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Local Consistency Combinatorial Problem Solving (CPS) Enric Rodr - PowerPoint PPT Presentation

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Local Consistency Combinatorial Problem Solving (CPS) Enric Rodr guez-Carbonell (based on materials by Javier Larrosa) February 28, 2020 Interaction Graph The interaction graph of a CSP is an undirected graph G = ( V, E ) s.t.: there

  1. Arc Consistency The AC name comes from the arcs (constraints) of the interaction graph ■ Example of enforcing AC. ■ Consider the CSP with variables x, y, z , domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z Recall nodes are labelled with variables (here, also with their domains) ■ Recall edges are labelled with constraints ■ y 2 x < y y < z z x 3 1 11 / 43

  2. Arc Consistency Another example of enforcing AC. ■ Consider the CSP with variables x, y, z , domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 1 x > z 12 / 43

  3. Arc Consistency Another example of enforcing AC. ■ Consider the CSP with variables x, y, z , domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 1 x > z 12 / 43

  4. Arc Consistency Another example of enforcing AC. ■ Consider the CSP with variables x, y, z , domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 x > z 12 / 43

  5. Arc Consistency Another example of enforcing AC. ■ Consider the CSP with variables x, y, z , domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 x > z The domain of x has become empty! ■ So the CSP cannot have any solution ■ 12 / 43

  6. Arc Consistency If while enforcing AC some domain becomes empty, ■ then we know the original CSP has no solution Else, when there are no more arc-inconsistent values, ■ we have a smaller equivalent CSP (no solution is lost) Uniqueness: the order of removal of arc-inconsistent values is irrelevant ■ Now let us see algorithms for effectively enforcing AC ■ 13 / 43

  7. AC-1: Revise( i, j ) The simplest algorithm to enforce AC is called AC-1 Based on function Revise ( i, j ), which removes values from d i without support in d j . Returns true if some value is removed function Revise ( i, j ) change := false for each a ∈ d i do if ∀ b ∈ d j ¬ c ij ( a, b ) then change := true remove a from d i return change The time complexity of Revise ( i, j ) is O ( | d i | · | d j | ) (we assume that evaluating a binary constraint takes constant time) 14 / 43

  8. AC-1 procedure AC-1 ( X, D, C ) repeat change := false for each c ij ∈ C do change := change ∨ Revise ( i, j ) ∨ Revise ( j, i ) until ¬ change The time complexity of AC-1 is O ( e · n · m 3 ) , ■ with n = | X | , m = max i {| d i |} and e = | C | (note e = O ( n 2 ) ) Whenever a value has been removed, ■ the domain should be checked if empty (not done here for simplicity) 15 / 43

  9. AC-3 A more efficient algorithm is AC-3, ■ which only revises constraints with a chance of filtering domains AC-3 uses a set of pairs Q s.t. ■ if ( i, j ) ∈ Q then we can’t ensure that all values in d i have support in x j procedure AC-3 ( X, D, C ) Q := { ( i, j ) , ( j, i ) | c ij ∈ C } // each pair is added twice while Q � = ∅ do ( i, j ) := Fetch ( Q ) // selects and removes from Q if Revise ( i, j ) then Q := Q ∪ { ( k, i ) | c ki ∈ C, k � = j } Space complexity: O ( e ) ■ Time complexity: O ( e · m 3 ) ■ 16 / 43

  10. Complexity of AC-3 Q := { ( i, j ) , ( j, i ) | c ij ∈ C } // each pair is added twice while Q � = ∅ do ( i, j ) := Fetch ( Q ) // selects and removes from Q if Revise ( i, j ) then Q := Q ∪ { ( k, i ) | c ki ∈ C, k � = j } ( i, j ) is in Q because d j has been pruned. ■ Therefore, it will be in Q at most m times. Consequently, the loop iterates at most O ( e · m ) times Without the red part, the cost of AC-3 would be O ( e · m 3 ) ■ For a given i , Revise ( i, j ) will be true at most m times. ■ Aggregated cost of red part due to i is O ( |{ k | c ki ∈ C }| · m ) = O (deg( i ) · m ) So aggregated cost of red part due to all vars is O ( e · m ) So the total cost in time is O ( e · m 3 ) ■ 17 / 43

