Hi, I'm trying to get a plan working to solve a predefined 8 Puzzle. The issue i'm having is that it will only solve if I remove the is_movable and at(empty,X) constraint. Otherwise it just seems to loop continuously.
Heres the code:
% Simple Prolog Planner for the 8 Puzzle Problem
% This predicate initialises the problem states. The first argument
% of solve/3 is the initial state, the 2nd the goal state, and the
% third the plan that will be produced.
test(Plan):-
write('Initial state:'),nl,
Init= [at(tile4,1), at(tile3,2), at(tile8,3), at(empty,4), at(tile2,5), at(tile6,6), at(tile5,7), at(tile1,8), at(tile7,9)],
write_sol(Init),
Goal= [at(tile1,1), at(tile2,2), at(tile3,3), at(tile4,4), at(empty,5), at(tile5,6), at(tile6,7), at(tile7,8), at(tile8,9)],
nl,write('Goal state:'),nl,
write(Goal),nl,nl,
solve(Init,Goal,Plan).
solve(State, Goal, Plan):-
solve(State, Goal, [], Plan).
%Determines whether Current and Destination tiles are a valid move.
is_movable(X1,Y1) :- (1 is X1 - Y1) ; (-1 is X1 - Y1) ; (3 is X1 - Y1) ; (-3 is X1 - Y1).
% This predicate produces the plan. Once the Goal list is a subset
% of the current State the plan is complete and it is written to
% the screen using write_sol/1.
solve(State, Goal, Plan, Plan):-
is_subset(Goal, State), nl,
write_sol(Plan).
solve(State, Goal, Sofar, Plan):-
act(Action, Preconditions, Delete, Add),
is_subset(Preconditions, State),
\+ member(Action, Sofar),
delete_list(Delete, State, Remainder),
append(Add, Remainder, NewState),
solve(NewState, Goal, [Action|Sofar], Plan).
% The problem has three operators.
% 1st arg = name
% 2nd arg = preconditions
% 3rd arg = delete list
% 4th arg = add list.
% Tile can move to new position only if the destination tile is empty & Manhattan distance = 1
act(move(X,Y,Z),
[at(X,Y), at(empty,Z), is_movable(Y,Z)],
[at(X,Y), at(empty,Z)],
[at(X,Z), at(empty,Y)]).
% Utility predicates.
% Check is first list is a subset of the second
is_subset([H|T], Set):-
member(H, Set),
is_subset(T, Set).
is_subset([], _).
% Remove all elements of 1st list from second to create third.
delete_list([H|T], Curstate, Newstate):-
remove(H, Curstate, Remainder),
delete_list(T, Remainder, Newstate).
delete_list([], Curstate, Curstate).
remove(X, [X|T], T).
remove(X, [H|T], [H|R]):-
remove(X, T, R).
write_sol([]).
write_sol([H|T]):-
write_sol(T),
write(H), nl.
append([H|T], L1, [H|L2]):-
append(T, L1, L2).
append([], L, L).
member(X, [X|_]).
member(X, [_|T]):-
member(X, T).
Heres the code:
% Simple Prolog Planner for the 8 Puzzle Problem
% This predicate initialises the problem states. The first argument
% of solve/3 is the initial state, the 2nd the goal state, and the
% third the plan that will be produced.
test(Plan):-
write('Initial state:'),nl,
Init= [at(tile4,1), at(tile3,2), at(tile8,3), at(empty,4), at(tile2,5), at(tile6,6), at(tile5,7), at(tile1,8), at(tile7,9)],
write_sol(Init),
Goal= [at(tile1,1), at(tile2,2), at(tile3,3), at(tile4,4), at(empty,5), at(tile5,6), at(tile6,7), at(tile7,8), at(tile8,9)],
nl,write('Goal state:'),nl,
write(Goal),nl,nl,
solve(Init,Goal,Plan).
solve(State, Goal, Plan):-
solve(State, Goal, [], Plan).
%Determines whether Current and Destination tiles are a valid move.
is_movable(X1,Y1) :- (1 is X1 - Y1) ; (-1 is X1 - Y1) ; (3 is X1 - Y1) ; (-3 is X1 - Y1).
% This predicate produces the plan. Once the Goal list is a subset
% of the current State the plan is complete and it is written to
% the screen using write_sol/1.
solve(State, Goal, Plan, Plan):-
is_subset(Goal, State), nl,
write_sol(Plan).
solve(State, Goal, Sofar, Plan):-
act(Action, Preconditions, Delete, Add),
is_subset(Preconditions, State),
\+ member(Action, Sofar),
delete_list(Delete, State, Remainder),
append(Add, Remainder, NewState),
solve(NewState, Goal, [Action|Sofar], Plan).
% The problem has three operators.
% 1st arg = name
% 2nd arg = preconditions
% 3rd arg = delete list
% 4th arg = add list.
% Tile can move to new position only if the destination tile is empty & Manhattan distance = 1
act(move(X,Y,Z),
[at(X,Y), at(empty,Z), is_movable(Y,Z)],
[at(X,Y), at(empty,Z)],
[at(X,Z), at(empty,Y)]).
% Utility predicates.
% Check is first list is a subset of the second
is_subset([H|T], Set):-
member(H, Set),
is_subset(T, Set).
is_subset([], _).
% Remove all elements of 1st list from second to create third.
delete_list([H|T], Curstate, Newstate):-
remove(H, Curstate, Remainder),
delete_list(T, Remainder, Newstate).
delete_list([], Curstate, Curstate).
remove(X, [X|T], T).
remove(X, [H|T], [H|R]):-
remove(X, T, R).
write_sol([]).
write_sol([H|T]):-
write_sol(T),
write(H), nl.
append([H|T], L1, [H|L2]):-
append(T, L1, L2).
append([], L, L).
member(X, [X|_]).
member(X, [_|T]):-
member(X, T).