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How can we show that the Hamming distance

  1. Nov 10, 2008 #1
    Hi,

    can anybody help me to show that the Hamming disatance d(x,y) is invarian to a shift of both x and y i.e d(x+z, y+z)=d(x,y)

    Thanks a lot

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

    Mark44

    Staff: Mentor

    Please include the formula for Hamming distance.
     
  4. Nov 10, 2008 #3
    The Hamming distance of two length-N words x, y, denoted as d(x,y), is defined as the number of components (symbols9 of x and y tha are different.

    we can writte as;

    d(x,y)=[tex]\sum[/tex]I{x[tex]\neq[/tex]y}

    thanks a lot
     
  5. Nov 10, 2008 #4

    Mark44

    Staff: Mentor

    Your summation doesn't make much sense to me. What does I in this summation mean?
    [tex]\sum I \{x \neq y\}[/tex]

    Also, it would help to give an example. For example, is d("eat", "eats") = 1?
     
  6. Nov 10, 2008 #5
    as an example;

    consider two binary words x=(1010110) and y=(1001010)

    the hamming distance between the binary words is d(x,y)=3

    because they change in 3 bits.
     
  7. Nov 10, 2008 #6

    Mark44

    Staff: Mentor

    How about my question about the summation?
     
  8. Nov 10, 2008 #7

    Mark44

    Staff: Mentor

    Assuming that z is another string of bits, it's intuitively obvious that if d(x, y) = n, say, then creating two longer strings x + z and y + z will not change the number of bits that are different. IOW d(x + z, y + z) = n as well.

    To formalize this, I would probably need the formula for Hamming distance, which you've provided, but I would also need to understand what this formula means, which I don't.
     
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