Prolog maze solving algorithm

Question

I want to implement a maze solving algorithm in Prolog. Therefore i searched for some maze solving algorithms and found the following: http://www.cs.bu.edu/teaching/alg/maze/

FIND-PATH(x, y):

if (x,y outside maze) return false
if (x,y is goal) return true
if (x,y not open) return false
mark x,y as part of solution path
if (FIND-PATH(North of x,y) == true) return true
if (FIND-PATH(East of x,y) == true) return true
if (FIND-PATH(South of x,y) == true) return true
if (FIND-PATH(West of x,y) == true) return true
unmark x,y as part of solution path
return false 

I already build a matrix in prolog, which represents a maze and where 0 is open and 1 is the wall, for example (starting position would be (2|1) and the goal is located at (4|1)):

11111
10001
10101

Further more i defined a clause named mazeDataAt(Coord_X, Coord_Y, MazeData, Result), which gives me the value of the matrix on a certain position.

So far. But now i have a problem implementing that algorithm in prolog. I already tried "the dirty way" (translate it one by one by use of nested if statements), but that escalated complexity and i don't think it's the way you do it in prolog.

So i tried this:

isNotGoal(X, Y) :- 
    X = 19, Y = 2.

notOpen(X, Y, MazeData) :-
    mazeDataAt(X, Y, MazeData, 1). 
    
findPath(X, Y, MazeData) :- 
    isNotGoal(X, Y),
    notOpen(X, Y, MazeData),
    increase(Y, Y_New),
    findPath(X, Y_New, MazeData),
    increase(X, X_New),
    findPath(X_New, Y, MazeData),
    decrease(Y, Y_New),
    findPath(X, Y_New, MazeData),
    decrease(X, X_New),
    findPath(X, Y_New, MazeData).

But this attempt didn't work like expected.

Actually, is this a correct prolog implementation of the algorithm above? How can i see if this approach really finds a path through the maze? Therefore how can i record the path or get the solution path (what is done by marking / unmarking the path in the algorithm above)?

Thank you very much for your help!

//UPDATE

Thanks to your answers! I adopted a more prolog like solution (see here) to solve my problem. So i now have:

d([2,1], [2,2]).
d([2,2], [1,2]).
d([2,2], [2,3]).

go(From, To, Path) :-
go(From, To, [], Path).

go(P, P, T, T).
go(P1, P2, T, NT) :-
    (d(P1, P3) ; d(P3, P2)),
    \+ member(P3, T),
    go(P3, P2, [P3|T], NT).

So far, this works. And i think i understand why the prolog way is much better. But now i have a small problem left.

I want my knowledge base be "dynamic". I can't define all the edges for every single waypoint in the maze. Therefore i wrote a clause named is_adjacent([X1, Y1], [X2, Y2]) which is true when [X1, Y1] is a neighbor of [X2, Y2].

I also have a list Waypoints = [[2, 1], [2, 2]| ...] which contains all possible waypoints in my maze.

Now the question: How can i use this to make my knowledge base "dynamic"? So that i can use it in the go clause for finding the path?

Thanks for your help!

//UPDATE 2

Ok, now i got all waypoints as facts:

w(2, 1).
w(2, 2).
...

I took the solution from Boris in one of his answers:

d(X0, Y0, X , Y) :-
    w(X0, Y0),
    next_w(X0, Y0, X, Y),
    w(X, Y).

next_w(X0, Y0, X0, Y) :- Y is Y0 + 1.
next_w(X0, Y0, X0, Y) :- Y is Y0 - 1.
next_w(X0, Y0, X, Y0) :- X is X0 + 1.
next_w(X0, Y0, X, Y0) :- X is X0 - 1.

After that, I updated the go clause, so that it fits:

go(X1, Y1, X2, Y2, Path) :-
go(X1, Y1, X2, Y2, [], Path).

go(X, Y, X, Y, T, T).
go(X1, Y1, X2, Y2, T, NT) :-
   (d(X1, Y1, X3, Y3) ; d(X3, Y3, X1, Y1)),
\+ member([X3, Y3], T),
go(X3, Y3, X2, Y2, [[X3, Y3]|T], NT).

But if i try to ask go(2, 1, 19, 2, R) prolog enters an infinite loop. If i try something easier like go(2, 1, 3, 8, R) it works and i get the solution path in R.

What am i doing wrong? What did i forget?

Answer 1

(this answer uses the same path finding algorithm as this answer)

EDIT 2

Indeed, if your input is just which cells of the rectangular matrix are not walls, you would need to somehow translate this to rules of the kind "you can get from A to B". If your waypoints are then:

w(2,1).
w(2,2).

etc, then you can translate the algorithm you originally pointed to into a Prolog rule like this:

% it is possible to move from (X0,Y0) to (X,Y)
d(X0,Y0,X,Y) :-
    w(X0,X0), % you can skip this check if you know for sure
              % that your starting point is a valid waypoint
              % or if you want to be able to start from inside
              % a wall :)
    next_w(X0,Y0,X,Y),
    w(X,Y).
% neighboring waypoints
next_w(X0,Y0,X0,Y) :- Y is Y0+1. % go up I guess
next_w(X0,Y0,X0,Y) :- Y is Y0-1. % go down
next_w(X0,Y0,X,Y0) :- X is X0+1. % go left
next_w(X0,Y0,X,Y0) :- X is X0-1. % go right

Note two things:

1. I am using a 4-argument rule for the possible moves from a square (so adjust accordingly)
2. The magic happens in next_w. When d is called, it uses next_w to generate the four possible neighbor squares (assuming you can only go up/down/left/right) and then checks whether this square is indeed a waypoint. You would not need to check both ways any more.

ANOTHER EDIT: Full code

w(0,0).
w(0,1). w(1,1). w(2,1). w(3,1). w(4,1). w(5,1).
        w(1,2).         w(3,2).         w(5,2).
        w(1,3).         w(3,3).         w(5,3).
w(0,4). w(1,4). w(2,4).         w(4,4). w(5,4).
                w(2,5). w(3,5). w(4,5).

d(X0,Y0,X,Y) :- next_w(X0,Y0,X,Y), w(X,Y).
next_w(X0,Y0,X0,Y) :- Y is Y0+1.
next_w(X0,Y0,X,Y0) :- X is X0+1.
next_w(X0,Y0,X0,Y) :- Y is Y0-1.
next_w(X0,Y0,X,Y0) :- X is X0-1.

go(X,Y,X,Y,Path,Path).
go(X0,Y0,X,Y,SoFar,Path) :-
    d(X0,Y0,X1,Y1),
    \+ memberchk( w(X1,Y1), SoFar ),
    go(X1,Y1,X,Y,[w(X1,Y1)|SoFar],Path).

You can call it with

? go(0,0,5,4,[],Path).

and you should get the two possible solutions.

In other words, I think your problem is the semicolon; it is no longer necessary, because you explicitly create all possible moves.