在Matlab中将值分配给索引子矩阵的子矩阵

问题描述 投票:0回答:2

我不确定我是否习惯纠正标题中的措辞来描述问题。请随意编辑它以反映下面的描述。

假设我有一个数独求解器程序,让我们说输入矩阵如下,

A = randi(10,[9,9])-1;

我将3x3子矩阵从1列到9列进行索引。假设代表此索引的变量nSubMat可以取1到9之间的任何值。

我用以下方式索引子矩阵,

SubMat(nSubMat) = A((1:3)+(3*floor((nSubMat-1)/3)),(1:3)+(3*mod(nSubMat-1,3)));

现在,我想访问和修改SubMat的(2x3)位置的值,而不必首先创建SubMat(比如说避免不必要的副本)。

详细说明,如果我有一个函数submatrix()来实现上述,我的语句将如下所示,

submatrix(A((1:3)+(3*floor((nSubMat-1)/3)),(1:3)+(3*mod(nSubMat-1,3))),[2,3]) = 5;

甚至,

submatrix(A((1:3)+(3*floor((nSubMat-1)/3)),(1:3)+(3*mod(nSubMat-1,3))),[2:3,2:3]) = [1 2;3 4];

我知道Matlab解释器会自动优化LHS = RHS类型的速度分配,但上述矩阵操作的重要性更多(算法上),而不仅仅是减少副本和加速代码,我不会在这里讨论。我在一个名为Armadillo的C ++库中看到了所需的语法,但我不确定是否可以用MATLAB完成相同的操作。

arrays matlab submatrix
2个回答
1
投票

您可以使用简单的线性索引来完成此操作。以下代码不言自明。

matrixRows=9;
matrixCols=9;
blockRows=3;
blockCols=3;
accessRow=2;
accessCol=3;

A = randi(10,[matrixRows,matrixCols])-1;
allPos=allcomb(accessRow:blockRows:matrixRows,accessCol:blockCols:matrixCols);
linPos=sub2ind(size(A),allPos(:,1),allPos(:,2));

% access them as usual and put any value
A(linPos)=-100;

结果:

A =

 8     9     7     3     6     4     1     6     8
 9     1     9     6     3     4     4     8     2
 1     9     6     1     9     6     9     9     9
 9     9     0     7     0     7     3     5     3
 6     4     8     0     4     7     5     1     1
 0     8     9     2     3     2     2     1     2
 2     1     6     0     7     6     7     2     6
 5     4     7     0     7     6     2     8     4
 9     9     7     8     1     1     5     2     3

运行上面的代码后:

A =

 8     9     7     3     6     4     1     6     8
 9     1  -100     6     3  -100     4     8  -100
 1     9     6     1     9     6     9     9     9
 9     9     0     7     0     7     3     5     3
 6     4  -100     0     4  -100     5     1  -100
 0     8     9     2     3     2     2     1     2
 2     1     6     0     7     6     7     2     6
 5     4  -100     0     7  -100     2     8  -100
 9     9     7     8     1     1     5     2     3

注意:allcomb生成输入参数的所有可能组合。你也可以使用比this更快的allcomb(根据答案)。


1
投票

这是任何大小数独谜题的一般功能。

function index = SudukoIndex(varargin)
%%%SudukoIndex provides the index or indicies of a sub-block of any n*n
%%%Suduko puzzle

%Possible inputs
%   One (1) number (i) between 1 and n will provide a sqrt(n) * sqrt(n) set of
%      indicies from the i-th block. Note this is counted using Matlab
%      syntax going from top to bottom then left to right, a simple check
%      for this can be found by typing the command 'reshape(1:n, sqrt(n),
%      sqrt(n))' into the command window.
%   Two (2) numbers between 1 and n will provide the single number index of
%       the row, column combination
%   Three (3) numbers the first (i) between 1 and n and the second (j) and 
%       third (k) between 1 and sqrt(n) will provide the index of the (j,k)
%       point in the i-th cell
n = 9;

if nargin == 1
    sM = varargin{1};
    majorColumn = floor((sM-1)/sqrt(n)) + 1;
    majorRow = mod(sM-1, sqrt(n)) + 1;

    x = (1:sqrt(n))';
    y = (1:n:n^1.5);
    [x, y] = meshgrid(x, y);
    m = (x + y - 1)';

    index = (n^1.5)*(majorColumn - 1) + sqrt(n)*(majorRow - 1) + m;
elseif nargin == 2
    index = n*(varargin{2} - 1) + varargin{1};
elseif nargin == 3
    m = nsm(varargin{1});
    index = m(varargin{2}, varargin{3});
end
end
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