from __future__ import print_function, division

from sympy.concrete.expr_with_limits import AddWithLimits
from sympy.concrete.expr_with_intlimits import ExprWithIntLimits
from sympy.core.basic import C
from sympy.core.containers import Tuple
from sympy.core.expr import Expr
from sympy.core.function import Derivative
from sympy.core.relational import Eq
from sympy.core.singleton import S
from sympy.core.symbol import (Dummy, Wild)
from sympy.core.sympify import sympify
from sympy.concrete.gosper import gosper_sum
from sympy.functions.elementary.piecewise import piecewise_fold, Piecewise
from sympy.polys import apart, PolynomialError
from sympy.solvers import solve
from sympy.core.compatibility import xrange


class Sum(AddWithLimits,ExprWithIntLimits):
    r"""Represents unevaluated summation.

    ``Sum`` represents a finite or infinite series, with the first argument
    being the general form of terms in the series, and the second argument
    being ``(dummy_variable, start, end)``, with ``dummy_variable`` taking
    all integer values from ``start`` through ``end``. In accordance with
    long-standing mathematical convention, the end term is included in the
    summation.

    Finite sums
    ===========

    For finite sums (and sums with symbolic limits assumed to be finite) we
    follow the summation convention described by Karr [1], especially
    definition 3 of section 1.4. The sum:

    .. math::

        \sum_{m \leq i < n} f(i)

    has *the obvious meaning* for `m < n`, namely:

    .. math::

        \sum_{m \leq i < n} f(i) = f(m) + f(m+1) + \ldots + f(n-2) + f(n-1)

    with the upper limit value `f(n)` excluded. The sum over an empty set is
    zero if and only if `m = n`:

    .. math::

        \sum_{m \leq i < n} f(i) = 0  \quad \mathrm{for} \quad  m = n

    Finally, for all other sums over empty sets we assume the following
    definition:

    .. math::

        \sum_{m \leq i < n} f(i) = - \sum_{n \leq i < m} f(i)  \quad \mathrm{for} \quad  m > n

    It is important to note that Karr defines all sums with the upper
    limit being exclusive. This is in contrast to the usual mathematical notation,
    but does not affect the summation convention. Indeed we have:

    .. math::

        \sum_{m \leq i < n} f(i) = \sum_{i = m}^{n - 1} f(i)

    where the difference in notation is intentional to emphasize the meaning,
    with limits typeset on the top being inclusive.

    Examples
    ========

    >>> from sympy.abc import i, k, m, n, x
    >>> from sympy import Sum, factorial, oo
    >>> Sum(k,(k,1,m))
    Sum(k, (k, 1, m))
    >>> Sum(k,(k,1,m)).doit()
    m**2/2 + m/2
    >>> Sum(k**2,(k,1,m))
    Sum(k**2, (k, 1, m))
    >>> Sum(k**2,(k,1,m)).doit()
    m**3/3 + m**2/2 + m/6
    >>> Sum(x**k,(k,0,oo))
    Sum(x**k, (k, 0, oo))
    >>> Sum(x**k,(k,0,oo)).doit()
    Piecewise((1/(-x + 1), Abs(x) < 1), (Sum(x**k, (k, 0, oo)), True))
    >>> Sum(x**k/factorial(k),(k,0,oo)).doit()
    exp(x)

    An example showing that the symbolic result of a summation is still
    valid for seemingly nonsensical values of the limits. Then the Karr
    convention allows us to give a perfectly valid interpretation to
    those sums by interchanging the limits according to the above rules:

    >>> S = Sum(i, (i,1,n)).doit()
    >>> S
    n**2/2 + n/2
    >>> S.subs(n, -4)
    6
    >>> Sum(i, (i, 1, -4)).doit()
    6
    >>> Sum(-i, (i, -3, 0)).doit()
    6

    An explicit example of the Karr summation convention:

    >>> S1 = Sum(i**2, (i, m, m+n-1)).doit()
    >>> S1
    m**2*n + m*n**2 - m*n + n**3/3 - n**2/2 + n/6
    >>> S2 = Sum(i**2, (i, m+n, m-1)).doit()
    >>> S2
    -m**2*n - m*n**2 + m*n - n**3/3 + n**2/2 - n/6
    >>> S1 + S2
    0
    >>> S3 = Sum(i, (i, m, m-1)).doit()
    >>> S3
    0

