The vanilla factorial challenge

Given a non-negative integer $$\n\$$, evaluate the factorial $$\n!\$$.

The factorial is defined as follows:

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

Rules

• All default I/O methods are allowed.
• Standard loopholes are forbidden.
• Built-ins are allowed.
• There is no time or memory limit.
• Giving imprecise or incorrect results for large inputs due to the limit of the native number format is fine, as long as the underlying algorithm is correct. Specifically, it is not allowed to abuse the native number type to trivialize the challenge, which is one of the standard loopholes.
• This is . Shortest code in bytes wins, but feel free to participate in various esolangs (especially the ones hindered by the restrictions of the former challenge).

Test cases

0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600

Note: We already have the old factorial challenge, but it has some restrictions on the domain, performance, and banning built-ins. As the consensus here was to create a separate challenge without those restrictions so that more esolangs can participate, here it goes.

Also, we discussed whether we should close the old one as a duplicate of this, and we decided to leave it open.

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GolfScript, 9 bytes

1\~,{)*}/

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1\         # Puts 1 under the input, this will be the acumulator
~,       # Makes an array with numbers from 0 to (n-1)
{)*}   # This block goes to the top of the stack without being executed, when executed it increments and multiplies, this avoids multiplying by 0 and also multiplies by n
/  # Executes the previous block for each number in the array
• Welcome to Code Golf! Sep 15 '20 at 22:41

Flurry-nii, 14 bytes

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

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Takes single number from the stack and prints its factorial from the return value.

The online interpreter implements certain arithmetic shortcuts, so it computes and prints 125! in an instant.

How it works

Iterate through 1 to n using the stack height and multiply all of them to the starting value of 1.

// n is the only content of the stack at program start
// 1 is popped from empty stack
main = pop push-mul pop
= n push-mul 1

// height yields 1 to n, since it is called after a push
// <a b> = a * b (where a, b are Church numerals)
push-mul = \x. <(push x) height>

Labyrinth, 16 bytes

?+1#*
(: (;
@!;

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Utilizes two loops, one to duplicate and decrement the input until zero, then one to multiply all the items on the stack.

Assembly (MIPS, SPIM), 78 bytes, 6*9 = 54 assembled bytes

main:li$2,5 syscall li$4,1
f:beqz$2,g mul$4,$4,$2
sub$2,1 b f g:li$2,1
syscall

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This is a significant optimization of this answer, but the account was deleted.

Specifically, this makes the following changes:

• It compares once at the top of the loop, using an unconditional branch at the bottom
• It uses the numerical register names which are shorter
• It removes spaces between mnemonics, as SPIM accepts this
• It removes the second operand from sub

Otherwise, the logic is identical.

