Code Golf: Prime Factors of a Number [closed]

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What is the shortest way, by character count, to find prime factors in any number?

Example Input: 1806046

Example Output: 2x11x11x17x439

Example Calculator

回答1:

C#, 69

x is input number

int i=2;while(x>1)if(x%i++==0){x/=--i;Console.Write(i+(x>1?"x":""));};

With includes:

using system;
namespace nameSP
{
   class Program
   {
     static void Main(string[] args)
     { 
        int i=2;while(x>1)if(x%i++==0){x/=--i;Console.Write(i+(x>1?"x":""));};
     }
   }
}

回答2:

Obligatory J answer (2 characters):

q:

回答3:

ANSI C, 79 characters

main(d,i){for(d+=scanf("%d",&i);i>1;i%d?++d:printf("%d%c",d,(i/=d)>1?'x':10));}

回答4:

Mathematica (15 chars including brackets):

FactorInteger

Example:

FactorInteger[42]

{{2, 1}, {3, 1}, {7, 1}}

回答5:

Python: 77 chars with input and output

d,s,n=2,'',input()
while n>1:
 if n%d:d+=1
 else:s+='%dx'%d;n/=d
print s[:-1]

回答6:

Haskell, 53 chars: (including 3 newlines)

a%1=[]
a%n|mod n a<1=a:p(div n a)|1>0=(a+1)%n
p=(2%)

Example:

*Main> p 1806046
[2,11,11,17,439]

回答7:

Python (228 chars without I/O, 340 with):

import sys

def primeFactors(n):
    l = []
    while n > 1:
        for i in xrange(2,n+1):
            if n % i == 0:
                l.append(i)
                n = n // i
                break
    return l if len(l) > 0 else [n]

n = int(sys.argv[1])
print '%d: %s' % (n, 'x'.join(map(lambda x: str(x), primeFactors(n))))

Can be compressed to 120 chars:

import sys
n,l=int(sys.argv[1]),[]
while n>1:
 for i in range(2,n+1):
    if n%i==0:l+=[str(i)];n/=i;break
print'x'.join(l)

Note: That's a tab character before the if, not four spaces. It works as another level of indentation and only costs one character instead of two.

回答8:

F#

81 chars

let rec f n=if n=1 then[]else let a=[2..n]|>List.find(fun x->n%x=0)in a::f(n/a)

It's terribly inefficient, but since the aim is undoubtedly to write the shortest code possible, I've neglected that matter.

Readable form (using #light syntax):

let rec factorise n =
    if n = 1 then [] else
    let a = [2 .. n] |> List.find (fun x -> n % x = 0)
    a :: factorise (n / a)

回答9:

GNU bc, 47 chars, including collecting input (need the GNU extensions for print, else and read):

x=read();for(i=2;x>1;)if(x%i){i+=1}else{x/=i;i}

If you really want the x characters in the output, it's 64 chars:

x=read();for(i=2;x>1;)if(x%i){i+=1}else{x/=i;print i;if(x>1)"x"}

Also, note that using bc allows this to process numbers of arbitrary length.

回答10:

11 characters in APL

Excluding function header and newlines

factors←{2÷/∪⌽∧\⍵∨⍳⍵}

回答11:

Erlang, the core is 122 chars and 152 for the whole module:

-module(pf).
-export([f/1]).

f(N) -> f(N,2,[]).
f(1,_,L) -> lists:reverse(L);
f(N,P,L) when N rem P == 0 -> f(N div P,P,[P|L]);
f(N,P,L) -> f(N,P+1,L).

To call from console:

70> string:join([integer_to_list(X) || X <- pf:f(1806046)], "x").
"2x11x11x17x439"

回答12:

A Mathematica answer that actually produces the specified output:

Print@@Riffle[Join@@ConstantArray@@@FactorInteger[n],x]

55 characters. Assumes n is the input number and x doesn't have a value assigned to it.

