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What is the complexity class of the general problem of integer programming feasibility?

The sources I've looked at are, in my opinion, very confusing. Some say NP-hard, some say NP-complete. Some do not distinguish between the general problem and the binary case. Some do not distinguish between the optimization problem and the feasibility problem. As far as I can tell, there appears to be no correlation between these three factors. I have only seen an actual proof of any claim in the binary feasibility case, which is NP-complete.

If I understand correctly, NP-completeness is a stronger condition than NP-hardness, since NP-hard problems do not need to be NP (or decision problems, but I can ignore that because I'm only interested in feasibility). So I am not surprised that the two are used interchangeably around for the binary case.

A reference would also be greatly appreciated.

Eric Stucky
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1 Answers1

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Edit 29 March 2015: As D.W. points out in a comment, Sahni's "P-hard" and "P-complete" do not mean what one might expect from modern usage. His terms are (roughly) conditional on P=NP. His section 1.1, Definition 4 is a clear example: "A problem $L$ is P-Hard iff a polynomial algorithm for $L$ implies P = NP." As D.W. asserts, this eliminates the discrepancy between Sahni and G&J: the problem is NP-hard/-complete, depending on the signs of the constraint coefficients.

P-hard/-complete, depending on the signs of the constraint coefficients. See Sahni, S., "Computationally Related Problems" SIAM J. Computing vol. 3, pp. 262-279, 1974. Also here (Section 2.5, I1 and I2).

Referenced as "A6 MP1" at Garey and Johnson, "Computers and Intractability: A Guide to the Theory of NP-Completeness", 1979. However, there must be an error in this reference, G&J assert NP-hard/-complete, but Sahni asserts P-Hard (positive or negative constraint coefficients) and P-complete (positive constraints).

Eric Towers
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  • a musical answer! – GEdgar Jun 24 '14 at 18:26
  • Thank you for your answer, but, ah, what? I have no idea where this fits. I see that if P$=$NP then we can conclude nothing and if P$\neq$NP then P-hard means not P. It must therefore be either NP$\smallsetminus$P or NP-hard, but we cannot conclude which one? But if it is the first then it is at least NP-complete, so I feel like I don't know more now than I did before. I guess I mean that P-hard feels no harder than NP-complete, but we already knew that it is at least NP-complete from the binary case. – Eric Stucky Jun 24 '14 at 18:41
  • P-hard is no worse than NP-hard (a deterministic polynomial time algorithm may be made NP by adding precisely zero nondeterminism, requiring no change). Similarly for P-complete. (However, as a practical matter, both are better than NP-[adjective] since they have polynomial time solutions.) – Eric Towers Jun 25 '14 at 14:24
  • @EricTowers: Re. "However…": No? P-hard is defined as "If polynomial time solution then P=NP" and I'm pretty sure that question is still open? ;) I feel like I must be missing something... – Eric Stucky Jun 25 '14 at 18:07
  • No. If every problem in P may be reduced to a given problem schema (using a logtime, logspace, or NC reduction), then that problem schema is P-hard. Note that this is not the same as P-complete since we do not require that the problem schema also be in P. It is already known that there are P-complete problems ("circuit evaluation", e.g.), so there are P-hard problems and so far no one has claimed that P=NP as a result. – Eric Towers Jun 27 '14 at 07:11
  • Thanks for the answer, but it is a bit misleading. It cites a 1974 paper that is using old terminology. What that paper actually proves is that the problem is NP-hard/NP-complete. (What that paper calls "P-complete" matches the modern meaning of the term NP-complete. The standard meaning of the term P-complete, in modern terminology, is something different than what that paper is referring to. See the definitions in Section 1.1.) So, there's no contradiction and no apparent error. – D.W. Mar 02 '15 at 22:25
  • @D.W. : Wow. Sahni's notation is conditional on P=NP. I had not noticed that (and am a little baffled by that choice). Rather than just fix the answer, I'm inclined to elaborate with a warning... – Eric Towers Mar 29 '15 at 23:49
  • @D.W. : What we need is a time machine so that all the mismatched notation and terminology can be in the present rather than the past. – Eric Towers Mar 29 '15 at 23:57