How to do the following with Matrix in MatLab

Hi! to all...am new to this site and this is my first thread... Warm regards..

I want to achieve the following:

First I want to create this matrix:

1 2 3

1 0 0 0

2 0 0 0

3 0 0 0

Then I want to create a Base Matrix(A matrix with predefined fixed values):

1 2 3

1 0 1 0

2 2 0 0

3 0 0 0

Then I want Random matrix from the Base matrix by adding +1 to +4 . but my constraint is that none of the number should be more than 4 so if the value is 0 then +4 is accepted but if the value is 2 then the accepted values should be +1 or +2..

Example of a randomly generated matrix from the base matrix will be:

1 2 3

1 0 1 1

2 4 0 1

3 0 2 0

now the twist here is that i also want to randomize the addresses along with the randomization of the values,

have a look

from this

1 2 3

1

2

3

to like this

3 2 1

3

2

1

and this

1 3 2

1

3

2

and so on...

now the last and my main program is to interchange the addresses of two matrix by keeping their values constant.

For example consider the following two matrix:

Matrix A:

3 2 1

3 0 1 1

2 4 0 1

1 0 2 0

Matrix B:

1 2 3

1 0 1 2

2 1 0 3

3 1 4 0

now the interchange of addresses will be:

Matrix B:

3 2 1

3 0 1 2

2 1 0 3

1 1 4 0

Matrix A:

1 2 3

1 0 1 1

2 4 0 1

3 0 2 0

Notice that the Values of Matrix B & A are constant, only the Addresses are interchanged..

so I want these. Kind help me to achieve it in MatLab... Thanks in advance.

[Walter Roberson]

in part, fliplr(flipud(A))

Answer 1

[Rick Rosson]

Here is a start:

   A = zeros(3,3);

   B = zeros(3,3);
   B(2,1) = 2;
   B(1,2) = 1;

   [M,N] = size(B);
   Q = 4
   C = B;
   for k = 0:Q-1
       imax = Q - k;
       C = C + (B == k) .* randi(imax,M,N);
   end

For more information:

   >> doc randi
   >> doc zeros
   >> doc size

[Image Analyst]

Don't you have to add

C = min(C, 4);

for the constraint that no number, after addition of the random numbers, exceeds 4?

[Rick Rosson]

Although he didn't specify, my assumption is that he wants equiprobable outcomes. If you simply take the min, the result will be biased too heavily to a value of 4. I modified my solution to account correctly for this constraint.