What does L_n is the bit length of the group order n states need for my calculation in ECDSA algorithm?

Question

I am doing the program of implementing ECDSA for which I am trying to solve the equation scenario. In ECDSA that signature generation algorithm which states as hash value from the SHA-1 and where $l$ denotes the leftmost bits of the hash.

For example of hash value de9f2c7fd25e1b3afad3e85a0bd17d9b100db4b3. In that hash value what is the leftmost bits represent in the z value?
Is it d or the last bit of binary of d such as 1101 which implies 1 in leftmost?

Let $z$ be the $L_n$ leftmost bits of $e$, where $L_n$ is the bit length of the group order $n$.

In my scenario the $n$ length is 17. I guess it is the $n=17$.

What does will have? Binary or hex value or decimal which is used in the calculation of $s$ in step 4?

Answer 1

In binary, your hash value is:

11011110100111110010....

And so the leftmost 17 bits would be 11011110100111110 or 1BD3E in hex.

Normally, for curves we actually use for ECDSA, $n$ is a multiple of 8 (and so we just use the leftmost $n/8$ bytes); in your (toy) case, we need to be a bit more careful (and end up not preserving byte boundaries).