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Heuristics & Informed Search Hill-climbing bold & informed, heuristics, potential dangers, simulated annealing Best first search tentative & uninformed, BFS + hill-climbing Algorithm A optimal cost solutions, admissibility, A*

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Search strategies so far… bold & uninformed: STUPID tentative & uninformed: DFS (and variants), BFS bold & informed: hill-climbing essentially, DFS utilizing a "measure of closeness" to the goal (a.k.a. a heuristic function) from a given state, pick the best successor state i.e., the state with highest heuristic value stop if no successor is better than the current state EXAMPLE: travel problem ( loc(omaha) loc(los_angeles) ) heuristic(State) = -(crow-flies distance from L.A.)

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Hill-climbing example move(loc(omaha), loc(chicago)). move(loc(omaha), loc(denver)). move(loc(chicago), loc(denver)). move(loc(chicago), loc(los_angeles)). move(loc(chicago), loc(omaha)). move(loc(denver), loc(los_angeles)). move(loc(denver), loc(omaha)). move(loc(los_angeles), loc(chicago)). move(loc(los_angeles), loc(denver)). heuristic(loc(omaha)) = -1700 heuristic(loc(chicago)) = -2200 heuristic(loc(denver)) = -1400 heuristic(loc(los_angeles)) = 0 the heuristic guides the search, always moving closer to the goal what if the flight from Denver to L.A. were cancelled?

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8-puzzle example start state has 5 tiles in place 123 86 754 hval = 5 457 368 21 457 68 321 57 468 321 hval = 4hval = 6 457 628 31 457 68 321 47 658 321 hval = 4 hval = 5 of the 3 possible moves, 2 improve the situation if assume left-to-right preference, move leads to a dead-end (no successor state improves the situation) so STOP! heuristic(State) = # of tiles in place, including the space

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Intuition behind hill-climbing if you think of the state space as a topographical map (heuristic value = elevation), then hill-climbing selects the steepest gradient to move up towards the goal potential dangers plateau: successor states have same values, no way to choose foothill: local maximum, can get stuck on minor peak ridge: foothill where N-step lookahead might help

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Hill-climbing variants could generalize hill-climbing to continue even if the successor states look worse always choose best successor don't stop unless reach the goal or no successors dead-ends are still possible (and likely if the heuristic is not perfect) simulated annealing allow moves in the wrong direction on a probabilistic basis decrease the probability of a backward move as the search continues idea: early in the search, when far from the goal, heuristic may not be good heuristic should improve as you get closer to the goal approach is based on a metallurgical technique

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Best first search hill-climbing is dependent on a near-perfect heuristic since bold tentative & informed: best first search like breadth first search, keep track of all paths searched like hill-climbing, use heuristics to guide the search always expand the "most promising" path i.e., the path ending in a state with highest heuristic value

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Best first example 123 86 754 hval = 5 457 628 31 457 68 321 47 658 321 hval = 4 hval = 5 457 68 321 57 468 321 hval = 7 567 48 321 57 468 321 hval = infinityhval = 4 hval = 6 457 368 21 457 68 321 57 468 321

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Breadth first search vs. best first search procedural description of breadth first search must keep track of all current paths, initially: [ [Start] ] look at first path in the list of current paths if the path ends in a goal, then DONE otherwise, remove the first path generate all paths by extending to each possible move add the newly extended paths to the end of the list recurse procedural description of best first search must keep track of all current paths w/ hvals, initially: [ Hval:[Start] ] look at first path in the list of current paths if the path ends in a goal, then DONE otherwise, remove the first path generate all paths by extending to each possible move add the newly extended paths to the list, sorted by (decreasing) hval recurse

