Informed search algorithms Chapter 4Outline Best-first search Greedy best-first search A* search Heuristics Memory Bounded A* SearchBest-first search Idea: use an evaluation function f(n) for each node f(n) provides an estimate for the total cost. Expand the node n with smallest f(n). Implementation: Order the nodes in fringe increasing order of cost. Special cases: greedy best-first search A* searchRomania with straight-line dist.Greedy best-first search f(n) = estimate of cost from n to goal e.g., f(n) = straight-line distance from n to Bucharest Greedy best-first search expands the node that appears to be closest to goal.Greedy best-first search exampleGreedy best-first search exampleGreedy best-first search exampleGreedy best-first search exampleGBFS is not complete gcbadstart state goal state f(n) = straightline distanceProperties of greedy best-first search Complete? No – can get stuck in loops. Time? O(bm), but a good heuristic can give dramatic improvement Space? O(bm) - keeps all nodes in memory Optimal? No e.g. AradSibiuRimnicu VireaPitestiBucharest is shorter!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 Best First search has f(n)=h(n) Uniform Cost search has f(n)=g(n)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: hSLD(n) (never overestimates the actual road distance) Theorem: If h(n) is admissible, A* using TREE-SEARCH is optimalAdmissible heuristics E.g., for the 8-puzzle: h1(n) = number of misplaced tiles h2(n) = total Manhattan distance (i.e., no. of squares from desired location of each tile) h1(S) = ? h2(S) = ?Admissible heuristics E.g., for the 8-puzzle: h1(n) = number of misplaced tiles h2(n) = total Manhattan distance (i.e., no. of squares from desired location of each tile) h1(S) = ? 8 h2(S) = ? 3+1+2+2+2+3+3+2 = 18Dominance If h2(n) ≥ h1(n) for all n (both admissible) then h2 dominates h1 h2 is better for search: it is guaranteed to expand less or equal nr of nodes. Typical search costs (average number of nodes expanded): d=12 IDS = 3,644,035 nodes A*(h1) = 227 nodes A*(h2) = 73 nodes d=24 IDS = too many nodes A*(h1) = 39,135 nodes A*(h2) = 1,641 nodesRelaxed 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 h1(n) gives the shortest solution If the rules are relaxed so that a tile can move to any adjacent square, then h2(n) gives the shortest solutionConsistent heuristics A heuristic is consistent if for every node n, every successor n' of n generated by any action a, h(n) ≤ c(n,a,n') + h(n') If h is consistent, we have f(n’) = g(n’) + h(n’) (by def.) = g(n) + c(n,a,n') + h(n’) (g(n’)=g(n)+c(n.a.n’)) ≥ g(n) + h(n) = f(n) (consistency) f(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 It’s the triangle inequality ! keeps all checked nodes in memory to avoid repeated statesA* search exampleA* search exampleA* search exampleA* search exampleA* search exampleA* search exampleProperties of A* Complete? Yes (unless there are infinitely many nodes with f ≤ f(G) , i.e. step-cost > ε) Time/Space? Exponential except if: Optimal? Yes Optimally Efficient: Yes (no algorithm with the same heuristic is guaranteed to expand fewer nodes)Optimality of A* (proof) Suppose some suboptimal goal G2 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(G2) = g(G2) since h(G2) = 0 f(G) = g(G) since h(G) = 0 g(G2) > g(G) since G2 is suboptimal f(G2) > f(G) from above h(n) ≤ h*(n) since h is admissible (under-estimate) g(n) + h(n) ≤ g(n) + h*(n) from above f(n) ≤ f(G) since g(n)+h(n)=f(n) & g(n)+h*(n)=f(G) f(n) < f(G2) from We want to prove: f(n) < f(G2) (then A* will prefer n over G2)SEARCH TREE 1) Consider the search tree to the right. There are 2 goal states, G1 and G2. The numbers on the edges represent step-costs. You also know the following heuristic estimates: h(BG2) = 9, h(DG2)=10, h(AG1)=2, h(CG1)=1 a) In what order will A* search visit the nodes? Explain your answer by indicating the value of the evaluation function for those nodes that the algorithm considers. G1 G2 C D A B R 1 2 10 9 1 1 ExerciseS B A D E C F G 1 20 2 3 4 8 6 1 1 straight-line distances h(S-G)=10 h(A-G)=7 h(D-G)=1 h(F-G)=1 h(B-G)=10 h(E-G)=8 h(C-G)=20 The graph above shows the step-costs for different paths going from the start (S) to the goal (G). On the right you find the straight-line distances. 1. Draw the search tree for this problem. Avoid repeated states. 2. Give the order in which the tree is searched (e.g. S-C-B...-G) for A* search. Use the straight-line dist. as a heuristic function, i.e. h=SLD, and indicate for each node visited what the value for the evaluation function, f, is. try yourselfMemory Bounded Heuristic Search: Recursive BFS How can we solve the memory problem for A* search? Idea: Try something like depth first search, but let’s not forget everything about the branches we have partially explored. We remember the best f-value we have found so far in the branch we are deleting.RBFS: RBFS changes its mind very often in practice. This is because the f=g+h become more accurate (less optimistic) as we approach the
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