Prolog 翻译成DCG文本不起作用

因为它使用列表，所以我想使用DCG解决它。在这个过程中，我意识到可以使用文本。（）

最初的问题是返回列表中项目的计数，但如果两个相同的项目相邻，则不增加计数

虽然我的代码对某些测试用例有效，但对其他测试用例无效。这只是一个失败的条款。在使用调试器查看代码时，第二个状态变量returned list似乎是在调用子句时绑定的，而我认为它应该被解除绑定编辑解决了下面这部分问题

我正在使用SWI Prolog 8.0

引起问题的条款：

count_dcg(N0,N),[C2] -->
    [C1,C2],
    { N is N0 + 1 }.

注意：C1被标记为

单一变量：[C1]

通常，我会将

C1

更改为

。

，但在这种情况下，我需要指出当前正在处理的第一个和第二个项目是不同的。换言之，它是以统一失败为挡箭牌

使用清单/1查看DCG会发现使用

\uuu

，这可能是问题所在，但不确定

count_dcg(C, B, A, E) :-
    A=[_, F|D],
    B is C+1,
    G=D,
    E=[F|G].

使用DCG解决问题的正确方法是什么

见下文

---

当前源代码

% empty list
% No semicontext (push back) needed because last item in list.
count_dcg(N,N) --> [].

% single item in list
% No semicontext (push back) needed because only one item removed from list.
count_dcg(N0,N) -->
    [_],
    \+ [_],
    { N is N0 + 1 }.

% Semicontext (push back) needed because two items removed from list.
% Need second item to stay on list.
count_dcg(N,N),[C] -->
    [C,C].

% Semicontext (push back) needed because two items removed from list.
% Need second item to stay on list.
count_dcg(N0,N),[C2] -->
    [C1,C2],
    { N is N0 + 1 }.

count(L,N) :-
    DCG = count_dcg(0,N),
    phrase(DCG,L).

---

测试用例

:- begin_tests(count).

test(1,[nondet]) :-
    count([],N),
    assertion( N == 0 ).

test(2,[nondet]) :-
    count([a],N),
    assertion( N == 1 ).

test(3,[nondet]) :-
    count([a,a],N),
    assertion( N == 1 ).

test(4,[nondet]) :-
    count([a,b],N),
    assertion( N == 2 ).

test(5,[nondet]) :-
    count([b,a],N),
    assertion( N == 2 ).

test(6,[nondet]) :-
    count([a,a,b],N),
    assertion( N == 2 ).

test(7,[nondet]) :-
    count([a,b,a],N),
    assertion( N == 3 ).

test(8,[nondet]) :-
    count([b,a,a],N),
    assertion( N == 2 ).

:- end_tests(count).

---

测试运行

?- run_tests.
% PL-Unit: count ..
ERROR: c:/question_110.pl:80:
        test 3: failed

ERROR: c:/question_110.pl:84:
        test 4: failed

ERROR: c:/question_110.pl:88:
        test 5: failed

ERROR: c:/question_110.pl:92:
        test 6: failed

ERROR: c:/question_110.pl:96:
        test 7: failed

ERROR: c:/question_110.pl:100:
        test 8: failed

 done
% 6 tests failed
% 2 tests passed
false.

?- run_tests.
% PL-Unit: count ........ done
% All 8 tests passed
true.

---

编辑1

意识到需要对其中两个谓词进行尾部调用

% Semicontext (push back) needed because two items removed from list.
% Need second item to stay on list.
count_dcg(N0,N),[C] -->
    [C,C],
    count_dcg(N0,N).

% Semicontext (push back) needed because two items removed from list.
% Need second item to stay on list.
count_dcg(N0,N),[C2] -->
    [C1,C2],
    { 
        C1 \== C2,
        N1 is N0 + 1 
    },
    count_dcg(N1,N).

代码仍然不起作用，但这解释了为什么状态变量被绑定，而我期望它被解除绑定

---

编辑2

虽然没有像我所希望的那样使用DCG Semicondext，而是使用Semicondext的一个变体作为前瞻，但代码仍然有效。不将此作为答案发布，因为我希望答案要么显示DCG代码与子句标题上的文本一起工作，要么解释为什么这是错误的

lookahead(C),[C] -->
    [C].

% empty list
% No lookahead needed because last item in list.
count_3_dcg(N,N) --> [].

% single item in list
% No lookahead  needed because only one in list.
count_3_dcg(N0,N) -->
    [_],
    \+ [_],
    { N is N0 + 1 }.

% Lookahead needed because two items in list and
% only want to remove first item.
count_3_dcg(N0,N) -->
    [C1],
    lookahead(C2),
    { C1 == C2 },
    count_3_dcg(N0,N).

% Lookahead needed because two items in list and
% only want to remove first item.
count_3_dcg(N0,N) -->
    [C1],
    lookahead(C2),
    {
        C1 \== C2,
        N1 is N0 + 1
    },
    count_3_dcg(N1,N).

count(L,N) :-
    DCG = count_3_dcg(0,N),
    phrase(DCG,L).

测试运行

?- run_tests.
% PL-Unit: count ..
ERROR: c:/question_110.pl:80:
        test 3: failed

ERROR: c:/question_110.pl:84:
        test 4: failed

ERROR: c:/question_110.pl:88:
        test 5: failed

ERROR: c:/question_110.pl:92:
        test 6: failed

ERROR: c:/question_110.pl:96:
        test 7: failed

ERROR: c:/question_110.pl:100:
        test 8: failed

 done
% 6 tests failed
% 2 tests passed
false.

?- run_tests.
% PL-Unit: count ........ done
% All 8 tests passed
true.

---

不需要推送列表或前瞻的替代解决方案：

count_dcg(N0,N) -->
    [C], {N1 is N0 + 1}, count_dcg(N1,N,C).
count_dcg(N,N) -->
    [].

count_dcg(N0,N,C) -->
    [C],
    count_dcg(N0,N,C).
count_dcg(N0,N,C) -->
    [C1],
    {C \== C1, N1 is N0 + 1},
    count_dcg(N1,N,C1).
count_dcg(N,N,_) -->
    [].

count(L,N) :-
    phrase(count_dcg(0,N),L).

---

self注意：有

count\u dcg/2

和

count\u dcg/3

@GuyCoder：通常，在同一谓词或非谓词的子句之间避免空行-terminal@false谢谢，我一直在修改我的代码来实现这一点。这个评论是因为起初我没有看到参数数量上的差异。有一次我看到对代码的理解是显而易见的。