Constraint Satisfaction Problems (CSP) (Where we postpone making difficult

Transcription

1 Constraint Satisfaction Problems (CSP) (Where we postpone making difficult decisions until they become easy to make) R&N: Chap. What we will try to do... Search techniques make choices in an often arbitrary order. Often little information is available to make each of them In many yp problems, the same states can be reached independent of the order in which choices are made ( commutative actions) Can we solve such problems more efficiently by picking the order appropriately? Can we even avoid making any choice? Place a queen in a square Remove the attacked squares from future consideration Count the number of non-attacked squares in every row and column Place a queen in a row or column with minimum number Remove the attacked squares from future consideration Repeat 6

2 What do we need? More than just a successor function and a goal test We also need: A means to propagate the constraints imposed by one queen s position on the positions of the other queens An early failure test Explicit representation of constraints Constraint propagation algorithms

3 Constraint Satisfaction Problem (CSP) Set of variables {,,, X n } Each variable X i has a domain D i of possible values. Usually, D i is finite Set of constraints {C, C,, C p } Each constraint relates a subset of variables by specifying the valid combinations of their values Goal: Assign a value to every variable such that all constraints are satisfied Map Coloring N 7 variables {,N,,,,,} Each variable has the same domain: {red, green, blue} No two adjacent variables have the same value: N,, N, N,,,,, 8-ueen Problem 8 variables X i, i = to 8 he domain of each variable is: {,,,8} Constraints are of the forms: X i = k X j k for all j = to 8, j i Similar constraints for diagonals All constraints are binary Street Puzzle N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {ea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, iolinist, Doctor} A i = {Dog, Snails, Fox, Horse, Zebra} he Englishman lives in the Red house he Spaniard has a Dog he Japanese is a Painter he Italian drinks ea he Norwegian lives in the first house on the left he owner of the Green house drinks Coffee he Green house is on the right of the White house he Sculptor breeds Snails he Diplomat lives in the Yellow house he owner of the middle house drinks Milk he Norwegian lives next door to the Blue house he iolinist drinks Fruit juice he Fox is in the house next to the Doctor s he Horse is next to the Diplomat s Who owns the Zebra? Who drinks Water? 6 Street Puzzle N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {ea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, iolinist, Doctor} A i = {Dog, Snails, Fox, Horse, Zebra} i,j [,], i j, N i N j he Englishman lives in the Red house he Spaniard has a Dog i,j [,], i j, C i C j he Japanese is a Painter... he Italian drinks ea he Norwegian lives in the first house on the left he owner of the Green house drinks Coffee he Green house is on the right of the White house he Sculptor breeds Snails he Diplomat lives in the Yellow house he owner of the middle house drinks Milk he Norwegian lives next door to the Blue house he iolinist drinks Fruit juice he Fox is in the house next to the Doctor s he Horse is next to the Diplomat s 7 Street Puzzle N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {ea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, iolinist, Doctor} A i = {Dog, Snails, Fox, Horse, Zebra} he Englishman lives in the Red house (N i = English) (C i = Red) he Spaniard has a Dog he Japanese is a Painter (N i = Japanese) (J i = Painter) he Italian drinks ea he Norwegian lives in the first house on the left (N = Norwegian) he owner of the Green house drinks Coffee he Green house is on the right of the White house he Sculptor breeds Snails he Diplomat lives in the Yellow house he owner of the middle house drinks Milk (C i = White) (C i+ = Green) (C White) he Norwegian lives next door to the Blue house (C Green) he iolinist drinks Fruit juice he Fox is in the house next to the Doctor s he Horse is next to the Diplomat s left as an exercise 8

