- #1
Askhwhelp
- 86
- 0
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?
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: