如何使用最小冲突启发式避免陷入8皇后的局部最小值中

Question

我编写了以下代码来解决n皇后问题：

(defun solve (board max-steps)

    (enforce-one-queen-per-column board)

    (dotimes (step max-steps) ; repeat for a max amount of times, 
        (if (eql (get-threatened-queens board) 0) ; if we have solved the board, return it
            (progn
                (format t "Solved!")
                (return board)
            )
        )

        (let ((threatened-queens (get-threatened-queens board)) ; get all threatened queens
              (queen nil))

             (setf queen (nth (random (length threatened-queens)) threatened-queens)) ; choose random threatened queen

             ; find which row in its column where it would be least threatened
             (let ((row-with-min-threats nil) ; (row_index, num_threats set to a high number)
                   (col (car queen))
                   (row (cdr queen)))

                  (format t "Dealing with threatened queen at (~A, ~A)...~%" col row)

                  (dotimes (i (array-dimension board 0)) ; for each row, find num of threats
                      (let ((num-threats (get-num-threats board col i)))
                           (print (car row-with-min-threats))
                           (format t "Checking (~A, ~A)~%" col i)
                           (format t "Threatened by ~A queens...~%" num-threats)

                            ; if row-with-min-threats has not yet been initialized
                           ; or if the row's threat is smaller than the tracked row with min threats

                            ; take first row as min threat so far
                            (if (eql row-with-min-threats nil) ; 
                                (setf row-with-min-threats (cons i num-threats))
                                (if (< num-threats (cdr row-with-min-threats)) ; if not first row and current min threats is smaller than tracked min
                                    (setf row-with-min-threats (cons i num-threats))
                                    (if (and (eql num-threats (cdr row-with-min-threats)) 
                                             (eql (random 2) 1))                            ; if current cell's & min cell's threats are equal, we randomly decide which cell to assign
                                        (progn 
                                            (format t "Randomly chose (~A, ~A)...~%" col i)
                                            (setf row-with-min-threats (cons i num-threats)))
                                        )

                                    )
                                )
                        )
                 )

                  (format t "Least threatened cell is (~A, ~A)...~%" col (car row-with-min-threats))

                  (if (not (eql row (car row-with-min-threats))) ; if its least threatened position isn't where it currently is
                      (progn
                          (setf (aref board (car row-with-min-threats) col) 1) ; move queen
                          (setf (aref board row col) 0) 
                          (format t "Moved queen to (~A, ~A)...~%" col (car row-with-min-threats))
                      )
                  )



             )
        )
    )

    nil
)

我正在尝试解决8皇后问题。问题出在solve函数中，但是我不确定自己做错了什么。由于我使用的是最小冲突启发式方法，因此我感觉自己陷入了局部最小值。我试图通过重新启动原始电路板的问题来解决此问题，但是无论我重新启动多少次，它似乎都无法正常工作，因为即使它们仍然存在冲突，但女王/王后都卡在最小冲突的位置上。我如何改进solve才能成功地将8个皇后放在不会互相威胁的牢房中？

运行程序：

(setf board (make-board 8))
(solve board 10)

其中10代表solve在原始板上被调用的次数。

我没有包含我的功能，因为它们是不言自明的，但是如果它们会有所帮助，我很乐意将它们包含在内。

Answer 1

我试图通过重新启动原始电路板来解决此问题，但是无论我重启多少次，它似乎都无法正常工作，因为皇后都卡在了最小冲突的位置，即使它们仍然冲突

我相信这只是意味着您无法通过启发式方法来“修复”您给定的初始位置。在这种情况下，您需要一个新的“种子”，您可以通过在所有8列中随机选择一行，然后这样做（如果事实证明它是不好的种子），直到获得一个好的“种子”。由于解空间渐近增加，因此在n * n板上，“ n”越高，“好的”初始位置/种子越多。

这里是一个具有以下考虑因素的Java类（主要在main和shuffle方法中：https://github.com/danielbmancini/JHTP8_JCP8_Sol._Comentadas/blob/master/7/src/EightQueensHeuristic.java

我希望这对您有帮助。如果我有任何错误，欢迎提出批评。