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Best first implementation %% best.pro Dave Reed 3/15/02 %% Best first search with cycle checking %% best(Current, Goal, Path): Path is a list of states that %% lead from Current to Goal with no duplicate states. %% %% best_help(ListOfPaths, Goal, Path): Path is a list %% of states (with associated hval) that lead from one of %% the paths in ListOfPaths (a list of lists) to Goal with %% no duplicate states. %% %% extend(Path, Goal, ListOfPaths): ListOfPaths is the list %% of all possible paths (with associated hvals) obtainable %% by extending Path (at the head) with no duplicate states. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% best(State, Goal, Path) :- best_help([0:[State]], Goal, RevPath), reverse(RevPath, Path). best_help([_:[Goal|Path]|_], Goal, [Goal|Path]). best_help([_:Path|RestPaths], Goal, SolnPath) :- extend(Path, Goal, NewPaths), append(RestPaths, NewPaths, TotalPaths), sort(TotalPaths, SortedPaths), reverse(SortedPaths,RevPaths), best_help(RevPaths, Goal, SolnPath). extend([State|Path], Goal, NewPaths) :- bagof(H:[NextState,State|Path], (move(State, NextState), not(member(NextState, [State|Path])), heuristic(NextState,Goal,H)), NewPaths), !. extend(_, _, []). differences from BFS associate heuristic value with each path H:Path since extend needs to know heuristic values for paths, must pass Goal to extend once the new paths have been appended, they must be sorted (in decreasing order by heuristic value)

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Travel example %% heuristic(Loc, Goal, Value) : Value is -distance from Goal %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% heuristic(loc(omaha), loc(los_angeles), -1700). heuristic(loc(chicago), loc(los_angeles), -2200). heuristic(loc(denver), loc(los_angeles), -1400). heuristic(loc(los_angeles), loc(los_angeles), 0). must define the heuristic function for this problem ?- best(loc(omaha), loc(los_angeles), Path). Path = [loc(omaha), loc(denver), loc(los_angeles)] Yes best first search is able to find a path

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8-puzzle example %% heuristic(Board, Goal, Value) : Value is the # of tiles in place %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% heuristic(tiles(Row1,Row2,Row3),tiles(Goal1,Goal2,Goal3), Value) :- same_count(Row1, Goal1, V1), same_count(Row2, Goal2, V2), same_count(Row3, Goal3, V3), Value is V1 + V2 + V3. same_count([], [], 0). same_count([H1|T1], [H2|T2], Count) :- same_count(T1, T2, TCount), (H1 = H2, Count is TCount + 1 ; H1 \= H2, Count is TCount). must define the heuristic function for this problem ?- best(tiles([1,2,3],[8,6,space],[7,5,4]), tiles([1,2,3],[8,space,4],[7,6,5]), Path). Path = [tiles([1, 2, 3], [8, 6, space], [7, 5, 4]), tiles([1, 2, 3], [8, 6, 4], [7, 5, space]), tiles([1, 2, 3], [8, 6, 4], [7, space, 5]), tiles([1, 2, 3], [8, space, 4], [7, 6, 5])] Yes ?- best(tiles([2,3,4],[1,8,space],[7,6,5]), tiles([1,2,3],[8,space,4],[7,6,5]), Path). Path = [tiles([2, 3, 4], [1, 8, space], [7, 6, 5]), tiles([2, 3, space], [1, 8, 4], [7, 6, 5]), tiles([2, space, 3], [1, 8, 4], [7, 6, 5]), tiles([space, 2, 3], [1, 8, 4], [7, 6, 5]), tiles([1, 2, 3], [space, 8, 4], [7, 6, 5]), tiles([1, 2|...], [8, space|...], [7, 6|...])] Yes best first search is able to find a path

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Other examples move(jugs(J1,J2), jugs(4,J2)) :- J1 < 4. move(jugs(J1,J2), jugs(J1,3)) :- J2 < 3. move(jugs(J1,J2), jugs(0,J2)) :- J1 > 0. move(jugs(J1,J2), jugs(J1,0)) :- J2 > 0. move(jugs(J1,J2), jugs(4,N)) :- J2 > 0, J1 + J2 >= 4, N is J2 - (4 - J1). move(jugs(J1,J2), jugs(N,3)) :- J1 > 0, J1 + J2 >= 3, N is J1 - (3 - J2). move(jugs(J1,J2), jugs(N,0)) :- J2 > 0, J1 + J2 < 4, N is J1 + J2. move(jugs(J1,J2), jugs(0,N)) :- J1 > 0, J1 + J2 < 3, N is J1 + J2. heuristic function for the Water Jug Problem? heuristic function for Missionaries & Cannibals?

