Question

忍受我，我知道这显然是超级简单的。我似乎无法绕过序幕。我有这个逻辑问题：

在最近的节日中，对100米高温进行了密切监控。

每位参赛者必须参加两场比赛才能确定平均场地。

两场比赛中只有一名选手在同一个地方完成比赛。

艾伦永远不会持久。查尔斯总是击败达伦。布莱恩至少有一个第一名。艾伦在至少一场比赛中获得第三名。达伦都是查尔斯获得了第二名。这两个结果是什么？

答案：第1场比赛：Brian，Charles，Alan，Darren。第2场比赛：查尔斯，达伦，艾伦，布赖恩。

这是我到目前为止所提出的，但我无法弄清楚＆＃34;至少＆＃34;和其他更复杂的条件。

place(one).
place(two).
place(three).
place(four).

/* Place (constants): one, two, three, four.

Names (variables): Alan_1, Brian_1, Charles_1, Darren_1
Names (variables): Alan_2, Brian_2, Charles_2, Darren_2
*/

higher(one,two).
higher(one,three).
higher(one,four).
higher(two,three).
higher(two,four).
higher(three,four).

is_higher(X,Y):- higher(X,Y).
is_higher(X,Y):- higher(X,Z), is_higher(Z,Y).


is_different(S,T,U,V,W,X,Y,Z):- W\=X,W\=Y,W\=Z,X\=Y,X\=Z,Y\=Z,W\=S,W\=T,
            W\=U,W\=V,S\=T,T\=U,U\=V,S\=U,S\=V,T\=V,
            T\=W,T\=X,T\=Y,T\=Z,U\=W,U\=X,U\=Y,U\=Z,
            V\=X,V\=Y,V\=Z.


solution(Alan_1, Brian_1, Charles_1, Darren_1, Alan_2, Brian_2, Charles_2, 
Darren_2):-
place(Alan_1), place(Brian_1), place(Charles_1), place(Darren_1), 
place(Alan_2), place(Brian_2), place(Charles_2), place(Darren_2),
Alan_1\=four, Alan_2\=four, higher(Charles_1, Darren_1), higher(Charles_2, 
Darren_2),

is_different(Alan_1, Brian_1, Charles_1, Darren_1, Alan_2, Brian_2, 
Charles_2, Darren_2).

/*  Query */

% solution(Alan_1, Brian_1, Charles_1, Darren_1, Alan_2, Brian_2, Charles_2, 
Darren_2).

任何帮助或智慧的话都会受到超级赞赏。

Answer 1

我用SWI + CLPFD解决了。为了表示＆＃34;至少＆＃34;，我使用了#\/。



:-use_module(library(clpfd)).

solve(RaceResult):-
    RaceResult = [Alan1,Alan2,Chales1,Chales2,Brian1,Brian2,Darren1,Darren2],
    RaceResult ins 1..4,

    all_different([Alan1,Chales1,Brian1,Darren1]),
    all_different([Alan2,Chales2,Brian2,Darren2]),

    % Only one runner finished in the same place in both races.
    (Alan1 #= Alan2) #<==> AlanSame,
    (Chales1 #= Chales2) #<==> ChalesSame,
    (Brian1 #= Brian2) #<==> BrianSame,
    (Darren1 #= Darren2) #<==> DarrenSame,
    sum([AlanSame,ChalesSame,BrianSame,DarrenSame], #=, 1),

    % Alan was never last. 
    Alan1 #\= 4,
    Alan2 #\= 4,

    % Alan finished third in at least one of the races. 
    (Alan1 #= 3 ) #\/ (Alan2 #= 3),

    % Charles always beat Darren. 
    Chales1 #< Darren1,
    Chales2 #< Darren2,

    % Brian had at least one first place. 
    (Brian1 #= 1 ) #\/ (Brian2 #= 1),

    % Both Darren and Charles had a second place. 
    (Darren1 #= 2) #\/ (Darren2 #= 2),
    (Chales1 #= 2) #\/ (Chales2 #= 2),

    labeling([ffc],RaceResult).


