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It also decomposes + Polynomials. For only Polynomial decomposition see ``decompose`` in polys. + + Examples + ======== + + >>> from sympy.abc import x + >>> from sympy import decompogen, sqrt, sin, cos + >>> decompogen(sin(cos(x)), x) + [sin(x), cos(x)] + >>> decompogen(sin(x)**2 + sin(x) + 1, x) + [x**2 + x + 1, sin(x)] + >>> decompogen(sqrt(6*x**2 - 5), x) + [sqrt(x), 6*x**2 - 5] + >>> decompogen(sin(sqrt(cos(x**2 + 1))), x) + [sin(x), sqrt(x), cos(x), x**2 + 1] + >>> decompogen(x**4 + 2*x**3 - x - 1, x) + [x**2 - x - 1, x**2 + x] + + """ + f = sympify(f) + if not isinstance(f, Expr) or isinstance(f, Relational): + raise TypeError('expecting Expr but got: `%s`' % func_name(f)) + if symbol not in f.free_symbols: + return [f] + + + # ===== Simple Functions ===== # + if isinstance(f, (Function, Pow)): + if f.is_Pow and f.base == S.Exp1: + arg = f.exp + else: + arg = f.args[0] + if arg == symbol: + return [f] + return [f.subs(arg, symbol)] + decompogen(arg, symbol) + + # ===== Min/Max Functions ===== # + if isinstance(f, (Min, Max)): + args = list(f.args) + d0 = None + for i, a in enumerate(args): + if not a.has_free(symbol): + continue + d = decompogen(a, symbol) + if len(d) == 1: + d = [symbol] + d + if d0 is None: + d0 = d[1:] + elif d[1:] != d0: + # decomposition is not the same for each arg: + # mark as having no decomposition + d = [symbol] + break + args[i] = d[0] + if d[0] == symbol: + return [f] + return [f.func(*args)] + d0 + + # ===== Convert to Polynomial ===== # + fp = Poly(f) + gens = list(filter(lambda x: symbol in x.free_symbols, fp.gens)) + + if len(gens) == 1 and gens[0] != symbol: + f1 = f.subs(gens[0], symbol) + f2 = gens[0] + return [f1] + decompogen(f2, symbol) + + # ===== Polynomial decompose() ====== # + try: + return decompose(f) + except ValueError: + return [f] + + +def compogen(g_s, symbol): + """ + Returns the composition of functions. + Given a list of functions ``g_s``, returns their composition ``f``, + where: + f = g_1 o g_2 o .. o g_n + + Note: This is a General composition function. It also composes Polynomials. + For only Polynomial composition see ``compose`` in polys. + + Examples + ======== + + >>> from sympy.solvers.decompogen import compogen + >>> from sympy.abc import x + >>> from sympy import sqrt, sin, cos + >>> compogen([sin(x), cos(x)], x) + sin(cos(x)) + >>> compogen([x**2 + x + 1, sin(x)], x) + sin(x)**2 + sin(x) + 1 + >>> compogen([sqrt(x), 6*x**2 - 5], x) + sqrt(6*x**2 - 5) + >>> compogen([sin(x), sqrt(x), cos(x), x**2 + 1], x) + sin(sqrt(cos(x**2 + 1))) + >>> compogen([x**2 - x - 1, x**2 + x], x) + -x**2 - x + (x**2 + x)**2 - 1 + """ + if len(g_s) == 1: + return g_s[0] + + foo = g_s[0].subs(symbol, g_s[1]) + + if len(g_s) == 2: + return foo + + return compogen([foo] + g_s[2:], symbol) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/deutils.py b/pllava/lib/python3.10/site-packages/sympy/solvers/deutils.py new file mode 100644 index 0000000000000000000000000000000000000000..c968b65c8d518b82ed308ba932ea9297a3fe3808 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/deutils.py @@ -0,0 +1,273 @@ +"""Utility functions for classifying and solving +ordinary and partial differential equations. + +Contains +======== +_preprocess +ode_order +_desolve + +""" +from sympy.core import Pow +from sympy.core.function import Derivative, AppliedUndef +from sympy.core.relational import Equality +from sympy.core.symbol import Wild + +def _preprocess(expr, func=None, hint='_Integral'): + """Prepare expr for solving by making sure that differentiation + is done so that only func remains in unevaluated derivatives and + (if hint does not end with _Integral) that doit is applied to all + other derivatives. If hint is None, do not do any differentiation. + (Currently this may cause some simple differential equations to + fail.) + + In case func is None, an attempt will be made to autodetect the + function to be solved for. + + >>> from sympy.solvers.deutils import _preprocess + >>> from sympy import Derivative, Function + >>> from sympy.abc import x, y, z + >>> f, g = map(Function, 'fg') + + If f(x)**p == 0 and p>0 then we can solve for f(x)=0 + >>> _preprocess((f(x).diff(x)-4)**5, f(x)) + (Derivative(f(x), x) - 4, f(x)) + + Apply doit to derivatives that contain more than the function + of interest: + + >>> _preprocess(Derivative(f(x) + x, x)) + (Derivative(f(x), x) + 1, f(x)) + + Do others if the differentiation variable(s) intersect with those + of the function of interest or contain the function of interest: + + >>> _preprocess(Derivative(g(x), y, z), f(y)) + (0, f(y)) + >>> _preprocess(Derivative(f(y), z), f(y)) + (0, f(y)) + + Do others if the hint does not end in '_Integral' (the default + assumes that it does): + + >>> _preprocess(Derivative(g(x), y), f(x)) + (Derivative(g(x), y), f(x)) + >>> _preprocess(Derivative(f(x), y), f(x), hint='') + (0, f(x)) + + Do not do any derivatives if hint is None: + + >>> eq = Derivative(f(x) + 1, x) + Derivative(f(x), y) + >>> _preprocess(eq, f(x), hint=None) + (Derivative(f(x) + 1, x) + Derivative(f(x), y), f(x)) + + If it's not clear what the function of interest is, it must be given: + + >>> eq = Derivative(f(x) + g(x), x) + >>> _preprocess(eq, g(x)) + (Derivative(f(x), x) + Derivative(g(x), x), g(x)) + >>> try: _preprocess(eq) + ... except ValueError: print("A ValueError was raised.") + A ValueError was raised. + + """ + if isinstance(expr, Pow): + # if f(x)**p=0 then f(x)=0 (p>0) + if (expr.exp).is_positive: + expr = expr.base + derivs = expr.atoms(Derivative) + if not func: + funcs = set().union(*[d.atoms(AppliedUndef) for d in derivs]) + if len(funcs) != 1: + raise ValueError('The function cannot be ' + 'automatically detected for %s.' % expr) + func = funcs.pop() + fvars = set(func.args) + if hint is None: + return expr, func + reps = [(d, d.doit()) for d in derivs if not hint.endswith('_Integral') or + d.has(func) or set(d.variables) & fvars] + eq = expr.subs(reps) + return eq, func + + +def ode_order(expr, func): + """ + Returns the order of a given differential + equation with respect to func. + + This function is implemented recursively. + + Examples + ======== + + >>> from sympy import Function + >>> from sympy.solvers.deutils import ode_order + >>> from sympy.abc import x + >>> f, g = map(Function, ['f', 'g']) + >>> ode_order(f(x).diff(x, 2) + f(x).diff(x)**2 + + ... f(x).diff(x), f(x)) + 2 + >>> ode_order(f(x).diff(x, 2) + g(x).diff(x, 3), f(x)) + 2 + >>> ode_order(f(x).diff(x, 2) + g(x).diff(x, 3), g(x)) + 3 + + """ + a = Wild('a', exclude=[func]) + if expr.match(a): + return 0 + + if isinstance(expr, Derivative): + if expr.args[0] == func: + return len(expr.variables) + else: + args = expr.args[0].args + rv = len(expr.variables) + if args: + rv += max(ode_order(_, func) for _ in args) + return rv + else: + return max(ode_order(_, func) for _ in expr.args) if expr.args else 0 + + +def _desolve(eq, func=None, hint="default", ics=None, simplify=True, *, prep=True, **kwargs): + """This is a helper function to dsolve and pdsolve in the ode + and pde modules. + + If the hint provided to the function is "default", then a dict with + the following keys are returned + + 'func' - It provides the function for which the differential equation + has to be solved. This is useful when the expression has + more than one function in it. + + 'default' - The default key as returned by classifier functions in ode + and pde.py + + 'hint' - The hint given by the user for which the differential equation + is to be solved. If the hint given by the user is 'default', + then the value of 'hint' and 'default' is the same. + + 'order' - The order of the function as returned by ode_order + + 'match' - It returns the match as given by the classifier functions, for + the default hint. + + If the hint provided to the function is not "default" and is not in + ('all', 'all_Integral', 'best'), then a dict with the above mentioned keys + is returned along with the keys which are returned when dict in + classify_ode or classify_pde is set True + + If the hint given is in ('all', 'all_Integral', 'best'), then this function + returns a nested dict, with the keys, being the set of classified hints + returned by classifier functions, and the values being the dict of form + as mentioned above. + + Key 'eq' is a common key to all the above mentioned hints which returns an + expression if eq given by user is an Equality. + + See Also + ======== + classify_ode(ode.py) + classify_pde(pde.py) + """ + if isinstance(eq, Equality): + eq = eq.lhs - eq.rhs + + # preprocess the equation and find func if not given + if prep or func is None: + eq, func = _preprocess(eq, func) + prep = False + + # type is an argument passed by the solve functions in ode and pde.py + # that identifies whether the function caller is an ordinary + # or partial differential equation. Accordingly corresponding + # changes are made in the function. + type = kwargs.get('type', None) + xi = kwargs.get('xi') + eta = kwargs.get('eta') + x0 = kwargs.get('x0', 0) + terms = kwargs.get('n') + + if type == 'ode': + from sympy.solvers.ode import classify_ode, allhints + classifier = classify_ode + string = 'ODE ' + dummy = '' + + elif type == 'pde': + from sympy.solvers.pde import classify_pde, allhints + classifier = classify_pde + string = 'PDE ' + dummy = 'p' + + # Magic that should only be used internally. Prevents classify_ode from + # being called more than it needs to be by passing its results through + # recursive calls. + if kwargs.get('classify', True): + hints = classifier(eq, func, dict=True, ics=ics, xi=xi, eta=eta, + n=terms, x0=x0, hint=hint, prep=prep) + + else: + # Here is what all this means: + # + # hint: The hint method given to _desolve() by the user. + # hints: The dictionary of hints that match the DE, along with other + # information (including the internal pass-through magic). + # default: The default hint to return, the first hint from allhints + # that matches the hint; obtained from classify_ode(). + # match: Dictionary containing the match dictionary for each hint + # (the parts of the DE for solving). When going through the + # hints in "all", this holds the match string for the current + # hint. + # order: The order of the DE, as determined by ode_order(). + hints = kwargs.get('hint', + {'default': hint, + hint: kwargs['match'], + 'order': kwargs['order']}) + if not hints['default']: + # classify_ode will set hints['default'] to None if no hints match. + if hint not in allhints and hint != 'default': + raise ValueError("Hint not recognized: " + hint) + elif hint not in hints['ordered_hints'] and hint != 'default': + raise ValueError(string + str(eq) + " does not match hint " + hint) + # If dsolve can't solve the purely algebraic equation then dsolve will raise + # ValueError + elif hints['order'] == 0: + raise ValueError( + str(eq) + " is not a solvable differential equation in " + str(func)) + else: + raise NotImplementedError(dummy + "solve" + ": Cannot solve " + str(eq)) + if hint == 'default': + return _desolve(eq, func, ics=ics, hint=hints['default'], simplify=simplify, + prep=prep, x0=x0, classify=False, order=hints['order'], + match=hints[hints['default']], xi=xi, eta=eta, n=terms, type=type) + elif hint in ('all', 'all_Integral', 'best'): + retdict = {} + gethints = set(hints) - {'order', 'default', 'ordered_hints'} + if hint == 'all_Integral': + for i in hints: + if i.endswith('_Integral'): + gethints.remove(i[:-len('_Integral')]) + # special cases + for k in ["1st_homogeneous_coeff_best", "1st_power_series", + "lie_group", "2nd_power_series_ordinary", "2nd_power_series_regular"]: + if k in gethints: + gethints.remove(k) + for i in gethints: + sol = _desolve(eq, func, ics=ics, hint=i, x0=x0, simplify=simplify, prep=prep, + classify=False, n=terms, order=hints['order'], match=hints[i], type=type) + retdict[i] = sol + retdict['all'] = True + retdict['eq'] = eq + return retdict + elif hint not in allhints: # and hint not in ('default', 'ordered_hints'): + raise ValueError("Hint not recognized: " + hint) + elif hint not in hints: + raise ValueError(string + str(eq) + " does not match hint " + hint) + else: + # Key added to identify the hint needed to solve the equation + hints['hint'] = hint + hints.update({'func': func, 'eq': eq}) + return hints diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/__init__.py b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/__init__.py new file mode 100644 index 0000000000000000000000000000000000000000..23c21242208d6f520c130250ecdce43382b9d868 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/__init__.py @@ -0,0 +1,5 @@ +from .diophantine import diophantine, classify_diop, diop_solve + +__all__ = [ + 'diophantine', 'classify_diop', 'diop_solve' +] diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/__pycache__/__init__.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/__pycache__/__init__.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..f5ea4496d035fb9e077779fc32268f1b4d5125e0 Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/__pycache__/__init__.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/diophantine.py b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/diophantine.py new file mode 100644 index 0000000000000000000000000000000000000000..3df4fe9b0df137828233a9243d2e1e604af309fd --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/diophantine.py @@ -0,0 +1,3960 @@ +from __future__ import annotations + +from sympy.core.add import Add +from sympy.core.assumptions import check_assumptions +from sympy.core.containers import Tuple +from sympy.core.exprtools import factor_terms +from sympy.core.function import _mexpand +from sympy.core.mul import Mul +from sympy.core.numbers import Rational, int_valued +from sympy.core.intfunc import igcdex, ilcm, igcd, integer_nthroot, isqrt +from sympy.core.relational import Eq +from sympy.core.singleton import S +from sympy.core.sorting import default_sort_key, ordered +from sympy.core.symbol import Symbol, symbols +from sympy.core.sympify import _sympify +from sympy.external.gmpy import jacobi, remove, invert, iroot +from sympy.functions.elementary.complexes import sign +from sympy.functions.elementary.integers import floor +from sympy.functions.elementary.miscellaneous import sqrt +from sympy.matrices.dense import MutableDenseMatrix as Matrix +from sympy.ntheory.factor_ import divisors, factorint, perfect_power +from sympy.ntheory.generate import nextprime +from sympy.ntheory.primetest import is_square, isprime +from sympy.ntheory.modular import symmetric_residue +from sympy.ntheory.residue_ntheory import sqrt_mod, sqrt_mod_iter +from sympy.polys.polyerrors import GeneratorsNeeded +from sympy.polys.polytools import Poly, factor_list +from sympy.simplify.simplify import signsimp +from sympy.solvers.solveset import solveset_real +from sympy.utilities import numbered_symbols +from sympy.utilities.misc import as_int, filldedent +from sympy.utilities.iterables import (is_sequence, subsets, permute_signs, + signed_permutations, ordered_partitions) + + +# these are imported with 'from sympy.solvers.diophantine import * +__all__ = ['diophantine', 'classify_diop'] + + +class DiophantineSolutionSet(set): + """ + Container for a set of solutions to a particular diophantine equation. + + The base representation is a set of tuples representing each of the solutions. + + Parameters + ========== + + symbols : list + List of free symbols in the original equation. + parameters: list + List of parameters to be used in the solution. + + Examples + ======== + + Adding solutions: + + >>> from sympy.solvers.diophantine.diophantine import DiophantineSolutionSet + >>> from sympy.abc import x, y, t, u + >>> s1 = DiophantineSolutionSet([x, y], [t, u]) + >>> s1 + set() + >>> s1.add((2, 3)) + >>> s1.add((-1, u)) + >>> s1 + {(-1, u), (2, 3)} + >>> s2 = DiophantineSolutionSet([x, y], [t, u]) + >>> s2.add((3, 4)) + >>> s1.update(*s2) + >>> s1 + {(-1, u), (2, 3), (3, 4)} + + Conversion of solutions into dicts: + + >>> list(s1.dict_iterator()) + [{x: -1, y: u}, {x: 2, y: 3}, {x: 3, y: 4}] + + Substituting values: + + >>> s3 = DiophantineSolutionSet([x, y], [t, u]) + >>> s3.add((t**2, t + u)) + >>> s3 + {(t**2, t + u)} + >>> s3.subs({t: 2, u: 3}) + {(4, 5)} + >>> s3.subs(t, -1) + {(1, u - 1)} + >>> s3.subs(t, 3) + {(9, u + 3)} + + Evaluation at specific values. Positional arguments are given in the same order as the parameters: + + >>> s3(-2, 3) + {(4, 1)} + >>> s3(5) + {(25, u + 5)} + >>> s3(None, 2) + {(t**2, t + 2)} + """ + + def __init__(self, symbols_seq, parameters): + super().__init__() + + if not is_sequence(symbols_seq): + raise ValueError("Symbols must be given as a sequence.") + + if not is_sequence(parameters): + raise ValueError("Parameters must be given as a sequence.") + + self.symbols = tuple(symbols_seq) + self.parameters = tuple(parameters) + + def add(self, solution): + if len(solution) != len(self.symbols): + raise ValueError("Solution should have a length of %s, not %s" % (len(self.symbols), len(solution))) + # make solution canonical wrt sign (i.e. no -x unless x is also present as an arg) + args = set(solution) + for i in range(len(solution)): + x = solution[i] + if not type(x) is int and (-x).is_Symbol and -x not in args: + solution = [_.subs(-x, x) for _ in solution] + super().add(Tuple(*solution)) + + def update(self, *solutions): + for solution in solutions: + self.add(solution) + + def dict_iterator(self): + for solution in ordered(self): + yield dict(zip(self.symbols, solution)) + + def subs(self, *args, **kwargs): + result = DiophantineSolutionSet(self.symbols, self.parameters) + for solution in self: + result.add(solution.subs(*args, **kwargs)) + return result + + def __call__(self, *args): + if len(args) > len(self.parameters): + raise ValueError("Evaluation should have at most %s values, not %s" % (len(self.parameters), len(args))) + rep = {p: v for p, v in zip(self.parameters, args) if v is not None} + return self.subs(rep) + + +class DiophantineEquationType: + """ + Internal representation of a particular diophantine equation type. + + Parameters + ========== + + equation : + The diophantine equation that is being solved. + free_symbols : list (optional) + The symbols being solved for. + + Attributes + ========== + + total_degree : + The maximum of the degrees of all terms in the equation + homogeneous : + Does the equation contain a term of degree 0 + homogeneous_order : + Does the equation contain any coefficient that is in the symbols being solved for + dimension : + The number of symbols being solved for + """ + name = None # type: str + + def __init__(self, equation, free_symbols=None): + self.equation = _sympify(equation).expand(force=True) + + if free_symbols is not None: + self.free_symbols = free_symbols + else: + self.free_symbols = list(self.equation.free_symbols) + self.free_symbols.sort(key=default_sort_key) + + if not self.free_symbols: + raise ValueError('equation should have 1 or more free symbols') + + self.coeff = self.equation.as_coefficients_dict() + if not all(int_valued(c) for c in self.coeff.values()): + raise TypeError("Coefficients should be Integers") + + self.total_degree = Poly(self.equation).total_degree() + self.homogeneous = 1 not in self.coeff + self.homogeneous_order = not (set(self.coeff) & set(self.free_symbols)) + self.dimension = len(self.free_symbols) + self._parameters = None + + def matches(self): + """ + Determine whether the given equation can be matched to the particular equation type. + """ + return False + + @property + def n_parameters(self): + return self.dimension + + @property + def parameters(self): + if self._parameters is None: + self._parameters = symbols('t_:%i' % (self.n_parameters,), integer=True) + return self._parameters + + def solve(self, parameters=None, limit=None) -> DiophantineSolutionSet: + raise NotImplementedError('No solver has been written for %s.' % self.name) + + def pre_solve(self, parameters=None): + if not self.matches(): + raise ValueError("This equation does not match the %s equation type." % self.name) + + if parameters is not None: + if len(parameters) != self.n_parameters: + raise ValueError("Expected %s parameter(s) but got %s" % (self.n_parameters, len(parameters))) + + self._parameters = parameters + + +class Univariate(DiophantineEquationType): + """ + Representation of a univariate diophantine equation. + + A univariate diophantine equation is an equation of the form + `a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are + integer constants and `x` is an integer variable. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import Univariate + >>> from sympy.abc import x + >>> Univariate((x - 2)*(x - 3)**2).solve() # solves equation (x - 2)*(x - 3)**2 == 0 + {(2,), (3,)} + + """ + + name = 'univariate' + + def matches(self): + return self.dimension == 1 + + def solve(self, parameters=None, limit=None): + self.pre_solve(parameters) + + result = DiophantineSolutionSet(self.free_symbols, parameters=self.parameters) + for i in solveset_real(self.equation, self.free_symbols[0]).intersect(S.Integers): + result.add((i,)) + return result + + +class Linear(DiophantineEquationType): + """ + Representation of a linear diophantine equation. + + A linear diophantine equation is an equation of the form `a_{1}x_{1} + + a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are + integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import Linear + >>> from sympy.abc import x, y, z + >>> l1 = Linear(2*x - 3*y - 5) + >>> l1.matches() # is this equation linear + True + >>> l1.solve() # solves equation 2*x - 3*y - 5 == 0 + {(3*t_0 - 5, 2*t_0 - 5)} + + Here x = -3*t_0 - 5 and y = -2*t_0 - 5 + + >>> Linear(2*x - 3*y - 4*z -3).solve() + {(t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3)} + + """ + + name = 'linear' + + def matches(self): + return self.total_degree == 1 + + def solve(self, parameters=None, limit=None): + self.pre_solve(parameters) + + coeff = self.coeff + var = self.free_symbols + + if 1 in coeff: + # negate coeff[] because input is of the form: ax + by + c == 0 + # but is used as: ax + by == -c + c = -coeff[1] + else: + c = 0 + + result = DiophantineSolutionSet(var, parameters=self.parameters) + params = result.parameters + + if len(var) == 1: + q, r = divmod(c, coeff[var[0]]) + if not r: + result.add((q,)) + return result + + ''' + base_solution_linear() can solve diophantine equations of the form: + + a*x + b*y == c + + We break down multivariate linear diophantine equations into a + series of bivariate linear diophantine equations which can then + be solved individually by base_solution_linear(). + + Consider the following: + + a_0*x_0 + a_1*x_1 + a_2*x_2 == c + + which can be re-written as: + + a_0*x_0 + g_0*y_0 == c + + where + + g_0 == gcd(a_1, a_2) + + and + + y == (a_1*x_1)/g_0 + (a_2*x_2)/g_0 + + This leaves us with two binary linear diophantine equations. + For the first equation: + + a == a_0 + b == g_0 + c == c + + For the second: + + a == a_1/g_0 + b == a_2/g_0 + c == the solution we find for y_0 in the first equation. + + The arrays A and B are the arrays of integers used for + 'a' and 'b' in each of the n-1 bivariate equations we solve. + ''' + + A = [coeff[v] for v in var] + B = [] + if len(var) > 2: + B.append(igcd(A[-2], A[-1])) + A[-2] = A[-2] // B[0] + A[-1] = A[-1] // B[0] + for i in range(len(A) - 3, 0, -1): + gcd = igcd(B[0], A[i]) + B[0] = B[0] // gcd + A[i] = A[i] // gcd + B.insert(0, gcd) + B.append(A[-1]) + + ''' + Consider the trivariate linear equation: + + 4*x_0 + 6*x_1 + 3*x_2 == 2 + + This can be re-written as: + + 4*x_0 + 3*y_0 == 2 + + where + + y_0 == 2*x_1 + x_2 + (Note that gcd(3, 6) == 3) + + The complete integral solution to this equation is: + + x_0 == 2 + 3*t_0 + y_0 == -2 - 4*t_0 + + where 't_0' is any integer. + + Now that we have a solution for 'x_0', find 'x_1' and 'x_2': + + 2*x_1 + x_2 == -2 - 4*t_0 + + We can then solve for '-2' and '-4' independently, + and combine the results: + + 2*x_1a + x_2a == -2 + x_1a == 0 + t_0 + x_2a == -2 - 2*t_0 + + 2*x_1b + x_2b == -4*t_0 + x_1b == 0*t_0 + t_1 + x_2b == -4*t_0 - 2*t_1 + + ==> + + x_1 == t_0 + t_1 + x_2 == -2 - 6*t_0 - 2*t_1 + + where 't_0' and 't_1' are any integers. + + Note that: + + 4*(2 + 3*t_0) + 6*(t_0 + t_1) + 3*(-2 - 6*t_0 - 2*t_1) == 2 + + for any integral values of 't_0', 't_1'; as required. + + This method is generalised for many variables, below. + + ''' + solutions = [] + for Ai, Bi in zip(A, B): + tot_x, tot_y = [], [] + + for arg in Add.make_args(c): + if arg.is_Integer: + # example: 5 -> k = 5 + k, p = arg, S.One + pnew = params[0] + else: # arg is a Mul or Symbol + # example: 3*t_1 -> k = 3 + # example: t_0 -> k = 1 + k, p = arg.as_coeff_Mul() + pnew = params[params.index(p) + 1] + + sol = sol_x, sol_y = base_solution_linear(k, Ai, Bi, pnew) + + if p is S.One: + if None in sol: + return result + else: + # convert a + b*pnew -> a*p + b*pnew + if isinstance(sol_x, Add): + sol_x = sol_x.args[0]*p + sol_x.args[1] + if isinstance(sol_y, Add): + sol_y = sol_y.args[0]*p + sol_y.args[1] + + tot_x.append(sol_x) + tot_y.append(sol_y) + + solutions.append(Add(*tot_x)) + c = Add(*tot_y) + + solutions.append(c) + result.add(solutions) + return result + + +class BinaryQuadratic(DiophantineEquationType): + """ + Representation of a binary quadratic diophantine equation. + + A binary quadratic diophantine equation is an equation of the + form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`, where `A, B, C, D, E, + F` are integer constants and `x` and `y` are integer variables. + + Examples + ======== + + >>> from sympy.abc import x, y + >>> from sympy.solvers.diophantine.diophantine import BinaryQuadratic + >>> b1 = BinaryQuadratic(x**3 + y**2 + 1) + >>> b1.matches() + False + >>> b2 = BinaryQuadratic(x**2 + y**2 + 2*x + 2*y + 2) + >>> b2.matches() + True + >>> b2.solve() + {(-1, -1)} + + References + ========== + + .. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], + Available: https://www.alpertron.com.ar/METHODS.HTM + .. