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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 339 / 2 / a / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 339 2 a
Created 2026-10-03 Updated 2026-10-06  0 By others on same topic  0 Discussions Create my own version
Split the sign vector as x=(pT,qT)T. Multiplication of the two off-diagonal blocks gives
xTAx=21​(pTSq+qTSTp)=pTSq,
(1)
because the two terms are the same real scalar. The map (p,q)↦x is a bijection between the feasible sign choices. Therefore
p,qmax​pTSq=xmax​xTAx​.
(2)
This is the symmetric lifting of bipartite binary quadratic optimization to binary quadratic optimization. The factor 1/2 is necessary: otherwise the two blocks would double the objective. No positive semidefiniteness of A is assumed; generally its off-diagonal structure makes its quadratic form indefinite.

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