    See Also
    ========

    summation
    Product, product

    References
    ==========

    .. [1] Michael Karr, "Summation in Finite Terms", Journal of the ACM,
           Volume 28 Issue 2, April 1981, Pages 305-350
           http://dl.acm.org/citation.cfm?doid=322248.322255
    .. [2] http://en.wikipedia.org/wiki/Summation#Capital-sigma_notation
    .. [3] http://en.wikipedia.org/wiki/Empty_sum
    """

    __slots__ = ['is_commutative']

    def __new__(cls, function, *symbols, **assumptions):
        obj = AddWithLimits.__new__(cls, function, *symbols, **assumptions)
        if not hasattr(obj, 'limits'):
            return obj
        if any(len(l) != 3 or None in l for l in obj.limits):
            raise ValueError('Sum requires values for lower and upper bounds.')

        return obj

    def _eval_is_zero(self):
        # a Sum is only zero if its function is zero or if all terms
        # cancel out. This only answers whether the summand is zero; if
        # not then None is returned since we don't analyze whether all
        # terms cancel out.
        if self.function.is_zero:
            return True

    def doit(self, **hints):
        if hints.get('deep', True):
            f = self.function.doit(**hints)
        else:
            f = self.function

        for n, limit in enumerate(self.limits):
            i, a, b = limit
            dif = b - a
            if dif.is_integer and (dif < 0) == True:
                a, b = b + 1, a - 1
                f = -f

            newf = eval_sum(f, (i, a, b))
            if newf is None:
                if f == self.function:
                    return self
                else:
                    return self.func(f, *self.limits[n:])
            f = newf

        if hints.get('deep', True):
            # eval_sum could return partially unevaluated
            # result with Piecewise.  In this case we won't
            # doit() recursively.
            if not isinstance(f, Piecewise):
                return f.doit(**hints)

        return f

    def _eval_derivative(self, x):
        """
        Differentiate wrt x as long as x is not in the free symbols of any of
        the upper or lower limits.

        Sum(a*b*x, (x, 1, a)) can be differentiated wrt x or b but not `a`
        since the value of the sum is discontinuous in `a`. In a case
        involving a limit variable, the unevaluated derivative is returned.
        """

        # diff already confirmed that x is in the free symbols of self, but we
        # don't want to differentiate wrt any free symbol in the upper or lower
        # limits
        # XXX remove this test for free_symbols when the default _eval_derivative is in
        if x not in self.free_symbols:
            return S.Zero

        # get limits and the function
        f, limits = self.function, list(self.limits)

        limit = limits.pop(-1)

        if limits:  # f is the argument to a Sum
            f = self.func(f, *limits)

        if len(limit) == 3:
            _, a, b = limit
            if x in a.free_symbols or x in b.free_symbols:
                return None
            df = Derivative(f, x, evaluate=True)
            rv = self.func(df, limit)
            if limit[0] not in df.free_symbols:
                rv = rv.doit()
            return rv
        else:
            return NotImplementedError('Lower and upper bound expected.')

    def _eval_simplify(self, ratio, measure):
        from sympy.simplify.simplify import sum_simplify
        return sum_simplify(self)

    def _eval_summation(self, f, x):
        return None

    def euler_maclaurin(self, m=0, n=0, eps=0, eval_integral=True):
        """
        Return an Euler-Maclaurin approximation of self, where m is the
        number of leading terms to sum directly and n is the number of
        terms in the tail.

        With m = n = 0, this is simply the corresponding integral
        plus a first-order endpoint correction.

