local consistency
play

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

Nov 05, 2022 •434 likes •1.55k views

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

Recommend

Consistency - Chapter 5 Introduce several notions of Local Consistency: arc consistency,

Consistency - Chapter 5 Introduce several notions of Local Consistency: arc consistency,

Constraints Course - Justin Pearson - Consistency 1 Consistency - Chapter 5 Introduce several notions of Local Consistency: arc consistency, hyper-arc consistency, k-consistency and strong k-consistency. Local

728 views • 37 slides
Web Cache Consistency Web Cache Consistency Web Cache Consistency Web Cache Consistency

Web Cache Consistency Web Cache Consistency Web Cache Consistency Web Cache Consistency

Web Cache Consistency Web Cache Consistency Web Cache Consistency Web Cache Consistency Requirements of performance, availability, and disconnected operation require us to relax the goal of semantic transparency. - HTTP 1.1 specification

598 views • 13 slides
Constraint Programming - An overview Node-consistency Arc-consistency Path-consistency

Constraint Programming - An overview Node-consistency Arc-consistency Path-consistency

Constraint Programming - An overview Node-consistency Arc-consistency Path-consistency I-consistency Generalised Arc-consistency Bounds-Consistency Propagators: The COMET example 7 November 2011 Constraint Propagation

615 views • 60 slides
Unifying Local Consistency and MAX SAT Relaxations for Scalable Inference with Rounding

Unifying Local Consistency and MAX SAT Relaxations for Scalable Inference with Rounding

Unifying Local Consistency and MAX SAT Relaxations for Scalable Inference with Rounding Guarantees Stephen H. Bach Bert Huang Lise Getoor Maryland Virginia Tech UC Santa Cruz AISTATS

715 views • 35 slides
TREE-TRANSVERSE MATCHINGS AND LOCAL CONSISTENCY Ross Churchley Simon Fraser University 13 June

TREE-TRANSVERSE MATCHINGS AND LOCAL CONSISTENCY Ross Churchley Simon Fraser University 13 June

Introduction Small tree-transverse matchings Local Consistency Open Questions TREE-TRANSVERSE MATCHINGS AND LOCAL CONSISTENCY Ross Churchley Simon Fraser University 13 June 2013 Introduction Small tree-transverse matchings Local

779 views • 57 slides
Constraint sa*sfac*on problems II CS171, Fall 2016 Introduc*on to Ar*ficial Intelligence Prof.

Constraint sa*sfac*on problems II CS171, Fall 2016 Introduc*on to Ar*ficial Intelligence Prof.

Constraint sa*sfac*on problems II CS171, Fall 2016 Introduc*on to Ar*ficial Intelligence Prof. Alexander Ihler You Should Know Node consistency, arc consistency, path consistency, K- consistency (6.2) Forward checking (6.3.2) Local

1.2k views • 82 slides
Consistent Storage or Scalable Storage Why Not Both? CONSISTENCY Strong Consistency

Consistent Storage or Scalable Storage Why Not Both? CONSISTENCY Strong Consistency

Consistent Storage or Scalable Storage Why Not Both? CONSISTENCY Strong Consistency Eventual Consistency &quot;Consistency in database systems refers to the requirement that any given database transaction must change affected data only in

668 views • 54 slides
Just-Right As available as possible, consistent when necessary Consistency The CAP theorem

Just-Right As available as possible, consistent when necessary Consistency The CAP theorem

Just-Right Consistency: Just-Right As available as possible, consistent when necessary Consistency The CAP theorem forces a choice In contrast, checking a data between strong consistency (CP) precondition on partitioned state is and

567 views • 10 slides
Lecture 10: Nonparametric Regression (2) Applied Statistics 2015 1 / 18 Consistency of

Lecture 10: Nonparametric Regression (2) Applied Statistics 2015 1 / 18 Consistency of

Consistency of Nadaraya-Watson Estimator Local linear regression Assignments Lecture 10: Nonparametric Regression (2) Applied Statistics 2015 1 / 18 Consistency of Nadaraya-Watson Estimator Local linear regression Assignments Consistency of

207 views • 19 slides
Eventual Consistency In the real world or Why You Already Know Eventual Consistency or

Eventual Consistency In the real world or Why You Already Know Eventual Consistency or

Eventual Consistency In the real world or Why You Already Know Eventual Consistency or Eventual Consistency is better* than Eventual Availability *depending on the use case @roder Matt Heitzenroder We &lt;3 Distributed Systems Basho

423 views • 38 slides
BRAND CONSISTENCY presented by Index Introduction What is Brand Consistency? Why is Brand

BRAND CONSISTENCY presented by Index Introduction What is Brand Consistency? Why is Brand

BRAND CONSISTENCY presented by Index Introduction What is Brand Consistency? Why is Brand Consistency Important? Easy Ways to Keep You Consistent Case Study: Concordia University of Edmonton Summary Contact Information Introduction Who

513 views • 17 slides
This talk is about Data consistency in 3D Understanding consistency (Its the invariants,

This talk is about Data consistency in 3D Understanding consistency (Its the invariants,