  11. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z 18 / 43

  12. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 1) : 1 ◆ 18 / 43

  13. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 2) : 2 ◆ 18 / 43

  14. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 3) : 3 ◆ 18 / 43

  15. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 4) : 4 ◆ 18 / 43

  16. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, x ) ■ Finding support for ( y, 1) : 1 ◆ 18 / 43

  17. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, x ) ■ Finding support for ( y, 2) : 1 ◆ 18 / 43

  18. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, x ) ■ Finding support for ( y, 3) : 1 ◆ 18 / 43

  19. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, x ) ■ Finding support for ( y, 4) : 1 ◆ 18 / 43

  20. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, z ) ■ Finding support for ( y, 1) : 1 ◆ 18 / 43

  21. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, z ) ■ Finding support for ( y, 2) : 1 ◆ 18 / 43

  22. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, z ) ■ Finding support for ( y, 3) : 1 ◆ 18 / 43

  23. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( y, z ) ■ Finding support for ( y, 4) : 1 ◆ 18 / 43

  24. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( z, y ) ■ Finding support for ( z, 3) : 1 ◆ 18 / 43

  25. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 1) : 1 ◆ 18 / 43

  26. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 2) : 2 ◆ 18 / 43

  27. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 3) : 3 ◆ 18 / 43

  28. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 4 4 x y z Let us count the number of constraint checks of Revise ( x, y ) ■ Finding support for ( x, 4) : 3 ◆ 18 / 43

  29. Example of AC-3 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z x ≤ y y � = z 1 1 2 2 3 3 4 4 x y z Altogether 10 + 4 + 4 + 1 + 9 = 28 constraint checks ■ From last 9, the only new check is (( x, 3) , ( y, 4)) ! ■ Still a lot of work is repeated over and over again ■ 18 / 43

  30. AC-4 AC-4 is an even more efficient algorithm. It uses: ■ N [ i, a, j ] = “number of supports that a ∈ d i has in d j ” ◆ S [ j, b ] = “set of pairs ( i, a ) supported by b ∈ d j ” ◆ ( i, a ) ∈ Q means that a has been pruned from d i and ◆ its effect has not been updated on the N array yet procedure AC-4 ( X, D, C ) // N is constructed full of 0’s and S is constructed full of ∅ ’s // Initialization phase for each c ij ∈ C , a ∈ d i , b ∈ d j such that c ij ( a, b ) do // Value b in d j is a support for value a ∈ d i N [ i, a, j ] + + S [ j, b ] := S [ j, b ] ∪ { ( i, a ) } for each c ij ∈ C , a ∈ d i such that N [ i, a, j ] = 0 do remove a from d i Q := Q ∪ ( i, a ) ... 19 / 43

  31. AC-4 ... // Propagation phase while Q � = ∅ do ( j, b ) := Fetch ( Q ) for each ( i, a ) ∈ S [ j, b ] such that a ∈ d i do N [ i, a, j ] − − if N [ i, a, j ] = 0 then remove a from d i Q := Q ∪ ( i, a ) Time complexity of AC-4: O ( e · m 2 ) (provable optimal!) ■ the initialization phase has cost O ( e · m 2 ) ◆ the propagation phase has cost O ( e · m 2 ) ◆ Space complexity of AC-4: O ( e · m 2 ) ■ 20 / 43

  32. Example of AC-4 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z During initialization, AC-4 performs all possible constraint checks for ■ every value in each domain For c 1 : 4 · 4 = 16 constraint checks ◆ For c 2 : 4 · 1 = 4 constraint checks ◆ In total 20 constraint checks ◆ 21 / 43

