1 Introduction and Examples

Transcription

1 1 Introduction and Examples Sequencing Problems Definition A sequencing problem is one that involves finding a sequence of steps that transforms an initial system state to a pre-defined goal state for that system. Sequencing Problem Example: Rubik s Cube Initial State of Rubik s Cube Figure 1: Rubik s Cube Initial State Sequencing Problem Example: Rubik s Cube Rubik s Cube Goal State Sequencing Problem Example: 8-Puzzle 8-Puzzle Initial and Goal State Sequencing Problem Example: Maze Navigation Initial State: Maze Entrance, Goal State: Maze Exit 1

2 Figure 2: Rubik s Cube Goal State Figure 3: 8-Puzzle Sequencing Problem Example: Blocks World Blocks World: Raising, Translating, and Setting Boxes 2 Modeling Sequencing Problems State Models State Model Definition A state model is a constraint model SM = (V, D, C) whose model solutions are called (legal) states. We let S denote the set of legal states for some state model. State Model Example: 8-Puzzle 2

3 Figure 4: Maze Navigation Figure 5: Blocks World Variables. x ij, i = 1, 2, 3, j = 1, 2, 3, denotes the tile located at row i and column j. Domains. dom(x ij ) = {1,..., 8, e}, where e denotes empty space. Constraints. alldifferent(x). State Graph Models State Graph Model Definition Let SM = (V, D, C) be a state model, and S the set of legal model states. A state-graph model for S has the following components. a designated initial state s 0 S 3

4 Figure 6: State s Figure 7: State 1 of τ(s) one or more designated goal states s g G S a state transition function τ : S subset(s) that maps each state to a subset of next states. An environmental factor that produces a transition from a state s to a state in τ(s) is called an action. Optional: each state transition may have an associated cost. The Purpose of Using State Graph Models Find a solution path (preferably the shortest or least costly) from s 0 to a goal state s g G. State Transition Function Example: 8-Puzzle Constraint Models for Sequencing Problems Finding a Solution for SM = (V, D, C) and (S, τ, s 0, G) Assume the solution path consists of a sequence of states s 0,..., s d 1 of length d Figure 8: State 2 of τ(s) 4

5 Figure 9: State 3 of τ(s) Variables. d copies of V : V 0, V 1,..., V d 1, where an assignment over V i encodes state s i of the solution path. Domains. d copies of D i, i = 1,..., m: D i0, D i1,..., D i(d 1), where D ij is the domain of the i th variable of V j. State Constraints. d copies of C: C 0, C 1,..., C d 1, where C i ensures that an assignment over V i encodes a valid state s i. Initial Constraint. The assignment over V 0 encodes the initial state s 0. Transition Constraints. For all i = 1,..., d 1, s i τ(s i 1 ). Goal Constraints. s i G for some i = 0,..., d 1. Termination Constraints. For all i = 0,..., d 2, if s i G, then s i+1 = s i. 3 A Search Proximity Heuristics Proximity Heuristic Definition Let (S, τ, s 0, G) be a state-transition model. A proximity heuristic is a function h : S R, where h(s) estimates the distance (in terms of number of remaining transitions or in terms of the sum of the transition costs of each remaining transition) from s to a goal state. We use the term heuristic since h is not guaranteed to be accurate. Informed Search We say that a search is informed if it uses additional problem information, such as a proximity heuristic, to assist in finding a solution. Enabling computers to learn such heuristics in an unsupervised setting is an ongoing area of research. This is similar to the problem of learning implied constraints within a constraintprogramming setting. 5

6 Proximity Heuristic Example: 8-Puzzle Two Heuristics h 1 and h 2 h 1 (s) counts the number of tiles that have been moved from their goalstate position. h 2 (s) = 8 m i, where m i is the number of moves needed to get tile i back i=1 to its goal-state position. Figure 10: h 1 (s) = 7, and h 2 (s) = = 21 Best First Search Using Proximity Heuristic h Mark s 0 as having been visited. Initialize priority queue Q = {s 0 } to contain the initial state. While Q is nonempty: Remove from Q the state s for which h(s) is minimum. Let C = τ(s) M, where M is the set of already-marked states. If C contains a goal state s g, then return the path from s 0 to s g Update h(ŝ) for states of ŝ M, if necessary. Mark each state of C as being visited and add C to Q. Mark s as having been explored. Return null since no goal state was reached. Best First Search Example s 0 = a, s g = i, h(s) = measures the alphabetical distance from the alphabet label of s to the letter i. For example, h(s 0 ) = 8 since a is 8 letters away from i. Q = {(a, 8)} Q = {(d, 5), (b, 7)} Q = {(h, 2), (e, 4), (b, 7)} Q = {(g, 1), (e, 4), (b, 7)} Q = {(e, 4), (b, 7)} 6

