Definition: Hamming weight
Consider any $z\in\mathbb Z$. The hamming weight of $binary(z)$ is then defined as the hamming distance [1] to the zero string of same length. That means (using the language of coding-theory) every set bit increases the weight. Otherwise defined by linear algebra as: Let $z=\sum_{i=0}^n z_i 2^i$ then the hamming weight $h(z)=\sum_{i=0}^n z_i$.
This is the definition of Hamming weight, that I know.
Question
How could it be, that then $u=-2^{77} +2^{50} + 2^{33}$ has $h(u)=3$ or $v=2^{35}-2^{32}-2^{18}+2^8+1$ has $h(v)=5$? The negative sign in the beginning is confusing me. That tells me, that I have to consider 3 conditions, instead of 2?
Binary representation of u and v
$u=-0b11111111111111111111111111011111111111111111000000000000000000000000000000000$
and
$v=0b11011111111111111000000000100000001$
That shows, that the given definition does not hold in case, the given hamming weight by [2] is right.
The Hamming weight does not take the value of the coefficient into account, it simply adds $1$ if the coefficient is $\ne 0$ :
$$u=-2^{77} +2^{50} + 2^{33} \Rightarrow h(u) = 1+1+1 = 3$$
$$v=2^{35}-2^{32}-2^{18}+2^8+1 \Rightarrow h(v) = 1+1+1+1+1 = 5 $$
The general formula is :
$$z = \sum_{i=0}^n z_i2^i \Rightarrow h(z) = \sum_{i=0\\z_i\ne 0}^n 1$$