Lecture 21 ( Oct . 24 ) : Max Cut SDP Gap and Max 2 - SAT

There is a simple example for Max Cut that shows our analysis is nearly tight. Consider G = C5, the cycle on 5 nodes {0, 1, 2, 3, 4}, with all edge weights being 1. The optimum solution is 4 with an optimum cut being, say, S = {1, 3} with δ(S) = {(0, 1), (1, 2), (2, 3), (3, 4)} (we can do no better because C5 is not bipartite). Next we define a feasible SDP solution with value greater than 4. Let θ = 4π/5. For 0 ≤ i ≤ 4, let vi = (cos(iθ), sin(iθ), 0, 0, 0). Note that vi ◦ vi = 1 for all i, since cos(iθ) + sin(iθ) = 1. So, this is a feasible solution to MAXCUT-SDP. The angle between any two nodes corresponding to an edge (i, i+ 1) is θ, so vi ◦ vi+1 = ||vi|| · ||vi+1|| · cos(θ) = cos(θ). Thus, OPTSDP ≥ 4 ∑