Announcements. Solution to Assignment 3 is posted Assignment 4 is available. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 1 / 25

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1 Announcements Solution to Assignment 3 is posted Assignment 4 is available c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 1 / 25

2 Review: So far... Constraint satisfaction problems defined in terms of variables, domains, constraints Constraint satisfactions problems can be solved with search An arc X, c is arc consistent if for every value for X there are values for the other variables that satisfies c. Arc consistency can be used to simplify the search space Domain splitting can be used to find solutions. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 2 / 25

3 Posing a Constraint Satisfaction Problem A CSP is characterized by A set of variables V 1, V 2,..., V n. Each variable V i has a domain dom(v i ) the set of possible values for V i. (We assume domains are finite.) A possible world or total assignment is an assignment of a value to each variable. A scope is a subset of the variables A hard constraint on a scope specifies which combination of values of the variables in the scope are legal. It is a function from the scope into {true, false}. A solution to CSP (a model) is possible world that satisfies all the constraints. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 3 / 25

4 Clicker Question Suppose there were 100 variables, each with domain size 17. How many possible worlds are there? A 1700 B 117 C D E None of the above c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 4 / 25

5 Arc Consistency An arc X, r(x, Y ) is arc consistent if, for each value x dom(x ), there is some value y dom(y ) such that r(x, y) is satisfied. A network is arc consistent if all its arcs are arc consistent. What if arc X, r(x, Y ) is not arc consistent? All values of X in dom(x ) for which there is no corresponding value in dom(y ) can be deleted from dom(x ) to make the arc X, r(x, Y ) consistent. What is the new domain of X? {x dom(x ) : y dom(y ) such that r(x, y)} c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 5 / 25

6 Clicker Question dom(x ) = {1, 3, 5} dom(y ) = {2, 3, 4} Making the arc X, X < Y arc consistent: A does nothing because it is already arc consistent B results in just 5 being removed from the domain of X C results in 3 and 5 being removed from the domain of X D results in 3 and 5 being removed from the domain of X, and 2 and 3 removed from the domain of Y E results in 3 being removed from both domains c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 6 / 25

7 Clicker Question dom(x ) = {1, 3, 5} dom(y ) = {2, 3, 4} Making the arc Y, X < Y arc consistent: A does nothing because it is already arc consistent B results in just 2 being removed from the domain of Y C results in 2 and 3 being removed from the domain of Y D results in the domain of Y becoming empty E results in 3 being removed from each domain c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 7 / 25

8 Clicker Question dom(x ) = {1, 3, 5} dom(y ) = {2, 3, 4} Making the arc X, X Y arc consistent: A does nothing because it is already arc consistent B results in just 3 being removed from the domain of X C results in 1 and 5 being removed from the domain of X D results in 1, 3 and 5 being removed from the domain of X E results in both domains becoming empty c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 8 / 25

9 Clicker Question dom(1-across) = {ant, big, bus, car, has} dom(2-down) = {ginger, search, symbol, yogurt} For the constraint C: 3rd letter of 1-across = 1st letter of 2-down. After making 1-across, C arc consistent, the domain of 1-across is A {ant, big, bus, car, has} B {big, bus, car, has} C {big, bus, has} D {ant, big, car} E {ant, car} c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 9 / 25

10 Clicker Question dom(1-across) = {ant, big, bus, car, has} dom(2-down) = {ginger, search, symbol, yogurt} For the constraint C: 3rd letter of 1-across = 1st letter of 2-down. After making 2-down, C arc consistent, the domain of 2-down is A {ginger, search, symbol, yogurt} B {ginger, yogurt} C {ginger, search, symbol} D {yogurt} E {} c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 10 / 25

11 Arc Consistency Algorithm The arcs can be considered in turn making each arc consistent. When an arc has been made arc consistent, does it ever need to be checked again? An arc X, r(x, Y ) needs to be revisited if the domain of Y is reduced. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 11 / 25