        Returns (s, e) where s is the Euler-Maclaurin approximation
        and e is the estimated error (taken to be the magnitude of
        the first omitted term in the tail):

            >>> from sympy.abc import k, a, b
            >>> from sympy import Sum
            >>> Sum(1/k, (k, 2, 5)).doit().evalf()
            1.28333333333333
            >>> s, e = Sum(1/k, (k, 2, 5)).euler_maclaurin()
            >>> s
            -log(2) + 7/20 + log(5)
            >>> from sympy import sstr
            >>> print(sstr((s.evalf(), e.evalf()), full_prec=True))
            (1.26629073187415, 0.0175000000000000)

        The endpoints may be symbolic:

            >>> s, e = Sum(1/k, (k, a, b)).euler_maclaurin()
            >>> s
            -log(a) + log(b) + 1/(2*b) + 1/(2*a)
            >>> e
            Abs(1/(12*b**2) - 1/(12*a**2))

        If the function is a polynomial of degree at most 2n+1, the
        Euler-Maclaurin formula becomes exact (and e = 0 is returned):

            >>> Sum(k, (k, 2, b)).euler_maclaurin()
            (b**2/2 + b/2 - 1, 0)
            >>> Sum(k, (k, 2, b)).doit()
            b**2/2 + b/2 - 1

        With a nonzero eps specified, the summation is ended
        as soon as the remainder term is less than the epsilon.
        """
        m = int(m)
        n = int(n)
        f = self.function
        if len(self.limits) != 1:
            raise ValueError("More than 1 limit")
        i, a, b = self.limits[0]
        if (a > b) == True:
            if a - b == 1:
                return S.Zero,S.Zero
            a, b = b + 1, a - 1
            f = -f
        s = S.Zero
        if m:
            if b.is_Integer and a.is_Integer:
                m = min(m, b - a + 1)
            if not eps:
                for k in range(m):
                    s += f.subs(i, a + k)
            else:
                term = f.subs(i, a)
                if term:
                    test = abs(term.evalf(3)) < eps
                    if test == True:
                        return s, abs(term)
                    elif not (test == False):
                        # a symbolic Relational class, can't go further
                        return term, S.Zero
                s += term
                for k in range(1, m):
                    term = f.subs(i, a + k)
                    if abs(term.evalf(3)) < eps:
                        return s, abs(term)
                    s += term
            if b - a + 1 == m:
                return s, S.Zero
            a += m
        x = Dummy('x')
        I = C.Integral(f.subs(i, x), (x, a, b))
        if eval_integral:
            I = I.doit()
        s += I

        def fpoint(expr):
            if b is S.Infinity:
                return expr.subs(i, a), 0
            return expr.subs(i, a), expr.subs(i, b)
        fa, fb = fpoint(f)
        iterm = (fa + fb)/2
        g = f.diff(i)
        for k in xrange(1, n + 2):
            ga, gb = fpoint(g)
            term = C.bernoulli(2*k)/C.factorial(2*k)*(gb - ga)
            if (eps and term and abs(term.evalf(3)) < eps) or (k > n):
                break
            s += term
            g = g.diff(i, 2, simplify=False)
        return s + iterm, abs(term)


    def reverse_order(self, *indices):
        """
        Reverse the order of a limit in a Sum.

        Usage
        =====

        ``reverse_order(self, *indices)`` reverses some limits in the expression
        ``self`` which can be either a ``Sum`` or a ``Product``. The selectors in
        the argument ``indices`` specify some indices whose limits get reversed.
        These selectors are either variable names or numerical indices counted
        starting from the inner-most limit tuple.

        Examples
        ========

        >>> from sympy import Sum
        >>> from sympy.abc import x, y, a, b, c, d

        >>> Sum(x, (x, 0, 3)).reverse_order(x)
        Sum(-x, (x, 4, -1))
        >>> Sum(x*y, (x, 1, 5), (y, 0, 6)).reverse_order(x, y)
        Sum(x*y, (x, 6, 0), (y, 7, -1))
        >>> Sum(x, (x, a, b)).reverse_order(x)
        Sum(-x, (x, b + 1, a - 1))
        >>> Sum(x, (x, a, b)).reverse_order(0)
        Sum(-x, (x, b + 1, a - 1))

        While one should prefer variable names when specifying which limits
        to reverse, the index counting notation comes in handy in case there
        are several symbols with the same name.