回答13:

Ruby 39B 71B (via STDIN)

#!ruby -nrmathn
p$_.to_i.prime_division.map{|d,c|[d]*c}.flatten.join"x"

回答14:

Best Perl answer yet - 70 characters, and no extra modules (unless you count special features of 5.10):

perl -nE'sub f{($a)=@_;$a%$_||return$_,f($a/$_)for 2..$a}$,=x;say f$_'

Doesn't work for 1 or 0, but works fine for everything else. If you don't like using say, or are using an earlier version of Perl, here's an 81 character version:

perl -ne'sub f{($a)=@_;$a%$_||return$_,f($a/$_)for 2..$a;}$,=x;$/="\n";print f$_'

回答15:

Wow, you guys aren't very good at optimizing. I can do it in Perl in 63 characters, or 79 if you insist on including a #!/usr/bin/perl at the top:

use Math::Big::Factors;
@f=factor_wheel($ARGV[0],1);
print @f;

(Don't look at me that way. Committed programmers are lazy programmers.)

回答16:

While it's not my best work, here's me answer in Haskell, 83 characters.

f n = s [2..n] n
s [] _ = []
s (p:z) n = p:s [x | x<-z, mod x p /= 0, mod n x == 0] n

I'm sure there's more that could be done, but for now it's good.

Edit: Rearranged things to shave off a character, less efficient, but smaller.

回答17:

Perl, 223 characters

perl -ne'f($o=$_,2);sub f{($v,$f)=@_;$d=$v/$f;if(!($d-int($d))){print"$f ";if(!p($d)){print"$d ";return(0);}else{f($d,$f);}}else{while(p(++$f)){}f($v,$f);}}sub p{for($i=2;$i<=sqrt($_[0]);$i++){if($_[0]%$i==0){return(1);}}}'

回答18:

VB6/VBA - 190 chars

Public Function P(N As Long) As String
Dim I As Long, O As String
Do While N > 1: For I = 2 To N
If N Mod I = 0 Then
O = O & " " & I: N = N / I: Exit For: End If: Next: Loop: P = O: End Function

回答19:

Perl, 70 char

$y=<>;for($i=2;$i<=$y;){next if$y%$i++;$y/=--$i;push@x,$i}print@{$,=x}

回答20:

Euphoria: 106 characters

procedure f(atom a)atom x=2
loop do
while remainder(a,x)do
x+=1
end while
?x
a/=x
until a=1
end procedure

回答21:

VB6/VBA - 147 chars

I'm not allowed to leave comments , but it is possible to shorten the previous answer somewhat by not having Option Explicit. Taking advantage of some of the more dangerous features of VB6/VBA you can use the one below. No need to declare what the variable is and also the function doesn't need to be declared public either if going for ultimate shortness! Also the End If is not needed if it is on the same line.

Function P(N As Long)
    Dim I, O
    Do While N > 1: For I = 2 To N
    If N Mod I = 0 Then O = O & " " & I: N = N / I: Exit For: 
    Next: Loop: P = O
End Function

This can be tested by :

Public Sub TestP()
    Dim s: s = P(1806046)
    Debug.Print s
End Sub

回答22:

The Go programming language, 100 characters:

package main;func main(){n:=42;c:="x";for i:=2;n>1;i++{if n%i<1{n/=i;if(n<2){c=""};print(i,c);i--}}}

My program, with the correct indentation:

package main

func main() {
 n := 42 // or whichever input number you like
 c := "x" // separating printed integers
 for i:=2 ; n>1; i++ {
  if n%i<1 { // n%i==0
   n /= i
   if(n<2) { c = "" } // n==1
   print(i, c)
   i--
  }
 }
}

回答23:

74 75 Characters in Python

a=input();b=2
while b*b<=a:
    if a%b==0:print b;a/=b;b=1
    b+=1
print a

Derived from my TI-BASIC code for prime factorization.

Since I'm talking about TI-Basic...

