Question

所以我的挑战是制作一个MxM零点矩阵，但其中大约15％的值为1。所有这些1的“总和”需要尽可能接近15％，同时也使它成为一个另一个函数的全局变量，但由于某种原因，它有时会陷入无限循环。有谁知道为什么会这样？

function [ board ] = randomking1( a,b )
clc
global sum
sum = 0; %initalizes sum to zero

kings = ceil(0.15*a*b); %finds number of kings needed for board size

 board = randi([0,1],a,b); %creates game board of random zeros and ones.
                           % ones being the kings on the board.


 for I = 1:a
        for J = 1:b

             if board(I,J) == 1
                     sum = sum + 1;  %this set of for loops counts the
                                     %number of kings on the board
             end

          end
 end


  while sum > kings || sum < kings-1 %if number of kings on the board is greater than
                       %number needed, the number of ones is reduced. 
       for I = 1:a
         for J = 1:b
               if sum

               if  board(I,J) == 1 %if a board value =1

                      board(I,J) = randi([0,1],1) %randomize 0 or 1

                      if board(I,J) == 0 %if the value becomes zero, subtract from the sum
                   sum = sum - 1
                      end
               end

           end
         end

       end

  disp(sum)
end

Answer 1

我建议不要试图使用你的暴力方法找到解决方案，而是建议在董事会中放置15％的国王，然后将其随机化。

function [ board ] = randomking1( a,b )

board_size = a*b;
% finds number of kings needed for board size
kings = ceil(0.15*board_size); 

% creates game board (as a linear vector) of ones and zeros
% ones being the kings on the board.
board = [ones(1,kings), zeros(1, board_size-kings)];

% shuffle the board
board = board(randperm(board_size));

% create the a x b matrix out of the linear vector
board = reshape(board,[a,b]);
end

示例运行：randomking1(3,3)：

board_size =

     9


kings =

     2


board =

     1     1     0     0     0     0     0     0     0


board =

     0     1     0     0     0     1     0     0     0


board =

     0     0     0
     1     0     0
     0     1     0

而循环无限循环

1 个答案: