• 设为首页
  • 点击收藏
  • 手机版
    手机扫一扫访问
    迪恩网络手机版
  • 关注官方公众号
    微信扫一扫关注
    迪恩网络公众号

Python polys.gcd函数代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

本文整理汇总了Python中sympy.polys.gcd函数的典型用法代码示例。如果您正苦于以下问题:Python gcd函数的具体用法?Python gcd怎么用?Python gcd使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。



在下文中一共展示了gcd函数的16个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。

示例1: rrs_simple

def rrs_simple(n):
    """
    @type n: integer > 0
    @rtype: list

    @return: a reduced residue system modulo n
    """
    return [x for x in range(n) if gcd(x, n) == 1]
开发者ID:wangjiezhe,项目名称:ent_note,代码行数:8,代码来源:residue_system.py


示例2: weak_normalizer

def weak_normalizer(a, d, DE, z=None):
    """
    Weak normalization.

    Given a derivation D on k[t] and f == a/d in k(t), return q in k[t]
    such that f - Dq/q is weakly normalized with respect to t.

    f in k(t) is said to be "weakly normalized" with respect to t if
    residue_p(f) is not a positive integer for any normal irreducible p
    in k[t] such that f is in R_p (Definition 6.1.1).  If f has an
    elementary integral, this is equivalent to no logarithm of
    integral(f) whose argument depends on t has a positive integer
    coefficient, where the arguments of the logarithms not in k(t) are
    in k[t].

    Returns (q, f - Dq/q)
    """
    z = z or Dummy('z')
    dn, ds = splitfactor(d, DE)

    # Compute d1, where dn == d1*d2**2*...*dn**n is a square-free
    # factorization of d.
    g = gcd(dn, dn.diff(DE.t))
    d_sqf_part = dn.quo(g)
    d1 = d_sqf_part.quo(gcd(d_sqf_part, g))

    a1, b = gcdex_diophantine(d.quo(d1).as_poly(DE.t), d1.as_poly(DE.t),
        a.as_poly(DE.t))
    r = (a - Poly(z, DE.t)*derivation(d1, DE)).as_poly(DE.t).resultant(
        d1.as_poly(DE.t))
    r = Poly(r, z)

    if not r.has(z):
        return (Poly(1, DE.t), (a, d))

    N = [i for i in r.real_roots() if i in ZZ and i > 0]

    q = reduce(mul, [gcd(a - Poly(n, DE.t)*derivation(d1, DE), d1) for n in N],
        Poly(1, DE.t))

    dq = derivation(q, DE)
    sn = q*a - d*dq
    sd = q*d
    sn, sd = sn.cancel(sd, include=True)

    return (q, (sn, sd))
开发者ID:Abhityagi16,项目名称:sympy,代码行数:46,代码来源:rde.py


示例3: _deflation

    def _deflation(p):
        for y in V:
            if not p.has(y):
                continue

            if _derivation(p) is not S.Zero:
                c, q = p.as_poly(y).primitive()
                return _deflation(c)*gcd(q, q.diff(y)).as_expr()
        else:
            return p
开发者ID:AALEKH,项目名称:sympy,代码行数:10,代码来源:heurisch.py


示例4: deflation

    def deflation(p):
        for y in V:
            if not p.has_any_symbols(y):
                continue

            if derivation(p) is not S.Zero:
                c, q = p.as_poly(y).as_primitive()
                return deflation(c)*gcd(q, q.diff(y))
        else:
            return p
开发者ID:jcockayne,项目名称:sympy-rkern,代码行数:10,代码来源:risch.py


示例5: extract

                def extract(f):
                    p = f.args[0]

                    for q in f.args[1:]:
                        p = gcd(p, q, *symbols)

                        if p.is_number:
                            return S.One, f

                    return p, Add(*[quo(g, p, *symbols) for g in f.args])
开发者ID:jcockayne,项目名称:sympy-rkern,代码行数:10,代码来源:rewrite.py


