How to change 0 to 1 and update a matrix

clc
clear all
close all
b1 = [10; 20 ; 30 ; 40];
A = [0 20 0 10 ; 20 0 30 0 ; 0 30 0 10 ; 20 0 10 0];
s = size(b1);
x = randi([0 1],s)
G = [];
[ n m ] = size(x);
for i = 1 : n
    for j = 1 : m
    G(:,i) = x(:,j)
    a = find(G(: , i) == 0 , 1 , 'first')
    [x y] = size(G(:,i))
    for k = 1:y
        G(: , k) = 0;
        G(a , k) = 1
    end
    end
    i = i + 1;
    j = j + 1;
end

This prob wil work only the first time. I want the matrix to take a column, find where the first 0 is and change it to 1. Then proceed to the next line and do the same but keep the first column and not change it to 0s. Final solution must look like

x =
     0
     1
     0
     0

(x changes randomly with 0 and 1 )

G =
     1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0     1

I would appreciate any answer :)

[Chinmay Anand]

Can you provide some example matrix x with the solution you expect ( matrix G ).

[Antonio Grivas]

x1 =

1

0

0

1

ans =

1 0 0 0

0 1 0 0

0 0 1 0

1 0 0 0

__________________________________________________________

x1 =

0

1

0

1

ans =

0 1 0 0

1 0 0 0

0 0 1 0

1 0 0 0

__________________________________________________________

x1 =

0

0

1

0

ans =

0 1 0 0

0 0 1 0

1 0 0 0

0 0 0 1

__________________________________________________________

x1 =

0

1

1

1

ans =

0 1 0 0

1 0 0 0

1 0 0 0

1 0 0 0

[Antonio Grivas]

I can do it in a 4x4 matrix by copying the same code 4 times but what if a matrix is 500x500 !!! final matrix G is always squared.

[Stephen23]

@Antonio Grivas: Whenever I run your code it always returns this:

G =
   0   4   4   4
   0   4   4   4
   0   4   4   4
   0   4   4   4

It is unclear how this output relates to your examples.

Please explain the logic of what you are trying to achieve.

[Antonio Grivas]

It is completely wrong. I don't know how to make it change the final result whick would be

1

0

0

0

1 0

0 1

0 0

0 0

1 0 0

0 1 0

0 0 1

0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

I send the (without the wrong loop) answer where the program runs correctly only the first time

I suppose that i need a

for i = 1:size(whatever)

whatever(: , i) = 0

whatever(: , a) = 1

end

that sould be put somewhere to keep the previous results and not change it back to

0

0

0

0

and change the position of the 1 after adding a column

If you run it just check only the 1st time before it changes it back to

0

0

1

0

becomes

0 1

0 0

1 0

0 0

Then for some reason that i cannot see does not continue but gives

0 4

0 4

0 4

0 4

I am trying to be precise if i can to help you

Thank you

Answer 1

[Andrei Bobrov]

n = numel(x1);
y1 = ~x1;

% if MATLAB >= R2016b
k = y1.*eye(n);
% if MATLAB <= R2016a
k = bsxfun(@times,y1,eye(n));

out = [x1, zeros(n,n-1)];
p = find(y1);
out(:,2:numel(p)+1) = k(:,p);

[Chinmay Anand]

Taking x to be a 4x4 matrix you can use the above code to get your result as follows.

[row,col] = size(x);
for i = 1:col
    
    x1 = x(:,i);
    
    n = numel(x1);
    y1 = ~x1;
    k = y1.*eye(n);
    out = [x1, zeros(n,n-1)];
    p = find(y1);    
    out(:,2:numel(p)+1) = k(:,p)  % It will print result for each column of x 

end

[Antonio Grivas]

I will run this program as soon as i finish from work and go home.

Thanks in advance

[Antonio Grivas]

Problem with matrix dimensions at k = y1.*eye(n);

[Andrei Bobrov]

You use old MATLAB.

Replace this expression with the following:

k = bsxfun(@times,y1,eye(n));

Answer 2

[Antonio Grivas]

Thank you very much

Matlab version 2016a (that was the problem)