- x is in A if and only if f(x) is in B (f reduces A to B),
- f is a bijection, i.e., f is 1-1 and onto,
- f is polynomial-time computable, and
- f-1 is polynomial-time computable.
The Isomorphic Relation between sets is an equivalence relation. The Berman-Hartmanis conjecture is equivalent to saying that every NP-complete set is isomorphic to SAT.
The conjecture is still open though it has generated a considerable amount of research in computational complexity. But for now let me just explain why this question is interesting.
Berman and Hartmanis showed that all of the natural NP-complete sets at the time, for example all of the problems listed in Garey & Johnson, are isomorphic. They established this fact by proving that every paddable NP-complete set is isomorphic to SAT. A set A is paddable if there is a polynomial-time computable length-increasing function g such that for all strings x and y, x is in A if and only if g(x,y) is in A.
Most NP-complete sets are easily seen to be paddable. Consider the clique problem. Given a graph G and a string y, we can create a new graph H that encodes y by adding disjoint edges to G while keeping the clique size of H the same as the clique size of G.
The isomorphism conjecture implies P≠NP, since if P=NP then there would be finite NP-complete sets which cannot not be isomorphic to the infinite set SAT. There was a time when Hartmanis was pushing on the idea of using the conjecture to prove P≠NP but most complexity theorists now believe the isomorphism conjecture is false.
More on the isomorphism conjecture in future posts.