This is the first question I ask in this forum concerning the NP-hardness proof of a decision problem X when some input to X is already exponentially long.
Let's say we have a decision problem DECIDE and it takes as its input a boolean algebra along with others. Relative to the input n, there will be up to 2^n different choices, but deciding if a given path is an answer is decidable in O(p(n)) for some polynomial p(n). Let's say that DECIDE is in NP and that we like to show that it is also NP-hard so we can conclude that DECIDE is NP-complete.
Here, let's also say that any instance of 3-SAT, provided that the input to it is as long as a string necessary to store a boolean algebra containing all the propositional variables occurring in the input, is many-one reducible to DECIDE. When an instance of 3-SAT is not as long as that, however, just producing such a boolean algebra (appears to) require longer than polynomially-bounded steps (relative to the size of the input to 3-SAT).
My question is: do we (or can we) still conclude that DECIDE is NP-hard?
Thank you! I have about 50 days-experience in complexity theory. (Before that, I did not have a clear view of what co-NP was.) I started to learn the theory with "Complexity Theory: Exploring the Limits of Efficient Algorithms" by Ingo Wegener, and now I am trying to prove \Pi_2-hardness of some decision problem.
