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256 lines
7.5 KiB
256 lines
7.5 KiB
from functools import reduce
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from math import prod
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from sympy.core.numbers import igcdex, igcd
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from sympy.ntheory.primetest import isprime
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from sympy.polys.domains import ZZ
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from sympy.polys.galoistools import gf_crt, gf_crt1, gf_crt2
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from sympy.utilities.misc import as_int
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def symmetric_residue(a, m):
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"""Return the residual mod m such that it is within half of the modulus.
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>>> from sympy.ntheory.modular import symmetric_residue
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>>> symmetric_residue(1, 6)
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1
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>>> symmetric_residue(4, 6)
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-2
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"""
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if a <= m // 2:
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return a
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return a - m
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def crt(m, v, symmetric=False, check=True):
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r"""Chinese Remainder Theorem.
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The moduli in m are assumed to be pairwise coprime. The output
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is then an integer f, such that f = v_i mod m_i for each pair out
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of v and m. If ``symmetric`` is False a positive integer will be
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returned, else \|f\| will be less than or equal to the LCM of the
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moduli, and thus f may be negative.
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If the moduli are not co-prime the correct result will be returned
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if/when the test of the result is found to be incorrect. This result
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will be None if there is no solution.
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The keyword ``check`` can be set to False if it is known that the moduli
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are coprime.
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Examples
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========
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As an example consider a set of residues ``U = [49, 76, 65]``
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and a set of moduli ``M = [99, 97, 95]``. Then we have::
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>>> from sympy.ntheory.modular import crt
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>>> crt([99, 97, 95], [49, 76, 65])
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(639985, 912285)
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This is the correct result because::
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>>> [639985 % m for m in [99, 97, 95]]
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[49, 76, 65]
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If the moduli are not co-prime, you may receive an incorrect result
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if you use ``check=False``:
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>>> crt([12, 6, 17], [3, 4, 2], check=False)
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(954, 1224)
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>>> [954 % m for m in [12, 6, 17]]
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[6, 0, 2]
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>>> crt([12, 6, 17], [3, 4, 2]) is None
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True
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>>> crt([3, 6], [2, 5])
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(5, 6)
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Note: the order of gf_crt's arguments is reversed relative to crt,
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and that solve_congruence takes residue, modulus pairs.
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Programmer's note: rather than checking that all pairs of moduli share
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no GCD (an O(n**2) test) and rather than factoring all moduli and seeing
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that there is no factor in common, a check that the result gives the
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indicated residuals is performed -- an O(n) operation.
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See Also
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========
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solve_congruence
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sympy.polys.galoistools.gf_crt : low level crt routine used by this routine
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"""
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if check:
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m = list(map(as_int, m))
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v = list(map(as_int, v))
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result = gf_crt(v, m, ZZ)
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mm = prod(m)
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if check:
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if not all(v % m == result % m for v, m in zip(v, m)):
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result = solve_congruence(*list(zip(v, m)),
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check=False, symmetric=symmetric)
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if result is None:
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return result
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result, mm = result
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if symmetric:
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return symmetric_residue(result, mm), mm
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return result, mm
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def crt1(m):
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"""First part of Chinese Remainder Theorem, for multiple application.
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Examples
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========
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>>> from sympy.ntheory.modular import crt1
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>>> crt1([18, 42, 6])
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(4536, [252, 108, 756], [0, 2, 0])
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"""
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return gf_crt1(m, ZZ)
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def crt2(m, v, mm, e, s, symmetric=False):
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"""Second part of Chinese Remainder Theorem, for multiple application.
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Examples
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========
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>>> from sympy.ntheory.modular import crt1, crt2
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>>> mm, e, s = crt1([18, 42, 6])
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>>> crt2([18, 42, 6], [0, 0, 0], mm, e, s)
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(0, 4536)
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"""
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result = gf_crt2(v, m, mm, e, s, ZZ)
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if symmetric:
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return symmetric_residue(result, mm), mm
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return result, mm
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def solve_congruence(*remainder_modulus_pairs, **hint):
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"""Compute the integer ``n`` that has the residual ``ai`` when it is
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divided by ``mi`` where the ``ai`` and ``mi`` are given as pairs to
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this function: ((a1, m1), (a2, m2), ...). If there is no solution,
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return None. Otherwise return ``n`` and its modulus.
