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我正在尝试解决 Prolog 中的一个难题，该难题涉及取一个数字的平方（数字列表的列表）并返回从顶部开始并逐行向下的最大数字组合列表。每次移动都必须向下、向右向下或向左向下。

我一直在尝试这样做一段时间，有人有我可以开始的地方吗？

例如，在板上

 [[0, 2, 1, 0],
  [0, 1, 1, 0],
  [0,10,20,30]]

最好的举动是[1, 2, 3]33分。

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3 回答 3

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所以这就是你如何做到的。我知道这有点罗嗦，那可能是因为我对 Prolog 也不是很流利......

% Lookup a value in a list by it's index.
% this should be built into prolog?
at(0, [H|_], H).
at(N, [_|T], X) :- 
    N > 0,
    N1 is N - 1,
    at(N1, T, X).

% like Haskell's maximumBy; takes a predicate, a 
% list and an initial maximum value, finds the
% maximum value in a list
maxby(_, [], M, M).
maxby(P, [H|T], M0, M) :-
    call(P, H, M0, M1),
    maxby(P, T, M1, M).

% which of two paths has the bigger score?
maxval(path(C,  I), path(C1, _), path(C, I)) :- C >= C1.
maxval(path(C0, _), path(C, I),  path(C, I)) :- C0 < C.

% generate N empty paths as a starting value for
% our search
initpaths(N, Ps) :- 
    findall(path(0, []), 
            between(0, N, _), 
        Ps). 

% given the known best paths to all indexes in the previous
% line and and index I in the current line, select the best
% path leading to I.
select(Ps, I, N, P) :-
    I0 is I-1,
    I1 is I+1,
    select(Ps, I0, N, path(-1, []), P0),
    select(Ps, I,  N, P0, P1),
    select(Ps, I1, N, P1, P).

% given the known best paths to the previous line (Ps),
% an index I and a preliminary choice P0, select the path 
% leading to the index I (in the previous line) if I is within 
% the range 0..N and its score is greater than the preliminary 
% choice. Stay with the latter otherwise.
select(_, I, _, P0, P0) :- I < 0.
select(_, I, N, P0, P0) :- I > N.
select(Ps, I, _, P0, P) :- 
    at(I, Ps, P1),
    maxby(maxval, [P0], P1, P).

% given the known best paths to the previous line (P1),
% and a Row, which is the current line, extend P1 to a 
% new list of paths P indicating the best paths to the
% current line.
update(P1, P, Row, N) :-
    findall(path(C, [X|Is]), 
            ( between(0, N, X)
            , select(P1, X, N, path(C0, Is))
            , at(X, Row, C1)
            , C is C0 + C1),
        P).

% solve the puzzle by starting with a list of empty paths
% and updating it as long as there are still more rows in 
% the square.
solve(Rows, Score, Path) :-
    Rows = [R|_],
    length(R, N0),
    N is N0 - 1,
    initpaths(N, IP),
    solve(N, Rows, IP, Score, Path).
solve(_, [], P, Score, Path) :- 
    maxby(maxval, P, path(-1, []), path(Score, Is0)),
    reverse(Is0, Path).
solve(N, [R|Rows], P0, Score, Path) :-
    update(P0, P1, R, N),
    solve(N, Rows, P1, Score, Path).

我们试试看？以下是您的示例：

?- solve([[0,2,1,0], [0,1,1,0], [0,10,20,30]], Score, Path).
Score = 33,
Path = [1, 2, 3] ;
false.

?- solve([[0,1,1], [0,2,1], [10,0,0]], Score, Path).
Score = 13,
Path = [1, 1, 0] ;
false.
于 2012-04-04T13:17:54.923 回答
0

我的序言有点不稳定。事实上，关于 prolog，我只记得它是声明性的。

这是一些用于查找最大路径值的haskell代码。找到跟踪应该是一个简单的下一步，但我想编码起来有点复杂。我想一个非常优雅的跟踪解决方案是使用单子。

maxValue :: [ [ Int ] ] -> Int
maxValue p = maximum $ maxValueHelper p
maxValueHelper :: [ [ Int ] ] -> [ Int ]
maxValueHelper [ row ] = row
maxValueHelper ( row : restOfRows ) = combine row ( maxValueHelper restOfRows )
combine :: [ Int ] -> [ Int ]-> [ Int ]
combine [ x ] [ y ] = [ x + y ]
combine ( x1 : x2 : lx ) ( y1 : y2 : ly ) =
   let ( z2 : lz ) = combine ( x2 : lx ) ( y2 : ly )
   in
   ( max ( x1 + y1 ) ( x1 + y2 ) : max ( x2 + y1 ) z2 : lz )

main :: IO()
main = print $ maxValue [[0,2,1,0], [0,1,1,0], [0,10,20,30]]
于 2012-04-03T21:49:23.817 回答
0 
?- best_path_score([[0, 2, 1, 0],[0, 1, 1, 0],[0,10,20,30]], P, S).
P = [1, 2, 3],
S = 33.

有了这个定义

best_path_score(Rs, BestPath, BestScore) :-
    aggregate_all(max(Score, Path), a_path(Rs, Path, Score), max(BestScore, BestPath)).

a_path([R|Rs], [P|Ps], Score) :-
    nth0(P, R, S0),
    a_path(Rs, P, Ps, S),
    Score is S0 + S.

a_path([], _, [], 0).
a_path([R|Rs], P, [Q|Qs], T) :-
    ( Q is P - 1 ; Q is P ; Q is P + 1 ),
    nth0(Q, R, S0),
    a_path(Rs, Q, Qs, S),
    T is S0 + S.
于 2013-02-21T16:51:31.760 回答