Find the Factorial!

76

15

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.

11How 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)? – Ami – 2011-02-06T22:43:29.687

Completes under a minute for factorials under 125? – Ming-Tang – 2011-02-06T23:08:48.193

@SHiNKiROU, most languages should be able to accomplish that in well under a minute. – Kevin Brown – 2011-02-06T23:10:32.560

@Ami, there is nothing restricting the format of the output, just as long as it is correct. – Kevin Brown – 2011-02-06T23:11:37.207

6@SHiNKiROU, even golfscript can manage 125! less than 1/10th of a second and it's and interpreted interpreted language! – gnibbler – 2011-02-08T03:21:05.863

5@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. – Michael Stern – 2014-01-07T03:18:06.343

4Completes in under a minute seems a very hardware-dependent requirement. Completes in under a minute on what hardware? – sergiol – 2017-08-24T18:05:55.893

4@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. – Kevin Brown – 2017-08-24T21:20:53.340

1Do you need an exact return or float? – l4m2 – 2017-12-25T13:16:55.193

Why aren't built-ins allowed? You haven't specified what built-ins are, and if you said that it was up to a "reasonable person" to decide (which is completely subjective, but ignoring that), you still say that any form of eval is a built-in for the factorial, even though it evaluates code, not the factorial of a given number. – MilkyWay90 – 2019-05-07T02:02:10.477

– Jared Updike – 2011-09-11T23:52:55.293

1

Looks like an exact duplicate of http://stackoverflow.com/questions/237496/code-golf-factorials which has a 2-char winner.

– ugoren – 2012-01-13T07:04:31.257

68

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.

8Interesting, the seems to be a bug in the spoiler markdown when you use code in a spoiler... Anyone cares to mention this on Meta? – Ivo Flipse – 2011-02-07T11:55:55.280

5I find it interesting how golfscript - a golfing language - allows multi-letter variable names and "punishes" you for using 1 letter with necessary whitespace – Cyoce – 2016-02-04T15:45:52.307

45

f n=product[1..n]


1I like Haskell very much. Such mathematical tasks are usually extremely short in it. – FUZxxl – 2011-02-06T17:42:41.207

FUZxxl: If it weren't for the horribly long name product ... – Joey – 2011-02-07T10:42:21.350

2I don't know Haskell... But Will this calculate factorial for 0 – The King – 2011-02-07T12:11:48.627

11@The King: yes it will. [1..0] ==> [] and product [] ==> 1 – J B – 2011-02-07T12:12:55.303

@TheKing product = foldl (*) 1, so product [] == foldl (*) 1 [], which is 1 (foldls initial value) – YoYoYonnY – 2016-02-06T15:58:36.677

@JB Might be worth adding this one too: (f 0=1 <NL> f n=n*f$n-1) – YoYoYonnY – 2016-02-06T16:02:09.670 2@YoYoYonnY: I count 17 characters as well, for less (subjective) readability. IMHO it's fine in the comments. – J B – 2016-02-08T13:32:31.617 5I would argue this uses the "built-in library" that the problem prohibits. Still, the other method f 0=1;f n=n*f$n-1 is 17 characters as well. – eternalmatt – 2011-07-26T01:20:24.557

5@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? – J B – 2011-07-27T09:28:29.783

@J B: Good point. I'm sure this has been debated somewhere else. I suppose in my naive mind, I'd like to set aside (*) and (-) operators and pretend that they execute straight to hardware, but I guess that's not true. – eternalmatt – 2011-07-31T04:36:15.927

4

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.

@Cyoce Can you explain? – YoYoYonnY – 2016-02-06T16:08:25.320

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. – Shaggy – 2019-07-29T08:29:42.320

3take out the whitespace and you can get it down to 36 chars – ratchet freak – 2011-07-03T13:36:53.347

4

Naïve:

filter f{if($_){$_*(--$_|f}else{1}}  Test: > 0,5,125|f 1 120 1,88267717688893E+209  4 Racket (scheme) 4035 29 bytes Computes 0! to be 1, and computes 125! in 0 seconds according to timer. Regular recursive approach (define(f n)(if(= n 0)1(* n(f(- n 1)))))  New version to beat common lisp: multiplies all elements of a list (same as that Haskell solution) (λ(n)(apply *(build-list n add1)))  Newer version to beat the other scheme solution and math the other racket solution by using foldl instead of apply and using range instead of buildlist (λ(n)(foldl * n(range 1 n)))  3 Befunge-932422 20 bytes Edit: Found a better way of generating the range of values 1&0>-#1:__\#0:#*_$.@


Try it online!

How it works:

1& Initialises the stack with 1 (for the special case of 0!) and the inputted factorial
0>-#1:_ Adds the current factorial level-1 to the stack until it reaches 0
_\# :# _ Pops another 0 and swaps the two top values of the stack. Duplicate the top value, and check whether it is 0.
_ #0 #*_ If not, multiply the top two values of the stack and add an extra 0 so the underscore will point right. Repeat the above section.
$.@ If so, pop the extra 0, print the value and exit the program  3 Jelly, 2 bytes RP  Try it online! Alternatively, a builtin answer is one byte: !  Try it online! 3 Kona (11 6) */1.+!  K works right-to-left (for the most part), so we enumerate x (make a list/array of numbers from 0 to x-1), add 1 to it (list ranges 0 to x), then multiply all numbers together. If it weren't a requirement to compute 125!, I could save 1 more byte by eliminating . next to the 1. In any event, 125! is computed in mere milliseconds:  */1.+!125. 1.882677e+209  You don't need a lot of this. K has currying, so the entire answer becomes */1.+!: 6 bytes. – kirbyfan64sos – 2015-06-25T14:43:59.390 @kirbyfan64sos: True & I'll edit it in. I think when I wrote this ~18 months ago, I was still stuck on everything must be callable (i.e., function). – Kyle Kanos – 2015-06-25T15:00:02.157 3 Aheui (esotope), 9390 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. ㅏ is left-to-right, ㅓ is right-to-left, ㅗ is down-to-up, ㅜ is up-to-down, ㅛ is down-to-up, with skipping one character in two characters. ㅐ is 'nothing' : keep same speed and direction. 박밴내  ㅂ commend is store given number in current stack, ㄴ commend is divide upmost two number in current stack. Both 박 and 백 store 2, so 박밴내 store 1 in current stack(default 아 or nothing stack) 색뱅  ㅅ commend change current stack. 색 change stack to 악 (or ㄱ) stack. ㅂ commend with ㅇ (like 뱅 or 방) get a number from STDIN. So 색뱅 get a number and store it in stack 악(ㄱ). 뿌 처  ㅃ commend duplicate upmost value in current stack, and ㅊ commend pop value from current stack and see if it is 0. If it is, it go to opposite direction from ㅓ indicate : in here right-to-left. If it is not, it go to direction where ㅓ 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 소 commend) First, change current stack to nothing(아). ㅁ commend is pop, and if used with ㅇ it print value as number. 희 halts program. So it print 1 and halt. 숙쌕빼서빼 타뿌싼때산쌕꾸 매차뽀요애애어  Look at this image for help. Here is main loop. ㅌ is subtraction, and ㄸ is multiply. ㅆ 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. 몽  Print number, then pointer go to 희 : halt. In one image : 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. 3 J - 6 characters */>:i.  Does this count? I know it is very similar to the earlier J example, but it is a little shorter :) I'm a beginner with J, but it's a lot of fun so far! 3 BrainFuck, 125 / CompressedFuck, 47 ,[>+>+>>>+<<<<<-]>>-<[[>[->+>+<<]>>[-<<+>>]<<<-]>>>>[<<<<+>>>>-]<<<<->[-]>[<+>-]<<[>>>>+>+<<<<<-]>>>>>[<<<<<+>>>>>-]<<<<<-]>.  In 8-bit text encodings the program had 1000 bits. However, any BrainFuck program could be stored with a 3-bit encoding. 125*3=375 375 bits / 8 = 47 bytes EDIT: In the CompressedFuck format it has 47 bytes :) Also I forgot to mention that this program only works with infinite-sized cells Still impressive. – timmyRS – 2016-02-05T11:05:10.463 Take a look at https://esolangs.org/wiki/CompressedFuck. But code-golf is more about competing with other programs in the same language than it is about competing with the rest, and 125 characters for brainfuck is pretty impressive :) – YoYoYonnY – 2016-02-06T16:15:30.877 @YoYoYonnY thanks, I updated the answer – KeksArmee – 2016-02-08T12:40:48.820 3 Befunge - 2x20 = 40 characters 0\:#v_# 1#<\$v *\<
>:1-:#^_$>\:#^_$


This is a function in that it is a standalone block of code not utilising the wraparound. You have to place the argument on the top of the stack then enter from the top-left going right, the function will exit from the bottom-right going right with the result on the top of the stack.

E.g. to calculate the factorial of 125

555**   0\:#v_# 1#<\$v *\< >:1-:#^_$>\:#^_$.@  Testing 0 0 0\:#v_# 1#<\$v *\<
>:1-:#^_$>\:#^_$    .@


I know this is pretty old, but I think this is somewhat shorter and quicker : &:!#@_>:# 1# -# :# _$>\# :#* _$.@ (where & should be replaced by the input). It's 32 chars/bytes – FliiFe – 2016-04-07T11:08:13.907

3

Python, 35 bytes

def f(n):return n and n*f(n-1) or 1


or

def f(n):return n*f(n-1) if n else 1


3I posted a shorter similar solution. No need to repeat it you don't have an extra insight. – Alexandru – 2011-02-07T12:33:11.447

@Alexandru if he wrote this without referencing your answer, I don't see why he shouldn't be allowed to post it just because it's longer. All Runners in a race get to post their score regardless of being first or last. – akozi – 2019-02-20T14:49:52.993

3

In C (23 Characters)

This abuses the GCC "feature" that makes the last assignment count as a return if no return is specified.

f(a){a=a>0?f(a-1)*a:1;}


In proper C, 28 characters

f(a){return a>0?f(a-1)*a:1;}


+1 for the GCC "feature". I think GCC even allows a block return value (Can remember doing something like this) 0 == ({printf("Hello, world!"); 0;}); – YoYoYonnY – 2016-02-06T16:11:45.767

2

PHP, 39 bytes

<?=array_product(range(1,$argv[1]))?:1;  breakdown <?= // 4. print result array_product( // 2. get product of the elements - special: 0 range(1,$argv[1])    // 1. build array from 1 to N - special: [1,0]
)
?:1                      // 3. special: if falsy, return 1
;


for($p=1;$i++<$argn;)$p*=$i;echo$p; is shorter and <?=array_product($argn?range(1,$argn):[]); is a more interesting way – Jörg Hülsermann – 2017-07-09T00:41:56.087

2

Brainfuck, 56 bytes

+>,[[>+>+<<-]>[-<<[->+<<+>]<[->+<]>>>]<<[-]>[->+<]>>-]<.


