4 ms·
The "too much time has passed" bound on each iteration guarantees that you don't end up running an iteration forever, and makes the algorithm O(2^N), since the
by xjtian 13y ago
The "too much time has passed" bound on each iteration guarantees that you don't end up running an iteration forever, and makes the algorithm O(2^N), since the number of programs up to N bits long is 2^(N+1) - 1.