The Deviation Constraint - UniversitÃ© catholique de Louvain

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An assignment A is said to be down-centered when:{= Xmaxif XA(X)max ≤ s(A)/n≥ s(A)/n otherwiseIn other words, each variable with maximum domain value smaller than the meanof the assigned values takes its maximum domain value and the other variablestake values larger than the mean of the assigned values.Example 2. Considering the variables and domains of Example 1, the followingassignment is up-centered with mean 17/4:A(X 1 ) = 8, A(X 2 ) = 4, A(X 3 ) = 2, A(X 4 ) = 3.Theorem 3. An assignment is an optimal solution to the problem of finding Dif and only if it is a down-centered assignment or an up-centered assignment ofmean m.Proof. (if) Given an assignment A of mean m i.e. s(A) = n.m. The only wayto decrease the sum of deviations while conserving the mean m is to find a pairof variables X i , X j such that A(X i ) > m, A(X i ) > Xi min , A(X j ) < m, A(X j ) m and A(X i ) > Ximin (violation of up-centered) and one withA(X j ) < m and A(X j ) < Xj max (violation of down-centered). Let define δ =min ( A(X i ) − max(Ximin , m), min(Xjmax , m) − A(X j ) ) . The assignment A(X) isnot optimal since the sum of deviations can be decreased by 2δ by modifyingthe assignment on X i and X j : A ′ (X i ) = A(X i ) − δ and A ′ (X j ) = A(X j ) + δ. Theorem 4. If DEVIATION is consistent thenD = 2. max(LD (X , m), RD(X , m)) .Proof. Assume LD ≥ RD, then it is possible to build a down-centered assignment∑A of mean m and which is optimal by Theorem 3. For this assignment∑ A(X)m(X − m) =X
Theorem 5. Computing D is N P-complete.Proof. It is possible to reduce the subset sum problem [3] to the problem ofcomputing D (see Definition 3). This problem is not more difficult than theparticular case where m = 0:such that :D = maxn∑|X i |i=1n∑X i = 0i=1Ximin≤ X i ≤ X maxi , 1 ≤ i ≤ n.Given a set of n positive values {b 1 , ..., b n−1 , T }, the subset sum problem consistsin finding if there exists a set of binary values {y 1 , ..., y n−1 }, y i ∈ {0, 1}, 1 ≤i < n such that ∑ n−1i=1 y i.b i = T . The reduction is the following:• X miniand Xmax i= − biP2 n−1i=1 bi2− T .= bi2for 1 ≤ i < n.• X n =P n−1Pi=1 n−1• There is a solution to the subset sum problem if D ≥bi i=1∣2+ ∣bi2− T ∣.This constraint on the optimal value ensures that the optimal solution is suchthat X i ∈ {− bi2 , bi2}. The solution to the subset sum problem is then givenby y i = 1 if X i = bi2 and y i = 0 if X i = −bi2 . Unless P = N P, the problem of computing D is exponential (Theorem 5).As explained in Section 2, in order to be effective, filtering algorithms should beas efficient as possible because they are applied many times during the searchprocess. This is why we prefer to find efficiently a good upper bound D ↑ for Dthan to find its exact value. An upper bound which can be computed in O(n) isD ↑ =n∑i=1max ( |X maxiThe filtering on Dom(D) becomes:− m|, |X mini − m| ) .Dom(D) ←− Dom(D) ∩ [D, D ↑ ]5 Filtering on XTwo propagators could be considered to filter the domain of X :1. from D min and m to X and2. from D max and m to X .9
Page 1 and 2: The Deviation ConstraintPierre Scha
Page 3 and 4: time. The maximization is proved to
Page 5 and 6: • RD(X , m) = ∑ X∈Xrd(X, m) a
Page 7: X 1 after a bound-consistent filter
Page 11 and 12: If both cases are considered togeth
Page 13 and 14: X 2 = ... = X n . Consequently at t
Page 15: failspread5000 10000 15000 20000Det

The Deviation Constraint - UniversitÃ© catholique de Louvain

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