filling a matrix with a loop

Hi community,

I want to create a large matrix 400x400, In this matrix I want it to have the form

A= [1 1 0 1 000000000000000000........;0 1 1 0 1 00000000000000......; 0 0 1 1 0 1 00000000000000000] and so on till it is a 400x400 matrix.

I could not find a way yet to easiliy do this. As doing this manually is way too much work.

Does anyone know how to do this?

[Adam Danz]

What's the pattern? Are you just shifting [1101] down by 1 index to the right and padding the rest of the array with 0s?

[bus14]

Yes exactely this

Answer 1

[Stephen23]

"Does anyone know how to do this?"

This is simple and efficient with toeplitz:

>> C = [1,zeros(1,399)];
>> R = [1,1,0,1,zeros(1,396)];
>> M = toeplitz(C,R);
>> size(M)
ans =
   400   400
>> M(1:9,1:20) % first rows and columns
ans =
   1   1   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   1   1   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   1   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   1   1   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   0   1   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   1   1   0   1   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   1   1   0   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   1   1   0   1   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   1   1   0   1   0   0   0   0   0   0   0   0
>> M(392:400,381:400) % last rows and columns
ans =
   0   0   0   0   0   0   0   0   0   0   0   1   1   0   1   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   1   1   0   1   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   0   1   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   0   1   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   0   1   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   0   1
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1

[Adam Danz]

Interesting... didn't know about that one.

[Alex Mcaulley]

Really nice

[bus14]

This worked perfectly for my described situation. However, I found out that I was using the wrong matrix..

Should be a 800 x 400 matrix which looks like

     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     1     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     1     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0

so like these two 2 colums which repeat itself every time.

Tried fixing it with Toeplitz again but toeplitz cannot handle 2 input columns.

would there be some sort of other matrix creating function that can do this?

[Stephen23]

Use blkdiag and a comma-separated list:

C = {[1,0;1,1;0,1;1,0]};
M = blkdiag(C{ones(1,200)});

Checking:

>> size(M)
ans =
   800   400
>> M(1:12,1:16)
ans =
   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   1   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   1   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0

Answer 2

[Alex Mcaulley]

A = repmat([1 1 0 1 zeros(1,396)],400,1);
A = cell2mat(arrayfun(@(i) circshift(A(i,:),i-1) , 1:size(A,1), 'UniformOutput',false)');

Answer 3

[Jos (10584)]

% clever indexing trick
A= [1 1 0 1]
N = 10 ; % smaller example! 400 in your case
X = triu(toeplitz(1:N)) ;
X(X > numel(A)) = 0 ;
tf = X > 0 ;
X(tf) = A(X(tf))

Answer 4

[Andrei Bobrov]

out = full(spdiags(ones(400,3),[0,1,3],400,400));