How to binary sum columns in a binary matrix

Hello guys,

I hope you can help me, I'm currently studying the Hamming code in college, but it's only my second time using MatLab so i'm having some trouble answering this question:

"Calculate the minimum number of rows that you need to sum (XOR actually) on the Parity Test matrix so that you obtain a null vector, also indicate the row's index of the first set you found."

I made a rough translation of the question, but from my understading I need to sum(XOR) the values of each column, until I obtain a null vector.

In terms of coding this is what I have:

%2.6 - Parity Test Matrix
n = 255;
k = 247;
[H,G,n,k] = hammgen(8);
H=H'

And this gives me a huge binary matrix, let's assume that these few lines:

[1 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0

...

1 1 1 0 0 0 0 0]

The sum of these 4 vectors should give me:

[0 0 0 0 0 0 0 0]

I have no idea how to this, can you guys please lend me a hand ? I hope the question is not confusing but i'm totally lost. I tried to use the sum() fuction but i got [128 128 128 128] or something in return.

Thanks in advance.

Answer 1

[James Tursa]

Just doing xor between succesive rows:

r = some row number
result = xor(H(r,:),H(r+1,:);

Can you wrap some code around this to get what you want?

[Redfoxxie]

Hey James,

Thanks for your input, I was thinking of something close to that, but i'm guessing i need to include those lines in a for statement and I need the loop to stop when result = [0 0 0 0 0 0 0 0] (am I thinking correctly?), also I don't know how to do both of these in MatLab.

[James Tursa]

Yes, the intent was for you to wrap that in a for loop to get the result. E.g., something like this:

result = H(1,:);
for k=2,size(H,1)
    if( some condition is met )  <-- you fill in this condition
        break;
    end
    result = xor(result,H(k,:));
end
% add code here to see if you met the condition

[Redfoxxie]

Actually there's no restraint on how to solve it, so anything goes. Is there a more direct ou simple approach ?

[James Tursa]

The above code is straightforward and simple to understand, so I suggest you use this method first. There are ways to vectorize this to avoid the loop using functions like cumsum( ) and mod( ) etc, but I see no compelling reason for you to go that route at this point.

[Redfoxxie]

James, I tried the following:

n = 255;
k = 247;
[H,G,n,k] = hammgen(8);
H=H';
result = H(1,:)
for count=2:size(H,1)
    result = xor(result,H(count,:));
    if( all(result == 0))
        disp('Im here')
        break;
    end
end
r = result
n_rows = count

This is the output:

result = 1 0 0 0 0 0 0 0

Im here

r = 0 0 0 0 0 0 0 0

count = 255

I'm guessing the code is running properly now, result has the proper vector for the first line of H, it enters the if condition when result has all 0 and the count was made up to 255 which is the total number of rows of the H matrix.

This is probably a characteristic of an Hamming Code, not sure though, but it seems the code is running as it should be, so thanks a lot James! If you notice any mistake on my part please let me know. Again, thanks.

[James Tursa]

For readability, I would avoid using l for a variable name (looks too much like 1).