a RG5d@sddlmZddlmZddlmZmZddlmZddl m Z ddl m Z m Z mZddlmZdd Zdd d ZddZdddZddZdS))reduce)prod)igcdexigcdisprime)ZZ)gf_crtgf_crt1gf_crt2as_intcCs||dkr|S||S)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 )amrrQ/var/www/html/django/DPS/env/lib/python3.9/site-packages/sympy/ntheory/modular.pysymmetric_residue s rFTcs|r ttt|}ttt|}t||tt|}|rtfddt||Dsttt||d|ddurzS\}|rt ||fS|fS)akChinese 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 c3s"|]\}}|||kVqdSNr).0vrresultrr [zcrt..F)check symmetricN) listmapr r rrallzipsolve_congruencer)rrrrmmrrrcrts : r#cCs t|tS)zFirst 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]) )r r)rrrrcrt1gs r$cCs,t|||||t}|r$t|||fS||fS)zSecond 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) )r rr)rrr"esrrrrrcrt2us r'c Os dd}|}|dd}|ddrdd|D}i}|D]4\}}||;}||vrf|||kr:d Sq:|||<q:d d|D}~td d |Drtt|\}}t|||dd Sd}|D],} ||| }|d urq|\} }| |} q|rt| ||fS| |fSd S)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 c s|\}}|\}}||||}}}tt|||gfdd|||fD\}}}|dkr~t||\} } dkrvdS|| 9}|||||}} || fS)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 csg|] }|qSrr)rigrr rz5solve_congruence..combine..N)rrr) c1c2a1m1a2m2rbcinv_a_rrr)rcombines z!solve_congruence..combinerFrTcSs g|]\}}t|t|fqSrr rrrrrrr+rz$solve_congruence..NcSsg|]\}}||fqSrr)rrr9rrrr+rcss|]\}}t|VqdSrrr8rrrrrz#solve_congruence..)rr)rr,)getitemsrrr r#r) Zremainder_modulus_pairshintr7Zrmruniqr9rrvZrminrrrr!s84        r!N)FT)F) functoolsrmathrsympy.core.numbersrrsympy.ntheory.primetestrsympy.polys.domainsrsympy.polys.galoistoolsr r r sympy.utilities.miscr rr#r$r'r!rrrrs      N