        >>> S = Sum(x**2, (x, a, b), (x, c, d))
        >>> S
        Sum(x**2, (x, a, b), (x, c, d))
        >>> S0 = S.reverse_order( 0)
        >>> S0
        Sum(-x**2, (x, b + 1, a - 1), (x, c, d))
        >>> S1 = S0.reverse_order( 1)
        >>> S1
        Sum(x**2, (x, b + 1, a - 1), (x, d + 1, c - 1))

        Of course we can mix both notations:

        >>> Sum(x*y, (x, a, b), (y, 2, 5)).reverse_order( x, 1)
        Sum(x*y, (x, b + 1, a - 1), (y, 6, 1))
        >>> Sum(x*y, (x, a, b), (y, 2, 5)).reverse_order( y, x)
        Sum(x*y, (x, b + 1, a - 1), (y, 6, 1))

        See Also
        ========

        index, reorder_limit, reorder

        References
        ==========

        .. [1] Michael Karr, "Summation in Finite Terms", Journal of the ACM,
               Volume 28 Issue 2, April 1981, Pages 305-350
               http://dl.acm.org/citation.cfm?doid=322248.322255
        """
        l_indices = list(indices)

        for i, indx in enumerate(l_indices):
            if not isinstance(indx, int):
                l_indices[i] = self.index(indx)

        e = 1
        limits = []
        for i, limit in enumerate(self.limits):
            l = limit
            if i in l_indices:
                e = -e
                l = (limit[0], limit[2] + 1 , limit[1] - 1)
            limits.append(l)

        return Sum(e * self.function, *limits)

def summation(f, *symbols, **kwargs):
    r"""
    Compute the summation of f with respect to symbols.

    The notation for symbols is similar to the notation used in Integral.
    summation(f, (i, a, b)) computes the sum of f with respect to i from a to b,
    i.e.,

    ::

                                    b
                                  ____
                                  \   `
        summation(f, (i, a, b)) =  )    f
                                  /___,
                                  i = a

    If it cannot compute the sum, it returns an unevaluated Sum object.
    Repeated sums can be computed by introducing additional symbols tuples::

    >>> from sympy import summation, oo, symbols, log
    >>> i, n, m = symbols('i n m', integer=True)

    >>> summation(2*i - 1, (i, 1, n))
    n**2
    >>> summation(1/2**i, (i, 0, oo))
    2
    >>> summation(1/log(n)**n, (n, 2, oo))
    Sum(log(n)**(-n), (n, 2, oo))
    >>> summation(i, (i, 0, n), (n, 0, m))
    m**3/6 + m**2/2 + m/3

    >>> from sympy.abc import x
    >>> from sympy import factorial
    >>> summation(x**n/factorial(n), (n, 0, oo))
    exp(x)

    See Also
    ========

    Sum
    Product, product

    """
    return Sum(f, *symbols, **kwargs).doit(deep=False)


def telescopic_direct(L, R, n, limits):
    """Returns the direct summation of the terms of a telescopic sum

    L is the term with lower index
    R is the term with higher index
    n difference between the indexes of L and R

    For example:

    >>> from sympy.concrete.summations import telescopic_direct
    >>> from sympy.abc import k, a, b
    >>> telescopic_direct(1/k, -1/(k+2), 2, (k, a, b))
    -1/(b + 2) - 1/(b + 1) + 1/(a + 1) + 1/a

    """
    (i, a, b) = limits
    s = 0
    for m in xrange(n):
        s += L.subs(i, a + m) + R.subs(i, b - m)
    return s


def telescopic(L, R, limits):
    '''Tries to perform the summation using the telescopic property

    return None if not possible
    '''
    (i, a, b) = limits
    if L.is_Add or R.is_Add:
        return None

    # We want to solve(L.subs(i, i + m) + R, m)
    # First we try a simple match since this does things that
    # solve doesn't do, e.g. solve(f(k+m)-f(k), m) fails

    k = Wild("k")
    sol = (-R).match(L.subs(i, i + k))
    s = None
    if sol and k in sol:
        s = sol[k]
        if not (s.is_Integer and L.subs(i, i + s) == -R):
            #sometimes match fail(f(x+2).match(-f(x+k))->{k: -2 - 2x}))
            s = None

