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Informed search A* algorithm. 2013/10/17. Outline. Informed = use problem-specific knowledge Which search strategies? Best-first search and its variants Heuristic functions? How to invent them Local search and optimization Hill climbing, simulated annealing, beam search,…

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Informed searchA* algorithm2013/10/17OutlineInformed = use problem-specific knowledge Which search strategies? Best-first search and its variants Heuristic functions? How to invent them Local search and optimization Hill climbing, simulated annealing, beam search,… Evolution algorithms Previously: tree-searchfunction TREE-SEARCH(problem,fringe) return a solution or failurefringe INSERT(MAKE-NODE(INITIAL-STATE[problem]), fringe)loop doif EMPTY?(fringe) then return failurenode REMOVE-FIRST(fringe)if GOAL-TEST[problem] applied to STATE[node] succeedsthen return SOLUTION(node)fringe INSERT-ALL(EXPAND(node, problem), fringe)A strategy is defined by picking the order of node expansionBest-first search General approach of informed search: Best-first search: node is selected for expansion based on an evaluation functionf(n) Idea: evaluation function measures distance to the goal. Choose node which appears best Implementation: fringe is queue sorted in decreasing order of desirability. Special cases: greedy search, A* search A heuristic function[dictionary]“A rule of thumb, simplification, or educated guess that reduces or limits the search for solutions in domains that are difficult and poorly understood.” h(n)= estimated cost of the cheapest path from node n to goal node. If n is goal thenh(n)=0 More information later.Romania with step costs in kmhSLD=straight-line distance heuristic. In this example f(n)=h(n) Expand node that is closest to goal = Greedy best-first searchGreedy search exampleArad (366)Assume that we want to use greedy search to solve the problem of travelling from Arad to Bucharest. The initial state=Arad Greedy search exampleAradZerind(374)Sibiu(253)Timisoara(329)The first expansion step produces: Sibiu, Timisoara and Zerind Greedy best-first will select Sibiu. Greedy search exampleAradSibiuArad(366)Rimnicu Vilcea(193)Fagaras(176)Oradea(380)If Sibiu is expanded we get: Arad, Fagaras, Oradea and Rimnicu Vilcea Greedy best-first search will select: Fagaras Greedy search exampleAradSibiuFagarasSibiu(253)Bucharest(0)If Fagaras is expanded we get: Sibiu and Bucharest Goal reached !! Yet not optimal (see Arad, Sibiu, Rimnicu Vilcea, Pitesti) Greedy search, evaluationCompleteness: NO (cfr. DF-search) Check on repeated states Minimizing h(n) can result in false starts, e.g. Iasi to Fagaras. Greedy search, evaluationCompleteness: NO (cfr. DF-search) Time complexity? Cfr. Worst-case DF-search (with m is maximum depth of search space)Good heuristic can give dramatic improvement. Greedy search, evaluationCompleteness: NO (cfr. DF-search) Time complexity: Space complexity: Keeps all nodes in memory Greedy search, evaluationCompleteness: NO (cfr. DF-search) Time complexity: Space complexity: Optimality? NO Same as DF-search A* searchBest-known form of best-first search. Idea: avoid expanding paths that are already expensive. Evaluation function f(n)=g(n) + h(n) g(n) the cost (so far) to reach the node. h(n) estimated cost to get from the node to the goal. f(n) estimated total cost of path through n to goal. A* searchA* search uses an admissible heuristic A heuristic is admissible if it never overestimates the cost to reach the goal Are optimistic Formally: 1. h(n) <= h*(n) where h*(n) is the true cost from n 2. h(n) >= 0 so h(G)=0 for any goal G.e.g. hSLD(n) never overestimates the actual road distanceRomania exampleA* search exampleFind Bucharest starting at Arad f(Arad) = c(??,Arad)+h(Arad)=0+366=366 A* search exampleExpand Arrad and determine f(n) for each node f(Sibiu)=c(Arad,Sibiu)+h(Sibiu)=140+253=393 f(Timisoara)=c(Arad,Timisoara)+h(Timisoara)=118+329=447 f(Zerind)=c(Arad,Zerind)+h(Zerind)=75+374=449 Best choice is Sibiu A* search exampleExpand Sibiu and determine f(n) for each node f(Arad)=c(Sibiu,Arad)+h(Arad)=280+366=646 f(Fagaras)=c(Sibiu,Fagaras)+h(Fagaras)=239+179=415 f(Oradea)=c(Sibiu,Oradea)+h(Oradea)=291+380=671 f(RimnicuVilcea)=c(Sibiu,RimnicuVilcea)+ h(RimnicuVilcea)=220+192=413Best choice is RimnicuVilcea A* search exampleExpand Rimnicu Vilcea and determine f(n) for each node f(Craiova)=c(Rimnicu Vilcea, Craiova)+h(Craiova)=360+160=526 f(Pitesti)=c(Rimnicu Vilcea, Pitesti)+h(Pitesti)=317+100=417 f(Sibiu)=c(Rimnicu Vilcea,Sibiu)+h(Sibiu)=300+253=553 Best choice is Fagaras A* search exampleExpand Fagaras and determine f(n) for each node f(Sibiu)=c(Fagaras, Sibiu)+h(Sibiu)=338+253=591 