""" This module contain solvers for all kinds of equations:

    - algebraic, use solve()

    - recurrence, use rsolve()

    - differential, use dsolve()

    - transcendental, use tsolve()

    - nonlinear (numerically), use nsolve()
      (you will need a good starting point)

"""

from sympy.core.sympify import sympify
from sympy.core import S, Mul, Add, Pow, Symbol, Wild, Equality, Dummy, Basic
from sympy.core.numbers import ilcm

from sympy.functions import log, exp, LambertW
from sympy.simplify import simplify, collect, powsimp
from sympy.matrices import Matrix, zeros
from sympy.polys import roots, cancel, Poly, together
from sympy.functions.elementary.piecewise import piecewise_fold

from sympy.utilities import any, all
from sympy.utilities.iterables import iff
from sympy.utilities.lambdify import lambdify
from sympy.mpmath import findroot

from sympy.solvers.polysys import solve_poly_system
from sympy.solvers.inequalities import reduce_inequalities

from warnings import warn

def denoms(eq, x=None):
    """Return (recursively) set of all denominators that appear in eq
    that contain any symbol in x; if x is None (default) then all
    denominators with symbols will be returned."""
    from sympy.utilities.iterables import preorder_traversal

    if x is None:
        x = eq.free_symbols
    dens = set()
    pt = preorder_traversal(eq)
    for e in pt:
        if e.is_Pow or e.func is exp:
            n, d = e.as_numer_denom()
            if d in dens:
                pt.skip()
            elif d.has(*x):
                dens.add(d.as_base_exp()[0])
    return dens

def checksol(f, symbol, sol=None, **flags):
    """Checks whether sol is a solution of equation f == 0.

    Input can be either a single symbol and corresponding value
    or a dictionary of symbols and values.

    Examples:
    ---------

       >>> from sympy import symbols
       >>> from sympy.solvers import checksol
       >>> x, y = symbols('x,y')
       >>> checksol(x**4-1, x, 1)
       True
       >>> checksol(x**4-1, x, 0)
       False
       >>> checksol(x**2 + y**2 - 5**2, {x:3, y: 4})
       True

       None is returned if checksol() could not conclude.

       flags:
           'numerical=True (default)'
               do a fast numerical check if f has only one symbol.
           'minimal=True (default is False)'
               a very fast, minimal testing.
           'warning=True (default is False)'
               print a warning if checksol() could not conclude.
           'simplified=True (default)'
               solution should be simplified before substituting into function
               and function should be simplified after making substitution.
           'force=True (default is False)'
               make positive all symbols without assumptions regarding sign.
    """

    if sol is not None:
        sol = {symbol: sol}
    elif isinstance(symbol, dict):
        sol = symbol
    else:
        msg = 'Expecting sym, val or {sym: val}, None but got %s, %s'
        raise ValueError(msg % (symbol, sol))

    if hasattr(f, '__iter__') and hasattr(f, '__len__'):
        if not f:
            raise ValueError('no functions to check')
        rv = set()
        for fi in f:
            check = checksol(fi, sol, **flags)
            if check is False:
                return False
            rv.add(check)
        if None in rv: # rv might contain True and/or None
            return None
        assert len(rv) == 1 # True
        return True

    if isinstance(f, Poly):
        f = f.as_expr()
    elif isinstance(f, Equality):
        f = f.lhs - f.rhs

    if not f:
        return True

    if not f.has(*sol.keys()):
        return False

    attempt = -1
    numerical = flags.get('numerical', True)
    while 1:
        attempt += 1
        if attempt == 0:
            val = f.subs(sol)
        elif attempt == 1:
            if not val.atoms(Symbol) and numerical:
                # val is a constant, so a fast numerical test may suffice
                if val not in [S.Infinity, S.NegativeInfinity]:
                    # issue 2088 shows that +/-oo chops to 0
                    val = val.evalf(36).n(30, chop=True)
        elif attempt == 2:
            if flags.get('minimal', False):
                return
            # the flag 'simplified=False' is used in solve to avoid
            # simplifying the solution. So if it is set to False there
            # the simplification will not be attempted here, either. But
            # if the simplification is done here then the flag should be
            # set to False so it isn't done again there.
            if flags.get('simplified', True):
                for k in sol:
                    sol[k] = simplify(sympify(sol[k]))
                flags['simplified'] = False
                val = simplify(f.subs(sol))
            if flags.get('force', False):
                val = posify(val)[0]
        elif attempt == 3:
            val = powsimp(val)
        elif attempt == 4:
            val = cancel(val)
        elif attempt == 5:
            val = val.expand()
        elif attempt == 6:
            val = together(val)
        elif attempt == 7:
            val = powsimp(val)
        else:
            break
        if val.is_zero:
            return True
        elif attempt > 0 and numerical and val.is_nonzero:
            return False

    if flags.get('warning', False):
        print("Warning: could not verify solution %s." % sol)
    # returns None if it can't conclude
    # TODO: improve solution testing

