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Informed search algorithms Chapter 4

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Outline Best-first search Greedy best-first search A * search Heuristics Local search algorithms Hill-climbing search

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Best-first search Idea: use an evaluation function f(n) for each node –estimate of "desirability" Expand most desirable unexpanded node Implementation: Order the nodes in fringe in decreasing order of desirability Special cases: –greedy best-first search –A * search

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Romania with step costs in km

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Greedy best-first search Evaluation function f(n) = h(n) (heuristic) = estimate of cost from n to goal e.g., h SLD (n) = straight-line distance from n to Bucharest Greedy best-first search expands the node that appears to be closest to goal

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Greedy best-first search example

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Properties of greedy best-first search Complete? No – can get stuck in loops, e.g., Iasi Neamt Iasi Neamt Time? O(b m ), but a good heuristic can give dramatic improvement Space? O(b m ) -- keeps all nodes in memory Optimal? No

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A * search Idea: avoid expanding paths that are already expensive Evaluation function f(n) = g(n) + h(n) g(n) = cost so far to reach n h(n) = estimated cost from n to goal f(n) = estimated total cost of path through n to goal

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A * search example

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Admissible heuristics A heuristic h(n) is admissible if for every node n, h(n) h * (n), where h * (n) is the true cost to reach the goal state from n. An admissible heuristic never overestimates the cost to reach the goal, i.e., it is optimistic Example: h SLD (n) (never overestimates the actual road distance) Theorem: If h(n) is admissible, A * using TREE-SEARCH is optimal

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Optimality of A * (proof) Suppose some suboptimal goal G 2 has been generated and is in the fringe. Let n be an unexpanded node in the fringe such that n is on a shortest path to an optimal goal G. f(G 2 ) = g(G 2 )since h(G 2 ) = 0 (G 2 is a goal) > g(G) since G 2 is sub-optimal = g(n) + d(n,G)since there is only one path from root to G (tree) g(n) + h(n)since h is admissible = f(n)by definition of f Hence f(G 2 ) > f(n), and A * will never select G 2 for expansion. Recall that it is only when a node is picked for expansion that we check if it is goal.

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Properties of A* Complete?? Yes Optimal?? Yes Let C* be the cost of the optimal solution. A* expands all nodes with f(n) < C* A* expands some nodes with f(n) = C* A* expands no nodes with f(n) > C*

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A* using Graph-Search A* using Graph-Search can produce sub-optimal solutions, even if the heuristic function is admissible. Sub-optimal solutions can be returned because Graph- Search can discard the optimal path to a repeated state if it is not the first one generated.

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Consistent heuristics A heuristic is consistent if for every node n, every successor n' of n generated by any action a, c(n,a,n') + h(n') h(n) If h is consistent, we have f(n') = g(n') + h(n') = g(n) + c(n,a,n') + h(n') g(n) + h(n) = f(n) i.e., f(n) is non-decreasing along any path. Theorem: If h(n) is consistent, A* using GRAPH-SEARCH is optimal

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Optimality under consistency Recall: If h is consistent, then A* expands nodes in order of increasing f value. First goal node selected for expansion must be optimal. –Recall that only when a node is selected for expansion only then it is tested whether it is a goal state or not. Why? Because if a goal node G 2 is selected later for expansion than G 1 this means that: –f(G 1 ) f(G 2 ) –g(G 1 ) + h(G 1 ) g(G 2 ) + h(G 2 ) recall h(G 1 ) = h(G 2 )=0 since G 1 and G 2 are goal, hence –g(G 1 ) g(G 2 ) which means that G 2 is sub-optimal solution.

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Straight-Line Distance Heuristic is Consistent Why? We know that the general triangle inequality is satisfied when each side is measured by the straight-line. And c(n,a,n) is greater or equal to the straight-line distance between n and n. –d(n.state, goalstate) d(n.state, goalstate) + d(n.state, n.state) By Euclidian distance triangle property Also recall that –d(n.state, goalstate) = h(n) and –d(n.state, goalstate) = h(n), and –d(n.state, n.state) c(n,a,n) for any action a. Hence, –h(n) h(n) + d(n.state, n.state) h(n) + c(n,a,n) I.e. h is consistent.

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Admissible heuristics E.g., for the 8-puzzle: h 1 (n) = number of misplaced tiles h 2 (n) = total Manhattan distance |x 1 -x 2 |+|y 1 +y 2 | (i.e., no. of squares from desired location of each tile) h 1 (S) = ? h 2 (S) = ?

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Admissible heuristics E.g., for the 8-puzzle: h 1 (n) = number of misplaced tiles h 2 (n) = total Manhattan distance |x 1 -x 2 |+|y 1 +y 2 | (i.e., no. of squares from desired location of each tile) h 1 (S) = ? 8 h 2 (S) = ? = 18

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Why are they admissible? Misplaced tiles: No move can get more than one misplaced tile into place,so this measure is a guaranteed underestimate and hence admissible. Manhattan: In fact, each move can at best decrease by one the rectilinear distance of a tile from its goal.

