Is there any function that does not suffers birthday problem? I am eager to know that if there is any function that does not suffer birthday problem and how to prove it formally that the function is not suffering the birthday problem.
If $f:\{0,1\}^m\rightarrow \{0,1\}^n$ with $n\geq m,$ then of course there are, the set of one-to-one functions, but such a function is not a cryptographic hash function, since it lacks the compression property. If $n&lt;m,$ (or more generally if $|X|&gt;|Y|$ for $f:X\rightarrow Y$), collisions will happen.