Find the Factorial!

Question

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.

Answer 1

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.

Answer 2

Haskell, 17

f n=product[1..n]

Answer 3

Python - 27

Just simply:

f=lambda x:0**x or x*f(x-1)

Answer 4

APL (4)

×/∘⍳

Works as an anonymous function:

    ×/∘⍳ 5
120

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

f←×/∘⍳

Answer 5

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

Answer 6

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

Answer 7

Perl 6: 13 chars

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

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

Answer 8

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!

Answer 9

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.

Answer 10

Python, 28 bytes

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

(based off Alexandru's solution)

Answer 11

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

Answer 12

Java, 85 Chars

BigInteger f(int n){return n<2?BigInteger.ONE:new BigInteger(""+n).multiply(f(n-1));}

Answer 13

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.

Answer 14

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}

Answer 15

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

Answer 16

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!

Answer 17

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/

Answer 18

Ruby - 30 29 characters

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

Test

f(0) -> 1
f(5) -> 120

Answer 19

C (39 chars)

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

Answer 20

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!

Answer 21

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.

Answer 22

Javascript, ES6 17

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

ES6:

• Arrow function

Answer 23

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

Answer 24

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.

Answer 25

PowerShell – 36

Naïve:

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

Test:

> 0,5,125|f
1
120
1,88267717688893E+209

Answer 26

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)))

Answer 27

Befunge-93 24 22 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

Answer 28

Jelly, 2 bytes

RP

Try it online!

Alternatively, a builtin answer is one byte:

!

Try it online!

Answer 29

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

Answer 30

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. 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.

Answer 31

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!

Answer 32

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

Answer 33

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-:#^_$>\:#^_$    .@

Answer 34

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

Answer 35

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;}

Answer 36

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
;

Answer 37

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.

Answer 38

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!

Answer 39

Mathematica, 17

f = Times@@Range@#&

works more or less instantaneously for f[125].

Answer 40

Ruby, 35 characters,

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

Test:

f(5)

=> 120

Answer 41

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.)

Answer 42

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)).

Answer 43

Bash+coreutils, 21 bytes

seq $1|paste -sd\*|bc

Answer 44

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

Answer 45

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.

Answer 46

C#: 37

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

Answer 47

Pico, 23

f(x):if(x=0,1,x*f(x-1))

but Pico max out at 12:

>f(12)
479001600

Answer 48

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

Answer 49

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)

Answer 50

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

Answer 51

Ruby, 22 bytes

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

Answer 52

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.

Answer 53

Golfscript, 10 chars:

~,{)}%{*}*

Answer 54

Haskell, 20

Gee, I sure hope folds don't count as built in functions...

f n=foldl(*)1[1..n]

Answer 55

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

Answer 56

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.

Answer 57

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

Answer 58

Python, 29 bytes

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

Answer 59

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.

Answer 60

JavaScript, 52 bytes

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

Answer 61

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

Answer 62

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, 17 13 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!

Answer 63

Python - 33

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

Answer 64

K, 9 bytes

f:*/1f+!:

k) f 125 
1.882677e+209

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

Answer 65

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.

Answer 66

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.

Answer 67

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

Answer 68

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!

Answer 69

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.

Answer 70

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).

Answer 71

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

Answer 72

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

Answer 73

Pyth, 7 Bytes

u*GhHQ1

Try it online!

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

Answer 74

Pyth, 3 bytes

*FS

Try it here.

Answer 75

K (oK), 5 bytes

Solution:

*/1+!

Try it online!

Explanation:

Interpretted right-to-left:

*/1+!   / solution
    !   / til, performs range, 0..n
  1+    / adds 1 (vectorised)
*/      / 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

Answer 76

dc, 23 22 bytes

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

Try it online or verify 0-125!

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
                     +  # add:                    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
    +   # add:                    1

Answer 77

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

Answer 78

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;};

Answer 79

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.

Answer 80

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.

Answer 81

PHP, 41

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

Answer 82

AsciiDots, 80 57 53 49 48 47 43 bytes

/{*}<$#&
\<^\*#1\
 \1#~*{-}-\
/#1/\*.>#?)
.

Outgolfs the sample by 34 57 61 65 66 67 71 bytes. Try it online!

Answer 83

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  :
<  *

Answer 84

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)

Answer 85

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

Explanation

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

Answer 86

PARI/GP, 16 bytes

n->prod(i=2,n,i)

The shortest answer would be the native ! which is disallowed.

