在 Prolog 中解決文本邏輯難題 - 查找生日和月份

Question

我正在閱讀“7 天中的 7 種語言”一書，並且已經到了 Prolog 章節。 作為學習練習，我試圖解決一些文本邏輯難題。 謎底如下：

五姐妹都在不同的月份過生日，而且每個人都在一周中的不同日子過生日。 使用下面的線索，確定每個姐妹生日的月份和日期。

1. 寶拉出生於三月，但不是周六。 阿比蓋爾的生日不是星期五或星期三。
2. 生日在星期一的女孩比布倫達和瑪麗早於這一年出生。
3. Tara 不是在二月出生的，她的生日是在周末。
4. 瑪麗不是十二月出生的，她的生日也不是工作日。 那個生日在六月的女孩是星期天出生的。
5. Tara 比 Brenda 早出生，Brenda 的生日不在周五。 瑪麗不是七月出生的。

對於有經驗的 Prolog 程序員來說，我當前的實現可能看起來像一個笑話。 代碼粘貼在下面。

我希望就如何解決問題以及如何使代碼既清晰又密集。

IE：

1. 我怎樣才能避免輸入限制，說 Days 必須是唯一的。
2. 我怎樣才能避免輸入限制，說月份必須是唯一的。
3. 添加關於生日排序的限制。

is_day(Day) :-
    member(Day, [sunday, monday, wednesday, friday, saturday]).

is_month(Month) :-
    member(Month, [february, march, june, july, december]).

solve(S) :-

    S = [[Name1, Month1, Day1],
         [Name2, Month2, Day2],
         [Name3, Month3, Day3],
         [Name4, Month4, Day4],
         [Name5, Month5, Day5]],

    % Five girls; Abigail, Brenda, Mary, Paula, Tara    
    Name1 = abigail,
    Name2 = brenda,
    Name3 = mary,
    Name4 = paula,
    Name5 = tara,

    is_day(Day1), is_day(Day2), is_day(Day3), is_day(Day4), is_day(Day5),
    Day1 \== Day2, Day1 \== Day3, Day1 \== Day4, Day1 \== Day5,
    Day2 \== Day1, Day2 \== Day3, Day2 \== Day4, Day2 \== Day5,
    Day3 \== Day1, Day3 \== Day2, Day3 \== Day4, Day3 \== Day5,
    Day4 \== Day1, Day4 \== Day2, Day4 \== Day3, Day4 \== Day5,

    is_month(Month1), is_month(Month2), is_month(Month3), is_month(Month4), is_month(Month5),
    Month1 \== Month2, Month1 \== Month3, Month1 \== Month4, Month1 \== Month5,
    Month2 \== Month1, Month2 \== Month3, Month2 \== Month4, Month2 \== Month5,
    Month3 \== Month1, Month3 \== Month2, Month3 \== Month4, Month3 \== Month5,
    Month4 \== Month1, Month4 \== Month2, Month4 \== Month3, Month4 \== Month5,

    % Paula was born in March but not on Saturday.  
    member([paula, march, _], S),
    Day4 \== sunday,

    % Abigail's birthday was not on Friday or Wednesday.    
    Day1 \== friday,
    Day1 \== wednesday,

    % The girl whose birthday is on Monday was born
    % earlier in the year than Brenda and Mary.

    % Tara wasn't born in February, and 
    % her birthday was on the weekend.
    Month5 \== february,
    Day5 \== monday, Day5 \== wednesday, Day5 \== friday,   

    % Mary was not born in December nor was her
    % birthday on a weekday.
    Month3 \== december,
    Day3 \== monday, Day3 \== wednesday, Day3 \== friday,

    % The girl whose birthday was in June was 
    % born on Sunday.
    member([_, june, sunday], S),

    % Tara was born before Brenda, whose birthday
    % wasn't on Friday.
    Day2 \== friday,

    % Mary wasn't born in July.
    Month3 \== july.

更新根據chac的回答，我能夠解決這個難題。 按照同樣的方法，我們（工作中的編程語言能力小組）也能夠解決第二個難題。 我已經在 GitHub 上發布了完整的實現和示例輸出作為要點。

Answer 1

使用 maplist/2 將大大縮短您的代碼。 例如：

maplist(is_month, [Month1,Month2,Month3,Month4,Month5]).