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], + Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf + + """ + + name = 'binary_quadratic' + + def matches(self): + return self.total_degree == 2 and self.dimension == 2 + + def solve(self, parameters=None, limit=None) -> DiophantineSolutionSet: + self.pre_solve(parameters) + + var = self.free_symbols + coeff = self.coeff + + x, y = var + + A = coeff[x**2] + B = coeff[x*y] + C = coeff[y**2] + D = coeff[x] + E = coeff[y] + F = coeff[S.One] + + A, B, C, D, E, F = [as_int(i) for i in _remove_gcd(A, B, C, D, E, F)] + + # (1) Simple-Hyperbolic case: A = C = 0, B != 0 + # In this case equation can be converted to (Bx + E)(By + D) = DE - BF + # We consider two cases; DE - BF = 0 and DE - BF != 0 + # More details, https://www.alpertron.com.ar/METHODS.HTM#SHyperb + + result = DiophantineSolutionSet(var, self.parameters) + t, u = result.parameters + + discr = B**2 - 4*A*C + if A == 0 and C == 0 and B != 0: + + if D*E - B*F == 0: + q, r = divmod(E, B) + if not r: + result.add((-q, t)) + q, r = divmod(D, B) + if not r: + result.add((t, -q)) + else: + div = divisors(D*E - B*F) + div = div + [-term for term in div] + for d in div: + x0, r = divmod(d - E, B) + if not r: + q, r = divmod(D*E - B*F, d) + if not r: + y0, r = divmod(q - D, B) + if not r: + result.add((x0, y0)) + + # (2) Parabolic case: B**2 - 4*A*C = 0 + # There are two subcases to be considered in this case. + # sqrt(c)D - sqrt(a)E = 0 and sqrt(c)D - sqrt(a)E != 0 + # More Details, https://www.alpertron.com.ar/METHODS.HTM#Parabol + + elif discr == 0: + + if A == 0: + s = BinaryQuadratic(self.equation, free_symbols=[y, x]).solve(parameters=[t, u]) + for soln in s: + result.add((soln[1], soln[0])) + + else: + g = sign(A)*igcd(A, C) + a = A // g + c = C // g + e = sign(B / A) + + sqa = isqrt(a) + sqc = isqrt(c) + _c = e*sqc*D - sqa*E + if not _c: + z = Symbol("z", real=True) + eq = sqa*g*z**2 + D*z + sqa*F + roots = solveset_real(eq, z).intersect(S.Integers) + for root in roots: + ans = diop_solve(sqa*x + e*sqc*y - root) + result.add((ans[0], ans[1])) + + elif int_valued(c): + solve_x = lambda u: -e*sqc*g*_c*t**2 - (E + 2*e*sqc*g*u)*t \ + - (e*sqc*g*u**2 + E*u + e*sqc*F) // _c + + solve_y = lambda u: sqa*g*_c*t**2 + (D + 2*sqa*g*u)*t \ + + (sqa*g*u**2 + D*u + sqa*F) // _c + + for z0 in range(0, abs(_c)): + # Check if the coefficients of y and x obtained are integers or not + if (divisible(sqa*g*z0**2 + D*z0 + sqa*F, _c) and + divisible(e*sqc*g*z0**2 + E*z0 + e*sqc*F, _c)): + result.add((solve_x(z0), solve_y(z0))) + + # (3) Method used when B**2 - 4*A*C is a square, is described in p. 6 of the below paper + # by John P. Robertson. + # https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf + + elif is_square(discr): + if A != 0: + r = sqrt(discr) + u, v = symbols("u, v", integer=True) + eq = _mexpand( + 4*A*r*u*v + 4*A*D*(B*v + r*u + r*v - B*u) + + 2*A*4*A*E*(u - v) + 4*A*r*4*A*F) + + solution = diop_solve(eq, t) + + for s0, t0 in solution: + + num = B*t0 + r*s0 + r*t0 - B*s0 + x_0 = S(num) / (4*A*r) + y_0 = S(s0 - t0) / (2*r) + if isinstance(s0, Symbol) or isinstance(t0, Symbol): + if len(check_param(x_0, y_0, 4*A*r, parameters)) > 0: + ans = check_param(x_0, y_0, 4*A*r, parameters) + result.update(*ans) + elif x_0.is_Integer and y_0.is_Integer: + if is_solution_quad(var, coeff, x_0, y_0): + result.add((x_0, y_0)) + + else: + s = BinaryQuadratic(self.equation, free_symbols=var[::-1]).solve(parameters=[t, u]) # Interchange x and y + while s: + result.add(s.pop()[::-1]) # and solution <--------+ + + # (4) B**2 - 4*A*C > 0 and B**2 - 4*A*C not a square or B**2 - 4*A*C < 0 + + else: + + P, Q = _transformation_to_DN(var, coeff) + D, N = _find_DN(var, coeff) + solns_pell = diop_DN(D, N) + + if D < 0: + for x0, y0 in solns_pell: + for x in [-x0, x0]: + for y in [-y0, y0]: + s = P*Matrix([x, y]) + Q + try: + result.add([as_int(_) for _ in s]) + except ValueError: + pass + else: + # In this case equation can be transformed into a Pell equation + + solns_pell = set(solns_pell) + solns_pell.update((-X, -Y) for X, Y in list(solns_pell)) + + a = diop_DN(D, 1) + T = a[0][0] + U = a[0][1] + + if all(int_valued(_) for _ in P[:4] + Q[:2]): + for r, s in solns_pell: + _a = (r + s*sqrt(D))*(T + U*sqrt(D))**t + _b = (r - s*sqrt(D))*(T - U*sqrt(D))**t + x_n = _mexpand(S(_a + _b) / 2) + y_n = _mexpand(S(_a - _b) / (2*sqrt(D))) + s = P*Matrix([x_n, y_n]) + Q + result.add(s) + + else: + L = ilcm(*[_.q for _ in P[:4] + Q[:2]]) + + k = 1 + + T_k = T + U_k = U + + while (T_k - 1) % L != 0 or U_k % L != 0: + T_k, U_k = T_k*T + D*U_k*U, T_k*U + U_k*T + k += 1 + + for X, Y in solns_pell: + + for i in range(k): + if all(int_valued(_) for _ in P*Matrix([X, Y]) + Q): + _a = (X + sqrt(D)*Y)*(T_k + sqrt(D)*U_k)**t + _b = (X - sqrt(D)*Y)*(T_k - sqrt(D)*U_k)**t + Xt = S(_a + _b) / 2 + Yt = S(_a - _b) / (2*sqrt(D)) + s = P*Matrix([Xt, Yt]) + Q + result.add(s) + + X, Y = X*T + D*U*Y, X*U + Y*T + + return result + + +class InhomogeneousTernaryQuadratic(DiophantineEquationType): + """ + + Representation of an inhomogeneous ternary quadratic. + + No solver is currently implemented for this equation type. + + """ + + name = 'inhomogeneous_ternary_quadratic' + + def matches(self): + if not (self.total_degree == 2 and self.dimension == 3): + return False + if not self.homogeneous: + return False + return not self.homogeneous_order + + +class HomogeneousTernaryQuadraticNormal(DiophantineEquationType): + """ + Representation of a homogeneous ternary quadratic normal diophantine equation. + + Examples + ======== + + >>> from sympy.abc import x, y, z + >>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadraticNormal + >>> HomogeneousTernaryQuadraticNormal(4*x**2 - 5*y**2 + z**2).solve() + {(1, 2, 4)} + + """ + + name = 'homogeneous_ternary_quadratic_normal' + + def matches(self): + if not (self.total_degree == 2 and self.dimension == 3): + return False + if not self.homogeneous: + return False + if not self.homogeneous_order: + return False + + nonzero = [k for k in self.coeff if self.coeff[k]] + return len(nonzero) == 3 and all(i**2 in nonzero for i in self.free_symbols) + + def solve(self, parameters=None, limit=None) -> DiophantineSolutionSet: + self.pre_solve(parameters) + + var = self.free_symbols + coeff = self.coeff + + x, y, z = var + + a = coeff[x**2] + b = coeff[y**2] + c = coeff[z**2] + + (sqf_of_a, sqf_of_b, sqf_of_c), (a_1, b_1, c_1), (a_2, b_2, c_2) = \ + sqf_normal(a, b, c, steps=True) + + A = -a_2*c_2 + B = -b_2*c_2 + + result = DiophantineSolutionSet(var, parameters=self.parameters) + + # If following two conditions are satisfied then there are no solutions + if A < 0 and B < 0: + return result + + if ( + sqrt_mod(-b_2*c_2, a_2) is None or + sqrt_mod(-c_2*a_2, b_2) is None or + sqrt_mod(-a_2*b_2, c_2) is None): + return result + + z_0, x_0, y_0 = descent(A, B) + + z_0, q = _rational_pq(z_0, abs(c_2)) + x_0 *= q + y_0 *= q + + x_0, y_0, z_0 = _remove_gcd(x_0, y_0, z_0) + + # Holzer reduction + if sign(a) == sign(b): + x_0, y_0, z_0 = holzer(x_0, y_0, z_0, abs(a_2), abs(b_2), abs(c_2)) + elif sign(a) == sign(c): + x_0, z_0, y_0 = holzer(x_0, z_0, y_0, abs(a_2), abs(c_2), abs(b_2)) + else: + y_0, z_0, x_0 = holzer(y_0, z_0, x_0, abs(b_2), abs(c_2), abs(a_2)) + + x_0 = reconstruct(b_1, c_1, x_0) + y_0 = reconstruct(a_1, c_1, y_0) + z_0 = reconstruct(a_1, b_1, z_0) + + sq_lcm = ilcm(sqf_of_a, sqf_of_b, sqf_of_c) + + x_0 = abs(x_0*sq_lcm // sqf_of_a) + y_0 = abs(y_0*sq_lcm // sqf_of_b) + z_0 = abs(z_0*sq_lcm // sqf_of_c) + + result.add(_remove_gcd(x_0, y_0, z_0)) + return result + + +class HomogeneousTernaryQuadratic(DiophantineEquationType): + """ + Representation of a homogeneous ternary quadratic diophantine equation. + + Examples + ======== + + >>> from sympy.abc import x, y, z + >>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadratic + >>> HomogeneousTernaryQuadratic(x**2 + y**2 - 3*z**2 + x*y).solve() + {(-1, 2, 1)} + >>> HomogeneousTernaryQuadratic(3*x**2 + y**2 - 3*z**2 + 5*x*y + y*z).solve() + {(3, 12, 13)} + + """ + + name = 'homogeneous_ternary_quadratic' + + def matches(self): + if not (self.total_degree == 2 and self.dimension == 3): + return False + if not self.homogeneous: + return False + if not self.homogeneous_order: + return False + + nonzero = [k for k in self.coeff if self.coeff[k]] + return not (len(nonzero) == 3 and all(i**2 in nonzero for i in self.free_symbols)) + + def solve(self, parameters=None, limit=None): + self.pre_solve(parameters) + + _var = self.free_symbols + coeff = self.coeff + + x, y, z = _var + var = [x, y, z] + + # Equations of the form B*x*y + C*z*x + E*y*z = 0 and At least two of the + # coefficients A, B, C are non-zero. + # There are infinitely many solutions for the equation. + # Ex: (0, 0, t), (0, t, 0), (t, 0, 0) + # Equation can be re-written as y*(B*x + E*z) = -C*x*z and we can find rather + # unobvious solutions. Set y = -C and B*x + E*z = x*z. The latter can be solved by + # using methods for binary quadratic diophantine equations. Let's select the + # solution which minimizes |x| + |z| + + result = DiophantineSolutionSet(var, parameters=self.parameters) + + def unpack_sol(sol): + if len(sol) > 0: + return list(sol)[0] + return None, None, None + + if not any(coeff[i**2] for i in var): + if coeff[x*z]: + sols = diophantine(coeff[x*y]*x + coeff[y*z]*z - x*z) + s = min(sols, key=lambda r: abs(r[0]) + abs(r[1])) + result.add(_remove_gcd(s[0], -coeff[x*z], s[1])) + return result + + var[0], var[1] = _var[1], _var[0] + y_0, x_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, coeff)) + if x_0 is not None: + result.add((x_0, y_0, z_0)) + return result + + if coeff[x**2] == 0: + # If the coefficient of x is zero change the variables + if coeff[y**2] == 0: + var[0], var[2] = _var[2], _var[0] + z_0, y_0, x_0 = unpack_sol(_diop_ternary_quadratic(var, coeff)) + + else: + var[0], var[1] = _var[1], _var[0] + y_0, x_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, coeff)) + + else: + if coeff[x*y] or coeff[x*z]: + # Apply the transformation x --> X - (B*y + C*z)/(2*A) + A = coeff[x**2] + B = coeff[x*y] + C = coeff[x*z] + D = coeff[y**2] + E = coeff[y*z] + F = coeff[z**2] + + _coeff = {} + + _coeff[x**2] = 4*A**2 + _coeff[y**2] = 4*A*D - B**2 + _coeff[z**2] = 4*A*F - C**2 + _coeff[y*z] = 4*A*E - 2*B*C + _coeff[x*y] = 0 + _coeff[x*z] = 0 + + x_0, y_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, _coeff)) + + if x_0 is None: + return result + + p, q = _rational_pq(B*y_0 + C*z_0, 2*A) + x_0, y_0, z_0 = x_0*q - p, y_0*q, z_0*q + + elif coeff[z*y] != 0: + if coeff[y**2] == 0: + if coeff[z**2] == 0: + # Equations of the form A*x**2 + E*yz = 0. + A = coeff[x**2] + E = coeff[y*z] + + b, a = _rational_pq(-E, A) + + x_0, y_0, z_0 = b, a, b + + else: + # Ax**2 + E*y*z + F*z**2 = 0 + var[0], var[2] = _var[2], _var[0] + z_0, y_0, x_0 = unpack_sol(_diop_ternary_quadratic(var, coeff)) + + else: + # A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, C may be zero + var[0], var[1] = _var[1], _var[0] + y_0, x_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, coeff)) + + else: + # Ax**2 + D*y**2 + F*z**2 = 0, C may be zero + x_0, y_0, z_0 = unpack_sol(_diop_ternary_quadratic_normal(var, coeff)) + + if x_0 is None: + return result + + result.add(_remove_gcd(x_0, y_0, z_0)) + return result + + +class InhomogeneousGeneralQuadratic(DiophantineEquationType): + """ + + Representation of an inhomogeneous general quadratic. + + No solver is currently implemented for this equation type. + + """ + + name = 'inhomogeneous_general_quadratic' + + def matches(self): + if not (self.total_degree == 2 and self.dimension >= 3): + return False + if not self.homogeneous_order: + return True + # there may be Pow keys like x**2 or Mul keys like x*y + return any(k.is_Mul for k in self.coeff) and not self.homogeneous + + +class HomogeneousGeneralQuadratic(DiophantineEquationType): + """ + + Representation of a homogeneous general quadratic. + + No solver is currently implemented for this equation type. + + """ + + name = 'homogeneous_general_quadratic' + + def matches(self): + if not (self.total_degree == 2 and self.dimension >= 3): + return False + if not self.homogeneous_order: + return False + # there may be Pow keys like x**2 or Mul keys like x*y + return any(k.is_Mul for k in self.coeff) and self.homogeneous + + +class GeneralSumOfSquares(DiophantineEquationType): + r""" + Representation of the diophantine equation + + `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. + + Details + ======= + + When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be + no solutions. Refer [1]_ for more details. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import GeneralSumOfSquares + >>> from sympy.abc import a, b, c, d, e + >>> GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve() + {(15, 22, 22, 24, 24)} + + By default only 1 solution is returned. Use the `limit` keyword for more: + + >>> sorted(GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve(limit=3)) + [(15, 22, 22, 24, 24), (16, 19, 24, 24, 24), (16, 20, 22, 23, 26)] + + References + ========== + + .. [1] Representing an integer as a sum of three squares, [online], + Available: + https://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares + """ + + name = 'general_sum_of_squares' + + def matches(self): + if not (self.total_degree == 2 and self.dimension >= 3): + return False + if not self.homogeneous_order: + return False + if any(k.is_Mul for k in self.coeff): + return False + return all(self.coeff[k] == 1 for k in self.coeff if k != 1) + + def solve(self, parameters=None, limit=1): + self.pre_solve(parameters) + + var = self.free_symbols + k = -int(self.coeff[1]) + n = self.dimension + + result = DiophantineSolutionSet(var, parameters=self.parameters) + + if k < 0 or limit < 1: + return result + + signs = [-1 if x.is_nonpositive else 1 for x in var] + negs = signs.count(-1) != 0 + + took = 0 + for t in sum_of_squares(k, n, zeros=True): + if negs: + result.add([signs[i]*j for i, j in enumerate(t)]) + else: + result.add(t) + took += 1 + if took == limit: + break + return result + + +class GeneralPythagorean(DiophantineEquationType): + """ + Representation of the general pythagorean equation, + `a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import GeneralPythagorean + >>> from sympy.abc import a, b, c, d, e, x, y, z, t + >>> GeneralPythagorean(a**2 + b**2 + c**2 - d**2).solve() + {(t_0**2 + t_1**2 - t_2**2, 2*t_0*t_2, 2*t_1*t_2, t_0**2 + t_1**2 + t_2**2)} + >>> GeneralPythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2).solve(parameters=[x, y, z, t]) + {(-10*t**2 + 10*x**2 + 10*y**2 + 10*z**2, 15*t**2 + 15*x**2 + 15*y**2 + 15*z**2, 15*t*x, 12*t*y, 60*t*z)} + """ + + name = 'general_pythagorean' + + def matches(self): + if not (self.total_degree == 2 and self.dimension >= 3): + return False + if not self.homogeneous_order: + return False + if any(k.is_Mul for k in self.coeff): + return False + if all(self.coeff[k] == 1 for k in self.coeff if k != 1): + return False + if not all(is_square(abs(self.coeff[k])) for k in self.coeff): + return False + # all but one has the same sign + # e.g. 4*x**2 + y**2 - 4*z**2 + return abs(sum(sign(self.coeff[k]) for k in self.coeff)) == self.dimension - 2 + + @property + def n_parameters(self): + return self.dimension - 1 + + def solve(self, parameters=None, limit=1): + self.pre_solve(parameters) + + coeff = self.coeff + var = self.free_symbols + n = self.dimension + + if sign(coeff[var[0] ** 2]) + sign(coeff[var[1] ** 2]) + sign(coeff[var[2] ** 2]) < 0: + for key in coeff.keys(): + coeff[key] = -coeff[key] + + result = DiophantineSolutionSet(var, parameters=self.parameters) + + index = 0 + + for i, v in enumerate(var): + if sign(coeff[v ** 2]) == -1: + index = i + + m = result.parameters + + ith = sum(m_i ** 2 for m_i in m) + L = [ith - 2 * m[n - 2] ** 2] + L.extend([2 * m[i] * m[n - 2] for i in range(n - 2)]) + sol = L[:index] + [ith] + L[index:] + + lcm = 1 + for i, v in enumerate(var): + if i == index or (index > 0 and i == 0) or (index == 0 and i == 1): + lcm = ilcm(lcm, sqrt(abs(coeff[v ** 2]))) + else: + s = sqrt(coeff[v ** 2]) + lcm = ilcm(lcm, s if _odd(s) else s // 2) + + for i, v in enumerate(var): + sol[i] = (lcm * sol[i]) / sqrt(abs(coeff[v ** 2])) + + result.add(sol) + return result + + +class CubicThue(DiophantineEquationType): + """ + Representation of a cubic Thue diophantine equation. + + A cubic Thue diophantine equation is a polynomial of the form + `f(x, y) = r` of degree 3, where `x` and `y` are integers + and `r` is a rational number. + + No solver is currently implemented for this equation type. + + Examples + ======== + + >>> from sympy.abc import x, y + >>> from sympy.solvers.diophantine.diophantine import CubicThue + >>> c1 = CubicThue(x**3 + y**2 + 1) + >>> c1.matches() + True + + """ + + name = 'cubic_thue' + + def matches(self): + return self.total_degree == 3 and self.dimension == 2 + + +class GeneralSumOfEvenPowers(DiophantineEquationType): + """ + Representation of the diophantine equation + + `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0` + + where `e` is an even, integer power. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import GeneralSumOfEvenPowers + >>> from sympy.abc import a, b + >>> GeneralSumOfEvenPowers(a**4 + b**4 - (2**4 + 3**4)).solve() + {(2, 3)} + + """ + + name = 'general_sum_of_even_powers' + + def matches(self): + if not self.total_degree > 3: + return False + if self.total_degree % 2 != 0: + return False + if not all(k.is_Pow and k.exp == self.total_degree for k in self.coeff if k != 1): + return False + return all(self.coeff[k] == 1 for k in self.coeff if k != 1) + + def solve(self, parameters=None, limit=1): + self.pre_solve(parameters) + + var = self.free_symbols + coeff = self.coeff + + p = None + for q in coeff.keys(): + if q.is_Pow and coeff[q]: + p = q.exp + + k = len(var) + n = -coeff[1] + + result = DiophantineSolutionSet(var, parameters=self.parameters) + + if n < 0 or limit < 1: + return result + + sign = [-1 if x.is_nonpositive else 1 for x in var] + negs = sign.count(-1) != 0 + + took = 0 + for t in power_representation(n, p, k): + if negs: + result.add([sign[i]*j for i, j in enumerate(t)]) + else: + result.add(t) + took += 1 + if took == limit: + break + return result + +# these types are known (but not necessarily handled) +# note that order is important here (in the current solver state) +all_diop_classes = [ + Linear, + Univariate, + BinaryQuadratic, + InhomogeneousTernaryQuadratic, + HomogeneousTernaryQuadraticNormal, + HomogeneousTernaryQuadratic, + InhomogeneousGeneralQuadratic, + HomogeneousGeneralQuadratic, + GeneralSumOfSquares, + GeneralPythagorean, + CubicThue, + GeneralSumOfEvenPowers, +] + +diop_known = {diop_class.name for diop_class in all_diop_classes} + + +def _remove_gcd(*x): + try: + g = igcd(*x) + except ValueError: + fx = list(filter(None, x)) + if len(fx) < 2: + return x + g = igcd(*[i.as_content_primitive()[0] for i in fx]) + except TypeError: + raise TypeError('_remove_gcd(a,b,c) or _remove_gcd(*container)') + if g == 1: + return x + return tuple([i//g for i in x]) + + +def _rational_pq(a, b): + # return `(numer, denom)` for a/b; sign in numer and gcd removed + return _remove_gcd(sign(b)*a, abs(b)) + + +def _nint_or_floor(p, q): + # return nearest int to p/q; in case of tie return floor(p/q) + w, r = divmod(p, q) + if abs(r) <= abs(q)//2: + return w + return w + 1 + + +def _odd(i): + return i % 2 != 0 + + +def _even(i): + return i % 2 == 0 + + +def diophantine(eq, param=symbols("t", integer=True), syms=None, + permute=False): + """ + Simplify the solution procedure of diophantine equation ``eq`` by + converting it into a product of terms which should equal zero. + + Explanation + =========== + + For example, when solving, `x^2 - y^2 = 0` this is treated as + `(x + y)(x - y) = 0` and `x + y = 0` and `x - y = 0` are solved + independently and combined. Each term is solved by calling + ``diop_solve()``. (Although it is possible to call ``diop_solve()`` + directly, one must be careful to pass an equation in the correct + form and to interpret the output correctly; ``diophantine()`` is + the public-facing function to use in general.) + + Output of ``diophantine()`` is a set of tuples. The elements of the + tuple are the solutions for each variable in the equation and + are arranged according to the alphabetic ordering of the variables. + e.g. For an equation with two variables, `a` and `b`, the first + element of the tuple is the solution for `a` and the second for `b`. + + Usage + ===== + + ``diophantine(eq, t, syms)``: Solve the diophantine + equation ``eq``. + ``t`` is the optional parameter to be used by ``diop_solve()``. + ``syms`` is an optional list of symbols which determines the + order of the elements in the returned tuple. + + By default, only the base solution is returned. If ``permute`` is set to + True then permutations of the base solution and/or permutations of the + signs of the values will be returned when applicable. + + Details + ======= + + ``eq`` should be an expression which is assumed to be zero. + ``t`` is the parameter to be used in the solution. + + Examples + ======== + + >>> from sympy import diophantine + >>> from sympy.abc import a, b + >>> eq = a**4 + b**4 - (2**4 + 3**4) + >>> diophantine(eq) + {(2, 3)} + >>> diophantine(eq, permute=True) + {(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)} + + >>> from sympy.abc import x, y, z + >>> diophantine(x**2 - y**2) + {(t_0, -t_0), (t_0, t_0)} + + >>> diophantine(x*(2*x + 3*y - z)) + {(0, n1, n2), (t_0, t_1, 2*t_0 + 3*t_1)} + >>> diophantine(x**2 + 3*x*y + 4*x) + {(0, n1), (-3*t_0 - 4, t_0)} + + See Also + ======== + + diop_solve + sympy.utilities.iterables.permute_signs + sympy.utilities.iterables.signed_permutations + """ + + eq = _sympify(eq) + + if isinstance(eq, Eq): + eq = eq.lhs - eq.rhs + + try: + var = list(eq.expand(force=True).free_symbols) + var.sort(key=default_sort_key) + if syms: + if not is_sequence(syms): + raise TypeError( + 'syms should be given as a sequence, e.g. a list') + syms = [i for i in syms if i in var] + if syms != var: + dict_sym_index = dict(zip(syms, range(len(syms)))) + return {tuple([t[dict_sym_index[i]] for i in var]) + for t in diophantine(eq, param, permute=permute)} + n, d = eq.as_numer_denom() + if n.is_number: + return set() + if not d.is_number: + dsol = diophantine(d) + good = diophantine(n) - dsol + return {s for s in good if _mexpand(d.subs(zip(var, s)))} + eq = factor_terms(n) + assert not eq.is_number + eq = eq.as_independent(*var, as_Add=False)[1] + p = Poly(eq) + assert not any(g.is_number for g in p.gens) + eq = p.as_expr() + assert eq.is_polynomial() + except (GeneratorsNeeded, AssertionError): + raise TypeError(filldedent(''' + Equation should be a polynomial with Rational coefficients.''')) + + # permute only sign + do_permute_signs = False + # permute sign and values + do_permute_signs_var = False + # permute few signs + permute_few_signs = False + try: + # if we know that factoring should not be attempted, skip + # the factoring step + v, c, t = classify_diop(eq) + + # check for permute sign + if permute: + len_var = len(v) + permute_signs_for = [ + GeneralSumOfSquares.name, + GeneralSumOfEvenPowers.name] + permute_signs_check = [ + HomogeneousTernaryQuadratic.name, + HomogeneousTernaryQuadraticNormal.name, + BinaryQuadratic.name] + if t in permute_signs_for: + do_permute_signs_var = True + elif t in permute_signs_check: + # if all the variables in eq have even powers + # then do_permute_sign = True + if len_var == 3: + var_mul = list(subsets(v, 2)) + # here var_mul is like [(x, y), (x, z), (y, z)] + xy_coeff = True + x_coeff = True + var1_mul_var2 = (a[0]*a[1] for a in var_mul) + # if coeff(y*z), coeff(y*x), coeff(x*z) is not 0 then + # `xy_coeff` => True and do_permute_sign => False. + # Means no permuted solution. + for v1_mul_v2 in var1_mul_var2: + try: + coeff = c[v1_mul_v2] + except KeyError: + coeff = 0 + xy_coeff = bool(xy_coeff) and bool(coeff) + var_mul = list(subsets(v, 1)) + # here var_mul is like [(x,), (y, )] + for v1 in var_mul: + try: + coeff = c[v1[0]] + except KeyError: + coeff = 0 + x_coeff = bool(x_coeff) and bool(coeff) + if not any((xy_coeff, x_coeff)): + # means only x**2, y**2, z**2, const is present + do_permute_signs = True + elif not x_coeff: + permute_few_signs = True + elif len_var == 2: + var_mul = list(subsets(v, 2)) + # here var_mul is like [(x, y)] + xy_coeff = True + x_coeff = True + var1_mul_var2 = (x[0]*x[1] for x in var_mul) + for v1_mul_v2 in var1_mul_var2: + try: + coeff = c[v1_mul_v2] + except KeyError: + coeff = 0 + xy_coeff = bool(xy_coeff) and bool(coeff) + var_mul = list(subsets(v, 1)) + # here var_mul is like [(x,), (y, )] + for v1 in var_mul: + try: + coeff = c[v1[0]] + except KeyError: + coeff = 0 + x_coeff = bool(x_coeff) and bool(coeff) + if not any((xy_coeff, x_coeff)): + # means only x**2, y**2 and const is present + # so we can get more soln by permuting this soln. + do_permute_signs = True + elif not x_coeff: + # when coeff(x), coeff(y) is not present then signs of + # x, y can be permuted such that their sign are same + # as sign of x*y. + # e.g 1. (x_val,y_val)=> (x_val,y_val), (-x_val,-y_val) + # 2. (-x_vall, y_val)=> (-x_val,y_val), (x_val,-y_val) + permute_few_signs = True + if t == 'general_sum_of_squares': + # trying to factor such expressions will sometimes hang + terms = [(eq, 1)] + else: + raise TypeError + except (TypeError, NotImplementedError): + fl = factor_list(eq) + if fl[0].is_Rational and fl[0] != 1: + return diophantine(eq/fl[0], param=param, syms=syms, permute=permute) + terms = fl[1] + + sols = set() + + for term in terms: + + base, _ = term + var_t, _, eq_type = classify_diop(base, _dict=False) + _, base = signsimp(base, evaluate=False).as_coeff_Mul() + solution = diop_solve(base, param) + + if eq_type in [ + Linear.name, + HomogeneousTernaryQuadratic.name, + HomogeneousTernaryQuadraticNormal.name, + GeneralPythagorean.name]: + sols.add(merge_solution(var, var_t, solution)) + + elif eq_type in [ + BinaryQuadratic.name, + GeneralSumOfSquares.name, + GeneralSumOfEvenPowers.name, + Univariate.name]: + sols.update(merge_solution(var, var_t, sol) for