4 Street Puzzle N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {ea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, iolinist, Doctor} A i = {Dog, Snails, Fox, Horse, Zebra} he Englishman lives in the Red house (N i = English) (C i = Red) he Spaniard has a Dog he Japanese is a Painter (N i = Japanese) (J i = Painter) he Italian drinks ea he Norwegian lives in the first house on the left (N = Norwegian) he owner of the Green house drinks Coffee he Green house is on the right of the White house he Sculptor breeds Snails he Diplomat lives in the Yellow house he owner of the middle house drinks Milk (C i = White) (C i+ = Green) (C White) he Norwegian lives next door to the Blue house (C Green) he iolinist drinks Fruit juice he Fox is in the house next to the Doctor s he Horse is next to the Diplomat s unary constraints 9 Street Puzzle N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {ea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, iolinist, Doctor} A i = {Dog, Snails, Fox, Horse, Zebra} he Englishman lives in the Red house he Spaniard has a Dog he Japanese is a Painter he Italian drinks ea he Norwegian lives in the first house on the left N = Norwegian he owner of the Green house drinks Coffee he Green house is on the right of the White house he Sculptor breeds Snails he Diplomat lives in the Yellow house he owner of the middle house drinks Milk D = Milk he Norwegian lives next door to the Blue house he iolinist drinks Fruit juice he Fox is in the house next to the Doctor s he Horse is next to the Diplomat s 0 Street Puzzle N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {ea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, iolinist, Doctor} A i = {Dog, Snails, Fox, Horse, Zebra} he Englishman lives in the Red house C Red he Spaniard has a Dog A Dog he Japanese is a Painter he Italian drinks ea he Norwegian lives in the first house on the left N = Norwegian he owner of the Green house drinks Coffee he Green house is on the right of the White house he Sculptor breeds Snails he Diplomat lives in the Yellow house he owner of the middle house drinks Milk D = Milk he Norwegian lives next door to the Blue house he iolinist drinks Fruit juice J iolinist he Fox is in the house next to the Doctor s he Horse is next to the Diplomat s ask Scheduling Four tasks,,, and are related by time constraints: must be done during must be achieved before starts must overlap with must start after is complete Are the constraints compatible? What are the possible time relations between two tasks? What if the tasks use resources in limited supply? How to formulate this problem as a CSP? - Finite vs. Infinite CSP n Boolean variables u,..., u n p constraints of the form u i * u j * u k *= where u* stands for either u or u Known to be NP-complete Finite CSP: each variable has a finite domain of values Infinite CSP: some or all variables have an infinite domain E.g., linear programming problems over the reals: for i =,,..., p : ai,x +ai,x +...+ai,nx n = ai,0 for j =,,..., q : bj,x +bj,x +...+bj,nx n bj,0 We will only consider finite CSP

5 CSP as a Search Problem n variables,..., X n alid assignment: {X i v i,..., X ik v ik }, 0 k n, such that the values v i,..., v ik satisfy all constraints relating the variables X i,..., X ik Complete assignment: one where k = n [if all variable domains have size d, there are O(d n ) complete assignments] States: valid assignments Initial state: empty assignment {}, i.e. k = 0 Successor of a state: {X i v i,..., X ik v ik } {X i v i,..., X ik v ik, X ik+ v ik+ } Goal test: k = n {X i v i,..., X ik v ik, X ik+ v ik+} {X i v i,..., X ik v ik } r = n k variables with s values r s branching factor 6 A Key property of CSP: Commutativity he order in which variables are assigned values has no impact on the reachable complete valid assignments {X i v i,..., X ik v ik } Hence: ) One can expand a node N by first selecting one variable X not in the assignment A associated with N and then assigning every value v in the domain of X [ big reduction in branching factor] 7 {X i v i,..., X ik v ik, X ik+ v ik+} r = n k n-k variables with s values r s s branching factor he depth of the solutions in the search tree is un-changed (n) 8 variables,..., X Let the valid assignment of N be: A = { v, v } For example pick variable X Let the domain of X be {v,, v,, v,} he successors of A are all the valid assignments among: { v, v, X v, } { v, v, X v, } { v, v, X v, } 9 A Key property of CSP: Commutativity he order in which variables are assigned values has no impact on the reachable complete valid assignments Hence: ) One can expand a node N by first selecting one variable X not in the assignment A associated with N and then assigning every value v in the domain of X [ big reduction in branching factor] ) One need not store the path to a node 0 Backtracking search algorithm

6 Essentially a simplified depth-first algorithm using recursion Assignment = {} v Assignment = {(, )} Assignment = {(, ), (,v )} v hen, the search algorithm backtracks to the previous variable ( ) and tries another value v Assume that no value of X leads to a valid assignment Assignment = {(, ), (,v )} Assignment = {(, ), (, )} 6 6

7 v he search algorithm backtracks to the previous variable ( ) and tries another value. But assume that t has only two possible values. he algorithm backtracks to v v Assume again that no value of leads to a valid assignment Assignment = {(, ), (, )} 7 Assignment = {(,v )} 8 v v v v v v he algorithm need not consider the variables in the same order in this sub-tree as in the other Assignment = {(,v ), (,v )} 9 Assignment = {(,v ), (,v )} 0 v v v v v v he algorithm need not consider the values of in the same order in this sub-tree Assignment = {(,v ), (,v ), (, )} Assignment = {(,v ), (,v ), (, )} 7

