Source code for sympy.ntheory.modular

from sympy.core.numbers import igcdex, igcd
from sympy.core.mul import prod
from sympy.core.compatibility import as_int
from sympy.ntheory.primetest import isprime
from sympy.polys.domains import ZZ
from sympy.polys.galoistools import gf_crt, gf_crt1, gf_crt2

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