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Comparing best first search unlike hill-climbing, best first search can handle "dips" in the search not so dependent on a perfect heuristic depth first search and breadth first search may be seen as special cases of best first search DFS: heuristic value is distance (number of moves) from start state BFS: heuristic value is inverse of distance (number of moves) from start state or, procedurally: DFS: assign all states the same heuristic value when adding new paths, add equal heuristic values at the front BFS: assign all states the same heuristic value when adding new paths, add equal heuristic values at the end

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Optimizing costs consider a related problem: instead of finding the shortest path (i.e., fewest moves) to a solution, suppose we want to minimize some cost EXAMPLE: airline travel problem could associate costs with each flight, try to find the cheapest route could associate distances with each flight, try to find the shortest route we could use a strategy similar to breadth first search repeatedly extend the minimal cost path search is guided by the cost of the path so far but such a strategy ignores heuristic information would like to utilize a best first approach, but not directly applicable search is guided by the remaining cost of the path IDEAL: combine the intelligence of both strategies cost-so-far component of breadth first search (to optimize actual cost) cost-remaining component of best first search (to make use of heuristics)

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Travel problem revisited %% travelcost.pro Dave Reed 3/15/02 %% %% This file contains the state space definition %% for air travel, with costs (distances) associated %% with each flight and heuristics providing %% as-the-crow-flies distance. %% %% loc(City): defines the state where the person %% is in the specified City. %%%%%%%%%%%%%%%%%%%%%%%%% move(loc(omaha), loc(chicago), 500). move(loc(omaha), loc(denver), 600). move(loc(chicago), loc(denver), 1000). move(loc(chicago), loc(los_angeles), 2200). move(loc(chicago), loc(omaha), 500). move(loc(denver), loc(los_angeles), 1400). move(loc(denver), loc(omaha), 600). move(loc(los_angeles), loc(chicago), 2200). move(loc(los_angeles), loc(denver), 1400). heuristic(loc(omaha), loc(los_angeles), 1700). heuristic(loc(chicago), loc(los_angeles), 2200). heuristic(loc(denver), loc(los_angeles), 1400). heuristic(loc(los_angeles), loc(los_angeles), 0). add the actual cost (number of miles per flight) to each move heuristic is crow-flies distance (note: no longer negative, since want estimate of remaining cost) want to minimize the total flight distance from Omaha to L.A.

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Algorithm A associate 2 costs with a path gactual cost of the path so far hheuristic estimate of the remaining cost to the goal f = g + hcombined heuristic cost estimate Algorithm A (for trees – note that usual formulation is for graphs) best first search using f as the heuristic

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Algorithm A implementation %% Algorithm A (for trees) %% atree(Current, Goal, Path): Path is a list of states %% that lead from Current to Goal with no duplicate states. %% %% atree_help(ListOfPaths, Goal, Path): Path is a list of %% states (with associated F and G values) that lead from one %% of the paths in ListOfPaths (a list of lists) to Goal with %% no duplicate states. %% %% extend(G:Path, Goal, ListOfPaths): ListOfPaths is the list %% of all possible paths (with associated F and G values) obtainable %% by extending Path (at the head) with no duplicate states. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% atree(State, Goal, G:Path) :- atree_help([0:0:[State]], Goal, G:RevPath), reverse(RevPath, Path). atree_help([_:G:[Goal|Path]|_], Goal, G:[Goal|Path]). atree_help([_:G:Path|RestPaths], Goal, SolnPath) :- extend(G:Path, Goal, NewPaths), append(RestPaths, NewPaths, TotalPaths), sort(TotalPaths, SortedPaths), atree_help(SortedPaths, Goal, SolnPath). extend(G:[State|Path], Goal, NewPaths) :- bagof(NewF:NewG:[NextState,State|Path], Cost^H^(move(State, NextState, Cost), not(member(NextState, [State|Path])), heuristic(NextState,Goal,H), NewG is G+Cost, NewF is NewG+H), NewPaths), !. extend(_, _, []). differences from best associate two values with each path (F is total estimated cost, G is actual cost so far) F:G:Path since extend needs to know of current path, must pass G new feature of bagof: if a variable appears only in the 2 nd arg, must identify it as backtrackable

