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我试图制作一个程序来解决 3 个任意大小的水壶问题。我写了这么多

refil(Size,_,_,Size).
refil(FirstSize,SecondSize,ThirdSize,Goal):-
    move(FirstSize,SecondSize,ThirdSize,FirstSize,0,0,Goal). 
move(_,_,_,Goal,_,_,Goal).
move(FirstSize,SecondSize,ThirdSize,New1,New2,New3,Goal):-
    (
            reduce(New1,TempVal1,-(New2-SecondSize),Goal)-> NextVal1 is Goal, NextVal2 is TempVal1,NextVal3 is New3, write('From 1st to 2nd'),nl;
            reduce(New1,TempVal1,-(New3-ThirdSize),Goal)-> NextVal1 is Goal,NextVal3 is TempVal1,NextVal2 is New2, write('From 1st to 3nd'),nl;
            reduce1(New1,NextVal1,-(New2-SecondSize))-> NextVal2 is New2 - (New2-SecondSize),NextVal3 is New3, write('From 1st to 2nd'),nl;
            reduce1(New1,NextVal1,-(New3-ThirdSize))-> NextVal3 is New3 - (New3-ThirdSize), NextVal2 is New2, write('From 1st to 3rd'),nl;
            reduce1(New2,NextVal2,-(New3-ThirdSize))-> NextVal3 is 0, NextVal1 is New1, write('From 2nd to 3rd'),nl;
            reduce1(New3,NextVal3,-(New2-SecondSize))-> NextVal2 is 0, NextVal1 is New1, write('From 3rd to 2nd'),nl;
            reduce2(New1, NextVal1, New2, SecondSize)-> NextVal2 is 0, NextVal3 is New3, write('From 2nd to 1st'),nl;
            reduce2(New1, NextVal1, New3, ThirdSize)-> NextVal3 is 0, NextVal2 is New2, write('From 3rd to 1st'),nl
    ),
    write('1st Jug: '), write(NextVal1),nl,
    write('2nd Jug: '), write(NextVal2),nl,
    write('3rd Jug: '), write(NextVal3),nl,
    move(FirstSize,SecondSize,ThirdSize,NextVal1,NextVal2,NextVal3,Goal).

reduce(D,Dn,Size,Goal):-
not(D mod 2 is 0),
X is D - Goal,
X > 0,
X < Size,
Dn is X.

reduce1(D, Dn, Size):-
X is Size, not(X is 0),
D - Size >= 0,
Dn is D - Size.

reduce2(D, Dn, Size, Jug):-
X is Size,
Jug - D >= X,
reduce1(D, Dn, Size).

使用此代码，我得到了我需要的结果。在不划分第一个水罐容量的情况下，我还有什么其他方法可以解决这个问题，而是将第一个水罐倒在任何其他水罐上，直到我装满它们，然后计算目标女巫是第一个水罐中的必要量？

PS我希望你能理解我蹩脚的英语，今天想不通

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