示例6: rsa

def rsa(bits):
    p = nextprime(randrange(2**(bits // 2 + 1)))
    q = nextprime(randrange(2**(bits // 2 + 1)))
    n = p * q
    phi_n = (p - 1) * (q - 1)
    while True:
        e = randrange(1, phi_n)
        if gcd(e, phi_n) == 1:
            break
    d = inv_mod(e, phi_n)
    return e, d, n
开发者ID:wangjiezhe,项目名称:ent_note,代码行数:11,代码来源:rsa.py


示例7: _splitter

    def _splitter(p):
        for y in V:
            if not p.has(y):
                continue

            if _derivation(y) is not S.Zero:
                c, q = p.as_poly(y).primitive()

                q = q.as_expr()

                h = gcd(q, _derivation(q), y)
                s = quo(h, gcd(q, q.diff(y), y), y)

                c_split = _splitter(c)

                if s.as_poly(y).degree() == 0:
                    return (c_split[0], q * c_split[1])

                q_split = _splitter(cancel(q / s))

                return (c_split[0]*q_split[0]*s, c_split[1]*q_split[1])
        else:
            return (S.One, p)
开发者ID:AALEKH,项目名称:sympy,代码行数:23,代码来源:heurisch.py


示例8: splitter

    def splitter(p):
        for y in V:
            if not p.has_any_symbols(y):
                continue

            if derivation(y) is not S.Zero:
                c, q = p.as_poly(y).as_primitive()

                q = q.as_basic()

                h = gcd(q, derivation(q), y)
                s = quo(h, gcd(q, q.diff(y), y), y)

                c_split = splitter(c)

                if s.as_poly(y).degree == 0:
                    return (c_split[0], q * c_split[1])

                q_split = splitter(Poly.cancel((q, s), *V))

                return (c_split[0]*q_split[0]*s, c_split[1]*q_split[1])
        else:
            return (S.One, p)
开发者ID:jcockayne,项目名称:sympy-rkern,代码行数:23,代码来源:risch.py


示例9: crack_given_decrypt

def crack_given_decrypt(n, m):
    n = int(n)
    m = int(m)

    while True:
        if m % 2 != 0:
            break
        divide_out = True
        for _ in range(5):
            a = randrange(1, n)
            if gcd(a, n) == 1:
                if pow(a, m // 2, n) != 1:
                    divide_out = False
                    break
        if divide_out:
            m //= 2
        else:
            break

    while True:
        a = randrange(1, n)
        g = gcd(pow(a, m // 2, n) - 1, n)
        if g != 1 and g != n:
            return g
开发者ID:wangjiezhe,项目名称:ent_note,代码行数:24,代码来源:rsa.py


示例10: _split_gcd

def _split_gcd(*a):
    """
    split the list of integers ``a`` into a list of integers, ``a1`` having
    ``g = gcd(a1)``, and a list ``a2`` whose elements are not divisible by
    ``g``.  Returns ``g, a1, a2``

    Examples
    ========

    >>> from sympy.simplify.radsimp import _split_gcd
    >>> _split_gcd(55, 35, 22, 14, 77, 10)
    (5, [55, 35, 10], [22, 14, 77])
    """
    g = a[0]
    b1 = [g]
    b2 = []
    for x in a[1:]:
        g1 = gcd(g, x)
        if g1 == 1:
            b2.append(x)
        else:
            g = g1
            b1.append(x)
    return g, b1, b2
开发者ID:tachycline,项目名称:sympy,代码行数:24,代码来源:radsimp.py


示例11: factor_nc

def factor_nc(expr):
    """Return the factored form of ``expr`` while handling non-commutative
    expressions.

    **examples**
    >>> from sympy.core.exprtools import factor_nc
    >>> from sympy import Symbol
    >>> from sympy.abc import x
    >>> A = Symbol('A', commutative=False)
    >>> B = Symbol('B', commutative=False)
    >>> factor_nc((x**2 + 2*A*x + A**2).expand())
    (x + A)**2
    >>> factor_nc(((x + A)*(x + B)).expand())
    (x + A)*(x + B)
    """
    from sympy.simplify.simplify import _mexpand
    from sympy.polys import gcd, factor

    expr = sympify(expr)
    if not isinstance(expr, Expr) or not expr.args:
        return expr
    if not expr.is_Add:
        return expr.func(*[factor_nc(a) for a in expr.args])