Like the other Brainfuck answers, this assumes the IO directly inputs and outputs the number in/from the cell and the interpreter has infinite cells and infinite cell size. Add a byte if you want to avoid negative cells. Add another byte if you want to do it in place.

Note: In CompressedFuck, this is only ~21 bytes.

2

Brain-Flak, 52 bytes

Came up with the solution independently, thanks JoKing for telling me that it's possible to get 52 bytes.

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


Try it online!

Ungolfed:

<>(())<>		# push 1 on the other stack
{			# while x:
(
({})			# copy x to the 3rd stack
<(			# push the
{			# running total of
<>({})<>		# top of the other stack
<({}[()])>		# (while decrementing x)
}
{}			# pop redundant 0
)<>{}>
[()]
)			# push x-1
}
<>


Try it online!

1

It's interesting that you don't always end up on the same stack at the end. Here's my own 52 byte answer if anyone else is interested.

– Jo King – 2018-05-22T06:26:01.733

2

Mathematica, 17

f = Times@@Range@#&


works more or less instantaneously for f[125].

1##& is 1 byte shorter than Times – LLlAMnYP – 2017-01-19T10:01:47.127

2

Ruby, 35 characters,

def f(x);p 1.upto(x).inject(:*);end

Test:

f(5)

=> 120

2

Keg, 16 bytes

(:|:1-)_(!1-|*).


This takes a top-of-stack item and returns [0..top]. Then, it discards the 0. After that, it multiplies everything in the stack, returning the factorial. (This is indeed too long.)

Nicely golfed, I gave my own shot at this but was only able to get 18 bytes, +1

– EdgyNerd – 2019-08-10T11:38:14.447

I'm getting an error running that code on TIO. – pppery – 2019-09-25T20:38:57.290

Run this in an old version. (Works for most of my Keg answers) – None – 2019-09-25T22:42:37.413

2

Julia - 14 characters (19 with non-arbitrary-precision input)

f(n)=prod(1:n)


If you want it to work all the way up to n=125, precision becomes an issue. If requiring the input value to be "big" to match the output is unacceptable, then an extra 5 characters can be used to overcome the problem:

g(n)=prod(1:big(n))


big(n) converts n to an arbitrary precision integer, and the code remains in arbitrary precision from there. The alternative is, with the 14 character code above, making the input arbitrary precision - for instance, calling f(big(125)).

And it works fast enough even for ridiculously high n. For n=100,000 my old i5-2410 laptop needs only 10.3 seconds. Displaying the 456,574 digits of the result is kind of a problem, though ;)

At first I didn’t see that there already was a Julia solution. I would have almost made a double post. – M L – 2015-06-28T07:44:26.333

2

Bash+coreutils, 21 bytes

seq $1|paste -sd\*|bc  2 Sage, 19 bytes For some reason, Guido hates prod(). But, Sage supports it: f=lambda n:prod(1..n)  edit: just had a statement previously, not a function 2 PowerShell, 34 {($p=1)..$_-ge1|%{$p*=$_};$p}


Creates a list of numbers from one to the argument, selects those greater than or equal to one and multiplies those. For 0 the list 1, 0 will be created where then only 1 remains, yielding the correct answer.

To test:

> &{($p=1).."$args"-ge1|%{$p*=$_};$p} 125 1,88267717688893E+209  It's just a scriptblock; i.e. a function without a name. 2 C#: 37 int f(int n){return n>0?n*f(n-1):1;} 2This function returns 1 always. – Alexandru – 2011-07-04T11:15:41.327 whoopsies, you're right. – Origamiguy – 2011-07-10T12:38:07.400 3int is way too small to hold 125!, which is something like 1.88e+209. – Justin Morgan – 2011-07-12T19:05:14.647 2 Pico, 23 f(x):if(x=0,1,x*f(x-1))  but Pico max out at 12: >f(12) 479001600  Does it work with 125? – user unknown – 2012-01-11T02:45:06.683 1@userunknown No, it doesn't as Pico's number type has the same limits as C's int. and fixing it so that it could would require implementing big integer multiplication and that would require at least 500-4000 characters. – Dan D. – 2012-01-11T09:55:04.753 2 Scheme - 33 characters Improved answer using an unnamed procedure and the λ symbol. (λ(n)(if(= 0 n)1(* n(!(- n 1)))))  Old 40 character answer below (define(! n)(if(= 0 n)1(* n(!(- n 1)))))  The white-space requirement is almost as much of a problem as the brackets for bloating things in scheme. Testing: > ((λ(n) (if(= 0 n)1(* n(!(- n 1))))) 0) 1 > ((λ(n) (if(= 0 n)1(* n(!(- n 1))))) 125) 188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000  what about removing the space between define and (! n) – jkabrg – 2015-10-13T21:49:11.520 Thanks NaN - can actually remove a lot of spaces so 47 down to 40. – Penguino – 2015-10-13T22:12:06.060 Doesn't work for me (the lambda one) - ! is undefined (running in DrRacket 6.4) – kronicmage – 2016-06-09T13:30:53.127 2 R, 22 (9 w/o function def; 35 w/ recursion) Simply f=function(n)prod(1:n)  Or, without defining a function: prod(1:n)  Or, recursive: f=function(n)if(n<2)1 else n*f(n-1)  1Only the last will work entirely (i.e. give 1 for n=0). A shorter solution would be : ifelse((n=scan())>0,prod(1:n),1), with 32 bytes. – Frédéric – 2016-08-26T16:11:06.597 @Frédéric or better yet "if"((n=scan())>0,prod(1:n),1) for 30 bytes. – Giuseppe – 2017-08-24T17:34:19.113 1@Giuseppe 0^(n=scan())+prod(1:n) is 22 bytes. – Robin Ryder – 2019-10-31T07:33:13.383 2 CJAM 9 I'm pretty sure this mmets all requirements. It ran on the online compiler for 125 is far less than a second. 1ri,{)*}%  It works as follows: 1 puts 1 on stack ri accepts input as integer , creates list of all non negative integers less than input { start block increments integer by 1 multiplies current product by integer, current product starts with 1 } repeat block for each element in list  Can you remove the -? It's throwing off the leaderboard snippet, thinks you're at negative 9 bytes. – Pavel – 2017-01-19T02:10:20.647 ri,1f+:* is even shorter – kaine – 2014-11-04T22:01:01.583 2 Ruby, 22 bytes n.downto(1).inject(:*)  n.downto(1).inject(1,:*) – histocrat – 2015-07-24T17:12:13.063 Or in fact (1..n).inject 1,:* seems to work fine. – histocrat – 2015-07-24T17:14:07.243 Unfortunately, this doesn't work for 0, so you probably have to add a ternary operator. – Martin Ender – 2014-11-29T20:05:38.383 2 Python, 30 bytes f=lambda n:n*f(n-1)if n else 1  Saves some characters by using lambda syntax and a ternary if-else. This is very similar to the Python answer on the first page, and doesn't really add anything. – lirtosiast – 2015-06-22T14:51:37.017 Firstly, there are many Python answers. Secondly, which answers are on which page is dependent on how you sort the answers. Thirdly, even if my answer doesn't add anything super cool or unique, it's still different enough for me to post it as my own. Because it IS my own. I created it without reading the other answers first. – mbomb007 – 2015-06-22T18:32:37.457 1There are five Python answers; two of them are exactly yours except that the authors used and/or rather than ternary or forgot to use lambda. If I had this solution, I would post it as an improvement comment on those answers due to similarity, or not post if there is no improvement. – lirtosiast – 2015-06-22T18:46:08.137 2Using and/or instead of ternary is pretty different in Python for this challenge. Your feedback is appreciated, but I'm not removing my answer. This answer was posted 5 months ago and was fine. – mbomb007 – 2015-06-22T18:51:31.753 2 Golfscript, 10 chars: ~,{)}%{*}*  2 Haskell, 20 Gee, I sure hope folds don't count as built in functions... f n=foldl(*)1[1..n]  product is probably shorter – Mega Man – 2019-08-05T18:15:07.277 2 Swift: 43 var f:Double->Double;f={$0<1 ?0:f($0-1)*$0}


Swift is definitely not made for conciseness, but I though let's give it a try anyways. This solution is obviously recursive.

There's also the more native Swift way to do it which is a bit longer (57 characters):

let f={stride(from:1,through:$0,by:1.0).reduce(1){$0*$1}}  If it would be allowed to add an additional rule for very typey languages: • You may add functions with the same behaviour as functions in the standard library for the purpose of shorter names Then I would declare these two: func ...(lhs: Double, rhs: Double) -> StrideThrough<Double> { return stride(from: lhs, through: rhs, by: 1) } extension SequenceType { func r<T>(initial: T, @noescape _ combine: (T, Self.Generator.Element) -> T) -> T { return reduce(initial, combine: combine) } }  which redeclares stride(from: a, through: b, to:1.0) to a...b which I even think should be in the standard library, and reduce(a, combine: f) becomes r(a, f). This would one enable to do this: let f={(1...$0).r(1,*)}


which would be 23 characters.

I'm even thinking about creating a Code Golf Swift extension, which just redeclares all the standard methods to something more concise.