  33. Example of AC-4 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z After initialization: ■ N [ x, 1 , y ] = 4 N [ y, 1 , x ] = 1 N [ y, 1 , z ] = 1 N [ z, 3 , y ] = 3 N [ x, 2 , y ] = 3 N [ y, 2 , x ] = 2 N [ y, 2 , z ] = 1 N [ x, 3 , y ] = 2 N [ y, 3 , x ] = 3 N [ y, 3 , z ] = 0 N [ x, 4 , y ] = 1 N [ y, 4 , x ] = 4 N [ y, 4 , z ] = 1 S [ x, 1] = { ( y, 1) , ( y, 2) , ( y, 3) , ( y, 4) } S [ y, 1] = { ( z, 3) , ( x, 1) } S [ x, 2] = { ( y, 2) , ( y, 3) , ( y, 4) } S [ y, 2] = { ( z, 3) , ( x, 1) , ( x, 2) } S [ x, 3] = { ( y, 3) , ( y, 4) } S [ y, 3] = { ( x, 1) , ( x, 2) , ( x, 3) } S [ x, 3] = { ( y, 4) } S [ y, 4] = { ( z, 3) , ( x, 1) , ( x, 2) , ( x, 3) , ( x, 4) } S [ z, 3] = { ( y, 1) , ( y, 2) , ( y, 4) } 21 / 43

  34. Example of AC-4 The only counter equal to zero is N [ y, 3 , z ] ■ ( y, 3) is removed, propagation loop starts with ( y, 3) in Q ■ When ( y, 3) is picked, traverse S [ y, 3] = { ( x, 1) , ( x, 2) , ( x, 3) } ■ N [ x, 1 , y ] , N [ x, 2 , y ] , N [ x, 3 , y ] are decremented ◆ No counter becomes zero, so nothing is added to Q ■ Propagation did not require any extra constraint check! ■ However ■ Space complexity is very high ◆ The initialization phase can be prohibitive ◆ (AC-4 has optimal worst-case complexity ... but always reaches this worst case) 22 / 43

  35. AC-6 Motivation: keep the optimal worst-case of AC-4 ■ but instead of counting all supports that a value has, just ensure that there is at least one AC-6 keeps the smallest support for each ( x i , a ) on c ij ■ 23 / 43

  36. AC-6 Motivation: keep the optimal worst-case of AC-4 ■ but instead of counting all supports that a value has, just ensure that there is at least one AC-6 keeps the smallest support for each ( x i , a ) on c ij ■ Initialization phase : cheaper than in AC-4 ■ Propagation phase : if value b of var x j is removed ■ if it is not the current support of value a of var x i : ◆ no work due to constraint c ij has to be done if it is the current support of value a of var x i : ◆ a new support is sought, but starting from next value of b in d j instead of min { d j } 23 / 43

  37. AC-6 Motivation: keep the optimal worst-case of AC-4 ■ but instead of counting all supports that a value has, just ensure that there is at least one AC-6 keeps the smallest support for each ( x i , a ) on c ij ■ Initialization phase : cheaper than in AC-4 ■ Propagation phase : if value b of var x j is removed ■ if it is not the current support of value a of var x i : ◆ no work due to constraint c ij has to be done if it is the current support of value a of var x i : ◆ a new support is sought, but starting from next value of b in d j instead of min { d j } The algorithm uses: ■ S [ j, b ] = “set of ( i, a ) s.t. b is current support of value a of x i wrt. x j ” Q = “set of ( i, a ) s.t. a has been pruned from d i but its effect has not been updated on S yet” 23 / 43

  38. AC-6 procedure AC-6 ( X, D, C ) Q := ∅ ; S [ j, b ] := ∅ , ∀ x j ∈ X , ∀ b ∈ d j // Initialization phase for each x i ∈ X, c ij ∈ C , a ∈ d i do b := smallest value in d j s.t. c ij ( a, b ) if b � = ⊥ then add ( i, a ) to S [ j, b ] else remove a from d i and add ( i, a ) to Q // Propagation phase while Q � = ∅ do ( j, b ) := Fetch ( Q ) for each ( i, a ) ∈ S [ j, b ] such that a ∈ d i do c := smallest value b ′ ∈ d j s.t. b ′ > b ∧ c ij ( a, b ′ ) if c � = ⊥ then add ( i, a ) to S [ j, c ] else remove a from d i and add ( i, a ) to Q Time complexity: O ( e · m 2 ) ■ Space complexity: O ( e · m ) ■ 24 / 43

