There's already a known algorithm M you can implement right now that can find satisfying assignments to satisfiable SAT instances in polynomial time conditioned

There's already a known algorithm M you can implement right now that can find satisfying assignments to satisfiable SAT instances in polynomial time conditioned only on P=NP. Note: I'll be glossing over a few technicalities in this explanation, feel free to ask about them. Roughly, the algorithm M is as follows. Suppose you have input x of length n. For each integer c between 1 and n, encode c in binary and try to decode it in UTF-8. If it decodes, try to interpret it as a Python program, and run it on input x for 100n^c steps. If it executes successfully and outputs a valid assignment, return the assignment. Otherwise, continue onto the next c. If c=n and you haven't found an assignment yet, do a brute force search over assignments and return what you find. Now suppose P=NP, so some Python program finds SAT solutions in polynomial time. In fact, there are infinitely many such programs, since each valid program can be prefixed with as many "pass" expressions as we want. Interpreting the UTF-8 encoding of any such program p in binary gives us some integer c. There are only finitely many inputs of length ≤ c, so the slowest of them still runs in finite (and thus polynomial) time. For any input of length > c, the above algorithm M will spend at most 100(c-1)n^c time trying to execute the (c-1) different strings in Python for at most 100n^c steps each. If M happens on a solution before then, great! If it doesn't, then now we run p -- which by assumption does find solutions in polynomial time -- for 100n^c steps. If 100n^c is an upper bound on the true running time of p, then this necessarily returns a valid assignment, meaning that we can find assignments in time O(c n^c) time for some integer c, i.e. polynomial time. Otherwise, if 100n^c is too small, we just wait for the next copy p' of p (with a pass prefixed to it) that will have an even looser time bound of 100n^c' . This also may not be enough, but eventually we will find some p''''' for which the time bound does suffice (depending on the true running time of p). And so we find valid solutions in O(c'''''n^c''''') time. None of this required us to know what p is; effectively M makes the non-constructive constructive by trying every program out in a systematic way that the running time never exceeds some (also unknown) polynomial bound. Is it practical? Suppose that the shortest Python program finding SAT assignments is at least 100 bytes long. Then c > 256^100, and thus the running time is something like O(n^(256^100)). So be sure to get a good CPU.