Any of those can be called like:

f(0) // 0
f(120) // 6.689502913449124e+198
f(170) // 7.257415615307994e+306


All of them can go up to 170, where the result will be Double.infinity when above.

The times are as follows (for input 170):

recursive (43 chars): 0.00000101 s
native (rule-bend)  : 0.00000027 s
native              : 0.00000027 s


Why is it var f:Double->Double? Isn't that declaring the type twice? var seems to declare it to be of "variable" type, whereas Double->Double seems to declare it to be a function that accepts a double and returns a double. If this is not the case, can I get an explanation of the type signature for someone who doesn't know swift? – Cyoce – 2016-02-05T06:21:19.310

@Cyoce If I was using let instead, the compiler would complain because I'm using the function itself in its definition without it being declared beforehand. By using var we can trick the compiler into thinking that f is declared already. – Kametrixom – 2016-02-05T13:30:42.640

ok, thanks for the clarification – Cyoce – 2016-02-05T15:56:24.743

2

JavaScript ES6 21 chars

Note: I'm just extending Casey Chu's answer using ES6 with minor change

f=(n)=>n<2?1:n*f(n-1)


fiddle: Factorial

1) Does not use any built-in functions

2) Calculates factorial up to 170

3) Calculates factorial for 0 too

4) Execution time is less than a millisecond

Note: It will work only in browsers that supports ES6. FF(22+) and Chrome(45+) supports arrow functions as per MDN at the time of writing this answer.

You don't need parens around the argument. f=n=>… is fine. – Cyoce – 2016-01-27T03:34:53.053

2

Simplex v.0.5, 12 bytes

(The Docs page may be outdated; mainly, the * also increments the pointer and the J being the max of two elements.)

h*M{*LTRpM}]


This defines a macro that performs the factorial function on the current byte. It maintains the structure of the strip, but inserts an extra 0 at the next byte. You can delete this by adding another command p before the function ends.

This one works for inputs on a strip whose sole member is the input. I.e., a strip which looks like [N,/,/,...] (/ is the empty or null bit.) It clocks in at…

11 Bytes!!

This beats the GolfScript entry, FYI.

h{*M}pwT1J]


This is what it does:

h{*M}pwT1J]
h         ] ~~ define new macro
{  }       ~~ repeat inside until zero met
*         ~~ copy the current byte and increment pointer
M        ~~ decrement byte
p      ~~ remove trailing zero
wT    ~~ spreads T (multiplication) across strip backwards; sets pointer to after the result
1J  ~~ Takes the maximum of 1 and the current byte


Here the non-destructive version being used in an example code:

h*M{*LTRpM}p]ih0o


This defines the macro, asks for numeric input (i), calls the first macro (h0) and outputs the byte as a number (o).

Here is the pseudo-code I used:

Function factorial(N)
A = N - 1
While A > 1
N = A * N
A = A - 1
End While
Return N
End Function


This is the expanded explanation.

h    ~~ open macro, implicit [
*   ~~ A=N [N,A]
M   ~~ A=N-1 [N,A-1]
{   ~~ Loop until current byte is zero
*  ~~ [N,A-1,A-1]
LT ~~ [N*(A-1),0,A-1]
Rp ~~ [N*(A-1),A-1]
M  ~~ [N*(A-1),A-2]
}
p   ~~ [N!]
]    ~~ close macro


2

Python, 29 bytes

f=lambda x:x and x*f(x-1)or 1


2

Burlesque, 4 bytes

Burlesque has a built-in ?! to do that, but since that is forbidden by the rules we can just use ropd (runs in less than a fraction of a second):

blsq ) 125ropd
188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000
blsq ) 5ropd
120
blsq ) 125?!
188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000


Basically factorial is just the product of a list [1..N] and ro creates [1..N] and pd is the product of a list. Simple as that.

Does this account for 0!=1? – AdmBorkBork – 2015-11-23T20:16:40.280

1Yes, because pd (Product) is defined to be 1 for empty lists. – mroman – 2015-11-24T09:37:33.530

2

JavaScript, 52 bytes

function f(m){n=1;for(i=1;i<=m;i++){n*=i;}return n;}


You can shorten that too function f(m){for(n=i=1;i<=m;)n*=i++;return n}. It's 6 character in less. – HoLyVieR – 2011-02-07T03:57:01.773

2

JacobFck, noncompeting

43 bytes. This answer is noncompeting, because the language was invented after the challenge was posted.

Might be a bit late but this is too good to pass up.

<^0|=_s~$t$c:m^1^c|=_e-$c^t*$t_m:e^t>!:s^1>


Here is the commented and expanded: here

2

Detour (non-competing), 5 bytes

?1RP.


Try it online!

?1 means "if n is 0, set n to 1"
RP means product [1..n], . is output

Terminates in 6ms for 170 (the highest number whose factorial can be represented in JS) on my craptop 4-year-old macbook air with 2GB RAM.

Here's a 100% symbolic method:

Detour, 10 bytes

[{<]?1}&*.


Try it online!

Old recursive way:

Detour, 1713 11 bytes

<Q0\
.$;p>P  Try it online! This is non-competing, as I just finished the language today. There's no good way to explain it, the website will give a visualization of the data flow at runtime. It's a shame I have to handle 0!=1, or this could be a one-liner. Another 11-byte solution (faster): Detour, 11 bytes ?1[$Q<]x
P.


Try it online!

@CᴏɴᴏʀO'Bʀɪᴇɴ as a matter of fact I am working on non-essentials such as permalinks and speed sliders now. :) – Cyoce – 2016-01-21T04:21:19.713

– Conor O'Brien – 2016-01-21T04:30:31.703

@CᴏɴᴏʀO'Bʀɪᴇɴ done. – Cyoce – 2016-01-21T05:11:25.833

2

Python - 33

f=lambda n:n*(n and f(n-1))+(n<1)


– lirtosiast – 2016-01-24T23:25:31.940

2

K, 9 bytes

f:*/1f+!:

k) f 125
1.882677e+209


Computes 125! in under a millisecond; 15ms for 10k iterations

Can you add a link to the language? – Mego – 2016-01-25T14:48:25.993

@Mego - it's https://en.wikipedia.org/wiki/K_(programming_language) I've just realised this is equivalent to Kona...

– Simon Major – 2016-01-25T14:59:51.737

2

05AB1E, 4 bytes (non-competing)

Since the language postdates the challenge, this is non-competing

Code:

L0KP


Explanation:

L     # Create the list [1, ..., input]
0K   # Remove all occurencies of zero
P  # Calculate the product


I have no idea why this works for 0, but it does.

Why does the list in the range [1,N] for N=0 becomes [0].. I did know taking the product of [] becomes 1. Both seem to be the ideal situation for the 0 edge-case here.. xD – Kevin Cruijssen – 2018-08-09T14:04:55.223

1Ý¦P is one byte shorter. – Grimmy – 2019-05-29T13:28:58.953

2

Pyth, 8 bytes

It's a shame that, 5 years after the challenge has been posted, there is no pyth answer. So I'm doing it now, even if it's ridiculous :). BTW, this is non-competiting, since the language is newer than the challenge...

Lu*GHSb1

You call it with yx, where x is a number. Test it here !

Explanation

Lu*GHSb1
L          Defines a lambda 'y' with argument 'b'
Sb    Create a range from one to 'b' (function argument)
*GH      Lambda function that takes two arguments and multiply them
u     1   Reduce the range with the above lambda.


L*F+1Sb is a bit shorter. – FryAmTheEggman – 2016-04-09T19:08:25.697

@FryAmTheEggman What kind of sorcery is this ? – FliiFe – 2016-04-09T19:38:47.493

F is fold, basically a reduce over the list, it expands pretty much into what you did, but without the start at 1 thing. So I manually add a 1 to the list Sb creates, which of course won't change the product. – FryAmTheEggman – 2016-04-09T19:40:36.763

@FryAmTheEggman I still have a lot to learn... – FliiFe – 2016-04-09T20:20:24.977

2

Common Lisp, 38 bytes

Disclaimer: I'm not responsible for any traumatic effect caused by the extreme density of parentheses. It's the language specification's fault.

(defun a(b)(if(< b 2)1(*(a(- b 1))b)))


Ungolfed & explained:

(defun a (b)                               ;Define a function called "a". It has one parameter called "b"
(if (< b 2)                    ;If b is a number that is smaller than 2 (0 and 1 satisfy this)
1                   ;Return 1
(* (a (- b 1)) b)));Otherwise, return a(b-1) multiplied by b


Hi, your solution does not handle the input value '0'; you should modify the code as follows: (defun a(b)(if(= b 0)1(*(a(- b 1))b))). – PieCot – 2016-06-06T21:47:59.850

You can save a byte by using (1- b) instead of (- b 1) – djeis – 2017-04-12T16:40:13.397

1

Pushy, 3 bytes

Non-competing as the language postdates the challenge.

RP#


Explanation:

R  \ Push the inclusive range of the input
P  \ Push the product
#  \ Print


Try it online!

Factorial built-ins aren't allowed, sorry. – ETHproductions – 2017-01-19T20:16:00.360

Is that in the question? I must have missed it. – None – 2017-01-19T20:20:06.900

You can do RP# which gets the range and prints the product. – FlipTack – 2017-01-19T20:23:11.460

Thanks for using Pushy, but as it postdates this challenge you should mark the answer as non-competing (and maybe include a TIO link) – FlipTack – 2017-01-19T20:24:05.803

I'm a bit hazy on the time Pushy was made – None – 2017-01-19T20:31:11.713

@JackBates late 2016. definitely not around 2011, when this challenge was made. – FlipTack – 2017-01-19T22:00:41.993

1

Python, 25 bytes

f=lambda x:x<2or x*f(x-1)


Try it online!

This is a recursive lambda. It returns True if the factorial is 1 (inputs 1 and 0), but that's allowed by meta.

1

Alice, 13 bytes

/o
\i@/r.1~&*


Try it online!

Explanation

This is a basic framework for arithmetic programs to read and write integer I/O and process them in Cardinal mode:

/o
\i@/...