月/1 可能是比 is_month/1 更好的謂詞名稱。 要說明兩個術語不同，請使用約束 dif/2。 使用 maplist/2 和 dif/2，您可以描述列表包含成對不同的元素：

all_dif([]).
all_dif([L|Ls]) :-
        maplist(dif(L), Ls),
        all_dif(Ls).

例子：

?- all_dif([X,Y,Z]).
dif(X, Z),
dif(X, Y),
dif(Y, Z).

solve/1 是一個命令式名稱 - 您正在描述解決方案，因此最好將其稱為 solution/1。

Answer 2

也許謎語未詳細說明，或者您的解決方案不完整：測試您的代碼，我明白了

?- solve(X),maplist(writeln,X).
[abigail,february,monday]
[brenda,july,wednesday]
[mary,june,sunday]
[paula,march,friday]
[tara,december,saturday]
X = [[abigail, february, monday], [brenda, july, wednesday], [mary, june, sunday], [paula, march, friday], [tara, december, saturday]] ;
[abigail,february,monday]
[brenda,december,wednesday]
[mary,june,sunday]
[paula,march,friday]
[tara,july,saturday]
X = [[abigail, february, monday], [brenda, december, wednesday], [mary, june, sunday], [paula, march, friday], [tara, july, saturday]] 

還有更多的解決方案。 那么布倫達什么時候出生？

唯一性的“交易技巧”是使用select /3 謂詞，或簡單地排列/2。 使用這最后的代碼變得像

solve(S) :-

    S = [[Name1, Month1, Day1],
         [Name2, Month2, Day2],
         [Name3, Month3, Day3],
         [Name4, Month4, Day4],
         [Name5, Month5, Day5]],

    Girls =  [abigail, brenda, mary, paula, tara],
    Girls =  [Name1, Name2, Name3, Name4, Name5],

    Months = [february, march, june, july, december],
    Days =   [sunday, monday, wednesday, friday, saturday],
    permutation(Months, [Month1, Month2, Month3, Month4, Month5]),
    permutation(Days,   [Day1, Day2, Day3, Day4, Day5]),

    % Paula was born in March but not on Saturday.
    member([paula, march, C1], S), C1 \= saturday,
   ...

'before in year' 的關系可以這樣編碼：

    ...
    % The girl whose birthday is on Monday was born
    % earlier in the year than Brenda and Mary.
    member([_, C3, monday], S),
    member([brenda, C4, C10], S), before_in_year(C3, C4, Months),
    member([mary, C5, _], S), before_in_year(C3, C5, Months),
    ...

與服務謂詞

before_in_year(X, Y, Months) :-
    nth1(Xi, Months, X),
    nth1(Yi, Months, Y),
    Xi < Yi.

“周末出生”可以這樣編碼

...
% Tara wasn't born in February, and
% her birthday was on the weekend.
member([tara, C6, C7], S), C6 \= february, (C7 = saturday ; C7 = sunday),

% Mary was not born in December nor was her
% birthday on a weekday.
member([mary, C8, C9], S), C8 \= december, (C9 = saturday ; C9 = sunday),
...

等等。 重寫后，我得到了獨特的解決方案

?- solve(X),maplist(writeln,X).
[abigail,february,monday]
[brenda,december,wednesday]
[mary,june,sunday]
[paula,march,friday]
[tara,july,saturday]
X = [[abigail, february, monday], [brenda, december, wednesday], [mary, june, sunday], [paula, march, friday], [tara, july, saturday]] ;
false.

