结束视图拼图，Prolog [英] End View Puzzle, Prolog

问题描述

我想为End View Puzzles编写一个求解器，使用CLPFD（对于不熟悉的人，这里有一个简单的描述 http://www.funwithpuzzles.com/2009/12/abcd-end-view-a1.html ）。我正在处理约束，我将应用于每一行/列，并遇到一些麻烦。

I'm trying to write a solver for End View Puzzles, using CLPFD (for those who are unfamiliar, here's a simple description http://www.funwithpuzzles.com/2009/12/abcd-end-view-a1.html ). I'm working on the constraint that I'll apply to each row/column, and running into some trouble.

所以我认为它应该看起来像这样：

So I'm thinking it should look something like this:

% NxN board, numbers from 0 to M in the row, Left/Right are the clues
% corresponding to the row
endviews(N,M,List,Left,Right):- 
    length(List,M),
    domain(List,0,M),
    stop_repeats(List,M),
    left_value(List,M,Left),
    reverse(List,RList),
    left_value(RList,M,Right)
    .

所以前三个步骤非常简单：初始化板，设置域，该行不会重复除 0 （我写了 stop_repeats / 2 谓词）之外的数字。我遇到的问题是处理 left_value / 3 谓词;我不太确定如何在这里继续，因为大多数列表不受约束，我实际上不知道这个第一个积极因素的位置。
任何帮助将非常感谢！

So the first 3 steps are pretty straight forward: initialize the board, set a domain, and make sure that this row won't repeat numbers other than 0 (I've written the stop_repeats/2 predicate). The problem I'm encountering is dealing with the left_value/3 predicate; I'm not really sure how to proceed here, since most of the list is not constrained and I do not actually know the position of this first positive element. Any help would be greatly appreciated!

推荐答案

也许 automaton / 3 约束可以帮助这里？您可以构造一个有限自动机，接受（可选）零后跟指定的第一个字母，然后是任意序列的零和字母。

Maybe the automaton/3 constraint can help here? You can construct a finite automaton that accepts (optional) zero followed by the first letter as indicated, then an arbitrary sequence of zero and letters.

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