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

2. 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)

3. 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

4. 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).

5. 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

6. 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

7. 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

8. 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

9. 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*)

10. 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?

11. 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.

12. 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

13. 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?

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. Minimax example

21. In-class exercise

22. 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

23. { 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

24. 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

25. 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

26. larger example

27. 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)