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Arc consistency

Arc consistency. V4  V3. V4. V3. V1  V4  1. V4 + V2 = 5. V2  V3  6. V1. Di = {1,2,3,4,5}. V2. V1  V2. What can you infer?. AR33 figure 18, page 35. V4  V3. V4. V3. V1  V4  1. V4 + V2 = 5. V2  V3  6. V1. V2. V1  V2. Di = {1,2,3,4,5}. D1 = {1,2}

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Arc consistency

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  1. Arc consistency

  2. V4  V3 V4 V3 V1  V4  1 V4 + V2 = 5 V2  V3  6 V1 Di = {1,2,3,4,5} V2 V1  V2 What can you infer? AR33 figure 18, page 35

  3. V4  V3 V4 V3 V1  V4  1 V4 + V2 = 5 V2  V3  6 V1 V2 V1  V2 Di = {1,2,3,4,5} D1 = {1,2} D2 = {2,3} D3 = {4,5} D4 = {2,3} Was that easy? Do you agree?

  4. I wrote a program!

  5. Here’s the reasoning Di = {1,2,3,4,5} V1  V2 V4  V3 V2  V3  6 V1  V4  1 V4 + V2 = 5 V1 < V2 D1 = {1,2,3,4,5} D2 = {1,2,3,4,5} D1 := {1,2,3,4} V2 > V1 D1 = {2,3,4,5} D2 = {1,2,3,4,5} D2 := {2,3,4,5} V4 >= V1 + 1 D4 = {1,2,3,4,5} D1 = {1,2,3,4} D4 := {2,3,4,5} V1 <= V4 - 1 D1 = {1,2,3,4} D4 = {2,3,4,5} no change V2 + V3 > 6 D2 = {2,3,4,5} D3 = {1,2,3,4,5} no change V3 + V2 > 6 D3 = {1,2,3,4,5} D2 = {2,3,4,5} D3 := {2,3,4,5} V2 + V4 = 5 D2 = {2,3,4,5} V4 = {2,3,4,5} D2 = {2,3} V1 < V2 D1 = {1,2,3,4} D2 = {2,3} D1 = {1,2} V2 > V1 D2 = {2,3} D1 = {1,2} no change V4 >= V1 + 1 D4 = {2,3,4,5} D1 = {1,2} no change V3 + V2 > 6 D3 = {2,3,4,5} D2 = {2,3} D3 = {4,5} V2 + V3 > 6 D2 = {2,3} D3 = {4,5} no change V4 + V2 = 5 D4 = {2,3,4,5} D2 = {2,3} D4 = {2,3} V1 <= V4 - 1 D1 = {1,2} D4 = {2,3} no change V2 + V4 = 5 D2 = {2,3} D4 = {2,3} no change V4 < V3 D4 = {2,3} D3 = {4,5} no change V3 > V4 D3 = {4,5} D4 = {2,3} no change

  6. Arc consistency: so what’s that then?

  7. A constraint Cij is arc consistent if • for every value x in Di there exists a value y in Dj that supports x • i.e. if v[i] = x and v[j] = y then Cij holds • note: we are assuming Cij is a binary constraint • A csp (V,D.C) is arc consistent if • every constraint is arc consistent This is also called 2-consistency • If (V,D,C) is arc consistent then • I can choose any variable v[i] • assign it a value x from its domain Di • I can now choose any other variable v[j] • I can find a consistent instantiation for v[j] from Dj • NOTE: this is in isolation, where I have only 2 variables that I instantiate

  8. A constraint Cij is arc consistent if • for every value x in Di there exists a value y in Dj that supports x • i.e. if v[i] = x and v[j] = y then Cij holds • note: we are assuming Cij is a binary constraint • A csp (V,D.C) is arc consistent if • every constraint is arc consistent

  9. Just because a problem (V,D,C) is arc consistent does not mean that it has a solution! But if it is a tree ... Arc-consistency processes a problem and removes from the domains of variables values that CANNOT occur in any solution Arguably, it makes the resultant problem easier. Why? The arc-consistent problem has the same set of solutions as the original problem

  10. So, is there 1-consistency? • Yip • when we have unary constraints • example odd(V[i]) • 1-consistency, we weed out all odd values from Di • also called node-consistency (NC) 3-consistency? Given constraints Cij and Cjk, disallow all pairs (x,z) in the constraint Cik where there is no value y in Dj such that Cij(x,y) and Cjk(y,z) This adds nogood tuples to an existing constraints, or creates a new constraint! sometimes called path-consistency (PC) (Given 2 variables in isolation, we can instantiate those consistently, and pick any third variable and …)

