使用 Prolog 查找网格中同一行和同一列中的重复项

发布于 2024-11-17 01:14:04 字数 1305 浏览 1 评论 0原文

我是 Prolog 新手。我正在编写 Prolog 代码来解决一个难题,该难题需要我在二维数字网格中查找重复项。


我的输入类似于:

grid(1,1,1).
grid(2,1,2).
grid(2,1,3).
grid(5,1,4).
grid(3,2,1).
grid(4,2,2).
grid(5,2,3).
grid(9,2,4).
grid(6,3,1).
grid(7,3,2).
grid(2,3,3).
grid(8,3,4).
grid(4,4,1).
grid(5,4,2).
grid(1,4,3).
grid(3,4,4).


代表矩阵:

1 2 2 5
3 4 5 9
6 7 2 8
4 5 1 3


我尝试的代码类似于:

finddup(N,X,Y):-
        a is X, b is Y,
        print('Rule 1 \n'),
        ((grid(N,U,1), U \= a, print('U->'), write(U), print('\n'), print('X->'), write(a));
        (grid(N,1,U), U \= b, print('U->'), write(U), print('\n'), print('Y->'), write(b));
        (grid(N,U,4), U \= a, print('U->'), write(U), print('\n'), print('X->'), write(a));
        grid(N,4,U), U \= b, print('U->'), write(U), print('\n'), print('Y->'), write(b)));
        print('Rule 2 \n'),
        ((finddup(N,X-1,Y), X-1 >= 1, X-1 =< 4, Y >= 1, Y =< 4);
        (finddup(N,X+1,Y), X+1 >= 1, X+1 =< 4, Y >= 1, Y =< 4);
        (finddup(N,X,Y-1), X >= 1, X =< 4, Y-1 >= 1, Y-1 =< 4);
        (finddup(N,X,Y+1), X >= 1, X =< 4, Y+1 >= 1, Y+1 =< 4)).


请帮助我...我从一两周以来一直在尝试这个...


谢谢,
维卡斯

I am new to Prolog. I am writing Prolog code to solve a puzzle which requires me to find duplicates in a 2 dimensional grid of numbers.

My input is something like:

grid(1,1,1).
grid(2,1,2).
grid(2,1,3).
grid(5,1,4).
grid(3,2,1).
grid(4,2,2).
grid(5,2,3).
grid(9,2,4).
grid(6,3,1).
grid(7,3,2).
grid(2,3,3).
grid(8,3,4).
grid(4,4,1).
grid(5,4,2).
grid(1,4,3).
grid(3,4,4).

which represents the matrix:

1 2 2 5
3 4 5 9
6 7 2 8
4 5 1 3

The code I was trying was something like:

finddup(N,X,Y):-
        a is X, b is Y,
        print('Rule 1 \n'),
        ((grid(N,U,1), U \= a, print('U->'), write(U), print('\n'), print('X->'), write(a));
        (grid(N,1,U), U \= b, print('U->'), write(U), print('\n'), print('Y->'), write(b));
        (grid(N,U,4), U \= a, print('U->'), write(U), print('\n'), print('X->'), write(a));
        grid(N,4,U), U \= b, print('U->'), write(U), print('\n'), print('Y->'), write(b)));
        print('Rule 2 \n'),
        ((finddup(N,X-1,Y), X-1 >= 1, X-1 =< 4, Y >= 1, Y =< 4);
        (finddup(N,X+1,Y), X+1 >= 1, X+1 =< 4, Y >= 1, Y =< 4);
        (finddup(N,X,Y-1), X >= 1, X =< 4, Y-1 >= 1, Y-1 =< 4);
        (finddup(N,X,Y+1), X >= 1, X =< 4, Y+1 >= 1, Y+1 =< 4)).

Please help me out guys ... I have been trying this since a week or two ...

Thanks,

Vikas

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揽月 2024-11-24 01:14:04

您不需要尝试指定行索引和列索引的范围,因为它们已经在 grid(..) 事实中明确给出。只需让 Prolog 回溯这些值即可。

finddup(N,R,C) :-
    grid(N,R,C),
    grid(N,R,C2), C \= C2,
    grid(N,R2,C), R \= R2.

grid(N,R,C) 将验证给定值是否在网格中,或者如果 N、R、C 中的任何一个(或全部)没有值,它将回溯所有可能的组合。

grid(N,R,C2), C \= C2, 会在同一行中找到具有相同 N 的另一列。

对于 grid(N,R2,C), R 也是如此\= R2 - 但对于另一行。

这将为它找到的任何重复项返回 true,但之后可能返回 false,因为最后一次回溯可能会失败。例如:(

?- finddup(N,R,C).
N = 2,
R = 1,
C = 3 ;
false.

?- finddup(2,1,3).
true ;
false.

?- 

其中分号是用户输入)

您可以在末尾添加一个剪切,以在找到解决方案后停止任何进一步的回溯:

finddup(N,R,C) :-
    grid(N,R,C),
    grid(N,R,C2), C \= C2,
    grid(N,R2,C), R \= R2, !.

给出

?- finddup(N,R,C).
N = 2,
R = 1,
C = 3.

?- finddup(2,1,3).
true.

?- 

You don't need to try and specify the ranges of the row and column indices since they are already explicitly given in the grid(..) facts. Just let Prolog backtrack over the values.

finddup(N,R,C) :-
    grid(N,R,C),
    grid(N,R,C2), C \= C2,
    grid(N,R2,C), R \= R2.

grid(N,R,C) will verify that the given values are in the grid, or it will backtrack over all the possible combinations if any (or all) of N, R, C have no values.

grid(N,R,C2), C \= C2, will find another column in the same row that has the same N.

Likewise for grid(N,R2,C), R \= R2 - but for another row.

This will return a true for any duplicate it finds, but will probably return a false after that, since the last backtrack may fail. For example:

?- finddup(N,R,C).
N = 2,
R = 1,
C = 3 ;
false.

?- finddup(2,1,3).
true ;
false.

?- 

(where the semicolons are user input)

You can add a cut at the end to stop any further backtracking once a solution has been found:

finddup(N,R,C) :-
    grid(N,R,C),
    grid(N,R,C2), C \= C2,
    grid(N,R2,C), R \= R2, !.

gives

?- finddup(N,R,C).
N = 2,
R = 1,
C = 3.

?- finddup(2,1,3).
true.

?- 
~没有更多了~
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