Code Golf: Prime Factors of a Number [closed]

Answer 1

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]

Answer 2

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.

Answer 3

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.

Answer 4

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))

Answer 5

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.

Answer 6

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));}