As for the actual computation:

r    Range: Replace the input N with 0, 1, 2, ..., N.
.    Duplicate N.
1~   Put a 1 underneath the copy to initialise the product correctly
for N = 0.
&    Repeat the next command N times.
*    Multiply (N times, multiplying up the entire stack).


1

Powershell, 38 bytes

filter f{((1..$_-join'*'|iex),1)[!$_]}


I used some of the other answers here for inspiration, but I'm not lower than @Joey. Although, I'm not sure how their code knows to stop subtracting once it hits 0...

PS C:\> 0|f
1
PS C:\> 4|f
24
PS C:\> 125|f
1.88267717688893E+209


1

Pyth, 15

K1FNr1hQ=K*KN;K

Try it here!

If you only need one variable, it’s better to use K or J which don’t need an equals sign to be assigned to on their first use

1

Pyth, 7 Bytes

u*GhHQ1


Try it online!

This is pretty much simple. Pyth is a good choise for code golfing.

1

Pyth, 3 bytes

*FS


Try it here.

the S appears to no longer be necessary – hakr14 – 2018-09-12T18:20:33.457

1

K (oK), 5 bytes

Solution:

*/1+!


Try it online!

Explanation:

Interpretted right-to-left:

*/1+!   / solution
!   / til, performs range, 0..n
*/      / multiply-over elements in list


Notes:

Unlike the K/Kona/Q implementations, the input is treated as a float:

oK)@5
-9
q)type 5
-7h


1

dc, 23 22 bytes

1r[dk*K1-d0<a]dsax_1*+


Explanation

1r[dk*K1-d0<a]dsax_1*+  # input on stack, eg:     4
1                       # push 1:                 1 4
r                      # reverse top two:        4 1
[dk*K1-d0<a]          # push [string]:          [string] 4 1
dsa       # copy top to register a: [string] 4 1
x      # exec top*               0 24
_1    # push -1:                -1 0 24
*   # multiply:               0 24

# first exec of [string]:

[dk*K1-d0<a]  # stack:                4 1
d            # duplicate top:        4 4 1
k           # pop & set precision:  4 1
*          # multiply              4
K         # push precision        4 4
1        # push 1                1 4 4
-       # subtract              3 4
d0<a   # if top > 0 exec content of register a, namely [string]
# output on stack:      24


In case of 0 after the exec part (*), the stack will be -1 0:

x       # exec top*               -1 0
_1     # push -1:                -1 -1 0
*    # multiply:               1 0


1

Pyt, 1 2 bytes

řΠ


Try it online!

Implicit input
ř creates a range from 1 to input, taking care of the 0 edge case
Π multiplies everything together
Implicit output


Builtin is not allowed. – Weijun Zhou – 2018-02-22T03:16:00.870

@WeijunZhou Ah yes, I will update, thank you – FantaC – 2018-02-22T14:55:11.293

1@Weijun Zhou and downvoters, please retract, I have taken out the builtin – FantaC – 2018-02-22T15:01:25.783

1

C++11 (35 chars)

Here's the function version:

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


C++11 template version (103 chars)

And here's the template version:

template<int I>struct f{static const int v=I*f<I-1>::v;};template<>struct f<0>{static const int v=1;};


1

APL (13)

∇R←F X
R←×/ιX
∇


May need a ⎕IO←1 line to be sure ι starts at 1 - it's been awhile since I last used APL.

⎕IO←1 is default in many APLs. Also, you can save 3 bytes: Remove the ∇ and the last line break, giving the ⎕CR instead of the ⎕VR. Typo: * should be ×. The former is Power: n*2 = n². – Adám – 2016-01-25T14:35:31.690

1

Golfscript — 16

{.!+,{(}%{*}*}:f


The way I handle 0! is to do this trick: .!+:

• 0 + 0! = 0 + 1 = 1
• a + a! = a + 0 = a (for every a != 0)

or:

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


Here, I start of by increasing the argument by 1. But before I factor the array, I drop the last element.

1

PHP, 41

function f($i){return$i==1?:$i*f($i-1);}


1

AsciiDots, 805753494847 43 bytes

/{*}<$#& \<^\*#1\ \1#~*{-}-\ /#1/\*.>#?) .  Outgolfs the sample by 34 57 61 65 66 67 71 bytes. Try it online! 1 Flobnar, 19 bytes |\@<: 1&:: <>-* 1  Try it online! An interesting sibling to Befunge. This uses the -d flag to enable decimal input. Explanation: This is basically equivalent to the recursive function: def f(x): if x == 0: return 1 else: return x*f(x-1)  The program starts from the @, and evaluates to its left. The \ evaluates underneath itself and stores that in the call stack. Underneath it is the &, which gives decimal input. If EOF has been reached, it evaluates underneath itself. Underneath, the > pushes the pointer right and returns the top value of the call stack minus 1. That's this section:  \@ &: >- 1  The \ basically counts down from the input and stores the latest version in the call stack. After that, it hits the |, which evaluates further along, and returns either the value above it if the evaluated value is non-zero, else the value below it. The : returns the top of the call stack, so if it is zero, the | goes down and returns 1, else it goes up and wraps around the field. It hits the < which wraps it around again, and then multiplies the top of the call stack by the next iteration of the function. | <: 1 : < *  You beat me to this answer and probably outgolfed any attempt I might have made! Looks like the -d flag is misdocumented as -e - that should be fixed soon. – Esolanging Fruit – 2018-08-07T20:41:47.813 1 µ6, 14 bytes #+[#.[#/0[+/1]/1/2][+/0]/1]  Try it online! Explanation # -- primitive recursion with + -- | base case: successor (of 0) [ -- | compose #.[#/0[+/1]/1/2] -- | | multiplication [+/0] -- | | successor of first argument /1 -- | | second argument ] -- | : f n * f (n-1)  1 Powershell, 21 byte 1..$_-ge1-join'*'|iex


Test script:

$f = { 1..$_-ge1-join'*'|iex
}

@(
0,1,2,3,4,5,6,125
) | % {
&$f$_
}


Output:

1
1
2
6
24
120
720
1.88267717688893E+209


1. 1..$_ generates a sequence of integer from 1 to argument (sequence=(1,0) if argument equal to 0) 2. -ge1 leaves in the sequence numbers great or equal to 1 3. -join'*' converts the sequence to string with '*' between elements 4. |iex evaluates string as powershell expression 1 PARI/GP, 16 bytes n->prod(i=2,n,i)  The shortest answer would be the native ! which is disallowed. 1 Muriel, 128 bytes C:"\";@%(B+\");B:\\\"\"+B+\"*\"+$a+\"\\\";a:\"+$(a-1)+\";C:\\\"\"+|C+C),(a>0)*(&B+2),a*999+&B+2";@"B:\".$(1\";a:"+~+";C:\""+|C+C


Try it online!

This is a difficult for a number of reasons. First is Muriel itself, where the only way to do loops is to create a quine and eval it, passing a condition into the code. Next is that large numbers lose precision over time (so the output for $$\125!\$$ is 1.88267717688893e+209), but the program can't handle that format inside the program itself, so you can't pass large numbers to the next iteration of code. The last problem is that numbers will eventually be so large, the program just renders them as Inf, but thankfully that's far beyond $$\125!\$$, so we don't have to worry about that.

Explanation:

The first iteration doesn't have to be a complete quine, but can take shortcuts in constructing the actual loop.

C:"\";@%(B+\");B:\\\"\"+B+\"*\"+$a+\"\\\";a:\"+$(a-1)+\";C:\\\"\"+|C+C),(a>0)*(&B+2),a*999+&B+2";


This creates an string version of the loop. This string persists across all iterations of the loop.

@                   # Evaluate the next strings as a new program
"B:\".$(1\"; # Initialise B as the string ".$(1"
a:"+~              # Initialise a as the inputted number
+"C:\""+|C         # Initialise C as the persistent string C
+C                 # And add the executing part of the program


The resulting program with input 125 looks like (newlines added for clarity)

B:".$(1"; a:125; C:"\";@%(B+\");B:\\\"\"+B+\"*\"+$a+\"\\\";a:\"+$(a-1)+\";C:\\\"\"+|C+C),(a>0)*(&B+2),a*999+&B+2"; @%(B+");B:\""+B+"*"+$a+"\";a:"+$(a-1)+";C:\""+|C+C),(a>0)*(&B+2),a*999+&B+2  The eval then executes @ # Evaluate %( ... ),(a>0)*(&B+2),a*999+&B+2 # The substring B+"); # If a is 0, evaluate B # Otherwise B:\""+B+"*"+$a+"\";                      # Concatonate "*a" to the end of B
a:"+$(a-1)+"; # Decrement a C:\""+|C # Set C to C +C # And add the executing part of the program  Eventually, a will reach 0, and B will be evaluated. B will look like: . # Print$(       ...         )   # The string form of
1*125*124*123*...*1    # The factorial


1

CJam, 10 bytes

qi1{_(j*}j


Explanation:

qi1{_(j*}j
qi            e# Read input as integer
{    }j    e# Define a recursive function
1           e# Where the value for f(0) = 1
_         e# For f(i), duplicate i
(        e# Then subtract 1
j       e# Run the function for this number (i-1)
*      e# Multiply i and f(i-1) together
e# Implicit output


Try it online!

1

MathGolf, 3 bytes

╒ε*


Try it online!

Explanation

╒    create 1-based range [1, ..., input]
ε*  Reduce by multiplication


1

Clam, 9 7 bytes

p;#qB1Q


-2 bytes thanks to ASCII-only

Explanation

p;#qB1Q - Implicit Q = first input
p       - Print...
;      - Product of...
B1Q - Range(1...Q) OR Range(Q...1) if (Q < 1)
#q    - Where (q => q) ie (q != 0)


get it on tio or GH pages pls. also 7: p;#qB1Q. alternatively, p;#>q0B1Q – ASCII-only – 2019-05-03T06:47:35.297

@ASCII-only waiting for Dennis to pull the latest version – Skidsdev – 2019-05-03T13:03:46.607

argh bad paste, was p|;B1Q1 or something – ASCII-only – 2019-05-03T13:09:15.997

1

><>, 14 bytes

1$:@?!n$:1-@*!


