Block sub diagonals matrix

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剑豪戴 2022-2-20

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1654045-block-sub-diagonals-matrix

编辑：剑豪戴 2022-2-21

采纳的回答： DGM

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I have known how to build a block main diagonals matrix. But I also need to fill the sub diagonals.

For example,

A=rand(3,3,5);
B=rand(3,3,5);
C=rand(3,3,5);

and I want to build a matrix D like,

D=[A(:,:,1),B(:,:,1),0,...,0
   C(:,:,1),A(:,:,2),B(:,:,2),...,0
   C(:,:,2),A(:,:,3),B(:,:,3),...,0
   ...
    A(:,:,5)]

D is 15×15

And A(:,:,i) is main diagonal, B and C is sub diagonals.

Kindly help me with this.

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Answer 1

DGM 2022-2-20

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If my interpretation is correct, this should be one method:

bksize = [2 2]; % smaller for example
A = ones(bksize);
B = 11*A;
C = 111*A;
blocks = {zeros(size(A)) A B C}; % this makes the blocks indexable
map = toeplitz([2 4 1 1 1],[2 3 1 1 1]); % a map of the block locations
D = cell2mat(reshape(blocks(map),size(map))) % the block array
D = 10×10
     1     1    11    11     0     0     0     0     0     0
     1     1    11    11     0     0     0     0     0     0
   111   111     1     1    11    11     0     0     0     0
   111   111     1     1    11    11     0     0     0     0
     0     0   111   111     1     1    11    11     0     0
     0     0   111   111     1     1    11    11     0     0
     0     0     0     0   111   111     1     1    11    11
     0     0     0     0   111   111     1     1    11    11
     0     0     0     0     0     0   111   111     1     1
     0     0     0     0     0     0   111   111     1     1

If the blocks are 3D, that can be done without changing anything.

bksize = [2 2 3]; % 3D blocks
A = ones(bksize);
B = 11*A;
C = 111*A;
blocks = {zeros(size(A)) A B C};
map = toeplitz([2 4 1 1 1],[2 3 1 1 1]);
D = cell2mat(reshape(blocks(map),size(map)))
D = 
D(:,:,1) =

   1    11    11     0     0     0     0     0     0
   1    11    11     0     0     0     0     0     0
 111     1     1    11    11     0     0     0     0
 111     1     1    11    11     0     0     0     0
   0   111   111     1     1    11    11     0     0
   0   111   111     1     1    11    11     0     0
   0     0     0   111   111     1     1    11    11
   0     0     0   111   111     1     1    11    11
   0     0     0     0     0   111   111     1     1
   0     0     0     0     0   111   111     1     1


D(:,:,2) =

   1    11    11     0     0     0     0     0     0
   1    11    11     0     0     0     0     0     0
 111     1     1    11    11     0     0     0     0
 111     1     1    11    11     0     0     0     0
   0   111   111     1     1    11    11     0     0
   0   111   111     1     1    11    11     0     0
   0     0     0   111   111     1     1    11    11
   0     0     0   111   111     1     1    11    11
   0     0     0     0     0   111   111     1     1
   0     0     0     0     0   111   111     1     1


D(:,:,3) =

   1    11    11     0     0     0     0     0     0
   1    11    11     0     0     0     0     0     0
 111     1     1    11    11     0     0     0     0
 111     1     1    11    11     0     0     0     0
   0   111   111     1     1    11    11     0     0
   0   111   111     1     1    11    11     0     0
   0     0     0   111   111     1     1    11    11
   0     0     0   111   111     1     1    11    11
   0     0     0     0     0   111   111     1     1
   0     0     0     0     0   111   111     1     1

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剑豪戴 2022-2-20

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Thanks! Maybe my description of problem is not clear and correct. What I want is:

For 3D matrix A(3,3,5), A(:,:,i) is different from A(:,:,j). And the main diagonal of D will be [A(:,:,1), A(:,:,2), ... ,A(:,:,5)].

The dimension of D is (3*5)×(3*5)×1.

But I follow your interpretation, it work right for me! Thanks!

A = rand(2,2,3);
B = rand(2,2,3);
C = rand(2,2,3);
[k,l,m] = size(A);
A = mat2cell(A,k,l,ones(1,m));
B = mat2cell(B,k,l,ones(1,m));
C = mat2cell(C,k,l,ones(1,m));
blocks = [{zeros(k,l)},A(:)',B(:)',C(:)'];
map = diag(1:m,0)+diag(m+1:(2*m-1),1)+diag((2*m+2):3*m,-1)+1;
D = cell2mat(reshape(blocks(map),size(map)))
D = 6×6
    0.3994    0.3321    0.2983    0.3994         0         0
    0.8713    0.4807    0.3125    0.3380         0         0
    0.6651    0.4984    0.8404    0.2329    0.4791    0.0781
    0.4750    0.6980    0.2003    0.1860    0.3050    0.0632
         0         0    0.1076    0.2159    0.1666    0.3028
         0         0    0.6675    0.6244    0.7260    0.2880

But I worried this is inefficient. Cause the acutal dimension of matrix in my code is 1000×1000. I wonder if it could use some techniques of sparse matrix.

DGM 2022-2-20

编辑：DGM 2022-2-20

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Sorry about the misunderstanding.

I don't know about the efficiency of the method given, but I'd imagine it's fair enough. I don't work with sparse tools, so offhand I don't know if there would be significant advantages or at what scale those advantages manifest. The only way to know would be to test various methods with different size inputs.

EDIT:

This is the above method versus a slightly modified version. On my hardware, the modified version is faster for smaller arrays, but reaches equivalence for about a 6000x6000 output size.

% prepare inputs
A = rand(3,3,100);
B = A*11;
C = A*111;
% original method
a = timeit(@() testA(A,B,C))
a = 0.0095
% avoiding mat2cell calls
b = timeit(@() testB(A,B,C))
b = 0.0040
% time ratio
a/b
ans = 2.3716
function testA(A,B,C)
    [k,l,m] = size(A);
    
    A = mat2cell(A,k,l,ones(1,m));
    B = mat2cell(B,k,l,ones(1,m));
    C = mat2cell(C,k,l,ones(1,m));
    
    blocks = [{zeros(k,l)},A(:)',B(:)',C(:)'];
    
    map = diag(1:m,0)+diag(m+1:(2*m-1),1)+diag((2*m+2):3*m,-1)+1;
    D = cell2mat(reshape(blocks(map),size(map)));
end
function testB(A,B,C)
    [k,l,m] = size(A);
    
    A = reshape(A,k,[]);
    B = reshape(B,k,[]);
    C = reshape(C,k,[]);
    
    blocks = mat2cell([zeros(k,l) A B C],k,l*ones(1,m*3+1));
    
    map = diag(1:m,0)+diag(m+1:(2*m-1),1)+diag((2*m+2):3*m,-1)+1;
    D = cell2mat(reshape(blocks(map),size(map)));
end

剑豪戴 2022-2-21

编辑：剑豪戴 2022-2-21

Thanks! Your advanced method worked nice for me! But in the end, I need to do x=D\b in linear equation system. Maybe sparse will be better.

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Block sub diagonals matrix

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