# Codes for guess solve strategy
GS_POLY = 0
GS_RATIONAL = 1
GS_POLY_CV_1 = 2 # can be converted to a polynomial equation via the change of variable y -> x**a, a real
GS_POLY_CV_2 = 3 # can be converted to a polynomial equation multiplying on both sides by x**m
                 # for example, x + 1/x == 0. Multiplying by x yields x**2 + x == 0
GS_RATIONAL_CV_1 = 4 # can be converted to a rational equation via the change of variable y -> x**n
GS_PIECEWISE = 5
GS_TRANSCENDENTAL = 6

def guess_solve_strategy(expr, symbol):
    """
    Tries to guess what approach should be used to solve a specific equation

    Returns
    =======
       - -1: could not guess
       - integer > 0: code representing certain type of equation. See GS_* fields
         on this module for a complete list

    Examples
    ========
    >>> from sympy import Symbol, Rational
    >>> from sympy.solvers.solvers import guess_solve_strategy
    >>> from sympy.abc import x
    >>> guess_solve_strategy(x**2 + 1, x)
    0
    >>> guess_solve_strategy(x**Rational(1,2) + 1, x)
    2

    """
    eq_type = -1
    if expr.is_Add:
        return max([guess_solve_strategy(i, symbol) for i in expr.args])

    elif expr.is_Mul:
        # check for rational functions
        num, denom = expr.as_numer_denom()
        if denom.has(symbol):
            #we have a quotient
            m = max(guess_solve_strategy(num, symbol), guess_solve_strategy(denom, symbol))
            if m == GS_POLY:
                return GS_RATIONAL
            elif m == GS_POLY_CV_1:
                return GS_RATIONAL_CV_1
            else:
                raise NotImplementedError
        else:
            return max([guess_solve_strategy(i, symbol) for i in expr.args])

    elif expr.is_Symbol:
        return GS_POLY

    elif expr.is_Pow:
        if expr.exp.has(symbol):
            return GS_TRANSCENDENTAL
        elif not expr.exp.has(symbol) and expr.base.has(symbol):
            if expr.exp.is_Integer and expr.exp > 0:
                eq_type = max(eq_type, GS_POLY)
            elif expr.exp.is_Integer and expr.exp < 0:
                eq_type = max(eq_type, GS_POLY_CV_2)
            elif expr.exp.is_Rational:
                eq_type = max(eq_type, GS_POLY_CV_1)
            else:
                return GS_TRANSCENDENTAL

    elif expr.is_Piecewise:
        return GS_PIECEWISE

    elif expr.is_Function and expr.has(symbol):
        return GS_TRANSCENDENTAL

    elif not expr.has(symbol):
        return GS_POLY

    return eq_type

def solve(f, *symbols, **flags):
    """
    A preprocessor to _solve.
    """

    solution = _solve(f, *symbols, **flags)
    return solution

def _solve(f, *symbols, **flags):
    """Solves equations and systems of equations.

       Currently supported are univariate polynomial, transcendental
       equations, piecewise combinations thereof and systems of linear
       and polynomial equations.  Input is formed as a single expression
       or an equation,  or an iterable container in case of an equation
       system.  The type of output may vary and depends heavily on the
       input. For more details refer to more problem specific functions.

       By default all solutions are simplified to make the output more
       readable. If this is not the expected behavior (e.g., because of
       speed issues) set simplified=False in function arguments.

       To solve equations and systems of equations like recurrence relations
       or differential equations, use rsolve() or dsolve(), respectively.

       >>> from sympy import I, solve
       >>> from sympy.abc import x, y

       Solve a polynomial equation:

       >>> solve(x**4-1, x)
       [1, -1, -I, I]

       Solve a linear system:

       >>> solve((x+5*y-2, -3*x+6*y-15), x, y)
       {x: -3, y: 1}

    """

    def sympified_list(w):
        return map(sympify, iff(isinstance(w,(list, tuple, set)), w, [w]))
    # make f and symbols into lists of sympified quantities
    # keeping track of how f was passed since if it is a list
    # a dictionary of results will be returned.
    bare_f = not isinstance(f, (list, tuple, set))
    f, symbols = (sympified_list(w) for w in [f, symbols])

    for i, fi in enumerate(f):
        if isinstance(fi, Equality):
            f[i] = fi.lhs - fi.rhs
        elif isinstance(fi, Poly):
            f[i] = fi.as_expr()
        elif isinstance(fi, bool) or fi.is_Relational:
            return reduce_inequalities(f, assume=flags.get('assume'))

    if not symbols:
        #get symbols from equations or supply dummy symbols since
        #solve(3,x) returns []...though it seems that it should raise some sort of error TODO
        symbols = set([])
        for fi in f:
            symbols |= fi.free_symbols or set([Dummy('x')])
        symbols = list(symbols)
        symbols.sort(key=Basic.sort_key)

    if len(symbols) == 1:
        if isinstance(symbols[0], (list, tuple, set)):
            symbols = symbols[0]

    result = list()

    # Begin code handling for Function and Derivative instances
    # Basic idea:  store all the passed symbols in symbols_passed, check to see
    # if any of them are Function or Derivative types, if so, use a dummy
    # symbol in their place, and set symbol_swapped = True so that other parts
    # of the code can be aware of the swap.  Once all swapping is done, the
    # continue on with regular solving as usual, and swap back at the end of
    # the routine, so that whatever was passed in symbols is what is returned.
    symbols_new = []
    symbol_swapped = False

    symbols_passed = list(symbols)

    for i, s in enumerate(symbols):
        if s.is_Symbol:
            s_new = s
        elif s.is_Function:
            symbol_swapped = True
            s_new = Dummy('F%d' % i)
        elif s.is_Derivative:
            symbol_swapped = True
            s_new = Dummy('D%d' % i)
        else:
            raise TypeError('not a Symbol or a Function')
        symbols_new.append(s_new)

        if symbol_swapped:
            swap_back_dict = dict(zip(symbols_new, symbols))
    # End code for handling of Function and Derivative instances

    if bare_f:
        f = f[0]

        # Create a swap dictionary for storing the passed symbols to be solved
        # for, so that they may be swapped back.
        if symbol_swapped:
            swap_dict = zip(symbols, symbols_new)
            f = f.subs(swap_dict)
            symbols = symbols_new

        # Any embedded piecewise functions need to be brought out to the
        # top level so that the appropriate strategy gets selected.
        f = piecewise_fold(f)

        if len(symbols) != 1:
            soln = None
            free = f.free_symbols
            ex = free - set(symbols)
            if len(ex) == 1:
                ex = ex.pop()
                try:
                    # may come back as dict or list (if non-linear)
                    soln = solve_undetermined_coeffs(f, symbols, ex)
                except NotImplementedError:
                    pass
            if soln is None:
                n, d = solve_linear(f, x=symbols)
                if n.is_Symbol:
                    soln = {n: cancel(d)}
            if soln:
                if symbol_swapped and isinstance(soln, dict):
                    return dict([(swap_back_dict[k],
                                  v.subs(swap_back_dict))
                                  for k, v in soln.iteritems()])
                return soln

        symbol = symbols[0]

        # first see if it really depends on symbol and whether there
        # is a linear solution
        f_num, sol = solve_linear(f, x=symbols)
        if not symbol in f_num.free_symbols:
            return []
        elif f_num.is_Symbol:
            return [cancel(sol)]

        strategy = guess_solve_strategy(f, symbol)
        result = False # no solution was obtained

        if strategy == GS_POLY:
            poly = f.as_poly(symbol)
            if poly is None:
                msg = "Cannot solve equation %s for %s" % (f, symbol)
            else:
                # for cubics and quartics, if the flag wasn't set, DON'T do it
                # by default since the results are quite long. Perhaps one could
                # base this decision on a certain crtical length of the roots.
                if poly.degree() > 2:
                    flags['simplified'] = flags.get('simplified', False)
                result = roots(poly, cubics=True, quartics=True).keys()

        elif strategy == GS_RATIONAL:
            P, _ = f.as_numer_denom()
            dens = denoms(f, x=symbols)
            # reject any result that makes Q affirmatively 0;
            # if in doubt, keep it
            try:
                soln = _solve(P, symbol, **flags)
            except NotImplementedError:
                msg = "Cannot solve equation %s for %s" % (P, symbol)
                result = []
            else:
                if dens:
                    result = [s for s in soln if all(not checksol(den, {symbol: s}) for den in dens)]
                else:
                    result = soln

