Help with Matrix block multiplication

A = [1 1 0;1 1 0; 1 1 1];
B = [1,2,3;4,5,6;7,8,9;2,6,8;4,1,6;1,12,16;4,2,1;4,9,6;3,8,2;];
expected_result=[A*B(1:3,:);A*B(4:6,:);A*B(7:9,:)]
expected_result =
     5     7     9
     5     7     9
    12    15    18
     6     7    14
     6     7    14
     7    19    30
     8    11     7
     8    11     7
    11    19     9
    
[m,n]=size(A);
[r,c]=size(B);
AA=A.*ones(n,n,n);
BB=permute(reshape(B',c,c,[]),[2 1 3]);
C=permute(AA.*BB,[1 3 2]);
Your_result = reshape(C,[],size(A,2),1)
Your_result =
     1     2     0
     4     5     0
     7     8     9
     2     6     0
     4     1     0
     1    12    16
     4     2     0
     4     9     0
     3     8     2
     

@Stephan yes it's true, but that was an example, and it was achieved by manually feeding inputs to the resulting matrix. Usually in my alghorithm the B matrix's depth is unknown and thus i can't ( or at least don't know how) to concatenate the resulting matrix in an "automated" way. I already found how to do it with loops/arrayfun but was wondering if it was possible to achieve the result with just matrix manipulation.

Stephan 2019-1-27

Is A always of size n x n ?

Is B always of size (k*n) x n with k=[1,2,3...] ?

Eugenio Grabovic 2019-1-27

编辑：Eugenio Grabovic 2019-1-27

@Stephan yes the only thing that is unkown is the row dimension of B ( parameter k) but is always a multiple of n.

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Answer 1

madhan ravi 2019-1-28

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/441731-help-with-matrix-block-multiplication#answer_358307

编辑：madhan ravi 2019-1-28

在 MATLAB Online 中打开

A = [1 1 0;1 1 0; 1 1 1];
B = [1,2,3;4,5,6;7,8,9;2,6,8;4,1,6;1,12,16;4,2,1;4,9,6;3,8,2];
expected_result=[A*B(1:3,:);A*B(4:6,:);A*B(7:9,:)];
%code starts here
[m,n]=size(A);
[r,c]=size(B);
BB=permute(reshape(B',c,c,[]),[2 1 3]);
[a,b,c]=size(BB);
BBB=reshape(BB,[a b*c]);
A_times_B_slices=A*BBB;
C=permute(reshape(A_times_B_slices,b,a,[]),[2 1 3]);
My_Result=reshape(C,c,[],1)';
isequal(My_Result,expected_result) % to check both the results are the same

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Stephan 2019-1-28

Nice! +1

Eugenio Grabovic 2019-1-28

编辑：Eugenio Grabovic 2019-1-28

在 MATLAB Online 中打开

Thank you very much, your code actually helped me with other algorithms too.

If anyone is curious if it's worth avoiding a loop , here is my test result:

function perform_test
n=3;
k=1000;
A = rand(n,n);
B = rand(n*k,3);
tic
result1 = zeros(n*k,3);
for i = 1 : k
    
    result1(i*3-2:i*3,:) = A*B(i*3-2:i*3,:);
    
end
disp("loop time: " + toc)
tic
[m,n]=size(A);
[r,c]=size(B);
BB=permute(reshape(B',c,c,[]),[2 1 3]);
[a,b,c]=size(BB);
BBB=reshape(BB,[a b*c]);
A_times_B_slices=A*BBB;
C=permute(reshape(A_times_B_slices,b,a,[]),[2 1 3]);
result2 = C(:,:).'; %<<<-------------------------***had to change this line***
disp("matrix manipulation time: " + toc)
equality = isequal(result1,result2)
end

and the results:

>> perform_test
loop time: 0.0046915
matrix manipulation time: 0.0009893
equality =
  logical
   1

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