MATLAB: Use the nested loop to calculate the kronecker product for different size

kronecker prouduct

A = round((rand(2)*11-5.5)) B = round((rand(3,4)*21-10.5)) calculate the kronecker product of AB

Best Answer

A = round((rand(2)*11-5.5))
B = round((rand(3,4)*21-10.5))
[rA,cA] = size(A);
[rB,cB] = size(B);
finalanswer = [];
for i = 1:rA
    answer =[];
    for j = 1:cA
        temp = A(i,j) * B;
        answer = [answer,temp];
    end
 finalanswer = [finalanswer;answer]; 
end
finalanswer
z = kron(A,B)
isequal(z,finalanswer)

Related Solutions

MATLAB: Using nested loops to solve kronecker product

you can do it as follows(if you have square matrices of same size):

A = round(rand(10,10)*10-5);
B = round(rand(10,10)*20-10);
for i = 1:size(A, 1)
  for j = 1:size(A, 2)
      AB{i, j} = A(i, j) * B;
  end
end
AB = cell2mat(AB);
disp(AB)

Good luck!

MATLAB: How to get the index from a matrix

Assuming B is in ascending order, idx(i) contains the index value of B that is closest to but not less than A(i).

A = [2;7.5;9;10.5;15;14]; 
B = 5:5:50; 
[~, idx] = max(A(:)'<B(:),[],1)
idx = 1×6
     1     2     2     3     4     3
●

If it cannot be assumed that B is in ascending order,

A = [2;7.5;9;10.5;15;14]; 
B = [5, 15, 10, 20:5:50]; % note: not in ascending order
d = (A(:)'-B(:)); 
d(d>=0)= -inf;
[~, idx] = max(d,[],1)
idx = 1×6
     1     3     3     2     4     2
●

Both solutions assume that at least 1 value in B is less than A(i). If this is not the case, you append inf to the end of B and if idx contains the index equal to the location of the final inf, you can replace it with 0. For example,

A = [99 ;7.5;9;10.5;15;14]; % Note; starts with 90
B = 5:5:50; 
[~, idx] = max(A(:)'<[B(:);inf],[],1);
idx(idx==numel(B)+1) = 0
idx = 1×6
     0     2     2     3     4     3
●

Best Answer

Related Solutions

MATLAB: Using nested loops to solve kronecker product

MATLAB: How to get the index from a matrix

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