Optimal Satisfiability and Conflict-directed A* Image credit: NASA. Assignment

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1 /27/ Optimal Satisfiability and onflict-directed * Brian. Williams 6.4 / 6.43 October 27 th, 2 Image credit: NS. Brian. Williams, copyright 2 ssignment Remember: Problem Set #6 Propositional Logic, due Today. 6:43 Project Part : Sat-based ctivity Planner, due Wednesday, November 3 rd. Problem Set #7 iagnosis, onflict-directed * and RRTs, due Wednesday, November th. Reading Today: Brian. Williams, and Robert Ragno, "onflict-directed * and its Role in Model-based Embedded Systems," Special Issue on Theory and pplications of Satisfiability Testing, Journal of iscrete pplied Math, January 23. /2/9 copyright Brian Williams, 2-2

2 /27/ Model-based iagnosis as onflict-directed Best First Search When you have eliminated the impossible, whatever remains, however improbable, must be the truth. - Sherlock Holmes. The Sign of the Four.. enerate most likely Hypothesis. 2. Test Hypothesis. 3. If Inconsistent, learn reason for inconsistency (a onflict). 4. Use conflicts to leap over similarly infeasible options to next best hypothesis. ompare Most Likely Hypothesis to Observations Helium tank Oxidizer tank Flow = zero Pressure = nominal Fuel tank Pressure 2 = nominal cceleration = zero Main Engines It is most likely that all components are okay. 2

3 /27/ Isolate onflicting Information Helium tank Oxidizer tank Fuel tank Flow = zero Main Engines The red component modes conflict with the model and observations. Leap to the Next Most Likely Hypothesis that Resolves the onflict Helium tank Oxidizer tank Fuel tank Flow = zero Main Engines The next hypothesis must remove the conflict. 3

4 /27/ New Hypothesis Exposes dditional onflicts Helium tank Oxidizer tank Pressure = nominal Fuel tank Pressure 2 = nominal cceleration = zero Main Engines nother conflict, try removing both. Final Hypothesis Resolves all onflicts Helium tank Oxidizer tank Pressure = nominal Flow = zero Fuel tank Pressure 2 = nominal Flow 2 = positive cceleration = zero Main Engines Implementation: onflict-directed * search. 4

5 /27/ Outline Model-based iagnosis Optimal SPs Informed Search onflict-directed * onstraint Satisfaction Problem SP = <X, X,> variables X with domain X. onstraint (X): X {True, False}. Problem: Find X in X s.t. (X) is True. V R,,B ifferent-color constraint V 2 R, V 3 /27/ 5

6 /27/ OSP= <Y, g, SP> Optimal SP ecision variables Y with domain Y. Utility function g(y): Y R. SP over variables <X;Y>. Find leading arg max g(y) Y y s.t. X X s.t. (X,Y) is True. g: multi-attribute utility with preferential independence, value constraint, SP: propositional state logic, simple temporal problem, mixed logic-linear program, /27/ SPs re Frequently Encoded in Propositional State Logic (mode(e) = ok implies (thrust(e) = on if and only if flow(v) = on and flow(v2) = on)) and (mode(e) = ok or mode(e) = unknown) and not (mode(e) = ok and mode(e) = unknown) V V2 E /27/ 2 6

7 /27/ Multi ttribute Utility Functions g(y) = (g (y ), g 2 (y 2 ),...) where (u, u 2 u n ) = (u,(u 2 u n )) (u ) = (u, I ) Example: iagnosis g i (y i =mode ij ) = P(y i = mode ij ) (u,u 2 ) = u x u 2 I = /27/ 3 Mutual Preferential Independence (MPI) ssignment δ is preferred over δ2 if g(δ ) < g(δ 2 ). For any set of decision variables W Y, our preference between two assignments to W is independent of the assignment to the remaining variables W Y. /27/ 4 7

8 /27/ MPI Example: iagnosis If = is more likely than = U, then { =, 2 =, 3 = U, X =, X2 = } is preferred to { = U, 2 =, 3 = U, X =, X2 = }. /27/ 5 Outline Model-based iagnosis Optimal SPs Informed Search * Branch and Bound onflict-directed * 8

9 /27/ Informed Search Extend search tree nodes to include path length g S 2 B S B 5 g = 8 Problem: iven graph <V,E> with weight function w: E R, and vertices S, in V, Find a path S p with the shortest path length δ(s,) where w(p) = Σ w(v i-,v i ). δ(u,v) = min {w(p) : u p v } Brian Williams, Fall 5 7 Best-first Search with Uniform ost spreads evenly from the start. x x B goal start Brian Williams, Fall 5 8 9

