离散连续数学方程

Question

我想实现一个代码，对大小为 (bs x bs x bs) 的立方体内的信息进行洗牌，这样立方体 (x,y,z) 内的每个元素都唯一地映射到 (x_new, y_new,z ）。但是代码不能正常工作。

所附图像是我尝试离散化和实现的代码的实际数学定义。原始数学方程适用于单位立方体，而我有一个维度为 bs 的立方体。有人可以看看我的代码并提示假定的错误吗？

n=bs;
NEWI=zeros(bs,bs,bs);
for row= 1:bs
    for col=1:bs 
        for height=1:bs
            if (     1<=row && row<=(n/2) && 1<=col && col<=(n/2) )
                x_new= 2*(row-1) + 1;
                y_new= 2*(col-1) + 1;
                 z=floor(0.25*(height-mod(height-1,2)))+1; 
             end
for 
for
for

Now this is just the first line implementation of the equation given in the figure. But as we see the coorepondence of points 
             (1,1,1) goes to (1,1,1)
              (1,1,2) goes to (1,1,1)
              (1,1,3) goes to  (1,1,1)
               (1,1,4) goes to (1,1,1)
which clearly is not a unique mapping, whereas the function claims of giving unique images for every (x,y,z). So my question is clearly there has to be some adjustments to discretize this map. Can somebody suggest

Answer 1

您的某些情况尚不清楚：

• 如果行总是>=1，为什么要检查1<=row？
• 为什么要使用模数？ （正如@Welbog 所指出的）
• ...

但要回答您的问题，我建议您使用逻辑索引和meshgrid，这将使您的代码更具可读性。您将能够在一行中完成所有操作。

% size of your 3D cube
n = 5;
% With meshgrid we generate the 3D grid that we will use to create the logical index.
[x,y,z] = meshgrid(1:n,1:n,1:n);
% Create empty array
M = zeros(n,n,n);
%first condition
I      = x<(n/2) & y<(n/2);        %create a logical index according to the first condition
M(I)   = 2*x(I) + 2*y(I) + z(I)/4; %generate the associated value
%second condition
I      = ...