f(Bucharest)=c(Fagaras,Bucharest)+h(Bucharest)=450+0=450 Best choice is Pitesti !!! A* search exampleExpand Pitesti and determine f(n) for each node f(Bucharest)=c(Pitesti,Bucharest)+h(Bucharest)=418+0=418 Best choice is Bucharest !!! Optimal solution (only if h(n) is admissable) Note values along optimal path !! Optimality of A*(standard proof)Suppose suboptimal goal G2 in the queue. Let n be an unexpanded node on a shortest to optimal goal G. f(G2 ) = g(G2 ) since h(G2 )=0> g(G) since G2 is suboptimal>= f(n) since h is admissibleSince f(G2) > f(n), A* will never select G2 for expansionBUT … graph search Discards new paths to repeated state. Previous proof breaks down Solution: Add extra bookkeeping i.e. remove more expensive of two paths. Ensure that optimal path to any repeated state is always first followed. Extra requirement on h(n): consistency (monotonicity) Consistency A heuristic is consistent if If h is consistent, we have i.e. f(n) is nondecreasing along any path.Optimality of A*(more usefull)A* expands nodes in order of increasing f value Contours can be drawn in state space Uniform-cost search adds circles. F-contours are gradually Added: 1) nodes with f(n)<C*2) Some nodes on the goalContour (f(n)=C*).Contour I has allNodes with f=fi, wherefi < fi+1.A* search, evaluationCompleteness: YES Since bands of increasing f are added Unless there are infinitly many nodes with f<f(G) A* search, evaluationCompleteness: YES Time complexity: Number of nodes expanded is still exponential in the length of the solution. A* search, evaluationCompleteness: YES Time complexity: (exponential with path length) Space complexity: It keeps all generated nodes in memory Hence space is the major problem not time A* search, evaluationCompleteness: YES Time complexity: (exponential with path length) Space complexity:(all nodes are stored) Optimality: YES Cannot expand fi+1 until fi is finished. A* expands all nodes with f(n)< C* A* expands some nodes with f(n)=C* A* expands no nodes with f(n)>C* Heuristic functionsE.g for the 8-puzzle Avg. solution cost is about 22 steps (branching factor +/- 3) Exhaustive search to depth 22: 3.1 x 1010 states. A good heuristic function can reduce the search process. Heuristic functionsE.g for the 8-puzzle knows two commonly used heuristics h1 = the number of misplaced tiles h1(s)=8 h2 = the sum of the distances of the tiles from their goal positions (Manhattan distance). h2(s)=3+1+2+2+2+3+3+2=18 Heuristic qualityEffective branching factor b* Is the branching factor that a uniform tree of depth d would have in order to contain N+1 nodes. Measure is fairly constant for sufficiently hard problems. Can thus provide a good guide to the heuristic’s overall usefulness. A good value of b* is 1. Heuristic quality and dominance1200 random problems with solution lengths from 2 to 24. If h2(n) >= h1(n) for all n (both admissible) then h2 dominates h1 and is better for searchInventing admissible heuristicsAdmissible heuristics can be derived from the exact solution cost of a relaxed version of the problem: Relaxed 8-puzzle for h1 : a tile can move anywhere As a result, h1(n) gives the shortest solutionRelaxed 8-puzzle for h2 : a tile can move to any adjacent square. As a result, h2(n) gives the shortest solution.The optimal solution cost of a relaxed problem is no greater than the optimal solution cost of the real problem.Inventing admissible heuristicsAdmissible heuristics can also be derived from the solution cost of a subproblem of a given problem. This cost is a lower bound on the cost of the real problem. Pattern databases store the exact solution to for every possible subproblem instance. The complete heuristic is constructed using the patterns in the DB Inventing admissible heuristicsAnother way to find an admissible heuristic is through learning from experience: Experience = solving lots of 8-puzzles An inductive learning algorithm can be used to predict costs for other states that arise during search. Tree search algorithmfunction TREE-SEARCH(problem,fringe) return a solution or failurefringe INSERT(MAKE-NODE(INITIAL-STATE[problem]), fringe)loop doif EMPTY?(fringe) then return failurenode REMOVE-FIRST(fringe)if GOAL-TEST[problem] applied to STATE[node] succeedsthen return SOLUTION(node)fringe INSERT-ALL(EXPAND(node, problem), fringe)Tree search algorithm (2)function EXPAND(node,problem) return a set of nodessuccessors the empty setfor each <action, result> in SUCCESSOR-FN[problem](STATE[node]) dos a new NODE STATE[s] result PARENT-NODE[s] node ACTION[s] action PATH-COST[s] PATH-COST[node]+ STEP-COST(node, action,s) DEPTH[s] DEPTH[node]+1 add s to successorsreturnsuccessors

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