sol in solution) + + else: + raise NotImplementedError('unhandled type: %s' % eq_type) + + # remove null merge results + if () in sols: + sols.remove(()) + null = tuple([0]*len(var)) + # if there is no solution, return trivial solution + if not sols and eq.subs(zip(var, null)).is_zero: + sols.add(null) + final_soln = set() + for sol in sols: + if all(int_valued(s) for s in sol): + if do_permute_signs: + permuted_sign = set(permute_signs(sol)) + final_soln.update(permuted_sign) + elif permute_few_signs: + lst = list(permute_signs(sol)) + lst = list(filter(lambda x: x[0]*x[1] == sol[1]*sol[0], lst)) + permuted_sign = set(lst) + final_soln.update(permuted_sign) + elif do_permute_signs_var: + permuted_sign_var = set(signed_permutations(sol)) + final_soln.update(permuted_sign_var) + else: + final_soln.add(sol) + else: + final_soln.add(sol) + return final_soln + + +def merge_solution(var, var_t, solution): + """ + This is used to construct the full solution from the solutions of sub + equations. + + Explanation + =========== + + For example when solving the equation `(x - y)(x^2 + y^2 - z^2) = 0`, + solutions for each of the equations `x - y = 0` and `x^2 + y^2 - z^2` are + found independently. Solutions for `x - y = 0` are `(x, y) = (t, t)`. But + we should introduce a value for z when we output the solution for the + original equation. This function converts `(t, t)` into `(t, t, n_{1})` + where `n_{1}` is an integer parameter. + """ + sol = [] + + if None in solution: + return () + + solution = iter(solution) + params = numbered_symbols("n", integer=True, start=1) + for v in var: + if v in var_t: + sol.append(next(solution)) + else: + sol.append(next(params)) + + for val, symb in zip(sol, var): + if check_assumptions(val, **symb.assumptions0) is False: + return () + + return tuple(sol) + + +def _diop_solve(eq, params=None): + for diop_type in all_diop_classes: + if diop_type(eq).matches(): + return diop_type(eq).solve(parameters=params) + + +def diop_solve(eq, param=symbols("t", integer=True)): + """ + Solves the diophantine equation ``eq``. + + Explanation + =========== + + Unlike ``diophantine()``, factoring of ``eq`` is not attempted. Uses + ``classify_diop()`` to determine the type of the equation and calls + the appropriate solver function. + + Use of ``diophantine()`` is recommended over other helper functions. + ``diop_solve()`` can return either a set or a tuple depending on the + nature of the equation. + + Usage + ===== + + ``diop_solve(eq, t)``: Solve diophantine equation, ``eq`` using ``t`` + as a parameter if needed. + + Details + ======= + + ``eq`` should be an expression which is assumed to be zero. + ``t`` is a parameter to be used in the solution. + + Examples + ======== + + >>> from sympy.solvers.diophantine import diop_solve + >>> from sympy.abc import x, y, z, w + >>> diop_solve(2*x + 3*y - 5) + (3*t_0 - 5, 5 - 2*t_0) + >>> diop_solve(4*x + 3*y - 4*z + 5) + (t_0, 8*t_0 + 4*t_1 + 5, 7*t_0 + 3*t_1 + 5) + >>> diop_solve(x + 3*y - 4*z + w - 6) + (t_0, t_0 + t_1, 6*t_0 + 5*t_1 + 4*t_2 - 6, 5*t_0 + 4*t_1 + 3*t_2 - 6) + >>> diop_solve(x**2 + y**2 - 5) + {(-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)} + + + See Also + ======== + + diophantine() + """ + var, coeff, eq_type = classify_diop(eq, _dict=False) + + if eq_type == Linear.name: + return diop_linear(eq, param) + + elif eq_type == BinaryQuadratic.name: + return diop_quadratic(eq, param) + + elif eq_type == HomogeneousTernaryQuadratic.name: + return diop_ternary_quadratic(eq, parameterize=True) + + elif eq_type == HomogeneousTernaryQuadraticNormal.name: + return diop_ternary_quadratic_normal(eq, parameterize=True) + + elif eq_type == GeneralPythagorean.name: + return diop_general_pythagorean(eq, param) + + elif eq_type == Univariate.name: + return diop_univariate(eq) + + elif eq_type == GeneralSumOfSquares.name: + return diop_general_sum_of_squares(eq, limit=S.Infinity) + + elif eq_type == GeneralSumOfEvenPowers.name: + return diop_general_sum_of_even_powers(eq, limit=S.Infinity) + + if eq_type is not None and eq_type not in diop_known: + raise ValueError(filldedent(''' + Although this type of equation was identified, it is not yet + handled. It should, however, be listed in `diop_known` at the + top of this file. Developers should see comments at the end of + `classify_diop`. + ''')) # pragma: no cover + else: + raise NotImplementedError( + 'No solver has been written for %s.' % eq_type) + + +def classify_diop(eq, _dict=True): + # docstring supplied externally + + matched = False + diop_type = None + for diop_class in all_diop_classes: + diop_type = diop_class(eq) + if diop_type.matches(): + matched = True + break + + if matched: + return diop_type.free_symbols, dict(diop_type.coeff) if _dict else diop_type.coeff, diop_type.name + + # new diop type instructions + # -------------------------- + # if this error raises and the equation *can* be classified, + # * it should be identified in the if-block above + # * the type should be added to the diop_known + # if a solver can be written for it, + # * a dedicated handler should be written (e.g. diop_linear) + # * it should be passed to that handler in diop_solve + raise NotImplementedError(filldedent(''' + This equation is not yet recognized or else has not been + simplified sufficiently to put it in a form recognized by + diop_classify().''')) + + +classify_diop.func_doc = ( # type: ignore + ''' + Helper routine used by diop_solve() to find information about ``eq``. + + Explanation + =========== + + Returns a tuple containing the type of the diophantine equation + along with the variables (free symbols) and their coefficients. + Variables are returned as a list and coefficients are returned + as a dict with the key being the respective term and the constant + term is keyed to 1. The type is one of the following: + + * %s + + Usage + ===== + + ``classify_diop(eq)``: Return variables, coefficients and type of the + ``eq``. + + Details + ======= + + ``eq`` should be an expression which is assumed to be zero. + ``_dict`` is for internal use: when True (default) a dict is returned, + otherwise a defaultdict which supplies 0 for missing keys is returned. + + Examples + ======== + + >>> from sympy.solvers.diophantine import classify_diop + >>> from sympy.abc import x, y, z, w, t + >>> classify_diop(4*x + 6*y - 4) + ([x, y], {1: -4, x: 4, y: 6}, 'linear') + >>> classify_diop(x + 3*y -4*z + 5) + ([x, y, z], {1: 5, x: 1, y: 3, z: -4}, 'linear') + >>> classify_diop(x**2 + y**2 - x*y + x + 5) + ([x, y], {1: 5, x: 1, x**2: 1, y**2: 1, x*y: -1}, 'binary_quadratic') + ''' % ('\n * '.join(sorted(diop_known)))) + + +def diop_linear(eq, param=symbols("t", integer=True)): + """ + Solves linear diophantine equations. + + A linear diophantine equation is an equation of the form `a_{1}x_{1} + + a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are + integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables. + + Usage + ===== + + ``diop_linear(eq)``: Returns a tuple containing solutions to the + diophantine equation ``eq``. Values in the tuple is arranged in the same + order as the sorted variables. + + Details + ======= + + ``eq`` is a linear diophantine equation which is assumed to be zero. + ``param`` is the parameter to be used in the solution. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import diop_linear + >>> from sympy.abc import x, y, z + >>> diop_linear(2*x - 3*y - 5) # solves equation 2*x - 3*y - 5 == 0 + (3*t_0 - 5, 2*t_0 - 5) + + Here x = -3*t_0 - 5 and y = -2*t_0 - 5 + + >>> diop_linear(2*x - 3*y - 4*z -3) + (t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3) + + See Also + ======== + + diop_quadratic(), diop_ternary_quadratic(), diop_general_pythagorean(), + diop_general_sum_of_squares() + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type == Linear.name: + parameters = None + if param is not None: + parameters = symbols('%s_0:%i' % (param, len(var)), integer=True) + + result = Linear(eq).solve(parameters=parameters) + + if param is None: + result = result(*[0]*len(result.parameters)) + + if len(result) > 0: + return list(result)[0] + else: + return tuple([None]*len(result.parameters)) + + +def base_solution_linear(c, a, b, t=None): + """ + Return the base solution for the linear equation, `ax + by = c`. + + Explanation + =========== + + Used by ``diop_linear()`` to find the base solution of a linear + Diophantine equation. If ``t`` is given then the parametrized solution is + returned. + + Usage + ===== + + ``base_solution_linear(c, a, b, t)``: ``a``, ``b``, ``c`` are coefficients + in `ax + by = c` and ``t`` is the parameter to be used in the solution. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import base_solution_linear + >>> from sympy.abc import t + >>> base_solution_linear(5, 2, 3) # equation 2*x + 3*y = 5 + (-5, 5) + >>> base_solution_linear(0, 5, 7) # equation 5*x + 7*y = 0 + (0, 0) + >>> base_solution_linear(5, 2, 3, t) # equation 2*x + 3*y = 5 + (3*t - 5, 5 - 2*t) + >>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0 + (7*t, -5*t) + """ + a, b, c = _remove_gcd(a, b, c) + + if c == 0: + if t is None: + return (0, 0) + if b < 0: + t = -t + return (b*t, -a*t) + + x0, y0, d = igcdex(abs(a), abs(b)) + x0 *= sign(a) + y0 *= sign(b) + if c % d: + return (None, None) + if t is None: + return (c*x0, c*y0) + if b < 0: + t = -t + return (c*x0 + b*t, c*y0 - a*t) + + +def diop_univariate(eq): + """ + Solves a univariate diophantine equations. + + Explanation + =========== + + A univariate diophantine equation is an equation of the form + `a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are + integer constants and `x` is an integer variable. + + Usage + ===== + + ``diop_univariate(eq)``: Returns a set containing solutions to the + diophantine equation ``eq``. + + Details + ======= + + ``eq`` is a univariate diophantine equation which is assumed to be zero. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import diop_univariate + >>> from sympy.abc import x + >>> diop_univariate((x - 2)*(x - 3)**2) # solves equation (x - 2)*(x - 3)**2 == 0 + {(2,), (3,)} + + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type == Univariate.name: + return {(int(i),) for i in solveset_real( + eq, var[0]).intersect(S.Integers)} + + +def divisible(a, b): + """ + Returns `True` if ``a`` is divisible by ``b`` and `False` otherwise. + """ + return not a % b + + +def diop_quadratic(eq, param=symbols("t", integer=True)): + """ + Solves quadratic diophantine equations. + + i.e. equations of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`. Returns a + set containing the tuples `(x, y)` which contains the solutions. If there + are no solutions then `(None, None)` is returned. + + Usage + ===== + + ``diop_quadratic(eq, param)``: ``eq`` is a quadratic binary diophantine + equation. ``param`` is used to indicate the parameter to be used in the + solution. + + Details + ======= + + ``eq`` should be an expression which is assumed to be zero. + ``param`` is a parameter to be used in the solution. + + Examples + ======== + + >>> from sympy.abc import x, y, t + >>> from sympy.solvers.diophantine.diophantine import diop_quadratic + >>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t) + {(-1, -1)} + + References + ========== + + .. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], + Available: https://www.alpertron.com.ar/METHODS.HTM + .. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], + Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf + + See Also + ======== + + diop_linear(), diop_ternary_quadratic(), diop_general_sum_of_squares(), + diop_general_pythagorean() + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type == BinaryQuadratic.name: + if param is not None: + parameters = [param, Symbol("u", integer=True)] + else: + parameters = None + return set(BinaryQuadratic(eq).solve(parameters=parameters)) + + +def is_solution_quad(var, coeff, u, v): + """ + Check whether `(u, v)` is solution to the quadratic binary diophantine + equation with the variable list ``var`` and coefficient dictionary + ``coeff``. + + Not intended for use by normal users. + """ + reps = dict(zip(var, (u, v))) + eq = Add(*[j*i.xreplace(reps) for i, j in coeff.items()]) + return _mexpand(eq) == 0 + + +def diop_DN(D, N, t=symbols("t", integer=True)): + """ + Solves the equation `x^2 - Dy^2 = N`. + + Explanation + =========== + + Mainly concerned with the case `D > 0, D` is not a perfect square, + which is the same as the generalized Pell equation. The LMM + algorithm [1]_ is used to solve this equation. + + Returns one solution tuple, (`x, y)` for each class of the solutions. + Other solutions of the class can be constructed according to the + values of ``D`` and ``N``. + + Usage + ===== + + ``diop_DN(D, N, t)``: D and N are integers as in `x^2 - Dy^2 = N` and + ``t`` is the parameter to be used in the solutions. + + Details + ======= + + ``D`` and ``N`` correspond to D and N in the equation. + ``t`` is the parameter to be used in the solutions. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import diop_DN + >>> diop_DN(13, -4) # Solves equation x**2 - 13*y**2 = -4 + [(3, 1), (393, 109), (36, 10)] + + The output can be interpreted as follows: There are three fundamental + solutions to the equation `x^2 - 13y^2 = -4` given by (3, 1), (393, 109) + and (36, 10). Each tuple is in the form (x, y), i.e. solution (3, 1) means + that `x = 3` and `y = 1`. + + >>> diop_DN(986, 1) # Solves equation x**2 - 986*y**2 = 1 + [(49299, 1570)] + + See Also + ======== + + find_DN(), diop_bf_DN() + + References + ========== + + .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. + Robertson, July 31, 2004, Pages 16 - 17. [online], Available: + https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf + """ + if D < 0: + if N == 0: + return [(0, 0)] + if N < 0: + return [] + # N > 0: + sol = [] + for d in divisors(square_factor(N), generator=True): + for x, y in cornacchia(1, int(-D), int(N // d**2)): + sol.append((d*x, d*y)) + if D == -1: + sol.append((d*y, d*x)) + return sol + + if D == 0: + if N < 0: + return [] + if N == 0: + return [(0, t)] + sN, _exact = integer_nthroot(N, 2) + if _exact: + return [(sN, t)] + return [] + + # D > 0 + sD, _exact = integer_nthroot(D, 2) + if _exact: + if N == 0: + return [(sD*t, t)] + + sol = [] + for y in range(floor(sign(N)*(N - 1)/(2*sD)) + 1): + try: + sq, _exact = integer_nthroot(D*y**2 + N, 2) + except ValueError: + _exact = False + if _exact: + sol.append((sq, y)) + return sol + + if 1 < N**2 < D: + # It is much faster to call `_special_diop_DN`. + return _special_diop_DN(D, N) + + if N == 0: + return [(0, 0)] + + sol = [] + if abs(N) == 1: + pqa = PQa(0, 1, D) + *_, prev_B, prev_G = next(pqa) + for j, (*_, a, _, _B, _G) in enumerate(pqa): + if a == 2*sD: + break + prev_B, prev_G = _B, _G + if j % 2: + if N == 1: + sol.append((prev_G, prev_B)) + return sol + if N == -1: + return [(prev_G, prev_B)] + for _ in range(j): + *_, _B, _G = next(pqa) + return [(_G, _B)] + + for f in divisors(square_factor(N), generator=True): + m = N // f**2 + am = abs(m) + for sqm in sqrt_mod(D, am, all_roots=True): + z = symmetric_residue(sqm, am) + pqa = PQa(z, am, D) + *_, prev_B, prev_G = next(pqa) + for _ in range(length(z, am, D) - 1): + _, q, *_, _B, _G = next(pqa) + if abs(q) == 1: + if prev_G**2 - D*prev_B**2 == m: + sol.append((f*prev_G, f*prev_B)) + elif a := diop_DN(D, -1): + sol.append((f*(prev_G*a[0][0] + prev_B*D*a[0][1]), + f*(prev_G*a[0][1] + prev_B*a[0][0]))) + break + prev_B, prev_G = _B, _G + return sol + + +def _special_diop_DN(D, N): + """ + Solves the equation `x^2 - Dy^2 = N` for the special case where + `1 < N**2 < D` and `D` is not a perfect square. + It is better to call `diop_DN` rather than this function, as + the former checks the condition `1 < N**2 < D`, and calls the latter only + if appropriate. + + Usage + ===== + + WARNING: Internal method. Do not call directly! + + ``_special_diop_DN(D, N)``: D and N are integers as in `x^2 - Dy^2 = N`. + + Details + ======= + + ``D`` and ``N`` correspond to D and N in the equation. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import _special_diop_DN + >>> _special_diop_DN(13, -3) # Solves equation x**2 - 13*y**2 = -3 + [(7, 2), (137, 38)] + + The output can be interpreted as follows: There are two fundamental + solutions to the equation `x^2 - 13y^2 = -3` given by (7, 2) and + (137, 38). Each tuple is in the form (x, y), i.e. solution (7, 2) means + that `x = 7` and `y = 2`. + + >>> _special_diop_DN(2445, -20) # Solves equation x**2 - 2445*y**2 = -20 + [(445, 9), (17625560, 356454), (698095554475, 14118073569)] + + See Also + ======== + + diop_DN() + + References + ========== + + .. [1] Section 4.4.4 of the following book: + Quadratic Diophantine Equations, T. Andreescu and D. Andrica, + Springer, 2015. + """ + + # The following assertion was removed for efficiency, with the understanding + # that this method is not called directly. The parent method, `diop_DN` + # is responsible for performing the appropriate checks. + # + # assert (1 < N**2 < D) and (not integer_nthroot(D, 2)[1]) + + sqrt_D = isqrt(D) + F = {N // f**2: f for f in divisors(square_factor(abs(N)), generator=True)} + P = 0 + Q = 1 + G0, G1 = 0, 1 + B0, B1 = 1, 0 + + solutions = [] + while True: + for _ in range(2): + a = (P + sqrt_D) // Q + P = a*Q - P + Q = (D - P**2) // Q + G0, G1 = G1, a*G1 + G0 + B0, B1 = B1, a*B1 + B0 + if (s := G1**2 - D*B1**2) in F: + f = F[s] + solutions.append((f*G1, f*B1)) + if Q == 1: + break + return solutions + + +def cornacchia(a:int, b:int, m:int) -> set[tuple[int, int]]: + r""" + Solves `ax^2 + by^2 = m` where `\gcd(a, b) = 1 = gcd(a, m)` and `a, b > 0`. + + Explanation + =========== + + Uses the algorithm due to Cornacchia. The method only finds primitive + solutions, i.e. ones with `\gcd(x, y) = 1`. So this method cannot be used to + find the solutions of `x^2 + y^2 = 20` since the only solution to former is + `(x, y) = (4, 2)` and it is not primitive. When `a = b`, only the + solutions with `x \leq y` are found. For more details, see the References. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import cornacchia + >>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35 + {(2, 3), (4, 1)} + >>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25 + {(4, 3)} + + References + =========== + + .. [1] A. Nitaj, "L'algorithme de Cornacchia" + .. [2] Solving the diophantine equation ax**2 + by**2 = m by Cornacchia's + method, [online], Available: + http://www.numbertheory.org/php/cornacchia.html + + See Also + ======== + + sympy.utilities.iterables.signed_permutations + """ + # Assume gcd(a, b) = gcd(a, m) = 1 and a, b > 0 but no error checking + sols = set() + for t in sqrt_mod_iter(-b*invert(a, m), m): + if t < m // 2: + continue + u, r = m, t + while (m1 := m - a*r**2) <= 0: + u, r = r, u % r + m1, _r = divmod(m1, b) + if _r: + continue + s, _exact = iroot(m1, 2) + if _exact: + if a == b and r < s: + r, s = s, r + sols.add((int(r), int(s))) + return sols + + +def PQa(P_0, Q_0, D): + r""" + Returns useful information needed to solve the Pell equation. + + Explanation + =========== + + There are six sequences of integers defined related to the continued + fraction representation of `\\frac{P + \sqrt{D}}{Q}`, namely {`P_{i}`}, + {`Q_{i}`}, {`a_{i}`},{`A_{i}`}, {`B_{i}`}, {`G_{i}`}. ``PQa()`` Returns + these values as a 6-tuple in the same order as mentioned above. Refer [1]_ + for more detailed information. + + Usage + ===== + + ``PQa(P_0, Q_0, D)``: ``P_0``, ``Q_0`` and ``D`` are integers corresponding + to `P_{0}`, `Q_{0}` and `D` in the continued fraction + `\\frac{P_{0} + \sqrt{D}}{Q_{0}}`. + Also it's assumed that `P_{0}^2 == D mod(|Q_{0}|)` and `D` is square free. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import PQa + >>> pqa = PQa(13, 4, 5) # (13 + sqrt(5))/4 + >>> next(pqa) # (P_0, Q_0, a_0, A_0, B_0, G_0) + (13, 4, 3, 3, 1, -1) + >>> next(pqa) # (P_1, Q_1, a_1, A_1, B_1, G_1) + (-1, 1, 1, 4, 1, 3) + + References + ========== + + .. [1] Solving the generalized Pell equation x^2 - Dy^2 = N, John P. + Robertson, July 31, 2004, Pages 4 - 8. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf + """ + sqD = isqrt(D) + A2 = B1 = 0 + A1 = B2 = 1 + G1 = Q_0 + G2 = -P_0 + P_i = P_0 + Q_i = Q_0 + + while True: + a_i = (P_i + sqD) // Q_i + A1, A2 = a_i*A1 + A2, A1 + B1, B2 = a_i*B1 + B2, B1 + G1, G2 = a_i*G1 + G2, G1 + yield P_i, Q_i, a_i, A1, B1, G1 + + P_i = a_i*Q_i - P_i + Q_i = (D - P_i**2) // Q_i + + +def diop_bf_DN(D, N, t=symbols("t", integer=True)): + r""" + Uses brute force to solve the equation, `x^2 - Dy^2 = N`. + + Explanation + =========== + + Mainly concerned with the generalized Pell equation which is the case when + `D > 0, D` is not a perfect square. For more information on the case refer + [1]_. Let `(t, u)` be the minimal positive solution of the equation + `x^2 - Dy^2 = 1`. Then this method requires + `\sqrt{\\frac{\mid N \mid (t \pm 1)}{2D}}` to be small. + + Usage + ===== + + ``diop_bf_DN(D, N, t)``: ``D`` and ``N`` are coefficients in + `x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions. + + Details + ======= + + ``D`` and ``N`` correspond to D and N in the equation. + ``t`` is the parameter to be used in the solutions. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import diop_bf_DN + >>> diop_bf_DN(13, -4) + [(3, 1), (-3, 1), (36, 10)] + >>> diop_bf_DN(986, 1) + [(49299, 1570)] + + See Also + ======== + + diop_DN() + + References + ========== + + .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. + Robertson, July 31, 2004, Page 15. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf + """ + D = as_int(D) + N = as_int(N) + + sol = [] + a = diop_DN(D, 1) + u = a[0][0] + + if N == 0: + if D < 0: + return [(0, 0)] + if D == 0: + return [(0, t)] + sD, _exact = integer_nthroot(D, 2) + if _exact: + return [(sD*t, t), (-sD*t, t)] + return [(0, 0)] + + if abs(N) == 1: + return diop_DN(D, N) + + if N > 1: + L1 = 0 + L2 = integer_nthroot(int(N*(u - 1)/(2*D)), 2)[0] + 1 + else: # N < -1 + L1, _exact = integer_nthroot(-int(N/D), 2) + if not _exact: + L1 += 1 + L2 = integer_nthroot(-int(N*(u + 1)/(2*D)), 2)[0] + 1 + + for y in range(L1, L2): + try: + x, _exact = integer_nthroot(N + D*y**2, 2) + except ValueError: + _exact = False + if _exact: + sol.append((x, y)) + if not equivalent(x, y, -x, y, D, N): + sol.append((-x, y)) + + return sol + + +def equivalent(u, v, r, s, D, N): + """ + Returns True if two solutions `(u, v)` and `(r, s)` of `x^2 - Dy^2 = N` + belongs to the same equivalence class and False otherwise. + + Explanation + =========== + + Two solutions `(u, v)` and `(r, s)` to the above equation fall to the same + equivalence class iff both `(ur - Dvs)` and `(us - vr)` are divisible by + `N`. See reference [1]_. No test is performed to test whether `(u, v)` and + `(r, s)` are actually solutions to the equation. User should take care of + this. + + Usage + ===== + + ``equivalent(u, v, r, s, D, N)``: `(u, v)` and `(r, s)` are two solutions + of the equation `x^2 - Dy^2 = N` and all parameters involved are integers. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import equivalent + >>> equivalent(18, 5, -18, -5, 13, -1) + True + >>> equivalent(3, 1, -18, 393, 109, -4) + False + + References + ========== + + .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. + Robertson, July 31, 2004, Page 12. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf + + """ + return divisible(u*r - D*v*s, N) and divisible(u*s - v*r, N) + + +def length(P, Q, D): + r""" + Returns the (length of aperiodic part + length of periodic part) of + continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`. + + It is important to remember that this does NOT return the length of the + periodic part but the sum of the lengths of the two parts as mentioned + above. + + Usage + ===== + + ``length(P, Q, D)``: ``P``, ``Q`` and ``D`` are integers corresponding to + the continued fraction `\\frac{P + \sqrt{D}}{Q}`. + + Details + ======= + + ``P``, ``D`` and ``Q`` corresponds to P, D and Q in the continued fraction, + `\\frac{P + \sqrt{D}}{Q}`. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import length + >>> length(-2, 4, 5) # (-2 + sqrt(5))/4 + 3 + >>> length(-5, 4, 17) # (-5 + sqrt(17))/4 + 4 + + See Also + ======== + sympy.ntheory.continued_fraction.continued_fraction_periodic + """ + from sympy.ntheory.continued_fraction import continued_fraction_periodic + v = continued_fraction_periodic(P, Q, D) + if isinstance(v[-1], list): + rpt = len(v[-1]) + nonrpt = len(v) - 1 + else: + rpt = 0 + nonrpt = len(v) + return rpt + nonrpt + + +def transformation_to_DN(eq): + """ + This function transforms general quadratic, + `ax^2 + bxy + cy^2 + dx + ey + f = 0` + to more easy to deal with `X^2 - DY^2 = N` form. + + Explanation + =========== + + This is used to solve the general quadratic equation by transforming it to + the latter form. Refer to [1]_ for more detailed information on the + transformation. This function returns a tuple (A, B) where A is a 2 X 2 + matrix and B is a 2 X 1 matrix such that, + + Transpose([x y]) = A * Transpose([X Y]) + B + + Usage + ===== + + ``transformation_to_DN(eq)``: where ``eq`` is the quadratic to be + transformed. + + Examples + ======== + + >>> from sympy.abc import x, y + >>> from sympy.solvers.diophantine.diophantine import transformation_to_DN + >>> A, B = transformation_to_DN(x**2 - 3*x*y - y**2 - 2*y + 1) + >>> A + Matrix([ + [1/26, 3/26], + [ 0, 1/13]]) + >>> B + Matrix([ + [-6/13], + [-4/13]]) + + A, B returned are such that Transpose((x y)) = A * Transpose((X Y)) + B. + Substituting these values for `x` and `y` and a bit of simplifying work + will give an equation of the form `x^2 - Dy^2 = N`. + + >>> from sympy.abc import X, Y + >>> from sympy import Matrix, simplify + >>> u = (A*Matrix([X, Y]) + B)[0] # Transformation for x + >>> u + X/26 + 3*Y/26 - 6/13 + >>> v = (A*Matrix([X, Y]) + B)[1] # Transformation for y + >>> v + Y/13 - 4/13 + + Next we will substitute these formulas for `x` and `y` and do + ``simplify()``. + + >>> eq = simplify((x**2 - 3*x*y - y**2 - 2*y + 1).subs(zip((x, y), (u, v)))) + >>> eq + X**2/676 - Y**2/52 + 17/13 + + By multiplying the denominator appropriately, we can get a Pell equation + in the standard form. + + >>> eq * 676 + X**2 - 13*Y**2 + 884 + + If only the final equation is needed, ``find_DN()`` can be used. + + See Also + ======== + + find_DN() + + References + ========== + + .