Phooey, 15 bytes

=>&.[@<*>-1]<$i Try it online! =>&.[@<*>-1]<$i # stack tape
=               #   (0)  >a    0     acc = 1
>              #   (0)   a   >0     move left
&.            #   (0)   a   >n     read int
[      ]    #   (0)   a   >n     loop while n != 0
@          #    n    a   >n         push n
<*>       #   (0)   a*n >n         multiply acc by n (popping from stack)
-1     #   (0)   a*n >n-1       subtract 1 from n
<   #   (0)  >res 0      go right to the result
$i # (0) >res 0 print as integer Since Phooey uses int64_t, this supports up to 20. Factor + math.factorials, 6 bytes [ n! ] Try it online! Thanks to @Bubbler for -9 bytes Factor + math.factorials, 15 bytes [ factorial . ] Try it online! • There's n! which is a synonym of factorial, and you don't need to print the result, so simply n! is a valid built-in submission. (You don't need to wrap it in a quotation, as a built-in submission is scored by its name.) Mar 31 '21 at 3:34 Vyxal, 1 byte ¡ Try it Online! Built-in factorial. No built-in, 2 bytes ɾΠ Explanation: # Implicit input ɾ # Range [1, N] Π # Reduce by multiplication # Implicit output Try it Online! jq, 10 bytes .+1|tgamma Uses the gamma function Γ(n) = (n - 1)! Try it online! • I think |ceil is not necessary since floating point imprecision is allowed. Sep 2 '21 at 23:50 Keg, -hr, 4 bytes Ï_∑* Try it online! SNOBOL4 (CSNOBOL4), 65 bytes i =input p =1 i x =x + 1 p =p * x output =p le(i,x) :f(i) end Try it online! Io, 20 bytes method(\,\factorial) Try it online! Seriously, 2 bytes ,! Try it online! Desmos, 2 bytes n! Try it in Desmos Finally a competitive Desmos answer that isn't graphical-output! Wolfram Language (Mathematica), 9 3 bytes -6 bytes thanks to @att #!& Try it online! • 3 bytes – att Aug 25 '20 at 5:00 MathGolf, 1 byte ! Try it online. Explanation: ! # Get the factorial (aka gamma(n+1)) of the (implicit) input-integer # (output the entire stack joined together implicitly as result) Elixir, 30 bytes Recursive. def f(n),do: n>1&&n*(f n-1)||1 Try it online! Burlesque, 4 bytes ri?! Try it online! Explanation: ri # Read integer ?! # Calculate factorial Lua, 36 bytes x=1 for i=1,...do x=x*i end print(x) Try it online! On Lua <= 5.2 this will use double, allowing big inputs to work. On Lua 5.3+ (as currently on TIO) integers are used instead, making big inputs fail. This can be worked around at cost of one byte: Lua, 37 bytes x=1. for i=1,...do x=x*i end print(x) Try it online! cQuents, 5 bytes$0:$! Try it online! The first three bytes are for 0-indexing, since by default cQuents uses 1-indexing. C# (Visual C# Interactive Compiler), 30 bytes int f(int n)=>n==0?1:n*f(n-1); Try it online! • Suggest n<1 instead of n==0 Aug 31 '20 at 9:36 PHP, 38 bytes Recursive... function f($n){return$n?$n*f($n-1):1;} Try it online! Or functional... PHP, 39 bytes fn($n)=>$n?array_product(range($n,1)):1

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• If anyone knows if there's a way to do 7.4 arrow functions with recursion I'd be very interested to know! Aug 25 '20 at 19:24

Python 2, 46 bytes

lambda n:reduce(lambda a,b:a*b,range(1,n+1),1)

How it works:

the reduce function takes in the list that contains all the numbers from 1 to n and reduces it by the multiplication function (lambda a,b:a*b) to a single number. An optional initial parameter is set in the case that n is equal to 0.

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• You can do -~b to just do range(n). You've also got a trailing space in your submission
– Jo King
Aug 27 '20 at 6:08
• If you didn't notice yet, TIO has a Code Golf submission generator. Aug 28 '20 at 2:26

f n=product[1..n]

Bonus reference: "The Evolution of a Haskell Programmer" by Fritz Ruehr

Arn, 2 bytes

↓;

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Unpacked: .fact. Passes STDIN (_) into the factorial function via the . infix

3 byte solution:

A→.

Unpacked: *\~

\      Fold with
*        Multiplication
~    1-range
_  Variable initialized to STDIN; implied

C#, 42 bytes

Using the power of fresh and new C# 9 we can achieve a stunning 42 bytes!

int f(int n)=>n<2?1:n*f(n-1);return f(10);

for C# 8 and the online example we need to add 38 bytes for a total of 70 bytes

class P{static int Main(){int f(int n)=>n<2?1:n*f(n-1);return f(10);}}

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ink, 50 bytes

==function f(n)
{
-!n:~return 1
}
~return n*f(n-1)

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Had to make it a proper function instead of just a stitch, since I actually have to use the return value (so outputting by printing is not an option - or at least not a good one).

Π

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Python 3, 30 27 bytes

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

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Kotlin, 37 bytes

fun a(n:Int)=(1..n).fold(1){a,b->a*b}

Had to use fold(1){a,b->a*b}(surprisingly enough 1 less byte than something like fold(1,Int::times)) due to a lack of a product function in the stdlib.