12 Today: Learning Objectives Today: Constraint Satisfaction Problems At the end of the class you should be able to: implement and trace arc-consistency of a constraint graph show how domain splitting can solve constraint problems show how a CSP can be solved using local search c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 12 / 25

13 Arc Consistency Algorithm Three possible outcomes when all arcs are made arc consistent: One domain is empty = no solution Each domain has a single value = unique solution Some domains have more than one value = there may or may not be a solution c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 13 / 25

14 Complexity of Arc Consistency Consider binary constraints Each variable domain is of size d There are e arcs. Checking an arc takes time O(d 2 ) X, c(x, Y ) for each value for X, check each value for Y Each constraint needs to be checked at most d times. X, c(x, Y ) rechecked when a value for Y is removed. Thus the algorithm GAC takes time O(ed 3 ). Solving a CSP is an NP-complete problem where n the number of variables Give a solution it can be checked in polynomial time But it can be made arc consistent in polynomial time. How? Making the network arc consistent does not solve the problem. We need to search for a solution. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 14 / 25

15 Finding solutions with AC and domain splitting To solve a CSP: Simplify with arc-consistency If a domain is empty, return no solution If all domains have size 1, return solution found Else split a domain, and recursively solve each half. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 15 / 25

16 Finding one solutions with AC and domain splitting Solve one(csp, domains) : simplify CSP with arc-consistency if one domain is empty: return False else if all domains have one element: return solution of that element for each variable else: select variable X with domain D and D > 1 partition D into D 1 and D 2 return Solve one(csp, domains with dom(x ) = D 1 ) or Solve one(csp, domains with dom(x ) = D 2 ) c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 16 / 25

17 Finding set of all solutions with AC and domain splitting Solve all(csp, domains) : simplify CSP with arc-consistency if one domain is empty: return {} else if all domains have one element: return {solution of that element for each variable} else: select variable X with domain D and D > 1 partition D into D 1 and D 2 return Solve all(csp, domains with dom(x ) = D 1 ) Solve all(csp, domains with dom(x ) = D 2 ) c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 17 / 25

18 AC and domain splitting as search Domain splitting lads to search space Nodes: CSP with arc-consistent domains Neighbors of CSP: if all domains are non-empty: select variable X with domain D and D > 1 partition D into D 1 and D 2 neighbors are make AC(CSP dom(x ) = D 1 ) make AC(CSP dom(x ) = D 2 ) Goal: all domains have size 1 Start node: make AC(CSP) c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 18 / 25

19 Local Search Local Search: Maintain a complete assignment of a value to each variable. Start with random assignment or a good guess Repeat: Select a variable to change Select a new value for that variable Until a satisfying assignment is found c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 19 / 25

20 Local Search for CSPs Aim: find an assignment with zero unsatisfied constraints. Given an assignment of a value to each variable, a conflict is an unsatisfied constraint. The goal is an assignment with zero conflicts. Heuristic function to be minimized: the number of conflicts. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 20 / 25

21 Greedy Descent (2 stage) Local Search: Start with random assignment Repeat: Select a variable that participates in the most conflicts Select a value for that variable that minimizes the number of conflicts Until a satisfying assignment is found c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 21 / 25

22 Random Walk Local Search: Start with random assignment (for each variable, select a value for that variable at random) Repeat: Select a variable at random Select a different value for that variable at random Until a satisfying assignment is found c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 22 / 25

23 Random Sampling Local Search: Repeat: For each variable, select a value for that variable at random Until a satisfying assignment is found c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 23 / 25

24 Comparing Stochastic Algorithms How can you compare three algorithms when one solves the problem 30% of the time very quickly but doesn t halt for the other 70% of the cases one solves 60% of the cases reasonably quickly but doesn t solve the rest one solves the problem in 100% of the cases, but slowly? Summary statistics, such as mean run time, median run time, and mode run time don t make much sense. c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 24 / 25

25 Runtime Distribution x-axis runtime (or number of steps) y-axis the proportion (or number) of runs that are solved within that runtime c D. Poole and A. Mackworth 2017 CPSC 322 Lecture 8 25 / 25