    expr, rep, nc_symbols = _mask_nc(expr)
    if rep:
        return factor(expr).subs(rep)
    else:
        args = [a.args_cnc() for a in Add.make_args(expr)]
        c = g = l = r = S.One
        hit = False
        # find any commutative gcd term
        for i, a in enumerate(args):
            if i == 0:
                c = Mul._from_args(a[0])
            elif a[0]:
                c = gcd(c, Mul._from_args(a[0]))
            else:
                c = S.One
        if c is not S.One:
            hit = True
            c, g = c.as_coeff_Mul()
            if g is not S.One:
                for i, (cc, _) in enumerate(args):
                    cc = list(Mul.make_args(Mul._from_args(list(cc))/g))
                    args[i][0] = cc
            else:
                for i, (cc, _) in enumerate(args):
                    cc[0] = cc[0]/c
                    args[i][0] = cc
        # find any noncommutative common prefix
        for i, a in enumerate(args):
            if i == 0:
                n = a[1][:]
            else:
                n = common_prefix(n, a[1])
            if not n:
                # is there a power that can be extracted?
                if not args[0][1]:
                    break
                b, e = args[0][1][0].as_base_exp()
                ok = False
                if e.is_Integer:
                    for t in args:
                        if not t[1]:
                            break
                        bt, et = t[1][0].as_base_exp()
                        if et.is_Integer and bt == b:
                            e = min(e, et)
                        else:
                            break
                    else:
                        ok = hit = True
                        l = b**e
                        il = b**-e
                        for i, a in enumerate(args):
                            args[i][1][0] = il*args[i][1][0]
                        break
                if not ok:
                    break
        else:
            hit = True
            lenn = len(n)
            l = Mul(*n)
            for i, a in enumerate(args):
                args[i][1] = args[i][1][lenn:]
        # find any noncommutative common suffix
        for i, a in enumerate(args):
            if i == 0:
                n = a[1][:]
            else:
                n = common_suffix(n, a[1])
            if not n:
                # is there a power that can be extracted?
                if not args[0][1]:
                    break
                b, e = args[0][1][-1].as_base_exp()
                ok = False
                if e.is_Integer:
                    for t in args:
#.........这里部分代码省略.........
开发者ID:FireJade,项目名称:sympy,代码行数:101,代码来源:exprtools.py


示例12: rsolve_ratio

def rsolve_ratio(coeffs, f, n, **hints):
    """Given linear recurrence operator L of order 'k' with polynomial
       coefficients and inhomogeneous equation Ly = f, where 'f' is a
       polynomial, we seek for all rational solutions over field K of
       characteristic zero.

       This procedure accepts only polynomials, however if you are
       interested in solving recurrence with rational coefficients
       then use rsolve() which will pre-process the given equation
       and run this procedure with polynomial arguments.

       The algorithm performs two basic steps:

           (1) Compute polynomial v(n) which can be used as universal
               denominator of any rational solution of equation Ly = f.

           (2) Construct new linear difference equation by substitution
               y(n) = u(n)/v(n) and solve it for u(n) finding all its
               polynomial solutions. Return None if none were found.

       Algorithm implemented here is a revised version of the original
       Abramov's algorithm, developed in 1989. The new approach is much
       simpler to implement and has better overall efficiency. This
       method can be easily adapted to q-difference equations case.

       Besides finding rational solutions alone, this functions is
       an important part of Hyper algorithm were it is used to find
       particular solution of inhomogeneous part of a recurrence.

       For more information on the implemented algorithm refer to:

       [1] S. A. Abramov, Rational solutions of linear difference
           and q-difference equations with polynomial coefficients,
           in: T. Levelt, ed., Proc. ISSAC '95, ACM Press, New York,
           1995, 285-289

    """
    f = sympify(f)

    if not f.is_polynomial(n):
        return None

    coeffs = map(sympify, coeffs)

    r = len(coeffs)-1

    A, B = coeffs[r], coeffs[0]
    A = A.subs(n, n-r).expand()

    h = Symbol('h', dummy=True)

    res = resultant(A, B.subs(n, n+h), n)

    if not res.is_polynomial(h):
        p, q = res.as_numer_denom()
        res = exquo(p, q, h)

    nni_roots = roots(res, h, filter='Z',
        predicate=lambda r: r >= 0).keys()

    if not nni_roots:
        return rsolve_poly(coeffs, f, n, **hints)
    else:
        C, numers = S.One, [S.Zero]*(r+1)

        for i in xrange(int(max(nni_roots)), -1, -1):
            d = gcd(A, B.subs(n, n+i), n)

