaScript 1.8 (developer.mozilla.org/en/new_in_javascript_1.8). Full disclosure, it only works on Firefox! \$\endgroup\$
    – Casey Chu
    Jul 13 '11 at 3:53
  • 1
    \$\begingroup\$ Nice, I didn't know about leaving out the return statement for JavaScript 1.8. Also, you can guarantee 1 instead of true for the n=0 case with the same length code: function f(n)n?n*f(--n):1 \$\endgroup\$
    – Briguy37
    Aug 3 '11 at 21:50
  • 11
    \$\begingroup\$ ES6, 17: f=n=>!n||n*f(n-1) Take that, CoffeeScript! \$\endgroup\$
    – Ry-
    Dec 29 '13 at 3:03
7
\$\begingroup\$

PostScript, 26 chars

/f{1 exch -1 1{mul}for}def

Example:

GS> 0 f =
1
GS> 1 f =
1
GS> 8 f =
40320

The function itself takes only 21 characters; the rest is to bind it to a variable. To save a byte, one can also bind it to a digit, like so:

GS> 0{1 exch -1 1{mul}for}def
GS> 8 0 load exec =
40320
\$\endgroup\$
1
  • 1
    \$\begingroup\$ Ghostscript cannot handle 125!; anything beyond 34! comes out as 1.#INF. (I used stock GNU Ghostscript 9.0.7 compiled for x64 Windows.) \$\endgroup\$ Jan 7 '14 at 6:17
7
\$\begingroup\$

Brachylog, 7 6 bytes

By making a range and multiplying it

-1 byte tanks to ovs having the idea to use the max() function

;1⌉⟦₁×

Explanation

;1          --  If n<1, use n=1 instead (zero case)
  ⟦₁        --      Construct the range [1,n]
    ×       --      return the product of said range

Try it online!


Brachylog, 10 9 bytes

recursion

≤1|-₁↰;?×

Explanation

            --f(n):
≤1          --  if n ≤ 1: return 1
|           --  else:
 -₁↰        --      f(n-1)
    ;?×     --            *n

Try it online!

\$\endgroup\$
2
  • 1
    \$\begingroup\$ This works for 6 bytes. Taking input as a singleton is allowed by default. \$\endgroup\$
    – ovs
    Aug 22 '18 at 17:58
  • \$\begingroup\$ @ovs thanks. But using ; instead of , allows for just a regular numerical input. -1byte anyway \$\endgroup\$
    – Kroppeb
    Aug 22 '18 at 18:12
6
\$\begingroup\$

Ruby - 30 29 characters

def f(n)(1..n).inject 1,:*end

Test

f(0) -> 1
f(5) -> 120
\$\endgroup\$
5
  • 1
    \$\begingroup\$ You can put the end directly after :* without a newline or semicolon. \$\endgroup\$
    – sepp2k
    Feb 6 '11 at 17:50
  • 1
    \$\begingroup\$ There's no need to pass 1 to the #inject call. (1..10).inject :* #=> 3628800 \$\endgroup\$
    – Dogbert
    Feb 7 '11 at 10:56
  • 1
    \$\begingroup\$ @Dogbert, what about for f(0)? \$\endgroup\$
    – Nemo157
    Feb 7 '11 at 19:05
  • \$\begingroup\$ @Nemo157, ah! forgot about that. \$\endgroup\$
    – Dogbert
    Feb 8 '11 at 13:38
  • 4
    \$\begingroup\$ Shorter to use 1.9 lambda syntax: f=->n{(1..n).inject 1,:*}. Call it with f[n]. \$\endgroup\$ Aug 22 '11 at 22:04
6
\$\begingroup\$

C#, 20 or 39 characters depending on your point of view

As a traditional instance method (39 characters; tested here):

double f(int x){return 2>x?1:x*f(x-1);}

As a lambda expression (20 characters, but see disclaimer; tested here):

f=x=>2>x?1:x*f(x-1);

We have to use double because 125! == 1.88 * 10209, which is much higher than ulong.MaxValue.

Disclaimer about the lambda version's character count:

If you recursion in a C# lambda, you obviously have to store the lambda in a named variable so that it can call itself. But unlike (e.g.) JavaScript, a self-referencing lambda must have been declared and initialized on a previous line. You can't call the function in the same statement in which you declare and/or initialize the variable.

In other words, this doesn't work:

Func<int,double> f=x=>2>x?1:x*f(x-1); //Error: Use of unassigned local variable 'f'

But this does:

Func<int,double> f=null;            
f=x=>2>x?1:x*f(x-1);  

There's no good reason for this restriction, since f can't ever be unassigned at the time it runs. The necessity of the Func<int,double> f=null; line is a quirk of C#. Whether that makes it fair to ignore it in the character count is up to the reader.

