# Show the following properties of Hamming weight

1. Nov 10, 2008

### lenti

hi,

I have to show the following properties of the Hamming weight for binary words x and y of equal lenght:

a)w(x+y)=w(x)+w(y)-2w(x*y)

b)w(x+y)>=w(x)-w(y)

c) For w(y) even, w(x+y) is even iff w(x) is even

d) For w(y) odd, w(x+y) is odd iff w(x) is even

can anybody help me,

thanks

lenti

Last edited: Nov 10, 2008
2. Nov 10, 2008

### Staff: Mentor

You probably won't get much help if you don't at least provide us the formula for Hamming weight.

3. Nov 10, 2008

### lenti

Well the Hamming weight of a length-N word x denoted w(x) is defined as the number of components (symbols) of x that are nonzero.

Well there is no special formula about the hamming weight it can be formulated as

w(x)= $$\sum$$I{x$$\neq$$0}

where I{x$$\neq$$0}, the indicator of event {x$$\neq$$0}, is 1 if x$$\neq$$0 and 0 if x=0

thanks

Last edited: Nov 10, 2008
4. Nov 10, 2008

### Staff: Mentor

a) I'm not sure what x*y means, but I suspect it is the dot product of the two words treated as vectors. If so, w(x*y) gives a measure of how many bits in x are 1 at the same place they are in y. If each bit in x is different from the corresponding bit in y, w(x*y) = 0. Is this correct?

For example, if
x = 1011
y = 0111

then x*y = 0011.

Also, x + y seems to be a new word of the same length as x (and as y), where a given bit is 1 in x + y if the corresponding bit in x or in y (or in both) is set.

For example, if
x = 1011
y = 0111
then x + y = 1111.

If I'm on the right track here, it seems that the formula should be w(x + y) = w(x) + w(y) - w(x*y), without the factor of 2 that you showed. Of course, I might not be on the right track, since I'm not sure what w(x*y) means.