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Inverse of a number modulo 2**255 -19

asked 2018-09-24 12:26:43 -0500

updated 2018-11-04 12:56:35 -0500

slelievre gravatar image

I don't understand this code to solve the inverse of a number:

b = 256;
q = 2**255 - 19

def expmod(b,e,m): 
    if e == 0: return 1
    t = expmod(b,e/2,m)**2 % m
    if e & 1: t = (t*b) % m
    return t

def inv(x):
   return expmod(x,q-2,q)`

Finally, If I want to put: $\frac{2}{3}$ I can to do this: aux=2*inv(3)

What does the variable e mean?

Could you explain me this code, please?

Thank you so much.

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answered 2018-09-25 02:47:01 -0500

slelievre gravatar image

updated 2018-09-25 02:47:34 -0500

Since $q$ is a prime here, $\mathbb{Z}/q$ is a field, usually denoted by $F_q$.

All its nonzero elements are invertible, and form a group under multiplication. This group is called the multiplicative group of this field, or its group of units.

This group has order $q-1$ (since it contains all nonzero elements of $F_q$), so every nonzero element $x$ in $F_q$ satisfies $x^{q-1} = 1$, which can be written as $x^{q-2} \cdot x = 1$, so we get that for any nonzero $x$ in $ F_q$, $x^{q-2}$ is the multiplicative inverse of $x$.

The function expmod(b, e, m) (for "exponentiation modulo") computes $b^e$ modulo $m$. Here the arguments of this function are named to reflect their role (basis, exponent, modulus).

Finally, the function inv just computes $x^{q-2}$ modulo $q$.

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Asked: 2018-09-24 12:26:43 -0500

Seen: 114 times

Last updated: Nov 04 '18