How to resolve the algorithm Maze solving step by step in the Prolog programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Maze solving step by step in the Prolog programming language
Table of Contents
Problem Statement
For a maze generated by this task, write a function that finds (and displays) the shortest path between two cells.
Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths, and a simple depth-first tree search can be used.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Maze solving step by step in the Prolog programming language
Source code in the prolog programming language
:- dynamic cell/2.
:- dynamic maze/3.
:- dynamic path/1.
maze_solve(Lig,Col) :-
retractall(cell(_,_)),
retractall(maze(_,_,_)),
retractall(path(_)),
% initialisation of the neighbours of the cells
forall(between(0, Lig, I),
( forall(between(0, Col, J), assert(maze(I, J, []))))),
% creation of the window of the maze
new(D, window('Maze')),
forall(between(0,Lig, I),
(XL is 50, YL is I * 30 + 50,
XR is Col * 30 + 50,
new(L, line(XL, YL, XR, YL)),
send(D, display, L))),
forall(between(0,Col, I),
(XT is 50 + I * 30, YT is 50,
YB is Lig * 30 + 50,
new(L, line(XT, YT, XT, YB)),
send(D, display, L))),
SX is Col * 30 + 100,
SY is Lig * 30 + 100,
send(D, size, new(_, size(SX, SY))),
L0 is random(Lig),
C0 is random(Col),
assert(cell(L0, C0)),
\+search(D, Lig, Col, L0, C0),
send(D, open),
% we look for a path from cell(0, 0) to cell(Lig-1, Col-1)
% creation of the entrance
erase_line(D, -1, 0, 0, 0),
% creation of the exit
Lig1 is Lig-1,
Col1 is Col-1,
erase_line(D, Lig1, Col1, Lig, Col1),
% seraching the path
assert(path([[0, 0], [-1, 0]])),
walk(Lig, Col),
path(P),
display_path(D, P).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
walk(Lig, Col) :-
path([[L, C] | _R]),
L is Lig - 1,
C is Col - 1,
retract(path(P)),
assert(path([[Lig, C]|P])).
walk(Lig, Col) :-
retract(path([[L, C] | R])),
maze(L, C, Edge),
member([L1, C1], Edge),
\+member([L1, C1], R),
assert(path([[L1,C1], [L, C] | R])),
walk(Lig, Col).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
display_path(_, []).
display_path(D, [[L, C] | R]):-
new(B, box(10,10)),
send(B, fill_pattern, new(_, colour(@default, 0,0,0))),
X is C * 30 + 60,
Y is L * 30 + 60,
send(D, display, B, point(X,Y)),
display_path(D, R).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
search(D, Lig, Col, L, C) :-
Dir is random(4),
nextcell(Dir, Lig, Col, L, C, L1, C1),
assert(cell(L1,C1)),
assert(cur(L1,C1)),
retract(maze(L, C, Edge)),
assert(maze(L, C, [[L1, C1] | Edge])),
retract(maze(L1, C1, Edge1)),
assert(maze(L1, C1, [[L, C] | Edge1])),
erase_line(D, L, C, L1, C1),
search(D, Lig, Col, L1, C1).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
erase_line(D, L, C, L, C1) :-
( C < C1 -> C2 = C1; C2 = C),
XT is C2 * 30 + 50,
YT is L * 30 + 51, YR is (L+1) * 30 + 50,
new(Line, line(XT, YT, XT, YR)),
send(Line, colour, white),
send(D, display, Line).
erase_line(D, L, C, L1, C) :-
XT is 51 + C * 30, XR is 50 + (C + 1) * 30,
( L < L1 -> L2 is L1; L2 is L),
YT is L2 * 30 + 50,
new(Line, line(XT, YT, XR, YT)),
send(Line, colour, white),
send(D, display, Line).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
nextcell(Dir, Lig, Col, L, C, L1, C1) :-
next(Dir, Lig, Col, L, C, L1, C1);
( Dir1 is (Dir+3) mod 4,
next(Dir1, Lig, Col, L, C, L1, C1));
( Dir2 is (Dir+1) mod 4,
next(Dir2, Lig, Col, L, C, L1, C1));
( Dir3 is (Dir+2) mod 4,
next(Dir3, Lig, Col, L, C, L1, C1)).
% 0 => northward
next(0, _Lig, _Col, L, C, L1, C) :-
L > 0,
L1 is L - 1,
\+cell(L1, C).
% 1 => rightward
next(1, _Lig, Col, L, C, L, C1) :-
C < Col - 1,
C1 is C + 1,
\+cell(L, C1).
% 2 => southward
next(2, Lig, _Col, L, C, L1, C) :-
L < Lig - 1,
L1 is L + 1,
\+cell(L1, C).
% 3 => leftward
next(2, _Lig, _Col, L, C, L, C1) :-
C > 0,
C1 is C - 1,
\+cell(L, C1).
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