The vanilla factorial challenge

Question

Task

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 code-golf. 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

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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.

Answer 1

Piet + ascii-piet, 24 bytes (4×10=40 codels)

rmttlddqbDlbaltdmu?_ V??

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Similar to below, but realized that I can remove 1 >. Also, using B2->B1 = I (which is a no-op) gave A2->B2 = D (which safely removes a 0) and a color sharing for 3. Luck based golfing?

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Piet + ascii-piet, 27 bytes (5×11=55 codels)

rmttldqbdTl?dckmvfnnN TuDuu

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The logic is the same as Parcly Taxel's.

Main loop (A1 -> A10 -> B9 -> B1 -> A1):

D * 2 1 r       Mostly no-op, just pushes [2 1]
I               Take input [2 1 n]; becomes a no-op afterwards
                Invariant: [2 prod i]
d 1 -           [2 prod i i-1]
3 1 r d 1 > !   [2 i-1 prod i i<=1]
D               [2 i-1 prod i]; turn right at A2 if n <= 1
* 2 1 r         [2 i-1 prod*i]

After turn right at A2 (A2 -> E2):

x O <halt>      Discard i and print prod

The main trick here is to move the DP+ to the cell A2, and rotate the main loop accordingly. This works because most commands are no-op when there are not enough items on the stack yet.

Answer 2

TypeScript Type System, 82 bytes

type X<A,B=A>=A extends`1${infer R}`?B extends`1${infer O}`?`${X<R>}${X<A,O>}`:B:1

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Quite similar to, but independently derived from noodle man's answer. I also stole their test harness.

This is basically the same as the JavaScript code f = (a, b = a) => a ? b ? f(a, b - 1) + f(a-1) : 0 : 1. The idea here is that we have two accumulators, a and b, where b starts at the value of a. We repeatedly decrement b and add f(a-1) to the total, which results in the factorial of a.

Answer 3

Uiua, 5 bytes

/×+1⇡

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    ⇡  # create range [0..input)
  +1   # amend it to [1..input]
/×     # reduce by multiplication

Answer 4

Keg, -hr, 4 bytes

Ï_∑*

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Answer 5

Pyth, 2 bytes

*F

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Alternate solution:

.!

Answer 6

05AB1E, 1 byte

!

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Answer 7

Python 3.8, ²³ 21 bytes

Saved 2 bytes thanks to Bubbler!!!

import math
math.perm

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Answer 8

Japt, 1 byte

l

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Answer 9

Charcoal, 8 bytes

ＩΠ…·¹∨Ｎ¹

Try it online! Link is to verbose version of code. Explanation:

      Ｎ     `n` as a number
     ∨      Logical Or
       ¹    Literal `1`
  …·¹       Inclusive range from `1` to `n`
 Π          Take the product
Ｉ           Cast to string
            Implicitly print

Answer 10

Seriously, 2 bytes

,!

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Answer 11

Elixir, 30 bytes

Recursive.

def f(n),do: n>1&&n*(f n-1)||1

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Answer 12

T-SQL, 91 bytes

CREATE FUNCTION dbo.f(@ INT)RETURNS INT AS BEGIN SET @=IIF(@<=1,1,@*dbo.f(@-1))RETURN @ END

Demo

It seems you need to specify a schema (dbo.) in order to make a recursive function :( Perhaps other SQL dialects are shorter.

Answer 13

ReRegex, 47 bytes

#import math
\b1!/1/(\d+)!/($1-1)!*$1/
#input
!

Defines two regexes;

(\d+)!/($1-1)!*$1/ Looks for a base10 number followed by a !, and replaces it with (n-1)!*n.

\b1!/1/ Replaces any free standing 1!'s with just 1.

The math library then takes care of the subtractions and multiplications. ReRegex will continue attempting its regex operations until it can no-longer modify the text.

The final two lines put the input as n! ready for the operations.

Values greater than 5 work offline, but due to how long they take; do not work on TIO.

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Answer 14

Perl 5 -p, 20 bytes

$\*=$_ for++$\..$_}{

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Answer 15

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

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Moving another answer from the previous challenge, this time because the answer uses an eval function... in a language where that's the only way to loop.

