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# Dave Reed applications of informed search optimization problems

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Dave Reed applications of informed search optimization problems
Algorithm A, admissibility, A* zero-sum game playing minimax principle, alpha-beta pruning

Optimization problems
consider a related search 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)

Algorithm A associate 2 costs with a path
g actual cost of the path so far h heuristic estimate of the remaining cost to the goal* f = g + h combined heuristic cost estimate *note: the heuristic value is inverted relative to best first Algorithm A: best first search using f as the heuristic

Travel problem revisited
g: cost is actual distances per flight h: cost estimate is crow-flies distance %%% h(Loc, Goal, Value) : Value is crow-flies %%% distance from Loc to Goal %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% h(loc(omaha), loc(los_angeles), 1700). h(loc(chicago), loc(los_angeles), 2200). h(loc(denver), loc(los_angeles), 1400). h(loc(los_angeles), loc(los_angeles), 0).

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])), h(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 2nd arg, must identify it as backtrackable

Travel example %%% travelcost.pro Dave Reed /15/02 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% move(loc(omaha), loc(chicago), ). move(loc(omaha), loc(denver), ). move(loc(chicago), loc(denver), ). move(loc(chicago), loc(los_angeles), 2200). move(loc(chicago), loc(omaha), ). move(loc(denver), loc(los_angeles), 1400). move(loc(denver), loc(omaha), ). move(loc(los_angeles), loc(chicago), ). move(loc(los_angeles), loc(denver), ). %%% h(Loc, Goal, Value) : Value is crow-flies %%% distance from Loc to Goal %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% h(loc(omaha), loc(los_angeles), 1700). h(loc(chicago), loc(los_angeles), 2200). h(loc(denver), loc(los_angeles), 1400). h(loc(los_angeles), loc(los_angeles), 0). 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 miles (instead of 2200) OmahaChicagoDenverLA would be shorter than OmahaChicagoLA ?- 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

8-puzzle example g: actual cost of each move is 1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% h(Board, Goal, Value) : Value is the number of tiles out of place h(tiles(Row1,Row2,Row3),tiles(Goal1,Goal2,Goal3), Value) :- diff_count(Row1, Goal1, V1), diff_count(Row2, Goal2, V2), diff_count(Row3, Goal3, V3), Value is V1 + V2 + V3. diff_count([], [], 0). diff_count([H1|T1], [H2|T2], Count) :- diff_count(T1, T2, TCount), (H1 = H2, Count is TCount ; H1 \= H2, Count is TCount+1). g: actual cost of each move is 1 h: remaining cost estimate is # of tiles out of place ?- 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]), 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|...])] here, Algorithm A finds the same paths as best first search not surprising since g is trivial still, not guaranteed to be the case

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

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 never overestimates actual cost, then Alg. A is admissible (when admissible, Alg. A is commonly referred to as Alg. A*)

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

Cost of the search the closer h is to the actual cost function, the fewer states considered however, the cost of computing h tends to go up as it improves the best algorithm is one that minimizes the total cost of the solution 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 exceeds optimal by more than D.

Algorithm A (for graphs)
representing the search search as a tree is conceptually simpler but must store entire paths, which include duplicates of states more efficient to store the search space as a graph store states w/o duplicates, need only remember parent for each state (so that the solution path can be reconstructed at the end) IDEA: keep 2 lists of states (along with f & g values, parent pointer) OPEN: states reached by the search, but not yet expanded CLOSED: states reached and already expanded while more efficient, the graph implementation is trickier when a state moves to the CLOSED list, it may not be finished may have to revise values of states if new (better) paths are found

Flashlight example consider the flashlight puzzle discussed in class:
Four people are on one side of a bridge. They wish to cross to the other side, but the bridge can only take the weight of two people at a time. It is dark and they only have one flashlight, so they must share it in order to cross the bridge. Assuming each person moves at a different speed (able to cross in 1, 2, 5 and 10 minutes, respectively), find a series of crossings that gets all four across in the minimal amount of time. state representation? cost of a move? heuristic?

Flashlight implementation
state representation must identify the locations of each person and the flashlight bridge(SetOfPeopleOnLeft, SetOfPeopleOnRight, FlashlightLocation) note: can use a list to represent a set, but must be careful of permutations e.g., [1,2,5,10] = [1,5,2,10], so must make sure there is only one list repr. per set solution: maintain the lists in sorted order, so only one permutation is possible only 3 possible moves: if the flashlight is on left and only 1 person on left, then move that person to the right (cost is time it takes for that person) if flashlight is on left and at least 2 people on left, then select 2 people from left and move them to right (cost is max time of the two) if the flashlight is on right, then select a person from right and move them to left (cost is time for that person) heuristic: h(State, Goal) = number of people in wrong place