10 /27/ x x Best-first Search with Uniform ost spreads evenly from the start. reedy goes for the goal, but forgets its past. start * finds an optimal solution if h never over estimates. Then h is called admissible B goal * biases uniform cost towards the goal by using h. f = g + h g = distance from start. h = estimated distance to goal. Brian Williams, Fall 5 9 * Best-first Search with Q ordered by admissible f = g + h. 5 4 S B 8 2 S B 5 Heuristic Value h in Red Edge cost in reen Brian Williams, Fall 5 2

11 /27/ * Search: State of Search Problem: State Space Search Problem. Θ Initial State. Expand(node) hildren of Search Node = adjacent states. oal-test(node) True if search node at a goal-state. Nodes Search Nodes to be expanded. Expanded Search Nodes already expanded. Initialize Search starts at Θ, with no expanded nodes. g(state) h(state) ost to state dmissible Heuristic - Optimistic cost to go. Search Node: Node in the search tree. State State the search is at. Parent Parent in search tree. Nodes[Problem]: Enqueue(node, f ) dds node to those to be expanded. Remove-Best(f) Removes best cost queued node according to f. /27/ 2 * Search Function *(problem, h) returns the best solution or failure. Problem pre-initialized. f(x) g[problem](x) + h(x) loop do Expand Best-first node Remove-Best(Nodes[problem], f) new-nodes Expand(node, problem) for each new-node in new-nodes then Nodes[problem] Enqueue(Nodes[problem], new-node, f ) end /27/ 22

12 /27/ * Search Function *(problem, h) returns the best solution or failure. Problem pre-initialized. f(x) g[problem](x) + h(x) loop do if Nodes[problem] is empty then return failure node Remove-Best(Nodes[problem], f) Terminate when... new-nodes Expand(node, problem) for each new-node in new-nodes then Nodes[problem] Enqueue(Nodes[problem], new-node, f ) if oal-test[problem] applied to State(node) succeeds then return node end /27/ 23 Expand Vertices More Than Once path length S 2 4 S 2 4 edge cost Suppose we expanded only the first path that visits each vertex X? The shortest path from S to is ( S). is reached first using path ( S). Brian Williams, Fall

13 /27/ Expand Vertices More Than Once 3 path length S 2 5 Suppose we expanded only the first path that visits each vertex X? S 2 4 The shortest path from S to is ( S). edge cost is reached first using path ( S). This prevents path ( S) from being expanded. Brian Williams, Fall 5 25 Expand Vertices More Than Once 3 path length S 2 5 Suppose we expanded only the first path that visits each vertex X? S 2 4 The shortest path from S to is ( S). edge cost is reached first using path ( S). This prevents path ( S) from being expanded. Brian Williams, Fall

14 /27/ Expand Vertices More Than Once 3 path length S 2 5 Suppose we expanded only the first path that visits each vertex X? Eliminate the Visited List. Return solution when taken off queue. S Brian Williams, Fall The shortest path from S to is ( S). edge cost is reached first using path ( S). This prevents path ( S) from being expanded. The suboptimal path ( S) is returned. s Shortest Paths ontain Only Shortest Paths u x y Subpaths of shortest paths are shortest paths. s p v = <s, x, u, v> Shortest, then. s p u = <s, x, u> Shortest s p x = <s, x> Shortest x p v = <x, u, v> Shortest x p v = <x, u> Shortest u p v = <u, v> Shortest 7 v Brian Williams, Fall

15 /27/ Shortest Paths can be rown From Shortest Paths s 5 u v The length of shortest path s p u v is δ(s,v) = δ(s,u) + w(u,v) such that <u,v> E δ(s,v) δ(s,u) + w(u,v). x y ynamic Programming Principle: " iven the shortest path to U, don t extend other paths to U; delete them (expanded list). " When * dequeues the first partial path with head node U, this path is guaranteed to be the shortest path from S to U. Brian Williams, Fall 5 29 * Search Function *(problem, h) returns the best solution or failure. Problem pre-initialized. f(x) g[problem](x) + h(x) loop do if Nodes[problem] is empty then return failure node Remove-Best(Nodes[problem], f) state State(node) ynamic Programming Principle... remove any n from Nodes[problem] such that State(n) = state Expanded[problem] Expanded[problem] {state} new-nodes Expand(node, problem) for each new-node in new-nodes unless State(new-node) is in Expanded[problem] then Nodes[problem] Enqueue(Nodes[problem], new-node, f ) if oal-test[problem] applied to State(node) succeeds then return node end /27/ 3 5