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, + John P.Robertson, May 8, 2003, Page 7 - 11. + https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf + """ + + var, coeff, diop_type = classify_diop(eq, _dict=False) + if diop_type == BinaryQuadratic.name: + return _transformation_to_DN(var, coeff) + + +def _transformation_to_DN(var, coeff): + + x, y = var + + a = coeff[x**2] + b = coeff[x*y] + c = coeff[y**2] + d = coeff[x] + e = coeff[y] + f = coeff[1] + + a, b, c, d, e, f = [as_int(i) for i in _remove_gcd(a, b, c, d, e, f)] + + X, Y = symbols("X, Y", integer=True) + + if b: + B, C = _rational_pq(2*a, b) + A, T = _rational_pq(a, B**2) + + # eq_1 = A*B*X**2 + B*(c*T - A*C**2)*Y**2 + d*T*X + (B*e*T - d*T*C)*Y + f*T*B + coeff = {X**2: A*B, X*Y: 0, Y**2: B*(c*T - A*C**2), X: d*T, Y: B*e*T - d*T*C, 1: f*T*B} + A_0, B_0 = _transformation_to_DN([X, Y], coeff) + return Matrix(2, 2, [S.One/B, -S(C)/B, 0, 1])*A_0, Matrix(2, 2, [S.One/B, -S(C)/B, 0, 1])*B_0 + + if d: + B, C = _rational_pq(2*a, d) + A, T = _rational_pq(a, B**2) + + # eq_2 = A*X**2 + c*T*Y**2 + e*T*Y + f*T - A*C**2 + coeff = {X**2: A, X*Y: 0, Y**2: c*T, X: 0, Y: e*T, 1: f*T - A*C**2} + A_0, B_0 = _transformation_to_DN([X, Y], coeff) + return Matrix(2, 2, [S.One/B, 0, 0, 1])*A_0, Matrix(2, 2, [S.One/B, 0, 0, 1])*B_0 + Matrix([-S(C)/B, 0]) + + if e: + B, C = _rational_pq(2*c, e) + A, T = _rational_pq(c, B**2) + + # eq_3 = a*T*X**2 + A*Y**2 + f*T - A*C**2 + coeff = {X**2: a*T, X*Y: 0, Y**2: A, X: 0, Y: 0, 1: f*T - A*C**2} + A_0, B_0 = _transformation_to_DN([X, Y], coeff) + return Matrix(2, 2, [1, 0, 0, S.One/B])*A_0, Matrix(2, 2, [1, 0, 0, S.One/B])*B_0 + Matrix([0, -S(C)/B]) + + # TODO: pre-simplification: Not necessary but may simplify + # the equation. + return Matrix(2, 2, [S.One/a, 0, 0, 1]), Matrix([0, 0]) + + +def find_DN(eq): + """ + This function returns a tuple, `(D, N)` of the simplified form, + `x^2 - Dy^2 = N`, corresponding to the general quadratic, + `ax^2 + bxy + cy^2 + dx + ey + f = 0`. + + Solving the general quadratic is then equivalent to solving the equation + `X^2 - DY^2 = N` and transforming the solutions by using the transformation + matrices returned by ``transformation_to_DN()``. + + Usage + ===== + + ``find_DN(eq)``: where ``eq`` is the quadratic to be transformed. + + Examples + ======== + + >>> from sympy.abc import x, y + >>> from sympy.solvers.diophantine.diophantine import find_DN + >>> find_DN(x**2 - 3*x*y - y**2 - 2*y + 1) + (13, -884) + + Interpretation of the output is that we get `X^2 -13Y^2 = -884` after + transforming `x^2 - 3xy - y^2 - 2y + 1` using the transformation returned + by ``transformation_to_DN()``. + + See Also + ======== + + transformation_to_DN() + + References + ========== + + .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, + John P.Robertson, May 8, 2003, Page 7 - 11. + https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + if diop_type == BinaryQuadratic.name: + return _find_DN(var, coeff) + + +def _find_DN(var, coeff): + + x, y = var + X, Y = symbols("X, Y", integer=True) + A, B = _transformation_to_DN(var, coeff) + + u = (A*Matrix([X, Y]) + B)[0] + v = (A*Matrix([X, Y]) + B)[1] + eq = x**2*coeff[x**2] + x*y*coeff[x*y] + y**2*coeff[y**2] + x*coeff[x] + y*coeff[y] + coeff[1] + + simplified = _mexpand(eq.subs(zip((x, y), (u, v)))) + + coeff = simplified.as_coefficients_dict() + + return -coeff[Y**2]/coeff[X**2], -coeff[1]/coeff[X**2] + + +def check_param(x, y, a, params): + """ + If there is a number modulo ``a`` such that ``x`` and ``y`` are both + integers, then return a parametric representation for ``x`` and ``y`` + else return (None, None). + + Here ``x`` and ``y`` are functions of ``t``. + """ + from sympy.simplify.simplify import clear_coefficients + + if x.is_number and not x.is_Integer: + return DiophantineSolutionSet([x, y], parameters=params) + + if y.is_number and not y.is_Integer: + return DiophantineSolutionSet([x, y], parameters=params) + + m, n = symbols("m, n", integer=True) + c, p = (m*x + n*y).as_content_primitive() + if a % c.q: + return DiophantineSolutionSet([x, y], parameters=params) + + # clear_coefficients(mx + b, R)[1] -> (R - b)/m + eq = clear_coefficients(x, m)[1] - clear_coefficients(y, n)[1] + junk, eq = eq.as_content_primitive() + + return _diop_solve(eq, params=params) + + +def diop_ternary_quadratic(eq, parameterize=False): + """ + Solves the general quadratic ternary form, + `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. + + Returns a tuple `(x, y, z)` which is a base solution for the above + equation. If there are no solutions, `(None, None, None)` is returned. + + Usage + ===== + + ``diop_ternary_quadratic(eq)``: Return a tuple containing a basic solution + to ``eq``. + + Details + ======= + + ``eq`` should be an homogeneous expression of degree two in three variables + and it is assumed to be zero. + + Examples + ======== + + >>> from sympy.abc import x, y, z + >>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic + >>> diop_ternary_quadratic(x**2 + 3*y**2 - z**2) + (1, 0, 1) + >>> diop_ternary_quadratic(4*x**2 + 5*y**2 - z**2) + (1, 0, 2) + >>> diop_ternary_quadratic(45*x**2 - 7*y**2 - 8*x*y - z**2) + (28, 45, 105) + >>> diop_ternary_quadratic(x**2 - 49*y**2 - z**2 + 13*z*y -8*x*y) + (9, 1, 5) + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type in ( + HomogeneousTernaryQuadratic.name, + HomogeneousTernaryQuadraticNormal.name): + sol = _diop_ternary_quadratic(var, coeff) + if len(sol) > 0: + x_0, y_0, z_0 = list(sol)[0] + else: + x_0, y_0, z_0 = None, None, None + + if parameterize: + return _parametrize_ternary_quadratic( + (x_0, y_0, z_0), var, coeff) + return x_0, y_0, z_0 + + +def _diop_ternary_quadratic(_var, coeff): + eq = sum(i*coeff[i] for i in coeff) + if HomogeneousTernaryQuadratic(eq).matches(): + return HomogeneousTernaryQuadratic(eq, free_symbols=_var).solve() + elif HomogeneousTernaryQuadraticNormal(eq).matches(): + return HomogeneousTernaryQuadraticNormal(eq, free_symbols=_var).solve() + + +def transformation_to_normal(eq): + """ + Returns the transformation Matrix that converts a general ternary + quadratic equation ``eq`` (`ax^2 + by^2 + cz^2 + dxy + eyz + fxz`) + to a form without cross terms: `ax^2 + by^2 + cz^2 = 0`. This is + not used in solving ternary quadratics; it is only implemented for + the sake of completeness. + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type in ( + "homogeneous_ternary_quadratic", + "homogeneous_ternary_quadratic_normal"): + return _transformation_to_normal(var, coeff) + + +def _transformation_to_normal(var, coeff): + + _var = list(var) # copy + x, y, z = var + + if not any(coeff[i**2] for i in var): + # https://math.stackexchange.com/questions/448051/transform-quadratic-ternary-form-to-normal-form/448065#448065 + a = coeff[x*y] + b = coeff[y*z] + c = coeff[x*z] + swap = False + if not a: # b can't be 0 or else there aren't 3 vars + swap = True + a, b = b, a + T = Matrix(((1, 1, -b/a), (1, -1, -c/a), (0, 0, 1))) + if swap: + T.row_swap(0, 1) + T.col_swap(0, 1) + return T + + if coeff[x**2] == 0: + # If the coefficient of x is zero change the variables + if coeff[y**2] == 0: + _var[0], _var[2] = var[2], var[0] + T = _transformation_to_normal(_var, coeff) + T.row_swap(0, 2) + T.col_swap(0, 2) + return T + + _var[0], _var[1] = var[1], var[0] + T = _transformation_to_normal(_var, coeff) + T.row_swap(0, 1) + T.col_swap(0, 1) + return T + + # Apply the transformation x --> X - (B*Y + C*Z)/(2*A) + if coeff[x*y] != 0 or coeff[x*z] != 0: + A = coeff[x**2] + B = coeff[x*y] + C = coeff[x*z] + D = coeff[y**2] + E = coeff[y*z] + F = coeff[z**2] + + _coeff = {} + + _coeff[x**2] = 4*A**2 + _coeff[y**2] = 4*A*D - B**2 + _coeff[z**2] = 4*A*F - C**2 + _coeff[y*z] = 4*A*E - 2*B*C + _coeff[x*y] = 0 + _coeff[x*z] = 0 + + T_0 = _transformation_to_normal(_var, _coeff) + return Matrix(3, 3, [1, S(-B)/(2*A), S(-C)/(2*A), 0, 1, 0, 0, 0, 1])*T_0 + + elif coeff[y*z] != 0: + if coeff[y**2] == 0: + if coeff[z**2] == 0: + # Equations of the form A*x**2 + E*yz = 0. + # Apply transformation y -> Y + Z ans z -> Y - Z + return Matrix(3, 3, [1, 0, 0, 0, 1, 1, 0, 1, -1]) + + # Ax**2 + E*y*z + F*z**2 = 0 + _var[0], _var[2] = var[2], var[0] + T = _transformation_to_normal(_var, coeff) + T.row_swap(0, 2) + T.col_swap(0, 2) + return T + + # A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, F may be zero + _var[0], _var[1] = var[1], var[0] + T = _transformation_to_normal(_var, coeff) + T.row_swap(0, 1) + T.col_swap(0, 1) + return T + + return Matrix.eye(3) + + +def parametrize_ternary_quadratic(eq): + """ + Returns the parametrized general solution for the ternary quadratic + equation ``eq`` which has the form + `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. + + Examples + ======== + + >>> from sympy import Tuple, ordered + >>> from sympy.abc import x, y, z + >>> from sympy.solvers.diophantine.diophantine import parametrize_ternary_quadratic + + The parametrized solution may be returned with three parameters: + + >>> parametrize_ternary_quadratic(2*x**2 + y**2 - 2*z**2) + (p**2 - 2*q**2, -2*p**2 + 4*p*q - 4*p*r - 4*q**2, p**2 - 4*p*q + 2*q**2 - 4*q*r) + + There might also be only two parameters: + + >>> parametrize_ternary_quadratic(4*x**2 + 2*y**2 - 3*z**2) + (2*p**2 - 3*q**2, -4*p**2 + 12*p*q - 6*q**2, 4*p**2 - 8*p*q + 6*q**2) + + Notes + ===== + + Consider ``p`` and ``q`` in the previous 2-parameter + solution and observe that more than one solution can be represented + by a given pair of parameters. If `p` and ``q`` are not coprime, this is + trivially true since the common factor will also be a common factor of the + solution values. But it may also be true even when ``p`` and + ``q`` are coprime: + + >>> sol = Tuple(*_) + >>> p, q = ordered(sol.free_symbols) + >>> sol.subs([(p, 3), (q, 2)]) + (6, 12, 12) + >>> sol.subs([(q, 1), (p, 1)]) + (-1, 2, 2) + >>> sol.subs([(q, 0), (p, 1)]) + (2, -4, 4) + >>> sol.subs([(q, 1), (p, 0)]) + (-3, -6, 6) + + Except for sign and a common factor, these are equivalent to + the solution of (1, 2, 2). + + References + ========== + + .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, + London Mathematical Society Student Texts 41, Cambridge University + Press, Cambridge, 1998. + + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type in ( + "homogeneous_ternary_quadratic", + "homogeneous_ternary_quadratic_normal"): + x_0, y_0, z_0 = list(_diop_ternary_quadratic(var, coeff))[0] + return _parametrize_ternary_quadratic( + (x_0, y_0, z_0), var, coeff) + + +def _parametrize_ternary_quadratic(solution, _var, coeff): + # called for a*x**2 + b*y**2 + c*z**2 + d*x*y + e*y*z + f*x*z = 0 + assert 1 not in coeff + + x_0, y_0, z_0 = solution + + v = list(_var) # copy + + if x_0 is None: + return (None, None, None) + + if solution.count(0) >= 2: + # if there are 2 zeros the equation reduces + # to k*X**2 == 0 where X is x, y, or z so X must + # be zero, too. So there is only the trivial + # solution. + return (None, None, None) + + if x_0 == 0: + v[0], v[1] = v[1], v[0] + y_p, x_p, z_p = _parametrize_ternary_quadratic( + (y_0, x_0, z_0), v, coeff) + return x_p, y_p, z_p + + x, y, z = v + r, p, q = symbols("r, p, q", integer=True) + + eq = sum(k*v for k, v in coeff.items()) + eq_1 = _mexpand(eq.subs(zip( + (x, y, z), (r*x_0, r*y_0 + p, r*z_0 + q)))) + A, B = eq_1.as_independent(r, as_Add=True) + + + x = A*x_0 + y = (A*y_0 - _mexpand(B/r*p)) + z = (A*z_0 - _mexpand(B/r*q)) + + return _remove_gcd(x, y, z) + + +def diop_ternary_quadratic_normal(eq, parameterize=False): + """ + Solves the quadratic ternary diophantine equation, + `ax^2 + by^2 + cz^2 = 0`. + + Explanation + =========== + + Here the coefficients `a`, `b`, and `c` should be non zero. Otherwise the + equation will be a quadratic binary or univariate equation. If solvable, + returns a tuple `(x, y, z)` that satisfies the given equation. If the + equation does not have integer solutions, `(None, None, None)` is returned. + + Usage + ===== + + ``diop_ternary_quadratic_normal(eq)``: where ``eq`` is an equation of the form + `ax^2 + by^2 + cz^2 = 0`. + + Examples + ======== + + >>> from sympy.abc import x, y, z + >>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic_normal + >>> diop_ternary_quadratic_normal(x**2 + 3*y**2 - z**2) + (1, 0, 1) + >>> diop_ternary_quadratic_normal(4*x**2 + 5*y**2 - z**2) + (1, 0, 2) + >>> diop_ternary_quadratic_normal(34*x**2 - 3*y**2 - 301*z**2) + (4, 9, 1) + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + if diop_type == HomogeneousTernaryQuadraticNormal.name: + sol = _diop_ternary_quadratic_normal(var, coeff) + if len(sol) > 0: + x_0, y_0, z_0 = list(sol)[0] + else: + x_0, y_0, z_0 = None, None, None + if parameterize: + return _parametrize_ternary_quadratic( + (x_0, y_0, z_0), var, coeff) + return x_0, y_0, z_0 + + +def _diop_ternary_quadratic_normal(var, coeff): + eq = sum(i * coeff[i] for i in coeff) + return HomogeneousTernaryQuadraticNormal(eq, free_symbols=var).solve() + + +def sqf_normal(a, b, c, steps=False): + """ + Return `a', b', c'`, the coefficients of the square-free normal + form of `ax^2 + by^2 + cz^2 = 0`, where `a', b', c'` are pairwise + prime. If `steps` is True then also return three tuples: + `sq`, `sqf`, and `(a', b', c')` where `sq` contains the square + factors of `a`, `b` and `c` after removing the `gcd(a, b, c)`; + `sqf` contains the values of `a`, `b` and `c` after removing + both the `gcd(a, b, c)` and the square factors. + + The solutions for `ax^2 + by^2 + cz^2 = 0` can be + recovered from the solutions of `a'x^2 + b'y^2 + c'z^2 = 0`. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import sqf_normal + >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11) + (11, 1, 5) + >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11, True) + ((3, 1, 7), (5, 55, 11), (11, 1, 5)) + + References + ========== + + .. [1] Legendre's Theorem, Legrange's Descent, + https://public.csusm.edu/aitken_html/notes/legendre.pdf + + + See Also + ======== + + reconstruct() + """ + ABC = _remove_gcd(a, b, c) + sq = tuple(square_factor(i) for i in ABC) + sqf = A, B, C = tuple([i//j**2 for i,j in zip(ABC, sq)]) + pc = igcd(A, B) + A /= pc + B /= pc + pa = igcd(B, C) + B /= pa + C /= pa + pb = igcd(A, C) + A /= pb + B /= pb + + A *= pa + B *= pb + C *= pc + + if steps: + return (sq, sqf, (A, B, C)) + else: + return A, B, C + + +def square_factor(a): + r""" + Returns an integer `c` s.t. `a = c^2k, \ c,k \in Z`. Here `k` is square + free. `a` can be given as an integer or a dictionary of factors. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import square_factor + >>> square_factor(24) + 2 + >>> square_factor(-36*3) + 6 + >>> square_factor(1) + 1 + >>> square_factor({3: 2, 2: 1, -1: 1}) # -18 + 3 + + See Also + ======== + sympy.ntheory.factor_.core + """ + f = a if isinstance(a, dict) else factorint(a) + return Mul(*[p**(e//2) for p, e in f.items()]) + + +def reconstruct(A, B, z): + """ + Reconstruct the `z` value of an equivalent solution of `ax^2 + by^2 + cz^2` + from the `z` value of a solution of the square-free normal form of the + equation, `a'*x^2 + b'*y^2 + c'*z^2`, where `a'`, `b'` and `c'` are square + free and `gcd(a', b', c') == 1`. + """ + f = factorint(igcd(A, B)) + for p, e in f.items(): + if e != 1: + raise ValueError('a and b should be square-free') + z *= p + return z + + +def ldescent(A, B): + """ + Return a non-trivial solution to `w^2 = Ax^2 + By^2` using + Lagrange's method; return None if there is no such solution. + + Parameters + ========== + + A : Integer + B : Integer + non-zero integer + + Returns + ======= + + (int, int, int) | None : a tuple `(w_0, x_0, y_0)` which is a solution to the above equation. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import ldescent + >>> ldescent(1, 1) # w^2 = x^2 + y^2 + (1, 1, 0) + >>> ldescent(4, -7) # w^2 = 4x^2 - 7y^2 + (2, -1, 0) + + This means that `x = -1, y = 0` and `w = 2` is a solution to the equation + `w^2 = 4x^2 - 7y^2` + + >>> ldescent(5, -1) # w^2 = 5x^2 - y^2 + (2, 1, -1) + + References + ========== + + .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, + London Mathematical Society Student Texts 41, Cambridge University + Press, Cambridge, 1998. + .. [2] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. + Mathematics of Computation, 72(243), 1417-1441. + https://doi.org/10.1090/S0025-5718-02-01480-1 + """ + if A == 0 or B == 0: + raise ValueError("A and B must be non-zero integers") + if abs(A) > abs(B): + w, y, x = ldescent(B, A) + return w, x, y + if A == 1: + return (1, 1, 0) + if B == 1: + return (1, 0, 1) + if B == -1: # and A == -1 + return + + r = sqrt_mod(A, B) + if r is None: + return + Q = (r**2 - A) // B + if Q == 0: + return r, -1, 0 + for i in divisors(Q): + d, _exact = integer_nthroot(abs(Q) // i, 2) + if _exact: + B_0 = sign(Q)*i + W, X, Y = ldescent(A, B_0) + return _remove_gcd(-A*X + r*W, r*X - W, Y*B_0*d) + + +def descent(A, B): + """ + Returns a non-trivial solution, (x, y, z), to `x^2 = Ay^2 + Bz^2` + using Lagrange's descent method with lattice-reduction. `A` and `B` + are assumed to be valid for such a solution to exist. + + This is faster than the normal Lagrange's descent algorithm because + the Gaussian reduction is used. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import descent + >>> descent(3, 1) # x**2 = 3*y**2 + z**2 + (1, 0, 1) + + `(x, y, z) = (1, 0, 1)` is a solution to the above equation. + + >>> descent(41, -113) + (-16, -3, 1) + + References + ========== + + .. [1] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. + Mathematics of Computation, 72(243), 1417-1441. + https://doi.org/10.1090/S0025-5718-02-01480-1 + """ + if abs(A) > abs(B): + x, y, z = descent(B, A) + return x, z, y + + if B == 1: + return (1, 0, 1) + if A == 1: + return (1, 1, 0) + if B == -A: + return (0, 1, 1) + if B == A: + x, z, y = descent(-1, A) + return (A*y, z, x) + + w = sqrt_mod(A, B) + x_0, z_0 = gaussian_reduce(w, A, B) + + t = (x_0**2 - A*z_0**2) // B + t_2 = square_factor(t) + t_1 = t // t_2**2 + + x_1, z_1, y_1 = descent(A, t_1) + + return _remove_gcd(x_0*x_1 + A*z_0*z_1, z_0*x_1 + x_0*z_1, t_1*t_2*y_1) + + +def gaussian_reduce(w:int, a:int, b:int) -> tuple[int, int]: + r""" + Returns a reduced solution `(x, z)` to the congruence + `X^2 - aZ^2 \equiv 0 \pmod{b}` so that `x^2 + |a|z^2` is as small as possible. + Here ``w`` is a solution of the congruence `x^2 \equiv a \pmod{b}`. + + This function is intended to be used only for ``descent()``. + + Explanation + =========== + + The Gaussian reduction can find the shortest vector for any norm. + So we define the special norm for the vectors `u = (u_1, u_2)` and `v = (v_1, v_2)` as follows. + + .. math :: + u \cdot v := (wu_1 + bu_2)(wv_1 + bv_2) + |a|u_1v_1 + + Note that, given the mapping `f: (u_1, u_2) \to (wu_1 + bu_2, u_1)`, + `f((u_1,u_2))` is the solution to `X^2 - aZ^2 \equiv 0 \pmod{b}`. + In other words, finding the shortest vector in this norm will yield a solution with smaller `X^2 + |a|Z^2`. + The algorithm starts from basis vectors `(0, 1)` and `(1, 0)` + (corresponding to solutions `(b, 0)` and `(w, 1)`, respectively) and finds the shortest vector. + The shortest vector does not necessarily correspond to the smallest solution, + but since ``descent()`` only wants the smallest possible solution, it is sufficient. + + Parameters + ========== + + w : int + ``w`` s.t. `w^2 \equiv a \pmod{b}` + a : int + square-free nonzero integer + b : int + square-free nonzero integer + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import gaussian_reduce + >>> from sympy.ntheory.residue_ntheory import sqrt_mod + >>> a, b = 19, 101 + >>> gaussian_reduce(sqrt_mod(a, b), a, b) # 1**2 - 19*(-4)**2 = -303 + (1, -4) + >>> a, b = 11, 14 + >>> x, z = gaussian_reduce(sqrt_mod(a, b), a, b) + >>> (x**2 - a*z**2) % b == 0 + True + + It does not always return the smallest solution. + + >>> a, b = 6, 95 + >>> min_x, min_z = 1, 4 + >>> x, z = gaussian_reduce(sqrt_mod(a, b), a, b) + >>> (x**2 - a*z**2) % b == 0 and (min_x**2 - a*min_z**2) % b == 0 + True + >>> min_x**2 + abs(a)*min_z**2 < x**2 + abs(a)*z**2 + True + + References + ========== + + .. [1] Gaussian lattice Reduction [online]. Available: + https://web.archive.org/web/20201021115213/http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404 + .. [2] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. + Mathematics of Computation, 72(243), 1417-1441. + https://doi.org/10.1090/S0025-5718-02-01480-1 + """ + a = abs(a) + def _dot(u, v): + return u[0]*v[0] + a*u[1]*v[1] + + u = (b, 0) + v = (w, 1) if b*w >= 0 else (-w, -1) + # i.e., _dot(u, v) >= 0 + + if b**2 < w**2 + a: + u, v = v, u + # i.e., norm(u) >= norm(v), where norm(u) := sqrt(_dot(u, u)) + + while _dot(u, u) > (dv := _dot(v, v)): + k = _dot(u, v) // dv + u, v = v, (u[0] - k*v[0], u[1] - k*v[1]) + c = (v[0] - u[0], v[1] - u[1]) + if _dot(c, c) <= _dot(u, u) <= 2*_dot(u, v): + return c + return u + + +def holzer(x, y, z, a, b, c): + r""" + Simplify the solution `(x, y, z)` of the equation + `ax^2 + by^2 = cz^2` with `a, b, c > 0` and `z^2 \geq \mid ab \mid` to + a new reduced solution `(x', y', z')` such that `z'^2 \leq \mid ab \mid`. + + The algorithm is an interpretation of Mordell's reduction as described + on page 8 of Cremona and Rusin's paper [1]_ and the work of Mordell in + reference [2]_. + + References + ========== + + .. [1] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. + Mathematics of Computation, 72(243), 1417-1441. + https://doi.org/10.1090/S0025-5718-02-01480-1 + .. [2] Diophantine Equations, L. J. Mordell, page 48. + + """ + + if _odd(c): + k = 2*c + else: + k = c//2 + + small = a*b*c + step = 0 + while True: + t1, t2, t3 = a*x**2, b*y**2, c*z**2 + # check that it's a solution + if t1 + t2 != t3: + if step == 0: + raise ValueError('bad starting solution') + break + x_0, y_0, z_0 = x, y, z + if max(t1, t2, t3) <= small: + # Holzer condition + break + + uv = u, v = base_solution_linear(k, y_0, -x_0) + if None in uv: + break + + p, q = -(a*u*x_0 + b*v*y_0), c*z_0 + r = Rational(p, q) + if _even(c): + w = _nint_or_floor(p, q) + assert abs(w - r) <= S.Half + else: + w = p//q # floor + if _odd(a*u + b*v + c*w): + w += 1 + assert abs(w - r) <= S.One + + A = (a*u**2 + b*v**2 + c*w**2) + B = (a*u*x_0 + b*v*y_0 + c*w*z_0) + x = Rational(x_0*A - 2*u*B, k) + y = Rational(y_0*A - 2*v*B, k) + z = Rational(z_0*A - 2*w*B, k) + assert all(i.is_Integer for i in (x, y, z)) + step += 1 + + return tuple([int(i) for i in (x_0, y_0, z_0)]) + + +def diop_general_pythagorean(eq, param=symbols("m", integer=True)): + """ + Solves the general pythagorean equation, + `a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`. + + Returns a tuple which contains a parametrized solution to the equation, + sorted in the same order as the input variables. + + Usage + ===== + + ``diop_general_pythagorean(eq, param)``: where ``eq`` is a general + pythagorean equation which is assumed to be zero and ``param`` is the base + parameter used to construct other parameters by subscripting. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import diop_general_pythagorean + >>> from sympy.abc import a, b, c, d, e + >>> diop_general_pythagorean(a**2 + b**2 + c**2 - d**2) + (m1**2 + m2**2 - m3**2, 2*m1*m3, 2*m2*m3, m1**2 + m2**2 + m3**2) + >>> diop_general_pythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2) + (10*m1**2 + 10*m2**2 + 10*m3**2 - 10*m4**2, 15*m1**2 + 15*m2**2 + 15*m3**2 + 15*m4**2, 15*m1*m4, 12*m2*m4, 60*m3*m4) + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type == GeneralPythagorean.name: + if param is None: + params = None + else: + params = symbols('%s1:%i' % (param, len(var)), integer=True) + return list(GeneralPythagorean(eq).solve(parameters=params))[0] + + +def diop_general_sum_of_squares(eq, limit=1): + r""" + Solves the equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. + + Returns at most ``limit`` number of solutions. + + Usage + ===== + + ``general_sum_of_squares(eq, limit)`` : Here ``eq`` is an expression which + is assumed to be zero. Also, ``eq`` should be in the form, + `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. + + Details + ======= + + When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be + no solutions. Refer to [1]_ for more details. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_squares + >>> from sympy.abc import a, b, c, d, e + >>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345) + {(15, 22, 22, 24, 24)} + + Reference + ========= + + .. [1] Representing an integer as a sum of three squares, [online], + Available: + https://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type == GeneralSumOfSquares.name: + return set(GeneralSumOfSquares(eq).solve(limit=limit)) + + +def diop_general_sum_of_even_powers(eq, limit=1): + """ + Solves the equation `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0` + where `e` is an even, integer power. + + Returns at most ``limit`` number of solutions. + + Usage + ===== + + ``general_sum_of_even_powers(eq, limit)`` : Here ``eq`` is an expression which + is assumed to be zero. Also, ``eq`` should be in the form, + `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0`. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_even_powers + >>> from sympy.abc import a, b + >>> diop_general_sum_of_even_powers(a**4 + b**4 - (2**4 + 3**4)) + {(2, 3)} + + See Also + ======== + + power_representation + """ + var, coeff, diop_type = classify_diop(eq, _dict=False) + + if diop_type == GeneralSumOfEvenPowers.name: + return set(GeneralSumOfEvenPowers(eq).solve(limit=limit)) + + +## Functions below this comment can be more suitably grouped under +## an Additive number theory module rather than the Diophantine +## equation module. + + +def partition(n, k=None, zeros=False): + """ + Returns a generator that can be used to generate partitions of an integer + `n`. + + Explanation + =========== + + A partition of `n` is a set of positive integers which add up to `n`. For + example, partitions of 3 are 3, 1 + 2, 1 + 1 + 1. A partition is returned + as a tuple. If ``k`` equals None, then all possible partitions are returned + irrespective of their size, otherwise only the partitions of size ``k`` are + returned. If the ``zero`` parameter is set to True then a suitable + number of zeros are added at the end of every partition of size less than + ``k``. + + ``zero`` parameter is considered only if ``k`` is not None. When the + partitions are over, the last `next()` call throws the ``StopIteration`` + exception, so this function should always be used inside a try - except + block. + + Details + ======= + + ``partition(n, k)``: Here ``n`` is a positive integer and ``k`` is the size + of the partition which is also positive integer. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import partition + >>> f = partition(5) + >>> next(f) + (1, 1, 1, 1, 1) + >>> next(f) + (1, 1, 1, 2) + >>> g = partition(5, 3) + >>> next(g) + (1, 1, 3) + >>> next(g) + (1, 2, 2) + >>> g = partition(5, 3, zeros=True) + >>> next(g) + (0, 0, 5) + + """ + if not zeros or k is None: + for i in ordered_partitions(n, k): + yield tuple(i) + else: + for m in range(1, k + 1): + for i in ordered_partitions(n, m): + i = tuple(i) + yield (0,)*(k - len(i)) + i + + +def prime_as_sum_of_two_squares(p): + """ + Represent a prime `p` as a unique sum of two squares; this can + only be done if the prime is congruent to 1 mod 4. + + Parameters + ========== + + p : Integer + A prime that is congruent to 1 mod 4 + + Returns + ======= + + (int, int) | None : Pair of positive integers ``(x, y)`` satisfying ``x**2 + y**2 = p``. + None if ``p`` is not congruent to 1 mod 4. + + Raises + ====== + + ValueError + If ``p`` is not prime number + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import prime_as_sum_of_two_squares + >>> prime_as_sum_of_two_squares(7) # can't be done + >>> prime_as_sum_of_two_squares(5) + (1, 2) + + Reference + ========= + + .. [1] Representing a number as a sum of four squares, [online], + Available: https://schorn.ch/lagrange.html + + See Also + ======== + + sum_of_squares + + """ + p = as_int(p) + if p % 4 != 1: + return + if not isprime(p): + raise ValueError("p should be a prime number") + + if p % 8 == 5: + # Legendre symbol (2/p) == -1 if p % 8 in [3, 5] + b = 2 + elif p % 12 == 5: + # Legendre symbol (3/p) == -1 if p % 12 in [5, 7] + b = 3 + elif p % 5 in [2, 3]: + # Legendre symbol (5/p) == -1 if p % 5 in [2, 3] + b = 5 + else: + b = 7 + while jacobi(b, p) == 1: + b = nextprime(b) + + b = pow(b, p >> 2, p) + a = p + while b**2 > p: + a, b = b, a % b + return (int(a % b), int(b)) # convert from long + + +def sum_of_three_squares(n): + r""" + Returns a 3-tuple $(a, b, c)$ such that $a^2 + b^2 + c^2 = n$ and + $a, b, c \geq 0$. + + Returns None if $n = 4^a(8m + 7)$ for some `a, m \in \mathbb{Z}`. See + [1]_ for more details. + + Parameters + ========== + + n : Integer + non-negative integer + + Returns + ======= + + (int, int, int) | None : 3-tuple non-negative integers ``(a, b, c)`` satisfying ``a**2 + b**2 + c**2 = n``. + a,b,c are sorted in ascending order. ``None`` if no such ``(a,b,c)``. + + Raises + ====== + + ValueError + If ``n`` is a negative integer + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import sum_of_three_squares + >>> sum_of_three_squares(44542) + (18, 37, 207) + + References + ========== + + .. [1] Representing a number as a sum of three squares, [online], + Available: https://schorn.ch/lagrange.html + + See Also + ======== + + power_representation : + ``sum_of_three_squares(n)`` is one of the solutions output by ``power_representation(n, 2, 3, zeros=True)`` + + """ + # https://math.stackexchange.com/questions/483101/rabin-and-shallit-algorithm/651425#651425 + # discusses these numbers (except for 1, 2, 3) as the exceptions of H&L's conjecture that + # Every sufficiently large number n is either a square or the sum of a prime and a square. + special = {1: (0, 0, 1), 2: (0, 1, 1), 3: (1, 1, 1), 10: (0, 1, 3), 34: (3, 3, 4), + 58: (0, 3, 7), 85: (0, 6, 7), 130: (0, 3, 11), 214: (3, 6, 13), 226: (8, 9, 9), + 370: (8, 9, 15), 526: (6, 7, 21), 706: (15, 15, 16), 730: (0, 1, 27), + 1414: (6, 17, 33), 1906: (13, 21, 36), 2986: (21, 32, 39), 9634: (56, 57, 57)} + n = as_int(n) + if n < 0: + raise ValueError("n should be a non-negative integer") + if n == 0: + return (0, 0, 0) + n, v = remove(n, 4) + v = 1 << v + if n % 8 == 7: + return + if n in special: + return tuple([v*i for i in special[n]]) + + s, _exact = integer_nthroot(n, 2) + if _exact: + return (0, 0, v*s) + if n % 8 == 3: + if not s % 2: + s -= 1 + for x in range(s, -1, -2): + N = (n - x**2) // 2 + if isprime(N): + # n % 8 == 3 and x % 2 == 1 => N % 4 == 1 + y, z = prime_as_sum_of_two_squares(N) + return tuple(sorted([v*x, v*(y + z), v*abs(y - z)])) + # We will never reach this point because there must be a solution. + assert False + + # assert n % 4 in [1, 2] + if not((n % 2) ^ (s % 2)): + s -= 1 + for x in range(s, -1, -2): + N = n - x**2 + if isprime(N): + # assert N % 4 == 1 + y, z = prime_as_sum_of_two_squares(N) + return tuple(sorted([v*x, v*y, v*z])) + # We will never reach this point because there must be a solution. + assert False + + +def sum_of_four_squares(n): + r""" + Returns a 4-tuple `(a, b, c, d)` such that `a^2 + b^2 + c^2 + d^2 = n`. + Here `a, b, c, d \geq 0`. + + Parameters + ========== + + n : Integer + non-negative integer + + Returns + ======= + + (int, int, int, int) : 4-tuple non-negative integers ``(a, b, c, d)`` satisfying ``a**2 + b**2 + c**2 + d**2 = n``. + a,b,c,d are sorted in ascending order. + + Raises + ====== + + ValueError + If ``n`` is a negative integer + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import sum_of_four_squares + >>> sum_of_four_squares(3456) + (8, 8, 32, 48) + >>> sum_of_four_squares(1294585930293) + (0, 1234, 2161, 1137796) + + References + ========== + + .. [1] Representing a number as a sum of four squares, [online], + Available: https://schorn.ch/lagrange.html + + See Also + ======== + + power_representation : + ``sum_of_four_squares(n)`` is one of the solutions output by ``power_representation(n, 2, 4, zeros=True)`` + + """ + n = as_int(n) + if n < 0: + raise ValueError("n should be a non-negative integer") + if n == 0: + return (0, 0, 0, 0) + # remove factors of 4 since a solution in terms of 3 squares is + # going to be returned; this is also done in sum_of_three_squares, + # but it needs to be done here to select d + n, v = remove(n, 4) + v = 1 << v + if n % 8 == 7: + d = 2 + n = n - 4 + elif n % 8 in (2, 6): + d = 1 + n = n - 1 + else: + d = 0 + x, y, z = sum_of_three_squares(n) # sorted + return tuple(sorted([v*d, v*x, v*y, v*z])) + + +def power_representation(n, p, k, zeros=False): + r""" + Returns a generator for finding k-tuples of integers, + `(n_{1}, n_{2}, . . . n_{k})`, such that + `n = n_{1}^p + n_{2}^p + . . . n_{k}^p`. + + Usage + ===== + + ``power_representation(n, p, k, zeros)``: Represent non-negative number + ``n`` as a sum of ``k`` ``p``\ th powers. If ``zeros`` is true, then the + solutions is allowed to contain zeros. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import power_representation + + Represent 1729 as a sum of two cubes: + + >>> f = power_representation(1729, 3, 2) + >>> next(f) + (9, 10) + >>> next(f) + (1, 12) + + If the flag `zeros` is True, the solution may contain tuples with + zeros; any such solutions will be generated after the solutions + without zeros: + + >>> list(power_representation(125, 2, 3, zeros=True)) + [(5, 6, 8), (3, 4, 10), (0, 5, 10), (0, 2, 11)] + + For even `p` the `permute_sign` function can be used to get all + signed values: + + >>> from sympy.utilities.iterables import permute_signs + >>> list(permute_signs((1, 12))) + [(1, 12), (-1, 12), (1, -12), (-1, -12)] + + All possible signed permutations can also be obtained: + + >>> from sympy.utilities.iterables import signed_permutations + >>> list(signed_permutations((1, 12))) + [(1, 12), (-1, 12), (1, -12), (-1, -12), (12, 1), (-12, 1), (12, -1), (-12, -1)] + """ + n, p, k = [as_int(i) for i in (n, p, k)] + + if n < 0: + if p % 2: + for t in power_representation(-n, p, k, zeros): + yield tuple(-i for i in t) + return + + if p < 1 or k < 1: + raise ValueError(filldedent(''' + Expecting positive integers for `(p, k)`, but got `(%s, %s)`''' + % (p, k))) + + if n == 0: + if zeros: + yield (0,)*k + return + + if k == 1: + if p == 1: + yield (n,) + elif n == 1: + yield (1,) + else: + be = perfect_power(n) + if be: + b, e = be + d, r = divmod(e, p) + if not r: + yield (b**d,) + return + + if p == 1: + yield from partition(n, k, zeros=zeros) + return + + if p == 2: + if k == 3: + n, v = remove(n, 4) + if v: + v = 1 << v + for t in power_representation(n, p, k, zeros): + yield tuple(i*v for i in t) + return + feasible = _can_do_sum_of_squares(n, k) + if not feasible: + return + if not zeros: + if n > 33 and k >= 5 and k <= n and n - k in ( + 13, 10, 7, 5, 4, 2, 1): + '''Todd G. Will, "When Is n^2 a Sum of k Squares?", [online]. + Available: https://www.maa.org/sites/default/files/Will-MMz-201037918.pdf''' + return + # quick tests since feasibility includes the possiblity of 0 + if k == 4 and (n in (1, 3, 5, 9, 11, 17, 29, 41) or remove(n, 4)[0] in (2, 6, 14)): + # A000534 + return + if k == 3 and n in (1, 2, 5, 10, 13, 25, 37, 58, 85, 130): # or n = some number >= 5*10**10 + # A051952 + return + if feasible is not True: # it's prime and k == 2 + yield prime_as_sum_of_two_squares(n) + return + + if k == 2 and p > 2: + be = perfect_power(n) + if be and be[1] % p == 0: + return # Fermat: a**n + b**n = c**n has no solution for n > 2 + + if n >= k: + a = integer_nthroot(n - (k - 1), p)[0] + for t in pow_rep_recursive(a, k, n, [], p): + yield tuple(reversed(t)) + + if zeros: + a = integer_nthroot(n, p)[0] + for i in range(1, k): + for t in pow_rep_recursive(a, i, n, [], p): + yield tuple(reversed(t + (0,)*(k - i))) + + +sum_of_powers = power_representation + + +def pow_rep_recursive(n_i, k, n_remaining, terms, p): + # Invalid arguments + if n_i <= 0 or k <= 0: + return + + # No solutions may exist + if n_remaining < k: + return + if k * pow(n_i, p) < n_remaining: + return + + if k == 0 and n_remaining == 0: + yield tuple(terms) + + elif k == 1: + # next_term^p must equal to n_remaining + next_term, exact = integer_nthroot(n_remaining, p) + if exact and next_term <= n_i: + yield tuple(terms + [next_term]) + return + + else: + # TODO: Fall back to diop_DN when k = 2 + if n_i >= 1 and k > 0: + for next_term in range(1, n_i + 1): + residual = n_remaining - pow(next_term, p) + if residual < 0: + break + yield from pow_rep_recursive(next_term, k - 1, residual, terms + [next_term], p) + + +def sum_of_squares(n, k, zeros=False): + """Return a generator that yields the k-tuples of nonnegative + values, the squares of which sum to n. If zeros is False (default) + then the solution will not contain zeros. The nonnegative + elements of a tuple are sorted. + + * If k == 1 and n is square, (n,) is returned. + + * If k == 2 then n can only be written as a sum of squares if + every prime in the factorization of n that has the form + 4*k + 3 has an even multiplicity. If n is prime then + it can only be written as a sum of two squares if it is + in the form 4*k + 1. + + * if k == 3 then n can be written as a sum of squares if it does + not have the form 4**m*(8*k + 7). + + * all integers can be written as the sum of 4 squares. + + * if k > 4 then n can be partitioned and each partition can + be written as a sum of 4 squares; if n is not evenly divisible + by 4 then n can be written as a sum of squares only if the + an additional partition can be written as sum of squares. + For example, if k = 6 then n is partitioned into two parts, + the first being written as a sum of 4 squares and the second + being written as a sum of 2 squares -- which can only be + done if the condition above for k = 2 can be met, so this will + automatically reject certain partitions of n. + + Examples + ======== + + >>> from sympy.solvers.diophantine.diophantine import sum_of_squares + >>> list(sum_of_squares(25, 2)) + [(3, 4)] + >>> list(sum_of_squares(25, 2, True)) + [(3, 4), (0, 5)] + >>> list(sum_of_squares(25, 4)) + [(1, 2, 2, 4)] + + See Also + ======== + + sympy.utilities.iterables.signed_permutations + """ + yield from power_representation(n, 2, k, zeros) + + +def _can_do_sum_of_squares(n, k): + """Return True if n can be written as the sum of k squares, + False if it cannot, or 1 if ``k == 2`` and ``n`` is prime (in which + case it *can* be written as a sum of two squares). A False + is returned only if it cannot be written as ``k``-squares, even + if 0s are allowed. + """ + if k < 1: + return False + if n < 0: + return False + if n == 0: + return True + if k == 1: + return is_square(n) + if k == 2: + if n in (1, 2): + return True + if isprime(n): + if n % 4 == 1: + return 1 # signal that it was prime + return False + # n is a composite number + # we can proceed iff no prime factor in the form 4*k + 3 + # has an odd multiplicity + return all(p % 4 !=3 or m % 2 == 0 for p, m in factorint(n).items()) + if k == 3: + return remove(n, 4)[0] % 8 != 7 + # every number can be written as a sum of 4 squares; for k > 4 partitions + # can be 0 + return True diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__init__.py b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__init__.py new file mode 100644 index 0000000000000000000000000000000000000000..e69de29bb2d1d6434b8b29ae775ad8c2e48c5391 diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__pycache__/__init__.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__pycache__/__init__.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..be5c49147ac62c9e56110ab7328ba1f5e52441d3 Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__pycache__/__init__.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__pycache__/test_diophantine.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__pycache__/test_diophantine.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..d27fc083eb39be30f44e9cfc10ed297ffa751a79 Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/__pycache__/test_diophantine.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/test_diophantine.py b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/test_diophantine.py new file mode 100644 index 0000000000000000000000000000000000000000..094770b7bba795aef306ea71831388414fde935e --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/diophantine/tests/test_diophantine.py @@ -0,0 +1,1051 @@ +from sympy.core.add import Add +from sympy.core.mul import Mul +from sympy.core.numbers import (Rational, oo, pi) +from sympy.core.relational import Eq +from sympy.core.singleton import S +from sympy.core.symbol import symbols +from sympy.matrices.dense import Matrix +from sympy.ntheory.factor_ import factorint +from sympy.simplify.powsimp import powsimp +from sympy.core.function import _mexpand +from sympy.core.sorting import default_sort_key, ordered +from sympy.functions.elementary.trigonometric import sin +from sympy.solvers.diophantine import diophantine +from sympy.solvers.diophantine.diophantine import (diop_DN, + diop_solve, diop_ternary_quadratic_normal, + diop_general_pythagorean, diop_ternary_quadratic, diop_linear, + diop_quadratic, diop_general_sum_of_squares, diop_general_sum_of_even_powers, + descent, diop_bf_DN, divisible, equivalent, find_DN, ldescent, length, + reconstruct, partition, power_representation, + prime_as_sum_of_two_squares, square_factor, sum_of_four_squares, + sum_of_three_squares, transformation_to_DN, transformation_to_normal, + classify_diop, base_solution_linear, cornacchia, sqf_normal, gaussian_reduce, holzer, + check_param, parametrize_ternary_quadratic, sum_of_powers, sum_of_squares, + _diop_ternary_quadratic_normal, _nint_or_floor, + _odd, _even, _remove_gcd, _can_do_sum_of_squares, DiophantineSolutionSet, GeneralPythagorean, + BinaryQuadratic) + +from sympy.testing.pytest import slow, raises, XFAIL +from sympy.utilities.iterables import ( + signed_permutations) + +a, b, c, d, p, q, x, y, z, w, t, u, v, X, Y, Z = symbols( + "a, b, c, d, p, q, x, y, z, w, t, u, v, X, Y, Z", integer=True) +t_0, t_1, t_2, t_3, t_4, t_5, t_6 = symbols("t_:7", integer=True) +m1, m2, m3 = symbols('m1:4', integer=True) +n1 = symbols('n1', integer=True) + + +def diop_simplify(eq): + return _mexpand(powsimp(_mexpand(eq))) + + +def test_input_format(): + raises(TypeError, lambda: diophantine(sin(x))) + raises(TypeError, lambda: diophantine(x/pi - 3)) + + +def test_nosols(): + # diophantine should sympify eq so that these are equivalent + assert diophantine(3) == set() + assert diophantine(S(3)) == set() + + +def test_univariate(): + assert diop_solve((x - 1)*(x - 2)**2) == {(1,), (2,)} + assert diop_solve((x - 1)*(x - 2)) == {(1,), (2,)} + + +def test_classify_diop(): + raises(TypeError, lambda: classify_diop(x**2/3 - 1)) + raises(ValueError, lambda: classify_diop(1)) + raises(NotImplementedError, lambda: classify_diop(w*x*y*z - 1)) + raises(NotImplementedError, lambda: classify_diop(x**3 + y**3 + z**4 - 90)) + assert classify_diop(14*x**2 + 15*x - 42) == ( + [x], {1: -42, x: 15, x**2: 14}, 'univariate') + assert classify_diop(x*y + z) == ( + [x, y, z], {x*y: 1, z: 1}, 'inhomogeneous_ternary_quadratic') + assert classify_diop(x*y + z + w + x**2) == ( + [w, x, y, z], {x*y: 1, w: 1, x**2: 1, z: 1}, 'inhomogeneous_general_quadratic') + assert classify_diop(x*y + x*z + x**2 + 1) == ( + [x, y, z], {x*y: 1, x*z: 1, x**2: 1, 1: 1}, 'inhomogeneous_general_quadratic') + assert classify_diop(x*y + z + w + 42) == ( + [w, x, y, z], {x*y: 1, w: 1, 1: 42, z: 1}, 'inhomogeneous_general_quadratic') + assert classify_diop(x*y + z*w) == ( + [w, x, y, z], {x*y: 1, w*z: 1}, 'homogeneous_general_quadratic') + assert classify_diop(x*y**2 + 1) == ( + [x, y], {x*y**2: 1, 1: 1}, 'cubic_thue') + assert classify_diop(x**4 + y**4 + z**4 - (1 + 16 + 81)) == ( + [x, y, z], {1: -98, x**4: 1, z**4: 1, y**4: 1}, 'general_sum_of_even_powers') + assert classify_diop(x**2 + y**2 + z**2) == ( + [x, y, z], {x**2: 1, y**2: 1, z**2: 1}, 'homogeneous_ternary_quadratic_normal') + + +def test_linear(): + assert diop_solve(x) == (0,) + assert diop_solve(1*x) == (0,) + assert diop_solve(3*x) == (0,) + assert diop_solve(x + 1) == (-1,) + assert diop_solve(2*x + 1) == (None,) + assert diop_solve(2*x + 4) == (-2,) + assert diop_solve(y + x) == (t_0, -t_0) + assert diop_solve(y + x + 0) == (t_0, -t_0) + assert diop_solve(y + x - 0) == (t_0, -t_0) + assert diop_solve(0*x - y - 5) == (-5,) + assert diop_solve(3*y + 2*x - 5) == (3*t_0 - 5, -2*t_0 + 5) + assert diop_solve(2*x - 3*y - 5) == (3*t_0 - 5, 2*t_0 - 5) + assert diop_solve(-2*x - 3*y - 5) == (3*t_0 + 5, -2*t_0 - 5) + assert diop_solve(7*x + 5*y) == (5*t_0, -7*t_0) + assert diop_solve(2*x + 4*y) == (-2*t_0, t_0) + assert diop_solve(4*x + 6*y - 4) == (3*t_0 - 2, -2*t_0 + 2) + assert diop_solve(4*x + 6*y - 3) == (None, None) + assert diop_solve(0*x + 3*y - 4*z + 5) == (4*t_0 + 5, 3*t_0 + 5) + assert diop_solve(4*x + 3*y - 4*z + 5) == (t_0, 8*t_0 + 4*t_1 + 5, 7*t_0 + 3*t_1 + 5) + assert diop_solve(4*x + 3*y - 4*z + 5, None) == (0, 5, 5) + assert diop_solve(4*x + 2*y + 8*z - 5) == (None, None, None) + assert diop_solve(5*x + 7*y - 2*z - 6) == (t_0, -3*t_0 + 2*t_1 + 6, -8*t_0 + 7*t_1 + 18) + assert diop_solve(3*x - 6*y + 12*z - 9) == (2*t_0 + 3, t_0 + 2*t_1, t_1) + assert diop_solve(6*w + 9*x + 20*y - z) == (t_0, t_1, t_1 + t_2, 6*t_0 + 29*t_1 + 20*t_2) + + # to ignore constant factors, use diophantine + raises(TypeError, lambda: diop_solve(x/2)) + + +def test_quadratic_simple_hyperbolic_case(): + # Simple Hyperbolic case: A = C = 0 and B != 0 + assert diop_solve(3*x*y + 34*x - 12*y + 1) == \ + {(-133, -11), (5, -57)} + assert diop_solve(6*x*y + 2*x + 3*y + 1) == set() + assert diop_solve(-13*x*y + 2*x - 4*y - 54) == {(27, 0)} + assert diop_solve(-27*x*y - 30*x - 12*y - 54) == {(-14, -1)} + assert diop_solve(2*x*y + 5*x + 56*y + 7) == {(-161, -3), (-47, -6), (-35, -12), + (-29, -69), (-27, 64), (-21, 7), + (-9, 1), (105, -2)} + assert diop_solve(6*x*y + 9*x + 2*y + 3) == set() + assert diop_solve(x*y + x + y + 1) == {(-1, t), (t, -1)} + assert diophantine(48*x*y) + + +def test_quadratic_elliptical_case(): + # Elliptical case: B**2 - 4AC < 0 + + assert diop_solve(42*x**2 + 8*x*y + 15*y**2 + 23*x + 17*y - 4915) == {(-11, -1)} + assert diop_solve(4*x**2 + 3*y**2 + 5*x - 11*y + 12) == set() + assert diop_solve(x**2 + y**2 + 2*x + 2*y + 2) == {(-1, -1)} + assert diop_solve(15*x**2 - 9*x*y + 14*y**2 - 23*x - 14*y - 4950) == {(-15, 6)} + assert diop_solve(10*x**2 + 12*x*y + 12*y**2 - 34) == \ + {(-1, -1), (-1, 2), (1, -2), (1, 1)} + + +def test_quadratic_parabolic_case(): + # Parabolic case: B**2 - 4AC = 0 + assert check_solutions(8*x**2 - 24*x*y + 18*y**2 + 5*x + 7*y + 16) + assert check_solutions(8*x**2 - 24*x*y + 18*y**2 + 6*x + 12*y - 6) + assert check_solutions(8*x**2 + 24*x*y + 18*y**2 + 4*x + 6*y - 7) + assert check_solutions(-4*x**2 + 4*x*y - y**2 + 2*x - 3) + assert check_solutions(x**2 + 2*x*y + y**2 + 2*x + 2*y + 1) + assert check_solutions(x**2 - 2*x*y + y**2 + 2*x + 2*y + 1) + assert check_solutions(y**2 - 41*x + 40) + + +def test_quadratic_perfect_square(): + # B**2 - 4*A*C > 0 + # B**2 - 4*A*C is a perfect square + assert check_solutions(48*x*y) + assert check_solutions(4*x**2 - 5*x*y + y**2 + 2) + assert check_solutions(-2*x**2 - 3*x*y + 2*y**2 -2*x - 17*y + 25) + assert check_solutions(12*x**2 + 13*x*y + 3*y**2 - 2*x + 3*y - 12) + assert check_solutions(8*x**2 + 10*x*y + 2*y**2 - 32*x - 13*y - 23) + assert check_solutions(4*x**2 - 4*x*y - 3*y- 8*x - 3) + assert check_solutions(- 4*x*y - 4*y**2 - 3*y- 5*x - 10) + assert check_solutions(x**2 - y**2 - 2*x - 2*y) + assert check_solutions(x**2 - 9*y**2 - 2*x - 6*y) + assert check_solutions(4*x**2 - 9*y**2 - 4*x - 12*y - 3) + + +def test_quadratic_non_perfect_square(): + # B**2 - 4*A*C is not a perfect square + # Used check_solutions() since the solutions are complex expressions involving + # square roots and exponents + assert check_solutions(x**2 - 2*x - 5*y**2) + assert check_solutions(3*x**2 - 2*y**2 - 2*x - 2*y) + assert check_solutions(x**2 - x*y - y**2 - 3*y) + assert check_solutions(x**2 - 9*y**2 - 2*x - 6*y) + assert BinaryQuadratic(x**2 + y**2 + 2*x + 2*y + 2).solve() == {(-1, -1)} + + +def test_issue_9106(): + eq = -48 - 2*x*(3*x - 1) + y*(3*y - 1) + v = (x, y) + for sol in diophantine(eq): + assert not diop_simplify(eq.xreplace(dict(zip(v, sol)))) + + +def test_issue_18138(): + eq = x**2 - x - y**2 + v = (x, y) + for sol in diophantine(eq): + assert not diop_simplify(eq.xreplace(dict(zip(v, sol)))) + + +@slow +def test_quadratic_non_perfect_slow(): + assert check_solutions(8*x**2 + 10*x*y - 2*y**2 - 32*x - 13*y - 23) + # This leads to very large numbers. + # assert check_solutions(5*x**2 - 13*x*y + y**2 - 4*x - 4*y - 15) + assert check_solutions(-3*x**2 - 2*x*y + 7*y**2 - 5*x - 7) + assert check_solutions(-4 - x + 4*x**2 - y - 3*x*y - 4*y**2) + assert check_solutions(1 + 2*x + 2*x**2 + 2*y + x*y - 2*y**2) + + +def test_DN(): + # Most of the test cases were adapted from, + # Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004. + # https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf + # others are verified using Wolfram Alpha. + + # Covers cases where D <= 0 or D > 0 and D is a square or N = 0 + # Solutions are straightforward in these cases. + assert diop_DN(3, 0) == [(0, 0)] + assert diop_DN(-17, -5) == [] + assert diop_DN(-19, 23) == [(2, 1)] + assert diop_DN(-13, 17) == [(2, 1)] + assert diop_DN(-15, 13) == [] + assert diop_DN(0, 5) == [] + assert diop_DN(0, 9) == [(3, t)] + assert diop_DN(9, 0) == [(3*t, t)] + assert diop_DN(16, 24) == [] + assert diop_DN(9, 180) == [(18, 4)] + assert diop_DN(9, -180) == [(12, 6)] + assert diop_DN(7, 0) == [(0, 0)] + + # When equation is x**2 + y**2 = N + # Solutions are interchangeable + assert diop_DN(-1, 5) == [(2, 1), (1, 2)] + assert diop_DN(-1, 169) == [(12, 5), (5, 12), (13, 0), (0, 13)] + + # D > 0 and D is not a square + + # N = 1 + assert diop_DN(13, 1) == [(649, 180)] + assert diop_DN(980, 1) == [(51841, 1656)] + assert diop_DN(981, 1) == [(158070671986249, 5046808151700)] + assert diop_DN(986, 1) == [(49299, 1570)] + assert diop_DN(991, 1) == [(379516400906811930638014896080, 12055735790331359447442538767)] + assert diop_DN(17, 1) == [(33, 8)] + assert diop_DN(19, 1) == [(170, 39)] + + # N = -1 + assert diop_DN(13, -1) == [(18, 5)] + assert diop_DN(991, -1) == [] + assert diop_DN(41, -1) == [(32, 5)] + assert diop_DN(290, -1) == [(17, 1)] + assert diop_DN(21257, -1) == [(13913102721304, 95427381109)] + assert diop_DN(32, -1) == [] + + # |N| > 1 + # Some tests were created using calculator at + # http://www.numbertheory.org/php/patz.html + + assert diop_DN(13, -4) == [(3, 1), (393, 109), (36, 10)] + # Source I referred returned (3, 1), (393, 109) and (-3, 1) as fundamental solutions + # So (-3, 1) and (393, 109) should be in the same equivalent class + assert equivalent(-3, 1, 393, 109, 13, -4) == True + + assert diop_DN(13, 27) == [(220, 61), (40, 11), (768, 213), (12, 3)] + assert set(diop_DN(157, 12)) == {(13, 1), (10663, 851), (579160, 46222), + (483790960, 38610722), (26277068347, 2097138361), + (21950079635497, 1751807067011)} + assert diop_DN(13, 25) == [(3245, 900)] + assert diop_DN(192, 18) == [] + assert diop_DN(23, 13) == [(-6, 1), (6, 1)] + assert diop_DN(167, 2) == [(13, 1)] + assert diop_DN(167, -2) == [] + + assert diop_DN(123, -2) == [(11, 1)] + # One calculator returned [(11, 1), (-11, 1)] but both of these are in + # the same equivalence class + assert equivalent(11, 1, -11, 1, 123, -2) + + assert diop_DN(123, -23) == [(-10, 1), (10, 1)] + + assert diop_DN(0, 0, t) == [(0, t)] + assert diop_DN(0, -1, t) == [] + + +def test_bf_pell(): + assert diop_bf_DN(13, -4) == [(3, 1), (-3, 1), (36, 10)] + assert diop_bf_DN(13, 27) == [(12, 3), (-12, 3), (40, 11), (-40, 11)] + assert diop_bf_DN(167, -2) == [] + assert diop_bf_DN(1729, 