\h!~SSKJr SSKJrJr SSKJr SSKJr SSK J r J r J r SSK Jr SrS SjrS rSS jrS rg ))prodgcdgcdextisprime)ZZ)gf_crtgf_crt1gf_crt2)as_intcXS-::aU$X- $)zReturn the residual mod m such that it is within half of the modulus. >>> from sympy.ntheory.modular import symmetric_residue >>> symmetric_residue(1, 6) 1 >>> symmetric_residue(4, 6) -2 )ams M/var/www/auris/envauris/lib/python3.13/site-packages/sympy/ntheory/modular.pysymmetric_residuer s F{ 5Lc^U(a2[[[U55n[[[U55n[X[5m[ U5nU(aK[ U4Sj[X555(d([[[X55SUS.6mTcT$TumnU(a [[TU55[U54$[T5[U54$)aChinese 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. Examples ======== 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 >>> 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 c3<># UHupX-TU-:Hv M g7fNr).0vrresults r crt..Zs=94115FQJ&9sF)check symmetric) listmapr r r rallzipsolve_congruenceintr)rrrrmmrs @rcrtr'st VQ  VQ  A" F aB =3q9===%tCI96F~ JFB$VR013r7:: v;B rc"[U[5$)aFirst part of Chinese Remainder Theorem, for multiple application. Examples ======== >>> from sympy.ntheory.modular import crt, crt1, crt2 >>> m = [99, 97, 95] >>> v = [49, 76, 65] The following two codes have the same result. >>> crt(m, v) (639985, 912285) >>> mm, e, s = crt1(m) >>> crt2(m, v, mm, e, s) (639985, 912285) However, it is faster when we want to fix ``m`` and compute for multiple ``v``, i.e. the following cases: >>> mm, e, s = crt1(m) >>> vs = [[52, 21, 37], [19, 46, 76]] >>> for v in vs: ... print(crt2(m, v, mm, e, s)) (397042, 912285) (803206, 912285) See Also ======== sympy.polys.galoistools.gf_crt1 : low level crt routine used by this routine sympy.ntheory.modular.crt sympy.ntheory.modular.crt2 )r r )rs rcrt1r)fsL 1b>rc[XX#U[5nU(a[[Xb55[U54$[U5[U54$)aSecond part of Chinese Remainder Theorem, for multiple application. See ``crt1`` for usage. 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) See Also ======== sympy.polys.galoistools.gf_crt2 : low level crt routine used by this routine sympy.ntheory.modular.crt sympy.ntheory.modular.crt1 )r r r%r)rrr&esrrs rcrt2r-sD,Q2!R (F$V013r7:: v;B rc>SnUnURSS5nURSS5(aUVVs/sHupV[U5[U54PM nnn0nUHupVXV-nXg;a XWU:wa gMXWU'M UR5VVs/sHupeXV4PM nnnA[SU55(a[ [ U65upV[ XeUSS9$S nUHn U"X5nUc gUupX-n M U(a[W W5U4$W W4$s snnfs snnf) aCompute 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 cUup#UupEX4U- Upn[XgU5n XgU4V s/sHoU -PM sn upgnUS:wa[Xh5upn U S:wagX{-nX#U--X8-pXm4$s sn f)zReturn the tuple (a, m) which satisfies the requirement that n = a + i*m satisfy n = a1 + j*m1 and n = a2 = k*m2. References ========== .. [1] https://en.wikipedia.org/wiki/Method_of_successive_substitution Nr)c1c2a1m1a2m2rbcgiinv_a_rs rcombine!solve_congruence..combines2gra aL"#+Aa4+a 6 ,KAaAv JAqDy"$1t ,sA"rFrTNc3<# UHup[U5v M g7frr)rrrs rr#solve_congruence..s)bdawqzzbs)rr)rr0)getr itemsr"r r#r'r) remainder_modulus_pairshintr=rmrr@runiqrvrmins rr$r$s6h, !Be,I xx13 4vay&)$ 4DA FAyQ<G "& .qf .  )b) ) )R>DAqy> > B R  :  E  $Q*A- -!t S5*/s "DDN)FT)F)mathrsympy.external.gmpyrrsympy.ntheory.primetestrsympy.polys.domainsr sympy.polys.galoistoolsr r r sympy.utilities.miscr rr'r)r-r$rrrrQs7++"<<' K \&R :wr