Python mul.prod函数代码示例

def eval_levicivita(*args):
    """Evaluate Levi-Civita symbol."""
    from sympy import factorial
    n = len(args)
    return prod(
        prod(args[j] - args[i] for j in xrange(i + 1, n))
        / factorial(i) for i in xrange(n))

def denester(nested):
    """
    Denests a list of expressions that contain nested square roots.
    This method should not be called directly - use 'sqrtdenest' instead.
    This algorithm is based on <http://www.almaden.ibm.com/cs/people/fagin/symb85.pdf>.

    It is assumed that all of the elements of 'nested' share the same
    bottom-level radicand. (This is stated in the paper, on page 177, in
    the paragraph immediately preceding the algorithm.)

    When evaluating all of the arguments in parallel, the bottom-level
    radicand only needs to be denested once. This means that calling
    denester with x arguments results in a recursive invocation with x+1
    arguments; hence denester has polynomial complexity.

    However, if the arguments were evaluated separately, each call would
    result in two recursive invocations, and the algorithm would have
    exponential complexity.

    This is discussed in the paper in the middle paragraph of page 179.
    """
    if all((n**2).is_Number for n in nested): #If none of the arguments are nested
        for f in subsets(len(nested)): #Test subset 'f' of nested
            p = prod(nested[i]**2 for i in range(len(f)) if f[i]).expand()
            if 1 in f and f.count(1) > 1 and f[-1]:
                p = -p
            if sqrt(p).is_Number:
                return sqrt(p), f #If we got a perfect square, return its square root.
        return nested[-1], [0]*len(nested) #Otherwise, return the radicand from the previous invocation.
    else:
        a, b, r, R = Wild('a'), Wild('b'), Wild('r'), None
        values = [expr.match(sqrt(a + b * sqrt(r))) for expr in nested]
        if any(v is None for v in values): # this pattern is not recognized
            return nested[-1], [0]*len(nested) #Otherwise, return the radicand from the previous invocation.
        for v in values:
            if r in v: #Since if b=0, r is not defined
                if R is not None:
                    assert R == v[r] #All the 'r's should be the same.
                else:
                    R = v[r]
        if R is None:
            return nested[-1], [0]*len(nested) #return the radicand from the previous invocation.
        d, f = denester([sqrt((v[a]**2).expand()-(R*v[b]**2).expand()) for v in values] + [sqrt(R)])
        if all(fi == 0 for fi in f):
            v = values[-1]
            return sqrt(v[a] + v[b]*d), f
        else:
            v = prod(nested[i]**2 for i in range(len(nested)) if f[i]).expand().match(a+b*sqrt(r))
            if 1 in f and f.index(1) < len(nested) - 1 and f[len(nested)-1]:
                v[a] = -1 * v[a]
                v[b] = -1 * v[b]
            if not f[len(nested)]: #Solution denests with square roots
                vad = (v[a] + d).expand()
                if not vad:
                    return nested[-1], [0]*len(nested) #Otherwise, return the radicand from the previous invocation.
                return (sqrt(vad/2) + sign(v[b])*sqrt((v[b]**2*R/(2*vad)).expand())).expand(), f
            else: #Solution requires a fourth root
                FR, s = (R.expand()**Rational(1,4)), sqrt((v[b]*R).expand()+d)
                return (s/(sqrt(2)*FR) + v[a]*FR/(sqrt(2)*s)).expand(), f

 def _terms(e):
     # return the number of terms of this expression
     # when multiplied out -- assuming no joining of terms
     if e.is_Add:
         return sum([_terms(ai) for ai in e.args])
     if e.is_Mul:
         return prod([_terms(mi) for mi in e.args])
     return 1

def _nP(n, k=None, replacement=False):
    from sympy.functions.combinatorial.factorials import factorial
    from sympy.core.mul import prod