1) == [(44611924489705, 1072885712316)] + assert diop_bf_DN(89, -8) == [(9, 1), (-9, 1)] + assert diop_bf_DN(21257, -1) == [(13913102721304, 95427381109)] + assert diop_bf_DN(340, -4) == [(756, 41)] + assert diop_bf_DN(-1, 0, t) == [(0, 0)] + assert diop_bf_DN(0, 0, t) == [(0, t)] + assert diop_bf_DN(4, 0, t) == [(2*t, t), (-2*t, t)] + assert diop_bf_DN(3, 0, t) == [(0, 0)] + assert diop_bf_DN(1, -2, t) == [] + + +def test_length(): + assert length(2, 1, 0) == 1 + assert length(-2, 4, 5) == 3 + assert length(-5, 4, 17) == 4 + assert length(0, 4, 13) == 6 + assert length(7, 13, 11) == 23 + assert length(1, 6, 4) == 2 + + +def is_pell_transformation_ok(eq): + """ + Test whether X*Y, X, or Y terms are present in the equation + after transforming the equation using the transformation returned + by transformation_to_pell(). If they are not present we are good. + Moreover, coefficient of X**2 should be a divisor of coefficient of + Y**2 and the constant term. + """ + A, B = transformation_to_DN(eq) + u = (A*Matrix([X, Y]) + B)[0] + v = (A*Matrix([X, Y]) + B)[1] + simplified = diop_simplify(eq.subs(zip((x, y), (u, v)))) + + coeff = dict([reversed(t.as_independent(*[X, Y])) for t in simplified.args]) + + for term in [X*Y, X, Y]: + if term in coeff.keys(): + return False + + for term in [X**2, Y**2, 1]: + if term not in coeff.keys(): + coeff[term] = 0 + + if coeff[X**2] != 0: + return divisible(coeff[Y**2], coeff[X**2]) and \ + divisible(coeff[1], coeff[X**2]) + + return True + + +def test_transformation_to_pell(): + assert is_pell_transformation_ok(-13*x**2 - 7*x*y + y**2 + 2*x - 2*y - 14) + assert is_pell_transformation_ok(-17*x**2 + 19*x*y - 7*y**2 - 5*x - 13*y - 23) + assert is_pell_transformation_ok(x**2 - y**2 + 17) + assert is_pell_transformation_ok(-x**2 + 7*y**2 - 23) + assert is_pell_transformation_ok(25*x**2 - 45*x*y + 5*y**2 - 5*x - 10*y + 5) + assert is_pell_transformation_ok(190*x**2 + 30*x*y + y**2 - 3*y - 170*x - 130) + assert is_pell_transformation_ok(x**2 - 2*x*y -190*y**2 - 7*y - 23*x - 89) + assert is_pell_transformation_ok(15*x**2 - 9*x*y + 14*y**2 - 23*x - 14*y - 4950) + + +def test_find_DN(): + assert find_DN(x**2 - 2*x - y**2) == (1, 1) + assert find_DN(x**2 - 3*y**2 - 5) == (3, 5) + assert find_DN(x**2 - 2*x*y - 4*y**2 - 7) == (5, 7) + assert find_DN(4*x**2 - 8*x*y - y**2 - 9) == (20, 36) + assert find_DN(7*x**2 - 2*x*y - y**2 - 12) == (8, 84) + assert find_DN(-3*x**2 + 4*x*y -y**2) == (1, 0) + assert find_DN(-13*x**2 - 7*x*y + y**2 + 2*x - 2*y -14) == (101, -7825480) + + +def test_ldescent(): + # Equations which have solutions + u = ([(13, 23), (3, -11), (41, -113), (4, -7), (-7, 4), (91, -3), (1, 1), (1, -1), + (4, 32), (17, 13), (123689, 1), (19, -570)]) + for a, b in u: + w, x, y = ldescent(a, b) + assert a*x**2 + b*y**2 == w**2 + assert ldescent(-1, -1) is None + assert ldescent(2, 6) is None + + +def test_diop_ternary_quadratic_normal(): + assert check_solutions(234*x**2 - 65601*y**2 - z**2) + assert check_solutions(23*x**2 + 616*y**2 - z**2) + assert check_solutions(5*x**2 + 4*y**2 - z**2) + assert check_solutions(3*x**2 + 6*y**2 - 3*z**2) + assert check_solutions(x**2 + 3*y**2 - z**2) + assert check_solutions(4*x**2 + 5*y**2 - z**2) + assert check_solutions(x**2 + y**2 - z**2) + assert check_solutions(16*x**2 + y**2 - 25*z**2) + assert check_solutions(6*x**2 - y**2 + 10*z**2) + assert check_solutions(213*x**2 + 12*y**2 - 9*z**2) + assert check_solutions(34*x**2 - 3*y**2 - 301*z**2) + assert check_solutions(124*x**2 - 30*y**2 - 7729*z**2) + + +def is_normal_transformation_ok(eq): + A = transformation_to_normal(eq) + X, Y, Z = A*Matrix([x, y, z]) + simplified = diop_simplify(eq.subs(zip((x, y, z), (X, Y, Z)))) + + coeff = dict([reversed(t.as_independent(*[X, Y, Z])) for t in simplified.args]) + for term in [X*Y, Y*Z, X*Z]: + if term in coeff.keys(): + return False + + return True + + +def test_transformation_to_normal(): + assert is_normal_transformation_ok(x**2 + 3*y**2 + z**2 - 13*x*y - 16*y*z + 12*x*z) + assert is_normal_transformation_ok(x**2 + 3*y**2 - 100*z**2) + assert is_normal_transformation_ok(x**2 + 23*y*z) + assert is_normal_transformation_ok(3*y**2 - 100*z**2 - 12*x*y) + assert is_normal_transformation_ok(x**2 + 23*x*y - 34*y*z + 12*x*z) + assert is_normal_transformation_ok(z**2 + 34*x*y - 23*y*z + x*z) + assert is_normal_transformation_ok(x**2 + y**2 + z**2 - x*y - y*z - x*z) + assert is_normal_transformation_ok(x**2 + 2*y*z + 3*z**2) + assert is_normal_transformation_ok(x*y + 2*x*z + 3*y*z) + assert is_normal_transformation_ok(2*x*z + 3*y*z) + + +def test_diop_ternary_quadratic(): + assert check_solutions(2*x**2 + z**2 + y**2 - 4*x*y) + assert check_solutions(x**2 - y**2 - z**2 - x*y - y*z) + assert check_solutions(3*x**2 - x*y - y*z - x*z) + assert check_solutions(x**2 - y*z - x*z) + assert check_solutions(5*x**2 - 3*x*y - x*z) + assert check_solutions(4*x**2 - 5*y**2 - x*z) + assert check_solutions(3*x**2 + 2*y**2 - z**2 - 2*x*y + 5*y*z - 7*y*z) + assert check_solutions(8*x**2 - 12*y*z) + assert check_solutions(45*x**2 - 7*y**2 - 8*x*y - z**2) + assert check_solutions(x**2 - 49*y**2 - z**2 + 13*z*y -8*x*y) + assert check_solutions(90*x**2 + 3*y**2 + 5*x*y + 2*z*y + 5*x*z) + assert check_solutions(x**2 + 3*y**2 + z**2 - x*y - 17*y*z) + assert check_solutions(x**2 + 3*y**2 + z**2 - x*y - 16*y*z + 12*x*z) + assert check_solutions(x**2 + 3*y**2 + z**2 - 13*x*y - 16*y*z + 12*x*z) + assert check_solutions(x*y - 7*y*z + 13*x*z) + + assert diop_ternary_quadratic_normal(x**2 + y**2 + z**2) == (None, None, None) + assert diop_ternary_quadratic_normal(x**2 + y**2) is None + raises(ValueError, lambda: + _diop_ternary_quadratic_normal((x, y, z), + {x*y: 1, x**2: 2, y**2: 3, z**2: 0})) + eq = -2*x*y - 6*x*z + 7*y**2 - 3*y*z + 4*z**2 + assert diop_ternary_quadratic(eq) == (7, 2, 0) + assert diop_ternary_quadratic_normal(4*x**2 + 5*y**2 - z**2) == \ + (1, 0, 2) + assert diop_ternary_quadratic(x*y + 2*y*z) == \ + (-2, 0, n1) + eq = -5*x*y - 8*x*z - 3*y*z + 8*z**2 + assert parametrize_ternary_quadratic(eq) == \ + (8*p**2 - 3*p*q, -8*p*q + 8*q**2, 5*p*q) + # this cannot be tested with diophantine because it will + # factor into a product + assert diop_solve(x*y + 2*y*z) == (-2*p*q, -n1*p**2 + p**2, p*q) + + +def test_square_factor(): + assert square_factor(1) == square_factor(-1) == 1 + assert square_factor(0) == 1 + assert square_factor(5) == square_factor(-5) == 1 + assert square_factor(4) == square_factor(-4) == 2 + assert square_factor(12) == square_factor(-12) == 2 + assert square_factor(6) == 1 + assert square_factor(18) == 3 + assert square_factor(52) == 2 + assert square_factor(49) == 7 + assert square_factor(392) == 14 + assert square_factor(factorint(-12)) == 2 + + +def test_parametrize_ternary_quadratic(): + assert check_solutions(x**2 + y**2 - z**2) + assert check_solutions(x**2 + 2*x*y + z**2) + assert check_solutions(234*x**2 - 65601*y**2 - z**2) + assert check_solutions(3*x**2 + 2*y**2 - z**2 - 2*x*y + 5*y*z - 7*y*z) + assert check_solutions(x**2 - y**2 - z**2) + assert check_solutions(x**2 - 49*y**2 - z**2 + 13*z*y - 8*x*y) + assert check_solutions(8*x*y + z**2) + assert check_solutions(124*x**2 - 30*y**2 - 7729*z**2) + assert check_solutions(236*x**2 - 225*y**2 - 11*x*y - 13*y*z - 17*x*z) + assert check_solutions(90*x**2 + 3*y**2 + 5*x*y + 2*z*y + 5*x*z) + assert check_solutions(124*x**2 - 30*y**2 - 7729*z**2) + + +def test_no_square_ternary_quadratic(): + assert check_solutions(2*x*y + y*z - 3*x*z) + assert check_solutions(189*x*y - 345*y*z - 12*x*z) + assert check_solutions(23*x*y + 34*y*z) + assert check_solutions(x*y + y*z + z*x) + assert check_solutions(23*x*y + 23*y*z + 23*x*z) + + +def test_descent(): + + u = ([(13, 23), (3, -11), (41, -113), (91, -3), (1, 1), (1, -1), (17, 13), (123689, 1), (19, -570)]) + for a, b in u: + w, x, y = descent(a, b) + assert a*x**2 + b*y**2 == w**2 + # the docstring warns against bad input, so these are expected results + # - can't both be negative + raises(TypeError, lambda: descent(-1, -3)) + # A can't be zero unless B != 1 + raises(ZeroDivisionError, lambda: descent(0, 3)) + # supposed to be square-free + raises(TypeError, lambda: descent(4, 3)) + + +def test_diophantine(): + assert check_solutions((x - y)*(y - z)*(z - x)) + assert check_solutions((x - y)*(x**2 + y**2 - z**2)) + assert check_solutions((x - 3*y + 7*z)*(x**2 + y**2 - z**2)) + assert check_solutions(x**2 - 3*y**2 - 1) + assert check_solutions(y**2 + 7*x*y) + assert check_solutions(x**2 - 3*x*y + y**2) + assert check_solutions(z*(x**2 - y**2 - 15)) + assert check_solutions(x*(2*y - 2*z + 5)) + assert check_solutions((x**2 - 3*y**2 - 1)*(x**2 - y**2 - 15)) + assert check_solutions((x**2 - 3*y**2 - 1)*(y - 7*z)) + assert check_solutions((x**2 + y**2 - z**2)*(x - 7*y - 3*z + 4*w)) + # Following test case caused problems in parametric representation + # But this can be solved by factoring out y. + # No need to use methods for ternary quadratic equations. + assert check_solutions(y**2 - 7*x*y + 4*y*z) + assert check_solutions(x**2 - 2*x + 1) + + assert diophantine(x - y) == diophantine(Eq(x, y)) + # 18196 + eq = x**4 + y**4 - 97 + assert diophantine(eq, permute=True) == diophantine(-eq, permute=True) + assert diophantine(3*x*pi - 2*y*pi) == {(2*t_0, 3*t_0)} + eq = x**2 + y**2 + z**2 - 14 + base_sol = {(1, 2, 3)} + assert diophantine(eq) == base_sol + complete_soln = set(signed_permutations(base_sol.pop())) + assert diophantine(eq, permute=True) == complete_soln + + assert diophantine(x**2 + x*Rational(15, 14) - 3) == set() + # test issue 11049 + eq = 92*x**2 - 99*y**2 - z**2 + coeff = eq.as_coefficients_dict() + assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \ + {(9, 7, 51)} + assert diophantine(eq) == {( + 891*p**2 + 9*q**2, -693*p**2 - 102*p*q + 7*q**2, + 5049*p**2 - 1386*p*q - 51*q**2)} + eq = 2*x**2 + 2*y**2 - z**2 + coeff = eq.as_coefficients_dict() + assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \ + {(1, 1, 2)} + assert diophantine(eq) == {( + 2*p**2 - q**2, -2*p**2 + 4*p*q - q**2, + 4*p**2 - 4*p*q + 2*q**2)} + eq = 411*x**2+57*y**2-221*z**2 + coeff = eq.as_coefficients_dict() + assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \ + {(2021, 2645, 3066)} + assert diophantine(eq) == \ + {(115197*p**2 - 446641*q**2, -150765*p**2 + 1355172*p*q - + 584545*q**2, 174762*p**2 - 301530*p*q + 677586*q**2)} + eq = 573*x**2+267*y**2-984*z**2 + coeff = eq.as_coefficients_dict() + assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \ + {(49, 233, 127)} + assert diophantine(eq) == \ + {(4361*p**2 - 16072*q**2, -20737*p**2 + 83312*p*q - 76424*q**2, + 11303*p**2 - 41474*p*q + 41656*q**2)} + # this produces factors during reconstruction + eq = x**2 + 3*y**2 - 12*z**2 + coeff = eq.as_coefficients_dict() + assert _diop_ternary_quadratic_normal((x, y, z), coeff) == \ + {(0, 2, 1)} + assert diophantine(eq) == \ + {(24*p*q, 2*p**2 - 24*q**2, p**2 + 12*q**2)} + # solvers have not been written for every type + raises(NotImplementedError, lambda: diophantine(x*y**2 + 1)) + + # rational expressions + assert diophantine(1/x) == set() + assert diophantine(1/x + 1/y - S.Half) == {(6, 3), (-2, 1), (4, 4), (1, -2), (3, 6)} + assert diophantine(x**2 + y**2 +3*x- 5, permute=True) == \ + {(-1, 1), (-4, -1), (1, -1), (1, 1), (-4, 1), (-1, -1), (4, 1), (4, -1)} + + + #test issue 18186 + assert diophantine(y**4 + x**4 - 2**4 - 3**4, syms=(x, y), permute=True) == \ + {(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)} + assert diophantine(y**4 + x**4 - 2**4 - 3**4, syms=(y, x), permute=True) == \ + {(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)} + + # issue 18122 + assert check_solutions(x**2 - y) + assert check_solutions(y**2 - x) + assert diophantine((x**2 - y), t) == {(t, t**2)} + assert diophantine((y**2 - x), t) == {(t**2, t)} + + +def test_general_pythagorean(): + from sympy.abc import a, b, c, d, e + + assert check_solutions(a**2 + b**2 + c**2 - d**2) + assert check_solutions(a**2 + 4*b**2 + 4*c**2 - d**2) + assert check_solutions(9*a**2 + 4*b**2 + 4*c**2 - d**2) + assert check_solutions(9*a**2 + 4*b**2 - 25*d**2 + 4*c**2 ) + assert check_solutions(9*a**2 - 16*d**2 + 4*b**2 + 4*c**2) + assert check_solutions(-e**2 + 9*a**2 + 4*b**2 + 4*c**2 + 25*d**2) + assert check_solutions(16*a**2 - b**2 + 9*c**2 + d**2 + 25*e**2) + + assert GeneralPythagorean(a**2 + b**2 + c**2 - d**2).solve(parameters=[x, y, z]) == \ + {(x**2 + y**2 - z**2, 2*x*z, 2*y*z, x**2 + y**2 + z**2)} + + +def test_diop_general_sum_of_squares_quick(): + for i in range(3, 10): + assert check_solutions(sum(i**2 for i in symbols(':%i' % i)) - i) + + assert diop_general_sum_of_squares(x**2 + y**2 - 2) is None + assert diop_general_sum_of_squares(x**2 + y**2 + z**2 + 2) == set() + eq = x**2 + y**2 + z**2 - (1 + 4 + 9) + assert diop_general_sum_of_squares(eq) == \ + {(1, 2, 3)} + eq = u**2 + v**2 + x**2 + y**2 + z**2 - 1313 + assert len(diop_general_sum_of_squares(eq, 3)) == 3 + # issue 11016 + var = symbols(':5') + (symbols('6', negative=True),) + eq = Add(*[i**2 for i in var]) - 112 + + base_soln = {(0, 1, 1, 5, 6, -7), (1, 1, 1, 3, 6, -8), (2, 3, 3, 4, 5, -7), (0, 1, 1, 1, 3, -10), + (0, 0, 4, 4, 4, -8), (1, 2, 3, 3, 5, -8), (0, 1, 2, 3, 7, -7), (2, 2, 4, 4, 6, -6), + (1, 1, 3, 4, 6, -7), (0, 2, 3, 3, 3, -9), (0, 0, 2, 2, 2, -10), (1, 1, 2, 3, 4, -9), + (0, 1, 1, 2, 5, -9), (0, 0, 2, 6, 6, -6), (1, 3, 4, 5, 5, -6), (0, 2, 2, 2, 6, -8), + (0, 3, 3, 3, 6, -7), (0, 2, 3, 5, 5, -7), (0, 1, 5, 5, 5, -6)} + assert diophantine(eq) == base_soln + assert len(diophantine(eq, permute=True)) == 196800 + + # handle negated squares with signsimp + assert diophantine(12 - x**2 - y**2 - z**2) == {(2, 2, 2)} + # diophantine handles simplification, so classify_diop should + # not have to look for additional patterns that are removed + # by diophantine + eq = a**2 + b**2 + c**2 + d**2 - 4 + raises(NotImplementedError, lambda: classify_diop(-eq)) + + +def test_issue_23807(): + # fixes recursion error + eq = x**2 + y**2 + z**2 - 1000000 + base_soln = {(0, 0, 1000), (0, 352, 936), (480, 600, 640), (24, 640, 768), (192, 640, 744), + (192, 480, 856), (168, 224, 960), (0, 600, 800), (280, 576, 768), (152, 480, 864), + (0, 280, 960), (352, 360, 864), (424, 480, 768), (360, 480, 800), (224, 600, 768), + (96, 360, 928), (168, 576, 800), (96, 480, 872)} + + assert diophantine(eq) == base_soln + + +def test_diop_partition(): + for n in [8, 10]: + for k in range(1, 8): + for p in partition(n, k): + assert len(p) == k + assert list(partition(3, 5)) == [] + assert [list(p) for p in partition(3, 5, 1)] == [ + [0, 0, 0, 0, 3], [0, 0, 0, 1, 2], [0, 0, 1, 1, 1]] + assert list(partition(0)) == [()] + assert list(partition(1, 0)) == [()] + assert [list(i) for i in partition(3)] == [[1, 1, 1], [1, 2], [3]] + + +def test_prime_as_sum_of_two_squares(): + for i in [5, 13, 17, 29, 37, 41, 2341, 3557, 34841, 64601]: + a, b = prime_as_sum_of_two_squares(i) + assert a**2 + b**2 == i + assert prime_as_sum_of_two_squares(7) is None + ans = prime_as_sum_of_two_squares(800029) + assert ans == (450, 773) and type(ans[0]) is int + + +def test_sum_of_three_squares(): + for i in [0, 1, 2, 34, 123, 34304595905, 34304595905394941, 343045959052344, + 800, 801, 802, 803, 804, 805, 806]: + a, b, c = sum_of_three_squares(i) + assert a**2 + b**2 + c**2 == i + assert a >= 0 + + # error + raises(ValueError, lambda: sum_of_three_squares(-1)) + + assert sum_of_three_squares(7) is None + assert sum_of_three_squares((4**5)*15) is None + # if there are two zeros, there might be a solution + # with only one zero, e.g. 25 => (0, 3, 4) or + # with no zeros, e.g. 49 => (2, 3, 6) + assert sum_of_three_squares(25) == (0, 0, 5) + assert sum_of_three_squares(4) == (0, 0, 2) + + +def test_sum_of_four_squares(): + from sympy.core.random import randint + + # this should never fail + n = randint(1, 100000000000000) + assert sum(i**2 for i in sum_of_four_squares(n)) == n + + # error + raises(ValueError, lambda: sum_of_four_squares(-1)) + + for n in range(1000): + result = sum_of_four_squares(n) + assert len(result) == 4 + assert all(r >= 0 for r in result) + assert sum(r**2 for r in result) == n + assert list(result) == sorted(result) + + +def test_power_representation(): + tests = [(1729, 3, 2), (234, 2, 4), (2, 1, 2), (3, 1, 3), (5, 2, 2), (12352, 2, 4), + (32760, 2, 3)] + + for test in tests: + n, p, k = test + f = power_representation(n, p, k) + + while True: + try: + l = next(f) + assert len(l) == k + + chk_sum = 0 + for l_i in l: + chk_sum = chk_sum + l_i**p + assert chk_sum == n + + except StopIteration: + break + + assert list(power_representation(20, 2, 4, True)) == \ + [(1, 1, 3, 3), (0, 0, 2, 4)] + raises(ValueError, lambda: list(power_representation(1.2, 2, 2))) + raises(ValueError, lambda: list(power_representation(2, 0, 2))) + raises(ValueError, lambda: list(power_representation(2, 2, 0))) + assert list(power_representation(-1, 2, 2)) == [] + assert list(power_representation(1, 1, 1)) == [(1,)] + assert list(power_representation(3, 2, 1)) == [] + assert list(power_representation(4, 2, 1)) == [(2,)] + assert list(power_representation(3**4, 4, 6, zeros=True)) == \ + [(1, 2, 2, 2, 2, 2), (0, 0, 0, 0, 0, 3)] + assert list(power_representation(3**4, 4, 5, zeros=False)) == [] + assert list(power_representation(-2, 3, 2)) == [(-1, -1)] + assert list(power_representation(-2, 4, 2)) == [] + assert list(power_representation(0, 3, 2, True)) == [(0, 0)] + assert list(power_representation(0, 3, 2, False)) == [] + # when we are dealing with squares, do feasibility checks + assert len(list(power_representation(4**10*(8*10 + 7), 2, 3))) == 0 + # there will be a recursion error if these aren't recognized + big = 2**30 + for i in [13, 10, 7, 5, 4, 2, 1]: + assert list(sum_of_powers(big, 2, big - i)) == [] + + +def test_assumptions(): + """ + Test whether diophantine respects the assumptions. + """ + #Test case taken from the below so question regarding assumptions in diophantine module + #https://stackoverflow.com/questions/23301941/how-can-i-declare-natural-symbols-with-sympy + m, n = symbols('m n', integer=True, positive=True) + diof = diophantine(n**2 + m*n - 500) + assert diof == {(5, 20), (40, 10), (95, 5), (121, 4), (248, 2), (499, 1)} + + a, b = symbols('a b', integer=True, positive=False) + diof = diophantine(a*b + 2*a + 3*b - 6) + assert diof == {(-15, -3), (-9, -4), (-7, -5), (-6, -6), (-5, -8), (-4, -14)} + + +def check_solutions(eq): + """ + Determines whether solutions returned by diophantine() satisfy the original + equation. Hope to generalize this so we can remove functions like check_ternay_quadratic, + check_solutions_normal, check_solutions() + """ + s = diophantine(eq) + + factors = Mul.make_args(eq) + + var = list(eq.free_symbols) + var.sort(key=default_sort_key) + + while s: + solution = s.pop() + for f in factors: + if diop_simplify(f.subs(zip(var, solution))) == 0: + break + else: + return False + return True + + +def test_diopcoverage(): + eq = (2*x + y + 1)**2 + assert diop_solve(eq) == {(t_0, -2*t_0 - 1)} + eq = 2*x**2 + 6*x*y + 12*x + 4*y**2 + 18*y + 18 + assert diop_solve(eq) == {(t, -t - 3), (-2*t - 3, t)} + assert diop_quadratic(x + y**2 - 3) == {(-t**2 + 3, t)} + + assert diop_linear(x + y - 3) == (t_0, 3 - t_0) + + assert base_solution_linear(0, 1, 2, t=None) == (0, 0) + ans = (3*t - 1, -2*t + 1) + assert base_solution_linear(4, 8, 12, t) == ans + assert base_solution_linear(4, 8, 12, t=None) == tuple(_.subs(t, 0) for _ in ans) + + assert cornacchia(1, 1, 20) == set() + assert cornacchia(1, 1, 5) == {(2, 1)} + assert cornacchia(1, 2, 17) == {(3, 2)} + + raises(ValueError, lambda: reconstruct(4, 20, 1)) + + assert gaussian_reduce(4, 1, 3) == (1, 1) + eq = -w**2 - x**2 - y**2 + z**2 + + assert diop_general_pythagorean(eq) == \ + diop_general_pythagorean(-eq) == \ + (m1**2 + m2**2 - m3**2, 2*m1*m3, + 2*m2*m3, m1**2 + m2**2 + m3**2) + + assert len(check_param(S(3) + x/3, S(4) + x/2, S(2), [x])) == 0 + assert len(check_param(Rational(3, 2), S(4) + x, S(2), [x])) == 0 + assert len(check_param(S(4) + x, Rational(3, 2), S(2), [x])) == 0 + + assert _nint_or_floor(16, 10) == 2 + assert _odd(1) == (not _even(1)) == True + assert _odd(0) == (not _even(0)) == False + assert _remove_gcd(2, 4, 6) == (1, 2, 3) + raises(TypeError, lambda: _remove_gcd((2, 4, 6))) + assert sqf_normal(2*3**2*5, 2*5*11, 2*7**2*11) == \ + (11, 1, 5) + + # it's ok if these pass some day when the solvers are implemented + raises(NotImplementedError, lambda: diophantine(x**2 + y**2 + x*y + 2*y*z - 12)) + raises(NotImplementedError, lambda: diophantine(x**3 + y**2)) + assert diop_quadratic(x**2 + y**2 - 1**2 - 3**4) == \ + {(-9, -1), (-9, 1), (-1, -9), (-1, 9), (1, -9), (1, 9), (9, -1), (9, 1)} + + +def test_holzer(): + # if the input is good, don't let it diverge in holzer() + # (but see test_fail_holzer below) + assert holzer(2, 7, 13, 4, 79, 23) == (2, 7, 13) + + # None in uv condition met; solution is not Holzer reduced + # so this will hopefully change but is here for coverage + assert holzer(2, 6, 2, 1, 1, 10) == (2, 6, 2) + + raises(ValueError, lambda: holzer(2, 7, 14, 4, 79, 23)) + + +@XFAIL +def test_fail_holzer(): + eq = lambda x, y, z: a*x**2 + b*y**2 - c*z**2 + a, b, c = 4, 79, 23 + x, y, z = xyz = 26, 1, 11 + X, Y, Z = ans = 2, 7, 13 + assert eq(*xyz) == 0 + assert eq(*ans) == 0 + assert max(a*x**2, b*y**2, c*z**2) <= a*b*c + assert max(a*X**2, b*Y**2, c*Z**2) <= a*b*c + h = holzer(x, y, z, a, b, c) + assert h == ans # it would be nice to get the smaller soln + + +def test_issue_9539(): + assert diophantine(6*w + 9*y + 20*x - z) == \ + {(t_0, t_1, t_1 + t_2, 6*t_0 + 29*t_1 + 9*t_2)} + + +def test_issue_8943(): + assert diophantine( + 3*(x**2 + y**2 + z**2) - 14*(x*y + y*z + z*x)) == \ + {(0, 0, 0)} + + +def test_diop_sum_of_even_powers(): + eq = x**4 + y**4 + z**4 - 2673 + assert diop_solve(eq) == {(3, 6, 6), (2, 4, 7)} + assert diop_general_sum_of_even_powers(eq, 2) == {(3, 6, 6), (2, 4, 7)} + raises(NotImplementedError, lambda: diop_general_sum_of_even_powers(-eq, 2)) + neg = symbols('neg', negative=True) + eq = x**4 + y**4 + neg**4 - 2673 + assert diop_general_sum_of_even_powers(eq) == {(-3, 6, 6)} + assert diophantine(x**4 + y**4 + 2) == set() + assert diop_general_sum_of_even_powers(x**4 + y**4 - 2, limit=0) == set() + + +def test_sum_of_squares_powers(): + tru = {(0, 0, 1, 1, 11), (0, 0, 5, 7, 7), (0, 1, 3, 7, 8), (0, 1, 4, 5, 9), (0, 3, 4, 7, 7), (0, 3, 5, 5, 8), + (1, 1, 2, 6, 9), (1, 1, 6, 6, 7), (1, 2, 3, 3, 10), (1, 3, 4, 4, 9), (1, 5, 5, 6, 6), (2, 2, 3, 5, 9), + (2, 3, 5, 6, 7), (3, 3, 4, 5, 8)} + eq = u**2 + v**2 + x**2 + y**2 + z**2 - 123 + ans = diop_general_sum_of_squares(eq, oo) # allow oo to be used + assert len(ans) == 14 + assert ans == tru + + raises(ValueError, lambda: list(sum_of_squares(10, -1))) + assert list(sum_of_squares(1, 1)) == [(1,)] + assert list(sum_of_squares(1, 2)) == [] + assert list(sum_of_squares(1, 2, True)) == [(0, 1)] + assert list(sum_of_squares(-10, 2)) == [] + assert list(sum_of_squares(2, 3)) == [] + assert list(sum_of_squares(0, 3, True)) == [(0, 0, 0)] + assert list(sum_of_squares(0, 3)) == [] + assert list(sum_of_squares(4, 1)) == [(2,)] + assert list(sum_of_squares(5, 1)) == [] + assert list(sum_of_squares(50, 2)) == [(5, 5), (1, 7)] + assert list(sum_of_squares(11, 5, True)) == [ + (1, 1, 1, 2, 2), (0, 0, 1, 1, 3)] + assert list(sum_of_squares(8, 8)) == [(1, 1, 1, 1, 1, 1, 1, 1)] + + assert [len(list(sum_of_squares(i, 5, True))) for i in range(30)] == [ + 1, 1, 1, 1, 2, + 2, 1, 1, 2, 2, + 2, 2, 2, 3, 2, + 1, 3, 3, 3, 3, + 4, 3, 3, 2, 2, + 4, 4, 4, 4, 5] + assert [len(list(sum_of_squares(i, 5))) for i in range(30)] == [ + 0, 0, 0, 0, 0, + 1, 0, 0, 1, 0, + 0, 1, 0, 1, 1, + 0, 1, 1, 0, 1, + 2, 1, 1, 1, 1, + 1, 1, 1, 1, 3] + for i in range(30): + s1 = set(sum_of_squares(i, 5, True)) + assert not s1 or all(sum(j**2 for j in t) == i for t in s1) + s2 = set(sum_of_squares(i, 5)) + assert all(sum(j**2 for j in t) == i for t in s2) + + raises(ValueError, lambda: list(sum_of_powers(2, -1, 1))) + raises(ValueError, lambda: list(sum_of_powers(2, 1, -1))) + assert list(sum_of_powers(-2, 3, 2)) == [(-1, -1)] + assert list(sum_of_powers(-2, 4, 2)) == [] + assert list(sum_of_powers(2, 1, 1)) == [(2,)] + assert list(sum_of_powers(2, 1, 3, True)) == [(0, 0, 2), (0, 1, 1)] + assert list(sum_of_powers(5, 1, 2, True)) == [(0, 5), (1, 4), (2, 3)] + assert list(sum_of_powers(6, 2, 2)) == [] + assert list(sum_of_powers(3**5, 3, 1)) == [] + assert list(sum_of_powers(3**6, 3, 1)) == [(9,)] and (9**3 == 3**6) + assert list(sum_of_powers(2**1000, 5, 2)) == [] + + +def test__can_do_sum_of_squares(): + assert _can_do_sum_of_squares(3, -1) is False + assert _can_do_sum_of_squares(-3, 1) is False + assert _can_do_sum_of_squares(0, 1) + assert _can_do_sum_of_squares(4, 1) + assert _can_do_sum_of_squares(1, 2) + assert _can_do_sum_of_squares(2, 2) + assert _can_do_sum_of_squares(3, 2) is False + + +def test_diophantine_permute_sign(): + from sympy.abc import a, b, c, d, e + eq = a**4 + b**4 - (2**4 + 3**4) + base_sol = {(2, 3)} + assert diophantine(eq) == base_sol + complete_soln = set(signed_permutations(base_sol.pop())) + assert diophantine(eq, permute=True) == complete_soln + + eq = a**2 + b**2 + c**2 + d**2 + e**2 - 234 + assert len(diophantine(eq)) == 35 + assert len(diophantine(eq, permute=True)) == 62000 + soln = {(-1, -1), (-1, 2), (1, -2), (1, 1)} + assert diophantine(10*x**2 + 12*x*y + 12*y**2 - 34, permute=True) == soln + + +@XFAIL +def test_not_implemented(): + eq = x**2 + y**4 - 1**2 - 3**4 + assert diophantine(eq, syms=[x, y]) == {(9, 1), (1, 3)} + + +def test_issue_9538(): + eq = x - 3*y + 2 + assert diophantine(eq, syms=[y,x]) == {(t_0, 3*t_0 - 2)} + raises(TypeError, lambda: diophantine(eq, syms={y, x})) + + +def test_ternary_quadratic(): + # solution with 3 parameters + s = diophantine(2*x**2 + y**2 - 2*z**2) + p, q, r = ordered(S(s).free_symbols) + assert s == {( + p**2 - 2*q**2, + -2*p**2 + 4*p*q - 4*p*r - 4*q**2, + p**2 - 4*p*q + 2*q**2 - 4*q*r)} + # solution with Mul in solution + s = diophantine(x**2 + 2*y**2 - 2*z**2) + assert s == {(4*p*q, p**2 - 2*q**2, p**2 + 2*q**2)} + # solution with no Mul in solution + s = diophantine(2*x**2 + 