16 /27/ Outline Model-based iagnosis Optimal SPs Informed Search * Branch and Bound onflict-directed * Branch and Bound Maintain the best solution found thus far (incumbent). Prune all subtrees worse than the incumbent S B 8 2 S B 5 Incumbent: cost U =, path P = (), 8 (S ) Heuristic Value h in Red Edge cost in reen Brian Williams, Fall

17 /27/ Branch and Bound Maintain the best solution found thus far (incumbent). Prune all subtrees worse than the incumbent. ny search order allowed (FS, Reverse-FS, BFS, Hill w BT ) Incumbent: cost U =, path P = (), S, (S B ) B 8 8 (S ) 2 S 2 Brian Williams, Fall B 5 Heuristic Values of g in Red Edge cost in reen Simple Optimal Search Using Branch and Bound Let <V,E> be a raph Let Q be a list of simple partial paths in <V,E> Let S be the start vertex in <V,E> and Let be a oal vertex in <V,E>. Let f = g + h be an admissible heuristic function. U and P are the cost and path of the best solution thus far (Incumbent).. Initialize Q with partial path (S); Incumbent U =, P = (); 2. If Q is empty, return Incumbent U and P, Else, remove a partial path N from Q; 3. If f(n) >= U, o to Step If head(n) =, then U = f(n) and P = N (a better path to the goal) 5. (Else) Find all children of head(n) (its neighbors in <V,E>) and create one-step extensions from N to each child. 6. dd extended paths to Q. 7. o to Step 2. Brian Williams, Fall

18 /27/ Outline Model-based iagnosis Optimal SPs Informed Search onflict-directed * * Increasing ost Infeasible Feasible /27/ 36 8

19 /27/ onflict-directed * Increasing ost Infeasible Feasible /27/ 37 onflict-directed * Increasing ost onflict Infeasible Feasible /27/ 38 9

20 /27/ onflict-directed * Increasing ost onflict Infeasible Feasible /27/ 39 onflict-directed * Increasing ost onflict Infeasible onflict 2 Feasible /27/ 4 2

21 /27/ onflict-directed * Increasing ost onflict Infeasible onflict 2 Feasible /27/ 4 onflict-directed * Increasing ost onflict Infeasible onflict 2 Feasible onflict 3 /27/ 42 2

22 /27/ onflict-directed * Increasing ost onflict Infeasible onflict 2 Feasible onflict 3 /27/ 43 Solving Optimal SPs Through enerate and Test Leading andidates Based on ost enerate andidate onflict-directed * one Yes Keep (Optional) Update ost Below Threshold? Yes No Test andidate onsistent? onstraint Solver No Extract onflict /27/ 44 22

23 /27/ onflict-directed * Function onflict-directed-*(osp) returns the leading minimal cost solutions. onflicts[osp] {} OSP Initialize-Best-Kernels(OSP) Solutions[OSP] {} loop do decision-state Next-Best-State-Resolving-onflicts(OSP) onflict-guided Expansion new-conflicts Extract-onflicts(SP[OSP], decision-state) onflicts[osp] Eliminate-Redundant-onflicts(onflicts[OSP] new-conflicts) end /27/ 45 onflict-directed * Function onflict-directed-*(osp) returns the leading minimal cost solutions. onflicts[osp] {} OSP Initialize-Best-Kernels(OSP) Solutions[OSP] {} loop do decision-state Next-Best-State-Resolving-onflicts(OSP) if no decision-state returned or Terminate?(OSP) then return Solutions[OSP] if onsistent?(sp[osp ], decision-state) then add decision-state to Solutions[OSP] new-conflicts Extract-onflicts(SP[OSP], decision-state) onflicts[osp] Eliminate-Redundant-onflicts(onflicts[OSP] new-conflicts) end /27/ 46 23

24 /27/ Increasing ost onflict-directed * Each feasible subregion described by a kernel assignment. pproach: Use conflicts to search for kernel assignment containing the best cost candidate. onflict Infeasible onflict 2 onflict 3 Kernel 3 Kernel 2 Feasible Kernel Recall: Mapping onflicts to Kernels B E 2 3 X Y Z X X2 onflict i : set of decision variable assignments that are inconsistent with constraints Φ. F i ^ Φ is inconsistent # # Φ entails i onstituent Kernel: n assignment a that resolves a conflict i. B E?? 3 X Y Z X? F a entails i Kernel: minimal set of decision variable assignments that resolves all known conflicts. entails i for all i in /27/ 48 24

