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Jean-Charles REGIN Michel RUEHER regin@ilog.fr rueher@essi.fr ILOG Sophia Antipolis Université de Nice – Sophia Antipolis A global constraint combining a sum constraint and binary inequalities 2 August, 2000 1
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Global constraint IS Combination of: A sum constraint: y=x i and binary Inequalities: x i x j + c ij Example: x 1 + x 2 = y & x 1 x 2 + 1 2 August, 2000 2
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Motivations to improve “back” propagation when solving Optimization Problems in the CP Framework 2 August, 2000 3
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Optimization Problems Objective function : a sum z=x i CP Framework solving a sequence of decision problems (B&B) where z < z* (each solution must be better than the previous one) Local consistency algorithms on each constraint poorly propagate (not very effective) Global constraint IS : to improve the back propagation when the bounds of z are modified 2 August, 2000 4
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Minimizing mean flow time F=1/n j=1…n (C j -r j ) Minimizing tardiness D=1/n j=1…n D j Where D j =max(C j -d j,0) C j, r j, d j : completion time, ready time, and duration of task T j D ifference constraints : precedence/distance constraints between the tasks 2 August, 2000 5 Applications in deterministic scheduling problems
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D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[1,13] C 1 : x 1 + x 2 =y C 2 : x 1 x 2 - 1 This system is arc consistent : each value belongs to a solution of every constraint If the lower bound of y is set to 6 this system remains arc consistent no pruning can be achieved 2 August, 2000 6 Example 1
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D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[6,13] C 1 : x 1 + x 2 =y C 2 : x 1 x 2 - 1 However, when x 2 belongs to [1,3] C 1 cannot be satisfied Global constraint IS will delete [1,3] from the domain D (x 2 ) 2 August, 2000 7 Example 1 (continued)
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Constraint S um : y= j=1…n x j Inequalities I neq : { x j - x j c ji (i,j [1,n])} Domain constraints D om : { l j x j u j (i [1,n])} Global constraint IS = {S um } D om I neq Algorithm (scheme): If a bound of x is modified Filtering D om I neq by interval consistency If a bound of some y is modified Filtering S um by interval consistency Updating the bounds of every x j with respect to IS 2 August, 2000 8 Summary of our framework
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Filtering S um by interval consistency : 0(n) Filtering I neq by interval consistency : 0(mn) Filtering IS by interval consistency : O(n(m + nlogn)) 2 August, 2000 9 Summary of our framework Contribution
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Sum constraint y=x i is interval consistent iff: (1)min(y) j=1…n min(x j ) (2)max(y) j=1…n max(x j ) (3) x i : min(x i ) min(y) - j i max(x j ) (4) x i : max(x i ) max(y) - j i min(x j ) 2 August, 2000 10
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Sum constraint (continued) Checking interval consistency of y=x i is in 0(n) (3) x i : min(x i ) min(y) - j i max(x j ) j i max(x j )= j=1…n max(x j )- max(x i ) 2 August, 2000 11
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Binary inequalities Filtering by AC complexity depends on the size of the domains Filtering by interval consistency can be achieved in O(mn) simple temporal CSP (Dechter et al) 2 August, 2000 12
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Simple Temporal CSP Shortest path : (s,x j ) (s,x i ) + c (x i, x j ) Distance Graph G=(N,E) N: source node s with D (s) ={0} + one node for each variable E: x i x j + c ji arc (x j, x i ) with cost c ji D (x) =[min x, max x ] s x – min(x) arc (x,s) with cost -min(x) x s + max(x) arc (s,x) with cost max(x) 2 August, 2000 13
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Example D (x i ) =[1,6] D (x j ) =[2,5] x j x i - 3 s i j 6 5 -2 -3 2 August, 2000 14 Distance graph (continued)
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x i is interval consistant if D (x i ) =[- (x i, s), (s,x i ) ] Example D (x i ) =[1 5,6] D (x j ) =[2,5 3 ] x j x i - 3 s i j -1 -5 6 5 3 -2 -3 2 August, 2000 15 Distance graph (continued) Computing shortest path : 0(nm) but after ONE computation we can use reduced costs 0(m+n log( n ))
