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Chapter 8 Advanced Consistency Methods

Chapter 8 Advanced Consistency Methods. Yaling Zheng CSCE 990-06: Advanced Constraint Processing. Outline: consistency methods. Relational consistency (Section 8.1) Relational arc-consistency Relational path-consistency Relational m-consistency

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Chapter 8 Advanced Consistency Methods

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  1. Chapter 8 Advanced Consistency Methods Yaling Zheng CSCE 990-06: Advanced Constraint Processing

  2. Outline: consistency methods • Relational consistency (Section 8.1) • Relational arc-consistency • Relational path-consistency • Relational m-consistency • Relational (i, m)-consistency • Directional relational-consistency (Section 8.2) • Global consistency (i.e., BT-free) • Domain tightness (Section 8.3.1) • Constraint tightness (Section 8.3.2) • Special types of networks • Boolean theories (Section 8.4) • Row convex constraints (Section 8.5) • Linear inequalities (Section 8.6)

  3. Arc-Consistency & Relational Arc-Consistency x y x<y • Constraint network 1 is arc-consistent iff x is arc-consistent relative to y and y is arc-consistent relative to x. • How should we express constraint network 2 is consistent? x+y =z y x Constraint network 1 Constraint network 2 z 3

  4. x { 1, 2 } x<y { 1, 2 } y Arc-Consistency Dx = { 1, 2 } Dy = { 1, 2 } Rxy : x<y y is arc-consistent relative to x Dy = { 2 } X is arc-consistent relative to y Dx = { 1 } Arc-consistency: y is arc-consistent relative to x iff for every value in Dy there is a value in Dx such that Cxy is satisfied Update domain using a binary relation 4

  5. Relational Arc-Consistency Dx= { 1, 2 } Dy= { 1, 2 } Dz = { -2, -1, 0, 1, 2, 3, 4, 5 } Rxyz : x+y =z Rxyz is relational arc-consistent relative to z Dz= { 2, 3, 4 } y { 1, 2 } x { 1, 2 } z {-2, -1, 0, 1, 2, 3, 4, 5} Similarly, update domain using one relation Except that relation is not binary, its arity > 2  Relational arc-consistency 5

  6. Notes for Relational arc-consistency • Relations not necessarily binary • Three levels of relational arc-consistency: • A relation (e.g,Rxyzt)is relational arc-consistent relative to a variablex • A relation (e.g, Rxyzt ) is relational arc-consistent • A network is relational arc-consistent 6

  7. Relational Arc-Consistency • Let R be a constraint network • over a set of variables X = {x1 , x2 , , xn}, • having domains D1 , , Dn • (A) is the set of all consistent instantiations of the subnetwork defined by A, which is a set of variables • Let Rs be a relation in R defined over a scope S, e.g. S = {x2, x3, x4}

  8. Relational Arc-Consistency • Rs in R is relational-arc-consistent relative to a variable x  S iff any consistent instantiation of the variables in S-{x} has an extension to a value in Dx that satisfies Rs . That is, iff  ( S-{x} )  s-{x} (R  Dx) • A relation Rs in R is relational-arc-consistent if it is relational arc-consistent relative to every variable in S • A network is relational-arc-consistent iff every relation in it is relational-arc-consistent

  9. Example of Relational Arc-consistency A network with Dx = Dy = Dz = {(a, b, c)} one constraint Rxyz = { (a, a, a), (a, b, c), (b, b, c), (d, a ,b)} Question • Is relation Rxyz relational arc-consistent relative to {x}, {y}, {z} ? • If we add a new constraint Ryz = {(a, a), (b, c)}, is relation Rxyz relational arc-consistent relative to {x} ? 9

  10. Example of Relational Arc-consistency A network with Dx[-2, 3] Dy[-5, 7] Rxyz:x+y=zRztf:z+t=f Question • Is this network relational arc-consistent ? • What constraints should we add to make this network relational arc-consistent? Hint Consider Rxyz should be relational-arc-consistent relative to x, y, z and relation Rztf should be relational-arc-consistent relative to z, t, f 10

