how to create this matrix without using for loop?

Question

I want to create a N*N matrix A.

when n = 4
2  0 -2  0
2  0  2  0
0  2  0 -2
0  2  0  2

when n = 8
2  0  0  0 -2  0  0  0
2  0  0  0  2  0  0  0
0  2  0  0  0 -2  0  0
0  2  0  0  0  2  0  0
0  0  2  0  0  0 -2  0
0  0  2  0  0  0  2  0
0  0  0  2  0  0  0 -2
0  0  0  2  0  0  0  2

I can create this using nested for loop, but how to achieve it more efficiently? Are there any methods without for loop?

Thanks

Check the diag function, this one can create off-diagonal elements as well. — Adriaan
– Adriaan, Commented Sep 16, 2015 at 8:18

Divakar · Accepted Answer · 2015-09-16 14:37:29Z

1

Here's one way with bsxfun -

A = zeros(n);
idx = bsxfun(@plus,[0:(n/2)-1]*((n+3)-1),[1:2].');
A(idx) = 2;
A(idx+numel(A)/2) = -2;

Sample runs -

Case #1 :

>> n = 4;
>> A
A =
     2     0    -2     0
     2     0    -2     0
     0     2     0    -2
     0     2     0    -2

Case #2 :

>> n = 8;
>> A
A =
     2     0     0     0    -2     0     0     0
     2     0     0     0    -2     0     0     0
     0     2     0     0     0    -2     0     0
     0     2     0     0     0    -2     0     0
     0     0     2     0     0     0    -2     0
     0     0     2     0     0     0    -2     0
     0     0     0     2     0     0     0    -2
     0     0     0     2     0     0     0    -2

answered Sep 16, 2015 at 14:37

Divakar

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3 Comments

kaniz Over a year ago

Thank you so much for your help... :) it works perfectly.

Luis Mendo Over a year ago

You will like this: kron in recent Matlab versions (from R2014b, maybe earlier) uses bsxfun internally!

Divakar Over a year ago

@LuisMendo Yes! I think first chappjc told me about this, or he commented on some answer from Ray. Seems like bsxfun is spreading like a good disease all over! ;)

Wauzl · Accepted Answer · 2015-09-16 08:31:55Z

0

You can do it like that:

[reshape([repmat([ 2;2;zeros(n,1)],n/2-1,1); 2;2],n,n/2)   ...
 reshape([repmat([-2;2;zeros(n,1)],n/2-1,1);-2;2],n,n/2) ]

This only will work, if n is a power of two, obviously.

[EDIT] It might be faster to use

[reshape([repmat([ 2;2;zeros(n,1)],n/2-1,1); 2;2;       ...
          repmat([-2;2;zeros(n,1)],n/2-1,1);-2;2] ,n,n) ]

[EDIT2] This is only a good idea, if you have n of moderate size. If you need really big matrices, you should use sparse matrices. In this case a loop is what you want.

edited Sep 16, 2015 at 8:31

answered Sep 16, 2015 at 8:25

Wauzl

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1 Comment

kaniz Over a year ago

thanks for the help... :) I learn how to use reshape and repmat. and it works perfect for the problem.

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