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Choose an arbitrary codeword **c** of **C**. Counting the number of vectors within radius 1, we have **c** itself and vectors (since we have options for each of the possible errors). We have shown these spheres are mutally disjoint, therefore exactly
vectors of **V** are covered by the spheres of **C**. Thus for a linear 1-error correcting code, **C** is perfect if and only if
, since that's how many vectors are in **V**.

Thus to show our Hamming code is perfect, we use the fact that the size of **C** is . Plugging this into our equation from the above theorem we have

.

, where is an integer.

Which is how is chosen by definition of our Hamming code.
QED

Bill Cherowitzo
2001-12-11