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Homework Help: Show the following properties of Hamming weight

  1. Nov 10, 2008 #1
    hi,

    I have to show the following properties of the Hamming weight for binary words x and y of equal lenght:

    a)w(x+y)=w(x)+w(y)-2w(x*y)

    b)w(x+y)>=w(x)-w(y)

    c) For w(y) even, w(x+y) is even iff w(x) is even

    d) For w(y) odd, w(x+y) is odd iff w(x) is even

    can anybody help me,

    thanks

    lenti
     
    Last edited: Nov 10, 2008
  2. jcsd
  3. Nov 10, 2008 #2

    Mark44

    Staff: Mentor

    You probably won't get much help if you don't at least provide us the formula for Hamming weight.
     
  4. Nov 10, 2008 #3
    Well the Hamming weight of a length-N word x denoted w(x) is defined as the number of components (symbols) of x that are nonzero.

    Well there is no special formula about the hamming weight it can be formulated as


    w(x)= [tex]\sum[/tex]I{x[tex]\neq[/tex]0}

    where I{x[tex]\neq[/tex]0}, the indicator of event {x[tex]\neq[/tex]0}, is 1 if x[tex]\neq[/tex]0 and 0 if x=0

    thanks
     
    Last edited: Nov 10, 2008
  5. Nov 10, 2008 #4

    Mark44

    Staff: Mentor

    a) I'm not sure what x*y means, but I suspect it is the dot product of the two words treated as vectors. If so, w(x*y) gives a measure of how many bits in x are 1 at the same place they are in y. If each bit in x is different from the corresponding bit in y, w(x*y) = 0. Is this correct?

    For example, if
    x = 1011
    y = 0111

    then x*y = 0011.

    Also, x + y seems to be a new word of the same length as x (and as y), where a given bit is 1 in x + y if the corresponding bit in x or in y (or in both) is set.

    For example, if
    x = 1011
    y = 0111
    then x + y = 1111.

    If I'm on the right track here, it seems that the formula should be w(x + y) = w(x) + w(y) - w(x*y), without the factor of 2 that you showed. Of course, I might not be on the right track, since I'm not sure what w(x*y) means.
     
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