  11. Solution to Example 8.1.3 To make a network: Dx[-2, 3] Dy[-5, 7] Rxyz:x+y=zRztf:z+t=f relational arc-consistent, Rxyz should be relational arc-consistent relative to x, y, z yz  yz (Rxyz  Dx)  Ryz: z – y [-2, 3] xz  xz (Rxyz  Dy)  Rxz: z – x [-5, 7] yz  xy (Rxyz Dz)  Rxy: x + y [-7, 10]  Dz[-7, 10] Rztf should be relational arc-consistent relative to z, t, f tf  tf (Rztf Dz)  Rtf: f – t [-7, 10] zf  zf (Rztf Dt)  zt  zt (Rztf Df)  11

  12. Path Consistency Dx= { 0, 1, 2 } Dy= { 0, 1, 2 } Dz= { 0, 1, 2 } Rxy: x< y Ryz : y< z Rxz is path consistent relative to y  Rxz : x<z y < x < < z Path of length 2 consistency (Dechter, pp. 72) x, z (Rxz) is path consistent relative to y iff for every pair (ax, az)  Rxz , there is a value ay  Dy s.t. (ax , ay)  Rxy and (ay , az)  Ryz Infer a new relation from 2 existing binary relations 12

  13. Relational Path-Consistency Dx[-2, 3] Dy[-5, 7 ] Rxyz: x + y = z Rztf : z + t = f Rxyz and Rztf is relational path-consistent to z  Rxytf : x+y+t=f x t y z f Similarly, infer new relation from 2 existing relations except that relations are not binary, their arity > 2  Relational path-consistency 13

  14. Relational Path-Consistency • Let R be a constraint network • over a set of variables X = {x1, …, xn} • having domains D1, …, Dn , • Let RS and RT be two constraints in R • with A = (S  T) – {x} •  (A)  s-{x}(RS RT  Dx)

  15. Relational Path-Consistency • RS and RT are relational-path-consistent relative to a variablex  S  T iff any consistent instantiation of the variables in A has an extension to a value in Dx,( the domain of x) that satisfies Rs and RT simultaneously; that is, iff  (A)  A(RS RT  Dx) • A pair of relations Rs and RT is relational-path-consistent iff it is relational-path-consistent relative to every variable in S  T

  16. Example of Relational Path-Consistency A network with • A network is relational-path-consistent iff every pair of its relations is relational-path-consistent Dx[-2, 3]Dy[-5, 7] Rxyz:x+y=zRztf:z+t=f Question • Is relation Rxyz and Rztf relational-path-consistency relative to z ? • What constraints should we added to made Rxyzand Rztf relational-path-consistency relative to z ? Hint: xytf  xytf(Rxyz Rztf  Dz)  Rxytf: x+y+t = f 16

  17. Relational –m – consistency • Let R be a constraint network • over a set of variables X = {x1, … ,xn} • having domains D1, …, Dn • Let Rs1, Rs2,…Rsm be m distinct relations in R • with Si  X • A =  i=1mSi – {x} (x is a common variable among s1, … sm) •  (A)  A(i=1mRsi Dx)

  18. Relational –m – consistency • Rs1, …, Rsmare relational-m-consistent relative to a variable x i=1m Si iff any consistent instantiation of the variables in A has an extension to a value in Dx, the domain of x, that satisfies Rs1, …, Rsm simultaneously; that is iff  (A)  A(i=1m Rsi Dx) • A set of relations {Rs1, … ,Rsm} is relational-m-consistent iff it is relational-m-consistent relative to every variable in i=1m Si • A network is relational-m-consistent iff every set of m relations is relational-m-consistent. • A network is strongly relational-m-consistent iff it is relational-i -consistent for every i  m.

  19. Example ofRelational-m -consistency • Consider the constraint network • over the set of variables {x1 , x2 , x3 , x4 , x5} • with Di = {a, b, c} , 1  i  5, and • relations are: R2,3,4,5 = {(a, a, a, a), (b, a, a, a), (a, b, a, a), (a, a, b, a), (a, a, a, b)} R1,2,5 = {(b, a, b), (c, b, c), (b, a, c)} • Question: Which constraints should be added to make the constraint network strongly relational-2-consistent? 19

  20. Solution to Example 8.1.7 For the constraint network • over the set of variables {x1 , x2 , x3 , x4 , x5} • with Di = {a, b, c} , 1  i  5, and • relations are: R2,3,4,5 = {(a, a, a, a), (b, a, a, a), (a, b, a, a), (a, a, b, a), (a, a, a, b)} R1,2,5 = {(b, a, b), (c, b, c), (b, a, c)} we should guarantee • R2,3,4,5relational arc-consistent relative to x2 , x3 , x4 , x5 • R1,2,5relational arc-consistent relative to x1 , x2 , x5 • R2,3,4,5and R1,2,5relational path-consistent relative to x2 , x5 20