        elif strategy == GS_POLY_CV_1:
            args = list(f.args)
            if isinstance(f, Pow):
                result = _solve(args[0], symbol, **flags)
            elif isinstance(f, Add):
                # we must search for a suitable change of variables
                # collect exponents
                exponents_denom = list()
                for arg in args:
                    if isinstance(arg, Pow):
                        exponents_denom.append(arg.exp.q)
                    elif isinstance(arg, Mul):
                        for mul_arg in arg.args:
                            if isinstance(mul_arg, Pow):
                                exponents_denom.append(mul_arg.exp.q)
                assert len(exponents_denom) > 0
                if len(exponents_denom) == 1:
                    m = exponents_denom[0]
                else:
                    # get the LCM of the denominators
                    m = reduce(ilcm, exponents_denom)
                # x -> y**m.
                # we assume positive for simplification purposes
                t = Dummy('t', positive=True)
                f_ = f.subs(symbol, t**m)
                if guess_solve_strategy(f_, t) != GS_POLY:
                    msg = "Could not convert to a polynomial equation: %s" % f_
                    result = []
                else:
                    soln = [s**m for s in _solve(f_, t)]
                    # we might have introduced solutions from another branch
                    # when changing variables; check and keep solutions
                    # unless they definitely aren't a solution
                    result = [s for s in soln if checksol(f, {symbol: s}) is not False]

            elif isinstance(f, Mul):
                result = []
                for m in f.args:
                    result.extend(_solve(m, symbol, **flags) or [])

        elif strategy == GS_POLY_CV_2:
            m = 0
            args = list(f.args)
            if isinstance(f, Add):
                for arg in args:
                    if isinstance(arg, Pow):
                        m = min(m, arg.exp)
                    elif isinstance(arg, Mul):
                        for mul_arg in arg.args:
                            if isinstance(mul_arg, Pow):
                                m = min(m, mul_arg.exp)
            elif isinstance(f, Mul):
                for mul_arg in args:
                    if isinstance(mul_arg, Pow):
                        m = min(m, mul_arg.exp)

            if m and m != 1:
                f_ = simplify(f*symbol**(-m))
                try:
                    sols = _solve(f_, symbol)
                except NotImplementedError:
                    msg = 'Could not solve %s for %s' % (f_, symbol)
                else:
                    # we might have introduced unwanted solutions
                    # when multiplying by x**-m; check and keep solutions
                    # unless they definitely aren't a solution
                    if sols:
                        result = [s for s in sols if checksol(f, {symbol: s}) is not False]
            else:
                msg = 'CV_2 calculated %d but it should have been other than 0 or 1' % m

        elif strategy == GS_PIECEWISE:
            result = set()
            for expr, cond in f.args:
                candidates = _solve(expr, *symbols)
                if isinstance(cond, bool) or cond.is_Number:
                    if not cond:
                        continue

                    # Only include solutions that do not match the condition
                    # of any of the other pieces.
                    for candidate in candidates:
                        matches_other_piece = False
                        for other_expr, other_cond in f.args:
                            if isinstance(other_cond, bool) \
                               or other_cond.is_Number:
                                continue
                            if bool(other_cond.subs(symbol, candidate)):
                                matches_other_piece = True
                                break
                        if not matches_other_piece:
                            result.add(candidate)
                else:
                    for candidate in candidates:
                        if bool(cond.subs(symbol, candidate)):
                            result.add(candidate)

            result = list(result)

        elif strategy == -1:
            raise ValueError('Could not parse expression %s' % f)

        # this is the fallback for not getting any other solution
        if result is False or strategy == GS_TRANSCENDENTAL:
            # reject any result that makes any dens affirmatively 0,
            # if in doubt, keep it
            soln = tsolve(f_num, symbol)
            dens = denoms(f, x=symbols)
            if not dens:
                result = soln
            else:
                result = [s for s in soln if all(not checksol(den, {symbol: s}) for den in dens)]

        if result is False:
            raise NotImplementedError(msg + "\nNo algorithms are implemented to solve equation %s" % f)

        if flags.get('simplified', True) and strategy != GS_RATIONAL:
            result = map(simplify, result)

