```
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
```