使用杰西解决河流拼图

Question

首先我解决了一个问题，即我有一个无限的循环，我通过在约束模块中添加规则来解决该问题。我已经考虑了所有的约束，但是似乎所有事实因某种原因被删除...到目前为止，这是我的代码：

;;MAIN Module
(deftemplate state
    (slot farmer-position)
    (slot fox-position)
    (slot goat-position)
    (slot cabbage-position)
    (slot id)
    (slot prev-state (default nil))
    (multislot move (default nil)))
(deftemplate finished
    (slot value))
(deffacts initial-facts
    (state (farmer-position s1) 
            (fox-position s1) 
            (goat-position s1) 
            (cabbage-position s1)
            (id 0))
    (opp s1 s2)
    (opp s2 s1))
;;CONSTARIN Modle
(defmodule CONSTRAIN)
(defrule CONSTRAIN::fox-goat
    (declare (auto-focus true))
    ?p<-(MAIN::state (fox-position ?f) (goat-position ?f) (farmer-position 
                                                                       ~?f))
    =>
    (retract ?p))
(defrule CONSTRAIN::goat-cabbge
    (declare (auto-focus true))
    ?p<-(MAIN::state (goat-position ?f) (cabbage-position ?f) (farmer-
                                                         position ~?f))
    =>
    (retract ?p))
(defrule CONSTRAIN::no-doubles
    (declare (auto-focus true)) 
    ?p1<-(MAIN::state (farmer-position ?s1) (fox-position ?s2) (goat-
                       position ?s3) (cabbage-position ?s4) (id ?id1))
    ?p2<-(MAIN::state (farmer-position ?s1) (fox-position ?s2) (goat-
          position ?s3) (cabbage-position ?s4) (id ?id2&:(> ?id2 ?id1)))
    =>
    (retract ?p2))
 (defrule CONSTRAIN::stop-exc
    (declare (auto-focus true)) 
    ?p1<-(MAIN::state (farmer-position s2) (fox-position s2) (goat-position 
                                                  s2) (cabbage-position s2))
    =>
    (assert (MAIN::finished (value yes))))
;;MOVE Module
(defmodule MOVE)
(defrule MOVE::move-fox
    ?p<-(MAIN::state (farmer-position ?old) (fox-position ?old) (id ?id))
    (not (MAIN::finished (value yes)))
    (opp ?old ?new)
    =>
    (duplicate ?p (farmer-position ?new)
        (fox-position ?new)
        (prev-state ?p)
        (id (+ ?id 1))
        (move fox ?new)))
(defrule MOVE::move-goat
    ?p<-(MAIN::state (farmer-position ?old) (goat-position ?old) (id ?id))
    (not (MAIN::finished (value yes)))
    (opp ?old ?new)
    => 
    (duplicate ?p (farmer-position ?new)
        (goat-position ?new)
        (prev-state ?p)
        (id (+ ?id 1))
        (move goat ?new)))
(defrule MOVE::move-cabbage
    ?p<-(MAIN::state (farmer-position ?old) (cabbage-position ?old) (id ?
                                                                       id))
    (not (MAIN::finished (value yes)))
    (opp ?old ?new)
    =>
    (duplicate ?p (farmer-position ?new)
       (cabbage-position ?new)
       (prev-state ?p)
       (id (+ ?id 1))
       (move cabbage ?new)))
 ;;RUN
 (reset)
 (watch all)
 (focus MOVE)
 (run)
 (facts)

这是我的输出：

<== Focus MAIN
 ==> Focus MOVE
 FIRE 1 MOVE::move-fox f-1,, f-2
  ==> f-4 (MAIN::state (farmer-position s2) (fox-position s2) (goat-position 
 s1) (cabbage-position s1) (id 1) (prev-state <Fact-1>) (move fox s2))
 ==> Activation: CONSTRAIN::goat-cabbge :  f-4
 ==> Activation: MOVE::move-fox :  f-4,, f-3
 <== Focus MOVE
  ==> Focus CONSTRAIN
 FIRE 2 CONSTRAIN::goat-cabbge f-4
 <== f-4 (MAIN::state (farmer-position s2) (fox-position s2) (goat-position 
 s1) (cabbage-position s1) (id 1) (prev-state <Fact-1>) (move fox s2))
 <== Activation: MOVE::move-fox :  f-4,, f-3
 <== Focus CONSTRAIN
 ==> Focus MOVE
 FIRE 3 MOVE::move-cabbage f-1,, f-2
  ==> f-5 (MAIN::state (farmer-position s2) (fox-position s1) (goat-position 
 s1) (cabbage-position s2) (id 1) (prev-state <Fact-1>) (move cabbage s2))
 ==> Activation: CONSTRAIN::fox-goat :  f-5
 ==> Activation: MOVE::move-cabbage :  f-5,, f-3
 <== Focus MOVE
  ==> Focus CONSTRAIN
 FIRE 4 CONSTRAIN::fox-goat f-5
 <== f-5 (MAIN::state (farmer-position s2) (fox-position s1) (goat-position 
 s1) (cabbage-position s2) (id 1) (prev-state <Fact-1>) (move cabbage s2))
 <== Activation: MOVE::move-cabbage :  f-5,, f-3
 <== Focus CONSTRAIN
 ==> Focus MOVE
FIRE 5 MOVE::move-goat f-1,, f-2
 ==> f-6 (MAIN::state (farmer-position s2) (fox-position s1) (goat-position 
s2) (cabbage-position s1) (id 1) (prev-state <Fact-1>) (move goat s2))
==> Activation: MOVE::move-goat :  f-6,, f-3
FIRE 6 MOVE::move-goat f-6,, f-3
 ==> f-7 (MAIN::state (farmer-position s1) (fox-position s1) (goat-position 
s1) (cabbage-position s1) (id 2) (prev-state <Fact-6>) (move goat s1))
==> Activation: CONSTRAIN::no-doubles :  f-1, f-7
==> Activation: MOVE::move-fox :  f-7,, f-2
==> Activation: MOVE::move-goat :  f-7,, f-2
==> Activation: MOVE::move-cabbage :  f-7,, f-2
<== Focus MOVE
==> Focus CONSTRAIN
FIRE 7 CONSTRAIN::no-doubles f-1, f-7
<== f-7 (MAIN::state (farmer-position s1) (fox-position s1) (goat-position 
s1) (cabbage-position s1) (id 2) (prev-state <Fact-6>) (move goat s1))
 <== Activation: MOVE::move-fox :  f-7,, f-2
 <== Activation: MOVE::move-goat :  f-7,, f-2
 <== Activation: MOVE::move-cabbage :  f-7,, f-2
 <== Focus CONSTRAIN
 ==> Focus MOVE
 <== Focus MOVE
 ==> Focus MAIN
 <== Focus MAIN
 For a total of 0 facts in module MOVE.

Answer 1

事实F-6未撤回。唯一有效的第一步是让农民用山羊过河，这是F-6代表的举动。您没有一个单独将农民移动到河流的规则，因此，从6号农民来说，唯一有效的举动是用山羊返回河流，将您带回最初的位置。由于由F-1表示的初始位置与F-7相同，因此NO倍缩回F-7，并且没有剩余的有效移动。

如果您从http://www.jessrules.com/jess/download.shtml下载Jess，则可以在file diremma.clp的示例目录中获得河流交叉示例的代码。

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