Help with Matrix block multiplication

Hi, i need help with block matrix multiplication. I think a practical example should explain what i'm looking for.

Given:

A = rand(3,3); B = rand(9,3);

so basically i have [A] nxn block (generalizing) and [B] (k*n)xn block.

I would like to achieve as a result the equivalent of the following:

[A*B(1:3,:);A*B(4:6,:);A*B(7:9,:)];

possibly without any loops and arrayfun/cellfun.

Thank you in advance.

[madhan ravi]

See if the below does what you want:

A = rand(3,3);
B = rand(9,3);
[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]);
CC = reshape(C,[],size(A,2),1)

[Eugenio Grabovic]

Not quite, it does the block multiplication but element wise, i needed scalar multiplication between the blocks.

[madhan ravi]

Give a short example of fixed A and B matrix and your expected result.

[Stephan]

What is the question? You gave the solution by yourself in the question:

A = rand(3,3)
B = rand(9,3)
result  = [A*B(1:3,:);A*B(4:6,:);A*B(7:9,:)]

The result is a concatenated block of the 3 results of matrix multiplication:

A =

    0.8143    0.3500    0.6160
    0.2435    0.1966    0.4733
    0.9293    0.2511    0.3517


B =

    0.8308    0.0759    0.3371
    0.5853    0.0540    0.1622
    0.5497    0.5308    0.7943
    0.9172    0.7792    0.3112
    0.2858    0.9340    0.5285
    0.7572    0.1299    0.1656
    0.7537    0.5688    0.6020
    0.3804    0.4694    0.2630
    0.5678    0.0119    0.6541


result =

    1.2200    0.4076    0.8206
    0.5776    0.2803    0.4899
    1.1123    0.2707    0.6333
    1.3134    1.0414    0.5404
    0.6379    0.4349    0.2581
    1.1904    1.0042    0.4802
    1.0967    0.6348    0.9852
    0.5271    0.2364    0.5079
    0.9956    0.6506    0.8554

[Eugenio Grabovic]

@madhan ravi

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]

Is A always of size n x n ?

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

[Eugenio Grabovic]

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

Answer 1

[madhan ravi]

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

[Stephan]

Nice! +1

[Eugenio Grabovic]

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