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Next: Proof Up: Basic Coding Topology for Previous: Application

Proof

First, it's clear that $ d$(v,w), which is simply a count of differing positions, is greater than or equal to zero. It's also clear that v and w are the same vector if and only if their count of differering positions equals zero. Third, it's clear that $ d$(v,w)$ =d$(w,v). What we need is a demonstration of the triangle inequality.

Suppose that we have three arbitrary vectors, u, v, and w, all of dimension $ n$. We wish to show that $ d$(u,w)$ \leq d$(u,v)$ +d$(v,w).

Suppose there are $ a$ positions in which u and w differ. Thus $ d$(u,w)$ =a$. If $ X$ represents identical elements and $ Y$ represents differing elements between u and w, a graphical representation might look like the following:

\begin{displaymath}
\begin{array}{ccc}
{\bf u:} & \begin{array}{cccccc} X & X & ...
... X & X \end{array} \\
& a=d({\bf u},{\bf w}) \\
\end{array}\end{displaymath}

\begin{displaymath}
\begin{array}{cccccc}
{\bf v:} & \underbrace{ \begin{array}{...
...egin{array}{cc} Y & Y \end{array} } \\
& b & & &c
\end{array}\end{displaymath}

Out of the $ a$ positions where u and v differ, there are $ b$ positions in which v's elements are identical to those of u, but not identical to those of w. And likewise, out of these $ a$ positions, there are $ a-b$ positions in which v's elements are identical to those of w, but not identical to those of u. And in the $ n-a$ positions in which u matches w, there are $ c$ positions in which the elements of v do not match either u or w.

We can now see that $ d$(v,w)$ =b+c$ and $ d$(u,v)$ =a-b+c$.

Adding $ d$(u,v) and $ d$(v,w) we have

$ d$(u,v)$ +d$(v,w) $ =(a-b+c)+(b+c)=a+2c\geq a=d$(u,w).
This is what we wished to show. Therefore, the Hamming distance is a metric.

Let v $ \epsilon$ V and let $ r \geq 0$. Then the Hamming sphere of radius r and centre v is defined by S$ _{r}$(v):={x $ \epsilon$ V : $ d({\bf x},{\bf v})\leq r\}$.

Let $ t \epsilon \mathbb{Z} $. A subset $ {\bf C} $ of $ \mathbb{V}:=\{0,1\}^{n} $ is called a t-error correcting code if any two distinct elements v,w $ \epsilon$ C satisfy $ d({\bf v},{\bf w})\geq 2t+1$.

If C is a $ t$-error correcting code, we define its minimum distance to be $ d$(C):= min{$ d$(c,c') : c,c' $ \epsilon$ C,c $ \ne$ c'}. Note that by Definition 1.4, the minimum distance for a C must be greater than or equal to $ 2t+1$.

If C is a $ t$-error correcting code and v $ \epsilon$ C such that $ {\bf v}=\{0,1\}^{n}$, then v is called a codeword of C.

Let C be a $ t$-error correcting code. Then for each v $ \epsilon$ V there exists at most one codeword c $ \epsilon$ C such that $ d({\bf v},{\bf c})\leq t$.


next up previous
Next: Proof Up: Basic Coding Topology for Previous: Application
Bill Cherowitzo 2001-12-11