v2007.09.13 - Convex Optimization
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v2007.09.13 - Convex Optimization

v2007.09.13 - Convex Optimization v2007.09.13 - Convex Optimization

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282 CHAPTER 4. SEMIDEFINITE PROGRAMMINGOur purpose here is to demonstrate how iteration of two simple convexproblems can quickly converge to an optimal solution of the max cutproblem with a 98% success rate, on average. 4.27 max cut is stated:maximizex∑a ij (1 − x i x j ) 1 21≤i

4.4. RANK-CONSTRAINED SEMIDEFINITE PROGRAM 283Because an estimate of upper bound to max cut is needed to ascertainconvergence when vector x has large cardinality, we digress to derive thedual problem because it is instructive: Directly from (679), the Lagrangianis [46,5.1.5] (1219)L(x, ν) = 1 4 〈xxT , δ(A1) − A〉 + 〈ν , δ(xx T ) − 1〉= 1 4 〈xxT , δ(A1) − A〉 + 〈δ(ν), xx T 〉 − 〈ν , 1〉= 1 4 〈xxT , δ(A1 + 4ν) − A〉 − 〈ν , 1〉(680)where quadratic x T (δ(A1+ 4ν)−A)x has supremum 0 if δ(A1+ 4ν)−A isnegative semidefinite, and has supremum ∞ otherwise. The finite supremumof dual functiong(ν) = sup L(x, ν) =x{ −〈ν , 1〉 , A − δ(A1 + 4ν) ≽ 0∞otherwise(681)is chosen to be the objective of minimization to dual convex problemminimize −ν T 1νsubject to A − δ(A1 + 4ν) ≽ 0(682)whose optimal value provides a least upper bound to max cut, but isnot tight (duality gap is nonzero). [107] In fact, we find that the bound’svariance is relatively large for this problem; thus ending our digression. 4.28To transform max cut to its convex equivalent, first definethen max cut (679) becomesX = xx T ∈ S n (687)maximize1X∈ S n 4〈X , δ(A1) − A〉subject to δ(X) = 1(X ≽ 0)rankX = 1(683)4.28 Taking the dual of (682) would provide (683) but without the rank constraint. [100]

282 CHAPTER 4. SEMIDEFINITE PROGRAMMINGOur purpose here is to demonstrate how iteration of two simple convexproblems can quickly converge to an optimal solution of the max cutproblem with a 98% success rate, on average. 4.27 max cut is stated:maximizex∑a ij (1 − x i x j ) 1 21≤i

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