        return result
    else:
        if not f:
            return []
        else:
            # Create a swap dictionary for storing the passed symbols to be
            # solved for, so that they may be swapped back.
            if symbol_swapped:
                swap_dict = zip(symbols, symbols_new)
                f = [fi.subs(swap_dict) for fi in f]
                symbols = symbols_new

            polys = []

            for g in f:

                poly = g.as_poly(*symbols, **{'extension':True})

                if poly is not None:
                    polys.append(poly)
                else:
                    raise NotImplementedError()

            if all(p.is_linear for p in polys):
                n, m = len(f), len(symbols)
                matrix = zeros((n, m + 1))

                for i, poly in enumerate(polys):
                    for monom, coeff in poly.terms():
                        try:
                            j = list(monom).index(1)
                            matrix[i, j] = coeff
                        except ValueError:
                            matrix[i, m] = -coeff

                # a dictionary of symbols: values or None
                soln = solve_linear_system(matrix, *symbols, **flags)
                # Use swap_dict to ensure we return the same type as what was
                # passed; this is not necessary in the poly-system case which
                # only supports zero-dimensional systems
                if symbol_swapped and soln:
                        soln = dict([(swap_back_dict[k],
                                      v.subs(swap_back_dict))
                                      for k, v in soln.iteritems()])
                return soln
            else:
                # a list of tuples, T, where T[i] [j] corresponds to the ith solution for symbols[j]
                return solve_poly_system(polys)

def solve_linear(lhs, rhs=0, x=[], exclude=[]):
    """ Return a tuple containing derived from f = lhs - rhs that is either:

        (numerator, denominator) of f; if this comes back as (0, 1) it means
            that f is independent of the symbols of x, e.g.
                y*cos(x)**2 + y*sin(x)**2 - y = y*(0) = 0
                cos(x)**2 + sin(x)**2 = 1
            If the numerator is not zero then the function is guaranteed
            to be dependent on a symbol in x.

        or

        (symbol, solution) where symbol appears linearly in the numerator of f,
            is in x (if given) and is not in exclude (if given).

        No simplification is done to f other than and mul=True expansion, so
        the solution will correspond strictly to a unique solution.

    Examples:

        >>> from sympy.solvers.solvers import solve_linear
        >>> from sympy.abc import x, y, z

    These are linear in x and 1/x:

        >>> solve_linear(x + y**2)
        (x, -y**2)
        >>> solve_linear(1/x - y**2)
        (x, y**(-2))

    When not linear in x or y then the numerator and denominator are returned.

        >>> solve_linear(x**2/y**2 - 3)
        (x**2 - 3*y**2, y**2)

    If x is allowed to cancel, then this appears linear, but this sort of
    cancellation is not done so the solution will always satisfy the original
    expression without causing a division by zero error.

        >>> solve_linear(x**2*(1/x - z**2/x))
        (x**2*(-x*z**2 + x), x**2)

    You can give a list of what you prefer for x candidates:

        >>> solve_linear(x + y + z, x=[y])
        (y, -x - z)

    You can also indicate what variables you don't want to consider:

        >>> solve_linear(x + y + z, exclude=[x, z])
        (y, -x - z)

    If only x was excluded then a solution for y or z might be obtained.

    """
    from sympy import expand_mul, Equality
    if isinstance(lhs, Equality):
        rhs += lhs.rhs
        lhs = lhs.lhs
    n, d = (lhs - rhs).as_numer_denom()
    ex = expand_mul(n)
    if not ex:
        return ex, S.One

    exclude = set(exclude)
    syms = ex.free_symbols
    if not x:
        x = syms
    else:
        x = syms.intersection(x)
    x = x.difference(exclude)
    d_free = d.free_symbols
    if x:
        all_zero = True
        for xi in x:
            dn = n.diff(xi)
            if dn:
                all_zero = False
                if not xi in dn.free_symbols:
                    vi = -(n.subs(xi, 0))/dn
                    if not checksol(d, {xi: vi}, minimal=True) is True:
                        return xi, vi

        if all_zero:
            return S.Zero, S.One
    return n, d # should we cancel now?

def solve_linear_system(system, *symbols, **flags):
    """Solve system of N linear equations with M variables, which means
       both Cramer and over defined systems are supported. The possible
       number of solutions is zero, one or infinite. Respectively this
       procedure will return None or dictionary with solutions. In the
       case of over-defined systems all arbitrary parameters are skipped.
       This may cause situation in which an empty dictionary is returned.
       In this case it means all symbols can be assigned arbitrary values.

