骑士之旅深度优先搜索回溯 [英] Knight's Tour Depth First Search Backtracking
问题描述
我正在使用DFS骑士的旅游咨询实施。
I'm working on a knight's tour implementation using DFS.
我的问题是，当我运行它，它工作正常到步骤20，但在此之后，该算法怪胎，并在一个5×5板输出这个（有一个5×5板开始溶液（0,0） ）：
My problem is, when I run it, it works fine up to step 20, but after that the algorithm freaks out and outputs this on a 5x5 board (there is a solution for a 5x5 board starting at (0,0)):
(1 10 5 16 24)
(4 15 2 11 20)
(9 6 17 23 22)
(14 3 8 19 12)
(7 18 13 21 25)
*合法继承人必须是0℃=行＆LT; n和0℃=列＆LT; n和不会是previous一步。
*Legal successors must be 0 <= row < n and 0 <= column < n and not be a previous step.
我的实现涉及生成使用genSuccessors功能*合法继承人，把它们入栈和堆栈作为新的当前位置的顶部递归运行算法的项目。我只递增STEP_COUNT（负责跟踪正方形的顺序骑士访问）如果下一个位置是没有采取之前的工序。当我不能产生任何更多的孩子，算法探讨在边境其他的替代品，直到空前沿（故障状态）或STEP_COUNT =＃广场上的板（胜利）。
My implementation involves generating *legal successors using the genSuccessors function, throwing them onto a stack and recursively running the algorithm with the item at the top of the stack as the new current position. I only increment the step_count (in charge of tracking the order of squares the knight visits) if the next position is a step not taken before. When I cannot generate any more children, the algorithm explores other alternatives in the frontier until frontier empty (fail condition) or the step_count = # squares on the board (win).
我觉得算法一般是健全的。
I think the algorithm in general is sound.
编辑：我认为问题是，当我不能产生更多的孩子，我去探索，我需要放弃一些目前旅游前沿的其余部分。我的问题是，我怎么知道我怎么早需要去？
edit: I think the problem is that when I can't generate more children, and I go to explore the rest of the frontier I need to scrap some of the current tour. My question is, how do I know how far back I need to go?
从图形上看，在树上，我想我需要去备份到了一个分支的未访问过的孩子最近的节点，并重新开始从那里报废的所有节点下降时，previous（错）分公司参观。它是否正确？我将如何跟踪，在我的code？
Graphically, in a tree I think I would need to go back up to the closest node that had a branch to an unvisited child and restart from there scrapping all the nodes visited when going down the previous (wrong) branch. Is this correct? How would I keep track of that in my code?
感谢您阅读这么长的职务;并感谢所有帮助你们可以给我。
Thanks for reading such a long post; and thanks for any help you guys can give me.
推荐答案
哎呀！您的code是真正可怕的。特别是：
Yikes! Your code is really scary. In particular:
1），它采用突变无处不在。 2），它试图模拟回报。 3）不具备的任意的测试用例。
1) It uses mutation everywhere. 2) It tries to model "return". 3) It doesn't have any test cases.
我将是一个傲慢，大便，在这里，简单地说此言的特性相结合，使得超硬调试的程序。
I'm going to be a snooty-poo, here, and simply remark that this combination of features makes for SUPER-hard-to-debug programs.
另外...为DFS，真的没有必要跟踪自己的堆栈;你可以用递归吧？
Also... for DFS, there's really no need to keep track of your own stack; you can just use recursion, right?
抱歉不能提供更多的帮助。
Sorry not to be more helpful.
下面是我会怎么写：
#lang racket
;; a position is (make-posn x y)
(struct posn (x y) #:transparent)
(define XDIM 5)
(define YDIM 5)
(define empty-board
(for*/set ([x XDIM]
[y YDIM])
(posn x y)))
(define (add-posn a b)
(posn (+ (posn-x a) (posn-x b))
(+ (posn-y a) (posn-y b))))
;; the legal moves, represented as posns:
(define moves
(list->set
(list (posn 1 2) (posn 2 1)
(posn -1 2) (posn 2 -1)
(posn -1 -2) (posn -2 -1)
(posn 1 -2) (posn -2 1))))
;; reachable knights moves from a given posn
(define (possible-moves from-posn)
(for/set ([m moves])
(add-posn from-posn m)))
;; search loop. invariant: elements of path-taken are not
;; in the remaining set. The path taken is given in reverse order.
(define (search-loop remaining path-taken)
(cond [(set-empty? remaining) path-taken]
[else (define possibilities (set-intersect (possible-moves
(first path-taken))
remaining))
(for/or ([p possibilities])
(search-loop (set-remove remaining p)
(cons p path-taken)))]))
(search-loop (set-remove empty-board (posn 0 0)) (list (posn 0 0)))
;; start at every possible posn:
#;(for/or ([p empty-board])
(search-loop (set-remove empty-board p) (list p)))
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