我编写了一个迭代加深算法,除非我添加循环检查,否则它会起作用,算法会返回一个比它应该更深的解决方案。但是当我没有检查周期时它确实可以正常工作,但这需要很长时间。任何人都可以发现这个错误吗?
(defun rec-depth-limited (problem node cutoff closed)
(if (= cutoff 0)
(if (funcall (problem-goalp problem) node)
node)
(if (visited-p node closed)
nil
(progn
;; when i remove the next line, it works correctly
(setf (gethash (node-state node) closed) t)
(loop for child in (expand node (problem-actions problem)) do
(let ((result (rec-depth-limited problem child (1- cutoff) closed)))
(if result
(return result))))))))
(defun iterative-deepening (problem)
"Iterative deepening search"
(let ((cutoff 0))
(loop
(format t "~%cut-off: ~A" cutoff)
(let ((solution (rec-depth-limited
problem
(make-node :state (problem-state problem))
cutoff
(make-hash-table :test #'equalp)))) ;solve problem up to cutoff
(if (null solution)
(incf cutoff);if solution is not found, increment the depth
(return solution))))))
(defun visited-p (node table)
"Checks if state in node was visited before by checking
if it exists in the table"
(nth-value 1 (gethash (node-state node) table)))
编辑:这是扩展功能
(defun expand (node actions)
"Expands a node, returns a list of the new nodes"
(remove-if #'null (apply-actions node actions)));apply all actions on all nodes
(defun apply-actions (node actions)
"Applies all actions to a state, returns a list of new states"
(mapcan #'(lambda (action)
(mapcar #'(lambda (tile) (funcall action tile node))
(node-state node)))
actions))
这是其中一项行动,除了微小改动外,它们都是相同的
(defun slide-right (tile node)
"slide the tile one cell to the right. returns nil if not possible,
otherwise returns a node with the new state"
(when (can-slide-right-p tile (node-state node));if can slide right
(and visualize (format t "~%slide ~A to the right" (tile-label tile)))
(let* ((newstate (mapcar #'copy-tile (node-state node)));copy the current state
(depth (node-depth node))
(newcol (incf (tile-col (find tile newstate :test #'equalp))));update state
(cost (1+ (node-cost node))))
(make-node :state newstate ;create new node with the new state
:parent node
:depth (1+ depth)
:action (concatenate 'string
"slide "
(tile-label tile)
" right" )
:cost cost))))
谓词
(defun can-slide-right-p (tile state)
"returns T if the specified tile can be sled one cell to the right"
(let ((row (tile-row tile))
(end (+ (tile-col tile) (tile-length tile))) ;col at which tile ends after being sled
(orient (tile-orientation tile)))
(and (equal orient 'H)
(or (tile-is-mouse tile) (< end *board-w*))
(empty-cell-p row end state))))
(defun spans-cell-p (row col tile)
"returns T if the specified tile spans the specified cell"
(if (equal (tile-orientation tile) 'H)
(horizontally-spans-cell-p row col tile)
(vertically-spans-cell-p row col tile)))
(defun horizontally-spans-cell-p (row col tile)
"Tests if the specified horizontal tile spans the specified cell"
(let ((tile-col (tile-col tile))
(tile-row (tile-row tile))
(tile-len (tile-length tile)))
(and (= tile-row row) (>= col tile-col) (< col (+ tile-col tile-len)))))
(defun vertically-spans-cell-p (row col tile)
"Tests if the specified vertical tile spans the specified cell"
(let ((tile-col (tile-col tile))
(tile-row (tile-row tile))
(tile-len (tile-length tile)))
(and (= tile-col col) (>= row tile-row) (< row (+ tile-row tile-len)))))
答案 0 :(得分:6)
当通向目标的第一条路径比包含相同状态的任何其他短路径更长时,带有周期检测的有限深度优先搜索可能会返回更长的路径。
让D成为目标状态:
A -- B -- C -- D
\
C -- D
深度限制为2时,如果首先访问顶部分支,将访问B和C并将其保存在哈希表中。访问底部分支时,它不会扩展到C,因为它被标记为已访问。
一种可能的解决方案是将哈希值设置为找到状态的最小深度。这使得状态被称为访问某个深度及更远,但如果访问的深度较小,则可以再次扩展它。
(defun visited-p (node table)
(let ((visited-depth (gethash (node-state node) table)))
(and visited-depth
(>= (node-depth node) visited-depth))))
(defun set-visited (node table)
(let ((visited-depth (gethash (node-state node) table)))
(setf (gethash (node-state node) table)
(if visited-depth
(min visited-depth (node-depth node))
(node-depth node)))))