25 /27/ Extracting a kernel s best state Select best utility value for unassigned variables (Why?). {X=U} =? 2=U 3=? X=? X2=? = 2=U 3= X= X2= /27/ 49 Next Best State Resolving onflicts function Next-Best-State-Resolving-onflicts(OSP) best-kernel Next-Best-Kernel(OSP) if best-kernel = failure then return failure else return kernel-best-state[problem](best-kernel) end function Kernel-Best-State(kernel) unassigned all variables not assigned in kernel return kernel {Best-ssignment(v) v unassigned} End function Terminate?(OSP) return True iff Solutions[OSP] is non-empty lgorithm for only finding the first solution, multiple later. /27/ 5 25

26 /27/ Example: iagnosis X B F X 2 Y X2 E 3 Z ssume Independent Failures: P (mi) >> P U(mi) P single >> P double P U(2) > P U() > P U(3) > P U(X) > P U(X2) /27/ 5 B E First Iteration X F X 2 Y X2 3 Z onflicts / onstituent Kernels none Best Kernel: {} Best andidate:? /27/ 52 26

27 /27/ Extracting the kernel s best state Select best value for unassigned variables { } =? 2=? 3=? X=? X2=? = 2= 3= X= 2= /27/ 53 B E First Iteration X F X 2 Y X2 3 Z onflicts / onstituent Kernels none Best Kernel: {} Best andidate: = 2= 3= X= X2=? /27/ 54 27

28 /27/ Test: = 2= 3= X= X2= B 2 X Y X X2 F E 3 Z /27/ 55 Test: = 2= 3= X= X2= B 2 X Y X X2 Symptom F E 3 Z Extract onflict and onstituent Kernels: [= 2= X=] =U 2=U X=U /27/ 56 28

29 /27/ Second Iteration P (mi) >> P U(mi) P single >> P double P U(2) > P U() > P U(3) > P U(X) > P U(X2) B E 2 3 X Y Z X X2 F onflicts onstituent Kernels =U 2=U X=U Best Kernel: 2=U (why?) Best andidate: = 2=U 3= X= X2= /27/ 57 Test: = 2=U 3= X= X2= B X Y X X2 F E 3 Z /27/ 58 29

30 /27/ Test: = 2=U 3= X= X2= B E 3 X Y Z X X2 F Extract onflict: [= 3= X= X2=] =U 3=U X=U X2=U /27/ 59 Third Iteration P (mi) >> P U(mi) P single >> P double P U(2) > P U() > P U(3) > P U(X) > P U(X2) B E 3 X Y Z X X2 F onflicts onstituent Kernels =U 2=U X=U =U 3=U X=U X2=U Best Kernel: =U Best andidate: =U 2= 3= X= X2= /27/ 6 3

31 /27/ Test: =U 2=U 3= X= X2= B 2 X Y X X2 F E 3 Z /27/ 6 Test: =U 2=U 3= X= X2= B E 2 3 X Y Z X X2 F onsistent! /27/ 62 3

32 /27/ Outline Model-based iagnosis Optimal SPs onflict-directed * enerating the Best Kernel Performance omparison enerating The Best Kernel of The Known onflicts onstituent Kernels X=U =U 2=U X=U, =U, 2=U X=U X2=U M=U 3=U X=U, X2=U, =U, 3=U X=U =U =U X2=U 2=U 3=U Insight: Kernels found by minimal set covering Minimal set covering is an instance of breadth first search. 32

33 /27/ Expanding a Node to Resolve a onflict onstituent kernels { } 2=U =U X=U 2=U =U X=U To Expand a Node: Select an unresolved onflict. Each child adds a constituent kernel of onflict. Prune any node that is Inconsistent, or superset of a known kernel. /27/ 65 enerating The Best Kernel of The Known onflicts onstituent Kernels X=U =U 2=U X=U, =U, 2=U X=U, X2=U, =U, 3=U =U Insight: Kernels found by minimal set covering Minimal set covering is an instance of breadth first search. To find the best kernel, expand tree in best first order. 33

