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Coding Theory, Finding Dual Code

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Let S = {0101, 1010, 1100}. From first principles, find a basis B for the dual code C orthogonal (couldn't find symbol)

    2. Relevant equations

    http://www.maths.uq.edu.au/courses/MATH3302/files/codingnotes.pdf [Broken]

    i'm using page 19,20 and 21

    3. The attempt at a solution

    so i did my matrix using the codes of S

    1. 0101
    A = 1010
    1100


    2. then i used REF on this getting

    1100 1100
    0110 => 0110 = G
    0101 0011

    So is this my generating matrix G? like shown on page 20
    The problem is the example on page 20 is in Reduced Row Echelon Form and not
    not just Row Echelon Form like above. So i don't get G = (I X) like the example.
    This is where i get stuck because i'm not getting the same form for G and so i can't
    get H which i need as the columns of H form a basis for C orthogonal.

    The solution to this problem i have i don't get either but maybe it might help you,
    It's a different approach to what i'm taking but i don't get it.

    Could someone please tell me if i got the correct G and how to get H because
    that's where i'm really stuck.



    Notes solution:

    4. Let x1x2x3x4 2 C?. Then by the defnition of C? we have
    (x1x2x3x4) x (0101) = 0 so x2 + x4 = 0
    (x1x2x3x4) x (1010) = 0 so x1 + x3 = 0
    (x1x2x3x4) x (1100) = 0 so x1 + x2 = 0

    Thus we have x1 = x2 = x3 = x4. Thus
    C orthogonal = {0000, 1111}

    Thanx for helping,

    Cheerz
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 11, 2009 #2

    Mark44

    Staff: Mentor

    I followed Ex. 2.14 to do this work. I got a different matrix G, but then I reduced A to completely reduced row echelon form (using paper and pen, BTW).

    My G looks like this:
    Code (Text):

    [1 0 0 1]
    [0 1 0 1]
    [0 0 1 1]
     
    From this I have [I3 X], where X is [1 1 1] (but as a column vector). From this I see that k = 3, and X is as above.

    My matrix H' is
    Code (Text):

    [1]
    [1]
    [1]
    [1]
     
    Note that the bottom entry is the 1 x 1 identity matrix. The columns in H' (all one of them) are the basis for Cperp.

    This makes at least some sense because the original code had three elements that are linearly independent, and the dimension of the code space is 4. So the space for the dual code has to be of dimension 1.

    The check on my work, which I didn't do, is to convince yourself that the original three codes plus the one I found are linearly independent.
     
    Last edited by a moderator: May 4, 2017
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