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given a LP half-integral solution from the minimum vertex cover problem with x € {0, 1/2 , 1} I'm looking for an algorithm that returns an integer solution with x € {0,1} that is optimal.

Obviously I can't just round up all x, since that would yield too many vertices in the minimum cover set and thus wouldn't be optimal.

So I have to decide which vertices with x=1/2 should be 0 and which should be 1.

I'm thinking about looking at the neighbours of a given vertex with x=1/2, so I can decide whether it's in the set or not but I think I'm missing something here.

Any hints are appreciated :)

moepmoep12
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  • If a vertex u has value 1/2 this means an adjacent vertex v also has value >= 1/2 to fulfill the requirement x[u] + x[v] >= 1. But doesn't that mean all vertices have value 1/2 since it's minimized in the LP? – moepmoep12 May 28 '18 at 16:37
  • What's the motivation/scope here? The question is probably off-topic here (not much programming, more math / CS). Given it's an NP-hard problem, assuming you can obtain your half-integral solution in polynomial-time (probably true), there is (for the general case) nothing you can do besides brute-force or some clever variant (backtracking). In practice, you probably would let the MIP-solver do it's job (branch and cut). – sascha May 28 '18 at 16:44
  • It's for educational purpose, I'm a student. The solution should be "trivial" but I'm stuck here. – moepmoep12 May 28 '18 at 16:46
  • I missunderstood the exercise. The exercise was to reduce x € ]0,1[, so that x' € {0, 1/2, 1}. – moepmoep12 May 28 '18 at 18:26
  • Have you seen this perhaps? Threshhold rounding. https://www2.cs.duke.edu/courses/fall15/compsci532/scribe_notes/lec13.pdf –  May 28 '18 at 22:05

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