編輯

我剛剛注意到我引入了一些冗余的 member/2 和自由變量，例如member([brenda, C4, C10], S),... 。 那些 C4、C10 obiouvsly 可以被綁定到 Brenda 的變量替換為 Month2、Day2，就像在原始代碼中一樣。

Answer 3

這是一個在問題空間上使用蠻力搜索的解決方案。 說我不為此感到自豪是遠遠不夠的。 當然，這個問題有一個更優雅的解決方案。

反正：

month(january).
month(february).
month(march).
month(april).
month(may).
month(june).
month(july).
month(august).
month(september).
month(october).
month(november).
month(december).

precedes(january, february).
precedes(february, march).
precedes(march, april).
precedes(april, may).
precedes(may, june).
precedes(june, july).
precedes(july, august).
precedes(august, september).
precedes(september, october).
precedes(october, november).
precedes(november, december).
earlier(M1, M2) :- precedes(M1, M2).
earlier(M1, M2) :- month(M1), month(M2), precedes(M1, X), month(X), earlier(X, M2).

weekday(monday).
weekday(tuesday).
weekday(wednesday).
weekday(thursday).
weekday(friday).
weekend(saturday).
weekend(sunday).

birthmonth(abigail, M) :- 
    month(M), 
    M \== march.
birthmonth(brenda, M) :- 
    month(M), 
    M \== march.
birthmonth(paula, march).
birthmonth(mary, M) :- 
    month(M), 
    M \== march, M \== december, M \== july.
birthmonth(tara, M) :- 
    month(M), 
    M \== march, 
    M \== february.

birthday(abigail, D) :- 
    weekday(D), 
    D \== friday, D \== wednesday.
birthday(brenda, D) :- 
    weekday(D), 
    D \== friday,
    D \== monday.
birthday(mary, D) :- weekend(D).
birthday(paula, D) :- weekday(D), D \==saturday.
birthday(tara, D) :- weekend(D).

answer(M, D):-
    candidate(M, D),
    member(june, M),
    member(sunday, D),
    nth(IM, M, june),
    nth(ID, D, sunday),
    IM =:= ID,
    nth(5, M, MTARA),
    nth(2, M, MBRENDA),
    earlier(MTARA, MBRENDA),
    nth(3, M, MMARY),
    nth(IMONDAY, D, monday),
    nth(IMONDAY, M, MMONDAY),
    earlier(MMONDAY, MBRENDA),
    earlier(MMONDAY, MMARY).


candidate([M1,M2,M3,M4,M5], [D1,D2,D3,D4,D5]):-
    birthday(abigail, D1),
    birthday(brenda, D2),
    D1 \== D2,
    birthday(mary, D3),
    D1 \== D3,
    D2 \== D3,
    birthday(paula, D4),
    D1 \== D4,
    D2 \== D4,
    D3 \== D4,
    birthday(tara, D5),
    D1 \== D5,
    D2 \== D5,
    D3 \== D5,
    D4 \== D5,
    birthmonth(abigail, M1), 
    birthmonth(brenda, M2), 
    M1 \== M2,
    birthmonth(mary, M3), 
    M1 \== M3, 
    M2 \== M3,
    birthmonth(paula, M4),
    M1 \== M4,
    M2 \== M4,
    M3 \== M4,
    birthmonth(tara, M5),
    M1 \== M5,   
    M2 \== M5,
    M3 \== M5,
    M4 \== M5.

更好的答案是將排序約束作為birthmonth/2或birthday/2子句的一部分。 到目前為止，我還無法讓它發揮作用。

candidate/2實現了相當於幾個嵌套for()循環的內容，您看不到這些循環，但 WAM（Prolog 的 Warren 抽象機）通過各種詭計來迭代值D1, D2, D3 ...等。

要查看可能的答案，請使用：

answer(M,D).

在 gprolog 中繼續按分號或“a”以查看所有答案。 每個列表的元素按字母順序對應於女孩。

Answer 4

獨特地- 從域中預先選擇所有實體允許輕松簡單，“既清晰又密集”的代碼。 使用數字域可以輕松進行比較：

day(   d(_,D,_), D).   
fname( d(N,_,_), N).   % first name
month( d(_,_,M), M).   

sistersP(X):-
    maplist( fname, X, ['Paula', 'Abigail', 'Brenda', 'Mary', 'Tara']),
    maplist( month, X, [PM, AM, BM, MM, TM]),
    maplist( day,   X, [PD, AD, BD, MD, TD]),
    permutation( [PM,AM,BM,MM,TM], [2,3,6,7,12]),            % months of year
    permutation( [PD,AD,BD,MD,TD], [sun,mon,wed,fri,sat]),   % days of week

    PM = 3, PD \== sat, AD \== fri, AD \== wed,              % the five rules,
    day(G,mon), member(G,X), month(G,GM), GM < BM, GM < MM,  %   one per line
    TM =\= 2, (TD == sat ; TD == sun),
    MM =\= 12, (MD == sat ; MD == sun), month(G2,6), day(G2,sun), member(G2,X),
    TM < BM, BD \== fri, MM =\= 7.