  21. Note for Relational-m - consistency For binary constraint, relational-m-consistency is identical to variable-based i –consistency where i =m+1.(Dechter, pp. 77, i-consistency) For general CSPs, 1.Given a set of constraints Rs1 ,…, Rsm, enforcing relational-m-consistency relative to a variable x can also be accomplished by enforcing r –consistency (r = |Si |) relative to x (prepocessing is needed here) 2.Enforcing i -consistency may be accomplished by relational-m-consistency where m ranges from 1 to 2i. 21

  22. Relational (i, m)- Consistency • A set of relations {Rs1 , …, Rsm} is relationally (i, m) -consistent iff for every subset of variables A of size i, A  j=1m Sj , any consistent assignment to A can be extended to an assignment to i=1m Si – A that satisfies Rs1 , …, Rsm simultaneously. • A network is relationally (i, m) -consistent iff every set of m relations is relationally (i, m) –consistent • A network is strong relational (i, m) -consistent iff it is relational (j, m) -consistent for every j  i. Question: relational-m-consistency  relational (m-1, m) consistency ?

  23. Algorithm Relation consistency(R, i, m) (RC(i, m)) Input: a network (X,D,C) Output: a relational-(i,m)- consistent network R‘, equivalent to R • Repeat • Q R • for every m relations Rs1,…,Rsm  Q and every • subset A of size i, A  i=1m Si • RA RA  A (i=1mRsi ) • If is RAthe empty relation • then exit and return the empty network • endif • endfor • Until Q = R

  24. Extended composition • The extended composition of relation Rs1, …, Rsm relative to a subset of variables A  i=1m Si , denoted by ECA(Rs1, …, Rsm), is defined by ECA(Rs1, …, Rsm) = A( i=1m Rsi) • When the operator is applied to m relations, it is called extended m-composition. • If the projection is restricted to subsets of sizes i, it is called extended (i, m)-composition.

  25. Two Practical relational-consistency algorithms • Generalized Arc-consistency (RC(1,1)) (a.k.a. domain propagation) if in relational-consistency algorithm, line 5 is replaced with Dx x (RsDx ) • Relational Arc-consistency (RC1) if line 5 is replaced with Rs– {x} s-{x}(RsDx )

  26. Directional-relational-consistency Given a network R an ordering d=(x1,,xn ) R is m-directionally relationally consistent iff for every subset of constraints {Rs1 , , Rsm } where the latest variable in any of the constraints’ scopes is Xl , and for every A, A{x1 , , xl-1 }, every consistent assignment to A can be extended to Xlwhile simultaneously satisfying all the relevant constraints in {Rs1, ,Rsm} .

  27. Algorithm:Directional-Relational-Consistency (DRCm) Input: a constraint network R=(X,D,C), an ordering d=(x1,…,xn ) Output: An m-directional-consistent network, along d 1. Build buckets. 2. Looping over buckets from n downto 1 3. Simplify all relations in current bucket 4. Instantiation if current bucket contains a unary constraint with xp=u then instantiate and propagate If no resulting relation is empty, then add them to appropriate bucket else exit and return the empty network 5. else (the general case) j  min {number of relations in current bucket, m } apply directional relation consistency If no resulting relation is empty, then add them to appropriate bucket else exit and return the empty network 9. Return all the buckets 27

  28. Example of DRC2 R{1,2,3,4,5} = {(H,O,S,E,S), (L,A,S,E,R), (S,H,E,E,T), (S,N,A,I,L), (S,T,E,E,R)} R{3,6,9,12} = {(A,L,S,O), (E,A,R,N), (H,I,K,E), (I,R,O,N), (S,A,M,E)} R{5,7,11} = {(E,A,T), (L,E,T), (R,U,N), (S,U,N), (T,E,N), (Y,E,S)} R{8,9,10,11} = {(A,L,S,O), (E,A,R,N), (H,I,K,E), (I,R,O,N), (S,A,M,E)} R{10,13} = {(B,E), (I,T), (N,O), (U,S)} R{12,13} = {(B,E), (I,T), (N,O), (U,S)} 28

