The Deviation Constraint - Université catholique de Louvain

1. RD(X , m) ≤ Dmax22. LD(X , m) ≤ Dmax23. RD(X , m) ≥ Dmin24. LD(X , m) ≥ Dmin25. [ LD(X , m), LD(X , m) ] ∩ [ RD(X , m), RD(X , m) ] ≠ φProof. 1. If RD(X , m) > Dmax2then ∑ DmaxX>m(X − m) >2(Property 1).Hence ∑ ni=1 |X i − m| > D max (by Theorem 1).2., 3. and 4. similar to 1.5. Direct consequence of Theorem 1 and Property 1. 3 Naive implementationThis section explains why a naive implementation of DEVIATION by decompositioninto more elementary constraints is not optimal in terms of filtering.As stated in Definition 1, DEVIATION(X , m, D) holds if and only if n.m =∑ ni=1 X i and D = ∑ ni=1 |X i −m|. This suggests a natural implementation of theconstraint by decomposing it into two SUM constraints. Figure 2 illustrates thatthe filtering obtained with the decomposition is not optimal.Dom X 2∣X 1−m∣∣X 2−m∣D maxD maxm , mX 1 X 2=2mDom X 1Fig. 2. Filtering of X 1 with decomposition and with DEVIATION.Assume two variables X 1 , X 2 with unbounded finite domains and the constraintDEVIATION({X 1 , X 2 }, m, D ∈ [0, D max ]).The diagonally shaded square (see Figure 2) delimits the set of points such that|X 1 − m| + |X 2 − m| ≤ D max . The diagonal line is the set of points such thatX 1 + X 2 = 2.m. The unbounded domains for X 1 and X 2 are bound-consistentfor the mean constraint. The vertically shaded rectangle defines the domain of6