    # But there are things that match doesn't do that solve
    # can do, e.g. determine that 1/(x + m) = 1/(1 - x) when m = 1

    if s is None:
        m = Dummy('m')
        try:
            sol = solve(L.subs(i, i + m) + R, m) or []
        except NotImplementedError:
            return None
        sol = [si for si in sol if si.is_Integer and
               (L.subs(i, i + si) + R).expand().is_zero]
        if len(sol) != 1:
            return None
        s = sol[0]

    if s < 0:
        return telescopic_direct(R, L, abs(s), (i, a, b))
    elif s > 0:
        return telescopic_direct(L, R, s, (i, a, b))


def eval_sum(f, limits):
    from sympy.concrete.delta import deltasummation, _has_simple_delta
    from sympy.functions import KroneckerDelta

    (i, a, b) = limits
    if f is S.Zero:
        return S.Zero
    if i not in f.free_symbols:
        return f*(b - a + 1)
    if a == b:
        return f.subs(i, a)
    if isinstance(f, Piecewise):
        if not any(i in arg.args[1].free_symbols for arg in f.args):
            # Piecewise conditions do not depend on the dummy summation variable,
            # therefore we can fold:     Sum(Piecewise((e, c), ...), limits)
            #                        --> Piecewise((Sum(e, limits), c), ...)
            newargs = []
            for arg in f.args:
                newexpr = eval_sum(arg.expr, limits)
                if newexpr is None:
                    return None
                newargs.append((newexpr, arg.cond))
            return f.func(*newargs)

    if f.has(KroneckerDelta) and _has_simple_delta(f, limits[0]):
        return deltasummation(f, limits)

    dif = b - a
    definite = dif.is_Integer
    # Doing it directly may be faster if there are very few terms.
    if definite and (dif < 100):
        return eval_sum_direct(f, (i, a, b))
    if isinstance(f, Piecewise):
        return None
    # Try to do it symbolically. Even when the number of terms is known,
    # this can save time when b-a is big.
    # We should try to transform to partial fractions
    value = eval_sum_symbolic(f.expand(), (i, a, b))
    if value is not None:
        return value
    # Do it directly
    if definite:
        return eval_sum_direct(f, (i, a, b))


def eval_sum_direct(expr, limits):
    (i, a, b) = limits

    dif = b - a
    return C.Add(*[expr.subs(i, a + j) for j in xrange(dif + 1)])


def eval_sum_symbolic(f, limits):
    f_orig = f
    (i, a, b) = limits
    if not f.has(i):
        return f*(b - a + 1)

    # Linearity
    if f.is_Mul:
        L, R = f.as_two_terms()

        if not L.has(i):
            sR = eval_sum_symbolic(R, (i, a, b))
            if sR:
                return L*sR

        if not R.has(i):
            sL = eval_sum_symbolic(L, (i, a, b))
            if sL:
                return R*sL

        try:
            f = apart(f, i)  # see if it becomes an Add
        except PolynomialError:
            pass

    if f.is_Add:
        L, R = f.as_two_terms()
        lrsum = telescopic(L, R, (i, a, b))

        if lrsum:
            return lrsum

        lsum = eval_sum_symbolic(L, (i, a, b))
        rsum = eval_sum_symbolic(R, (i, a, b))

        if None not in (lsum, rsum):
            r = lsum + rsum
            if not r is S.NaN:
                return r

    # Polynomial terms with Faulhaber's formula
    n = Wild('n')
    result = f.match(i**n)

    if result is not None:
        n = result[n]

        if n.is_Integer:
            if n >= 0:
                if (b is S.Infinity and not a is S.NegativeInfinity) or \
                   (a is S.NegativeInfinity and not b is S.Infinity):
                    return S.Infinity
                return ((C.bernoulli(n + 1, b + 1) - C.bernoulli(n + 1, a))/(n + 1)).expand()
            elif a.is_Integer and a >= 1:
                if n == -1:
                    return C.harmonic(b) - C.harmonic(a - 1)
                else:
                    return C.harmonic(b, abs(n)) - C.harmonic(a - 1, abs(n))

