Summary. Local Search and cycle cut set. Local Search Strategies for CSP. Greedy Local Search. Local Search. and Cycle Cut Set. Strategies for CSP

Transcription

1 Summary Greedy and cycle cut set

2 Greedy... in short G.B. Shaw A life spent doing mistakes is not only more honorable, but more useful than a life spent doing nothing Reminder: Hill-climbing Algorithm function Hill-Climbing( problem) returns a state that is a local maximum inputs: problem, a problem local variables: current, a node neighbor, a node current Make-Node(Initial-State[problem]) loop do end neighbor a highest-valued successor of current if Value[neighbor] Value[current] then return State[current] current neighbor

3 Greedy s Hill-climbing, simulated annealing, etc. typically work with complete states, i.e., all variables assigned To apply to CSPs: allow states with unsatised constraints operators reassign variable values Variable selection: randomly select any conicted variable Value selection by min-conicts heuristic choose value that violates the fewest constraints i.e., hillclimb with Value[State] = total number of violated constraints in State Example: 4-Queens States: 4 queens in 4 columns (4 4 = 256 states) Operators: move queen in column Goal test: no attacks Evaluation: Value(State) = h(state) = number of attacks in State

4 Reminder: Hill-climbing issues Example Example ( for 4-Queens) (1, 2, 3, 4) Conicts: 6 (1, 2, 4, 4) Conicts: 4 (1, 2, 4, 1) Conicts: 2 (3, 2, 4, 1) Conicts: 1 Local Minima

5 Performance of min-conicts Given random initial state, can solve n-queens in almost constant time for arbitrary n with high probability (e.g., n = 10,000,000) The same appears to be true for any randomly-generated CSP except in a narrow range of the ratio R = number of constraints number of variables Improvements to Hill Climbing Improving HC Trying to avoid local minima Plateau (aka Ramdom Sideway Moves) Constraint Weighting Tabu

6 Plateau Going Sideways Plateau: set of solutions that have same cost Local minima can be due to a plateau: HC stops as soon as a plateau is found Keep on changing as long as solution is no worse than current one We can still have local minima Example Example (Plateau ) Variables: x, y, z, k, s, r Domains: D i = {0, 1, 2, 3} Constraints: x < z, y < z, z < k, k < r, k < s Initial Assignment (0, 0, 1, 1, 2, 2) cost = 1 z < k No single variable change improves the cost but a solution exists (0, 0, 1, 2, 3, 3) Changing to a same cost assignment we can nd the solution

7 Example: Cycles in Plateau Example (Plateau s) Variables: x, y, z, k, m, r, s Domains: D i = {0, 1, 2, 3} Constraints: x = z, x = y, y = z, z < k, k < m, m = r, r = s, m = s Initial Assignment (0, 0, 0, 1, 1, 1, 1) cost = 1 k < m Modifying any variable in {x, y, z, m, r, s} results in at least two violated constraints Setting k = 0 cost is constant but now z < k is violated The only modication with constant cost is k = 1 cycle Constraint Weighting Breakout Method Cost function: F (ā) = i w i C i (ā) w i current cost weight, C i (ā) = 1 i ā violates constraint C i Find local modication that maximise the decrement of F When local minima increase by one violated constraints Current assignment is no longer a local minima and we can progress towards a solution In general is not complete but extremely good empirical results [Morris 93] If no solution exists we can still cycle through inconsistent solutions

8 Example: Constraint Weighting Example (Constraint Weighting) Variables: x, y, z, k, m, r, s Domains: D i = {0, 1, 2} Constraints: x = z, x = y, y = z, z < k, k < m, m = r, r = s, m = s Initial Assignment (0, 0, 0, 1, 1, 1, 1) cost = 1 k < m Increasing constraints by 1 each time a local minima is reached we can nd a solution Example: Inconsistent Problem Example (Constraint Weighting and Inconsistency) Variables: x, y, z Domains: D i = {R, B} Constraints: x! = y, x! = z, y! = z Initial Assignment (R, B, R) cost = 1 Lifting weights results in cycling over inconsistent states

9 Tabu Tabu Given an assignment ā, consider all other assignments that change only one variable (as in hill climbing) Avoid backward moves (i.e. assignments already visited) Build a queue of last n assignments <variable,value> Assignments in the queue are forbidden Forget the oldest assignment when the queue is full Choose the assignment that has lowest cost Example: Tabu search Example (Tabu ) Variables: x, y, z, k, m, r, s Domains: D i = {0, 1, 2} Constraints: x = z, x = y, y = z, z < k, k < m, m = r, r = s, m = s Initial Assignment (0, 0, 0, 1, 1, 1, 1) cost = 1 k < m Fix n = 4, rst three moves k = 0, k = 1, k = 2 with constant cost Then k can not be changed anymore and we can get out of the cycle e.g. m = 2 We could make a wrong choice z = 1 If list is long enough at some point we will change m If list is too long we can make good paths longer

