Let A and B be NP-complete problems. Suppose I have established reductions from problem A to problem B and vice versa. Now, considering a specific instance (or set of instances) of problem A that can be solved efficiently in polynomial time, does this imply that the instances reduced to problem B can also be solved in polynomial time? In other words, does the polynomial-time solvability of instances of problem A extend to instances of problem B through the established reductions?

  • $\begingroup$ I would say no, unless the particular reduction is a restriction or isomorphism that preserves the particular property of A which is “easier” to solve. Otherwise, there’s no guarantee that the corresponding subset of B after the reduction can be solved efficiently. It’s possible that what you’re describing is a separate complexity class (I do not know). I’m just kicking off the discussion. $\endgroup$ Commented Mar 31 at 15:54
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    $\begingroup$ One way to think about this is to define a problem A' that contains those instances of A that have the special structure that makes them efficiently solvable. Let problem B' contain those instances of B that, if given to the reduction from B to A, produce an instance of A'. Then B' is also polynomially solvable: both A' and B' are in P. What exactly B' will look like may or may not be easy to say. $\endgroup$
    – Neal Young
    Commented Mar 31 at 22:39
  • $\begingroup$ Trivially if you can reduce an instance of B to an efficiently solvable instance of A, then that instance of B is efficiently solvable just by using the reduction and the algorithm for A. $\endgroup$
    – rus9384
    Commented May 1 at 17:08

1 Answer 1


Yes, that would be the case, and here is my argument. Suppose there is a polynomial time reduction problem $A$ to problem $B$. So basically, we have a mapping function $f$ that converts any input of problem $A$ into an input of problem $B$ in polynomial time. Now we are to run a solver that can solve problem $B$, which is typically the bottleneck in cases of NP-hard problems. Now, as you said, some special class of instances of problem $B$ may indeed be solved in polynomial time. Thus, by virtue of the polynomial time mapping function $f$, we clearly have a polynomial time solvable subset (which may also be empty) of instances of problem $A$ that map to those special instances of problem $B$. Also, keep in mind that any reduction is a one-way path; you cannot comment on the other direction using that reduction.


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