EEC EECS 3401: Intr tro to to AI & LP Constr traint t Sati tisfacti tion Problems CSPs & Backtr tracking Search ● The search algorithms we discussed so far had ● Readings: R & N Chapter 6. no knowledge of the states representation (black box). So, we could not take advantage of domain-specific information. ● CSP are a special class of search problems with a uniform and simple state representation. ● This allows to design more efficient algorithms. 1 1 2 2 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance Constr traint t Sati tisfacti tion Problems Constr traint t Sati tisfacti tion Problems ● Sudoku: ● Many problems can be represented as a search for a vector of feature values. ■ 81 variables, the value in each cell. ■ Values: a fixed value for those cells that are already ■ k-features: variables. filled in, the values {1-9} for those cells that are ■ Each feature has a value. Domain of values for the empty. variables. ■ Solution: a value for each cell satisfying the ■ e.g., height = {short, average, tall}, weight = {light, constraints: average, heavy}. ● no cell in the same column can have the same value. ● In these problems the problem is to search for ● no cell in the same row can have the same value. a set of values for the features (variables) so ● no cell in the same sub-square can have the same that the values satisfy some conditions value. (constraints). 3 3 4 4 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance 1
Constr traint t Sati tisfacti tion Problems Constr traint t Sati tisfacti tion Problems ● Scheduling ● Variables: ■ Want to schedule a time and a space for each final ■ T1, …, Tm: Ti is a variable representing the exam so that scheduled time for the i-th final. ● No student is scheduled to take more than one final ● Assume domains are fixed to {MonAm, MonPm, …, at the same time. FriAm, FriPm}. ● The space allocated has to be available at the time ■ S1, …, Sm: Si is the space variable for the i-th final. set. ● Domain of Si are all rooms big enough to hold the ● The space has to be large enough to accommodate i-th final. all of the students taking the exam. 5 5 6 6 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance Constr traint t Sati tisfacti tion Problems Constr traint t Sati tisfacti tion Problems (CSP) ● Want to find an assignment of values to each ● More formally. variable (times, rooms for each final), subject ● A CSP consists of to the constraints: ■ a set of variables V1, …, Vn ■ For all pairs of finals i, j such that there is a student ■ for each variable a domain of possible values taking both: Dom[Vi]. ● Ti ≠ Tj ■ A set of constraints C1,…, Cm. ■ For all pairs of finals i, j ● Ti ≠ Tj or Si ≠ Sj ■ either i and j are not scheduled at the same time, or if they are they are not in the same space. 7 7 8 8 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance 2
Constr traint t Sati tisfacti tion Problems Constr traint t Sati tisfacti tion Problems ● Each variable be assigned any value from its ● A solution to a CSP is domain. ■ an assignment of a value to all of the variables such ● Vi = d where d ∈ Dom[Vi] that ● Each constraint C has ● every constraint is satisfied. ■ A set of variables it is over, called its scope: e.g., C(V1,V2,V4). ■ Is a boolean function that maps assignments to these variables to true/false. ● e.g. C(V1=a,V2=b,V4=c) = True ● this set of assignments satisfies the constraint. ● e.g. C(V1=b,V2=c,V4=c) = False ● this set of assignments falsifies the constraint. 9 9 10 10 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance Constr traint t Sati tisfacti tion Problems Constr traint t Sati tisfacti tion Problems ● Sudoku: ● Sudoku: ■ V 11 , V 12 , …, V 21 , V 22 , …, V 91 , …, V 99 ■ Each of these constraints is over 9 variables, and they ● Dom[V ij ] = {1-9} for empty cells are all the same constraint: ● Dom[V ij ] = {k} a fixed value k for filled cells. ■ Row constraints: ● Any assignment to these 9 variables such that each ● CR1(V 11 , V 12 , V 13 , …, V 19 ) variable has a unique value satisfies the constraint. ● CR2(V 21 , V 22 , V 23 , …, V 29 ) ● Any assignment where two or more variables have ● ...., CR9(V 91 , V 92 , …, V 99 ) ■ Column Constraints: the same value falsifies the constraint. ● CC1(V 11 , V 21 , V 31 , …, V 91 ) ■ Such constraints are often called ALL-DIFF ● CC2(V 21 , V 22 , V 13 , …, V 92 ) constraints. ● ...., CC9(V 19 , V 29 , …, V 99 ) ■ Sub-Square Constraints: ● CSS1(V 11 , V 12 , V 13 , V 21 , V 22 , V 23 , V 31 , V 32 , V 33 ) ● CSS1(V 14 , V 15 , V 16 ,…, V 34 , V 35 , V 36 ) 11 11 12 12 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance 3
