constantine/constantine/montgomery.nim
Mamy André-Ratsimbazafy 08e12df4ed
internal proc renaming
2019-04-28 14:05:13 +02:00

81 lines
3.3 KiB
Nim

# Constantine
# Copyright (c) 2018 Status Research & Development GmbH
# Licensed and distributed under either of
# * MIT license (license terms in the root directory or at http://opensource.org/licenses/MIT).
# * Apache v2 license (license terms in the root directory or at http://www.apache.org/licenses/LICENSE-2.0).
# at your option. This file may not be copied, modified, or distributed except according to those terms.
# ############################################################
#
# Montgomery domain primitives
#
# ############################################################
import
./word_types, ./bigints, ./field_fp
from bitops import fastLog2
# This will only be used at compile-time
# so no constant-time worries (it is constant-time if using the De Bruijn multiplication)
func montyMagic*(M: static BigInt): static Word =
## Returns the Montgomery domain magic number for the input modulus:
## -1/M[0] mod LimbSize
## M[0] is the least significant limb of M
## M must be odd and greater than 2.
# Test vectors: https://www.researchgate.net/publication/4107322_Montgomery_modular_multiplication_architecture_for_public_key_cryptosystems
# on p354
# Reference C impl: http://www.hackersdelight.org/hdcodetxt/mont64.c.txt
# ######################################################################
# Implementation of modular multiplication inverse
# Assuming 2 positive integers a and m the modulo
#
# We are looking for z that solves `az ≡ 1 mod m`
#
# References:
# - Knuth, The Art of Computer Programming, Vol2 p342
# - Menezes, Handbook of Applied Cryptography (HAC), p610
# http://cacr.uwaterloo.ca/hac/about/chap14.pdf
# Starting from the extended GCD formula (Bezout identity),
# `ax + by = gcd(x,y)` with input x,y and outputs a, b, gcd
# We assume a and m are coprimes, i.e. gcd is 1, otherwise no inverse
# `ax + my = 1` <=> `ax + my ≡ 1 mod m` <=> `ax ≡ 1 mod m`
# For Montgomery magic number, we are in a special case
# where a = M and m = 2^LimbSize.
# For a and m to be coprimes, a must be odd.
# M being a power of 2 greatly simplifies computation:
# - https://crypto.stackexchange.com/questions/47493/how-to-determine-the-multiplicative-inverse-modulo-64-or-other-power-of-two
# - http://groups.google.com/groups?selm=1994Apr6.093116.27805%40mnemosyne.cs.du.edu
# - https://mumble.net/~campbell/2015/01/21/inverse-mod-power-of-two
# - https://eprint.iacr.org/2017/411
# We have the following relation
# ax ≡ 1 (mod 2^k) <=> ax(2 - ax) ≡ 1 (mod 2^(2k))
#
# To get -1/M0 mod LimbSize
# we can either negate the resulting x of `ax(2 - ax) ≡ 1 (mod 2^(2k))`
# or do ax(2 + ax) ≡ 1 (mod 2^(2k))
const
M0 = M.limbs[0]
k = fastLog2(WordBitSize)
result = M0 # Start from an inverse of M0 modulo 2, M0 is odd and it's own inverse
for _ in static(0 ..< k):
result *= 2 + M * result # x' = x(2 + ax) (`+` to avoid negating at the end)
func toMonty*[P: static BigInt](a: Fp[P]): Montgomery[P] =
## Convert a big integer over Fp to it's montgomery representation
## over Fp.
## i.e. Does "a * (2^LimbSize)^W (mod p), where W is the number
## of words needed to represent p in base 2^LimbSize
result = a
for i in static(countdown(P.limbs.high, 0)):
shiftAdd(result, 0)