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Travel example ?- atree(loc(omaha), loc(los_angeles), Path). Path = 2000:[loc(omaha), loc(denver), loc(los_angeles)] ; Path = 2700:[loc(omaha), loc(chicago), loc(los_angeles)] ; Path = 2900:[loc(omaha), loc(chicago), loc(denver), loc(los_angeles)] ; No note: Algorithm A finds the path with least cost (here, distance) not necessarily the path with fewest steps suppose the flight from Chicago to L.A. was 2500 miles (instead of 2200) then Omaha Chicago Denver L.A. is shorter than Omaha Chicago L.A. can use semi-colon to see alternative paths in order

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8-puzzle example ?- atree(tiles([1,2,3],[8,6,space],[7,5,4]), tiles([1,2,3],[8,space,4],[7,6,5]), Path). Path = 3:[tiles([1, 2, 3], [8, 6, space], [7, 5, 4]), tiles([1, 2, 3], [8, 6, 4], [7, 5, space]), tiles([1, 2, 3], [8, 6, 4], [7, space, 5]), tiles([1, 2, 3], [8, space, 4], [7, 6, 5])] Yes ?- atree(tiles([2,3,4],[1,8,space],[7,6,5]), tiles([1,2,3],[8,space,4],[7,6,5]), Path). Path = 5:[tiles([2, 3, 4], [1, 8, space], [7, 6, 5]), tiles([2, 3, space], [1, 8, 4], [7, 6, 5]), tiles([2, space, 3], [1, 8, 4], [7, 6, 5]), tiles([space, 2, 3], [1, 8, 4], [7, 6, 5]), tiles([1, 2|...], [space, 8|...], [7, 6|...]), tiles([1|...], [8|...], [7|...])] Yes here, Algorithm A finds the same paths as best first search not surprising since actual cost component (g) is trivial still, not guaranteed to be the case (recall: best only cares about the future) could apply Alg. A to the 8-puzzle problem actual cost of each move (g) is 1 remaining cost estimate (h) is # of tiles out of place

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Algorithm A vs. hill-climbing if the cost estimate function h is perfect, then f is a perfect heuristic Algorithm A is deterministic if know actual costs for each state, Alg. A reduces to hill-climbing

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Admissibility in general, actual costs are unknown at start – must rely on heuristics if the heuristic is imperfect, Alg. A is NOT guaranteed to find an optimal solution if a control strategy is guaranteed to find an optimal solution (when a solution exists), we say it is admissible if cost estimate h is never overestimates actual cost, then Alg. A is admissible (when admissible, Alg. A is commonly referred to as Alg. A*)

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Admissible examples is our heuristic for the travel problem admissible? h (State) = crow-flies distance from L.A. is our heuristic for the 8-puzzle admissible? h (State) = number of tiles out of place, including the space is our heuristic for the Missionaries & Cannibals admissible?

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Cost of the search the closer h is to the actual cost, the fewer states considered however, the cost of computing h tends to go up as it improves also, admissibility is not always needed or desired Graceful Decay of Admissibility: If h rarely overestimates the actual cost by more than D, then Alg. A will rarely find a solution whose cost is more than D greater than optimal. the best algorithm is one that minimizes the total cost of the solution

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