[英]Square Puzzle Problem Solution with Constraint Programming
Question: Fill in the grid with squares (of any size) that do not touch or overlap, even at the corners.
问题:用不接触或不重叠的正方形(任何大小)填充网格,即使在角落也是如此。 The numbers below and at the right indicate the number of grid squares that are filled in the corresponding column / row.
下方和右侧的数字表示填充在相应列/行中的网格方块的数量。
To solve this problem, I applied the following constraints: the placed squares should be disjoint and to make sure the number of grid squares is correct, I constrained the sum of the length of the squares that intersect a given row/column to be equal to that row/column number.为了解决这个问题,我应用了以下约束:放置的正方形应该是不相交的,并且为了确保网格正方形的数量是正确的,我将与给定行/列相交的正方形的长度之和限制为等于该行/列号。
However, the outputed solution is [1, 0, 0, 1]
( [NumSquares, X, Y, SquareSize]
), a single square with length of value one in coordinates (0, 0) and it should be the one depicted in the right image (13 squares with different sizes and coordinates).但是,输出的解决方案是
[1, 0, 0, 1]
( [NumSquares, X, Y, SquareSize]
),一个在坐标 (0, 0) 中长度值为 1 的正方形,它应该是在正确的图像(13 个不同大小和坐标的正方形)。
:- use_module(library(clpfd)).
:- include('utils.pl').
solve(Rows, Columns, Vars) :-
% Domain and variables definition
length(Rows, Size),
MaxNumSquares is Size * Size,
NumSquares #>= 0,
NumSquares #< MaxNumSquares,
length(StartsX, NumSquares),
length(StartsY, NumSquares),
length(SquareSizes, NumSquares),
S is Size - 1,
domain(StartsX, 0, S),
domain(StartsY, 0, S),
domain(SquareSizes, 1, Size),
construct_squares(Size, StartsX, StartsY, SquareSizes, Squares),
% Constraints
disjoint2(Squares, [margin(0, 0, 1, 1)]),
lines_constraints(0, Rows, StartsX, SquareSizes),
lines_constraints(0, Columns, StartsY, SquareSizes),
% Solution search
VarsList = [NumSquares, StartsX, StartsY, SquareSizes],
flatten(VarsList, Vars),
labeling([], Vars).
construct_squares(_, [], [], [], []).
construct_squares(Size, [StartX|T1], [StartY|T2], [SquareSize|T3], [square(StartX, SquareSize, StartY, SquareSize)|T4]) :-
StartX + SquareSize #=< Size,
StartY + SquareSize #=< Size,
construct_squares(Size, T1, T2, T3, T4).
% Rows and columns NumFilledCells cells constraints
lines_constraints(_, [], _, _).
lines_constraints(Index, [NumFilledCells|T], Starts, SquareSizes) :-
line_constraints(Index, NumFilledCells, Starts, SquareSizes),
I is Index + 1,
lines_constraints(I, T, Starts, SquareSizes).
line_constraints(Index, NumFilledCells, Starts, SquareSizes) :-
findall(
SquareSize,
(
element(N, Starts, Start),
element(N, SquareSizes, SquareSize),
intersect(Index, Start, SquareSize)
),
Lines),
sum(Lines, #=, NumFilledCells).
% Check if a square intersects a row or column
intersect(Index, Start, SquareSize) :-
Start #=< Index,
Index #=< Start + SquareSize.
The issue is in your line_constraint/4
predicate.问题出在您的
line_constraint/4
谓词中。 In it, you are posting some clpfd constraints inside a findall/3
.在其中,您在
findall/3
中发布了一些 clpfd 约束。 This means that those constraints are only valid inside the findall/3
.这意味着这些约束仅在
findall/3
内有效。 Here is a way to rewrite your predicate that keeps the constraints posted (given that you are using SICStus, I use the do
loop style, which is just syntactic sugar around a recursive predicate):这是一种重写谓词以保持发布约束的方法(鉴于您使用的是 SICStus,我使用
do
循环样式,它只是递归谓词周围的语法糖):
line_constraints(Index, NumFilledCells, Starts, SquareSizes) :-
(
foreach(Start,Starts),
foreach(SquareSize,SquareSizes),
foreach(Usage,Usages),
param(Index)
do
Intersect #<=> ( Start #=< Index #/\ Index #< Start + SquareSize),
Usage #= Intersect * SquareSize
),
sum(Usages, #=, NumFilledCells).
(Note that I changed the second inequality to be a strict one: The end of the square is right before Start + SquareSize
.) (请注意,我将第二个不等式更改为严格的不等式:正方形的结束就在
Start + SquareSize
之前。)
As you will probably experience, this formulation is pretty weak in terms of reducing the search space.正如您可能会体验到的那样,这个公式在减少搜索空间方面非常薄弱。 One way to improve it (but I haven't tried it myself) would be to replace the
lines_constraints/4
by some cumulative constraints.改进它的一种方法(但我自己没有尝试过)是用一些累积约束替换
lines_constraints/4
。
Since the problem was in the number of squares, I fixed them to be the highest possible (total number of cells divided by four, since they must be disjoint), but allowed its width/height to be equal to zero, effectively not existing and then allowing for the number of squares to be bounded between zero and the maximum number of squares.由于问题在于正方形的数量,我将它们固定为尽可能高的(单元格总数除以四,因为它们必须不相交),但允许其宽度/高度等于零,实际上不存在并且然后允许平方数限制在零和最大平方数之间。
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