Question

我的矩阵大小为[n,n]。

我需要找到总和例如

1  2  3  4 
5  6  7  8 
9  10 11 12
13 14 15 16

sum = 3+4+7+8

我需要找到第一象限矩阵的元素总和

Answer 1

使用library(clpfd)，其中提供有用的sum/3和transpose/2：

:- use_module(library(clpfd)).

sum_first_quadrant(M, S) :-
    first_quadrant(M, Q),
    maplist(sum_, Q, Ss),
    sum_(Ss, S).

sum_(L, S) :-
    sum(L, #=, S).

first_quadrant(M, Q) :-
    transpose(M, T),
    reverse(T, RT),
    dichotomize(RT, RD),
    reverse(RD, D),
    transpose(D, TD),
    dichotomize(TD, Q).

dichotomize(M, D) :-
    length(M, L),
    X #= L//2,
    dichotomize_(M, X, D).

dichotomize_(_, 0, []).
dichotomize_([H|T], X, [H|T2]) :-
    X #> 0,
    Y #= X - 1,
    dichotomize_(T, Y, T2).

示例：

?- sum_first_quadrant([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]], Z).
Z = 22 ;
false.

注意

您可以使用library reif中的dichotomize_和if_/3摆脱(=)/3中无关的选择点：

dichotomize_(L, X, D) :-
    X #>= 0,
    if_(X = 0,
        D = [],
        (   Y #= X - 1,
            L = [H|T],
            D = [H|T2],
            dichotomize_(T, Y, T2)
        )
    ).

Answer 2

*> you may get more performance if you `REDEFINE` source-text as a `PIC X OCCURS length-of-text TIMES` - but I find this one more better to read and it shouldn't consume much more time...

MOVE 0 TO target-pointer
PERFORM VARYING source-pointer
        FROM 1 BY 1
        UNTIL   source-pointer > length-of-text
   IF source-text (source-pointer:1) = '#'
      *> a very good optimizer would calculate the constant on the
      *> right side, you may write it directly
      IF source-pointer + 4 > length-of-text
         ADD  1 TO target-pointer
         MOVE source-text (source-pointer:)
           TO target-text (target-pointer:)
         EXIT PERFORM
      END-IF
      IF source-text (source-pointer:4) = '#RT#' OR '#LT#' OR ...
         ADD 4 TO source-pointer
         EXIT PERFORM CYCLE
      END-IF
   END-IF
   ADD  1 TO target-pointer
   MOVE source-text (source-pointer:1)
     TO target-text (target-pointer:1)
END-PERFORM

Prolog如何找到元素矩阵之和

2 个答案:

注意