Nice! More explicitly: if the polynomial-time algorithm is at (constant) index K in our enumeration of all algorithms, we’d need about R*2^K steps of the meta-algorithm to run R steps of the algorithm K. Thus, if the algorithm K is bound by polynomial P(n) in problem size n, it’d take P(n)*2^K steps of the meta-algorithm (polynomial in n, K is a constant) to solve the problem of size n.
Nice! More explicitly: if the polynomial-time algorithm is at (constant) index K in our enumeration of all algorithms, we’d need about R*2^K steps of the meta-algorithm to run R steps of the algorithm K. Thus, if the algorithm K is bound by polynomial P(n) in problem size n, it’d take P(n)*2^K steps of the meta-algorithm (polynomial in n, K is a constant) to solve the problem of size n.