       Input to this functions is a Nx(M+1) matrix, which means it has
       to be in augmented form. If you prefer to enter N equations and M
       unknowns then use 'solve(Neqs, *Msymbols)' instead. Note: a local
       copy of the matrix is made by this routine so the matrix that is
       passed will not be modified.

       The algorithm used here is fraction-free Gaussian elimination,
       which results, after elimination, in an upper-triangular matrix.
       Then solutions are found using back-substitution. This approach
       is more efficient and compact than the Gauss-Jordan method.

       >>> from sympy import Matrix, solve_linear_system
       >>> from sympy.abc import x, y

       Solve the following system:

              x + 4 y ==  2
           -2 x +   y == 14

       >>> system = Matrix(( (1, 4, 2), (-2, 1, 14)))
       >>> solve_linear_system(system, x, y)
       {x: -6, y: 2}

    """
    matrix = system[:,:]
    syms = list(symbols)

    i, m = 0, matrix.cols-1  # don't count augmentation

    while i < matrix.rows:
        if i == m:
            # an overdetermined system
            if any(matrix[i:,m]):
                return None   # no solutions
            else:
                # remove trailing rows
                matrix = matrix[:i,:]
                break

        if not matrix[i, i]:
            # there is no pivot in current column
            # so try to find one in other columns
            for k in xrange(i+1, m):
                if matrix[i, k]:
                    break
            else:
                if matrix[i, m]:
                    return None   # no solutions
                else:
                    # zero row or was a linear combination of
                    # other rows so now we can safely skip it
                    matrix.row_del(i)
                    continue

            # we want to change the order of colums so
            # the order of variables must also change
            syms[i], syms[k] = syms[k], syms[i]
            matrix.col_swap(i, k)

        pivot_inv = S.One / matrix [i, i]

        # divide all elements in the current row by the pivot
        matrix.row(i, lambda x, _: x * pivot_inv)

        for k in xrange(i+1, matrix.rows):
            if matrix[k, i]:
                coeff = matrix[k, i]

                # subtract from the current row the row containing
                # pivot and multiplied by extracted coefficient
                matrix.row(k, lambda x, j: simplify(x - matrix[i, j]*coeff))

        i += 1

    # if there weren't any problems, augmented matrix is now
    # in row-echelon form so we can check how many solutions
    # there are and extract them using back substitution

    simplified = flags.get('simplified', True)

    if len(syms) == matrix.rows:
        # this system is Cramer equivalent so there is
        # exactly one solution to this system of equations
        k, solutions = i-1, {}

        while k >= 0:
            content = matrix[k, m]

            # run back-substitution for variables
            for j in xrange(k+1, m):
                content -= matrix[k, j]*solutions[syms[j]]

            if simplified:
                solutions[syms[k]] = simplify(content)
            else:
                solutions[syms[k]] = content

            k -= 1

        return solutions
    elif len(syms) > matrix.rows:
        # this system will have infinite number of solutions
        # dependent on exactly len(syms) - i parameters
        k, solutions = i-1, {}

        while k >= 0:
            content = matrix[k, m]

            # run back-substitution for variables
            for j in xrange(k+1, i):
                content -= matrix[k, j]*solutions[syms[j]]

            # run back-substitution for parameters
            for j in xrange(i, m):
                content -= matrix[k, j]*syms[j]

            if simplified:
                solutions[syms[k]] = simplify(content)
            else:
                solutions[syms[k]] = content

            k -= 1

        return solutions
    else:
        return None   # no solutions

def solve_undetermined_coeffs(equ, coeffs, sym, **flags):
    """Solve equation of a type p(x; a_1, ..., a_k) == q(x) where both
       p, q are univariate polynomials and f depends on k parameters.
       The result of this functions is a dictionary with symbolic
       values of those parameters with respect to coefficients in q.

       This functions accepts both Equations class instances and ordinary
       SymPy expressions. Specification of parameters and variable is
       obligatory for efficiency and simplicity reason.