  29. DRC2 for crossword puzzlewith d = (x13 , x12 ,  , x1) Bucket(x1) R1,2,3,4,5 Bucket(x2) H2,3,4,5 Bucket(x3) R3,6,9,12 H3,4,5 Bucket(x4) H4,5,6,9,12 Bucket(x5) R5,7,11 H5,6,9,12 Bucket(x6) H6,7,9,11,12 Bucket(x7) H7,9,11,12 Bucket(x8) R8,9,10,11 Bucket(x9) H9,10,11 H9,11,12 Bucket(x10) R10,13 H10,11,12 Bucket(x11) empty relation exit Bucket(x12) Bucket(x13) 29

  30. Complexity of directional relational consistency • Given a network R , the time complexity and size of the network generated by DRCm along ordering d is O(nm·(2mk2)(w*(d)+1) ) , • where w*(d) is induced width along ordering d, and n is the number of variables(a.k.a. number of buckets) • Consequently, the time complexity of DRC2 is O( n (4k2) (w*(d)+1) ) • The complexity of DRC1 is O(n · e2· t2) • where e is the number of input relations and • t bounds the number of tuples in each relation.

  31. Proof of Complexity of DRCm The number of variables in any bucket is at most w*(d )+1  The number of relations in a bucket is at most 2w*(d)+1  The number of subsets of size m is at most 2(w*(d)+1)m  The number of subsets of size m in n buckets is at most n 2(w*(d)+1)m The number of tuples in each relation is at most kw*(d)+1  join operations over m relations(pair-wise join) Complexity of each join operation O(m k2(w*(d)+1)) Join operation done for every subsets of m in n buckets  DRCm complexity = O(nm·(2mk2)(w*(d)+1) ) 31

  32. Proof of Complexity of DRC1 Facts: 1.The number of relations in a bucket is at most e 2. Complexity of simplication is O(e2t2) 3. complexity of instantiation is O(t) 4. complexity of “general case” is O(t) 5. The iteration time is n Complexity of DRC1 is O(n · e2· t2) 32

  33. Application of crossword puzzles Note: whenever we apply adaptive-consistency to the crossword puzzle, at most two relations will ever reside in a single bucket, no matter what variable ordering is selected. For the crossword puzzles, then, adaptive-consistency(Decther, pp.113 ) coincides with DRC2 . (Proposistion 8.2.5) • Crossword puzzles can be made directional backtrack-free by DRC2 (Decther, pp.113 , theorem 4.5.2 ) • The complexity of DRC2 applied to crossword puzzles is O( n·k 2(w*(d)+1) ) 33

  34. Domain and constraint tightness • Theorem 8.3.1 A strong relational 2-consistent constraint network over bi-valued domains is globally consistent. • Theorem 8.3.2 (domain tightness and local consistency) A strong relational k-consistent constraint network R=(X, D, C) over domains with at most k values is globally consistent

  35. Proof of Theorem 8.3.1 1. We can prove this theorem by proving that all instantation āi-1 = (a1 ,  , ai-1) (i>1) to variable x1 ,  , xi-1can be successfully extended to xi . 2. Assume that there is an instantation ēj-1= (e1 ,  , ei-1 ) can not successfully be extended to ej . Let denote A0 be the constraints violated by (ēj-1 , ej= 0) A1 be the constraints violated by (ēj-1 , ej = 1) 3. Let’s pick R1 A0, R2  A0 4. If R1=R2, which indicates in relation R1, (x1 ,  , xj-1) can not be successfully extended to xi, contradicting that the network is relational 1-consistent. 5. If R1R2, whichalso indicates (x1 ,  , xj-1) can not be successfully extended to xi , contradicting that the network is relational 2-consistent. 6. Thus our assumption is wrong. 7. So we can conclude: A strong relational 2-consistent network over bi-valued domains is globally consistent. 35

  36. m-tight • A constraint relation Rs of arity r is called m-tight if, for any variable xi S and any instantiation ā of the remaining r1 variables in S  {xi } , either there are at most m extensions of ā to xi that satisfy Rs , or there are exactly |Di | such extensions. Note: Constraints having domains of size 2 are 1-tight.

