1. 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*

2. 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.)

3. 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?

4. 8-puzzle example start state has 5 tiles in place
heuristic(State) = # of tiles in place, including the space 1 2 3 8 6 7 5 4
start state has 5 tiles in place
hval = 5 4 5 7 3 6 8 2 1
hval = 4
hval = 6
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! 4 5 7 6 2 8 3 1
hval = 4
hval = 5

5. 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

6. 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

7. 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

8. Best first example hval = 5 hval = 4 hval = 6 hval = 4 hval = 51 2 3 8 6 7 5 4
hval = 5
hval = 4
hval = 6 4 5 7 3 6 8 2 1 4 5 7 6 2 8 3 1
hval = 4
hval = 5 4 5 7 6 8 3 2 1
hval = 5
hval = 7 5 6 7 4 8 3 2 1
hval = infinity
hval = 4

9. 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

10. Best first implementation
%%% best.pro Dave Reed /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)

11. Travel example must define the heuristic function for this problem
%%% 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

12. 8-puzzle example must define the heuristic function for this problem
%%% 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]),
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|...])]
best first search is able to find a path

13. Other examples heuristic function for the Water Jug Problem?
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 Missionaries & Cannibals?

14. 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

15. 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)

16. Travel problem revisited
%%% travelcost.pro Dave Reed /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), ).
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), ).
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.

17. 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
Algorithm A (for trees – note that usual formulation is for graphs)
best first search using f as the heuristic

18. 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 2nd arg, must identify it as backtrackable

19. Travel example can use semi-colon to see alternative paths in order
?- 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 miles (instead of 2200)
then OmahaChicagoDenverL.A. is shorter than OmahaChicagoL.A.

20. 8-puzzle example 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
?- 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 actual cost component (g) is trivial
still, not guaranteed to be the case (recall: best only cares about the future)

21. 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

22. 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*)

23. 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?

24. 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
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 is more than D greater than optimal.