  39. Example of AC-6 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z In initialization, AC-6 performs the same number of constraint checks as ■ AC-3: 10+4 on c 1 and 4+1 on c 2 S [ x, 1] = { ( y, 1) , ( y, 2) , ( y, 3) , ( y, 4) } S [ y, 1] = { ( z, 3) , ( x, 1) } S [ x, 2] = {} S [ y, 2] = { ( x, 2) } S [ x, 3] = {} S [ y, 3] = { ( x, 3) } S [ x, 3] = {} S [ y, 4] = { ( x, 4) } S [ z, 3] = { ( y, 1) , ( y, 2) , ( y, 4) } 25 / 43

  40. Example of AC-6 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z In initialization, AC-6 performs the same number of constraint checks as ■ AC-3: 10+4 on c 1 and 4+1 on c 2 S [ x, 1] = { ( y, 1) , ( y, 2) , ( y, 3) , ( y, 4) } S [ y, 1] = { ( z, 3) , ( x, 1) } S [ x, 2] = {} S [ y, 2] = { ( x, 2) } S [ x, 3] = {} S [ y, 3] = { ( x, 3) } S [ x, 3] = {} S [ y, 4] = { ( x, 4) } S [ z, 3] = { ( y, 1) , ( y, 2) , ( y, 4) } Q contains ( y, 3) , and 3 has been removed from the domain of y ■ 25 / 43

  41. Example of AC-6 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z In initialization, AC-6 performs the same number of constraint checks as ■ AC-3: 10+4 on c 1 and 4+1 on c 2 S [ x, 1] = { ( y, 1) , ( y, 2) , ( y, 3) , ( y, 4) } S [ y, 1] = { ( z, 3) , ( x, 1) } S [ x, 2] = {} S [ y, 2] = { ( x, 2) } S [ x, 3] = {} S [ y, 3] = { ( x, 3) } S [ x, 3] = {} S [ y, 4] = { ( x, 4) } S [ z, 3] = { ( y, 1) , ( y, 2) , ( y, 4) } When AC-6 enters the propagation loop it pops ( y, 3) from Q , ■ S [ y, 3] = { ( x, 3) } is traversed and a new support greater than 3 is sought for ( x, 3) . 25 / 43

  42. Example of AC-6 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z In initialization, AC-6 performs the same number of constraint checks as ■ AC-3: 10+4 on c 1 and 4+1 on c 2 S [ x, 1] = { ( y, 1) , ( y, 2) , ( y, 3) , ( y, 4) } S [ y, 1] = { ( z, 3) , ( x, 1) } S [ x, 2] = {} S [ y, 2] = { ( x, 2) } S [ x, 3] = {} S [ y, 3] = { ( x, 3) } S [ x, 3] = {} S [ y, 4] = { ( x, 4) } S [ z, 3] = { ( y, 1) , ( y, 2) , ( y, 4) } Just 1 extra constraint check: ■ as c 1 (3 , 4) , we add ( x, 3) to S [ y, 4] 25 / 43

  43. Example of AC-6 Let x, y, z be vars with domains d x = d y = { 1 , 2 , 3 , 4 } , d z = { 3 } , and ■ c 1 ≡ x ≤ y , c 2 ≡ y � = z In initialization, AC-6 performs the same number of constraint checks as ■ AC-3: 10+4 on c 1 and 4+1 on c 2 S [ x, 1] = { ( y, 1) , ( y, 2) , ( y, 3) , ( y, 4) } S [ y, 1] = { ( z, 3) , ( x, 1) } S [ x, 2] = {} S [ y, 2] = { ( x, 2) } S [ x, 3] = {} S [ y, 3] = { ( x, 3) } S [ x, 3] = {} S [ y, 4] = { ( x, 4) } S [ z, 3] = { ( y, 1) , ( y, 2) , ( y, 4) } In total 20 constraint checks ■ 25 / 43

  44. Weaker than AC Directional AC (DAC) ■ Bounds Consistency (BC) ■ 26 / 43

  45. Directional Arc Consistency Given an order ≺ among variables, ■ each x i only needs supports with respect to greater variables in the order Consider a CSP P = ( X, D, C ) . ■ Constraint c ij ∈ C is directionally arc-consistent iff x i ≺ x j implies x i is arc-consistent with respect to x j The CSP P is directionally arc-consistent (DAC) ■ iff all its constraints are directionally arc-consistent DAC is weaker than AC but is enforced more efficiently in practice ■ 27 / 43