  37. Example of m-tight X= (x1 , x2 , x3 , x4) Di = {a, b, c} 1  i  4 Rs1 = {(a, a, a), (a, a, c), (a, b, c), (a, c, b), (b, a, c), (b, b, b), (b, c, a), (c, a, b), (c, b, a), (c, c, c)} Rs2 = {(a, b), (b, a), (b, c), (c, a), (a, c)} Rs3 = {(a, b), (a, c), (b, b), (c, a), (c, b)} Where s1={x1 , x2 , x3 } s2={x2 , x4 } s3={x3 , x4 } All the constraints are 2-tight. 37

  38. Lemma 8.3.1 constraint-tightness and local consistency Let Rs1 , … , Rsl be l relations that constrain a variable x , let d be the size of the domain of variable x , and let ā be an instantiation of all of the variables except for x that are constrained by the l relations (i.e. , ā is an instantiation of the variables in (S1 …  Sl) – x . If • Each relation is m-tight, for some 1m<d , and • For every subset of m+1 or fewer relations from {Rs1, …, Rsl }, there exists at least one extension of ā to x that satisfys each of the relations in the subset, Then there exists at least one extension of ā to x that satisfies all l relations.

  39. Proof of lemma 8.3.1 1.Let a1, a2 , … , ad be the d elements in the domain of x . We say that a relation allows an element ai if the extension (ā, ai ) of ā to x satisfies the relation. 2.Assume to the contrary that an extension of ā to x that satisfies all therelations does not exist. 3.for each element ai in the domain of x there must exist at least one relation that does not allow ai . Let ci denote one of the relations that does not allow ai . By construction. The set c={c1, c2 , … , cd } is a set of relations for which there does not exist an extension of ā to x that satisfies each of the relations in the set (every candicate ai is ruled out since ci does not allow ai). For every possible value of m , this leads to a contradiction. Case 1 (m = d  1): The contradiction is immediate as c={c1 , c2 , … , cd } is a set of relations of size m+1 for which there does not exist an extension to x that satisfy every relation in the set. This contradicts condition (2). 39

  40. Proof of lemma 8.3.1 (Cont.) Case 2 (m = d  2): The nominal size of the set c={c1 , c2 , … , cd } is m+2 . We can prove there must be a repetition in c . Assume to the contrary that ci cj for i  j . Recall ci is a relation that does not allow ai , i = 1, … , d and • consider {c1 , c2 , … , cd -1}. This is a set of m+1 relations so by condition (2) there must exist an ai that every relation in the set allows. The only possiblity is ad . • Consider {c1 , … , cd –2 , cd }. This is a set of m+1 relations so by condition (2) there must exist an ai that every relation in the set allows. The only possiblity is ad –1 • Simliar … Continuing in this manner , we an show that relation c1 must allow ad , ad-1 , … a2 ; that is , c1 must allow exactly m+1 extensions. This contradicts condition (1). Therefore, our assumption is false. So we can conclude that it must be the case that ci = cj for some i  j 40

  41. Proof of lemma 8.3.1 (Cont.) • case 3 (m = d 3), … , case d 1 (m =1) : • The remaining cases are similar. In each case we argue that • There are repetitions in the set c={c1, c2, … cd ), • The true size of the set c is m +1, • A contradiction is derived by appealing to condition (2) • Thus , there exists at least one extension of ā to x that satisfies all of the relations. •  41

  42. constraint-tightness and local consistency • A strongly relationally (m+1)-consistent constraint network that has m-tight constraints, is globally consistent. Reference P.van Beek and R.Dechter. Constraint tightness and looseness versus local and global consistency. Journal of the ACM(in press), in 1997

  43. Proof of theorem 8.3.5Constraint-tightness and local consistency 1.Let X = (x1 ,  , xk-1) be a set of k 1 variables, let ā be an instantation of the variables in X that is consistent with the constraint network, and let Xk be an additional variable 2. Using lemma 8.3.1 we show that there exists an extension of ā to xkthat is consistent with the constraint network. 3. Let Rs1, … , Rsl be l relations which are all and only the relations which constraint only xk and a subset of variables from X . To be consistent with the constraint network, the extension of ā to xk must satisfy each of the l relations. 4. Now , condition (1) of Lemma 8.3.1 is satisfied since each of the l relations is m-tight. 5. Further, condition (2) of Lemma 8.3.1 is satisfied since, by definiton, the requirement of strong relational (m+1)-consistency ensures that for any subset of m+1 or fewer relations, there exists an extension of ā to xk that is consistent with the constraint network.  43

  44. Questions • ? 44

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