CoffeeScript, 21 19 characters for real

f=(x)->+!x||x*f x-1

Tested here: http://jsfiddle.net/0xjdm971/

\$\endgroup\$
1
  • \$\begingroup\$ Wanted to point out that while inline declaration as a Func type doesn't work, an inline function does: double f(int x)=>2>x?1:x*f(x-1); \$\endgroup\$ Jun 8 '20 at 17:30
5
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C (39 chars)

double f(int n){return n<2?1:n*f(n-1);}
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2
  • 3
    \$\begingroup\$ Nice. But can save some characters: double f(n){return!n?1:n*f(n-1);} - 33 chars. \$\endgroup\$
    – ugoren
    Jan 12 '12 at 16:38
  • 2
    \$\begingroup\$ f(125) will overflow \$\endgroup\$
    – ogogmad
    Oct 14 '15 at 7:12
5
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Jelly, 2 bytes

RP

Try it online!

Alternatively, a builtin answer is one byte:

!

Try it online!

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5
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Aheui (esotope), 93 90 87 bytes

박밴내색뱅뿌뮹
숙쌕빼서빼처소
타뿌싼때산쑥희
매차뽀요@어몽

Try it online!

Nice, small, and fast code. Slightly golfed after writing explanation. I'll not change it, because it is just same code.

Explaination

Aheui is befunge-like language and (almost) every character of Aheui is operator. Part of character looks like ㅏ, ㅐ, ㅓ, ㅜ, ㅛ, ㅗ, ㅢ determines direction where next operator execute. is left-to-right, is right-to-left, is down-to-up, is up-to-down, is down-to-up, with skipping one character in two characters. is 'nothing' : keep same speed and direction.

박밴내

commend is store given number in current stack, commend is divide upmost two number in current stack. Both and store 2, so 박밴내 store 1 in current stack(default or nothing stack)

색뱅

commend change current stack. change stack to (or ) stack. commend with (like or ) get a number from STDIN. So 색뱅 get a number and store it in stack 악(ㄱ).

뿌
처

commend duplicate upmost value in current stack, and commend pop value from current stack and see if it is 0. If it is, it go to opposite direction from indicate : in here right-to-left. If it is not, it go to direction where indicate. So 뿌(\n)처 see if input is 0 or not, and go right if zero, and go left if not.

망희
소

If input is zero, here is evaluated. (from commend) First, change current stack to nothing(). commend is pop, and if used with it print value as number. halts program. So it print 1 and halt.

숙쌕빼서빼
타뿌싼때산쌕꾸
매차뽀요애애어

enter image description here

Look at this image for help. Here is main loop. is subtraction, and is multiply. move value from current stack to selected one.

Put it shortly, it get number from nothing stack(or get 1), subtract to find if it is zero, and if not zero multiply and restart loop. And if zero, go to rightmost place of code with popping one number.

Print number, then pointer go to : halt.

In one image :

AheuiChem image translated

SEL is select, MOV is move, DUP is duplicate. This image is produced by AheuiChem, Aheui development tool in Korean. Translated with paint tool of windows.

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5
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8088 / 8087 machine code, 13 bytes

D9 E8       FLD1                ; start with 1 
E3 08       JCXZ DONE           ; if N = 0, return 1 
        FACT_LOOP: 
51          PUSH CX             ; push current N onto stack
8B F4       MOV  SI, SP         ; SI to top of stack for N 
DE 0C       FIMUL WORD PTR[SI]  ; ST = ST * N 
59          POP  CX             ; remove N from stack 
E2 F8       LOOP FACT_LOOP      ; decrement N, loop until N = 0
        DONE: 
C3          RET                 ; return to caller

Input \$n\$ is in CX, output \${n!}\$ is in ST(0).

Test output displayed using Borland Turbo Debugger 3:

n = 1:

enter image description here

n = 15:

enter image description here

n = 50:

enter image description here

n = 125:

enter image description here

Notes:

  • Calculates \${125!}\$ in an imperceptible amount of time (difficult to accurately profile in DOS)
  • Language is not newer than the challenge
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1
  • 1
    \$\begingroup\$ "Language is not newer than the challenge" facepalm \$\endgroup\$
    – S.S. Anne
    Mar 18 '20 at 19:03
4
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D: 45 Characters

T f(T)(T n){return n < 2 ? 1 : n * f(n - 1);}

More legibly:

T f(T)(T n)
{
    return n < 2 ? 1 : n * f(n - 1);
}

A cooler (though longer version) is the templatized one which does it all at compile time (64 characters):

template F(int n){static if(n<2)enum F=1;else enum F=n*F!(n-1);}

More legibly:

template F(int n)
{
    static if(n < 2)
        enum F = 1;
    else
        enum F = n * F!(n - 1);
}

Eponymous templates are pretty verbose though, so you can't really use them in code golf very well. D's already verbose enough in terms of character count to be rather poor for code golf (though it actually does really well at reducing overall program size for larger programs). It's my favorite language though, so I figure that I might as well try and see how well I can get it to do at code golf, even if the likes of GolfScript are bound to cream it.

