#
# Copyright (c) 2008-2020 Stefan Krah. All rights reserved.
#
# Redistribution and use in source and binary forms, with or without
# modification, are permitted provided that the following conditions
# are met:
#
# 1. Redistributions of source code must retain the above copyright
# notice, this list of conditions and the following disclaimer.
#
# 2. Redistributions in binary form must reproduce the above copyright
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#
# THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS "AS IS" AND
# ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
# ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
# FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
# DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
# OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
# HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
# LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
# OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
# SUCH DAMAGE.
#
######################################################################
# This file lists and checks some of the constants and limits used #
# in libmpdec's Number Theoretic Transform. At the end of the file #
# there is an example function for the plain DFT transform. #
######################################################################
#
# Number theoretic transforms are done in subfields of F(p). P[i]
# are the primes, D[i] = P[i] - 1 are highly composite and w[i]
# are the respective primitive roots of F(p).
#
# The strategy is to convolute two coefficients modulo all three
# primes, then use the Chinese Remainder Theorem on the three
# result arrays to recover the result in the usual base RADIX
# form.
#
# ======================================================================
# Primitive roots
# ======================================================================
#
# Verify primitive roots:
#
# For a prime field, r is a primitive root if and only if for all prime
# factors f of p-1, r**((p-1)/f) =/= 1 (mod p).
#
def prod(F, E):
"""Check that the factorization of P-1 is correct. F is the list of
factors of P-1, E lists the number of occurrences of each factor."""
x = 1
for y, z in zip(F, E):
x *= y**z
return x
def is_primitive_root(r, p, factors, exponents):
"""Check if r is a primitive root of F(p)."""
if p != prod(factors, exponents) + 1:
return False
for f in factors:
q, control = divmod(p-1, f)
if control != 0:
return False
if pow(r, q, p) == 1:
return False
return True
# =================================================================
# Constants and limits for the 64-bit version
# =================================================================
RADIX = 10**19
# Primes P1, P2 and P3:
P = [2**64-2**32+1, 2**64-2**34+1, 2**64-2**40+1]
# P-1, highly composite. The transform length d is variable and
# must divide D = P-1. Since all D are divisible by 3 * 2**32,
# transform lengths can be 2**n or 3 * 2**n (where n <= 32).
D = [2**32 * 3 * (5 * 17 * 257 * 65537),
2**34 * 3**2 * (7 * 11 * 31 * 151 * 331),
2**40 * 3**2 * (5 * 7 * 13 * 17 * 241)]
# Prime factors of P-1 and their exponents:
F = [(2,3,5,17,257,65537), (2,3,7,11,31,151,331), (2,3,5,7,13,17,241)]
E = [(32,1,1,1,1,1), (34,2,1,1,1,1,1), (40,2,1,1,1,1,1)]
# Maximum transform length for 2**n. Above that only 3 * 2**31
# or 3 * 2**32 are possible.
MPD_MAXTRANSFORM_2N = 2**32
# Limits in the terminology of Pollard's paper:
m2 = (MPD_MAXTRANSFORM_2N * 3) // 2 # Maximum length of the smaller array.
M1 = M2 = RADIX-1 # Maximum value per single word.
L = m2 * M1 * M2
P[0] * P[1] * P[2] > 2 * L
# Primitive roots of F(P1), F(P2) and F(P3):
w = [7, 10, 19]
# The primitive roots are correct:
for i in range(3):
if not is_primitive_root(w[i], P[i], F[i], E[i]):
print("FAIL")
# =================================================================
# Constants and limits for the 32-bit version
# =================================================================
RADIX = 10**9
# Primes P1, P2 and P3:
P = [2113929217, 2013265921, 1811939329]
# P-1, highly composite. All D = P-1 are divisible by 3 * 2**25,
# allowing for transform lengths up to 3 * 2**25 words.
D = [2**25 * 3**2 * 7,
2**27 * 3 * 5,
2**26 * 3**3]
# Prime factors of P-1 and their exponents:
F = [(2,3,7), (2,3,5), (2,3)]
E = [(25,2,1), (27,1,1), (26,3)]
# Maximum transform length for 2**n. Above that only 3 * 2**24 or
# 3 * 2**25 are possible.
MPD_MAXTRANSFORM_2N = 2**25
# Limits in the terminology of Pollard's paper:
m2 = (MPD_MAXTRANSFORM_2N * 3) // 2 # Maximum length of the smaller array.
M1 = M2 = RADIX-1 # Maximum value per single word.
L = m2 * M1 * M2
P[0] * P[1] * P[2] > 2 * L
# Primitive roots of F(P1), F(P2) and F(P3):
w = [5, 31, 13]
# The primitive roots are correct:
for i in range(3):
if not is_primitive_root(w[i], P[i], F[i], E[i]):
print("FAIL")
# ======================================================================
# Example transform using a single prime
# ======================================================================
def ntt(lst, dir):
"""Perform a transform on the elements of lst. len(lst) must
be 2**n or 3 * 2**n, where n <= 25. This is the slow DFT."""
p = 2113929217 # prime
d = len(lst) # transform length
d_prime = pow(d, (p-2), p) # inverse of d
xi = (p-1)//d
w = 5 # primitive root of F(p)
r = pow(w, xi, p) # primitive root of the subfield
r_prime = pow(w, (p-1-xi), p) # inverse of r
if dir == 1: # forward transform
a = lst # input array
A = [0] * d # transformed values
for i in range(d):
s = 0
for j in range(d):
s += a[j] * pow(r, i*j, p)
A[i] = s % p
return A
elif dir == -1: # backward transform
A = lst # input array
a = [0] * d # transformed values
for j in range(d):
s = 0
for i in range(d):
s += A[i] * pow(r_prime, i*j, p)
a[j] = (d_prime * s) % p
return a
def ntt_convolute(a, b):
"""convolute arrays a and b."""
assert(len(a) == len(b))
x = ntt(a, 1)
y = ntt(b, 1)
for i in range(len(a)):
y[i] = y[i] * x[i]
r = ntt(y, -1)
return r
# Example: Two arrays representing 21 and 81 in little-endian:
a = [1, 2, 0, 0]
b = [1, 8, 0, 0]
assert(ntt_convolute(a, b) == [1, 10, 16, 0])
assert(21 * 81 == (1*10**0 + 10*10**1 + 16*10**2 + 0*10**3))
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