Lisp 如何使用最小冲突启发式避免陷入8皇后的局部极小值

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
)

其中，10表示原始板上调用

solve

的次数

我没有包括我的功能，因为它们是不言自明的，但如果它们有帮助，我将很高兴包括它们

我试图通过重新启动原始板来克服这个问题，但无论我重新启动多少次，它似乎都不起作用，因为皇后们都被困在最小冲突点，即使它们仍然冲突

我相信这仅仅意味着你给出的初始位置根本无法通过启发式“修复”。在这种情况下，您需要一个新的“种子”，您可以通过在所有8列中随机选择一行来获得，然后这样做（如果结果是一个坏种子），直到获得一个好种子。
由于解空间逐渐增大，在一个n*n板中，“n”越高，初始位置/种子的数量就越多

下面是一个Java类，其中包含以下注意事项（主要是main和shuffle方法）：

我希望这能对你有所帮助。如果我有任何错误，欢迎批评

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