Constr traint t Sati tisfacti tion Problems Constr traint t Sati tisfacti tion Problems ● Sudoku: ● Exam Scheduling ■ Thus Sudoku has 3x9 ALL-Diff constraints, one over ■ constraints: each set of variables in the same row, one over each ■ For all pairs of finals i, j such that there is a student set of variables in the same column, and one over taking both: each set of variables in the same sub-square. ● NEQ(Ti,Tj) ■ For all pairs of finals i, j ■ Note also that an ALL-Diff constraint over k variables ● C(Ti,Tj,Si,Sj) can be equivalently represented by k choose 2 not- ■ This constraint is satisfied equal constraints over each pair of these variables. ● by any set of assignments in which Ti ≠ Tj. ● e.g. CSS1(V 11 , V 12 , V 13 , V 21 , V 22 , V 23 , V 31 , V 32 , V 33 ) = ● any set of assignments in which Si ≠ Sj. NEQ(V 11 ,V 12 ), NEQ(V 11 ,V 13 ), NEQ(V 11 ,V 21 ) …, NEQ(V 32 ,V 33 ) ■ Falsified by any set of assignments in which Ti=Tj as well as Si=Sj. ■ NEQ is a not-equal constraint. 13 13 14 14 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance Solvi So ving ng CSPs CSPs Backtr tracking Search ● These ideas lead to the backtracking search algorithm ● CSPs can be solved by a specialized version of depth first search. Algorithm BT (Backtracking) BT(Level) If all variables assigned ● Key intuitions: PRINT Value of each Variable RETURN or EXIT (RETURN for more solutions) ■ We can build up to a solution by searching through (EXIT for only one solution) V := PickUnassignedVariable() the space of partial assignments. Variable[Level] := V ■ Order in which we assign the variables does not Assigned[V] := TRUE for d := each member of Domain(V) matter---eventually they all have to be assigned. Value[V] := d OK := TRUE ■ If during the process of building up a solution we for each constraint C such that V is a variable of C falsify a constraint, we can immediately reject all and all other variables of C are assigned. possible ways of extending the current partial if C is not not satisfied by the current set of assignments OK := FALSE assignment. if(OK) BT(Level+1) return 15 15 16 16 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance 4
Solvi So ving ng CSPs CSPs Backtr tracking Search ● The algorithm searches a tree of partial assignments. ● Heuristics are used to determine which variable to assign next “ PickUnassignedVariable ” . Children of a node are The root has the empty set all possible values of of assignments ● The choice can vary from branch to branch, some (any) unassigned Root {} variable e.g., ■ under the assignment V1=a we might choose to assign V4 next, while under V1=b we might choose Vi=a Vi=b Vi=c to assign V5 next. ● This “ dynamically ” chosen variable ordering has a tremendous impact on performance. Search stops Vj=1 Vj=2 descending if the assignments on path to the node Subtree violate a constraint 17 17 18 18 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance Ex Example. Example. Ex ● N-Queens. Place N Queens on an N X N chess board so ● 4X4 Queens that no Queen can attack any other Queen. ■ Variables, one per row. ● Value of Vi is the column the Queen in row i is place. ■ Constrants. ● Vi ≠ Vj for all i ≠ j (can put two Queens in same column) ● |Vi-Vj| ≠ i-j (Diagonal constraint) ■ (i.e., the difference in the values assigned to Vi and Vj can ’ t be equal to the difference between i and j. 19 19 20 20 EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance EECS 3401 Winter 2017 Fahiem Bacchus & Yves Lesperance 5
Recommend
More recommend