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The ``mi`` values need not be co-prime. If it is known that the moduli are
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not co-prime then the hint ``check`` can be set to False (default=True) and
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the check for a quicker solution via crt() (valid when the moduli are
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co-prime) will be skipped.
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If the hint ``symmetric`` is True (default is False), the value of ``n``
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will be within 1/2 of the modulus, possibly negative.
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Examples
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========
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>>> from sympy.ntheory.modular import solve_congruence
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What number is 2 mod 3, 3 mod 5 and 2 mod 7?
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>>> solve_congruence((2, 3), (3, 5), (2, 7))
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(23, 105)
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>>> [23 % m for m in [3, 5, 7]]
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[2, 3, 2]
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If you prefer to work with all remainder in one list and
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all moduli in another, send the arguments like this:
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>>> solve_congruence(*zip((2, 3, 2), (3, 5, 7)))
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(23, 105)
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The moduli need not be co-prime; in this case there may or
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may not be a solution:
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>>> solve_congruence((2, 3), (4, 6)) is None
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True
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>>> solve_congruence((2, 3), (5, 6))
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(5, 6)
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The symmetric flag will make the result be within 1/2 of the modulus:
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>>> solve_congruence((2, 3), (5, 6), symmetric=True)
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(-1, 6)
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See Also
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========
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crt : high level routine implementing the Chinese Remainder Theorem
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"""
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def combine(c1, c2):
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"""Return the tuple (a, m) which satisfies the requirement
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that n = a + i*m satisfy n = a1 + j*m1 and n = a2 = k*m2.
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References
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==========
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.. [1] https://en.wikipedia.org/wiki/Method_of_successive_substitution
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"""
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a1, m1 = c1
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a2, m2 = c2
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a, b, c = m1, a2 - a1, m2
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g = reduce(igcd, [a, b, c])
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a, b, c = [i//g for i in [a, b, c]]
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if a != 1:
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inv_a, _, g = igcdex(a, c)
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if g != 1:
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return None
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b *= inv_a
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a, m = a1 + m1*b, m1*c
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return a, m
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rm = remainder_modulus_pairs
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symmetric = hint.get('symmetric', False)
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if hint.get('check', True):
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rm = [(as_int(r), as_int(m)) for r, m in rm]
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# ignore redundant pairs but raise an error otherwise; also
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# make sure that a unique set of bases is sent to gf_crt if
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# they are all prime.
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#
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# The routine will work out less-trivial violations and
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# return None, e.g. for the pairs (1,3) and (14,42) there
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# is no answer because 14 mod 42 (having a gcd of 14) implies
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# (14/2) mod (42/2), (14/7) mod (42/7) and (14/14) mod (42/14)
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# which, being 0 mod 3, is inconsistent with 1 mod 3. But to
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# preprocess the input beyond checking of another pair with 42
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# or 3 as the modulus (for this example) is not necessary.
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uniq = {}
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for r, m in rm:
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r %= m
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if m in uniq:
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if r != uniq[m]:
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return None
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continue
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uniq[m] = r
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rm = [(r, m) for m, r in uniq.items()]
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del uniq
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# if the moduli are co-prime, the crt will be significantly faster;
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# checking all pairs for being co-prime gets to be slow but a prime
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# test is a good trade-off
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if all(isprime(m) for r, m in rm):
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r, m = list(zip(*rm))
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return crt(m, r, symmetric=symmetric, check=False)
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rv = (0, 1)
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for rmi in rm:
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rv = combine(rv, rmi)
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if rv is None:
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break
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n, m = rv
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n = n % m
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else:
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if symmetric:
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return symmetric_residue(n, m), m
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return n, m
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