*polynomial-time isomorphic*if there exists a function f such that

- 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.

**Conjecture (Berman-Hartmanis):**Every pair of NP-complete sets are isomorphic.

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.