8 v v Assignment = {(,v ), (,v ), (, )} v Since there are only three variables, the assignment is complete Backtracking Algorithm CSP-BACKRACKING(A). If assignment A is complete then return A. X select a variable not in A. D select an ordering on the domain of X. For each value v in D do a. Add (X v) to A b. If A is valid then i. result CSP-BACKRACKING(A) ii. If result failure then return result c. Remove (X v) from A. Return failure Call CSP-BACKRACKING({}) [his recursive algorithm keeps too much data in memory. An iterative version could save memory (left as an exercise)] Critical uestions for the Efficiency of CSP-Backtracking CSP-BACKRACKING(A). If assignment A is complete then return A. X select a variable not in A. D select an ordering on the domain of X. For each value v in D do a. Add (X v) to A b. If a is valid then i. result CSP-BACKRACKING(A) ii. If result failure then return result c. Remove (X v) from A. Return failure Critical uestions for the Efficiency of CSP-Backtracking ) Which variable X should be assigned a value next? he current assignment may not lead to any solution, but the algorithm still does know it. Selecting the right variable to which to assign a value may help discover the contradiction more quickly ) In which order should X s values be assigned? he current assignment may be part of a solution. Selecting the right value to assign to X may help discover this solution more quickly More on these questions in a short while... 6 Critical uestions for the Efficiency of CSP-Backtracking ) Which variable X should be assigned a value next? he current assignment may not lead to any solution, but the algorithm does not know it yet. Selecting the right variable X may help discover the contradiction more quickly ) In which order should X s values be assigned? he current assignment may be part of a solution. Selecting the right value to assign to X may help discover this solution more quickly Critical uestions for the Efficiency of CSP-Backtracking ) Which variable X should be assigned a value next? he current assignment may not lead to any solution, but the algorithm does not know it yet. Selecting the right variable X may help discover the contradiction more quickly ) In which order should X s values be assigned? he current assignment may be part of a solution. Selecting the right value to assign to X may help discover this solution more quickly More on these questions in a short while... 7 More on these questions in a short while

9 Critical uestions for the Efficiency of CSP-Backtracking ) Which variable X should be assigned a value next? he current assignment may not lead to any solution, but the algorithm does not know it yet. Selecting the right variable X may help discover the contradiction more quickly ) In which order should X s values be assigned? he current assignment may be part of a solution. Selecting the right value to assign to X may help discover this solution more quickly X X X 6 X 7 X 8 Forward Checking A simple constraint-propagation technique: Assigning the value to leads to removing values from the domains of,,..., X 8 More on these questions very soon Forward Checking in Map Coloring Forward Checking in Map Coloring N Constraint graph N N RGB RGB RGB RGB RGB RGB RGB N RGB RGB RGB RGB RGB RGB RGB R RGB RGB RGB RGB RGB RGB Forward checking removes the value Red of N and of Forward Checking in Map Coloring Forward Checking in Map Coloring N N N RGB RGB RGB RGB RGB RGB RGB R GB RGB RGB RGB GB RGB R GB G RGB RGB GB RGB N RGB RGB RGB RGB RGB RGB RGB R GB RGB RGB RGB GB RGB R B G RB RGB B RGB R B G RB B B RGB 9

10 Forward Checking in Map Coloring Empty set: the current assignment {( R), ( G), ( B)} does not lead to a solution N RGB RGB RGB RGB RGB RGB RGB R GB RGB RGB RGB GB RGB R B G RB RGB B RGB R B G RB B B RGB Forward Checking (General Form) Whenever a pair (X v) is added to assignment A do: For each variable Y not in A do: For every constraint C relating Y to the variables in A do: Remove all values from Y s domain that do not satisfy C 6 Modified Backtracking Algorithm CSP-BACKRACKING(A, var-domains). If assignment A is complete then return A. X select a variable not in A. D select an ordering on the domain of X. For each value v in D do a. Add (X v) to A b. var-domains forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result CSP-BACKRACKING(A, var-domains) (ii) If result failure then return result d. Remove (X v) from A. Return failure 7 Modified Backtracking Algorithm CSP-BACKRACKING(A, var-domains). If assignment A is complete then return A. X select a variable not in A. D select an ordering on the domain of X. For each value v in D do No need any more to a. Add (X v) to A verify that A is valid b. var-domains forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result CSP-BACKRACKING(A, var-domains) (ii) If result failure then return result d. Remove (X v) from A. Return failure 8 Modified Backtracking Algorithm CSP-BACKRACKING(A, var-domains). If assignment A is complete then return A. X select a variable not in A. D select an ordering on the domain of X. For each value v in D do a. Add (X v) to A b. var-domains forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result CSP-BACKRACKING(A, var-domains) (ii) If result failure then return result d. Remove (X v) from A. Return failure Need to pass down the updated variable domains 9 Modified Backtracking Algorithm CSP-BACKRACKING(A, var-domains). If assignment A is complete then return A. X select a variable not in A. D select an ordering on the domain of X. For each value v in D do a. Add (X v) to A b. var-domains forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result CSP-BACKRACKING(A, var-domains) (ii) If result failure then return result d. Remove (X v) from A. Return failure 60 0