34 /27/ dmissible h(α): ost of best state that extends partial assignment α f = g + h 2=U =? 3=? X=? X2=? P 2=u x P = x P 3= x P X= x P X2= Select best value of unassigned variables. /27/ 67 dmissible Heuristic h Let g = <,g i,y> describe a multi-attribute utility fn ssume the preference for one attribute x i is independent of another x k alled Mutual Preferential Independence: For all u, v Y If g i (u) g i (v) then for all w (g i (u),g k (w)) (g i (v),g k (w)) n dmissible h: iven a partial assignment, to X Y h selects the best value of each unassigned variable Z = X Y h(y) = ({g zi_max z i Z, max g zi (v ij ))}) v ij zi candidate always exists satisfying h(y). /27/ 68 34

35 /27/ Terminate when all conflicts resolved Function oal-test-kernel (node, problem) returns True IFF node is a complete decision state. if forall K in onstituent-kernels(onflicts[problem]), State[node] contains a kernel in K then return True else return False /27/ 69 Next Best Kernel of Known onflicts Function Next-Best-Kernel (OSP) returns the next best cost kernel of onflicts[osp]. f(x) [OSP] (g[osp](x), h[osp](x)) n instance loop do if Nodes[OSP] is empty then return failure of * node Remove-Best(Nodes[OSP], f) add State[node] to Visited[OSP] new-nodes Expand-onflict(node, OSP) for each new-node new-nodes unless n Nodes[OSP] such that State[new-node] = State[n] OR State[new-node] Visited[problem] then Nodes[OSP] Enqueue(Nodes[OSP], new-node, f ) if oal-test-kernel[osp] applied to State[node] succeeds Best-Kernels[OSP] dd-to-minimal-sets(best-kernels[osp], best-kernel) if best-kernel Best-Kernels[OSP] then return State[node] end /27/ 7 35

36 /27/ Outline Model-based iagnosis Optimal SPs onflict-directed * enerating the Best Kernel Performance omparison Problem Parameters om Size Performance: With and Without onflicts ec Vars lau -ses lau -se lngth onstraint-based * (no conflicts) Nodes Expande d Queue Size onflict-directed * Nodes Expand Queue Size onflicts used Mean -B Ratio Nodes Expanded Queue Size , % 5.6% ,36 3, % 3.5% ,27 6, %.% 6 3,79 3, %.% 3 6,43 5, % 5.8% , % 3.9% % 3.% % 5.4% %.% /27/ 72 36

37 /27/ Multiple Fault iagnosis of Systems with Novel Failures onsistency-based iagnosis: iven symptoms, find diagnoses that are consistent with symptoms. Suspending onstraints: Make no presumption about a component s faulty behavior. B E 2 3 X Y Z X X2 F Symptom Model-based iagnosis as onflict-directed Best First Search When you have eliminated the impossible, whatever remains, however improbable, must be the truth. - Sherlock Holmes. The Sign of the Four.. enerate most likely Hypothesis. 2. Test Hypothesis. 3. If Inconsistent, learn reason for inconsistency (a onflict). 4. Use conflicts to leap over similarly infeasible options to next best hypothesis. 37

38 /27/ Outline Model-based iagnosis Optimal SPs onflict-directed * enerating the Best Kernel Performance omparison ppendix: Intelligent Tree Expansion Extending to Multiple Solutions Review of * Expand Only Best hild & Sibling onstituent kernels { } 2=U =U X=U Order constituents by decreasing utility. 2=U =U > > X=U Traditionally all children expanded. Only need child containing best candidate. hild with best estimated cost f = g+h. /27/ 76 38

39 /27/ Expand Only Best hild & Sibling onstituent kernels { } 2=U =U X=U 2=U Order constituents by decreasing utility Traditionally all children expanded. Only need child with best candidate. hild with best estimated cost f = g+h. /27/ 77 When o We Expand The hild s Next Best Sibling? onstituent kernels { } 2=U =U X=U 2=U =U =U 3=U X=U X2=U =U When a best child has a subtree or leaf pruned, it may have lost its best candidate. One of the child s siblings might now contain the best candidate. Expand child s next best sibling: when expanding children to resolve another conflict. /27/ 78 39