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Are they consistent? c(n,a,n) = 1 for any action a Claim: h 1 (n) h 1 (n) + c(n,a,n) = h 1 (n) + 1 –Now, no move (action) can get more than one misplaced tile into place. –Also, no move can create more than one new misplaced tile. –Hence, the above follows. I.e. h 1 is consistent. Similar reasoning for h 2 as well.

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A* Graph-Search (without consistent heuristic) The problem is that sub-optimal solutions can be returned because Graph- Search can discard the optimal path to a repeated state if it is not the first one generated. When the heuristic function h is consistent, then we proved that the optimal path to any repeated state is the first one to be followed. When h is not consistent, we can still use Graph-Search if we discard the longer path. We store in the hashtable pairs (statekey, pathcost) Graph-Search(problem, fringe) node = MakeNode(problem.initialstate); Insert(fringe, node); do if ( Empty(fringe) ) return null; //failure node = Remove(fringe); if (problem.GoalTest(node.state)) return node; key_cost_pair = Find( closed, node.state.getKey() ); if (key_cost_pair == null || key_cost_pair.pathcost > node.pathcost) Insert (closed, (node.state.getKey(), node.pathcost) ); InsertAll (fringe, Expand(node, problem) ); while (true);

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A* using Graph-Search Lets try it here

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Dominance If h 2 (n) h 1 (n) for all n (both admissible) then h 2 dominates h 1 h 2 is better for search –Let C* be the cost of the optimal solution. –A* expands all nodes with f(n) < C* –A* expands some nodes with f(n) = C* –A* expands no nodes with f(n) > C* –Hence, we want f(n) (g(n)+h(n)) to be as big as possible. Since we cant do anything about g(n) we are interested in having h(n) as big as possible. Typical search costs (average number of nodes expanded): d=12IDS = 3,644,035 nodes A * (h 1 ) = 227 nodes A * (h 2 ) = 73 nodes d=24 IDS = too many nodes A * (h 1 ) = 39,135 nodes A * (h 2 ) = 1,641 nodes

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The max of heuristics If a collection of admissible heuristics h 1, …, h m is available for a problem, and none of them dominates any of the others, we create a compound heuristic as: –h(n) = max {h 1 (n), …, h m (n)} Is it admissible? Yes, because each component is admissible, so h wont over- estimate the distance to the goal. If h 1, …, h m are consistent, is h consistent as well? Yes. For any action (move) a: c(n,a,n)+h(n) = c(n,a,n)+ max {h 1 (n), …, h m (n)} = max {c(n,a,n)+h 1 (n), …, c(n,a,n)+h m (n)} (from consistency of h i ) max {h 1 (n), …, h m (n)} = h(n)

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Relaxed problems A problem with fewer restrictions on the actions is called a relaxed problem The cost of an optimal solution to a relaxed problem is an admissible heuristic for the original problem If the rules of the 8-puzzle are relaxed so that a tile can move anywhere, then h 1 (n) gives the shortest solution If the rules are relaxed so that a tile can move to any adjacent square, then h 2 (n) gives the shortest solution

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Local search algorithms In many optimization problems, the path to the goal is irrelevant; the goal state itself is the solution State space = set of "complete" configurations Find configuration satisfying constraints, e.g., n-queens In such cases, we can use local search algorithms, which keep a single "current" state, and try to improve it.

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Example: n-queens Put n queens on an n × n board with no two queens on the same row, column, or diagonal How many successors from each state? (n-1)*n

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8-queens problem h = number of pairs of queens that are attacking each other, either directly or indirectly h = 17 for the above state

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Hill-climbing search "Like climbing Everest in thick fog with amnesia" Hill-Climbing chooses randomly among the set of best successors, if there is more than one.

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Hill-climbing search Problem: depending on initial state, can get stuck in local maxima

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Hill-climbing search: 8-queens problem A local minimum with h = 1

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Local Minima Starting from a randomly generated state of the 8- queens, hill-climbing gets stuck 86% of the time, solving only 14% of problems. It works quickly, taking just 4 steps on average when it succeeds, and 3 when it gets stuck – not bad for a state space with million states. Memory is constant, since we keep only one state. Since it is so attractive, what can we do in order to not get stuck?

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Random-restart hill climbing Well known adage: If at first you dont succeed, try, try again. It conducts a series of hill-climbing searches from randomly generated initial states, stopping when a goal is found. If each hill-climbing search has a probability p of success, then the expected number of restarts is 1/p. For 8-queens, p=0.14, so we need roughly 7 iterations to find a goal, i.e. 22 steps (3 steps for failures and 4 for success)

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