How does wNAF work with prime finite fields?

Question

According to wikipedia, in the precomputation step of the w-ary non-adjacent form (wNAF) point multiplication method you do $d \bmod 2$ and, later, $d \gets \frac{d}2$.

The mod operation doesn't make a lot of sense in the context of a prime finite field since division over a finite field doesn't produce a remainder. Rather, it's the numerator multiplied by the modulo inverse of the denominator. But I guess in this context it can probably just be taken to be the same thing as "is d odd".

But what about $d \gets \frac{d}2$? Outside of the context of prime finite fields that would normally be the same thing as right bit shifting by 1 but in the context of a binary field idk that dividing by two is the same thing as right bit shifting by one. It could be "multiply d by the modulo inverse of 2".

The fact that the while loop is doing while (d > 0) do kinda makes me think it's doing bit shifting but idk.

Any ideas?

Answer 1

The scalar is a scalar, it is not uncommon to view it in $\mathbb{Z}$ instead of in a field defined by the order of the curve/point (note:it does never make sense to view it in a binary field).

So $d\mod2$ becomes the least significant bit of $d$ and $\frac{d}{2}$ becomes the same thing as $floor(\frac{d}{2})$ which is the same thing as a right shift by 1.

Answer 2

The point of wNAF is computing an exponentiation ($g^k$) or a elliptic point multiplication ($kP$). The fact that the order of $g$ or $P$ can be a prime number and thus $k$ can be interpreted as being an element of a prime field is irrelevant. The algorithm would work even if the order was a composite number.

So yes, when wNAF divides by two, it is a integer division by two e.g. right shift.