Answer 87

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

Answer 88

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!

Answer 89

MathGolf, 3 bytes

╒ε*

Try it online!

Explanation

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

Answer 90

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)

Answer 91

><>, 14 bytes

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

Try it online!

Answer 92

Julia 1.0, 12 bytes

n->prod(1:n)

Try it online!

Answer 93

Dreaderef, 47 bytes

"??"-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
10.  add   ?  28  -1

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

; N = N - 1
21.  add   *  -1  22

; Jump to start
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.

Answer 94

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
C3          RET                     ; return to caller
        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

Answer 95

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.

Answer 96

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!

Answer 97

@, 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

Answer 98

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.

Answer 99

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.

Answer 100

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.

Answer 101

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.

Answer 102

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

Answer 103

><>, 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 /

Answer 104

JavaScript, 41

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

or 39 if globals are okay.

Answer 105

Ruby: 22 characters

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

Answer 106

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);
}

Answer 107

Racket, 29

(λ(x)(apply * x(range 1 x)))

Answer 108

Scala, 39

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

Answer 109

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.

Answer 110

In Q (18 characters)

f:{(*/)9h$1+til x}

Computes in less than one millisecond.

q)\t f 125
0

Answer 111

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

Answer 112

Bash/coreutils/dc, 25

dc<<<"1 `seq -f%g* $1`p"

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.

Answer 113

PlatyPar, 8 bytes

c?1,_p\1

Try it online!

Explanation:

c?        ## if (n != 0)
  1,_p     ## product [1..n]
       \  ## else
        1  ## 1

Answer 114

, 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!

Answer 115

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.

Answer 116

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

Answer 117

Lua, 47 bytes

function f(n)return(n<1 and 1 or n*f(n-1))end

Answer 118

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

Answer 119

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}

Answer 120

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.

Answer 121

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>

or Try it online!

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.

Answer 122

Desmos, 18 bytes

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

Uses the formula for a! instead of a!

Answer 123

Joy, 13 bytes

[1][*]primrec

30 char requirement in codegolf?

Answer 124

><>, 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.

Answer 125

Maple, 17 bytes

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

Usage:

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

Answer 126

Mathematica

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

Answer 127

Fourier, 18 bytes

Non-competing: Fourier is newer than the challenge

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

Try it online!

Answer 128

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

Answer 129

Tcl, 41 37 35 bytes

proc f n {expr $n?($n)*\[f $n-1]:1}

Try it online!

Answer 130

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

Answer 131

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

Answer 132

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

Answer 133

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)

Answer 134

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

Answer 135

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)

Answer 136

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

• -5 bytes thanks to mgthomas99

Answer 137

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)

Answer 138

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]

Answer 139

MuPAD – 7

`*`($n)

Computes n!, no recursion.

Answer 140

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

Answer 141

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!

Answer 142

Clojure, 29

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

Answer 143

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).

Answer 144

Japt, 4 bytes

oÄ ×

Try it

---

Explanation

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

Answer 145

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.

Answer 146

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)

Answer 147

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!

Answer 148

Kotlin, 50 45 bytes

fun f(x:Int):Int=when(x){0->1;else->x*f(x-1)}

Thanks to @jonathan-frech

Answer 149

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.

Answer 150

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

Answer 151

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

Answer 152

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.

Answer 153

Pepe, 60 bytes

rrEERREeeeeeeeErEeREeEreEREErEEEEErEEEeeReererRrEEEEEEeEreEE

Try it online!

Answer 154

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.

Answer 155

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.

Answer 156

Keg, 9 5 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

Answer 157

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

Answer 158

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);
}

Answer 159

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.

Answer 160

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); 

Answer 161

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.

Answer 162

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.

Answer 163

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}

Answer 164

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.

Answer 165

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)

Answer 166

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!

Answer 167

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

Answer 168

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.

Answer 169

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.

Answer 170

Clojure/ClojureScript, 26 bytes

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

Answer 171

R 27 Bytes

function(n)prod(seq_len(n))

Answer 172

Y, 10 bytes

Try it here!

jC:t:tF|*p

A three-link program. Ungolfed:

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.

Answer 173

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

Answer 174

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..<>)

Answer 175

Pyke, 2 bytes (non-competing)

SB

Try it here!

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

Answer 176

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!

Answer 177

Mathcad, tbd "bytes"

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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.

Answer 178

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.

Answer 179

Neoscript, 31 bytes

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

Answer 180

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)

Answer 181

Logy, 47 bytes

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

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