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3
  • 3
    \$\begingroup\$ take out the whitespace and you can get it down to 36 chars \$\endgroup\$ Jul 3 '11 at 13:36
  • \$\begingroup\$ @Cyoce Can you explain? \$\endgroup\$
    – yyny
    Feb 6 '16 at 16:08
  • \$\begingroup\$ Welcome to the site, @user272735. Note that we don't edit people's solutions in order to make improvements here. Instead we leave comments suggesting those improvements, as ratchet freak did above. \$\endgroup\$
    – Shaggy
    Jul 29 '19 at 8:29
4
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PowerShell – 36

Naïve:

filter f{if($_){$_*(--$_|f}else{1}}

Test:

> 0,5,125|f
1
120
1,88267717688893E+209
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4
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Scala, 39 characters

def f(x:BigInt)=(BigInt(1)to x).product

Most of the characters are ensuring that BigInts are used so the requirement for values up to 125 is met.

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1
  • \$\begingroup\$ Some shorter options: (x:Int)=>(BigInt(1)to x).product def f(x:Int)=(BigInt(1)to x).product def f(x:BigInt)=(x.to(1,-1)).product def f(x:BigInt)=(-x to-1).product.abs \$\endgroup\$
    – LRLucena
    Jun 7 '16 at 14:32
4
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Javascript, ES6 17

f=n=>n?n*f(n-1):1

ES6:

  • Arrow function
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3
  • \$\begingroup\$ ES6 is younger than this challenge if I'm remembering correctly and therefore not eligible. \$\endgroup\$
    – lirtosiast
    Jun 22 '15 at 19:44
  • \$\begingroup\$ There is smth strange with conditional operator. Why there are two colons? \$\endgroup\$
    – Qwertiy
    Jun 25 '15 at 15:43
  • \$\begingroup\$ @Qwertiy You're right, that was a typo, thanks. \$\endgroup\$ Jun 25 '15 at 18:21
4
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PowerShell, 42 bytes

(saved 2 chars using filter instead of function)

filter f($x){if(!$x){1}else{$x*(f($x-1))}}

Output:

PS C:\> f 0
1
PS C:\> f 5
120
PS C:\> f 1
1
PS C:\> f 125
1.88267717688893E+209
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1
  • 1
    \$\begingroup\$ This is way old now, but... Can save 1 more character by reversing the if/else: filter f($x){if($x){$x*(f($x-1))}else{1}}. And it can be reduced further to 36 characters if it's called via pipeline since it's a filter (e.g. 125|f): filter f{if($_){$_*($_-1|f)}else{1}} \$\endgroup\$
    – Andrew
    Oct 14 '15 at 18:31
4
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Racket (scheme) 40 35 29 bytes

Computes 0! to be 1, and computes 125! in 0 seconds according to timer. Regular recursive approach

(define(f n)(if(= n 0)1(* n(f(- n 1)))))

New version to beat common lisp: multiplies all elements of a list (same as that Haskell solution)

(λ(n)(apply *(build-list n add1)))

Newer version to beat the other scheme solution and math the other racket solution by using foldl instead of apply and using range instead of buildlist

(λ(n)(foldl * n(range 1 n)))
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4
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Python 2, 45 bytes

lambda n:reduce(long.__mul__,range(1,n+1),1L)

Try it online! Not the shortest Python solution but I thought it would be interesting to post a solution with reduce and long.__mul__.

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4
  • \$\begingroup\$ Doesn't work past 22. \$\endgroup\$ Jun 22 '20 at 17:25
  • \$\begingroup\$ @Calculuswhiz you are right, it doesn't. Python 2's int has a max size. \$\endgroup\$
    – RGS
    Jun 22 '20 at 17:33
  • 1
    \$\begingroup\$ You can fix it at the cost of 2 bytes. \$\endgroup\$
    – Bubbler
    Aug 13 '20 at 0:49
  • \$\begingroup\$ @Bubbler indeed I can, thanks o/ \$\endgroup\$
    – RGS
    Aug 13 '20 at 7:11
1
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