How can we show that the Hamming distance

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Homework Help Overview

The discussion revolves around the properties of the Hamming distance, specifically its invariance under shifts of the input strings. Participants are exploring how to demonstrate that d(x+z, y+z) = d(x,y) for given strings x, y, and z.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the definition and implications of the Hamming distance, with some questioning the notation used in the summation formula. Others are providing examples to illustrate the concept.

Discussion Status

The discussion is active, with participants asking for clarifications on the formula and its components. Some have provided examples to aid understanding, while others are seeking further explanation of the underlying concepts.

Contextual Notes

There is a request for the formula for Hamming distance, and some participants express confusion regarding the notation and its meaning. Additionally, the discussion includes examples that may not fully align with the original question about invariance.

lenti
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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
 
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Please include the formula for Hamming distance.
 
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
 
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?
 
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.
 
How about my question about the summation?
 
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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