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IS constraint ( Sum & binary Inequalities ) (3) x i : min(x i ) min(y) - ji max (x j ) is too weak to enforce interval consistency on IS Example: D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[6,13] C 1 : x 1 + x 2 =y C 2 : x 1 x 2 - 1 min(x 2 ) min(y) - max (x 1 ) = 0 although IS cannot be satisfied when x 1 3 2 August, 2000 16
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IS constraint ( Sum & binary Inequalities ) Interval consistent iff: min(y) i=1…n min(x i ) max(y) i=1…n max(x i ) (3b) x i : min(x i ) min(y) - ji max x i min(x i ) (x j ) (4b) x i : max(x i ) max(y) - ji min x i max(x i ) (x j ) max x i min(x i ) (x j ) : maximum value of D (xj) which satisfies I neq when x i is set to min(x i ) 2 August, 2000 17
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IS constraint ( Sum & binary Inequalities ) (3b) x i : min(x i ) min(y) - ji max x i min(x i ) (x j ) Example: D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[6,13] C 1 : x 1 + x 2 =y C 2 : x 1 x 2 - 1 min(x 2 ) min(y) - max x 2 min(x 2 ) (x 1 ) = 4 2 August, 2000 18
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How to compute min(x i ) ? (3b) x i : min(x i ) min(y) - ji max x i min(x i ) (x j ) x i = min(y) - j i max x i x i (x j ) 2 August, 2000 19
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Computing x i max (x j )=min( (s,x j ), x i + ’ (x i,x j ) ) (1) xi xi where ’ (x i,x j ) ) is the shortest path from x i to x j in G-{s} and thus x i = min(y) - j i min( (s,x j ), x i + ’ (x i,x j ) ) (2) 2 August, 2000 20 max (x j ) depends only on the upper-bounds of the variables x k that belong to a shortest path from s to x j
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Computing min(x i ) (continued) x i = min(y) - j i min( (s,x j ), x i + ’ (x i,x j ) ) (2) 2 August, 2000 21 algorithm (1) Computing ’ (x i,x j ) ) for all j i and be (s,x j ) - (x i,x j ) = j (2) L sorted sorted list of j S { x j : j 0} S is the set of the x j ’s for which min( (s,x j ), x i + ’ (x i,x j ) ) = (s,x j ) Loop x i = min(y) - j i min( (s,x j ), x i + ’ (x i,x j ) ) S S { x j : j x i } until of S does no more change
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Computing min(x i ) (continued) 2 August, 2000 22 (x i,x j ) can be computed on the graph of reduced costs of G-{s} in O(nlogn) Identifying the shortest paths from x i to x j which go through s iteration step : O(nlogn) Filtering of IS by interval consistency : O(n(m + nlogn)) (no propagation step is required when min(x i ) is increased)
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Extensions Works also for i x i =y (3b) x i 1 (min(y) - ji j (x i,x j ) ) |X| Works also if all the variables do not occur in the sum constraint 2 August, 2000 23
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Conclusion An original combination and an efficient algorithm for a new global constraint to improve propagation in optimization problems Further work : implementation & experimentation 2 August, 2000 24
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Computing min(x i ) (continued) x i 1 (min(y) - ji (x i,x j ) ) |X| 2 August, 2000 25 (x i,x j ) can be computed on the graph of reduced costs of G-{s} Identifying the shortest paths from x i to x j which go through s iteration step : O(nlogn) Filtering of IS by interval consistency : O(n(m + nlogn)) (no propagation step is required when min(x i ) is increased)
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Computing x i max (x j )=min( (s,x j ), x i + (x i,x j ) ) (1) xi xi and thus max (x j ) x i + (x i,x j ) (2) xi xi and x i 1 (min(y) - ji (x i,x j ) ) (3) |X| 2 August, 2000 26 max (x j ) depends only on the upper-bounds of the variables x k that belong to a shortest path from s to x j
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Computing min(x i ) (continued) max (x j )= x i + (x i,x j ) (2) xi min(xi) 2 August, 2000 27 Proof (scheme) Two cases : 1. x i belongs to a shortest path from s to x j when x i is set to the value of x i we are searching for max xi min(xi) (x j )=x i + (x i,x j ) 2. x i does not belong to a shortest from s to x j (whatever value is assigned to x i ) x i =min(x i ) =- (x i,s) max xi min(xi) (x j )= (s,x j ) = x i + (x i,s) + (s,x j ) and s belongs to a shortest path from s to x j
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Computing min(x i ) (improvement) 3b) requires x i = 1 (min(y) - j i (x i,x j ) ) |X| 2 August, 2000 28 (x i,x j ) can be approximated by its lower bound max(x j )-max(x i ) Computation can be stopped earlier
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