2*y**2 - z**2) + assert s == {(2*p**2 - q**2, -2*p**2 + 4*p*q - q**2, + 4*p**2 - 4*p*q + 2*q**2)} + # reduced form when parametrized + s = diophantine(3*x**2 + 72*y**2 - 27*z**2) + assert s == {(24*p**2 - 9*q**2, 6*p*q, 8*p**2 + 3*q**2)} + assert parametrize_ternary_quadratic( + 3*x**2 + 2*y**2 - z**2 - 2*x*y + 5*y*z - 7*y*z) == ( + 2*p**2 - 2*p*q - q**2, 2*p**2 + 2*p*q - q**2, 2*p**2 - + 2*p*q + 3*q**2) + assert parametrize_ternary_quadratic( + 124*x**2 - 30*y**2 - 7729*z**2) == ( + -1410*p**2 - 363263*q**2, 2700*p**2 + 30916*p*q - + 695610*q**2, -60*p**2 + 5400*p*q + 15458*q**2) + + +def test_diophantine_solution_set(): + s1 = DiophantineSolutionSet([], []) + assert set(s1) == set() + assert s1.symbols == () + assert s1.parameters == () + raises(ValueError, lambda: s1.add((x,))) + assert list(s1.dict_iterator()) == [] + + s2 = DiophantineSolutionSet([x, y], [t, u]) + assert s2.symbols == (x, y) + assert s2.parameters == (t, u) + raises(ValueError, lambda: s2.add((1,))) + s2.add((3, 4)) + assert set(s2) == {(3, 4)} + s2.update((3, 4), (-1, u)) + assert set(s2) == {(3, 4), (-1, u)} + raises(ValueError, lambda: s1.update(s2)) + assert list(s2.dict_iterator()) == [{x: -1, y: u}, {x: 3, y: 4}] + + s3 = DiophantineSolutionSet([x, y, z], [t, u]) + assert len(s3.parameters) == 2 + s3.add((t**2 + u, t - u, 1)) + assert set(s3) == {(t**2 + u, t - u, 1)} + assert s3.subs(t, 2) == {(u + 4, 2 - u, 1)} + assert s3(2) == {(u + 4, 2 - u, 1)} + assert s3.subs({t: 7, u: 8}) == {(57, -1, 1)} + assert s3(7, 8) == {(57, -1, 1)} + assert s3.subs({t: 5}) == {(u + 25, 5 - u, 1)} + assert s3(5) == {(u + 25, 5 - u, 1)} + assert s3.subs(u, -3) == {(t**2 - 3, t + 3, 1)} + assert s3(None, -3) == {(t**2 - 3, t + 3, 1)} + assert s3.subs({t: 2, u: 8}) == {(12, -6, 1)} + assert s3(2, 8) == {(12, -6, 1)} + assert s3.subs({t: 5, u: -3}) == {(22, 8, 1)} + assert s3(5, -3) == {(22, 8, 1)} + raises(ValueError, lambda: s3.subs(x=1)) + raises(ValueError, lambda: s3.subs(1, 2, 3)) + raises(ValueError, lambda: s3.add(())) + raises(ValueError, lambda: s3.add((1, 2, 3, 4))) + raises(ValueError, lambda: s3.add((1, 2))) + raises(ValueError, lambda: s3(1, 2, 3)) + raises(TypeError, lambda: s3(t=1)) + + s4 = DiophantineSolutionSet([x, y], [t, u]) + s4.add((t, 11*t)) + s4.add((-t, 22*t)) + assert s4(0, 0) == {(0, 0)} + + +def test_quadratic_parameter_passing(): + eq = -33*x*y + 3*y**2 + solution = BinaryQuadratic(eq).solve(parameters=[t, u]) + # test that parameters are passed all the way to the final solution + assert solution == {(t, 11*t), (t, -22*t)} + assert solution(0, 0) == {(0, 0)} diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/inequalities.py b/pllava/lib/python3.10/site-packages/sympy/solvers/inequalities.py new file mode 100644 index 0000000000000000000000000000000000000000..9a5cf9737f7e66b98949aa93ffd3adb44c144b1a --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/inequalities.py @@ -0,0 +1,986 @@ +"""Tools for solving inequalities and systems of inequalities. """ +import itertools + +from sympy.calculus.util import (continuous_domain, periodicity, + function_range) +from sympy.core import sympify +from sympy.core.exprtools import factor_terms +from sympy.core.relational import Relational, Lt, Ge, Eq +from sympy.core.symbol import Symbol, Dummy +from sympy.sets.sets import Interval, FiniteSet, Union, Intersection +from sympy.core.singleton import S +from sympy.core.function import expand_mul +from sympy.functions.elementary.complexes import Abs +from sympy.logic import And +from sympy.polys import Poly, PolynomialError, parallel_poly_from_expr +from sympy.polys.polyutils import _nsort +from sympy.solvers.solveset import solvify, solveset +from sympy.utilities.iterables import sift, iterable +from sympy.utilities.misc import filldedent + + +def solve_poly_inequality(poly, rel): + """Solve a polynomial inequality with rational coefficients. + + Examples + ======== + + >>> from sympy import solve_poly_inequality, Poly + >>> from sympy.abc import x + + >>> solve_poly_inequality(Poly(x, x, domain='ZZ'), '==') + [{0}] + + >>> solve_poly_inequality(Poly(x**2 - 1, x, domain='ZZ'), '!=') + [Interval.open(-oo, -1), Interval.open(-1, 1), Interval.open(1, oo)] + + >>> solve_poly_inequality(Poly(x**2 - 1, x, domain='ZZ'), '==') + [{-1}, {1}] + + See Also + ======== + solve_poly_inequalities + """ + if not isinstance(poly, Poly): + raise ValueError( + 'For efficiency reasons, `poly` should be a Poly instance') + if poly.as_expr().is_number: + t = Relational(poly.as_expr(), 0, rel) + if t is S.true: + return [S.Reals] + elif t is S.false: + return [S.EmptySet] + else: + raise NotImplementedError( + "could not determine truth value of %s" % t) + + reals, intervals = poly.real_roots(multiple=False), [] + + if rel == '==': + for root, _ in reals: + interval = Interval(root, root) + intervals.append(interval) + elif rel == '!=': + left = S.NegativeInfinity + + for right, _ in reals + [(S.Infinity, 1)]: + interval = Interval(left, right, True, True) + intervals.append(interval) + left = right + else: + if poly.LC() > 0: + sign = +1 + else: + sign = -1 + + eq_sign, equal = None, False + + if rel == '>': + eq_sign = +1 + elif rel == '<': + eq_sign = -1 + elif rel == '>=': + eq_sign, equal = +1, True + elif rel == '<=': + eq_sign, equal = -1, True + else: + raise ValueError("'%s' is not a valid relation" % rel) + + right, right_open = S.Infinity, True + + for left, multiplicity in reversed(reals): + if multiplicity % 2: + if sign == eq_sign: + intervals.insert( + 0, Interval(left, right, not equal, right_open)) + + sign, right, right_open = -sign, left, not equal + else: + if sign == eq_sign and not equal: + intervals.insert( + 0, Interval(left, right, True, right_open)) + right, right_open = left, True + elif sign != eq_sign and equal: + intervals.insert(0, Interval(left, left)) + + if sign == eq_sign: + intervals.insert( + 0, Interval(S.NegativeInfinity, right, True, right_open)) + + return intervals + + +def solve_poly_inequalities(polys): + """Solve polynomial inequalities with rational coefficients. + + Examples + ======== + + >>> from sympy import Poly + >>> from sympy.solvers.inequalities import solve_poly_inequalities + >>> from sympy.abc import x + >>> solve_poly_inequalities((( + ... Poly(x**2 - 3), ">"), ( + ... Poly(-x**2 + 1), ">"))) + Union(Interval.open(-oo, -sqrt(3)), Interval.open(-1, 1), Interval.open(sqrt(3), oo)) + """ + return Union(*[s for p in polys for s in solve_poly_inequality(*p)]) + + +def solve_rational_inequalities(eqs): + """Solve a system of rational inequalities with rational coefficients. + + Examples + ======== + + >>> from sympy.abc import x + >>> from sympy import solve_rational_inequalities, Poly + + >>> solve_rational_inequalities([[ + ... ((Poly(-x + 1), Poly(1, x)), '>='), + ... ((Poly(-x + 1), Poly(1, x)), '<=')]]) + {1} + + >>> solve_rational_inequalities([[ + ... ((Poly(x), Poly(1, x)), '!='), + ... ((Poly(-x + 1), Poly(1, x)), '>=')]]) + Union(Interval.open(-oo, 0), Interval.Lopen(0, 1)) + + See Also + ======== + solve_poly_inequality + """ + result = S.EmptySet + + for _eqs in eqs: + if not _eqs: + continue + + global_intervals = [Interval(S.NegativeInfinity, S.Infinity)] + + for (numer, denom), rel in _eqs: + numer_intervals = solve_poly_inequality(numer*denom, rel) + denom_intervals = solve_poly_inequality(denom, '==') + + intervals = [] + + for numer_interval, global_interval in itertools.product( + numer_intervals, global_intervals): + interval = numer_interval.intersect(global_interval) + + if interval is not S.EmptySet: + intervals.append(interval) + + global_intervals = intervals + + intervals = [] + + for global_interval in global_intervals: + for denom_interval in denom_intervals: + global_interval -= denom_interval + + if global_interval is not S.EmptySet: + intervals.append(global_interval) + + global_intervals = intervals + + if not global_intervals: + break + + for interval in global_intervals: + result = result.union(interval) + + return result + + +def reduce_rational_inequalities(exprs, gen, relational=True): + """Reduce a system of rational inequalities with rational coefficients. + + Examples + ======== + + >>> from sympy import Symbol + >>> from sympy.solvers.inequalities import reduce_rational_inequalities + + >>> x = Symbol('x', real=True) + + >>> reduce_rational_inequalities([[x**2 <= 0]], x) + Eq(x, 0) + + >>> reduce_rational_inequalities([[x + 2 > 0]], x) + -2 < x + >>> reduce_rational_inequalities([[(x + 2, ">")]], x) + -2 < x + >>> reduce_rational_inequalities([[x + 2]], x) + Eq(x, -2) + + This function find the non-infinite solution set so if the unknown symbol + is declared as extended real rather than real then the result may include + finiteness conditions: + + >>> y = Symbol('y', extended_real=True) + >>> reduce_rational_inequalities([[y + 2 > 0]], y) + (-2 < y) & (y < oo) + """ + exact = True + eqs = [] + solution = S.EmptySet # add pieces for each group + for _exprs in exprs: + if not _exprs: + continue + _eqs = [] + _sol = S.Reals + for expr in _exprs: + if isinstance(expr, tuple): + expr, rel = expr + else: + if expr.is_Relational: + expr, rel = expr.lhs - expr.rhs, expr.rel_op + else: + expr, rel = expr, '==' + + if expr is S.true: + numer, denom, rel = S.Zero, S.One, '==' + elif expr is S.false: + numer, denom, rel = S.One, S.One, '==' + else: + numer, denom = expr.together().as_numer_denom() + + try: + (numer, denom), opt = parallel_poly_from_expr( + (numer, denom), gen) + except PolynomialError: + raise PolynomialError(filldedent(''' + only polynomials and rational functions are + supported in this context. + ''')) + + if not opt.domain.is_Exact: + numer, denom, exact = numer.to_exact(), denom.to_exact(), False + + domain = opt.domain.get_exact() + + if not (domain.is_ZZ or domain.is_QQ): + expr = numer/denom + expr = Relational(expr, 0, rel) + _sol &= solve_univariate_inequality(expr, gen, relational=False) + else: + _eqs.append(((numer, denom), rel)) + + if _eqs: + _sol &= solve_rational_inequalities([_eqs]) + exclude = solve_rational_inequalities([[((d, d.one), '==') + for i in eqs for ((n, d), _) in i if d.has(gen)]]) + _sol -= exclude + + solution |= _sol + + if not exact and solution: + solution = solution.evalf() + + if relational: + solution = solution.as_relational(gen) + + return solution + + +def reduce_abs_inequality(expr, rel, gen): + """Reduce an inequality with nested absolute values. + + Examples + ======== + + >>> from sympy import reduce_abs_inequality, Abs, Symbol + >>> x = Symbol('x', real=True) + + >>> reduce_abs_inequality(Abs(x - 5) - 3, '<', x) + (2 < x) & (x < 8) + + >>> reduce_abs_inequality(Abs(x + 2)*3 - 13, '<', x) + (-19/3 < x) & (x < 7/3) + + See Also + ======== + + reduce_abs_inequalities + """ + if gen.is_extended_real is False: + raise TypeError(filldedent(''' + Cannot solve inequalities with absolute values containing + non-real variables. + ''')) + + def _bottom_up_scan(expr): + exprs = [] + + if expr.is_Add or expr.is_Mul: + op = expr.func + + for arg in expr.args: + _exprs = _bottom_up_scan(arg) + + if not exprs: + exprs = _exprs + else: + exprs = [(op(expr, _expr), conds + _conds) for (expr, conds), (_expr, _conds) in + itertools.product(exprs, _exprs)] + elif expr.is_Pow: + n = expr.exp + if not n.is_Integer: + raise ValueError("Only Integer Powers are allowed on Abs.") + + exprs.extend((expr**n, conds) for expr, conds in _bottom_up_scan(expr.base)) + elif isinstance(expr, Abs): + _exprs = _bottom_up_scan(expr.args[0]) + + for expr, conds in _exprs: + exprs.append(( expr, conds + [Ge(expr, 0)])) + exprs.append((-expr, conds + [Lt(expr, 0)])) + else: + exprs = [(expr, [])] + + return exprs + + mapping = {'<': '>', '<=': '>='} + inequalities = [] + + for expr, conds in _bottom_up_scan(expr): + if rel not in mapping.keys(): + expr = Relational( expr, 0, rel) + else: + expr = Relational(-expr, 0, mapping[rel]) + + inequalities.append([expr] + conds) + + return reduce_rational_inequalities(inequalities, gen) + + +def reduce_abs_inequalities(exprs, gen): + """Reduce a system of inequalities with nested absolute values. + + Examples + ======== + + >>> from sympy import reduce_abs_inequalities, Abs, Symbol + >>> x = Symbol('x', extended_real=True) + + >>> reduce_abs_inequalities([(Abs(3*x - 5) - 7, '<'), + ... (Abs(x + 25) - 13, '>')], x) + (-2/3 < x) & (x < 4) & (((-oo < x) & (x < -38)) | ((-12 < x) & (x < oo))) + + >>> reduce_abs_inequalities([(Abs(x - 4) + Abs(3*x - 5) - 7, '<')], x) + (1/2 < x) & (x < 4) + + See Also + ======== + + reduce_abs_inequality + """ + return And(*[ reduce_abs_inequality(expr, rel, gen) + for expr, rel in exprs ]) + + +def solve_univariate_inequality(expr, gen, relational=True, domain=S.Reals, continuous=False): + """Solves a real univariate inequality. + + Parameters + ========== + + expr : Relational + The target inequality + gen : Symbol + The variable for which the inequality is solved + relational : bool + A Relational type output is expected or not + domain : Set + The domain over which the equation is solved + continuous: bool + True if expr is known to be continuous over the given domain + (and so continuous_domain() does not need to be called on it) + + Raises + ====== + + NotImplementedError + The solution of the inequality cannot be determined due to limitation + in :func:`sympy.solvers.solveset.solvify`. + + Notes + ===== + + Currently, we cannot solve all the inequalities due to limitations in + :func:`sympy.solvers.solveset.solvify`. Also, the solution returned for trigonometric inequalities + are restricted in its periodic interval. + + See Also + ======== + + sympy.solvers.solveset.solvify: solver returning solveset solutions with solve's output API + + Examples + ======== + + >>> from sympy import solve_univariate_inequality, Symbol, sin, Interval, S + >>> x = Symbol('x') + + >>> solve_univariate_inequality(x**2 >= 4, x) + ((2 <= x) & (x < oo)) | ((-oo < x) & (x <= -2)) + + >>> solve_univariate_inequality(x**2 >= 4, x, relational=False) + Union(Interval(-oo, -2), Interval(2, oo)) + + >>> domain = Interval(0, S.Infinity) + >>> solve_univariate_inequality(x**2 >= 4, x, False, domain) + Interval(2, oo) + + >>> solve_univariate_inequality(sin(x) > 0, x, relational=False) + Interval.open(0, pi) + + """ + from sympy.solvers.solvers import denoms + + if domain.is_subset(S.Reals) is False: + raise NotImplementedError(filldedent(''' + Inequalities in the complex domain are + not supported. Try the real domain by + setting domain=S.Reals''')) + elif domain is not S.Reals: + rv = solve_univariate_inequality( + expr, gen, relational=False, continuous=continuous).intersection(domain) + if relational: + rv = rv.as_relational(gen) + return rv + else: + pass # continue with attempt to solve in Real domain + + # This keeps the function independent of the assumptions about `gen`. + # `solveset` makes sure this function is called only when the domain is + # real. + _gen = gen + _domain = domain + if gen.is_extended_real is False: + rv = S.EmptySet + return rv if not relational else rv.as_relational(_gen) + elif gen.is_extended_real is None: + gen = Dummy('gen', extended_real=True) + try: + expr = expr.xreplace({_gen: gen}) + except TypeError: + raise TypeError(filldedent(''' + When gen is real, the relational has a complex part + which leads to an invalid comparison like I < 0. + ''')) + + rv = None + + if expr is S.true: + rv = domain + + elif expr is S.false: + rv = S.EmptySet + + else: + e = expr.lhs - expr.rhs + period = periodicity(e, gen) + if period == S.Zero: + e = expand_mul(e) + const = expr.func(e, 0) + if const is S.true: + rv = domain + elif const is S.false: + rv = S.EmptySet + elif period is not None: + frange = function_range(e, gen, domain) + + rel = expr.rel_op + if rel in ('<', '<='): + if expr.func(frange.sup, 0): + rv = domain + elif not expr.func(frange.inf, 0): + rv = S.EmptySet + + elif rel in ('>', '>='): + if expr.func(frange.inf, 0): + rv = domain + elif not expr.func(frange.sup, 0): + rv = S.EmptySet + + inf, sup = domain.inf, domain.sup + if sup - inf is S.Infinity: + domain = Interval(0, period, False, True).intersect(_domain) + _domain = domain + + if rv is None: + n, d = e.as_numer_denom() + try: + if gen not in n.free_symbols and len(e.free_symbols) > 1: + raise ValueError + # this might raise ValueError on its own + # or it might give None... + solns = solvify(e, gen, domain) + if solns is None: + # in which case we raise ValueError + raise ValueError + except (ValueError, NotImplementedError): + # replace gen with generic x since it's + # univariate anyway + raise NotImplementedError(filldedent(''' + The inequality, %s, cannot be solved using + solve_univariate_inequality. + ''' % expr.subs(gen, Symbol('x')))) + + expanded_e = expand_mul(e) + def valid(x): + # this is used to see if gen=x satisfies the + # relational by substituting it into the + # expanded form and testing against 0, e.g. + # if expr = x*(x + 1) < 2 then e = x*(x + 1) - 2 + # and expanded_e = x**2 + x - 2; the test is + # whether a given value of x satisfies + # x**2 + x - 2 < 0 + # + # expanded_e, expr and gen used from enclosing scope + v = expanded_e.subs(gen, expand_mul(x)) + try: + r = expr.func(v, 0) + except TypeError: + r = S.false + if r in (S.true, S.false): + return r + if v.is_extended_real is False: + return S.false + else: + v = v.n(2) + if v.is_comparable: + return expr.func(v, 0) + # not comparable or couldn't be evaluated + raise NotImplementedError( + 'relationship did not evaluate: %s' % r) + + singularities = [] + for d in denoms(expr, gen): + singularities.extend(solvify(d, gen, domain)) + if not continuous: + domain = continuous_domain(expanded_e, gen, domain) + + include_x = '=' in expr.rel_op and expr.rel_op != '!=' + + try: + discontinuities = set(domain.boundary - + FiniteSet(domain.inf, domain.sup)) + # remove points that are not between inf and sup of domain + critical_points = FiniteSet(*(solns + singularities + list( + discontinuities))).intersection( + Interval(domain.inf, domain.sup, + domain.inf not in domain, domain.sup not in domain)) + if all(r.is_number for r in critical_points): + reals = _nsort(critical_points, separated=True)[0] + else: + sifted = sift(critical_points, lambda x: x.is_extended_real) + if sifted[None]: + # there were some roots that weren't known + # to be real + raise NotImplementedError + try: + reals = sifted[True] + if len(reals) > 1: + reals = sorted(reals) + except TypeError: + raise NotImplementedError + except NotImplementedError: + raise NotImplementedError('sorting of these roots is not supported') + + # If expr contains imaginary coefficients, only take real + # values of x for which the imaginary part is 0 + make_real = S.Reals + if (coeffI := expanded_e.coeff(S.ImaginaryUnit)) != S.Zero: + check = True + im_sol = FiniteSet() + try: + a = solveset(coeffI, gen, domain) + if not isinstance(a, Interval): + for z in a: + if z not in singularities and valid(z) and z.is_extended_real: + im_sol += FiniteSet(z) + else: + start, end = a.inf, a.sup + for z in _nsort(critical_points + FiniteSet(end)): + valid_start = valid(start) + if start != end: + valid_z = valid(z) + pt = _pt(start, z) + if pt not in singularities and pt.is_extended_real and valid(pt): + if valid_start and valid_z: + im_sol += Interval(start, z) + elif valid_start: + im_sol += Interval.Ropen(start, z) + elif valid_z: + im_sol += Interval.Lopen(start, z) + else: + im_sol += Interval.open(start, z) + start = z + for s in singularities: + im_sol -= FiniteSet(s) + except (TypeError): + im_sol = S.Reals + check = False + + if im_sol is S.EmptySet: + raise ValueError(filldedent(''' + %s contains imaginary parts which cannot be + made 0 for any value of %s satisfying the + inequality, leading to relations like I < 0. + ''' % (expr.subs(gen, _gen), _gen))) + + make_real = make_real.intersect(im_sol) + + sol_sets = [S.EmptySet] + + start = domain.inf + if start in domain and valid(start) and start.is_finite: + sol_sets.append(FiniteSet(start)) + + for x in reals: + end = x + + if valid(_pt(start, end)): + sol_sets.append(Interval(start, end, True, True)) + + if x in singularities: + singularities.remove(x) + else: + if x in discontinuities: + discontinuities.remove(x) + _valid = valid(x) + else: # it's a solution + _valid = include_x + if _valid: + sol_sets.append(FiniteSet(x)) + + start = end + + end = domain.sup + if end in domain and valid(end) and end.is_finite: + sol_sets.append(FiniteSet(end)) + + if valid(_pt(start, end)): + sol_sets.append(Interval.open(start, end)) + + if coeffI != S.Zero and check: + rv = (make_real).intersect(_domain) + else: + rv = Intersection( + (Union(*sol_sets)), make_real, _domain).subs(gen, _gen) + + return rv if not relational else rv.as_relational(_gen) + + +def _pt(start, end): + """Return a point between start and end""" + if not start.is_infinite and not end.is_infinite: + pt = (start + end)/2 + elif start.is_infinite and end.is_infinite: + pt = S.Zero + else: + if (start.is_infinite and start.is_extended_positive is None or + end.is_infinite and end.is_extended_positive is None): + raise ValueError('cannot proceed with unsigned infinite values') + if (end.is_infinite and end.is_extended_negative or + start.is_infinite and start.is_extended_positive): + start, end = end, start + # if possible, use a multiple of self which has + # better behavior when checking assumptions than + # an expression obtained by adding or subtracting 1 + if end.is_infinite: + if start.is_extended_positive: + pt = start*2 + elif start.is_extended_negative: + pt = start*S.Half + else: + pt = start + 1 + elif start.is_infinite: + if end.is_extended_positive: + pt = end*S.Half + elif end.is_extended_negative: + pt = end*2 + else: + pt = end - 1 + return pt + + +def _solve_inequality(ie, s, linear=False): + """Return the inequality with s isolated on the left, if possible. + If the relationship is non-linear, a solution involving And or Or + may be returned. False or True are returned if the relationship + is never True or always True, respectively. + + If `linear` is True (default is False) an `s`-dependent expression + will be isolated on the left, if possible + but it will not be solved for `s` unless the expression is linear + in `s`. Furthermore, only "safe" operations which do not change the + sense of the relationship are applied: no division by an unsigned + value is attempted unless the relationship involves Eq or Ne and + no division by a value not known to be nonzero is ever attempted. + + Examples + ======== + + >>> from sympy import Eq, Symbol + >>> from sympy.solvers.inequalities import _solve_inequality as f + >>> from sympy.abc import x, y + + For linear expressions, the symbol can be isolated: + + >>> f(x - 2 < 0, x) + x < 2 + >>> f(-x - 6 < x, x) + x > -3 + + Sometimes nonlinear relationships will be False + + >>> f(x**2 + 4 < 0, x) + False + + Or they may involve more than one region of values: + + >>> f(x**2 - 4 < 0, x) + (-2 < x) & (x < 2) + + To restrict the solution to a relational, set linear=True + and only the x-dependent portion will be isolated on the left: + + >>> f(x**2 - 4 < 0, x, linear=True) + x**2 < 4 + + Division of only nonzero quantities is allowed, so x cannot + be isolated by dividing by y: + + >>> y.is_nonzero is None # it is unknown whether it is 0 or not + True + >>> f(x*y < 1, x) + x*y < 1 + + And while an equality (or inequality) still holds after dividing by a + non-zero quantity + + >>> nz = Symbol('nz', nonzero=True) + >>> f(Eq(x*nz, 1), x) + Eq(x, 1/nz) + + the sign must be known for other inequalities involving > or <: + + >>> f(x*nz <= 1, x) + nz*x <= 1 + >>> p = Symbol('p', positive=True) + >>> f(x*p <= 1, x) + x <= 1/p + + When there are denominators in the original expression that + are removed by expansion, conditions for them will be returned + as part of the result: + + >>> f(x < x*(2/x - 1), x) + (x < 1) & Ne(x, 0) + """ + from sympy.solvers.solvers import denoms + if s not in ie.free_symbols: + return ie + if ie.rhs == s: + ie = ie.reversed + if ie.lhs == s and s not in ie.rhs.free_symbols: + return ie + + def classify(ie, s, i): + # return True or False if ie evaluates when substituting s with + # i else None (if unevaluated) or NaN (when there is an error + # in evaluating) + try: + v = ie.subs(s, i) + if v is S.NaN: + return v + elif v not in (True, False): + return + return v + except TypeError: + return S.NaN + + rv = None + oo = S.Infinity + expr = ie.lhs - ie.rhs + try: + p = Poly(expr, s) + if p.degree() == 0: + rv = ie.func(p.as_expr(), 0) + elif not linear and p.degree() > 1: + # handle in except clause + raise NotImplementedError + except (PolynomialError, NotImplementedError): + if not linear: + try: + rv = reduce_rational_inequalities([[ie]], s) + except PolynomialError: + rv = solve_univariate_inequality(ie, s) + # remove restrictions wrt +/-oo that may have been + # applied when using sets to simplify the relationship + okoo = classify(ie, s, oo) + if okoo is S.true and classify(rv, s, oo) is S.false: + rv = rv.subs(s < oo, True) + oknoo = classify(ie, s, -oo) + if (oknoo is S.true and + classify(rv, s, -oo) is S.false): + rv = rv.subs(-oo < s, True) + rv = rv.subs(s > -oo, True) + if rv is S.true: + rv = (s <= oo) if okoo is S.true else (s < oo) + if oknoo is not S.true: + rv = And(-oo < s, rv) + else: + p = Poly(expr) + + conds = [] + if rv is None: + e = p.as_expr() # this is in expanded form + # Do a safe inversion of e, moving non-s terms + # to the rhs and dividing by a nonzero factor if + # the relational is Eq/Ne; for other relationals + # the sign must also be positive or negative + rhs = 0 + b, ax = e.as_independent(s, as_Add=True) + e -= b + rhs -= b + ef = factor_terms(e) + a, e = ef.as_independent(s, as_Add=False) + if (a.is_zero != False or # don't divide by potential 0 + a.is_negative == + a.is_positive is None and # if sign is not known then + ie.rel_op not in ('!