7 a b c d e f h g i Figure 11: Initial State a b c d e f h g i Figure 12: a marks b and d a b c d e f h g i Figure 13: d marks e and h 7

8 a b c d e f h g i Figure 14: h marks g a b c d e f h g i Figure 15: g marks i: Goal Reached! 8

9 A Search: a Balance of Past and Future Total Path Cost Heuristic A search refers to a best-first search in which the proximity heuristic h(n) is replaced by the total path cost heuristic f(n) = g(n) + h(n), where f(n) is the path cost from initial state to n, and h(n) is an estimate of the cost of reaching a goal state from n. Admissible and Consistent Heuristics h(n) is called admissible iff h(n) is no greater than the actual minimal cost of moving from n to a goal state. h(n) is called consistent or monotone iff h(n) cost(n, n ) + h(n ) for every successor node n τ(n). Examples of Admissible Heuristics 8-Puzzle h 1 (s) counts the number of tiles that have been moved from their goalstate position. h 2 (s) = 8 m i, where m i is the number of moves needed to get tile i back i=1 to its goal-state position. Both heuristics are monotone (why?). Finding Shortest Driving Distance from one Location to Another h(n) is the Euclidean distance between n and final destination. This heuristic is also monotone (why?). Consistent Heuristics are Admissible Theorem 1: Every consistent heuristic is admissible. Proof of Theorem 1. Suppose n is a goal state. Then we may assume that h(n) = 0, and hence the value is admissible. Now suppose h(n ) has an admissible value, for every node n that is within k or fewer steps (i.e. actions) from a goal state, for some k 0. Let n be a node that is k + 1 steps from a goal state, and let n be a successor of n along a path that is optimal from n to a goal state. Then by consistency we have h(n) cost(n, n ) + h(n ) cost(n, n ) + min cost of reaching goal from n = the minimum cost of reaching a goal state from n. Therefore, h(n) is admissible. 9

10 Admissible Heuristics Lead to Optimal Solutions Theorem 2 A best-first search that uses f(n) = g(n) + h(n) as ordering heuristic, where h(n) is admissible, will find a goal state along an optimal path, provided action costs are all positive-valued, and ties are broken according to g(n). Proof that Admissible Heuristics Lead to Optimal Solutions Proof of Theorem 2. It suffices to prove that, if P is an optimal path from initial state to a goal node n, then the parent of n is removed from Q before n. Let m n be the node of P that is furthest from the root, and is removed from Q before n is discovered. We know m exists, since the root node is the first node of P, and is the first node removed from Q. We must show that f(m) f(n), in which case m will be removed from the queue before n (since g(m) < g(n)). Hence the sucessors of m will be considered, and any successors that have not already been placed in the queue, will then be placed. But since m n is the furthest from the root that belongs to P and was placed in the queue, it follows that n must be a successor of m, and hence the optimal path to n will have been recorded. Proof that Admissible Heuristics Lead to Optimal Solutions Completing the Proof of Theorem 2: f(m) f(n) f(m) = g(m) + h(m) g(m) + (g(n) g(m)), since h(m) is admissible, and g(n) g(m) is the cost in moving from m to goal state n. But g(n) = f(n), since h(n) = 0 (n is a goal state). Therefore, f(m) = g(m) + h(m) g(m) + (g(n) g(m)) = g(n) = f(n). Informedness Informedness Definition Suppose h 1 and h 2 are two admissible heuristics. We say h 2 is more informed than h 1 iff h 2 (n) h 1 (n) for every node n S. Theorem 3 Given a search problem, let C denote the minimum cost from the initial state to a goal state. If admissible heuristic h 2 is more informed than admissible h 1, then A search using h 2 visits at most the same number of nodes n that are visited when using h 1, where f(n) < C. In other words, if an A search visits a node n using h 2 that was not visited when using h 1, then it must be the case that f(n) = g(n) + h 2 (n) = C. 10