77 Characters in TI-Basic

input a
2→b
while b²<a
a/b→c
if int(c)=c:then:disp b:c→a:1→b:end
b+1→b
end
disp a

回答24:

C# and LINQ, 241 Characters:

public IEnumerable<int> F(int n)
{
    return Enumerable.Range(2,n-1)
        .Where(x => (n%x)==0 && F(x).Count()==1)
        .Take(1)
        .SelectMany(x => new[]{x}.Concat(F(n/x)))
        .DefaultIfEmpty(n);
}

public string Factor(int n) {
    return F(n).Aggregate("", (s,i) => s+"x"+i).TrimStart('x'); 
}

Compressed:

int[] F(int n){return Enumerable.Range(2,n-1).Where(x=>(n%x)==0&&F(x).Length==1).Take(1).SelectMany(x=>new[]{x}.Concat(F(n/x))).DefaultIfEmpty(n).ToArray();}void G(int n){Console.WriteLine(F(n).Aggregate("",(s,i)=>s+"x"+i).TrimStart('x'));}

回答25:

C#, 366 characters

C# is not the most averbose language for something like this, but this is quite compact:

class P {
   static void Main(string[] a) {
      int i = int.Parse(a[0]);
      var p = new System.Collections.Generic.List<int>();
      for (int n = 2; i > 1; n++)
         if (p.Find(q => n % q == 0) == 0) {
            p.Add(n);
            while (i % n == 0) {
               System.Console.WriteLine(n);
               i /= n;
            }
         }
   }
}

Edit:
I saw that Noldorin used the List.Find method in his F# code, and realised that it would be a bit shorter than a foreach...

Edit:
Well, if it doesn't have to be a complete program...

C#, 181 characters

string f(int i) {
   var r = "";
   var p = new System.Collections.Generic.List<int>();
   for (int n = 2; i > 1; n++)
      if (p.Find(q => n % q == 0) == 0) {
         p.Add(n);
         while (i % n == 0) {
            r += "x" + n;
            i /= n;
         }
      }
   return r.Substring(1);
}

Compressed:

string f(int i){var r="";var p=new System.Collections.Generic.List<int>();for(int n=2;i>1;n++)if(p.Find(q=>n%q==0)==0){p.Add(n);while(i%n==0){r+="x"+n;i/=n;}}return r.Substring(1);}

回答26:

In a similar vein as Paxinum (Mathematica answer), here's one in bash:

$ factor 1806046
1806046: 2 11 11 17 439

7 characters the excluding number.

回答27:

Python recursive solution

99 characters (including spaces) 87 characters (without spaces)

def f(n,i=2,r=""):
    while n%i<1:r+="%dx"%i;n/=i
    return f(n,i+1,r)if n>1 else r
print f(input())[:-1]

Update: A completely recursive version

def f(n,i=2,x=""): return x if n<2 else f(n,i+1,x)if n%i else f(n/i,i,x+'%dx'%i)
print f(input())[:-1]

Both versions are prone to stack overflows for all but the smallest of inputs.

回答28:

In PARLANSE, this would do the trick (252 chars):

(action (procedure natural)
   (loop 
      (ifthen (== ? 1) (return))
      (do f i 2 ? 1
         (ifthen (== (modulo ? i) 0)
            (print ?)
            (= ? (/ ? i))
            (exit f)
         )ifthen
      )do
    )loop
)action

I'm sure there's a much smaller APL program to do this.

回答29:

Javascript, 56

f="";
for(i=2;i<n;i++)
    if(n%i==0){
        f+=i+"x";
        n/=i;i--
    }
f+=n;

(54 characters)

first declare n= the number to be factored (2 characters included)

then execute the code.

example:

> n=12345
  12345

> f="";for(i=2;i<n;i++)if(n%i==0){f+=i+"x";n/=i;i--}f+=n
  "3x5x823"

回答30:

Python 3 163

Without printing the result.

l,f,p=len,lambda n:list(filter(lambda b:n%b==0,range(2,n))),lambda n:l(f(n))==0;r=lambda n: n if p(n) else[x if p(x) else r(x) for x in [f(n)[0],f(n)[l(f(n))-1]]]

Use it as a function:

print(r(1806046))

来源：https://stackoverflow.com/questions/1304713/code-golf-prime-factors-of-a-number