# Explain hamming distance satisfies the triangle inequity

1. Nov 6, 2013

Hamming distance satisfies the triangle inequity that is for all x, y, u in c such that d(x,y) <= d(x,u) + d(u,y) where c is a code. Also when does the equality hold?

My approach is
Turning x to u by changing at most d(x,u) letters and turning u to y by changing at most d(u,y) letters. So turning x to t will change no more than d(x,u) + d(u,y) letters.

By looking at the following example, x = 001111, y = 111111, z = 011111. The equality holds when changing x to y, z is just a intermediate step, the equality holds.

Does this make sense?

If not, how should I approach this?

Last edited: Nov 6, 2013
2. Nov 7, 2013

### jbunniii

You have the right idea, but you might state it more precisely. If we define $x_n$, $y_n$, and $u_n$ to be the $n$'th letter of $x$, $y$, and $u$, respectively, and $N$ is the code length, then we have
$$d(x,y) = \sum_{n=1}^{N} |x_n - y_n|$$
and similarly for $d(y,u)$ and $d(x,u)$. Proving the triangle inequality is therefore equivalent to proving that
$$\sum_{n=1}^{N} |x_n -y_n| \leq \sum_{n=1}^{N} |x_n - u_n| + \sum_{n=1}^{N} |u_n - y_n|$$