  11. ac3: Mackworth 1977 Alan Mackworth presented ac1, ac2, and ac3 in 1977. Ac1 and ac2 were “straw men”

  12. ac3: revise a constraint (claire3 code) [revise(d:array<set>,c:tuple,i:integer,j:integer) : boolean ->let revised := false in (for x in copy(d[i]) (let supported := false in (for y in d[j] (if check(c,x,y) (supported := true, return())), if not(supported) (delete(d[i],x), revised := true))), CONSISTENT := d[i] != {}, revised)]

  13. The micro-structure of C23 1 1 support 2 2 3 3 4 4 5 5 revise searches for 1st support for a value Is it a bijection? What are implications of this?

  14. ac3 [ac3(d:array<set>,constraints:set) : boolean -> let Q := copy(constraints) in (while (Q != {} & CONSISTENT) let c := Q[1], i := c[2], j := c[3] in (delete(Q,c), if revise(d,c,i,j) Q := Q U {z in constraints | z[3] = i})), CONSISTENT]

  15. ac3 if revise(d,c,i,j) Q := Q U {z in constraints | z[3] = i} What?

  16. Note: ac3 has a queue of constraints that need revision because some values in the domains of the variables may be unsupported. Remember: ac3 processes a queue of constraints But forgive me, the queue might be treated as just a set

  17. Ac1 and ac2 (the straw men) essentially revised constraints over and over again, until no change … until reaching a fixed point

  18. [ac3(d:array<set>,constraints:set) : boolean -> let Q := copy(constraints) in (while (Q != {} & CONSISTENT) let c := Q[1], i := c[2], j := c[3] in (delete(Q,c), if revise(d,c,i,j) Q := Q U {z in constraints | z[3] = i})), CONSISTENT] e is number of constraints d is domain size Complexity of ac3 proved in 1985 by Mackworth & Freuder (AIJ 25)

  19. [ac3(d:array<set>,constraints:set) : boolean -> let Q := copy(constraints) in (while (Q != {} & CONSISTENT) let c := Q[1], i := c[2], j := c[3] in (delete(Q,c), if revise(d,c,i,j) Q := Q U {z in constraints | z[3] = i})), CONSISTENT] e is number of constraints d is domain size Prove it! Also look at paper by Zhang & Yap

  20. [ac3(d:array<set>,constraints:set) : boolean -> let Q := copy(constraints) in (while (Q != {} & CONSISTENT) let c := Q[1], i := c[2], j := c[3] in (delete(Q,c), if revise(d,c,i,j) Q := Q U {z in constraints | z[3] = i})), CONSISTENT] Notice similarity to depth first search (dfs) The order that we revise the constraints make no difference to the outcome It reaches the same fixed point, the same set of arc-consistent domains The order that we revise the constraints may make a difference to run time. … constraint ordering heuristic, anyone?

  21. Revise ignored any semantics of the constraint Is that dumb, or what? • Could we get round this? • Use OOP? • A class of constraint? • revise as a specialised method?

  22. AC4, AC6, AC7, …. “Optimal” support counting algorithms

  23. AC4, AC6, AC7, ... 1 1 2 2 3 3 4 4 5 5 • Associate with each value in Di • a counter support[x,i,j] • the number of values in Dj that support x • a boolean supports[x,y,j] • true if x supports y in Dj 1st stage of the algorithm builds up the support counts • 2nd stage • if support[x,i,j] = 0 (x has no support in Dj over constraint Cij) • delete(Di,x) • decrement support[y,k,i] (where supports[x,y,k] is true) • continue this till no change • i.e. propagate

  24. Best case and worst case performance of ac4 is the same • Ac6, 7, and 8 exploit symmetries, and lazy evaluation • if x supports y over constraint Cij then y supports x over Cji • find the 1st support for x, and only look for more when support is lost Ac3 worst case performance rarely occurs (experimental evidence due to Rick Wallace)

  25. Why is best case and worst case performance of ac4 ?

  26. AC4, AC6, AC7, ... 1 1 2 2 3 3 4 4 5 5 • Associate with each value in Di • a counter support[x,i,j] • the number of values in Dj that support x • a boolean supports[x,y,j] • true if x supports y in Dj 1st stage of the algorithm builds up the support counts • 2nd stage • if support[x,i,j] = 0 (x has no support in Dj over constraint Cij) • delete(Di,x) • decrement support[y,k,i] (where supports[x,y,k] is true) • continue this till no change • i.e. propagate

  27. History Lesson • ac4 Mohr & Henderson AIJ28 1986 • ac6 , 7, 8 due to Freuder, Bessiere, Regin, and others • in AIJ, IJCAI, etc Downside of these algorithms is “hard to code” ac3 is easy!