10 Exploiting Problem Structure in Cycle + Arc Consisteny on Tree Cycle + Tree Solving (Dynamic Programming on Tree) Cycle + Arc Consistency Hybrid approach 1 Find a cycle cutset 2 Fix an assignment for the cutset variables (this leaves a forest of unassigned subproblems) 3 Force arc consistency on each tree and propagate constraints. If solution found stop. 4 Otherwise stochastic local search on cutset variables only. 5 If improvement, change assignment and go to step 3 6 Otherwise stop

11 Example: Arc Consistency Example (Arc Consistency for Australia Map Coloring) Solve this map colouring problem by using Cycle and Arc Consistency Tree Solving based on cycle cutset Decomposing the problem Idea: given an assignment for cutset cycle variables nd an assignment of other variables that minimises number of violated constraints Partition X in {Z, Y } Y cutset variables, Z = X \ Y tree variables ā y current assignment of cycle cutset variables Divide the problem in rooted sub-trees Duplicate cutset variables for each neighbour and assign current value (ā y [y i ])

12 Tree Inference based on cycle cutset Propagating constraints C zi (a i, ā y ) number of violated conicts in the tree rooted at the tree variable z i C(ā z, ā y ) number of violated constraint for overall problem with assignments z and ȳ Want to compute C min = min Y =y min Z=z C(z, y) General Idea compute number of violated constraint from leaves to root choose assignment at root propagate assignment from root to leaves Tree Algorithm Tree Algorithm Algorithm 1 Tree Require: An arc consistent network R, a partition of X = {Z, Y }, an assignment for cutset variables ā y Ensure: An assignment ā z such that C(ā z, ā y ) = C T min (ā y ) Initialisation: C yi (ā y [y i ], ā y ) = 0 for all y i Y going from leaves to root on the tree compute: for all variable z i and every value a i D zi compute C zi (a i, ā y ) end for going from root to leaves on the tree compute: for all every tree variable z i Z let D zi its consistent values with its parent's value v pj do do compute best a i end for return {< z 1, a 1 >,, < z k, a k >}

13 Main Computation for Tree Algorithm Main Computation Computations performed in the algorithm C zi (a i, ā y ) = z j z j child of z i min aj D zj (C zj (a j, ā y ) + R zi,z j (a i, a j )) R zi,z j (a i, a j ) = 0 if < a i, a j > R zi,z j ; 1 Otherwise a i = arg min ai D zi C zi (a i, ā y ) + R zi,p j (a i, v pj ) p j parent of z i Example: Tree Algorithm Example (Tree Algorithm for Australia Map Colouring) Solve this map colouring problem by using Cycle and Tree Algorithm

14 HC with cycle cutset HC + CC-TREE Replace backtracking with local search Start random initial assignment Perform a given number of TRY within each TRY alternate between HC and CC-TREE: Fix one assignment for CC variables, perform TREE Fix the given assignment for tree variables and perform HC on CC HC with cycle cutset: performance Behaviour of HC and CC-TREE Empirical result on randomly generated instances: HC + CC-TREE Not always better than HC Empirical evidence show that behavior depends on ratio of CC variables Crossing point around 36% Not completely clear how general these results are

15 Dierences between HC+tree and CC+AC What does the tree solving phase give us? Tree solving phase provides a quantitative measure of how good a CC variable assignment is. Crucial to run the HC phase (or use heuristics in Backtracking) Example (Dierence between CC+AC and HC+tree) Solve the following problem using the CC+AC and HC+Tree Variables: x 1, x 2, x 3, x 4 Domain: D 1 = {G, R, B}, D 2 = D 3 = D 4 = {R, B} Constraints: x 1! = x 2, x 1! = x 3, x 1! = x 4, x 2! = x 3, x 3! = x 4 Cycle Cutset variables Y = {x 1 } Start with ā y = x 1 = R Properties of stochastic General Features Anytime: the longer they run the better the solution Local Minima: guaranteed to terminate at local minima Not complete: can not be used to prove inconsistency

16 Exercise Cycle and Tree Solving Solve the following problem using the Tree Solving approach Variables: x 1, x 2, x 3, x 4, x 5 Domain: R, B, Y Constraints: x 1! = x 2, x 1! = x 3, x 1! = x 4, x 1! = x 5, x 2! = x 3, x 3! = x 5, x 4! = x 5 Cycle Cutset variables Y = {x 3, x 5 } Assignment for cycle cutset variables ā y = x 3 = R, x 5 = B