    if not (a.has(S.Infinity, S.NegativeInfinity) or
            b.has(S.Infinity, S.NegativeInfinity)):
        # Geometric terms
        c1 = C.Wild('c1', exclude=[i])
        c2 = C.Wild('c2', exclude=[i])
        c3 = C.Wild('c3', exclude=[i])

        e = f.match(c1**(c2*i + c3))

        if e is not None:
            p = (c1**c3).subs(e)
            q = (c1**c2).subs(e)

            r = p*(q**a - q**(b + 1))/(1 - q)
            l = p*(b - a + 1)

            return Piecewise((l, Eq(q, S.One)), (r, True))

        r = gosper_sum(f, (i, a, b))

        if not r in (None, S.NaN):
            return r

    return eval_sum_hyper(f_orig, (i, a, b))


def _eval_sum_hyper(f, i, a):
    """ Returns (res, cond). Sums from a to oo. """
    from sympy.functions import hyper
    from sympy.simplify import hyperexpand, hypersimp, fraction, simplify
    from sympy.polys.polytools import Poly, factor

    if a != 0:
        return _eval_sum_hyper(f.subs(i, i + a), i, 0)

    if f.subs(i, 0) == 0:
        if simplify(f.subs(i, Dummy('i', integer=True, positive=True))) == 0:
            return S(0), True
        return _eval_sum_hyper(f.subs(i, i + 1), i, 0)

    hs = hypersimp(f, i)
    if hs is None:
        return None

    numer, denom = fraction(factor(hs))
    top, topl = numer.as_coeff_mul(i)
    bot, botl = denom.as_coeff_mul(i)
    ab = [top, bot]
    factors = [topl, botl]
    params = [[], []]
    for k in range(2):
        for fac in factors[k]:
            mul = 1
            if fac.is_Pow:
                mul = fac.exp
                fac = fac.base
                if not mul.is_Integer:
                    return None
            p = Poly(fac, i)
            if p.degree() != 1:
                return None
            m, n = p.all_coeffs()
            ab[k] *= m**mul
            params[k] += [n/m]*mul

    # Add "1" to numerator parameters, to account for implicit n! in
    # hypergeometric series.
    ap = params[0] + [1]
    bq = params[1]
    x = ab[0]/ab[1]
    h = hyper(ap, bq, x)

    return f.subs(i, 0)*hyperexpand(h), h.convergence_statement


def eval_sum_hyper(f, i_a_b):
    from sympy.logic.boolalg import And

    i, a, b = i_a_b

    if (b - a).is_Integer:
        # We are never going to do better than doing the sum in the obvious way
        return None

    old_sum = Sum(f, (i, a, b))

    if b != S.Infinity:
        if a == S.NegativeInfinity:
            res = _eval_sum_hyper(f.subs(i, -i), i, -b)
            if res is not None:
                return Piecewise(res, (old_sum, True))
        else:
            res1 = _eval_sum_hyper(f, i, a)
            res2 = _eval_sum_hyper(f, i, b + 1)
            if res1 is None or res2 is None:
                return None
            (res1, cond1), (res2, cond2) = res1, res2
            cond = And(cond1, cond2)
            if cond == False:
                return None
        return Piecewise((res1 - res2, cond), (old_sum, True))

    if a == S.NegativeInfinity:
        res1 = _eval_sum_hyper(f.subs(i, -i), i, 1)
        res2 = _eval_sum_hyper(f, i, 0)
        if res1 is None or res2 is None:
            return None
        res1, cond1 = res1
        res2, cond2 = res2
        cond = And(cond1, cond2)
        if cond == False:
            return None
        return Piecewise((res1 + res2, cond), (old_sum, True))

    # Now b == oo, a != -oo
    res = _eval_sum_hyper(f, i, a)
    if res is not None:
        r, c = res
        if c == False:
            if r.is_number:
                f = f.subs(i, Dummy('i', integer=True, positive=True) + a)
                if f.is_positive or f.is_zero:
                    return S.Infinity
                elif f.is_negative:
                    return S.NegativeInfinity
            return None
        return Piecewise(res, (old_sum, True))
