>>49
i'd also argue that it has more to do with the number of disjoint cycles than collisions
Each cycle is just the same as if you replaced the set of inputs
X with
Y, and reordered the set of all numbers
j, which can be denoted as
J.
The operations that come out of this, as noted in
>>46, can just be treated as the generation of an
n-dimensional vector, for
n cycles, and the problem of finding the ``cyclic quine'' equivalent to checking if the vector is a linear combination over the space. Assuming random distribution, since it's a hash function, and little to no collision, you can calculate the chance of finding one of these quines by the size of the space, and each cycle adding another dimension. The more dimensions, the higher the chance. The number of disjoint cycles cannot be easily computed, due to the fact that hash functions are made to resist such things. But you can just compare the number of elements to the dimension of the vector.