    if k == 0:
        return 1
    if isinstance(n, SYMPY_INTS):  # n different items
        # assert n >= 0
        if k is None:
            return sum(_nP(n, i, replacement) for i in range(n + 1))
        elif replacement:
            return n**k
        elif k > n:
            return 0
        elif k == n:
            return factorial(k)
        elif k == 1:
            return n
        else:
            # assert k >= 0
            return _product(n - k + 1, n)
    elif isinstance(n, _MultisetHistogram):
        if k is None:
            return sum(_nP(n, i, replacement) for i in range(n[_N] + 1))
        elif replacement:
            return n[_ITEMS]**k
        elif k == n[_N]:
            return factorial(k)/prod([factorial(i) for i in n[_M] if i > 1])
        elif k > n[_N]:
            return 0
        elif k == 1:
            return n[_ITEMS]
        else:
            # assert k >= 0
            tot = 0
            n = list(n)
            for i in range(len(n[_M])):
                if not n[i]:
                    continue
                n[_N] -= 1
                if n[i] == 1:
                    n[i] = 0
                    n[_ITEMS] -= 1
                    tot += _nP(_MultisetHistogram(n), k - 1)
                    n[_ITEMS] += 1
                    n[i] = 1
                else:
                    n[i] -= 1
                    tot += _nP(_MultisetHistogram(n), k - 1)
                    n[i] += 1
                n[_N] += 1
            return tot

def gf_crt1(M, K):
    """
    First part of the Chinese Remainder Theorem.

    Examples
    ========
    >>> from sympy.polys.domains import ZZ
    >>> from sympy.polys.galoistools import gf_crt1

    >>> gf_crt1([99, 97, 95], ZZ)
    (912285, [9215, 9405, 9603], [62, 24, 12])

    """
    E, S = [], []
    p = prod(M, start=K.one)

    for m in M:
        E.append(p // m)
        S.append(K.gcdex(E[-1], m)[0] % m)

    return p, E, S

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),
                    check=False, symmetric=symmetric)
            if result is None:
                return result
            result, mm = result

    if symmetric:
        return symmetric_residue(result, mm), mm
    return result, mm

def nC(n, k=None, replacement=False):
    """Return the number of combinations of ``n`` items taken ``k`` at a time.

    Possible values for ``n``::
        integer - set of length ``n``
        sequence - converted to a multiset internally
        multiset - {element: multiplicity}

    If ``k`` is None then the total of all combinations of length 0
    through the number of items represented in ``n`` will be returned.

    If ``replacement`` is True then a given item can appear more than once
    in the ``k`` items. (For example, for 'ab' sets of 2 would include 'aa',
    'ab', and 'bb'.) The multiplicity of elements in ``n`` is ignored when
    ``replacement`` is True but the total number of elements is considered
    since no element can appear more times than the number of elements in
    ``n``.

    Examples
    ========

    >>> from sympy.functions.combinatorial.numbers import nC
    >>> from sympy.utilities.iterables import multiset_combinations
    >>> nC(3, 2)
    3
    >>> nC('abc', 2)
    3
    >>> nC('aab', 2)
    2

    When ``replacement`` is True, each item can have multiplicity
    equal to the length represented by ``n``:

    >>> nC('aabc', replacement=True)
    35
    >>> [len(list(multiset_combinations('aaaabbbbcccc', i))) for i in range(5)]
    [1, 3, 6, 10, 15]
    >>> sum(_)
    35

    If there are ``k`` items with multiplicities ``m_1, m_2, ..., m_k``
    then the total of all combinations of length 0 hrough ``k`` is the
    product, ``(m_1 + 1)*(m_2 + 1)*...*(m_k + 1)``. When the multiplicity
    of each item is 1 (i.e., k unique items) then there are 2**k
    combinations. For example, if there are 4 unique items, the total number
    of combinations is 16:

    >>> sum(nC(4, i) for i in range(5))
    16

    References
    ==========

    .. [1] http://en.wikipedia.org/wiki/Combination
    .. [2] http://tinyurl.com/cep849r

    See Also
    ========
    sympy.utilities.iterables.multiset_combinations
    """
    from sympy.functions.combinatorial.factorials import binomial
    from sympy.core.mul import prod

    if isinstance(n, SYMPY_INTS):
        if k is None:
            if not replacement:
                return 2**n
            return sum(nC(n, i, replacement) for i in range(n + 1))
        if k < 0:
            raise ValueError("k cannot be negative")
        if replacement:
            return binomial(n + k - 1, k)
        return binomial(n, k)
    if isinstance(n, _MultisetHistogram):
        N = n[_N]
        if k is None:
            if not replacement:
                return prod(m + 1 for m in n[_M])
            return sum(nC(n, i, replacement) for i in range(N + 1))
        elif replacement:
            return nC(n[_ITEMS], k, replacement)
        # assert k >= 0
        elif k in (1, N - 1):
            return n[_ITEMS]
        elif k in (0, N):
            return 1
        return _AOP_product(tuple(n[_M]))[k]
    else:
        return nC(_multiset_histogram(n), k, replacement)