  28. AC5 A generic arc-consistency algorithm and its specializations AIJ 57 (2-3) October 1992 P. Van Hentenryck, Y. Deville, and C.M. Teng

  29. ac5 Ac5 is a “generic” ac algorithm and can be specialised for special constraints (i.e. made more efficient when we know something about the constraints) Ac5 is at the heart of constraint programming It can be “tweaked” to become ac3, or “tweaked” to become ac4 But, it can exploit the semantics of constraints, and there’s the win

  30. ac5 • ac5 processes a queue of tuples (i,j,w) • w has been removed from Dj • we now need to reconsider constraint Cij • to determine the unsupported values in Di • A crucial difference between ac3 and ac5 • ac3 • revise(i,j) because Dj has lost some values • seek support for each and every value in Di • ac5 • process(i,j,w) because Dj has lost the value w (very specific) • seek support, possibly, for select few (1?) value in Di Therefore, by having a queue of (i,j,w) we might be able to do things much more efficiently

  31. ac5, the algorithm [arcCons(d:array<set>,c:tuple) : set -> let s := {}, i := c[2], j := c[3] in (for x in d[i] let supported := false in (for y in d[j] (supported := check(c,x,y), if supported return()), if not(supported) s := s U set(x)), s)] This is just like revise, but delivers the set s of unsupported values in d[i] over the constraint C_i,j s is the set of disallowed values in d[i] remember c is a tuple <r,i,j>, the relation r between v[i] and v[j]

  32. ac5, the algorithm [localArcCons(d:array<set>,c:tuple,w:integer) : set -> arcCons(d,c)] // // opportunity to specialise this with resepect to c // localArcCons(d,c,w) - c is the tuple <r,i,j> - d[j] lost the value w - return the set of values unsupported in d[i] We can specialise this method if we know something about the semantics of c We might then do much better that arcCons in terms of complexity What we have above is the dummy’s version, or when we know nothing about the constraint (and we get AC3!)

  33. ac5, the algorithm [enqueue(i:integer,delta:set,q:set,constraints:set) : set -> for c in {z in constraints | z[3] = i} for w in delta q := q U set(tuple(c,w)), q] // // d[i] has lost the values delta // add to the q tuples (c,w) where // - the constraint c is the constraint C_k,i // - w is in delta // The domain d[i] has lost values delta Add to the queue of “things to do” the tuple (c,w) where - c is the constraint C_k,i and - w is in delta This is similar to the step in AC3 Q := Q U {z in constraints | z[3] = i}

  34. ac5, the algorithm [ac5(d:array<set>,constraints:set) : boolean -> let Q := {} in (for c in constraints let delta := arcCons(d,c), i := c[2] in (d[i] := difference(d[i],delta), Q := enqueue(i,delta,Q,constraints)), while (Q != {}) (let t := Q[1], c := t[1], w := t[2], i := c[2], delta := localArcCons(d,c,w) in (delete(Q,t), Q := enqueue(i,delta,Q,constraints), d[i] := difference(d[i],delta)))), CONSISTENT] ac5 has 2 phases - revise all the constraints and build up the Q, using arcCons - process the queue and propagate using localArcCons, exploiting knowledge on deleted values and the semantics of constraints

  35. Complexity of ac5 If arcCons is O(d2) and localArcCons is O(d) then ac5 is O(e.d2) If arcCons is O(d) and localArcCons is O() then ac5 is O(e.d) See AIJ 57 (2-3) page 299 Remember - ac3 is O(e.d3) - ac4 is O(e.d2)

  36. Functional constraints

  37. 1 1 2 2 3 3 4 4 5 5 If c is functional We can specialise arcCons and localArcCons for functional constraints • A constraint C_i,j is functional, with respect to a domainD, if and only if • for all x in D[i] • there exists at most one value y in D[j] such that • C_i,j(x,y) holds • and • for all y in D[j] • there exists at most one value in D[i] such that • C_j,i(y,x) holds • The relation is a bijection An example: 3.x = y

  38. 1 1 2 2 3 3 4 4 5 5 If c is functional Let f(x) = y and f’(y) = x (f’ is the inverse of f) [arcCons(d:array<set>,c:tuple) : set -> let s := {}, f := c[1], i := c[2], j := c[3] in (for x in d[i] if not(f(x) % d[j]) s := s U set(x), s)] // // functional arcCons // the relation is a bijection // The % operator is 