Try it online!

1

Julia 1.0, 12 bytes

n->prod(1:n)


Try it online!

1

"??"-14"?"-1"?"*-1" "-1" "

Try it online! Takes input from command-line arguments.

This file contains unprintables. A hexdump is provided below:

00000000: 2201 1604 043f 0703 3f22 2d31 3422 0b02  "....?..?"-14"..
00000010: 3f1c 222d 3122 0116 1303 013f 1202 222a  ?."-1".....?.."*
00000020: 2d31 2216 0120 222d 3122 0112 2005 22    -1".. "-1".. ."


Ungolfed, the program looks like this:

; Jump to 28 if N = 0
0.   deref 22 4
3.   bool  ?  7
6.   mul   ?  -14 11

; R = R * N
14.  deref 22 19
17.  mul   1  ?   18

; N = N - 1

25.  deref 32 -1

; Print R
28.  deref 18 32
31.  numo  ?


There are two variables: N, which is located at position 22 and initialized to the input integer, and R, which is located at position 18 and initialized to 1.

1

8088 / 8087 machine code, 14 bytes

More trivial iterative solution:

 D9 E8      FLD1                    ; start with 1
E3 0A      JCXZ +10                ; if 0, do nothing
51         PUSH CX                 ; push counter onto stack
8B F4      MOV  SI, SP             ; use stack memory for N
F_LOOP:
89 0C      MOV  WORD PTR[SI], CX   ; N = counter
DE 0C      FIMUL WORD PTR[SI]      ; ST =  ST * N
E2 FA      LOOP F_LOOP             ; keep looping
59         POP  CX                 ; restore stack


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

8088 / 8087 machine code, 28 bytes

Original recursive solution:

        FACT PROC
50          PUSH AX                 ; push input to top of stack
D9 E8       FLD1                    ; load initial 1
E8 0112     CALL FACT_H             ; start recursion
FACT_H:                     ; recursive helper
55          PUSH BP                 ; save BP
8B EC       MOV  BP, SP             ; point BP to top of stack
8B 46 04    MOV  AX, [BP+4]         ; AX = N
48          DEC  AX                 ; decrement N
7C 08       JL   DONE               ; if N = 0, end recursion
50          PUSH AX                 ; save N on stack
E8 0112     CALL FACT_H             ; recursive call
9B          FWAIT                   ; synchronize CPU and FPU
DE 4E 04    FIMUL WORD PTR[BP+4]    ; accumulate result in ST(0) * N
DONE:
5D          POP  BP                 ; restore BP
C2 0002     RET  2                  ; remove N from stack and return
FACT ENDP


A recursive solution (at the machine code level) using only the 8088 CPU, and the 8087 FPU to calculate the factorial sum at 80-bit double-extended precision.

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

Test output displayed using Borland Turbo Debugger 3:

n = 1:

n = 15:

n = 50:

n = 125:

Notes:

• Calculates $$\{125!}\$$ in an imperceptible amount of time (difficult to accurately profile in DOS)
• Specifically targeted to 8088 CPU (even includes FWAITs) to run on an original IBM PC (with FPU installed)
• Language is not newer than the challenge

1

Julia, 14 characters

f(n)=prod(1:n)


although I think recursive implementations are more interesting for this challenge. The best I could do there was 18 characters:

f(n)=n<1||n*f(n-1)


for n = 125, note that one has to use a BigInt like this:

julia> f(big(125))
188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000


Both complete in much less than a second.

1

Wren, 34 bytes

Generate range from a to 1 and then reduce it with product of this whole sequence.

Fn.new{|a|(a..1).reduce{|a,b|a*b}}


Try it online!

1

@, 8 bytes

#*¨1^ň


Explanation

     ň Input a single number from STDIN
^  Increment this number
¨1   Exclusive range from 1 to this number
#*     Fold multiplication over this vector


1

Julia - 17

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


This defines !n as !(n-1)*n if n>1, 1 otherwise. To make it work with big numbers you just need to make "n" a BigInt type (build in Julia).

And if its permitted (13 chars.):

!n=gamma(n+1)


with gamma equals to:

In the particular case that z its an integer the gamma function would be equal to:

Like its not a build in factorial it must not break the rules, but Im not posting it as solution just in case it does.

1

JavaScript (ES6) - 17 Characters

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


Or:

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


JavaScript - 17 Characters (not a function)

for(a=1;n;)a*=n--


Assumes that the variable n contains the number you want the factorial for and outputs the answer to the console and stores it in the variable a.

@WallyWest yes, JavaScript has only one numeric type, Number. It is not arbitrary precision, but it can hold up to 170! Before overflow, at which point it is said to be Infinity. JS is weird?, but it is actually helpful in this case. – Cyoce – 2016-02-05T06:51:55.143

Wow... I just looked at the similarities between my code and yours. Ours are basically the same. – Drew Christensen – 2016-06-08T19:30:51.097

But will this provide the full numeric value of 125!? – WallyWest – 2014-11-05T01:37:37.343

1

TI-BASIC, 65 14 bytes

For(I,0,Ans:IAns+not(Ans:End


Tricky tricky... +not(Ans and loop starting from zero should handle the special case. seq( isn't as viable of an option for that reason. TI-Basic only supports up to 10^100 which makes 70! and above fail, but it's easy to see that this solution would extend indefinitely.

What's wrong with For(? – Khuldraeseth na'Barya – 2017-09-09T02:01:56.120

@lirtosiast I apologize for the poor quality of the previous edit, which I fixed. – Timtech – 2017-09-09T13:33:44.547

@Scrooble Nothing, I just didn't understand TI-Basic as well in Apr '14. – Timtech – 2017-09-09T13:34:08.113

1

Mathematica – 46 characters

f[x_]:=Integrate[(x+1)^(t-1)Exp[-x-1],{t,0,∞}]


This is using the integral definition of the Gamma Function.

I like this solution! – lambruscoAcido – 2014-09-09T10:45:51.157

1

Python - 74

Python without any library calls. Not the shortest but calculate factorial of 125 in a flash (< 1 sec)

def factorial(n):
if n < 1:
return 1
else:
return n * factorial(n-1)


Output of 125 factorial

188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000


@Sieg Neither of your suggestions works. They both always output 1. – mbomb007 – 2015-01-21T20:39:05.167

@mbomb007 I goof when I'm tired. Try f=lambda n:n and f(n-1)*n or 1 – seequ – 2015-01-21T20:46:25.060

@Sieg I came up with my own in a new answer. It has the same length as yours. f=lambda n:n*f(n-1)if n else 1 – mbomb007 – 2015-01-21T20:47:55.930

Because the shorter version in the answer did not actually run, I rolled back the answer to a previous revision.

– mbomb007 – 2018-04-27T21:48:40.260

For some reason my indentation is not getting preserved. – Ricardo A – 2014-06-06T03:59:59.367

Someone indented it for you, and then I reduced the indentation to only a single space per indent. I also added a character count. – Rainbolt – 2014-06-06T21:45:10.393

1You can save characters with def factorial(n):return 1 if n<1 else n*factorial(n-1) – Rainbolt – 2014-06-06T21:46:57.857

Also, in Python, 0 is equivalent to false when used as a condition, so you could literally say if n: instead of if n < 1: – Rainbolt – 2014-06-06T21:49:44.447

1You seem to have missed the point of the challenge, as this does not seem the least bit golfed, i.e., written such that the code has as few characters as possible. – Wrzlprmft – 2014-06-06T22:11:46.233

I wrote it so it is readable as well. Sure I could have called my function n and since python relies on indents jamming it all on one line will not work. Is the point to make it as short as possible even if it will not work? Serious question. I am new at this. – Ricardo A – 2014-06-06T22:21:57.100

@Ricardo The point is to write the shortest working program/function. – seequ – 2014-06-06T23:20:54.737

Also, this can be written in one line: f=lambda n:f(n-1)if n else 1 or even f=lambda n:n and f(n-1)or 1 – seequ – 2014-06-06T23:24:11.600

Also use @nicknamehere to answer to someone. – seequ – 2014-06-06T23:26:22.033

@TheRare Thanks. Then your way would have been the shortest since python relies on the indents. – Ricardo A – 2014-06-11T00:33:09.117

Check my answers and you won't think so anymore. – seequ – 2014-06-11T07:17:42.983

1

><>, 18 22

Launch with -v number for inputting the argument, or put it before the one.

Now also handles 0, some more intelligent direction usage, and some more space for putting numbers up to ff* or 225:

   1&:?\&n;
:-1&*&:/?=0


Old version

 1&>:&*&\
;n&\?-1 /


1

JavaScript, 41

function(n,r){for(r=1;n;r*=n--);return r}


or 39 if globals are okay.

1

Ruby: 22 characters

n.downto(1).reduce(:*)


1

JAVA

I rarely see Java solutions here. Why is that?

    public static void main(String[] args)
{
int tot = 1;
for(int i = 1;i<=5;i++)
tot *= i;
System.out.println(tot);
}


2This is [tag:code-golf]. With barely any work at all, you can significantly reduce the length by removing unnecessary whitespace and using 1-letter variable names – Cyoce – 2016-02-05T06:30:30.950

Java's int is incapable of holding 125!, a requirement of the challenge. A Java program must use BigInteger or something similar to work. – None – 2017-03-15T17:22:28.267

Yes, and it can calculate the factorial for 0. Put the factorial value in the loop continuation condition. i.e 5 – Mob – 2011-08-06T11:17:20.503

Java's a pretty verbose language, so it's not great for getting the lowest character count. – Gareth – 2011-08-06T13:34:14.467

@Gareth Yeah, but Brain Fuck isn't right? – Mob – 2011-08-07T19:13:59.433

You asked why you rarely see Java solutions here - it's because Java's verbose and less likely to win at code-golf. That's not to say there are no Java solutions, or that people shouldn't post Java solutions - they're just rarer for that reason. – Gareth – 2011-08-07T20:06:41.820

You could try to transform your Javacode to Beanshell. – user unknown – 2012-01-11T02:46:59.613

1

Racket, 29

(λ(x)(apply * x(range 1 x)))


@overactor It creates an anonymous function, creates a list from 1 to the given number, and applies multiplication on all of it. – Mega Man – 2019-08-05T18:23:35.737

– EdgyNerd – 2019-11-26T15:57:04.103

Can you explain how this solution works? Not everyone is familiar with Racket. – overactor – 2014-09-09T10:15:46.073

1

Scala, 39

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


1

Ruby, 19

[1,*2..n].inject :*

The extra hardcoded 1 at the beginning makes it work for when n=0.
Ruby auto-converts to BigInt after a certain point, so it has 100% accuracy.

