Refactoring A* algo from Prolog Programming for Artificial Intelligence #2585
jjtolton
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Maybe start with aiming for a generic version, like this. |
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hmm I've read over this a few times and I'm not sure how that applies here, but I get the feeling it would be very elegant if I did |
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Adding 8-puzzle: :- use_module(library(time)).
:- use_module(library(lambda)).
:- use_module('/path/to/astar.pl').
/* Problem-specific procedures for the eight puzzle
Current situation is represented as a list of positions of the tiles, with first item in the
list corresponding to the empty square.
Example:
3
2
1
1 2 3
8 4
7 6 5
This position is represented by:
[2/2, 1/3, 2/3, 3/3, 3/2, 3/1, 2/1, 1/1, 1/2]
1 2 3
'Empty' can move to any of its neighbours, which means that 'empty' and its
neighbour interchange their positions.
*/
% D is I A-B
% s( Node, SuccessorNode, Cost)
s( [Empty | Tiles], [Tile | Tiles1], 1) :-
swap( Empty, Tile, Tiles, Tiles1).
swap( Empty, Tile, [Tile | Ts], [Empty | Ts]):-
mandist( Empty, Tile, 1).
swap( Empty, Tile, [T1 | Ts], [T1 | Ts1]) :-
swap( Empty, Tile, Ts, Ts1).
mandist( X/Y, X1/Y1, D) :-
D #= abs(X1-X)+abs(Y1-Y).
% Heuristic estimate h is the sum of distances of each tile
% from its 'home' square plus 3 times 'sequence' score
h( [_Empty | Tiles], H) :-
goal( [_Empty1 | GoalSquares]),
totdist( Tiles, GoalSquares, D),
seq_score( Tiles, S),
H #= D + 3*S.
totdist([],[],0).
totdist( [Tile | Tiles], [Square | Squares], D) :-
mandist( Tile, Square, D1),
totdist( Tiles, Squares, D2),
D #= D1 + D2.
% seq( TilePositions, Score): sequence score
% All arc costs are 1
% Swap Empty and Tile in Tiles
% Manhattan distance = 1
% D is Manh. dist. between two squares
% Total distance from home squares
% Sequence score
seq_score( [First | OtherTiles], S) :-
seq_score( [First | OtherTiles ], First, S).
seq_score( [Tile1, Tile2 | Tiles], First, S) :-
score( Tile1, Tile2, S1),
seq_score( [Tile2 | Tiles], First, S2),
S #= S1 + S2.
seq_score( [Last], First, S) :-
score( Last, First, S).
score(2/2,_,1).
score( 1/3, 2/3, 0).
score( 2/3, 3/3, 0).
score( 3/3, 3/2, 0).
score( 3/2, 3/1, 0).
score( 3/1, 2/1, 0).
score( 2/1, 1/1, 0).
score( 1/1, 1/2, 0).
score( 1/2, 1/3, 0).
score(_, _, 2).
goal( [2/2,1/3,2/3,3/3,3/2,3/1,2/1,1/1,1/2]).
% Display a solution path as a list of board positions
showsol( [ ]).
showsol( [P | L]) :-
showsol( L),
nl,write('---'),
showpos( P).
% Display a board position
showpos( [S0,S1,S2,S3,S4,S5,S6,S7,S8]) :-
( member( Y, [3,2,1]),
nl, member( X, [1,2,3]),
member( Tile-X/Y, [' '-S0,1-S1,2-S2,3-S3,4-S4,5-S5,6-S6,7-S7,8-S8]),
write( Tile),
fail
; true
).
% Starting positions for some puzzles
start1( [2/2,1/3,3/2,2/3,3/3,3/1,2/1,1/1,1/2]).
start2( [2/1,1/2,1/3,3/3,3/2,3/1,2/2,1/1,2/3]).
start3( [2/2,2/3,1/3,3/1,1/2,2/1,3/3,1/1,3/2]).
% Backtrack to next square
% All squares done
% Requires 4 steps
% Requires 5 steps
% Requires 18 steps
% An example query:
?- _+\(time((start1(Pos),
bestfirst( Pos, Sol),
showsol( Sol)))
).