這僅找到一種解決方案，僅使用拼圖中提到的一年中的幾個月和一周中的幾天：

?- sistersP(X).
X = [d('Paula', fri, 3), d('Abigail', mon, 2), d('Brenda', wed, 12), 
     d('Mary', sun, 6), d('Tara', sat, 7)] ;
No

?- time( sistersP(_) ).
% 19,537 inferences, 0.01 CPU in 0.01 seconds (100% CPU, 2624221 Lips)
Yes

?- time( (sistersP(_),fail;true) ).  % exhaust the search space
% 56,664 inferences, 0.03 CPU in 0.04 seconds (75% CPU, 2441285 Lips)
Yes

盡快進行測試，增量選擇，會產生更高效的代碼。 我喜歡使用我自己的select/2 ，它讓我從域中唯一地選擇列表的元素（即另一個列表，允許比第一個列表長，因此不能使用permutation/2 ）。

select([A|As],S):- select(A,S,S1),select(As,S1).
select([],_). 

sisters(X):-
    maplist(fname, X, ['Paula', 'Abigail', 'Brenda', 'Mary', 'Tara']),
    maplist(month, X, [PM, AM, BM, MM, TM]),
    maplist(day,   X, [PD, AD, BD, MD, TD]),
    Months = [2,3,6,7,12],           %%% [1,2,3,4,5,6,7,8,9,10,11,12],
    Days = [sun,mon,wed,fri,sat],    %%% [sun,mon,tue,wed,thu,fri,sat], 

    select(3,Months,M2),  PM = 3, 
    select(PD,Days,D2),   PD \== sat,              % 1a
    select(AD,D2,D3),     AD \== fri, AD \== wed,  % 1b
    select(TM,M2,M3),     TM =\= 2,                % 3a
    select(MM,M3,M4),     MM =\= 12,  MM =\= 7,    % 4a1 % 5c
    select(TD,D3,D4),  select([TD,MD],[sat,sun]),  % 3b  % 4a2
    month(G,6), day(G,sun), member(G,X),           % 4b
    select([MD,BD],D4),   BD \== fri,              % 5a
    select([BM,AM],M4),   TM < BM,                 % 5b
    day(G2,mon),          member(G2,X),
    month(G2,G2M),        G2M < BM, G2M < MM.      % 2

運行：

?- sisters(X).
X = [d('Paula', fri, 3), d('Abigail', mon, 2), d('Brenda', wed, 12), 
     d('Mary', sun, 6), d('Tara', sat, 7)] ;
No

?- time(sisters(_)).
% 2,071 inferences, 0.00 CPU in 0.00 seconds (?% CPU, Infinite Lips)
Yes

?- time( (sisters(_),fail;true) ).  % exhaust the search space
% 2,450 inferences, 0.00 CPU in 0.00 seconds (?% CPU, Infinite Lips)
Yes

使用一年中的所有 12 個月和一周中的 7 天（我一開始這樣做，不幸的是:)），有 4561 個解決方案，第二個代碼足夠快地找到（0.16 秒，424,600 次推理）。 第一個代碼使用select/2而不是permutation/2 ，需要180,400,000 次推理和 75 秒才能產生第一個答案，而第二個更快的代碼在 0.01 秒內需要 19,400 個 infs。

Answer 5

在這種問題中，我喜歡按照拼圖的文本（適用於 SWI Prolog 6.3.0）：

week_end(Day) :-
    member(Day, [saturday, sunday]).

day(Day) :-
    member(Day, [monday, wednesday, friday, saturday, sunday]).

month(Month) :-
    member(Month, [february, march, june, july, december]).