1

f:{(*/)9h$1+til x}  Computes in less than one millisecond. q)\t f 125 0  f:{prd 1f+til x} for 16. f:{prd 1f+(!)x} for 15. – streetster – 2017-09-13T07:44:28.163 1 Powershell, 31 $a=1;$args[0]..1|%{$a=$_*$a};$a  usage powershell -nologo .\fact125.ps1 0 0 powershell -nologo .\fact125.ps1 1 1 owershell -nologo .\fact125.ps1 5 120 powershell -nologo .\fact125.ps1 125 1.88267717688893E+209  This doesn't account for 0!=1. You can use the Invoke-Expression command to evaluate on-the-fly, and then use bool casting to select the appropriate answer -- try param($a)$b=1..$a-join'*'|iex;($b,1)[!$a] for 41 bytes – AdmBorkBork – 2015-11-23T20:13:33.323

1Actually, we can move the |iex and skip the $b entirely -- 37 Bytes for param($a)((1..$a-join'*'),1)[!$a]|iex – AdmBorkBork – 2015-11-23T20:38:24.827

1

Bash/coreutils/dc, 25

dc<<<"1 seq -f%g* $1p"  This forms a dc script and evaluates it. So ,with input of 5, we evaluate 1 1* 2* 3* 4* 5*p  It took my machine 2.05 seconds to compute 10000! here (that's factorial ten-thousand, with 36693 digits), so seems to scale reasonably well. For the zero case, seq produces no output, so the dc script is just 1 p which produces the correct output 1. 2 bytes shorter using only dc ;) – ბიმო – 2017-11-26T00:09:05.110 1 PlatyPar, 8 bytes c?1,_p\1  Explanation: c? ## if (n != 0) 1,_p ## product [1..n] \ ## else 1 ## 1  This does not seem to handle the special 0 case. – Mama Fun Roll – 2016-01-26T04:04:19.493 @ՊՓԼՃՐՊՃՈԲՍԼ oops, I completely forgot. Adding... – Cyoce – 2016-01-26T04:46:09.937 1 , 9 chars / 19 bytes (noncompetitive) +!ï⋎⨴⩤⁽1ï  Try it here (Firefox only). Ay, 19th byte! Great thing about this is that it also calculates factorials up to 171 instantly without returning Infinity. Bonus solution! +!ï⋎⨴МĂ⩤⁽1ï  Try it here (Firefox only). This one allows you to calculate past 171 without getting Infinity. Still superbly fast! Mind explaining how this gets around Infinity? :) – ETHproductions – 2016-01-27T18:30:53.780 МĂ is math.js's bignumber function. – Mama Fun Roll – 2016-01-27T23:07:52.900 1 JavaScript, 34 bytes function f(n){return n?n*f(n-1):1}  (or) function f(n){return n?n*f(--n):1}  Explanation Function takes in a value, returns itself multiplied by if n != 0: the same function on the number decreased by one if n == 0: 1 The final f(0) returns first with 1, times 1, times 2, etc. Terminator removed, may upset use strict. There were already some similar answers but uses ES6. – jimmy23013 – 2016-01-26T22:42:44.123 1 Japt, 8 bytes (non-competing) This answer is non-competing because Japt was created long after this challenge. UòJ ¤r*1  Test it online! How it works UòJ ¤r*1 // Implicit: U = input integer 5 UòJ // Create the inclusive range [-1..U]. [-1, 0, 1, 2, 3, 4, 5] ¤ // Slice off the first two items. [1, 2, 3, 4, 5] r*1 // Reduce by multiplication, starting at 1. 1*1=1*2=2*3=6*4=24*5=120 // Implicit output 120  This does not handle the zero case correctly (0! should return 1). – Mama Fun Roll – 2016-01-27T03:40:13.717 @ՊՓԼՃՐՊՃՈԲՍԼ Thanks, fixed now. – ETHproductions – 2016-01-27T23:21:31.173 I know this is old but, wouldn't á be enough? – Luis felipe De jesus Munoz – 2018-08-07T17:51:25.107 @LuisfelipeDejesusMunoz I think you mean l, but yes :-) – ETHproductions – 2018-08-08T15:30:45.797 1 Lua, 47 bytes function f(n)return(n<1 and 1 or n*f(n-1))end  1 PHP Short, 58 <?$r=$i=$argv[1];while($i>1){$i--;$r=$r*$i;}echo($r==0?1:$r);  Tests 0 -> 1 1 -> 1 5 -> 120 125 -> 1.8826771768889E+209 170 -> 7.257415615308E+306 171 -> INF Executes in microseconds. Ungolfed <?php$r = $i =$argv[1]; // Set $r and$i to Arg.
while($i > 1) // Calculate while$i bigger than 1
{
$i--; // Decrement$i (so it's not infinite)
$r =$r * $i; // Calculation the Factorial } echo ($r==0 ? 1: $r); // Output and make 0! = 1 ?>  Slighty Longer, 86 <?$r=$i=(isset($argv[1])?$argv[1]:0);while($i>1){$i--;$r=$r*$i;}echo($r==0?1:$r)."\n";


Improvements

• Output with \n
• Doesn't throw error if no arg defined

1

DUP, 19 bytes

[$[$1-a;!*][%1]?]a:


Try it here!

A recursive lambda that leaves result on the stack. Usage:

6[$[$1-a;!*][%1]?]a:a;!


Explanation

[               ]a: {set a to lambda}
${check if top of stack >0} [ ][ ]? {conditional}$1-a;!*          {if so, top of stack *a(top of stack -1)}
%1      {otherwise, replace top of stack with 1}


1

Hoon, 29 bytes

|=(@ (reel (gulf [1 +<]) mul)


Hoon's native number is a bignum, so it works fine with 125 (or even 2000). It also correctly gives 1 for 0.

It uses +< in order to access the sample of the gate. This is axis navigation syntax: It means to access the tail of the subject, and then the head, which is where the sample is stored in the binary tree model Hoon uses.

Urbit drops you into a shell and Hoon REPL when you start it, :dojo. To test this, simply enter %. 125 on one line and then the snippet for 125! Note there are two spaces between the dot and 1.

Hoon is a beautiful mystery. – Lynn – 2016-03-01T01:10:42.253

1It's a surprisingly nice language to code in! It takes a little bit to learn all the runes, but you don't even need to internalize them to read it since they belong in "families" based on the first symbol. The fact it's strongly typed and has a novel type system is just icing on the cake. – RenderSettings – 2016-03-01T01:36:34.347

1

Yup, 33 31 29 bytes

*{{:0e-}]~{~|~|0~--e~}~#\}0e#


Here's the github. Invoke like this:

node yup.js <location>.yup -n <input>


Or

node yup.js -l "*{{:0e-}]~{~|~|0~--e~}~#\}0e#" -n <input>


Examples:

λ node yup.js -l "*{{:0e-}]~{~|~|0~--e~}~#\}0e#" -n 5
120
λ node yup.js examples\factorial.yup -n 0
1


Explanation

*{{:0e-}]~{~|~|0~--e~}~#\}0e#
*                               take input
{                       }      while TOS -- if zero, we advance to the }
0e#   print number 1 (exp(0))
 otherwise (nonzero)
{    }                        while TOS is not zero
:                            duplicate TOS
0                           push 0
e                          pop 0, push exp(0) = 1
-                         subtract 1
 we eventually are at zero.
]                       we move that zero to the bottom of the stack
~                      switch top two for looping offset
{~        ~}          while STOS
|~|0~--e            multiply two elements (see further down)
~         switch the top zero with the result
#        print the result
\       exit program (so we don't print the final one)


Multiplication

In this program, I have multiplication defined as thus:

|~|0~--e


First, observe 0~--. This pushes a zero behind the TOS, and subtracts twice:

command | stack
| a b
0       | a b 0
~       | a 0 b
-       | a (-b)
-       | a - (-b) = a + b


This performs addition. Let's replace 0~-- with + for clarity:

|~|+e


Now, | is ln. So watch the stack:

command | stack
| a b
|       | a ln(b)
~       | ln(b) a
|       | ln(b) ln(a)
+       | (ln(b)+ln(a))
e       | exp(ln(b)+ln(a))


And, by the theorem of logarithms, this is multiplication.

1

Desmos, 18 bytes

a=1
\prod _{n=1}^an


Uses the formula for a! instead of a!

1

Joy, 13 bytes

[1][*]primrec


30 char requirement in codegolf?

@FryAmTheEggman To make a function you would actually have to write DEFINE f==[1][*]primerec.. I believe program arguments are thrown on the stack when the program starts, and that everything outside a define block is executed – BlackCap – 2016-06-08T20:12:34.643

1It'd be best if you knew for sure rather than just believing ;) Anyway, seems alright then, but you also wouldn't run in to that 30 character requirement if you gave some explanation of your code :P – FryAmTheEggman – 2016-06-08T20:16:02.580

1

><>, 17 16 bytes

1v;n
$<*}-1.!?::  -1 byte thanks to Jo King. Since the question asks for a function as opposed to a full program, I allowed myself to accept the input from the stack without counting an additional 3 bytes for using the -v option. This manages to be shorter than the other ><> answer because it jumps to the end-of-iteration code without having to hardcode the jump destination address : the current iteration counter (duplicated) is used as an address. The iteration stops when the counter is 0, and jumping to (0, 0) while the direction pointer points to the right will execute the n; code that is otherwise unreachable, displaying the result and stopping the execution. It handles 0! correctly and executes in 10*(n+1) ticks for n > 0 or 9 ticks for n = 0. You can try it online. If you run the bottom line to the left so that the pointer is moving left when it is transported, You remove the need for the second line. -1 byte 1v;n \n $<*}-1.!?:: – Jo King – 2017-12-20T05:41:28.663

Don’t forget you can remove the ; as well! – Jo King – 2017-12-21T12:41:11.507

1

Maple, 17 bytes

n->*(seq(1..n))


Usage:

> f:=n->*(seq(1..n));
> f(0);
1
> f(5);
120
> f(125);
188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000


1

Mathematica

f = If[# > 0, # f[# - 1], 1] &
f[125] = 188267.....