%@
%@ ---
%@ 134
%@ 8 2
%@ 765
%@ ---
%@ 134
%@ 82
%@ 765
%@ ---
%@ 13
%@ 824
%@ 765
%@ ---
%@ 1 3
%@ 824
%@ 765
%@ ---
%@ 123
%@ 8 4
%@ 765 % CPU time: 0.005s, 7_635 inferences
%@ true
%@ ; ... . |
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Monoton-ish version of A*: :- use_module(library(debug)).
:- use_module(library(lists)).
% Start( % bestfirst, Solution): Solution is a path from Start to a goal
bestfirst( Start, Solution) :-
expand( [], leaf( Start, 0/0), _, yes, Solution).
% expand( Path, Tree, Tree1, Solved, Solution):
% Path is path between start node of search and subtree Tree,
% if goal found then Solution is solution path and Solved = yes
% Case 1: goal leaf-node, construct a solution path
expand( P, leaf( N, _), _, yes, [N | P]) :-
goal(N).
% Case 2: leaf-node, f-value less than Bound
% Generate successors and expand them within Bound
expand( P, leaf(N, F/G), Tree1, Solved, Sol) :-
( bagof( M/C, ( s(N, M, C), maplist(dif(M), P)), Succ),
succlist( G, Succ, Ts), % Make subtrees Ts
bestf( Ts, F1), % f-value of best successor
expand( P, t(N, F1/G, Ts), Tree1, Solved, Sol)
; Solved=never
).
% Case 3: non-leaf, f-value less than Bound
% Expand the most promising subtree; depending on
% results, procedure continue will decide how to proceed
expand( P, t(N, F/G, [T | Ts]), Tree1, Solved, Sol) :-
bestf( Ts, BF),
expand( [N | P], T, T1, Solved1, Sol),
continue( P, t(N, F/G, [T1 | Ts]), Tree1, Solved1, Solved, Sol).
% Case 4: non-leaf with empty subtrees
% This is a dead end which will never be solved
expand( _, t(_, _, []), _, never, _) :- !.
% continue( Path, Tree, Bound, NewTree, SubtreeSolved, TreeSolved, Solution)
continue( _,_,_, yes, yes, _Sol).
continue( P, t(N, _F/G, [T1 | Ts]), Tree1, no, Solved, Sol) :-
insert( Ts, T1, NTs),
bestf( NTs, F1),
expand( P, t(N, F1/G, NTs), Tree1, Solved, Sol).
continue( P, t(N, _F/G, [_ | Ts]), Tree1, never, Solved, Sol) :-
bestf( Ts, F1),
expand( P, t(N, F1/G, Ts), Tree1, Solved, Sol).
%% succlist( G0, [Nodel/Cost1, ...], [leaf(BestNode, BestF/G),
% make list of search leaves ordered by their f-values
succlist( _,[],[]).
succlist( G0, [N/C | NCs], Ts) :-
G #= G0 + C,
h( N, H),
F #= G + H,
succlist( G0, NCs, Ts1),
insert(Ts1, leaf(N, F/G), Ts).
% Heuristic term h(N)
% Insert T into list of trees Ts preserving order with respect to f-values
insert([], T, [T]).
insert(Ts0, T, Ts) :-
f(T, F),
bestf(Ts0, F1),
if_(clpz_t(F #=< F1),
Ts=[T|Ts0],
( Ts0=[Tx|Ts0_],
Ts =[Tx|Ts_],
insert(Ts0_, T, Ts_)
)
).
f(leaf(_, F/_), F).
f(t(_, F/_,_),F).
bestf( [T | _], F) :-
f( T, F).
bestf( [],9999).
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I've adapted the A* program from Prolog Programming for Artificial Intelligence with very few modifications except switching to clpz. So far I have not been able to make a "pure" version, and every effort to do so has failed.
That being said, I think the approach taken here is quite interesting, I've never seen A* done using a literal tree before (as opposed to maintaining a priority queue of nodes). Encountered a lot of interesting things while exploring this problem, may highlight a few in the discussion below.
A*
Example problem: Task Scheduling
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