  46. Directional Arc Consistency Consider CSP with vars x ≺ y ≺ z , ■ domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 3 x < y y < z z x 0 2 3 1 2 3 x > z 28 / 43

  47. Directional Arc Consistency Consider CSP with vars x ≺ y ≺ z , ■ domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 3 x < y y < z z x 0 2 3 1 2 3 x > z 28 / 43

  48. Directional Arc Consistency Consider CSP with vars x ≺ y ≺ z , ■ domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 x < y y < z z x 0 2 3 1 2 3 x > z 28 / 43

  49. Directional Arc Consistency Consider CSP with vars x ≺ y ≺ z , ■ domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 x < y y < z z x 0 2 3 1 2 x > z 28 / 43

  50. Directional Arc Consistency Consider CSP with vars x ≺ y ≺ z , ■ domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 x < y y < z z x 0 2 3 1 x > z 28 / 43

  51. Directional Arc Consistency Consider CSP with vars x ≺ y ≺ z , ■ domains d x = d y = { 1 , 2 , 3 } and d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 x < y y < z z x 0 2 3 1 x > z Inconsistency is not detected by DAC! ■ 28 / 43

  52. DAC enforcing procedure DAC ( X, D, C ) for each i := n − 1 downto 1 do for each c ij s.t. x i ≺ x j do Revise ( i, j ) endprocedure Only one pass is required ■ Once x i is made arc-consistent with respect to x j ≻ x i , removing values ■ from x k ≺ x i does not destroy the arc-consistency of x i wrt. x j Time complexity: O ( e · m 2 ) ■ 29 / 43

  53. DAC and Acyclic CSP’s A CSP is acyclic if its interaction graph has no loops ■ Theorem. Acyclic CSP’s can be solved in time O ( e · m 2 ) ■ To prove this theorem we need some ingredients ■ Given an acyclic graph, i.e. a forest, ■ let us choose a root for each tree and orient edges away from the roots Given a directed acyclic graph G = ( V, E ) , ■ a topological ordering is a sequence of all vertices in V s.t. if ( u, v ) ∈ E then u comes before v in the sequence 30 / 43

  54. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 This CSP is acyclic, as its interaction graph is: x 3 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 ( x 3 , x 4 , x 5 , x 1 , x 2 ) is a topological ordering ◆ ( x 3 , x 2 , x 4 , x 1 , x 5 ) is a topological ordering ◆ ( x 3 , x 2 , x 1 , x 4 , x 5 ) is not a topological ordering ◆ 31 / 43

  55. DAC and Acyclic CSP’s Theorem. Acyclic CSP’s can be solved in time O ( e · m 2 ) ■ Proof. Consider the following algorithm: // Pre: The graph of the CSP ( X, D, C ) is a tree rooted at x 1 // ( x 1 , x 2 , . . . , x n ) is a topological ordering // Post: The algorithm returns a solution µ procedure AcyclicSolver ( X, D, C ) ( X, D, C ) := DAC ( X, D, C ) // enforce DAC wrt. the ordering a := any element from d 1 µ := ( x 1 �→ a ) for each i := 2 to n do // Any non-root node x i of the tree has a parent x j := parent( x i ) // x j already assigned due to topological ordering v := any support of µ ( x j ) from d i // ∃ because DAC µ := µ ◦ ( x i �→ v ) 32 / 43

  56. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ First we enforce DAC ■ x 3 1 2 3 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 5 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  57. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ First we enforce DAC ■ x 3 1 2 3 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 5 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  58. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ First we enforce DAC ■ x 3 1 2 3 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  59. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ First we enforce DAC ■ x 3 1 2 3 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  60. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ First we enforce DAC ■ x 3 1 2 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  61. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ First we enforce DAC ■ x 3 2 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  62. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ Second we build the assignment ■ x 3 2 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  63. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ Second we build the assignment ■ x 3 2 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  64. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ Second we build the assignment ■ x 3 2 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  65. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ Second we build the assignment ■ x 3 2 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  66. DAC and Acyclic CSP’s Consider a CSP with 5 integer vars with domain [1 , 5] , and constraints ■ | x 2 − x 3 | = 3 , x 3 > x 4 , x 4 + x 1 = 5 , x 4 = x 5 Let us take the topological ordering ( x 3 , x 4 , x 5 , x 1 , x 2 ) ■ Second we build the assignment ■ x 3 2 4 5 | x 2 − x 3 | = 3 x 3 > x 4 x 4 x 2 1 2 3 4 1 2 3 4 5 x 4 = x 5 x 4 + x 1 = 5 x 5 x 1 1 2 3 4 5 1 2 3 4 5 33 / 43