1

Fourier, 18 bytes

Non-competing: Fourier is newer than the challenge

1~NI(i^~iN*i~Ni)No


Try it online!

1

Oasis, 3 bytes

Try it online

n*1


Explanation:

n*1
n    Push n
*   Multiply the two items on the top of the stack
Because there is only one item on the stack, A(n - 1) is pushed
Implicit output
1  Special case A(0) = 1


0

proc f n {expr $n?($n)*\[f $n-1]:1}  Try it online! 0 Bash w/Core Utils & BC, 15 Bytes Saw 2 other bash solutions, but both longer than this one: seq -s* dd|bc  number to factorial for from stdin, factorial to stdout 0 Recursiva, 12 bytes =a0:1!*a#~a$


Try it online!

Explanation:

=a0:1!*a#~a$=a0:1 - If a==0 return 1 ! - Else * - Multiply a - a #~a$ - Call self but with parameter a-1


0

Axiom, 42 31 bytes

h x==(x=0=>1;product(i,i=1..x))


The 42 bytes one

s(x)==(x=0=>1;reduce(*,[j for j in 1..x]))


There could be even the 37 bytes

f(x)==(x=0=>1;reduce(*,expand(1..x)))


but there is one warning when I use it the first time

0

Brainfuck, 109 Bytes

>>>,[[<<<+>>+>>+<-]>-[<+>-]<<<<-[>>[>[>+>+<<-]>>[<<+>>-]<<<-]<<-[>+>+<<-]>[<+>-]>>[-]>[<+>-]<<<<]>>[->+<]<<]+


The factorial is left on the tape at position 4, user input is taken as an ASCII character. Meets all requirements except for completing in under a minute.

How it works:

>>>,


Gets user input and stores it in position 4. The rest of the code is in brackets to account for an input of zero. In that case, none of the following code is executed, and it skips to the last command.

[<<<+>>+>>+<-]


Move the input number to positions 1, 3, and 5.

>-[<+>-]


Decreases the number at 5, and moves it to 4.

<<<<-


Decrease the number at 1. This serves as a counter for how many times we need to multiply

This is where most of the work occurs:

[                | While 1 is non-zero (n-1 times)
>>[            | Move to 3, and while it's non-zero
>[           | Move to 4
>+>+<<-    | Copy it into 5 and 6
]>>          | Move to 6
[            |
<<+>>-     | Copy it to 4
]<<<-        | Move to 3, decrease it
]<<-           | Move to 1, decrease it (one multiplication has been done)
[              |
>+>+<<-      | Copy the value at 1 to 2 and 3
]>             | Move to 2
[              |
<+>-         | Put it back in 1
]>>            | Go to 4
[-]            | Zero it
>[             | Go to 5
<+>-         | Copy 5 to 4
]              |
<<<<]            | Go back to 1


For values >1, we could end here, but for 1:

>>[->+<]<<]


No multiplications will have been done, and we'll have a 1 in position 3. If this is the case, move it to 4. Return to 1.

]+


If the input was 0, put a one in the current position (4, since we never moved)

– Jo King – 2017-11-24T09:51:59.587

0

Javascript (No body version), 23 bytes

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

Saved 2 bytes with n?n, got the idea from Drew Christensen. This is what I had before

f=(n,t=0**n||n*f(n-1))=>t

0

4, 37 bytes

3.70060101602018002010100100000295014


Try it online!

If you question the input method, please visit first the numerical input and output may be given as a character code meta post.

Pseudo code

g_0 = input()
g_1 = 1
while g_0 != 0:
g_1 *= g_0
g_0 -= 1
print(g_1)


0

Java (JDK 10), 84 bytes

n->{var r=java.math.BigInteger.ONE;for(;n>0;)r=r.multiply(r.valueOf(n--));return r;}


Try it online!

Credits

1You can save 5 bytes by using Java 9's var and removing the r=null initialisation: n->{var r=java.math.BigInteger.ONE;for(;n>0;)r=r.multiply(r.valueOf(n--));return r;} – mgthomas99 – 2018-09-11T21:45:06.897

2@mgthomas99 Indeed! But it's Java 10's var, not Java 9's ;) and it wasn't released at the time I wrote that answer. But I have no issue switching to Java 10. – Olivier Grégoire – 2018-09-12T07:48:43.680

0

Forth (gforth), 31 bytes

: f 1e 0 ?do i 1+ s>f f* loop ;


Try it online!

Answer will be on the floating point stack as 125! exceeds the maximum double precision integer value (by a lot)

0

Stax, 4 bytes

┤c5ô


Run and debug it online

Unpacked version with 5 bytes:

!x+k*


Explanation

!x+      Compute logical not, then add to itself
The result is denoted m
k*    reduce with multiplication over the range [1..m]


0

*($n)  Computes n!, no recursion. 0 Gol><>, 8 bytes 1IFLP*|h  Try it online! How it works 1IFLP*|h 1 Push 1 I Take input (n) as int F | Repeat n times... LP* Multiply loop counter(L) + 1; L = 0..n-1 h Print the top as int and halt  Function form, 8 bytes Assuming that stack input and stack output are allowed for Gol><> functions. 1$FLP*|B


Try it online!

Takes the stack as input (assuming the stack has only one value), and leaves the factorial as the only content of the stack. $ is used to push 1 to the bottom, and B is the "return" command. Note that B inside an F (for) or W (while) loop acts as a "break" instead. Using the Gol><> function The following code uses the function above to output 0! to 125!. 1AG~FLGN|; 1$FLP*|B

1AG          Register row 1 as function G
~        126
F   |   Repeat from 0 to 125...
Stack is empty at this point
L        Push loop counter
G       Call G on the current stack
N      Pop and print as number, with newline
;  Halt


0

Python 3 - 52 characters

r=n=int(input())
for i in range(1,n):r=r*i
print(r)


This is my best try. My C++ solution (not posted) was over 100 characters even without #includes and whitespace!

2Can't you do r*=i instead of r=r*i ? – Cyoce – 2016-01-21T07:05:23.997

You may discard prompt string in input – AMK – 2014-01-07T00:23:02.877

0

Clojure, 29

#(reduce * (range 1 (inc %)))


0

dc, 22 bytes

[1q]sg[d2>gd1-d2<f*]sf


Try it online! (requires Javascript)

This is the obvious 12-byte implementation ([d1-d1<f*]sf) with an additional test to short-circuit the calculation for numbers less than 2 (which otherwise subtract too far, yielding a product of zero).

0

Japt, 4 bytes

oÄ ×


Try it

Explanation

o        :Range [0,input) (= empty array when input is 0)
×     :Reduce by multiplication, with an initial value of 1


0

Symbolic Python, 36 bytes

__('__=_==_'+';__*=_;_=~-_'*_)
_=+__


Try it online!

Can compute 125! with ease.

Works using an 'exec' pseudo-loop:

• First, we set __ to True, which is equivalent to 1.
• We multiply this by _ (the input), and then decrement _.
• The second step is repeated _ (input) times using string multiplication in the __ (exec) function's argument.

0

C#, just the relevant code, 59

(assuming the argument variable is called a)

Enumerable.Range(1,int.Parse(a[0])).Aggregate(1,(x,y)=>x*y)


With boilerplate, 122

using System.Linq;class A{static int Main(string[] a){return Enumerable.Range(1,int.Parse(a[0])).Aggregate(1,(x,y)=>x*y);}


(note that this solution returns the result)

