11112005 NP-Hardness and Turing Reducibility 5.1 Turing reducibility and NP-Hardness Problems Problems outside NP can be proved to be hard using the same technique. Any decision problem to which any NP-Complete problem can be reduced in polynomial time can be shown to be not solvable in polynomial time unless P=NP. We may say that such a problem is NP-Hard in a sense that they are at least as hard as NP-Complete problems. KTH LARGEST SUBSET INSTANCE: A finite set A, a size s(a) Z for each a A , and 2 non-negative integers B s(a) and K 2| A| . aA QUESTION: Are there at least K distinct subsets A A that satisfies s ( A) B ? It is shown by Lawler that this problem can be solved in pseudo-polynomial time, in fact in time bounded by a polynomial function of | A | K log( s (a)). Thus for any fixed value of K it can be solved in polynomial time. The question then arises, can it be solved in general in polynomial time? Not only does this problem appear not to be in P, it does not appear even to be in NP, since the natural way of solving it non-deterministically involves guessing K subsets of A, and there seems to be no way to write down such a guess using only a polynomial number of symbols in |A|logKlog(s(a)). On the other hand, no transformation from an NP-Complete problem to this problem is known. However, it has been shown that the NP-Complete partition problem can be Turing-reduced to Kth LARGEST SUBSET. TRAVELLING SALESMAN EXTENSION(TSE) INSTANCE: A finite set C {c1 , c2 ,..., cm } of cities, a distance d (ci , c j ) Z for each pair of cities ci , c j C, a bound B Z , and a partial tour c (1) , c (2) ,..., c ( m ) of K distinct cities from C ,1 K m QUESTION: Can be extended to a full tour c (1) , c (2) ,..., c ( K ) , c ( K 1) ,..., c ( m) having total length B or less? It is easy to see that this problem belongs to NP, and hence, by the definition of NPCompleteness, TSETS . Since a transformation is a special case of a Turing reduction, this in turn implies that TSET TS . Thus letting TSO stand for the traveling salesman optimization problem, all we need to show is that TSOT TS and by transitivity we will have that TSO is Turing reducible to TS.