Algorithm for the matrices

Hello dears,

I am a beginner and want to use the algorithm

Y(i+1)=((1/2).*(1/4)*(Y(i))+2*A1)+((1/4).*(1/6)*(Y(i))+2*A2)+((1/6).*(1/8)*(Y(i))+2*A3);

where

A1=[4/7 7/8 0; 7/8 4/7 0; 0 0 0], A2=[3/4 1 0; 1 3/4 0; 0 0 0], A3=[1/8 1 0; 1 1/8 0; 0 0 0]

and initial values for Y are zeros.

My code is

N=10;
nstep=0:10;
Y=zeros(3,3);
A1=[4/7 7/4 0; 7/4 4/7 0; 0 0 0];
A2=[3/4 1 0; 1 3/4 0; 0 0 0];
A3=[1/8 1 0; 1 1/8 0; 0 0 0];
for i=1:N
    Y(i+1)=(1./2).*(((1./4).*(Y(i)+A1))+(1./4).*(Y(i)+A1))+(1./4).*(((1./6).*(Y(i)+A2))+(1./6).*(Y(i)+A2))+((1./6).*(((1./8).*(Y(i)+A3))+((1./8).*(Y(i)+A3))))
end

But the code gives the following error

"In an assignment A(I) = B, the number of elements in B and I must be the same.

Error in Code (line 8)

Y(i+1)=(1./2).*(((1./4)*(Y(i)+A1))+(1./4).*(Y(i)+A1))+(1./4).*(((1./6)*(Y(i)+A2))+(1./6).*(Y(i)+A2))+(1./6).*(((1./8)*(Y(i)+A3))+(1./8).*(Y(i)+A3)); "

I don't know how to proceed. Any help will be appreciated. Thank you all in advance for your valuable time.

[Dyuman Joshi]

You are assigning an array as an element of another matrix, which is simply not possible.

"In an assignment A(I) = B, the number of elements in B and I must be the same"

This error is encountered when you try to place more elements into A that are specified by the I variable. For example if A is a simple vector, and I is a simple index, like 3, then trying to put 100 numbers into A(3) will obviously not work.

MATLAB FAQs

[Waseem Ahmad]

Dear Joshi, now i understood the cause of the error. Can you please help me with how to initiate the algorithm. As im a beginner and don't know much about metrics algorithms. Thank you

[Walter Roberson]

Each of your A variables is a 3 by 3 array. How big are you expecting each of your Y to be?

[Waseem Ahmad]

Dear Roberson, I want Y to be 3by3

[Image Analyst]

Then why is N 10? If you set it to 3, does this give you what you want:

N=3;
nstep=0:10;
A1=[4/7 7/4 0; 7/4 4/7 0; 0 0 0];
A2=[3/4 1 0; 1 3/4 0; 0 0 0];
A3=[1/8 1 0; 1 1/8 0; 0 0 0];
[rows, columns] = size(A1)
Y = zeros(rows, N);
% Loop over all rows in the A's:
for row = 1 : rows
	for i = 1 : N - 1
		term1 = (1/2).*(((1/4).*(Y(row, i)+A1(row,:)))+ (1/4).*(Y(row, i)+A1(row,:)))
		term2 = (1/4).*(((1/6).*(Y(row, i)+A2(row,:)))+ (1/6).*(Y(row, i)+A2(row,:)))
		term3 = ((1/6).*(((1/8).*(Y(row, i)+A3(row,:)))+ ((1/8).*(Y(row, i)+A3(row,:)))))
		Y(row, 1:length(term1)) = term1 + term2 + term3
	end
end

[Waseem Ahmad]

Thank you Image Analyst your code work for me

Answer 1

[Jan]

Either collect all Y in a 3D array:

N  = 10;
A1 = [4/7, 7/4, 0; 7/4, 4/7, 0; 0, 0, 0];
A2 = [3/4, 1,   0; 1,   3/4, 0; 0, 0, 0];
A3 = [1/8, 1,   0; 1,   1/8, 0; 0, 0, 0];
Y  = zeros(3, 3, N);
for i = 1:N
    % Original:
    % Y(:, :, i+1)=(1./2).*(((1./4).*(Y(:, :, i)+A1))+(1./4).*...
    %    (Y(:, :, i)+A1))+(1./4).*(((1./6).*(Y(:, :, i)+A2))+...
    %    (1./6).*(Y(:, :, i)+A2))+((1./6).*(((1./8).*...
    %    (Y(:, :, i)+A3))+((1./8).*(Y(:, :, i)+A3))))
    
    % Simplified 1:
    Y(:, :, i+1) = 0.5 * ((0.25 * (Y(:, :, i) + A1)) + ...
                           0.25 * (Y(:, :, i) + A1)) + ...
                   0.25 * (((1 / 6) * (Y(:, :, i) + A2)) + ...
                            (1 / 6) * (Y(:, :, i) + A2)) + ...
                   1/6 *  (((1 / 8) * (Y(:, :, i) + A3)) + ...
                           ((1 / 8) * (Y(:, :, i) + A3)));
    
    % Simplified 2:
    % Y(:, :, i+1) = 1 / 4  * (Y(:, :, i) + A1) + ...
    %                1 / 12 * (Y(:, :, i) + A2) + ...
    %                1 / 24 * (Y(:, :, i) + A3);  
end

I've simplified the code, because then it is much easier to modify and to debug.

Or omit the indexing for Y:

N  = 10;
A1 = [4/7, 7/4, 0; 7/4, 4/7, 0; 0, 0, 0];
A2 = [3/4, 1,   0; 1,   3/4, 0; 0, 0, 0];
A3 = [1/8, 1,   0; 1,   1/8, 0; 0, 0, 0];
Y  = zeros(3, 3);
for i = 1:N
    Y = (Y + A1) / 4  + ...
        (Y + A2) / 12 + ...
        (Y + A3) / 24;
end

Answer 2

[Image Analyst]

N=3;
nstep=0:10;
A1=[4/7 7/4 0; 7/4 4/7 0; 0 0 0];
A2=[3/4 1 0; 1 3/4 0; 0 0 0];
A3=[1/8 1 0; 1 1/8 0; 0 0 0];
[rows, columns] = size(A1)
Y = zeros(rows, N);
% Loop over all rows in the A's:
for row = 1 : rows
	for i = 1 : N - 1
		term1 = (1/2).*(((1/4).*(Y(row, i)+A1(row,:)))+ (1/4).*(Y(row, i)+A1(row,:)))
		term2 = (1/4).*(((1/6).*(Y(row, i)+A2(row,:)))+ (1/6).*(Y(row, i)+A2(row,:)))
		term3 = ((1/6).*(((1/8).*(Y(row, i)+A3(row,:)))+ ((1/8).*(Y(row, i)+A3(row,:)))))
		Y(row, 1:length(term1)) = term1 + term2 + term3
	end
end

[Jan]

It's interesting, how differently we understood the question.