0

PHP, 40 39 bytes

<?=array_product(range($argv[1]?:1,1));  Thanks to @JoKing for saving 1 byte and correcting the input Try it online! @JoKing changed the answer to reflect – EvanBarbour3 – 2019-02-19T12:05:21.493 0 Kotlin, 50 45 bytes fun f(x:Int):Int=when(x){0->1;else->x*f(x-1)}  Thanks to @jonathan-frech 1Welcome to PPCG. I think the 1->1; clause is superfluous. – Jonathan Frech – 2019-02-19T14:22:26.153 0 Excel Formula, 41 bytes The following should be entered as an array formula (Ctrl+Shift+Enter): =IFERROR(PRODUCT(ROW(OFFSET(A1,,,A1))),0)  Where A1 contains the value for n. The IFERROR is just there to handle n=0. 0 Python 2, 52 bytes I'm posting this purely because it uses reduce :) f=lambda x:x<1or reduce(lambda a,b:a*b,range(1,x+1))  Try it online Note: returns True for n<1. I assume this is ok because True == 1 0 NuStack, 55 bytes f(n:int):int{r:int=n;while(n>1){n=n-1;r=r*n;}return r;}  Naive while-based approach 0 Swift - 53 Characters func f(_ x:Double)->(Double){return(x<2 ?1:x*f(x-1))}  I'm sure it can be improved using closures...still investigating. 1 Welcome to PPCG. There is already a Swift answer in 43 bytes. (It doesn't mean you can't post. But you may want to have a look at the existing answers in the same language first.) – jimmy23013 – 2019-05-29T00:10:02.237 Welcome to the site! Since you are golfing in Swift you might want to check out our tips for golfing in swift question. There is one of these for most other languages as well. – Post Rock Garf Hunter – 2019-05-29T02:29:46.030 Thanks for the prompt feedback - much appreciated! – L Rettberg – 2019-05-30T18:07:04.780 0 Pepe, 60 bytes rrEERREeeeeeeeErEeREeEreEREErEEEEErEEEeeReererRrEEEEEEeEreEE  Try it online! 0 C, 37 bytes long double f(n){return!n?1:n*f(n-1);}  This program accepts one number from stdin and then prints the answer in floating point form. Hello, and welcome to Code Golf! This is a great answer, but to reduce the byte count even more, why not remove the extra spaces? – NoOneIsHere – 2019-08-03T22:23:28.753 Oh, so you only want the nonmain function. Alright. I will keep that in mind for next time. – T. Salim – 2019-08-03T22:33:49.053 You are right Jo, I fixed it. – T. Salim – 2019-08-04T00:05:55.890 Thank you A__, I removed the parentheses. – T. Salim – 2019-08-04T02:41:20.387 No, I tried to simply use long but the computer outputted 0 due to overflow, so I kept it as long double. – T. Salim – 2019-08-04T02:45:06.447 long double f(n){return!n?1:n*f(n-1);} returns the exactly same result. – None – 2019-08-04T02:47:07.803 Wow, how are you that good identifying these errors. Is there any site on codegolf that tutors people these clever shortcuts. Thanks for the advice once again. – T. Salim – 2019-08-04T02:49:24.043 There's the Tips for golfing in C page. You can also remove the (accounting spaces) part, since you're meant to count whitespace anyway. – Jo King – 2019-08-07T05:58:18.240 0 Pip, 13 bytes Fi,a{o*:i+1}o  Try it online! Recursive factorial is given in Pip documentation, so I did iterative. Also, it couldn't handle up to 125! anyway. 0 Keg, 95 4 bytes Ï⑨∑*  Try it online! -1 byte thanks to @A̲̲ Answer History 5 Bytes Ï_1∑*  Try it online! This: • Takes the (Ï)ota of the implicit input (pushes input, input - 1, input - 2 ... 0) • Pops the bottom 0 • Pushes an extra 1 (so that the 0 case works) • Multiplies the entire stack together 1Maybe simply increment the 0 of the stack? – None – 2019-11-24T04:10:43.337 0 W, 5 bytes *R:e+  Explanation *R % Reduce by the product :e % Is this item 0? + % Add this value to the product % We need to also calculate fact(0) % Implicit output  0 C 20 characters x(){while(n)f*=n--;}  Assuming f and n are global variables. Here is the entire program : double n=5,f=1; x(){while(n)f*=n--;} main(){ x(); printf("%f",f); }  0 C 37 characters double f(int n){return n?n*f(n-1):1;}  This returns the value but is slightly longer than my previous answer which used global variables. 0 the minimum solutions is already given using C# lamada. But just try to this another way.  var seq = Enumerable.Range(1, 5).ToList(); int O=1; //O will contain Factorial or ouput seq.ForEach(x=>O*=x);  0 C#, 30 double a=1;while(p>0){a*=p--;}  With spaces for ease of reading: double a = 1; while(p > 0) { a *= p--; }  a is the factorial result, while p is the number for which factorial is computed. Couldn't you just say "while(p)statement;" instead of "while(p>0){statement;}"? Unless C# behaves differently from C, it should work. – Glenn Randers-Pehrson – 2014-04-01T18:59:48.397 if p is bool, it would. Here I need to continue checking p with 0... – s3l1n – 2014-04-02T05:20:55.360 0 Python 2.7.5 - 29 characters f=lambda n:n and n*f(n-1)or 1  29 characters. It's still mathematically sound for a negative input value, since (-n)! = ∞ and therefore the program gives a Runtime Error maximum recursion depth exceeded. 0 F# based on cfern's 63 36 characters His didn't work on 125 for me. Adapted to use BigInteger let f n:BigInteger=Seq.fold(*)BigInteger.One{BigInteger.One..n}  Edit: I just realized that double works too. let f n:double=Seq.fold(*)1.{1.0..n}  0 Simplefunge, 87 chars including whitespace v v *&< > &V  & v < o ^H^ @ v>>!1-^ >iV >1o@  I don't actually have time to test this right now, but it should work. If it doesn't work, I'll fix it tomorrow. 0 Scala, 28 (39 w/ recursion) Solution: def f(n:Int)=(1 to n)product  Recursive solution: def f(n:Int):Int=if(n<2)1 else n*f(n-1)  0 Befunge-93, 24 22 0\0>-#1:_$> #\:#*_$:!+  As with the other Befunge answer(s?), this is a function in that you enter with your input on the top of the stack, start on the top-left with your pointer facing right, and exit still facing right with the result on top of the stack. The difference is that this one is shorter, and only one row. And it still doesn't use wraparound or self-modification. You can test it here. Testing six factorial: 6 0\0>-#1:_$> #\:#*_$:!+ .@  Outputs 720. Testing zero: 0 0\0>-#1:_$> #\:#*_$:!+ .@  Outputs 1. The :!+ (formerly :0!+) does at the end does nothing but check for zeroes. EDIT: A suggestion golfed two characters off. Thanks! 1You don't need the 0\ part,!will return the correct result directly. So instead your implementation should look like that0\0>-#1:$> #:#*$:!+ – FliiFe – 2016-04-07T11:14:22.020 Ah! Thank you. I'll change that now. – Kasran – 2016-04-09T17:10:21.253 0 R - 33 function(x) ifelse(x,prod(1:x),1) > (function(x) ifelse(x,prod(1:x),1))(0) [1] 1 > (function(x) ifelse(x,prod(1:x),1))(5) [1] 120 > (function(x) ifelse(x,prod(1:x),1))(120) [1] 6.689503e+198  0 PHP, 13 Might sound like cheating, but in PHP it's just: gmp_fact($n);

Will get you 100% precision, but it won't always be fast, especially for larger numbers.

Does not use any built-in libraries that can calculate the factorial – user unknown – 2012-01-11T02:44:24.313

3Ah, so it is cheating. – Mr. Llama – 2012-01-11T15:13:12.223

0

Python, 43 38

import math
f=lambda n:math.gamma(n+1)


Explanation: The gamma function is a very quickly-growing complex function which, at integer values, is equal to the factorial of one less than the number. So we add one to n and take the gamma function of it.

I hope this isn't considered cheating, since the gamma function is not technically able to directly calculate the factorial.

1To quote Wikipedia, "In mathematics, the gamma function is an extension of the factorial function, with its argument shifted down by 1, to real and complex numbers". Some would say it's "close enough" to a factorial function that it shouldn't count, I personally don't care because it's longer than the other Python answers. – Kevin Brown – 2015-10-14T17:49:18.440

0

Clojure/ClojureScript, 26 bytes

#(apply *(range 1(inc %)))


0

R 27 Bytes

function(n)prod(seq_len(n))


0

Y, 10 bytes

Try it here!

jC:t:tF|*p


j C :t :t F |* p


j takes numeric input, C begins a new link, :t duplicates and decrements to [n n-1]; the :t:t becomes [n n-1 n-2]; the last n-2 is popped for a zero-check by the F node; once a zero is found, we continue. |* folds the stack over multiplication, and p prints it.

0

ForceLang, 83 bytes

Noncompeting, language postdates the challenge

set a io.readnum()
set b 1
label l
set b b.mult a
set a a+-1
if a
goto l
io.write b


0

Perl 5, 31 bytes

$s=1;map{$s*=$_}(2..<>);print$s


Prints the result in scientific notation, and takes care of the 0 value as well.

Another way to do it, but without the 0 case, for 26 bytes :

print eval join'*',(1..<>)


26 bytes: $\=1;map$\*=$_,2..<>;print. Or 21 + 3 (flag -l61): map$\*=$_,2..<>;print. – Denis Ibaev – 2016-11-10T20:29:13.400 21 bytes: map$.*=$_,2..<>;say$. – Xcali – 2017-11-15T23:15:54.770

0

Pyke, 2 bytes (non-competing)

SB


Try it here!

   - implicit input
S  -  range(1, input+1)
B - product(^)


0

JavaScript ES6, 19 17 bytes

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


Saved two bytes by changing the conditional to n instead of n>1, because they are effectively equal.

Ungolfed

var factorial = function(n) {
if (n > 1) {
return f(n-1)*n;
} else {
return 1;
}
}


Works just like a standard factorial. Defines a function f which multiplies it's input n by f(n-1). If n is equal to 0 or 1, then the function returns 1.

See it in Action

Check it out on JSFiddle!

Yes, exactly the same as my answer from 2 years previously.

– MT0 – 2016-06-08T20:06:34.517

0

Mathcad byte-equivalence system yet to be determined. Some operators cannot be entered as text but have keyboard "shortcuts" instead (or can be picked from a toolbar). For example, the product operator is inserted using the key combination ctl-# ; this results in a capital Pi symbol being placed upon the Mathcad worksheet, together with 4 associated placeholders (black rectangles) which hold the iteration variable, starting value, final value and the expression for each element of the product. Balanced parentheses can be entered (in most cases) using the quote key. Typing : enters the assignment operator :=.

Mathcad has both a standard IEEE numerical floating point processing system and a longnum symbolic processing system that run in parallel over the same worksheet. = is the numeric evaluation operator whilst → is the symbolic evaluation operator.

0

Java, 51 bytes

class a{double A(double b){return b<1?1:b*A(b-1);}}


I'm actually glad Coding Ground can handle more stack frames than what is needed to overflow a double with this.

I'd +1 if the result was precise for n=125, but it is not (also, not that I reduced your byte count to 39).

– Olivier Grégoire – 2017-12-24T11:09:15.453

0

Neoscript, 31 bytes

{n|(0:(]:n):reduce({x y|x*y}1)}


0

Python (31 29 character)

f=lambda n:n and n*f(n-1)or 1


28 characters

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


As I see no Python solution present – AMK – 2012-11-24T13:49:55.977

The first page has a Python solution with only 28 characters. – Kevin Brown – 2012-11-24T15:36:04.910

0

Logy, 47 bytes

f[X]->include["@stdlib.logy"]~product[..[2,X]];


Trivial. Return the product of the range [2..n]`