Assigning values to a sub-matrix of a indexed sub-matrix in Matlab

Question

I'm not sure if I've used to correct wording in the Title to describe the problem. Please do feel free to edit it to reflect the description below.

Suppose I have a sudoku solver program and lets say the input matrix is the following,

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

I index the 3x3 sub-matrices columnwise from 1 to 9. Let's say the variable nSubMat representing this index can take any values between 1 to 9.

I index the submatrices in the following way,

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

Now, I want to access and modify the value in the (2x3) position of SubMat without having to create SubMat in the first place(say to avoid unnecessary copies).

To elaborate, if I were to have a function submatrix() which would implement the above, my statement would look something like the following,

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

or even,

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

I know that Matlab interpreter automatically optimizes LHS=RHS type assignments for speed, but the above matrix operation is important for more reasons (algorithmically) than just reducing copies and speeding up the code which I will not dwell into here. I have seen the required syntax in a C++ library called Armadillo, but I'm not sure if the same can be done with MATLAB.

Did you find any of the answer useful? If yes, then please accept it so that it gets off the unanswered list. — Autonomous, May 31 '15 at 21:28

score 1 · Accepted Answer · edited May 23 '17 at 12:30

You could do this using simple linear indexing. The following code is self-explanatory.

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;

Result:

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

After running above code:

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

Note: allcomb generates all possible combinations of the input arguments. You can also use this which is faster than allcomb (according to the answer).

An armadillo [submat views](http://arma.sourceforge.net/docs.html#submat) type access is what I was looking for, but linear indexing should do. Also, how efficient is allcomb() for large square matrices, larger access region or large number of iterations in sudoku? — Naveen, May 31 '15 at 23:29
I think it will be pretty efficient. As I mentioned, you can also test out the alternative which is apparently faster. — Autonomous, Jun 01 '15 at 03:12

user1543042 · Answer 2 · 2015-05-31T05:05:41.303

This is a general function for any size Sudoku puzzle.

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

Assigning values to a sub-matrix of a indexed sub-matrix in Matlab

2 Answers2