       >>> from sympy import Eq
       >>> from sympy.abc import a, b, c, x
       >>> from sympy.solvers import solve_undetermined_coeffs

       >>> solve_undetermined_coeffs(Eq(2*a*x + a+b, x), [a, b], x)
       {a: 1/2, b: -1/2}

       >>> solve_undetermined_coeffs(Eq(a*c*x + a+b, x), [a, b], x)
       {a: 1/c, b: -1/c}

    """
    if isinstance(equ, Equality):
        # got equation, so move all the
        # terms to the left hand side
        equ = equ.lhs - equ.rhs

    equ = cancel(equ).as_numer_denom()[0]

    system = collect(equ.expand(), sym, evaluate=False).values()

    if not any([ equ.has(sym) for equ in system ]):
        # consecutive powers in the input expressions have
        # been successfully collected, so solve remaining
        # system using Gaussian elimination algorithm
        return solve(system, *coeffs, **flags)
    else:
        return None # no solutions

def solve_linear_system_LU(matrix, syms):
    """ LU function works for invertible only """
    assert matrix.rows == matrix.cols-1
    A = matrix[:matrix.rows,:matrix.rows]
    b = matrix[:,matrix.cols-1:]
    soln = A.LUsolve(b)
    solutions = {}
    for i in range(soln.rows):
        solutions[syms[i]] = soln[i,0]
    return solutions

x = Dummy('x')
a,b,c,d,e,f,g,h = [Wild(t, exclude=[x]) for t in 'abcdefgh']
patterns = None

def _generate_patterns():
    """
    Generates patterns for transcendental equations.

    This is lazily calculated (called) in the tsolve() function and stored in
    the patterns global variable.
    """

    tmp1 = f ** (h-(c*g/b))
    tmp2 = (-e*tmp1/a)**(1/d)
    global patterns
    patterns = [
        (a*(b*x+c)**d + e   , ((-(e/a))**(1/d)-c)/b),
        (    b+c*exp(d*x+e) , (log(-b/c)-e)/d),
        (a*x+b+c*exp(d*x+e) , -b/a-LambertW(c*d*exp(e-b*d/a)/a)/d),
        (    b+c*f**(d*x+e) , (log(-b/c)-e*log(f))/d/log(f)),
        (a*x+b+c*f**(d*x+e) , -b/a-LambertW(c*d*f**(e-b*d/a)*log(f)/a)/d/log(f)),
        (    b+c*log(d*x+e) , (exp(-b/c)-e)/d),
        (a*x+b+c*log(d*x+e) , -e/d+c/a*LambertW(a/c/d*exp(-b/c+a*e/c/d))),
        (a*(b*x+c)**d + e*f**(g*x+h) , -c/b-d*LambertW(-tmp2*g*log(f)/b/d)/g/log(f))
    ]

def tsolve(eq, sym):
    """
    Solves a transcendental equation with respect to the given
    symbol. Various equations containing mixed linear terms, powers,
    and logarithms, can be solved.

    Only a single solution is returned. This solution is generally
    not unique. In some cases, a complex solution may be returned
    even though a real solution exists.

        >>> from sympy import tsolve, log
        >>> from sympy.abc import x

        >>> tsolve(3**(2*x+5)-4, x)
        [(-5*log(3) + 2*log(2))/(2*log(3))]

        >>> tsolve(log(x) + 2*x, x)
        [LambertW(2)/2]

    """
    if patterns is None:
        _generate_patterns()
    eq = sympify(eq)
    if isinstance(eq, Equality):
        eq = eq.lhs - eq.rhs
    sym = sympify(sym)
    eq2 = eq.subs(sym, x)
    # First see if the equation has a linear factor
    # In that case, the other factor can contain x in any way (as long as it
    # is finite), and we have a direct solution to which we add others that
    # may be found for the remaining portion.
    r = Wild('r')
    m = eq2.match((a*x+b)*r)
    if m and m[a]:
        return [(-b/a).subs(m).subs(x, sym)] + solve(m[r], x)
    for p, sol in patterns:
        m = eq2.match(p)
        if m:
            soln = sol.subs(m).subs(x, sym)
            if not(soln is S.NaN or
                   soln.has(S.Infinity) or
                   soln.has(S.NegativeInfinity) or
                   sym in soln.free_symbols):
                return [soln]

    # let's also try to inverse the equation
    lhs = eq
    rhs = S.Zero

    while True:
        indep, dep = lhs.as_independent(sym)

        # dep + indep == rhs
        if lhs.is_Add:
            # this indicates we have done it all
            if indep is S.Zero:
                break

            lhs = dep
            rhs-= indep

        # dep * indep == rhs
        else:
            # this indicates we have done it all
            if indep is S.One:
                break

            lhs = dep
            rhs/= indep

    #                    -1
    # f(x) = g  ->  x = f  (g)
    if lhs.is_Function and lhs.nargs==1 and hasattr(lhs, 'inverse'):
        rhs = lhs.inverse() (rhs)
        lhs = lhs.args[0]

        sol = solve(lhs-rhs, sym)
        return sol

    elif lhs.is_Add:
        # just a simple case - we do variable substitution for first function,
        # and if it removes all functions - let's call solve.
        #      x    -x                   -1
        # UC: e  + e   = y      ->  t + t   = y
        t = Dummy('t')
        terms = lhs.args

