我被要求为著名的“山羊、狼和白菜”场景编写解决方案。场景如下:
农夫想要将这三者都运过河。然而,如果:
- 山羊和白菜被单独留下,山羊会吃掉白菜
- 如果狼和山羊被单独留下,狼会吃掉山羊!
因此,解决这个问题的一种方法如下:
- 将山羊运过河,并在对岸放下
- 返回河对岸
- 拿起白菜或狼中的一个,带到对岸
- 放下狼,捡起山羊,返回另一边
- 放下山羊,捡起白菜,返回另一边
- 捡起山羊,瞧!三者都成功运输了。
然而,我在将其投射到PDDL上遇到了一些困难。我被给出了问题定义:
(define (problem boat1)(:domain boat); only needs two objects, namely representing; either banke side of the river, [w]est and [e]ast(:objects w e)(:INIT ; wolf, goat, cabbage, boat are all on ; the west side to start with (config w w w w) ; represent all valid states ; these two are the special case, ; representing that wolf and cabbage are ; safe together even if the boat is away (valid w e w e) (valid e w e w) ; these are all cases where two entities ; are always safe as long as the boat is ; with them. In other words, a single entity ; on the other side is also always safe ; for west side (valid w w w w) (valid w w e w) (valid w e w w) (valid e w w w) ; for east side (valid e e e e) (valid e e w e) (valid e w e e) (valid w e e e) ; these are all valid states that are ; ever allowed)(:goal (AND ; they all have to move to the east side (config e e e e) ))
最后,我们只被给出了一个谓词,并被告知可以用四个动作完成。Move_empty, move_goat, move_wolf, move_cabbage.
谓词是:
(config ?wolf ?goat ?cabbage ?boat)(valid ?wolf ?goat ?cabbage ?boat)
我尝试从move_empty开始:
(:action move_empty :parameters (?from ?to) :precondition (and (valid ?x ?y ?z ?w) (on_left ?from) (on_right ?to)) :effect (and (valid ?x ?y ?z ?w)))
我并不希望得到答案,只希望得到关于如何解决这个问题的帮助和建议,因为根据我所能找到的,关于PDDL的信息并不多。
回答:
注意: 我不知道pddl语言,这是我通过查看您的代码所取得的成果。
主要思路
在您的问题中,每个实体都被描述为在河的西岸或东岸,并且实体通过在谓词config和valid中的相对位置来识别。
每个动作都从一个给定的配置开始,并且必须结束于另一个配置。此外,我们必须要求结束的配置是有效的。
因此,从东岸到西岸的move_empty动作非常简单,如下所示:
(:action move_empty_ew :parameters (?x ?y ?z) :precondition (and (config ?x ?y ?z e) (valid ?x ?y ?z w)) :effect (and (not (config ?x ?y ?z e)) (config ?x ?y ?z w)) )
在这里,我们让其他所有实体的位置(狼、山羊和白菜)保持不确定,同时我们要求最初船在东岸,并且让船前往西岸而留下动物无人看管是有效的移动。如果所有这些条件都满足,那么我们就移动到所需的配置。
解决方案
请注意,我对动作的名称进行了改进,使它们更能反映实际执行的动作。
boat-domain.pddl
(define (domain boat) (:requirements :equality) (:predicates (config ?wolf ?goat ?cabbage ?boat) (valid ?wolf ?goat ?cabbage ?boat) ) (:action move_empty_ew :parameters (?x ?y ?z) :precondition (and (config ?x ?y ?z e) (valid ?x ?y ?z w)) :effect (and (not (config ?x ?y ?z e)) (config ?x ?y ?z w)) ) (:action move_empty_we :parameters (?x ?y ?z) :precondition (and (config ?x ?y ?z w) (valid ?x ?y ?z e)) :effect (and (not (config ?x ?y ?z w)) (config ?x ?y ?z e)) ) (:action move_wolf_ew :parameters (?y ?z) :precondition (and (config e ?y ?z e) (valid w ?y ?z w)) :effect (and (not (config e ?y ?z e)) (config w ?y ?z w)) ) (:action move_wolf_we :parameters (?y ?z) :precondition (and (config w ?y ?z w) (valid e ?y ?z e)) :effect (and (not (config w ?y ?z w)) (config e ?y ?z e)) ) (:action move_goat_ew :parameters (?x ?z) :precondition (and (config ?x e ?z e) (valid ?x w ?z w)) :effect (and (not (config ?x e ?z e)) (config ?x w ?z w)) ) (:action move_goat_we :parameters (?x ?z) :precondition (and (config ?x w ?z w) (valid ?x e ?z e)) :effect (and (not (config ?x w ?z w)) (config ?x e ?z e)) ) (:action move_cabbage_ew :parameters (?x ?y) :precondition (and (config ?x ?y e e) (valid ?x ?y w w)) :effect (and (not (config ?x ?y e e)) (config ?x ?y w w)) ) (:action move_cabbage_we :parameters (?x ?y) :precondition (and (config ?x ?y w w) (valid ?x ?y e e)) :effect (and (not (config ?x ?y w w)) (config ?x ?y e e)) ))
boat-prob.pddl
(define (problem boat) (:domain boat) (:objects w e) (:INIT (config w w w w) (valid w e w e) (valid e w e w) (valid w w w w) (valid w w e w) (valid w e w w) (valid e w w w) (valid e e e e) (valid e e w e) (valid e w e e) (valid w e e e) ) (:goal (config e e e e)))
我使用了fast-downward来寻找最短路径的解决方案:
~$ fast-downward.py --alias seq-opt-bjolp boat-domain.pddl boat-prob.pddl
实际的解决方案是:
move_goat_we w w (1)move_empty_ew w e w (1)move_cabbage_we w e (1)move_goat_ew w e (1)move_wolf_we w e (1)move_empty_ew e w e (1)move_goat_we e e (1)Plan length: 7 step(s).Plan cost: 7