Question

Prove that the elements of even Hamming weight in$\mathbf{B}\left(n\right)$ form an$\left(n,n-1\right)$ code.

Short answer

It is proved that the elements of even Hamming weight in$\mathrm{B}\left(\mathrm{n}\right)$ form an$\left(\mathrm{n},\mathrm{n}-1\right)$ code.

Step-by-step solution

Definition of humming weight

The Hamming weight of an element u of $\mathbf{B}\left(n\right)$. Is the number of nonzerocoordinates in u; it is denoted $\mathbf{Wt}\left(u\right)$.

Proving that even weight in B(n) form an (n,n-1) code

We know, that even humming implies the sum of digits is zero

Let us take an example of $\left(6,5\right)$parity-check code it is an example of$\left(\mathrm{n},\mathrm{n}-1\right)$code with $\mathrm{n}=6$

Now we need to prove that it is an even humming means the sum of all the digits is zero.

Now we have two cases

• Sum of the first five digits is 0.

Since the sum of five digits is zero, therefore the sixth digit must be zero. And we know the sixth digit is some of the first five digits.

Hence, we can construct a codeword in the$\left(6,5\right)$ parity-check code.

• Sum of the first five digits is 1.

Since the sum of five digits is 1, therefore the sixth digit must be 1 and, and we know the sixth digit is some of five digits and $1+1=0$.

Therefore, in both cases, some of the digits are zero, so it is an even humming code.

This implies, $\mathrm{B}\left(6\right)$is a codeword in the$\left(6,5\right)$ code if the sum of its digits is 0 or even.

Hence, if we generalize, we can conclude that the elements of even Hamming weight in data-custom-editor="chemistry" $\mathrm{B}\left(\mathrm{n}\right)$ form andata-custom-editor="chemistry" $\left(\mathrm{n},\mathrm{n}-1\right)$ code.