Is there any function that does not suffers 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.