  67. Bounds Consistency AC may be too costly when domains are very large ■ Idea: Establish an order on domain values and ■ require supports only for the extreme values (i.e., min { d i } and max { d i } ) Consider a CSP ( X, D, C ) with ordered domains ■ Variable x i ∈ X is bounds-consistent wrt. x j iff ◆ min { d i } and max { d i } have a support in x j Constraint c ij ∈ C is bounds-consistent iff ◆ x i is bounds-consistent wrt. x j , and x j wrt. x i The CSP is bounds-consistent iff ◆ all its constraints are bounds-consistent Notation: BC means bounds-consistent ■ BC weaker than AC, but can be enforced more efficiently in practice ■ 34 / 43

  68. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 3 x < y y < z z x 0 2 3 1 2 3 x > z 35 / 43

  69. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 3 x < y y < z z x 0 2 3 1 2 3 x > z 35 / 43

  70. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 1 2 3 x < y y < z z x 0 2 3 1 2 x > z 35 / 43

  71. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 3 x < y y < z z x 0 2 3 1 2 x > z 35 / 43

  72. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 0 2 3 1 2 x > z 35 / 43

  73. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 2 3 1 2 x > z 35 / 43

  74. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 1 2 x > z 35 / 43

  75. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 1 x > z 35 / 43

  76. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 x > z 35 / 43

  77. Bounds Consistency Let x, y ∈ X be integer variables with domains [1 , 10] . ■ Constraint | x − y | > 5 is BC (but not AC nor DAC) Consider CSP with vars x, y, z , domains d x = d y = { 1 , 2 , 3 } and ■ d z = { 0 , 2 , 3 } , and constraints x < y, y < z, x > z y 2 x < y y < z z x 3 x > z The examples show that DAC and BC are incomparable ■ (and both weaker than AC) 35 / 43

  78. BC-3: ReviseBounds( i , j ) Natural adaptation of AC-3 to bounds consistency: BC-3 ■ Based on function ReviseBounds ( i , j ), ■ which removes values from the extremes of d i without support in d j Returns true if some value is removed function ReviseBounds ( i, j ) change := false while | d i | � = ∅ ∧ ( ∀ b ∈ d j ¬ c ij (min { d i } , b )) do change := true remove min { d i } from d i while | d i | � = ∅ ∧ ∀ b ∈ d j ¬ c ij (max { d i } , b ) do change := true remove max { d i } from d i return change The time complexity of ReviseBounds ( i, j ) is O ( | d i | · | d j | ) ■ 36 / 43

  79. BC-3 // ( i, j ) ∈ Q means “cannot guarantee min { d i } , max { d i } have support in d j ” procedure BC3 ( X, D, C ) Q := { ( i, j ) , ( j, i ) | c ij ∈ C } // each constraint is added twice while Q � = ∅ do ( i, j ) := Fetch ( Q ) if ReviseBounds ( i, j ) then Q := Q ∪ { ( k, i ) | c ki ∈ C, k � = j } Space complexity: O ( e ) ■ Time complexity: O ( e · m 3 ) ■ Same asymptotic costs of AC-3, but BC-3 is more efficient in practice ■ 37 / 43

  80. Stronger than AC Singleton AC (SAC) ■ Neighborhood Inverse Consistency (NIC) ■ 38 / 43

  81. Singleton AC Let AC ( P ) denote the CSP resulting from enforcing AC on P ■ If AC ( P [ x i → a ]) has an empty domain, ■ then P [ x i → a ] does not have any solution, and a ∈ d i is unfeasible in P A CSP P = ( X, D, C ) is singleton arc-consistent (SAC) iff ■ ∀ d i ∈ D, ∀ a ∈ d i , problem AC ( P [ x i → a ]) has no empty domains 39 / 43

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