  39. Claire manual, section 4.5, inverse of a relation!

  40. 1 1 2 2 3 3 4 4 5 5 If c is functional Let inverse(f) deliver the f’, the inverse of function f [localArcCons(d:array<set>,c:tuple,w:integer) : set -> let f := c[1], i := c[2], j := c[3], g := inverse(f) in if (g(w) % d[i]) set(g(w)) else {}] // // assume inverse delivers the inverse of a function // localArcCons(d,c,w) - c is the tuple <r,i,j> - d[j] lost the value w - return the set of values unsupported in d[i] • In our picture, • assume d[j] looses the value 2 • g(2) = 3 • localArcCons(d,c12,2) = {3}

  41. Anti-functional constraints

  42. 1 1 2 2 3 3 4 4 5 5 If c is anti-functional The value w in d[j] conflicts with only one value in d[i] an example: x + 4 != y [localArcCons(d:array<set>,c:tuple,w:integer) : set -> let f := c[1], i := c[2], j := c[3], g := inverse(f) in if (size(d[i]) = 1 & g(w) = d[i][1]) set(d[i][1]) else {}] // // assume inverse delivers the inverse of a function // If D[i] has more than one value, no problem!

  43. Other kinds of constraints that can be specialised • monotonic •     • localArcCons tops and tails domains • but domains must be ordered • piecewise functional • piecewise anti-functional • piecewise monotonic See: AIJ 57 & AR33 section 7

  44. AIJ57 • The AIJ57 paper by van Hentenryck, Deville, and Teng is a remarkable paper • they tell us how to build a constraint programming language • they tell us all the data structures we need • for representing variables, their domains, etc • they introduce efficient algorithms for arc-consistency • they show us how to recognise special constraints and how we should process them • they show us how to incorporate ac5 into the search process • they explain all of this in easy to understand, well written English • It is a classic paper, in my book!

  45. AC2001 Taken from IJCAI01 “Making AC-3 an Optimal Algorithm” by Y Zhang and R. Yap and “Refining the basic constraint propagation algorithm” by Christian Bessiere & Jean-Charles Regin

  46. The complexity of ac3 A beautiful proof • A constraint C_i,j is revised iff it enters the Q • C_i,j enters the Q iff some value in d[j] is deleted • C_i,j can enter Q at most d times (the size of domain d[j]) • A constraint can be revised at most d times • There are e constraints in C (the set of constraints) • revise is therefore executed at most e.d times • the complexity of revise is O(d2) • the complexity of ac3 is then O(e.d3)

  47. When searching for a support for x in d[j] we always start from scratch This is naïve. [revise(d:array<set>,c:tuple,i:integer,j:integer) : boolean ->let revised := false in (for x in copy(d[i]) (let supported := false in (for y in d[j] // this is naive (if check(c,x,y) (supported := true, return())), if not(supported) (delete(d[i],x), revised := true))), CONSISTENT := d[i] != {}, revised)]

  48. Assume domains are totally ordered • Let resumePoint[i,j,x] = y • where y is the first value in d[j] that supports x in d[i] • i.e. C_i,j is satisfied by the pair (x,y) • Assume succ(y,d) delivers the successor of y in domain d • (and nil if no successor exists) • When we revise constraint C_i,j, we look for support for x in d[j] • get the resumePoint for x in domain d[j] over the constraint • call this value y • if y is in d[j] then x is supported! Do nothing! • If y is not in d[j] then • we search through the remainder of d[j] looking for support • we use a successor function to get next y in d[j] • if we find a y that satisfies the constraint set resumePoint[i,j,x] = y

  49. [succ(x:integer,l:list) : integer -> if (l = nil) -1 else if (l[1] > x) l[1] else succ(x,cdr(l))] // // find the successor of x in ordered list l // NOTE: in the ac2001 algorithms we assume // we can do this in O(1) //

  50. [existsY(d:array<set>,c:tuple,i:integer,j:integer,x:integer) : boolean -> let y := RP[i][j][x] in (if (y % d[j]) true else let supported := false in (y := succ(y,list!(d[j])), while (y > 0 & not(supported)) (if check(c,x,y) (RP[i][j][x] := y, supported := true) else y := succ(y,list!(d[j]))), supported))] // // find the first value y in d[j] that supports x in d[i] // if found return true otherwise false //

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