11 Proof of Theorem 3 We know that for both searches, the initial state s 0 is added to Q. Now suppose n is added to Q when using f 2. Suppose, by way of induction, that the parent m of n is added to Q in both searches. Then since f 1 (m) f 2 (m) < C, and since m is popped from Q using f 2, it will also be popped using f 1. Hence, n will be added to Q using f 1. Now, since best-first search is used, every node n for which f(n) < C will be popped from the queue before the goal node is popped. Moreover, if f 2 (n) = g(n) + h 2 (n) < C, then also f 1 (n) = g(n) + h 1 (n) < C, since h 1 (n) h 2 (n). Hence, if node n is popped when f 2 is used, then it will also be popped when f 1 is used. Therefore, if n is popped using f 2, but not using f 1, then we must have f 2 (n) = C. Methods for Generating Admissible Heuristics Problem Relaxations Problem P 2 is a relaxed version of problem P 1, provided both problems share the same initial state, state space and goal-testing function, but the action set for problem P 2 contains the action set for problem P 1. Thus, if the minimum cost of reaching a goal in the relaxed problem can be computed, then this cost serves as a lower bound for the cost in the original problem (since the original problem perhaps cannot use all of the actions that were used in the relaxed problem). Example of 8-Puzzle Relaxation. Any two adjacent tiles may be swapped (even if neither is the empty space). Methods for Generating Admissible Heuristics Other Methods Maximizing over a set of heuristics. If h 1,..., h k are all admissible heuristics, then so is max(h 1,..., h k )(n). Pattern Databases. These databases contain a set of state patterns, so that each state of the state space will fit at least one of the patterns. Then associated with each pattern is a value of the cost in moving from this pattern to a pattern that corresponds with a goal state. Used by Richard Korf of UCLA for finding optimal solutions to Rubik s Cube. 11

12 Exercises 1. Give an example that shows that not every admissible heuristic is monotone. 2. A robot is standing due north on square (1, 1) of an N N grid. The goal of the robot is to reach square (N, N). The robot has three possible actions: turn left, turn right, and move forward (either north, south, east, or west) to an adjacent square. However, throughout the grid there are walls that prevent the robot from moving forward from one square to an adjacent one (there are also walls around the entire grid boundary). Specify the state space for this problem, the goal-testing function, and the set of actions. What is the size of the state space? Provide a bound on the branching factor b, and a bound on the minimum goal depth d. 3. For the 4-puzzle, show that the state space can be divided into two sets of equal size, such that no state from one set is adjacent to a state from the other set. Note: this property also holds for the 8-puzzle, and shows that in general not all states from a space are reachable from an initial state. Extra credit: prove it is also true for the 8-puzzle. 4. For the n-queens problem, prove that the state space has at least (n!) 1 3 states. Hint: derive a lower bound on the branching factor by considering the maximum number of of squares that a queen can attack in any column. 5. Given three jugs, one of 12 gallons, 8 gallons, and 3 gallons, and a water faucet, you are allowed to fill the jugs and/or empty them into one another, or on to the ground. The goal is to measure out exactly one gallon. Hint: every action must have the effect of either entirely emptying or filling one of the jugs. For example, if you pour from the 8-gallon jug, then it must either empty, or it must fill at least one other jug. 6. Consider the state space consisting of the numbers 1 to 15. The two possible actions on a state i are to multiply it by 2, or multiply it by 2 and add 1. Assuming 1 is the initial state, and 11 is the goal state. Show the order of the search for breadth-first, bounded depth-first (with depth bound of 3), Assume the 2k action has precedence over the 2k + 1 action. 7. Provide a sequence of state spaces S 1,..., S n,... for which iterative deepening search finds the goal state in O(n 2 ) steps, while depth-first search finds the goal state in O(n) steps. 8. n checkers occupy squares (1, 1) through (1, n) of an n n grid (i.e. they are at the bottom row of the grid). The checkers must be moved to the top row, but in reverse order. In other words, checker i must be moved to square (n i + 1, n). On each step each checker can move one square up, down, left, right, or stay put. If a checker does not move, then it may be jumped (vertically or horizontally) by one other checker that is adjacent 12

13 to it. Two checkers cannot occupy the same square. Calculate the size of the state space as a function of n and estimate the branching factor as a function of n. 9. For the previous problem, provide a nontrivial admissible heuristic h i that estimates the number of steps needed for the i th checker to reach its goal square (n i + 1, n). Will n i=1 h i(n) be admissible? What about min i (h i (n))? max i (h i (n))? 10. In what way can breadth-first search be considered a special case of bestfirst search? Explain. 11. Prove that the set of nodes considered during an A search is a subset of nodes considered during a breadth-first search. 13