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 16

COW, 150 bytes

oomMOOMOoMMMmoOmoOmoOMMMMMMMoOmOomOomOoMOOmoOMMMmoOmoOMMMOOOmOoMMMmOoMOOmoOMMMMOOmoOMoOmOoMOomooMMMmOoMOomoomOoMOoMMMmoomoOmoOmoOOOMOOOMOomOOmooMoOOOM

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Commented

Four memory blocks are used. Block 1 stores a persistent counter. Blocks 2 and 3 store the temporary loop variables used for multiplication (by repeated addition). Block 4 stores the result.

oom                           # read input into Block 1 
MOO                           # if input is not 0
  MOoMMM                        # decrement and copy
  moOmoOmoOMMMMMMMoOmOomOomOo   # paste and increment to initialise Block 4
  MOO                           # Loop 1: from Block 1 value down to 1
    moOMMM                        # set Block 2 to Block 1 value
    moOmoOMMMOOOmOoMMM            # set Block 3 to Block 4 value and zero Block 4
    mOoMOO                        # Loop 2: from Block 2 value down to 1
      moOMMMMOO                     # Loop 3: as many times as Block 3 value
        moOMoO                        # increment Block 4
      mOoMOomooMMM                  # end Loop 3
    mOoMOomoo                     # end Loop 2
  mOoMOoMMMmoo                  # end Loop 1
  moOmoOmoOOOM                  # print Block 4 value
  OOOMOomOO                     # exit
moo                           # (only get here if input is 0)
MoOOOM                        # print 1

Answer 17

Scala 2.12, 14 bytes

1 to _ product

This is a function expression, where the underscore is the parameter. 1 to _ creates a range object, and product calculates the product of all numbers.

Answer 18

Forth (gforth), 30 bytes

: f 1 swap 0 ?do i 1+ * loop ;

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Code Explanation

: f          \ start a new word definition
  1 swap 0   \ create accumulator and set up loop parameters
  ?do        \ start a conditional counted loop from 0 to n
    i 1+ *   \ multiply accumulator by loop index + 1
  loop       \ end counted loop
;            \ end word definition 

Answer 19

F#, 26 bytes, by Laikoni

fun x->List.fold(*)1[1..x]

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F#, 35 bytes

let rec f=function|0->1|x->x*f(x-1)

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Answer 20

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

Answer 21

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>

Answer 22

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.

Answer 23

Husk, 1 byte

Π

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Answer 24

ARM Thumb-2, 12 bytes

2101 b110 4341 3801 d1fc 4770

Commented assembly:

        .syntax unified
        .arch armv6t2
        .thumb
        .globl factorial
        .thumb_func
        // input: r0
        // output: r1
factorial:
        // accumulator starts at 1
        movs    r1, #1
        // skip to end if r0 is zero
        cbz     r0, .Lend
.Lloop:
        // r1 *= r0
        muls    r1, r0
        // while (--r0)
        subs    r0, #1
        bne     .Lloop
.Lend:
        // return
        bx      lr

Standard iterative approach.

Takes a 32-bit integer in r0, returns a 32-bit integer in r1.

ARM Thumb-2, 20 bytes

2201 2300 b128 fba2 2100 fb03 1300 3801 d1f9 4770

Commented assembly:

        .globl factorial64
        .thumb_func
        // input: r0
        // output: {r2, r3}
factorial64:
        // accumulator starts at 1
        movs    r2, #1
        movs    r3, #0
        // skip to end if r0 is zero
        cbz     r0, .Lend64
.Lloop64:
        // {r2, r3} *= r0
        // tmp = (u64)r2 * r0
        umull   r2, r1, r0, r2
        // {r2, r3} = tmp + (r3 * r0 << 32)
        mla     r3, r3, r0, r1
        // while (--r0)
        subs    r0, #1
        bne     .Lloop64
.Lend64:
        // return
        bx      lr

This version uses 64-bit arithmetic, at the cost of 8 bytes. It is only 2 more instructions, but two of those instructions are now wide instructions. 😭

Takes a 32-bit integer in r0 and returns a 64-bit integer in {r2, r3} (easy to forward to printf)

Answer 25

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.

Answer 26

Phooey, 15 bytes

=>&.[@<*>-1]<$i

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=>&.[@<*>-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.

Answer 27

Factor + math.factorials, 6 bytes

[ n! ]

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Thanks to @Bubbler for -9 bytes

Factor + math.factorials, 15 bytes

[ factorial . ]

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Answer 28

jq, 10 bytes

.+1|tgamma

Uses the gamma function Γ(n) = (n - 1)!

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Answer 29

Cascade, 18 bytes

#1]*
\aa?
(\&;
a ?

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Golfed version by Jo King.

A program without @ starts at the top-left corner. The main difference with my version is that a is decremented under the second branch, not first:

  ?
  ] 
 a ?
  &;(
    a

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Cascade, 22 bytes

? @
]*#1
)(|a
a?\
&;a\

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Solved as part of LYAL 2021-09-29.

How it works

Cascade programs start the execution at the entry point @.

? @   The main program prints (#) the result of the huge loop
  #   starting at `?`
  | 
  \
   \

 ?    (grid rotated once for clarity)
 ]    On entry, the if-clause is run:
a (   Push to the stack "a" the decrement of...
  ?   If EOF, the top value of "a", otherwise the input from stdin
 &;a

 ?    If the above is not positive, return 1;
1 *   Otherwise return (increment of a) times
 ) |  the next invocation of the main loop
 a \
\

Answer 30

Python 3, 25 bytes

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

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