11 ) Which variable X i should be assigned a value next? Most-constrained-variable heuristic Most-constraining-variable heuristic ) In which order should its values be assigned? Least-constraining-value heuristic hese heuristics can be quite confusing Keep in mind that all variables must eventually get a value, while only one value from a domain must be assigned to each variable 6 Most-Constrained-ariable Heuristic ) Which variable X i should be assigned a value next? Select the variable with the smallest remaining domain [Rationale: Minimize the branching factor] 6 8-ueens 8-ueens Forward checking Forward checking New assignment New assignment Numbers of values for each un-assigned variable New numbers of values for each un-assigned variable 6 6 Map Coloring N s remaining domain has size (value Blue remaining) s remaining domain has size s, s, and s remaining domains have size Select 6 Most-Constraining-ariable Heuristic ) Which variable X i should be assigned a value next? Among the variables with the smallest remaining domains (ties with respect to the most-constrained-variable heuristic), select the one that appears in the largest number of constraints on variables not in the current assignment [Rationale: Increase future elimination of values, to reduce future branching factors] 66

12 Map Coloring N Before any value has been assigned, all variables have a domain of size, but is involved in more constraints () than any other variable Select and assign a value to it (e.g., Blue) 67 Least-Constraining-alue Heuristic ) In which order should X s values be assigned? Select the value of X that removes the smallest number of values from the domains of those variables which are not in the current assignment [Rationale: Since only one value will eventually be assigned to X, pick the least-constraining value first, since it is the most likely not to lead to an invalid assignment] [Note: Using this heuristic requires performing a forward-checking step for every value, not just for 68 the selected value] Map Coloring Map Coloring N N {} {Blue} s domain has two remaining values: Blue and Red Assigning Blue to would leave 0 value for, while assigning Red would leave value 69 s domain has two remaining values: Blue and Red Assigning Blue to would leave 0 value for, while assigning Red would leave value So, assign Red to 70 Modified Backtracking Algorithm CSP-BACKRACKING(A, var-domains). If assignment A is complete then return A. X select a variable not in A. D select an ordering on the domain of X. For each value v in D do a. Add (X v) to A ) Most-constrained-variable heuristic b. var-domains forward checking(var-domains, X, v, A) ) Most-constraining-variable heuristic c. If no variable has an empty domain then (i) result CSP-BACKRACKING(A, var-domains) (ii) If result failure then return result d. Remove (X v) from A ) Least-constraining-value heuristic. Return failure Applications of CSP CSP techniques are widely used Applications include: Crew assignments to flights Management of transportation fleet Flight/rail schedules Job shop scheduling ask scheduling in port operations Design, including spatial layout design Radiosurgical procedures 7 7

13 Radiosurgery Problem Minimally invasive procedure that uses a beam of radiation as an ablative surgical instrument to destroy tumors umor = bad Critical structures = good and sensitive Brain = good Burn tumor without damaging healthy tissue 7 7 he CyberKnife linear accelerator robot arm Inputs ) Regions of interest X-Ray cameras 7 76 Inputs Beam Sampling ) Dose constraints Dose to tumor Dose to critical structure Critical umor Falloff of dose around tumor Falloff of dose in critical structure 77 78

14 Constraints Case Results B B C B 000 < umor < < B + B < < B < < B + B < < B < < B + B + B < < B + B < < B + B + B < < B < < B + B < 00 B 000 umor B + B B B + B B B + B + B B + B B + B + B B B + B 00 0 Critical 00 0 B < umor < < B 000 < B B + B + B < 00 B + B + B < < B 79 0% Isodose Surface 80% Isodose Surface LINAC system Cyberknife 80 8