我试图使用CLIPS解决8个皇后问题。这个我的代码解决了它的一部分。只有水平和垂直检查，所以你可以说它解决了8个车问题。我无法得到的是如何进行对角线检查。 CLIPS不会允许任何检查事实，如（不是（女王？i-1？j-1）），（不是（女王？i-2？j-2））等......
(assert (cur 1 1))
(assert (placed 0))
(defrule placer "Placer"
?a <- (cur ?i ?j)
?b <- (placed ?placed)
(test (<= ?i 8))
(test (<= ?j 8))
(test (< ?placed 8))
(not (queen ?i 1|2|3|4|5|6|7|8))
(not (queen 1|2|3|4|5|6|7|8 ?j))
(not (queen (?i-1)| //diagonal check there
=>
(assert (queen ?i ?j))
(assert (placed (+ ?placed 1)))
(retract ?b)
(if (< ?j 8)
then (assert (cur ?i (+ ?j 1)))
else (assert (cur (+ ?i 1) 1)))
(retract ?a)
(printout t "Placing queen" crlf)
)
(defrule horiz "Check-hor"
?a <- (cur ?i ?j)
?b <- (placed ?placed)
(test (<= ?i 8))
(test (<= ?j 8))
(test (< ?placed 8))
(queen ?i 1|2|3|4|5|6|7|8)
=>
(if (< ?j 8)
then (assert (cur ?i (+ ?j 1)))
else (assert (cur (+ ?i 1) 1)))
(retract ?a)
(printout t "Horiz" crlf)
)
(defrule vert "Check-vert"
?a <- (cur ?i ?j)
?b <- (placed ?placed)
(test (<= ?i 8))
(test (<= ?j 8))
(test (< ?placed 8))
(queen 1|2|3|4|5|6|7|8 ?j)
=>
(if (< ?j 8)
then (assert (cur ?i (+ ?j 1)))
else (assert (cur (+ ?i 1) 1)))
(retract ?a)
(printout t "Vert" crlf)
)
答案 0 :(得分：1)
您想要使用返回值约束：
(queen =(- ?i 1) =(- ?j 1))
但是，我建议完全限制女王这样：
(queen ?i2 ?j2)
(test (or (eq ?i ?i2)
(eq ?j ?j2)
(= (abs (- ?i2 ?i)) (abs (- ?j2 ?j))))))
如果您修改规则，现在可以获得部分解决方案：
CLIPS>
(deffacts initial
(cur 1 1)
(placed 0))
CLIPS>
(defrule place
?a <- (cur ?i ?j)
?b <- (placed ?placed)
(test (<= ?i 8))
(test (<= ?j 8))
(test (< ?placed 8))
(not (and (queen ?i2 ?j2)
(test (or (eq ?i ?i2)
(eq ?j ?j2)
(= (abs (- ?i2 ?i)) (abs (- ?j2 ?j)))))))
=>
(assert (queen ?i ?j))
(assert (placed (+ ?placed 1)))
(retract ?b)
(if (< ?j 8)
then (assert (cur ?i (+ ?j 1)))
else (assert (cur (+ ?i 1) 1)))
(retract ?a)
(printout t "Placing queen " (+ ?placed 1) crlf))
CLIPS>
(defrule skip
?a <- (cur ?i ?j)
?b <- (placed ?placed)
(test (<= ?i 8))
(test (<= ?j 8))
(test (< ?placed 8))
(exists (queen ?i2 ?j2)
(test (or (eq ?i ?i2)
(eq ?j ?j2)
(= (abs (- ?i2 ?i)) (abs (- ?j2 ?j))))))
=>
(if (< ?j 8)
then (assert (cur ?i (+ ?j 1)))
else (assert (cur (+ ?i 1) 1)))
(retract ?a))
CLIPS> (reset)
CLIPS> (run)
Placing queen 1
Placing queen 2
Placing queen 3
Placing queen 4
Placing queen 5
CLIPS> (facts)
f-0 (initial-fact)
f-3 (queen 1 1)
f-15 (queen 2 3)
f-27 (queen 3 5)
f-34 (queen 4 2)
f-46 (queen 5 4)
f-47 (placed 5)
f-76 (cur 9 1)
For a total of 8 facts.
CLIPS>
只放置了5个女王。因为你没有在你的程序中实现任何回溯，所以可以有效地放置一个阻止所有8个皇后被放置的女王。
这是来自https://sites.google.com/site/drriggsnewsite/Home/half-baked-essays/composing-thoughts-an-introduction-to-rules/eight-queens-problem的N皇后计划，它实现了回溯：
CLIPS>
(deffunction shareDiag (?row1 ?col1 ?row2 ?col2)
(= (abs (- ?row1 ?row2))
(abs (- ?col1 ?col2))))
CLIPS>
(deffacts queens
(queen 1 1)
(queen 2 1)
(queen 3 1)
(queen 4 1)
(queen 5 1)
(queen 6 1)
(queen 7 1)
(queen 8 1))
CLIPS>
(defrule moveBadlyPlacedQueen
(queen ?row1 ?col1)
?f <- (queen ?row2 ?col2)
(test (< ?row1 ?row2))
(test (or (= ?col1 ?col2)
(shareDiag ?row1 ?col1 ?row2 ?col2)))
=>
(retract ?f)
(assert (queen ?row2 (+ 1 ?col2))))
CLIPS>
(defrule backTrack
(declare (salience 100))
?g <- (queen ?row1 ?col1)
?f <- (queen ?row2 9)
(test (= (+ ?row1 1) ?row2))
=>
(retract ?f ?g)
(assert (queen ?row1 (+ ?col1 1))
(queen ?row2 1)))
CLIPS> (deffacts otpt (print 1))
CLIPS>
(defrule visualize
(declare (salience -10))
?f <- (print ?row)
(queen ?row ?col)
=>
(retract ?f)
(printout t ?row ":")
(loop-for-count (- ?col 1) (printout t "|_"))
(printout t "|Q" )
(bind ?more (- 8 ?col))
(if (> ?more 0) then (loop-for-count ?more (printout t "|_")))
(printout t crlf)
(assert (print (+ ?row 1))))
CLIPS> (reset)
CLIPS> (run)
1:|Q|_|_|_|_|_|_|_
2:|_|_|_|_|_|Q|_|_
3:|_|_|_|_|_|_|_|Q
4:|_|_|Q|_|_|_|_|_
5:|_|_|_|_|_|_|Q|_
6:|_|_|_|Q|_|_|_|_
7:|_|Q|_|_|_|_|_|_
8:|_|_|_|_|Q|_|_|_
CLIPS>