            A = exquo(A, d, n)
            B = exquo(B, d.subs(n, n-i), n)

            C *= Mul(*[ d.subs(n, n-j) for j in xrange(0, i+1) ])

        denoms = [ C.subs(n, n+i) for i in range(0, r+1) ]

        for i in range(0, r+1):
            g = gcd(coeffs[i], denoms[i], n)

            numers[i] = exquo(coeffs[i], g, n)
            denoms[i] = exquo(denoms[i], g, n)

        for i in xrange(0, r+1):
            numers[i] *= Mul(*(denoms[:i] + denoms[i+1:]))

        result = rsolve_poly(numers, f * Mul(*denoms), n, **hints)

        if result is not None:
            if hints.get('symbols', False):
                return (simplify(result[0] / C), result[1])
            else:
                return simplify(result / C)
        else:
            return None
开发者ID:Sumith1896,项目名称:sympy-polys,代码行数:93,代码来源:recurr.py


示例13: rsolve_ratio

def rsolve_ratio(coeffs, f, n, **hints):
    """
    Given linear recurrence operator `\operatorname{L}` of order `k`
    with polynomial coefficients and inhomogeneous equation
    `\operatorname{L} y = f`, where `f` is a polynomial, we seek
    for all rational solutions over field `K` of characteristic zero.

    This procedure accepts only polynomials, however if you are
    interested in solving recurrence with rational coefficients
    then use ``rsolve`` which will pre-process the given equation
    and run this procedure with polynomial arguments.

    The algorithm performs two basic steps:

        (1) Compute polynomial `v(n)` which can be used as universal
            denominator of any rational solution of equation
            `\operatorname{L} y = f`.

        (2) Construct new linear difference equation by substitution
            `y(n) = u(n)/v(n)` and solve it for `u(n)` finding all its
            polynomial solutions. Return ``None`` if none were found.

    Algorithm implemented here is a revised version of the original
    Abramov's algorithm, developed in 1989. The new approach is much
    simpler to implement and has better overall efficiency. This
    method can be easily adapted to q-difference equations case.

    Besides finding rational solutions alone, this functions is
    an important part of Hyper algorithm were it is used to find
    particular solution of inhomogeneous part of a recurrence.

    Examples
    ========

    >>> from sympy.abc import x
    >>> from sympy.solvers.recurr import rsolve_ratio
    >>> rsolve_ratio([-2*x**3 + x**2 + 2*x - 1, 2*x**3 + x**2 - 6*x,
    ... - 2*x**3 - 11*x**2 - 18*x - 9, 2*x**3 + 13*x**2 + 22*x + 8], 0, x)
    C2*(2*x - 3)/(2*(x**2 - 1))

    References
    ==========

    .. [1] S. A. Abramov, Rational solutions of linear difference
           and q-difference equations with polynomial coefficients,
           in: T. Levelt, ed., Proc. ISSAC '95, ACM Press, New York,
           1995, 285-289

    See Also
    ========

    rsolve_hyper
    """
    f = sympify(f)

    if not f.is_polynomial(n):
        return None

    coeffs = map(sympify, coeffs)

    r = len(coeffs) - 1

    A, B = coeffs[r], coeffs[0]
    A = A.subs(n, n - r).expand()

    h = Dummy('h')

    res = resultant(A, B.subs(n, n + h), n)

    if not res.is_polynomial(h):
        p, q = res.as_numer_denom()
        res = quo(p, q, h)

    nni_roots = roots(res, h, filter='Z',
        predicate=lambda r: r >= 0).keys()

    if not nni_roots:
        return rsolve_poly(coeffs, f, n, **hints)
    else:
        C, numers = S.One, [S.Zero]*(r + 1)

        for i in xrange(int(max(nni_roots)), -1, -1):
            d = gcd(A, B.subs(n, n + i), n)

            A = quo(A, d, n)
            B = quo(B, d.subs(n, n - i), n)

