2

作为对提出问题的这个问题的后续行动

返回列表中项目的计数，但如果两个相同的项目彼此相邻，则不要增加计数。

这段代码是我最接近用 DCG 和 semicontext 解决这个问题的。

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

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

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

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

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

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

在子句头部使用带有半上下文的 DCG 解决问题的正确方法是什么？

想知道从句头上半上下文的变化是否可能。如果可能，则需要工作示例代码，如果不可能，则需要解释。

4

2 回答 2

1

怎么样：

:-use_module(library(clpfd)).

list([])     --> [].
list([L|Ls]) --> [L], list(Ls).

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

count_dcg(N,N) --> [].
count_dcg(N0,N) --> %last item.
    [_],
    list(R),
    {R = [], N #=N0+1}.
count_dcg(N0,N) -->
    [C1],
    lookahead(C1),
    count_dcg(N0,N).
count_dcg(N0,N) -->
    [C1],
    lookahead(C2),
    {
        dif(C1,C2),
        N1 #= N0 + 1
    },
    count_dcg(N1,N).

count(L,N) :-
    phrase(count_dcg(0,N),L).
于 2019-03-04T17:52:46.513 回答
1

我认为这是正确使用半上下文符号。我正在使用0,s(0),...

% Constraint Logic Programming
:- use_module(library(dif)).    % Sound inequality
:- use_module(library(clpfd)).  % Finite domain constraints

list([])     --> [].
list([L|Ls]) --> [L], list(Ls).

state(S), [state(S)] --> [state(S)].
state(S, s(S)), [state(s(S))] --> [state(S)].

keep_state(S,I),[state(S)] --> [state(S)],[I].
end_state(S)  -->[state(S)],[].

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

count_dcg(S,S) --> 
    state(S), %might not need this
    end_state(S).

/* Can be used get the length of a list
count_dcg(S,S2) --> 
    state(S,S1),
    keep_state(S1,_),
    count_dcg(S1,S2),
    {}.
*/

%last item.
count_dcg(S,S1) -->     
    state(S,S1),
    keep_state(S1,_C),
    list(R),
    {R = [state(_)]}.

%Two the same dont increase state 
count_dcg(S,S1) -->
    state(S), %might not need this
    keep_state(S,C1),
    lookahead(C1),
    count_dcg(S,S1).

%Two different increase state  
count_dcg(S,S2)  -->
    state(S,S1),
    keep_state(S1,C1),
    lookahead(C2),
    {
        dif(C1,C2)
    },
    count_dcg(S1,S2).

count(L,S) :-
    phrase(count_dcg(0,S),[state(0)|L]).

对于以下情况，这并不像我希望的那样有效：

65 ?- count([a,b,X,c],L).
X = b,
L = s(s(s(0))) ;
;
X = c,
L = s(s(s(0))) .

您可以使用以下方式转换 peano：

natsx_int(0, 0). 
natsx_int(s(N), I1) :- 
  I1 #> 0, 
  I2 #= I1 - 1, 
  natsx_int(N, I2).

或者您可以更改状态谓词：

state(S), [state(S)] --> [state(S)].
state(S, S2), [state(S2)] --> [state(S)],{S2#=S+1}.
于 2019-03-05T10:35:10.847 回答