Code Golf: Prime Factors of a Number [closed]

Answer 1

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);}}}'

Answer 2

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'));}

Answer 3

Ruby 39B 71B (via STDIN)

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

Answer 4

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.

Answer 5

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]

Answer 6

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)