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Proof

First, it's clear that (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 (v,w)(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 . We wish to show that (u,w)(u,v)(v,w).

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

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

We can now see that (v,w) and (u,v).

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

(u,v)(v,w) (u,w).
This is what we wished to show. Therefore, the Hamming distance is a metric.

Let v V and let . Then the Hamming sphere of radius r and centre v is defined by S(v):={x V : .

Let . A subset of is called a t-error correcting code if any two distinct elements v,w C satisfy .

If C is a -error correcting code, we define its minimum distance to be (C):= min{(c,c') : c,c' C,c c'}. Note that by Definition 1.4, the minimum distance for a C must be greater than or equal to .

If C is a -error correcting code and v C such that , then v is called a codeword of C.

Let C be a -error correcting code. Then for each v V there exists at most one codeword c C such that .

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