The problem has plenty of symmetry and there are ways to solve it other than naive brute force. For example, the proof that R(3,3)=6 is given in the first part of the article, and does not involve listing 2^15 graphs.

This also could apply to AES256 - if it has some mathematical structure that can be exploited, then brute forcing might not be needed to break it - see https://en.wikipedia.org/wiki/Advanced_Encryption_Standard#K... .