This talk is about Data consistency in 3D Understanding consistency (Its the invariants, stupid) Primitive consistency mechanisms How primitives compose models How models relate / differ What they cost Understanding

365 views • 13 slides
Non-Gaussianity Consistency Relation for Multi-Field Inflation for the local form Eiichiro

Non-Gaussianity Consistency Relation for Multi-Field Inflation for the local form Eiichiro

arXiv: 1101.3636 Non-Gaussianity Consistency Relation for Multi-Field Inflation for the local form Eiichiro Komatsu (Texas Cosmology Center, Univ. of Texas at Austin) Cosmological Non-Gaussianity at Univ. of Michigan, May 15, 2011 Nao

356 views • 32 slides
Weak Consistency Dan Ports, CSEP 552 CAP Theorem Cant have all three of consistency,

Weak Consistency Dan Ports, CSEP 552 CAP Theorem Cant have all three of consistency,

Weak Consistency Dan Ports, CSEP 552 CAP Theorem Cant have all three of consistency, availability, and tolerance to partitions (but the devil is in the details!) CAP Eric Brewer, 2000: conjecture on reliable

418 views • 20 slides
Local Search CPSC 322 CSPs 4 Textbook 4.8 Local Search CPSC 322 CSPs 4, Slide 1

Local Search CPSC 322 CSPs 4 Textbook 4.8 Local Search CPSC 322 CSPs 4, Slide 1

Recap Arc Consistency Example Local Search Hill Climbing Local Search CPSC 322 CSPs 4 Textbook 4.8 Local Search CPSC 322 CSPs 4, Slide 1 Recap Arc Consistency Example Local Search Hill Climbing Lecture Overview 1 Recap 2 Arc

705 views • 16 slides
Consistency The syntactic counterpart of satisfiability is consistency. Definition 1 is

Consistency The syntactic counterpart of satisfiability is consistency. Definition 1 is

Computational Logic, Spring 2006 Pete Manolios Consistency The syntactic counterpart of satisfiability is consistency. Definition 1 is consistent, written Con , iff there is no formula such that and . is

199 views • 8 slides
Replication and Consistency Setting: Concurrent threads accessing shared data Roland Meyer (TU

Replication and Consistency Setting: Concurrent threads accessing shared data Roland Meyer (TU

Replication and Consistency in and hardware general memory consistency models in particular Roland Meyer Technische Universit at Kaiserslautern Roland Meyer (TU KL) Replication and Consistency IFIP WG 2.2 09/2016 1 / 32

1.76k views • 150 slides
household recycling Mike Gardner A framework for Local Authority Account Manager greater

household recycling Mike Gardner A framework for Local Authority Account Manager greater

Consistency in household recycling Mike Gardner A framework for Local Authority Account Manager greater consistency in household 22 nd February 2018 recycling in England Within the Framework a 5 point action plan: Packaging:

322 views • 19 slides
The many faces of consistency Marcos K. Aguilera, Douglas B. Terry Presented by Ji Wang Outline

The many faces of consistency Marcos K. Aguilera, Douglas B. Terry Presented by Ji Wang Outline

The many faces of consistency Marcos K. Aguilera, Douglas B. Terry Presented by Ji Wang Outline Introduction: What is consistency and why important Abstract model and terminology Two types of consistency State consistency Operation

390 views • 15 slides
Consistency in key- value stores Monika Moser Consistency guarantees explain a system's

Consistency in key- value stores Monika Moser Consistency guarantees explain a system's

Consistency in key- value stores Monika Moser Consistency guarantees explain a system's behavior Where is my data I just updated? The consistency guarantees of a system influence the behavior perceived by a client. An ideal world The result

591 views • 31 slides
Seminar: Search and Optimization Directional Consistency Gabi R oger Universit at Basel

Seminar: Search and Optimization Directional Consistency Gabi R oger Universit at Basel

Seminar: Search and Optimization Directional Consistency Gabi R oger Universit at Basel November 6, 2014 Directional Arc-consistency Directional Path-consistency Adaptive Consistency Summary Directional Arc-consistency Directional

772 views • 76 slides
Eventual Consistency Eventual Consistency In the real world In the real world or Why you

Eventual Consistency Eventual Consistency In the real world In the real world or Why you

Eventual Consistency Eventual Consistency In the real world In the real world or Why you already know Eventual Consistency /usr/bin/whoami Chris Molozian Client Services Engineer, Basho EMEA Basho Technologies cmolozian@basho.com 03

371 views • 34 slides
Consistency Protocols Description describe an implementation of a specific

Consistency Protocols Description describe an implementation of a specific

Consistency Protocols Description describe an implementation of a specific consistency model Classification Distributed Systems (ICE 601) primary-based protocols remote-write protocols Replication &amp; Consistency

425 views • 4 slides
1 Applications ? Trading Consistency for Performance Applications ? Trading Consistency for

1 Applications ? Trading Consistency for Performance Applications ? Trading Consistency for

Consistency in Replication Consistency in Replication ! Replication comes with consistency cost: Continuous Consistency Continuous Consistency ! Reasons for replication: Better performance and availability and Availability and Availability !

280 views • 6 slides

More recommend