Flashlight code merge(L1,L2,L3):
%%% flashlight.pro Dave Reed /15/02 %%% %%% This file contains the state space definition %%% for the flashlight puzzle. %%% bridge(Left, Right, Loc): Left is a (sorted) list of people %%% on the left shore, Right is a (sorted) list of people on %%% the right shore, and Loc is the location of the flashlight. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% move(bridge([P], Right, left), bridge([], NewRight, right), P) :- merge([P], Right, NewRight). move(bridge(Left, Right, left), bridge(NewLeft, NewRight, right), Cost) :- length(Left, L), L >= 2, select(P1, Left, Remain), select(P2, Remain, NewLeft), merge([P1], Right, TempRight), merge([P2], TempRight, NewRight), Cost is max(P1, P2). move(bridge(Left, Right, right), bridge(NewLeft, NewRight, left), P) :- select(P, Right, NewRight), merge([P], Left, NewLeft). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% h(State, Goal, Value) : Value is # of people out of place h(bridge(Left, Right, _), bridge(GoalLeft, GoalRight, _), Value) :- diff_count(Left, GoalLeft, V1), diff_count(Right, GoalRight, V2), Value is V1 + V2. diff_count([], _, 0). diff_count([H|T], L, Count) :- diff_count(T, L, TCount), (member(H,L), !, Count is TCount ; Count is TCount+1). Flashlight code merge(L1,L2,L3): L3 is the result of merging sorted lists L1 & L2 select(X,L,R): R is the remains after removing X from list L

Flashlight answers Algorithm A finds optimal solutions to the puzzle
note: more than one solution is optimal can use ';' to enumerate solutions, from best to worst ?- atree(bridge([1,2,5,10],[],left), bridge([],[1,2,5,10],right), Path). Path = 17:[bridge([1, 2, 5, 10], [], left), bridge([5, 10], [1, 2], right), bridge([1, 5, 10], [2], left), bridge([1], [2, 5, 10], right), bridge([1, 2], [5, 10], left), bridge([], [1|...], right)] ; Path = 17:[bridge([1, 2, 5, 10], [], left), bridge([5, 10], [1, 2], right), bridge([2, 5, 10], [1], left), bridge([2], [1, 5, 10], right), bridge([1, 2], [5, 10], left), bridge([], [1|...], right)] ; Path = 19:[bridge([1, 2, 5, 10], [], left), bridge([2, 5], [1, 10], right), bridge([1, 2, 5], [10], left), bridge([2], [1, 5, 10], right), bridge([1, 2], [5, 10], left), bridge([], [1|...], right)] Yes

Search in game playing consider games involving:
2 players perfect information zero-sum (player's gain is opponent's loss) examples: tic-tac-toe, checkers, chess, othello, … non-examples: poker, backgammon, prisoner's dilemma, … von Neumann (the father of game theory) showed that for such games, there is always a "rational" strategy that is, can always determine a best move, assuming the opponent is equally rational

Game trees idea: model the game as a search tree
associate a value with each game state (possible since zero-sum) player 1 wants to maximize the state value (call him/her MAX) player 2 wants to minimize the state value (call him/her MIN) players alternate turns, so differentiate MAX and MIN levels in the tree the leaves of the tree will be end-of-game states

Minimax search minimax search:
at a MAX level, take the maximum of all possible moves at a MIN level, take the minimum of all possible moves can visualize the search bottom-up (start at leaves, work up to root) likewise, can search top-down using recursion

Minimax example

In-class exercise

Minimax in practice while Minimax Principle holds for all 2-party, perfect info, zero-sum games, an exhaustive search to find best move may be infeasible EXAMPLE: in an average chess game, ~100 moves with ~35 options/move  ~35100 states in the search tree! practical alternative: limit the search depth and use heuristics expand the search tree a limited number of levels (limited look-ahead) evaluate the "pseudo-leaves" using a heuristic high value  good for MAX low value  good for MIN back up the heuristic estimates to determine the best-looking move at MAX level, take minimum at MIN level, take maximum

{ Tic-tac-toe example 1000 if win for MAX (X)
heuristic(State) = if win for MIN (O) (#rows/cols/diags open for MAX – #rows/cols/diags open for MIN) otherwise suppose look-ahead of 2 moves

a-b bounds a-b technique: associate bonds with state in the search
sometimes, it isn't necessary to search the entire tree a-b technique: associate bonds with state in the search associate lower bound a with MAX: can increase associate upper bound b with MIN: can decrease

a-b pruning discontinue search below a MIN node if b value <= a value of ancestor discontinue search below a MAX node if a value >= b value of ancestor

larger example

tic-tac-toe example a-b vs. minimax:
worst case: a-b examines as many states as minimax best case: assuming branching factor B and depth D, a-b examines ~2bd/2 states (i.e., as many as minimax on a tree with half the depth)

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