# Creating a mxn matrix that is all zeros except for the middle row and middle column which are 1s

46 views (last 30 days)

Show older comments

Hi,

Given an 7x5 matrix, i am to return a mxn matrix M that is all zeros except for the middle row and middle column, which should be all 1s.

I have gotten as far as creating a zeros matrix using the code M=zeros(7,5), but am clueless on how to change the middle row and middle column to 1s.

I should also mention that the code should be a one-liner.

Thank you

##### 0 Comments

### Accepted Answer

Wan Ji
on 28 Aug 2021

Edited: Wan Ji
on 28 Aug 2021

M = [0,0,0,1,0,0,0; 0,0,0,1,0,0,0; 1,1,1,1,1,1,1; 0,0,0,1,0,0,0; 0,0,0,1,0,0,0];

Once for all

Or you can do

a = zeros(5,7);

a((1:numel(a))>size(a,1)*(size(a,2)-1)/2&1:numel(a)<=size(a,1)*(size(a,2)+1)/2|mod(1:numel(a),size(a,1))==(size(a,1)+1)/2) = 1

The result is

a =

0 0 0 1 0 0 0

0 0 0 1 0 0 0

1 1 1 1 1 1 1

0 0 0 1 0 0 0

0 0 0 1 0 0 0

We can extend it to any matrix with both odd columns and rows.

a = zeros(9,11);

a((1:numel(a))>size(a,1)*(size(a,2)-1)/2&1:numel(a)<=size(a,1)*(size(a,2)+1)/2|mod(1:numel(a),size(a,1))==(size(a,1)+1)/2) = 1

Then a becomes

a =

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

##### 0 Comments

### More Answers (3)

Awais Saeed
on 28 Aug 2021

M = zeros(7,5)

M(:,ceil(size(M,2)/2)) = 1

M(ceil(size(M,1)/2),:) = 1

##### 4 Comments

Awais Saeed
on 28 Aug 2021

John D'Errico
on 29 Mar 2023

Edited: John D'Errico
on 31 Mar 2023

Clearly a homework assignment. But now long since past, and with multiple answers, most of which are not in one line. However, simple enough to do, in one line in at least one way. Here is my first try:

A1 = (1:7)' == 4 | (1:5) == 3

Short. Sweet. Simple enough. I'm not sure I can beat that. It is logical, so if you needed it to be composed of doubles, that too is not difficult. Just use a unary plus. It forces me to add a parens though.

A2 = +((1:7)' == 4 | (1:5) == 3)

This next one uses a slight tweak on the diagonal, else there would be a 2 at the center element.

A3 = ((1:7)' == 4) + ((1:5) == 3) - ((1:7)' == 4) * ((1:5) == 3)

I'd bet I could do it in other ways too. How about this cute one?

A4 = dec2bin([4 4 4 31 4 4 4]) - '0'

This next one seems kind of silly, as you would never want to use it for a larger problem. But it works.

A5 = blkdiag(1,1,1,flip(eye(4)))*[zeros(6,4),ones(6,1);ones(1,5)]*blkdiag(1,1,flip(eye(3)))

I'm sure there are other ways too. Personally, I prefer A1 and A4. The point is, as you learn to use MATLAB, you can learn different ways to manipulate the elements of arrays. But as you look for other ways to solve the problem, it forces you to learn about other functions you may never have seen in that context.

##### 0 Comments

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!