?- solve([Alan1,Alan2,Chales1,Chales2,Brian1,Brian2,Darren1,Darren2]).
Alan1 = Alan2, Alan2 = 3,
Chales1 = Brian2, Brian2 = 1,
Chales2 = Darren1, Darren1 = 2,
Brian1 = Darren2, Darren2 = 4 ;
Alan1 = Alan2, Alan2 = 3,
Chales1 = Darren2, Darren2 = 2,
Chales2 = Brian1, Brian1 = 1,
Brian2 = Darren1, Darren1 = 4.

:-use_module(library(clpfd)).

solve(RaceResult):-
    RaceResult = [Alan1,Alan2,Chales1,Chales2,Brian1,Brian2,Darren1,Darren2],
    RaceResult ins 1..4,

    all_different([Alan1,Chales1,Brian1,Darren1]),
    all_different([Alan2,Chales2,Brian2,Darren2]),

    % Only one runner finished in the same place in both races.
    (Alan1 #= Alan2) #<==> AlanSame,
    (Chales1 #= Chales2) #<==> ChalesSame,
    (Brian1 #= Brian2) #<==> BrianSame,
    (Darren1 #= Darren2) #<==> DarrenSame,
    sum([AlanSame,ChalesSame,BrianSame,DarrenSame], #=, 1),

    % Alan was never last. 
    Alan1 #\= 4,
    Alan2 #\= 4,

    % Alan finished third in at least one of the races. 
    (Alan1 #= 3 ) #\/ (Alan2 #= 3),

    % Charles always beat Darren. 
    Chales1 #< Darren1,
    Chales2 #< Darren2,

    % Brian had at least one first place. 
    (Brian1 #= 1 ) #\/ (Brian2 #= 1),

    % Both Darren and Charles had a second place. 
    (Darren1 #= 2) #\/ (Darren2 #= 2),
    (Chales1 #= 2) #\/ (Chales2 #= 2),

    labeling([ffc],RaceResult).


?- solve([Alan1,Alan2,Chales1,Chales2,Brian1,Brian2,Darren1,Darren2]).
Alan1 = Alan2, Alan2 = 3,
Chales1 = Brian2, Brian2 = 1,
Chales2 = Darren1, Darren1 = 2,
Brian1 = Darren2, Darren2 = 4 ;
Alan1 = Alan2, Alan2 = 3,
Chales1 = Darren2, Darren2 = 2,
Chales2 = Brian1, Brian1 = 1,
Brian2 = Darren1, Darren1 = 4.

也许有另一种解决方案？

Race1： Chales 达伦艾伦布赖恩

RACE2：布赖恩 Chales 艾伦达伦

...抱歉只是替换了race1和race2。

Answer 2

我是这样做的：

puzzle :-
    Races = [[_,_,_,_],[_,_,_,_]],
    Races = [R1,R2],
    (first(brian,R1);first(brian,R2);(first(brian,R1),first(brian,R2))),
    (third(alan,R1);third(alan,R2);(third(alan,R1),third(alan,R2))),
    ((second(darren,R1),second(charles,R2));(second(darren,R2),second(charles,R1))),
    member(brian,R1),
    member(charles,R1),
    member(brian,R2),
    member(charles,R2),
    member(alan,R1),
    member(darren,R1),
    member(alan,R2),
    member(darren,R2),
    before(charles,darren,R1),
    before(charles,darren,R2),
    never_last(alan,Races),
    only_one_same(Races),
    write(Races),nl.

zip_equal([],[],[]).
zip_equal([R|R1s],[R|R2s],[R|Rs]) :- !, zip_equal(R1s,R2s,Rs).
zip_equal([_|R1s],[_|R2s],Rs) :- !, zip_equal(R1s,R2s,Rs).

never_last(X,[[A1,B1,C1,_],[A2,B2,C2,_]]) :- member(X,[A1,B1,C1]), member(X,[A2,B2,C2]).

before(X,Y,[X|Rs]) :- !, member(Y,Rs).
before(X,Y,[_|Rs]) :- before(X,Y,Rs).

first(X,[X,_,_,_]).
second(X,[_,X,_,_]).
third(X,[_,_,X,_]).

only_one_same([R1s,R2s]) :- zip_equal(R1s,R2s,[_]).

初学者Prolog逻辑拼图

2 个答案: