By Trevisan L.

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RANDOMIZED ROUNDING From which we derive the linear programming relaxation minimize subject to n i=1 xi xi ≥ 1 ∀v ∈ U xi ≤ 1 ∀i ∈ {1, . . , n} xi ≥ 0 ∀i ∈ {1, . . 1 In an earlier lecture, we noted that every instance G of the vertex cover problem can be translated into an instance (U, S1 , . . , Sn ) of the set cover problem in which every element of U belongs precisely to two sets. 2) of the resulting instance of set cover is identical to the linear programming relaxation of the vertex cover problem on the graph G.

We also set y(u,v) to min{c(u) − pu , c(v) − pv }. Here is the psedocode of the algorithm: • Input: undirected, unweighted, graph G = (V, E) • p = (0, · · · , 0) • y = (0, · · · , 0) • for each edge (u, v) ∈ E – if pu < c(u) and pv < c(v) then ∗ y(u,v) := min{c(u) − pu , c(v) − pv } ∗ pu := pu + min{c(u) − pu , c(v) − pv } ∗ pv := pv + min{c(u) − pu , c(v) − pv } 54 LECTURE 7. ROUNDING LINEAR PROGRAMS • S := {v : pv ≥ c(v)} • return S, y The algorithm outputs a correct vertex cover, because for each edge (u, v), the algorithm makes sure that at least one of pu = c(u) or pv = c(v) is true, and so at least one of u or v belongs to S at the end.

These are the only three edges that could have had (0, 1, 0) as an endpoint, 36 LECTURE 5. LINEAR PROGRAMMING and since (0, 1, 0) is a vertex of the above polytope, we have to conclude that the polytope has no edge, and so it is made of the single point (0, 1, 0). 3 The General Case In general, if we have a linear program with n variables x1 , . . , xn , we can think of every assignment to the variables as an n-dimensional point (x1 , . . , xn ) in Rn . Every inequality a1 x1 + . . an xn ≤ b divides the space Rn into the region that satisfies the inequality and the region that does not satisfy the inequality, with the hyperplane a1 x1 + · · · an xn = b being the boundary between the two regions.

### Combinatorial optimization: Exact and approximate algorithms by Trevisan L.

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