            C *= Mul(*[ d.subs(n, n - j) for j in xrange(0, i + 1) ])

        denoms = [ C.subs(n, n + i) for i in range(0, r + 1) ]

        for i in range(0, r + 1):
            g = gcd(coeffs[i], denoms[i], n)

            numers[i] = quo(coeffs[i], g, n)
            denoms[i] = quo(denoms[i], g, n)

        for i in xrange(0, r + 1):
            numers[i] *= Mul(*(denoms[:i] + denoms[i + 1:]))

#.........这里部分代码省略.........
开发者ID:alhirzel,项目名称:sympy,代码行数:101,代码来源:recurr.py


示例14: nc_gcd

 def nc_gcd(aa, bb):
     a, b = [i.as_coeff_Mul() for i in [aa, bb]]
     c = gcd(a[0], b[0]).as_numer_denom()[0]
     g = Mul(*(a[1].args_cnc(cset=True)[0] & b[1].args_cnc(cset=True)[0]))
     return _keep_coeff(c, g)
开发者ID:bjodah,项目名称:sympy,代码行数:5,代码来源:powsimp.py


示例15: gcd

        def gcd(f, g):
            from sympy.polys import gcd

            return f.__class__(gcd(f.ex, f.__class__(g).ex))
开发者ID:Carreau,项目名称:sympy,代码行数:4,代码来源:expressiondomain.py


示例16: normal

def normal(f, g, n=None):
    """Given relatively prime univariate polynomials 'f' and 'g',
       rewrite their quotient to a normal form defined as follows:

                       f(n)       A(n) C(n+1)
                       ----  =  Z -----------
                       g(n)       B(n)  C(n)

       where Z is arbitrary constant and A, B, C are monic
       polynomials in 'n' with following properties:

           (1) gcd(A(n), B(n+h)) = 1 for all 'h' in N
           (2) gcd(B(n), C(n+1)) = 1
           (3) gcd(A(n), C(n)) = 1

       This normal form, or rational factorization in other words,
       is crucial step in Gosper's algorithm and in difference
       equations solving. It can be also used to decide if two
       hypergeometric are similar or not.

       This procedure will return a tuple containing elements
       of this factorization in the form (Z*A, B, C). For example:

       >>> from sympy import Symbol, normal
       >>> n = Symbol('n', integer=True)

       >>> normal(4*n+5, 2*(4*n+1)*(2*n+3), n)
       (1/4, 3/2 + n, 1/4 + n)

    """
    f, g = map(sympify, (f, g))

    p = f.as_poly(n, field=True)
    q = g.as_poly(n, field=True)

    a, p = p.LC(), p.monic()
    b, q = q.LC(), q.monic()

    A = p.as_basic()
    B = q.as_basic()

    C, Z = S.One, a / b

    h = Dummy('h')

    res = resultant(A, B.subs(n, n+h), n)

    nni_roots = roots(res, h, filter='Z',
        predicate=lambda r: r >= 0).keys()

    if not nni_roots:
        return (f, g, S.One)
    else:
        for i in sorted(nni_roots):
            d = gcd(A, B.subs(n, n+i), n)

            A = quo(A, d, n)
            B = quo(B, d.subs(n, n-i), n)

            C *= Mul(*[ d.subs(n, n-j) for j in xrange(1, i+1) ])

        return (Z*A, B, C)
开发者ID:Aang,项目名称:sympy,代码行数:62,代码来源:gosper.py



注:本文中的sympy.polys.gcd函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。


鲜花

握手

雷人

路过

鸡蛋
该文章已有0人参与评论

请发表评论

全部评论

专题导读
上一篇:
Python polys.groebner函数代码示例发布时间:2022-05-27
下一篇:
Python polys.field函数代码示例发布时间:2022-05-27
热门推荐
阅读排行榜

扫描微信二维码

查看手机版网站

随时了解更新最新资讯

139-2527-9053

在线客服(服务时间 9:00~18:00)

在线QQ客服
地址:深圳市南山区西丽大学城创智工业园
电邮:jeky_zhao#qq.com
移动电话:139-2527-9053

Powered by 互联科技 X3.4© 2001-2213 极客世界.|Sitemap