The wikipedia page about sha-2 states:
Currently, the best public attacks break preimage resistance for 52 out of 64 rounds of SHA-256 or 57 out of 80 rounds of SHA-512
How exactly is the sha-256 algorithm with fewer rounds defined? I am wondering in particular about the message schedule array $W$, since in round $i$ of the $64$ round sha-256 algorithm, $W[i]$ is accessed. Does 52 round sha-256 simply ignore the elements $53..64$ of $W$?