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How Far From a Worst Solution a Random Solution of a k CSP Instance Can Be?

机译:距离最糟糕的解决方案有多远的K CSP实例的随机解决方案可以是?

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Given an instance I of an optimization constraint satisfaction problem (CSP), finding solutions with value at least the expected value of a random solution is easy. We wonder how good such solutions can be. Namely, we initiate the study of ratio ρE(I) = (E_X[v(I,X)] - wor(I))/(opt(I) - wor(I)) where opt(I), wor(I) and E_X [v(I, X)} refer to respectively the optimal, the worst, and the average solution values on I. We here focus on the case when the variables have a domain of size q ≥ 2 and the constraint arity is at most k ≥ 2, where k,q are two constant integers. Connecting this ratio to the highest frequency in orthogonal arrays with specified parameters, we prove that it is Ω(1/n~(k/2)) if q = 2, n(1/n~(k-1-「log_pk(k-1))」 where p~k is the smallest prime power such that p~k ≥ q otherwise, and Ω(1/q~k ) in (max{q, k} + 1})-partite instances.
机译:鉴于优化约束满足问题(CSP)的实例I,请至少使用值的随机解决方案的预期值查找解决方案。我们想知道这些解决方案有多好。即,我们启动了比率ρe(i)=(e_x [v(i,x)] - wor(i))/(选择(i) - wor(i)),其中opt(i),wor(i )和e_x [v(i,x)}分别参考最佳,最差和平均解决方案值I.我们在这里专注于变量具有大小Q≥2域的情况和约束arity的情况至多K≥2,其中k,q是两个恒定整数。将此比率与具有指定参数的正交阵列中的最高频率连接到,我们证明它是ω(1 / n〜(k / 2)),如果q = 2,n(1 / n〜(k-1-「log_pk( K-1))」其中P〜K是最小的主要功率,使得p〜k≥q否则,ω(1 / q〜k)(max {q,k} + 1}) - partite实例。

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