Is there an equivalent to the birthday paradox for more than 2 messages.
Solving Hash(x) = 0 takes $2^{bits}$ steps on average
Solving Hash(x) XOR Hash(y) = 0 takes around $2 ^{bits/2}$ due to the birthday paradox.
Is there an algorithm that can solve Hash(x) XOR Hash(y) XOR Hash(z) = 0 in $2^{bits/3}$ steps (or similar)?
If so, does it get easier to find a solution if you add more terms?
[edit] The second post by Squeamish Ossifrage is what I was looking for, thanks.
