Question

我一直在学习Matlab的业余爱好。这是我在堆栈溢出中的第二个问题。

这些天，我对Matlab的Sudoku感兴趣。我从互联网上获取了代码，并且正在研究此代码3星期。

我想添加表示“子功能（候选）”执行次数的功能。 ->函数可用于解决方案的次数。

我曾经使用cnt = 0和cnt = cnt + 1进行计数（在if语句和while语句下）但我意识到这段代码的结构是loop（??） *我尝试过，但尝试时将cnt重置了，我想我知道重置的原因但我不能写代码

谢谢您的帮助

function X = s2(X) 

% SUDOKU  Solve Sudoku using recursive backtracking. 
%   sudoku(X), expects a 9-by-9 array X. 
 % Fill in all ?singletons?. 
 % C is a cell array of candidate vectors for each cell. 
 % s is the first cell, if any, with one candidate. 
 % e is the first cell, if any, with no candidates. 
 [C,s,e] = candidates(X); 
 while ~isempty(s) && isempty(e) 
    X(s) = C{s}; 
    [C,s,e] = candidates(X); 
 end 
 % Return for impossible puzzles. 
 if ~isempty(e)

    return 
 end 
 % Recursive backtracking. 
 if any(X(:) == 0)


    Y = X; 
    z = find(X(:) == 0,1);    % The first unfilled cell. 
    for r = [C{z}]            % Iterate over candidates. 
       X = Y; 
       X(z) = r;              % Insert a tentative value. 
       X = s2(X);         % Recursive call. 
       if all(X(:) > 0)       % Found a solution. 
          return 
       end 
    end 

   end 
% ??????????????? 
   function [C,s,e] = candidates(X) 


      C = cell(9,9); 
      tri = @(k) 3*ceil(k/3-1) + (1:3); 
      for j = 1:9 
         for i = 1:9 
            if X(i,j)==0 
               z = 1:9; 
               z(nonzeros(X(i,:))) = 0; 
               z(nonzeros(X(:,j))) = 0; 
               z(nonzeros(X(tri(i),tri(j)))) = 0; 
               C{i,j} = nonzeros(z)'; 
            end 
         end 
      end 
 L = cellfun(@length,C);   % Number of candidates. 
 s = find(X==0 & L==1,1); 
 e = find(X==0 & L==0,1); 

 end % candidates 
end % s2

我使用了变量X

X=[4 0  0   0   2   0   0   0   0;
0   1   0   3   0   0   5   0   0;
0   0   9   0   0   8   0   6   0;
7   0   0   6   0   0   1   0   0;
0   2   0   0   0   0   0   0   9;
0   0   3   0   0   4   0   0   0;
6   0   0   7   0   0   0   2   0;
0   8   0   0   1   0   0   0   4;
0   0   0   0   0   9   3   0   0];

Answer 1

您还可以使用一个持久变量，该变量应在函数的多次调用中保留其值。

例如：

function [ result ] = myfactorial( num )
    % using persistent variable based on 
    % https://www.mathworks.com/help/matlab/ref/persistent.html
    persistent numcalls;
    if isempty(numcalls)
        numcalls = 0;
    end
    numcalls = numcalls+1;
    disp(strcat('numcalls is now: ', int2str(numcalls)));
    % factorial function based on
    % https://www.quora.com/How-can-I-create-a-factorial-function-in-MATLAB
    if num > 1
        result = myfactorial(num-1)*num;
    else
        result = 1;
    end
end

现在调用该函数。

clear all;
result = myfactorial(5);

输出为：
numcalls现在是：1
numcalls现在是：2
numcalls现在是：3
numcalls现在是：4
numcalls现在是：5

如何计算此代码中执行了多少个函数？

1 个答案: