Cryptographic hash golf (robbers)
Python, 109 bytes by Sp3000
Note that Martin cracked first, so I'm not sure if this deserves points. On the other hand, I did make a preimage attack rather than a simple collision - a much stronger result. This means that you can give it an arbitrary hash value, and it will construct an input that generates that hash value.
M = 2**128

# The hash to crack.
def jenkins(n):
    h = 42
    while n:
        h += n &amp; (M - 1)
        n &gt;&gt;= 128
        h *= 1025
        h ^= h &gt;&gt; 6
        h %= M

    h *= 9
    h ^= h &gt;&gt; 11
    h *= 32769

    return h % M

def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)

def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m

def invxorshift(h, s):
    r = h &gt;&gt; s
    while r:
        h ^= r
        r &gt;&gt;= s
    return h

def moddiv(a, b):
    return (a * modinv(b, M)) % M

def jenkins_crack(h):
    h = moddiv(h, 32769)
    h = invxorshift(h, 11)
    h = moddiv(h, 9)
    h = invxorshift(h, 6)
    h = moddiv(h, 1025)
    h -= 42
    return h

And to show that it works:
&gt;&gt;&gt; from crack import *
&gt;&gt;&gt; n = 2**128 + 1337
&gt;&gt;&gt; h = jenkins(n)
&gt;&gt;&gt; n2 = jenkins_crack(h)
&gt;&gt;&gt; h2 = jenkins(n2)
&gt;&gt;&gt; n != n2
True
&gt;&gt;&gt; h == h2
True

And to give a particular set of numbers that collide:
N: 2**128
M: 43617