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I don't understand why the following works. It initializes row 1 of the array c with values of 1 in roughly the first half of the columns and then places a zero in the exact middle column (half way) while also placing zeros in the right half of the columns. So, if the number of columns is 11, columns 1 through 5 are initialized with 1 and columns 6 through 11 with zero.

1) c(1,:)=[ones(1,(N-1)/2),zeros(1,(N+1)/2)]

I would have thought that the way to do this is to specify the zeros from columns (N+1)/2 to N. More like:

2) c(1,:)=[ones(1,(N-1)/2),zeros((N+1)/2,N)]

So if N = 11, the script, 2) would result in:

c(1,:)=[ones(1,(10)/2),zeros((11+1)/2,11)] or c(1,:)=[ones(1,5),zeros(6,11)]

So, Matlab just knows that in 1) above to continue counting from the column where it stopped putting ones to start putting in values of zeros for the remaining columns. So, rather than an absolute reference like I've shown in 2), it uses a relative reference to decide which columns to initialize?

David Hill
on 8 Jun 2021

You are concatenating two arrays. Look at the ones() and zeros() functions.

a=ones(1,5);

b=zeros(1,6);

c=[a,b];

David Hill
on 8 Jun 2021

because you are specifing:

c(1,:)=[a,b];%you are only doing it to the first row of c

% you could also do it the the 5th row

c(5,:)=[a,b];

%or you could do it to all odd rows of c

c(1:2:end,:)=[a,b];

Fangjun Jiang
on 8 Jun 2021

Edited: Fangjun Jiang
on 8 Jun 2021

You mis-understood it. The code involves built-in functions ones() and zeros(). The input arguments are size of the matrix, not the index of elements in a matrix.

The "[ ]" operator combines the two matrix (or vectors, five ones and six zeros) together.

Or you could do this

N=11

c(1,1:(N-1)/2)=1

c(1,(N+1)/2:N)=0

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