before(M1, M2) :-
    nth0(I1, [february, march, june, july, december], M1),
    nth0(I2, [february, march, june, july, december], M2),
    I1 < I2.

names([person(abigail, _, _),
       person(brenda, _, _),
       person(mary, _, _),
       person(paula, _, _),
       person(tara, _, _)]).


solve(L) :-
    maplist(\X^(X = person(_, Day, Month),
            day(Day),
            month(Month)),
        L),

    forall((select(X,L, L1), select(Y, L1, _)),
           (   X = person(_, D1, M1),
           Y = person(_, D2, M2),
           D1 \= D2,
           M1 \= M2)).

/*
1.Paula was born in March but not on Saturday. Abigail's birthday was not on Friday or Wednesday.
*/
rule_1(L) :-
    member(person(paula, D, march), L),
        D \== saturday,

    member(person(abigail, D1, _M), L),
    day(D1),
    \+ member(D1, [friday, wednesday]).


/*
2.The girl whose birthday is on Monday was born earlier in the year than Brenda and Mary.
*/
rule_2(L) :-
    member(person(_N, monday, M), L),
    member(person(brenda, _D1, M1), L),
    member(person(mary, _D2, M2), L),
    before(M, M1),
    before(M, M2).

/*
3.Tara wasn't born in February and her birthday was on the weekend.
*/

rule_3(L) :-
    member(person(tara, D, M), L),
    M \== february,
    week_end(D).

/*
4.Mary was not born in December nor was her birthday on a weekday. The girl whose birthday was in June was born on Sunday.
*/

rule_4(L) :-
    member(person(mary, D, M), L),
    week_end(D),
    M \== december,
    member(person(_N, sunday, june), L).

/*
5.Tara was born before Brenda, whose birthday wasn't on Friday. Mary wasn't born in July.
*/

rule_5(L) :-
    member(person(tara, _DT, MT), L),
    member(person(brenda, DB, MB), L),
    before(MT, MB),
    % DB \== friday,
    day(DB),
    DB \= friday,    
    member(person(mary, _D, M), L),
    M \== july.



puzzle :-
    names(L),
    rule_1(L),
    rule_2(L),
    rule_3(L),
    rule_4(L),
    rule_5(L),
    solve(L),
    maplist(writeln, L).

我得到：

 ?- time(puzzle).
person(abigail,monday,february)
person(brenda,wednesday,december)
person(mary,sunday,june)
person(paula,friday,march)
person(tara,saturday,july)
% 45,144 inferences, 0.016 CPU in 0.031 seconds (50% CPU, 3294080 Lips)
true .

Answer 6

#clpfd 方法序言：-

:-use_module(library(clpfd)).
puzzle(Sisters,Months,Days):-
Sisters=[Paula, Brenda, Abigail, Mary, Tara], Sisters ins 1..5,
Months=[Feburary, March, June, July, December], Months ins 1..5,
Days=[Monday, Wednesday, Friday, Saturday, Sunday], Days ins 1..5,

Paula#=March,
Paula#\=Saturday,
Abigail#\=Friday #\/ Abigail #\=Wednesday,
Tara#\=Feburary #/\ (Tara#=Saturday #\/ Tara#=Sunday),
Mary#\=December #/\ (Mary#\=Saturday #\/ Mary#\=Sunday),
Tara#=Brenda-1,
Brenda#\=Friday,
Mary#\=July,
June#=Sunday,
Brenda #\=Monday #/\ Mary #\=Monday,

all_different(Sisters),
all_different(Months),
all_different(Days),

labeling([], Sisters), labeling([],Months), labeling([], Days).

?-puzzle(Sisters,Months,Days).
OUTPUT:
Days = [1, 3, 4, 2, 5],
Months = [3, 1, 5, 2, 4],
Sisters = [1, 3, 4, 5, 2]
Days = [4, 3, 1, 2, 5],
Months = [3, 1, 5, 2, 4],
Sisters = [1, 3, 4, 5, 2]
Days = [1, 3, 4, 2, 5],
Months = [3, 1, 5, 4, 2],
Sisters = [1, 3, 4, 5, 2]
......