        # find first term which is Function
        for f1 in lhs.args:
            if f1.is_Function:
                break
        else:
            raise NotImplementedError("Unable to solve the equation" + \
                "(tsolve: at least one Function expected at this point")

        # perform the substitution
        lhs_ = lhs.subs(f1, t)

        # if no Functions left, we can proceed with usual solve
        if not (lhs_.is_Function or
                any(term.is_Function for term in lhs_.args)):
            cv_sols = solve(lhs_ - rhs, t)
            for sol in cv_sols:
                if sol.has(sym):
                    raise NotImplementedError("Unable to solve the equation")
            cv_inv = solve( t - f1, sym )[0]
            sols = list()
            for sol in cv_sols:
                sols.append(cv_inv.subs(t, sol))
            return sols


    raise NotImplementedError("Unable to solve the equation.")

# TODO: option for calculating J numerically
def nsolve(*args, **kwargs):
    """
    Solve a nonlinear equation system numerically.

    nsolve(f, [args,] x0, modules=['mpmath'], **kwargs)

    f is a vector function of symbolic expressions representing the system.
    args are the variables. If there is only one variable, this argument can be
    omitted.
    x0 is a starting vector close to a solution.

    Use the modules keyword to specify which modules should be used to evaluate
    the function and the Jacobian matrix. Make sure to use a module that
    supports matrices. For more information on the syntax, please see the
    docstring of lambdify.

    Overdetermined systems are supported.

    >>> from sympy import Symbol, nsolve
    >>> import sympy
    >>> sympy.mpmath.mp.dps = 15
    >>> x1 = Symbol('x1')
    >>> x2 = Symbol('x2')
    >>> f1 = 3 * x1**2 - 2 * x2**2 - 1
    >>> f2 = x1**2 - 2 * x1 + x2**2 + 2 * x2 - 8
    >>> print nsolve((f1, f2), (x1, x2), (-1, 1))
    [-1.19287309935246]
    [ 1.27844411169911]

    For one-dimensional functions the syntax is simplified:

    >>> from sympy import sin, nsolve
    >>> from sympy.abc import x
    >>> nsolve(sin(x), x, 2)
    3.14159265358979
    >>> nsolve(sin(x), 2)
    3.14159265358979

    mpmath.findroot is used, you can find there more extensive documentation,
    especially concerning keyword parameters and available solvers.
    """
    # interpret arguments
    if len(args) == 3:
        f = args[0]
        fargs = args[1]
        x0 = args[2]
    elif len(args) == 2:
        f = args[0]
        fargs = None
        x0 = args[1]
    elif len(args) < 2:
        raise TypeError('nsolve expected at least 2 arguments, got %i'
                        % len(args))
    else:
        raise TypeError('nsolve expected at most 3 arguments, got %i'
                        % len(args))
    modules = kwargs.get('modules', ['mpmath'])
    if isinstance(f,  (list,  tuple)):
        f = Matrix(f).T
    if not isinstance(f, Matrix):
        # assume it's a sympy expression
        if isinstance(f, Equality):
            f = f.lhs - f.rhs
        f = f.evalf()
        atoms = f.atoms(Symbol)
        if fargs is None:
            fargs = atoms.copy().pop()
        if not (len(atoms) == 1 and (fargs in atoms or fargs[0] in atoms)):
            raise ValueError('expected a one-dimensional and numerical function')

        # the function is much better behaved if there is no denominator
        f = f.as_numer_denom()[0]

        f = lambdify(fargs, f, modules)
        return findroot(f, x0, **kwargs)
    if len(fargs) > f.cols:
        raise NotImplementedError('need at least as many equations as variables')
    verbose = kwargs.get('verbose', False)
    if verbose:
        print 'f(x):'
        print f
    # derive Jacobian
    J = f.jacobian(fargs)
    if verbose:
        print 'J(x):'
        print J
    # create functions
    f = lambdify(fargs, f.T, modules)
    J = lambdify(fargs, J, modules)
    # solve the system numerically
    x = findroot(f, x0, J=J, **kwargs)
    return x