40 /27/ Expand Node to Resolve onflict function Expand-onflict(node, OSP) return Expand-onflict-Best-hild(node, OSP) Expand-Next-Best-Sibling (node, OSP) function Expand-onflict-Best-hild(node, OSP) if for all K γ in onstituent-kernels(γ[osp]) State[node] contains a kernel K γ then return {} else return Expand-onstituent-Kernel(node,OSP) function Expand-onstituent-Kernel(node, OSP) K γ = smallest uncovered set onstituent-kernels(γ[osp]) {y i = v ij {y i = v ij } in K γ, y i = v ij is consistent with State[node]} Sort such that for all i from to -, Better-Kernel?([i],[i+], OSP) is True hild-ssignments[node] y i = v ij [], which is the best kernel in K γ consistent with State[node] return {Make-Node({y i = v ij }, node)} /27/ 79 Expand Node to Resolve onflict function Expand-Next-Best-Sibling(node, OSP) if Root?[node] then return {} else {y i = v ij } ssignment[node] {y k = v kl } next best assignment in consistent child-assignments[parent[node]] after {y i = v ij } if no next assignment {y k = v kl } or Parent[node] already has a child with {y k = v kl } then return {} else return {Make-Node({y k = v kl }, Parent[node])} /27/ 8 4

41 /27/ Outline Model-based iagnosis Optimal SPs onflict-directed * enerating the Best Kernel Performance omparison ppendix: Intelligent Tree Expansion Extending to Multiple Solutions Review of * Multiple Solutions: Systematically Exploring Kernels onstituent Kernels X=U =U 2=U X=U, =U, 2=U X2=U 3=U X=U, X2=U, =U, 3=U X=U =U =U X2=U 2=U 3=U 4

42 /27/ hild Expansion For Finding Multiple Solutions onflict (2= = X=) { } 2=U =U X=U X2= X2=U If Unresolved onflicts: Select unresolved conflict. Each child adds a constituent kernel. If ll onflicts Resolved: Select unassigned variable y i. Each child adds an assignment from i. /27/ 83 Intelligent Expansion Below a Kernel {} Select Unassigned Variable. 2= 2=U Order assignments by decreasing utility. Expand best child. =U 2= 3= X= 2=U 3=U X=U ontinue expanding best descendents. When leaf visited,, expand all next best ancestors. (why?) X2= X2=U /27/ 84 42

43 /27/ Putting It Together: Expansion Of ny Search Node { } onstituent kernels 2=U =U X=U 2=U =U =U 3=U X=U X2=U =U 2= 2=U When a best child loses any candidate, expand child s next best sibling: If child has unresolved conflicts, expand sibling when child expands its next conflict. If child resolves all conflicts, expand sibling when child expands a leaf. 3= X= X2= 3=U =U 2=U /27/ 85 onflict-directed * When you have eliminated the impossible, whatever remains, however improbable, must be the truth. - Sherlock Holmes. The Sign of the Four.. enerate most likely hypothesis. 2. Test hypothesis. 3. If inconsistent, learn reason for inconsistency (a onflict). 4. Use conflicts to leap over similarly infeasible options to next best hypothesis. /27/ 86 43

44 /27/ Outline Using conflicts in backtrack search ependency-directed backtracking onflict learning onflict-directed backjumping /2/9 copyright Brian Williams, 2-87 Using onflicts to uide Search: ependency-directed Search [Stallman & Sussman, 978] Input: Output: onstraint satisfaction problem. Satisfying assignment. Repeat while a next candidate assignment exists. enerate candidate assignment c. heck candidate c against conflicts. If c is a superset of a conflict, Then loop to the next candidate. heck consistency of c. If inconsistent, Then extract and record a conflict from c. Else return c as a solution. Like a raphplan memo, but generalizes an inconsistent solution. /2/9 copyright Brian Williams,

45 /27/ Procedure ependency_directed_backtracking (<X,,>) Input: constraint network R = <X,, > Output: solution, or notification that the network is inconsistent. i ; a i = {}; conflicts = {} Initialize variable counter, assignments, i i ; opy domain of first variable. while i n instantiate x i Select-B-Value(); dd to assignments a i. if x i is null No value was returned, i i - ; then backtrack else i i + ; else step forward and i i ; copy domain of next variable end while if i = return inconsistent else return a i, the instantiated values of {x i,, x n } end procedure /2/9 copyright Brian Williams, 2-89 Procedure Select-B-Value() Output: value in i consistent with a i-, or null, if none. while i is not empty select an arbitrary element a i and remove a from i ; a i a i-, {x i = a}; if for every c in conflicts, not (a i superset c) if consistent(a i-, x i = a ) return a; else conflicts conflicts minimal inconsistent subset of a i- ; end while return null end procedure /2/9 copyright Brian Williams,

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