# 8 puzzle

## Description

Problem-specific procedures for the eight puzzle, to be used in best-first search

Source: PROLOG programming for artificial intelligence, 3rd Edition, Harlow, 2001, ISBN 0-201-40375-7.

Program source code: 8_puzzle.pl

## Listing

```% Figure 12.6  Problem-specific procedures for the eight
% puzzle.

:- op( 900, fy, not).

/* 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:

This position is represented by:
3        1    2    3
2        8         4       [2/2, 1/3, 2/3, 3/3, 3/2, 3/1, 2/1, 1/1, 1/2]
1        7    6    5

1    2    3

"Empty' can move to any of its neighbours which means
that "empty' and its neighbour interchange their positions.
*/

% s( Node, SuccessorNode, Cost)

s( [Empty | Tiles], [Tile | Tiles1], 1)  :-  % All arc costs are 1
swap( Empty, Tile, Tiles, Tiles1).         % Swap Empty and Tile in Tiles

swap( Empty, Tile, [Tile | Ts], [Empty | Ts] )  :-
mandist( Empty, Tile, 1).                  % Manhattan distance = 1

swap( Empty, Tile, [T1 | Ts], [T1 | Ts1] )  :-
swap( Empty, Tile, Ts, Ts1).

mandist( X/Y, X1/Y1, D)  :-          % D is Manhhattan dist. between two squares
dif( X, X1, Dx),
dif( Y, Y1, Dy),
D is Dx + Dy.

dif( A, B, D)  :-              % D is |A-B|
D is A-B, D >= 0, !
;
D is B-A.

% 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),      % Total distance from home squares
seq( Tiles, S),                       % Sequence score
H is D + 3*S.

totdist( [], [], 0).

totdist( [Tile | Tiles], [Square | Squares], D)  :-
mandist( Tile, Square, D1),
totdist( Tiles, Squares, D2),
D is D1 + D2.

% seq( TilePositions, Score): sequence score

seq( [First | OtherTiles], S)  :-
seq( [First | OtherTiles ], First, S).

seq( [Tile1, Tile2 | Tiles], First, S)  :-
score( Tile1, Tile2, S1),
seq( [Tile2 | Tiles], First, S2),
S is S1 + S2.

seq( [Last], First, S)  :-
score( Last, First, S).

score( 2/2, _, 1)  :-  !.              % Tile in centre scores 1

score( 1/3, 2/3, 0)  :-  !.            % Proper successor scores 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).                       % Tiles out of sequence score 2

goal( [2/2,1/3,2/3,3/3,3/2,3/1,2/1,1/1,1/2] ).  % Goal squares for tiles

% 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] ),                           % Order of Y-coordinates
nl, member( X, [1,2,3] ),                       % Order of X-coordinates
member( Tile-X/Y,                               % Tile on square X/Y
[' '-S0,1-S1,2-S2,3-S3,4-S4,5-S5,6-S6,7-S7,8-S8] ),
write( Tile),
fail                                            % Backtrack to next square
;
true.                                           % All squares done

% Starting positions for some puzzles

start1( [2/2,1/3,3/2,2/3,3/3,3/1,2/1,1/1,1/2] ).  % Requires 4 steps

start2( [2/1,1/2,1/3,3/3,3/2,3/1,2/2,1/1,2/3] ).  % Requires 5 steps

start3( [2/2,2/3,1/3,3/1,1/2,2/1,3/3,1/1,3/2] ).  % Requires 18 steps

% An example query: ?- start1( Pos), bestfirst( Pos, Sol), showsol( Sol).

% not Goal): negation as failure;
%   Note: This is often available as a built-in predicate,
%   often written as prefix operator "\+", e.g. \+ likes(mary,snakes)

not Goal  :-
Goal, !, fail
;
true.

% bestfirst( Start, Solution): Solution is a path from Start to a goal

bestfirst( Start, Solution) :-
expand( [], l( Start, 0/0),  9999, _, yes, Solution).
%  Assume 9999 is greater than any f-value

% expand( Path, Tree, Bound, Tree1, Solved, Solution):
%   Path is path between start node of search and subtree Tree,
%   Tree1 is Tree expanded within Bound,
%   if goal found then Solution is solution path and Solved = yes

%  Case 1: goal leaf-node, construct a solution path

expand( P, l( N, _), _, _, yes, [N|P])  :-
goal(N).

%  Case 2: leaf-node, f-value less than Bound
%  Generate successors and expand them within Bound.

expand( P, l(N,F/G), Bound, Tree1, Solved, Sol)  :-
F  =<  Bound,
(  bagof( M/C, ( s(N,M,C), not member(M,P) ), Succ),
!,                                    % Node N has successors
succlist( G, Succ, Ts),               % Make subtrees Ts
bestf( Ts, F1),                       % f-value of best successor
expand( P, t(N,F1/G,Ts), Bound, Tree1, Solved, Sol)
;
Solved = never                        % N has no successors - dead end
) .

%  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]), Bound, Tree1, Solved, Sol)  :-
F  =<  Bound,
bestf( Ts, BF), min( Bound, BF, Bound1),          % Bound1 = min(Bound,BF)
expand( [N|P], T, Bound1, T1, Solved1, Sol),
continue( P, t(N,F/G,[T1|Ts]), Bound, Tree1, Solved1, Solved, Sol).

%  Case 4: non-leaf with empty subtrees
%  This is a dead end which will never be solved

expand( _, t(_,_,[]), _, _, never, _) :- !.

%  Case 5:  f-value greater than Bound
%  Tree may not grow.

expand( _, Tree, Bound, Tree, no, _)  :-
f( Tree, F), F > Bound.

% continue( Path, Tree, Bound, NewTree, SubtreeSolved, TreeSolved, Solution)

continue( _, _, _, _, yes, yes, Sol).

continue( P, t(N,F/G,[T1|Ts]), Bound, Tree1, no, Solved, Sol)  :-
insert( T1, Ts, NTs),
bestf( NTs, F1),
expand( P, t(N,F1/G,NTs), Bound, Tree1, Solved, Sol).

continue( P, t(N,F/G,[_|Ts]), Bound, Tree1, never, Solved, Sol)  :-
bestf( Ts, F1),
expand( P, t(N,F1/G,Ts), Bound, Tree1, Solved, Sol).

% succlist( G0, [ Node1/Cost1, ...], [ l(BestNode,BestF/G), ...]):
%   make list of search leaves ordered by their F-values

succlist( _, [], []).

succlist( G0, [N/C | NCs], Ts)  :-
G is G0 + C,
h( N, H),                             % Heuristic term h(N)
F is G + H,
succlist( G0, NCs, Ts1),
insert( l(N,F/G), Ts1, Ts).

% Insert T into list of trees Ts preserving order w.r.t. f-values

insert( T, Ts, [T | Ts])  :-
f( T, F), bestf( Ts, F1),
F  =<  F1, !.

insert( T, [T1 | Ts], [T1 | Ts1])  :-
insert( T, Ts, Ts1).

% Extract f-value

f( l(_,F/_), F).        % f-value of a leaf

f( t(_,F/_,_), F).      % f-value of a tree

bestf( [T|_], F)  :-    % Best f-value of a list of trees
f( T, F).

bestf( [], 9999).       % No trees: bad f-value

min( X, Y, X)  :-
X  =<  Y, !.

min( X, Y, Y).```