=', '==')): # reject if not Eq/Ne + e = ef + a = S.One + rhs /= a + if a.is_positive: + rv = ie.func(e, rhs) + else: + rv = ie.reversed.func(e, rhs) + + # return conditions under which the value is + # valid, too. + beginning_denoms = denoms(ie.lhs) | denoms(ie.rhs) + current_denoms = denoms(rv) + for d in beginning_denoms - current_denoms: + c = _solve_inequality(Eq(d, 0), s, linear=linear) + if isinstance(c, Eq) and c.lhs == s: + if classify(rv, s, c.rhs) is S.true: + # rv is permitting this value but it shouldn't + conds.append(~c) + for i in (-oo, oo): + if (classify(rv, s, i) is S.true and + classify(ie, s, i) is not S.true): + conds.append(s < i if i is oo else i < s) + + conds.append(rv) + return And(*conds) + + +def _reduce_inequalities(inequalities, symbols): + # helper for reduce_inequalities + + poly_part, abs_part = {}, {} + other = [] + + for inequality in inequalities: + + expr, rel = inequality.lhs, inequality.rel_op # rhs is 0 + + # check for gens using atoms which is more strict than free_symbols to + # guard against EX domain which won't be handled by + # reduce_rational_inequalities + gens = expr.atoms(Symbol) + + if len(gens) == 1: + gen = gens.pop() + else: + common = expr.free_symbols & symbols + if len(common) == 1: + gen = common.pop() + other.append(_solve_inequality(Relational(expr, 0, rel), gen)) + continue + else: + raise NotImplementedError(filldedent(''' + inequality has more than one symbol of interest. + ''')) + + if expr.is_polynomial(gen): + poly_part.setdefault(gen, []).append((expr, rel)) + else: + components = expr.find(lambda u: + u.has(gen) and ( + u.is_Function or u.is_Pow and not u.exp.is_Integer)) + if components and all(isinstance(i, Abs) for i in components): + abs_part.setdefault(gen, []).append((expr, rel)) + else: + other.append(_solve_inequality(Relational(expr, 0, rel), gen)) + + poly_reduced = [reduce_rational_inequalities([exprs], gen) for gen, exprs in poly_part.items()] + abs_reduced = [reduce_abs_inequalities(exprs, gen) for gen, exprs in abs_part.items()] + + return And(*(poly_reduced + abs_reduced + other)) + + +def reduce_inequalities(inequalities, symbols=[]): + """Reduce a system of inequalities with rational coefficients. + + Examples + ======== + + >>> from sympy.abc import x, y + >>> from sympy import reduce_inequalities + + >>> reduce_inequalities(0 <= x + 3, []) + (-3 <= x) & (x < oo) + + >>> reduce_inequalities(0 <= x + y*2 - 1, [x]) + (x < oo) & (x >= 1 - 2*y) + """ + if not iterable(inequalities): + inequalities = [inequalities] + inequalities = [sympify(i) for i in inequalities] + + gens = set().union(*[i.free_symbols for i in inequalities]) + + if not iterable(symbols): + symbols = [symbols] + symbols = (set(symbols) or gens) & gens + if any(i.is_extended_real is False for i in symbols): + raise TypeError(filldedent(''' + inequalities cannot contain symbols that are not real. + ''')) + + # make vanilla symbol real + recast = {i: Dummy(i.name, extended_real=True) + for i in gens if i.is_extended_real is None} + inequalities = [i.xreplace(recast) for i in inequalities] + symbols = {i.xreplace(recast) for i in symbols} + + # prefilter + keep = [] + for i in inequalities: + if isinstance(i, Relational): + i = i.func(i.lhs.as_expr() - i.rhs.as_expr(), 0) + elif i not in (True, False): + i = Eq(i, 0) + if i == True: + continue + elif i == False: + return S.false + if i.lhs.is_number: + raise NotImplementedError( + "could not determine truth value of %s" % i) + keep.append(i) + inequalities = keep + del keep + + # solve system + rv = _reduce_inequalities(inequalities, symbols) + + # restore original symbols and return + return rv.xreplace({v: k for k, v in recast.items()}) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__init__.py b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__init__.py new file mode 100644 index 0000000000000000000000000000000000000000..2b543425251dea6380a1860279cb6d636f3dd629 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__init__.py @@ -0,0 +1,16 @@ +from .ode import (allhints, checkinfsol, classify_ode, + constantsimp, dsolve, homogeneous_order) + +from .lie_group import infinitesimals + +from .subscheck import checkodesol + +from .systems import (canonical_odes, linear_ode_to_matrix, + linodesolve) + + +__all__ = [ + 'allhints', 'checkinfsol', 'checkodesol', 'classify_ode', 'constantsimp', + 'dsolve', 'homogeneous_order', 'infinitesimals', 'canonical_odes', 'linear_ode_to_matrix', + 'linodesolve' +] diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__pycache__/__init__.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__pycache__/__init__.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..0741bd60ae664f28d441a158bbe5016d123c0209 Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__pycache__/__init__.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__pycache__/lie_group.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/__pycache__/lie_group.cpython-310.pyc new file mode 100644 index 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nth_linear_constant_coeff_undetermined_coefficients, +nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients, +nth_linear_constant_coeff_variation_of_parameters, +and nth_linear_euler_eq_nonhomogeneous_variation_of_parameters. + +All the functions in this file are used by more than one solvers so, instead of creating +instances in other classes for using them it is better to keep it here as separate helpers. + +""" +from collections import defaultdict +from sympy.core import Add, S +from sympy.core.function import diff, expand, _mexpand, expand_mul +from sympy.core.relational import Eq +from sympy.core.sorting import default_sort_key +from sympy.core.symbol import Dummy, Wild +from sympy.functions import exp, cos, cosh, im, log, re, sin, sinh, \ + atan2, conjugate +from sympy.integrals import Integral +from sympy.polys import (Poly, RootOf, rootof, roots) +from sympy.simplify import collect, simplify, separatevars, powsimp, trigsimp # type: ignore +from sympy.utilities import numbered_symbols +from sympy.solvers.solvers import solve +from sympy.matrices import wronskian +from .subscheck import sub_func_doit +from sympy.solvers.ode.ode import get_numbered_constants + + +def _test_term(coeff, func, order): + r""" + Linear Euler ODEs have the form K*x**order*diff(y(x), x, order) = F(x), + where K is independent of x and y(x), order>= 0. + So we need to check that for each term, coeff == K*x**order from + some K. We have a few cases, since coeff may have several + different types. + """ + x = func.args[0] + f = func.func + if order < 0: + raise ValueError("order should be greater than 0") + if coeff == 0: + return True + if order == 0: + if x in coeff.free_symbols: + return False + return True + if coeff.is_Mul: + if coeff.has(f(x)): + return False + return x**order in coeff.args + elif coeff.is_Pow: + return coeff.as_base_exp() == (x, order) + elif order == 1: + return x == coeff + return False + + +def _get_euler_characteristic_eq_sols(eq, func, match_obj): + r""" + Returns the solution of homogeneous part of the linear euler ODE and + the list of roots of characteristic equation. + + The parameter ``match_obj`` is a dict of order:coeff terms, where order is the order + of the derivative on each term, and coeff is the coefficient of that derivative. + + """ + x = func.args[0] + f = func.func + + # First, set up characteristic equation. + chareq, symbol = S.Zero, Dummy('x') + + for i in match_obj: + if i >= 0: + chareq += (match_obj[i]*diff(x**symbol, x, i)*x**-symbol).expand() + + chareq = Poly(chareq, symbol) + chareqroots = [rootof(chareq, k) for k in range(chareq.degree())] + collectterms = [] + + # A generator of constants + constants = list(get_numbered_constants(eq, num=chareq.degree()*2)) + constants.reverse() + + # Create a dict root: multiplicity or charroots + charroots = defaultdict(int) + for root in chareqroots: + charroots[root] += 1 + gsol = S.Zero + ln = log + for root, multiplicity in charroots.items(): + for i in range(multiplicity): + if isinstance(root, RootOf): + gsol += (x**root) * constants.pop() + if multiplicity != 1: + raise ValueError("Value should be 1") + collectterms = [(0, root, 0)] + collectterms + elif root.is_real: + gsol += ln(x)**i*(x**root) * constants.pop() + collectterms = [(i, root, 0)] + collectterms + else: + reroot = re(root) + imroot = im(root) + gsol += ln(x)**i * (x**reroot) * ( + constants.pop() * sin(abs(imroot)*ln(x)) + + constants.pop() * cos(imroot*ln(x))) + collectterms = [(i, reroot, imroot)] + collectterms + + gsol = Eq(f(x), gsol) + + gensols = [] + # Keep track of when to use sin or cos for nonzero imroot + for i, reroot, imroot in collectterms: + if imroot == 0: + gensols.append(ln(x)**i*x**reroot) + else: + sin_form = ln(x)**i*x**reroot*sin(abs(imroot)*ln(x)) + if sin_form in gensols: + cos_form = ln(x)**i*x**reroot*cos(imroot*ln(x)) + gensols.append(cos_form) + else: + gensols.append(sin_form) + return gsol, gensols + + +def _solve_variation_of_parameters(eq, func, roots, homogen_sol, order, match_obj, simplify_flag=True): + r""" + Helper function for the method of variation of parameters and nonhomogeneous euler eq. + + See the + :py:meth:`~sympy.solvers.ode.single.NthLinearConstantCoeffVariationOfParameters` + docstring for more information on this method. + + The parameter are ``match_obj`` should be a dictionary that has the following + keys: + + ``list`` + A list of solutions to the homogeneous equation. + + ``sol`` + The general solution. + + """ + f = func.func + x = func.args[0] + r = match_obj + psol = 0 + wr = wronskian(roots, x) + + if simplify_flag: + wr = simplify(wr) # We need much better simplification for + # some ODEs. See issue 4662, for example. + # To reduce commonly occurring sin(x)**2 + cos(x)**2 to 1 + wr = trigsimp(wr, deep=True, recursive=True) + if not wr: + # The wronskian will be 0 iff the solutions are not linearly + # independent. + raise NotImplementedError("Cannot find " + str(order) + + " solutions to the homogeneous equation necessary to apply " + + "variation of parameters to " + str(eq) + " (Wronskian == 0)") + if len(roots) != order: + raise NotImplementedError("Cannot find " + str(order) + + " solutions to the homogeneous equation necessary to apply " + + "variation of parameters to " + + str(eq) + " (number of terms != order)") + negoneterm = S.NegativeOne**(order) + for i in roots: + psol += negoneterm*Integral(wronskian([sol for sol in roots if sol != i], x)*r[-1]/wr, x)*i/r[order] + negoneterm *= -1 + + if simplify_flag: + psol = simplify(psol) + psol = trigsimp(psol, deep=True) + return Eq(f(x), homogen_sol.rhs + psol) + + +def _get_const_characteristic_eq_sols(r, func, order): + r""" + Returns the roots of characteristic equation of constant coefficient + linear ODE and list of collectterms which is later on used by simplification + to use collect on solution. + + The parameter `r` is a dict of order:coeff terms, where order is the order of the + derivative on each term, and coeff is the coefficient of that derivative. + + """ + x = func.args[0] + # First, set up characteristic equation. + chareq, symbol = S.Zero, Dummy('x') + + for i in r.keys(): + if isinstance(i, str) or i < 0: + pass + else: + chareq += r[i]*symbol**i + + chareq = Poly(chareq, symbol) + # Can't just call roots because it doesn't return rootof for unsolveable + # polynomials. + chareqroots = roots(chareq, multiple=True) + if len(chareqroots) != order: + chareqroots = [rootof(chareq, k) for k in range(chareq.degree())] + + chareq_is_complex = not all(i.is_real for i in chareq.all_coeffs()) + + # Create a dict root: multiplicity or charroots + charroots = defaultdict(int) + for root in chareqroots: + charroots[root] += 1 + # We need to keep track of terms so we can run collect() at the end. + # This is necessary for constantsimp to work properly. + collectterms = [] + gensols = [] + conjugate_roots = [] # used to prevent double-use of conjugate roots + # Loop over roots in theorder provided by roots/rootof... + for root in chareqroots: + # but don't repoeat multiple roots. + if root not in charroots: + continue + multiplicity = charroots.pop(root) + for i in range(multiplicity): + if chareq_is_complex: + gensols.append(x**i*exp(root*x)) + collectterms = [(i, root, 0)] + collectterms + continue + reroot = re(root) + imroot = im(root) + if imroot.has(atan2) and reroot.has(atan2): + # Remove this condition when re and im stop returning + # circular atan2 usages. + gensols.append(x**i*exp(root*x)) + collectterms = [(i, root, 0)] + collectterms + else: + if root in conjugate_roots: + collectterms = [(i, reroot, imroot)] + collectterms + continue + if imroot == 0: + gensols.append(x**i*exp(reroot*x)) + collectterms = [(i, reroot, 0)] + collectterms + continue + conjugate_roots.append(conjugate(root)) + gensols.append(x**i*exp(reroot*x) * sin(abs(imroot) * x)) + gensols.append(x**i*exp(reroot*x) * cos( imroot * x)) + + # This ordering is important + collectterms = [(i, reroot, imroot)] + collectterms + return gensols, collectterms + + +# Ideally these kind of simplification functions shouldn't be part of solvers. +# odesimp should be improved to handle these kind of specific simplifications. +def _get_simplified_sol(sol, func, collectterms): + r""" + Helper function which collects the solution on + collectterms. Ideally this should be handled by odesimp.It is used + only when the simplify is set to True in dsolve. + + The parameter ``collectterms`` is a list of tuple (i, reroot, imroot) where `i` is + the multiplicity of the root, reroot is real part and imroot being the imaginary part. + + """ + f = func.func + x = func.args[0] + collectterms.sort(key=default_sort_key) + collectterms.reverse() + assert len(sol) == 1 and sol[0].lhs == f(x) + sol = sol[0].rhs + sol = expand_mul(sol) + for i, reroot, imroot in collectterms: + sol = collect(sol, x**i*exp(reroot*x)*sin(abs(imroot)*x)) + sol = collect(sol, x**i*exp(reroot*x)*cos(imroot*x)) + for i, reroot, imroot in collectterms: + sol = collect(sol, x**i*exp(reroot*x)) + sol = powsimp(sol) + return Eq(f(x), sol) + + +def _undetermined_coefficients_match(expr, x, func=None, eq_homogeneous=S.Zero): + r""" + Returns a trial function match if undetermined coefficients can be applied + to ``expr``, and ``None`` otherwise. + + A trial expression can be found for an expression for use with the method + of undetermined coefficients if the expression is an + additive/multiplicative combination of constants, polynomials in `x` (the + independent variable of expr), `\sin(a x + b)`, `\cos(a x + b)`, and + `e^{a x}` terms (in other words, it has a finite number of linearly + independent derivatives). + + Note that you may still need to multiply each term returned here by + sufficient `x` to make it linearly independent with the solutions to the + homogeneous equation. + + This is intended for internal use by ``undetermined_coefficients`` hints. + + SymPy currently has no way to convert `\sin^n(x) \cos^m(y)` into a sum of + only `\sin(a x)` and `\cos(b x)` terms, so these are not implemented. So, + for example, you will need to manually convert `\sin^2(x)` into `[1 + + \cos(2 x)]/2` to properly apply the method of undetermined coefficients on + it. + + Examples + ======== + + >>> from sympy import log, exp + >>> from sympy.solvers.ode.nonhomogeneous import _undetermined_coefficients_match + >>> from sympy.abc import x + >>> _undetermined_coefficients_match(9*x*exp(x) + exp(-x), x) + {'test': True, 'trialset': {x*exp(x), exp(-x), exp(x)}} + >>> _undetermined_coefficients_match(log(x), x) + {'test': False} + + """ + a = Wild('a', exclude=[x]) + b = Wild('b', exclude=[x]) + expr = powsimp(expr, combine='exp') # exp(x)*exp(2*x + 1) => exp(3*x + 1) + retdict = {} + + def _test_term(expr, x): + r""" + Test if ``expr`` fits the proper form for undetermined coefficients. + """ + if not expr.has(x): + return True + elif expr.is_Add: + return all(_test_term(i, x) for i in expr.args) + elif expr.is_Mul: + if expr.has(sin, cos): + foundtrig = False + # Make sure that there is only one trig function in the args. + # See the docstring. + for i in expr.args: + if i.has(sin, cos): + if foundtrig: + return False + else: + foundtrig = True + return all(_test_term(i, x) for i in expr.args) + elif expr.is_Function: + if expr.func in (sin, cos, exp, sinh, cosh): + if expr.args[0].match(a*x + b): + return True + else: + return False + else: + return False + elif expr.is_Pow and expr.base.is_Symbol and expr.exp.is_Integer and \ + expr.exp >= 0: + return True + elif expr.is_Pow and expr.base.is_number: + if expr.exp.match(a*x + b): + return True + else: + return False + elif expr.is_Symbol or expr.is_number: + return True + else: + return False + + def _get_trial_set(expr, x, exprs=set()): + r""" + Returns a set of trial terms for undetermined coefficients. + + The idea behind undetermined coefficients is that the terms expression + repeat themselves after a finite number of derivatives, except for the + coefficients (they are linearly dependent). So if we collect these, + we should have the terms of our trial function. + """ + def _remove_coefficient(expr, x): + r""" + Returns the expression without a coefficient. + + Similar to expr.as_independent(x)[1], except it only works + multiplicatively. + """ + term = S.One + if expr.is_Mul: + for i in expr.args: + if i.has(x): + term *= i + elif expr.has(x): + term = expr + return term + + expr = expand_mul(expr) + if expr.is_Add: + for term in expr.args: + if _remove_coefficient(term, x) in exprs: + pass + else: + exprs.add(_remove_coefficient(term, x)) + exprs = exprs.union(_get_trial_set(term, x, exprs)) + else: + term = _remove_coefficient(expr, x) + tmpset = exprs.union({term}) + oldset = set() + while tmpset != oldset: + # If you get stuck in this loop, then _test_term is probably + # broken + oldset = tmpset.copy() + expr = expr.diff(x) + term = _remove_coefficient(expr, x) + if term.is_Add: + tmpset = tmpset.union(_get_trial_set(term, x, tmpset)) + else: + tmpset.add(term) + exprs = tmpset + return exprs + + def is_homogeneous_solution(term): + r""" This function checks whether the given trialset contains any root + of homogeneous equation""" + return expand(sub_func_doit(eq_homogeneous, func, term)).is_zero + + retdict['test'] = _test_term(expr, x) + if retdict['test']: + # Try to generate a list of trial solutions that will have the + # undetermined coefficients. Note that if any of these are not linearly + # independent with any of the solutions to the homogeneous equation, + # then they will need to be multiplied by sufficient x to make them so. + # This function DOES NOT do that (it doesn't even look at the + # homogeneous equation). + temp_set = set() + for i in Add.make_args(expr): + act = _get_trial_set(i, x) + if eq_homogeneous is not S.Zero: + while any(is_homogeneous_solution(ts) for ts in act): + act = {x*ts for ts in act} + temp_set = temp_set.union(act) + + retdict['trialset'] = temp_set + return retdict + + +def _solve_undetermined_coefficients(eq, func, order, match, trialset): + r""" + Helper function for the method of undetermined coefficients. + + See the + :py:meth:`~sympy.solvers.ode.single.NthLinearConstantCoeffUndeterminedCoefficients` + docstring for more information on this method. + + The parameter ``trialset`` is the set of trial functions as returned by + ``_undetermined_coefficients_match()['trialset']``. + + The parameter ``match`` should be a dictionary that has the following + keys: + + ``list`` + A list of solutions to the homogeneous equation. + + ``sol`` + The general solution. + + """ + r = match + coeffs = numbered_symbols('a', cls=Dummy) + coefflist = [] + gensols = r['list'] + gsol = r['sol'] + f = func.func + x = func.args[0] + + if len(gensols) != order: + raise NotImplementedError("Cannot find " + str(order) + + " solutions to the homogeneous equation necessary to apply" + + " undetermined coefficients to " + str(eq) + + " (number of terms != order)") + + trialfunc = 0 + for i in trialset: + c = next(coeffs) + coefflist.append(c) + trialfunc += c*i + + eqs = sub_func_doit(eq, f(x), trialfunc) + + coeffsdict = dict(list(zip(trialset, [0]*(len(trialset) + 1)))) + + eqs = _mexpand(eqs) + + for i in Add.make_args(eqs): + s = separatevars(i, dict=True, symbols=[x]) + if coeffsdict.get(s[x]): + coeffsdict[s[x]] += s['coeff'] + else: + coeffsdict[s[x]] = s['coeff'] + + coeffvals = solve(list(coeffsdict.values()), coefflist) + + if not coeffvals: + raise NotImplementedError( + "Could not solve `%s` using the " + "method of undetermined coefficients " + "(unable to solve for coefficients)." % eq) + + psol = trialfunc.subs(coeffvals) + + return Eq(f(x), gsol.rhs + psol) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/ode.py b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/ode.py new file mode 100644 index 0000000000000000000000000000000000000000..be82bb18a5c800ae3848605d636d9744a011b6bc --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/ode.py @@ -0,0 +1,3575 @@ +r""" +This module contains :py:meth:`~sympy.solvers.ode.dsolve` and different helper +functions that it uses. + +:py:meth:`~sympy.solvers.ode.dsolve` solves ordinary differential equations. +See the docstring on the various functions for their uses. Note that partial +differential equations support is in ``pde.py``. Note that hint functions +have docstrings describing their various methods, but they are intended for +internal use. Use ``dsolve(ode, func, hint=hint)`` to solve an ODE using a +specific hint. See also the docstring on +:py:meth:`~sympy.solvers.ode.dsolve`. + +**Functions in this module** + + These are the user functions in this module: + + - :py:meth:`~sympy.solvers.ode.dsolve` - Solves ODEs. + - :py:meth:`~sympy.solvers.ode.classify_ode` - Classifies ODEs into + possible hints for :py:meth:`~sympy.solvers.ode.dsolve`. + - :py:meth:`~sympy.solvers.ode.checkodesol` - Checks if an equation is the + solution to an ODE. + - :py:meth:`~sympy.solvers.ode.homogeneous_order` - Returns the + homogeneous order of an expression. + - :py:meth:`~sympy.solvers.ode.infinitesimals` - Returns the infinitesimals + of the Lie group of point transformations of an ODE, such that it is + invariant. + - :py:meth:`~sympy.solvers.ode.checkinfsol` - Checks if the given infinitesimals + are the actual infinitesimals of a first order ODE. + + These are the non-solver helper functions that are for internal use. The + user should use the various options to + :py:meth:`~sympy.solvers.ode.dsolve` to obtain the functionality provided + by these functions: + + - :py:meth:`~sympy.solvers.ode.ode.odesimp` - Does all forms of ODE + simplification. + - :py:meth:`~sympy.solvers.ode.ode.ode_sol_simplicity` - A key function for + comparing solutions by simplicity. + - :py:meth:`~sympy.solvers.ode.constantsimp` - Simplifies arbitrary + constants. + - :py:meth:`~sympy.solvers.ode.ode.constant_renumber` - Renumber arbitrary + constants. + - :py:meth:`~sympy.solvers.ode.ode._handle_Integral` - Evaluate unevaluated + Integrals. + + See also the docstrings of these functions. + +**Currently implemented solver methods** + +The following methods are implemented for solving ordinary differential +equations. See the docstrings of the various hint functions for more +information on each (run ``help(ode)``): + + - 1st order separable differential equations. + - 1st order differential equations whose coefficients or `dx` and `dy` are + functions homogeneous of the same order. + - 1st order exact differential equations. + - 1st order linear differential equations. + - 1st order Bernoulli differential equations. + - Power series solutions for first order differential equations. + - Lie Group method of solving first order differential equations. + - 2nd order Liouville differential equations. + - Power series solutions for second order differential equations + at ordinary and regular singular points. + - `n`\th order differential equation that can be solved with algebraic + rearrangement and integration. + - `n`\th order linear homogeneous differential equation with constant + coefficients. + - `n`\th order linear inhomogeneous differential equation with constant + coefficients using the method of undetermined coefficients. + - `n`\th order linear inhomogeneous differential equation with constant + coefficients using the method of variation of parameters. + +**Philosophy behind this module** + +This module is designed to make it easy to add new ODE solving methods without +having to mess with the solving code for other methods. The idea is that +there is a :py:meth:`~sympy.solvers.ode.classify_ode` function, which takes in +an ODE and tells you what hints, if any, will solve the ODE. It does this +without attempting to solve the ODE, so it is fast. Each solving method is a +hint, and it has its own function, named ``ode_``. That function takes +in the ODE and any match expression gathered by +:py:meth:`~sympy.solvers.ode.classify_ode` and returns a solved result. If +this result has any integrals in it, the hint function will return an +unevaluated :py:class:`~sympy.integrals.integrals.Integral` class. +:py:meth:`~sympy.solvers.ode.dsolve`, which is the user wrapper function +around all of this, will then call :py:meth:`~sympy.solvers.ode.ode.odesimp` on +the result, which, among other things, will attempt to solve the equation for +the dependent variable (the function we are solving for), simplify the +arbitrary constants in the expression, and evaluate any integrals, if the hint +allows it. + +**How to add new solution methods** + +If you have an ODE that you want :py:meth:`~sympy.solvers.ode.dsolve` to be +able to solve, try to avoid adding special case code here. Instead, try +finding a general method that will solve your ODE, as well as others. This +way, the :py:mod:`~sympy.solvers.ode` module will become more robust, and +unhindered by special case hacks. WolphramAlpha and Maple's +DETools[odeadvisor] function are two resources you can use to classify a +specific ODE. It is also better for a method to work with an `n`\th order ODE +instead of only with specific orders, if possible. + +To add a new method, there are a few things that you need to do. First, you +need a hint name for your method. Try to name your hint so that it is +unambiguous with all other methods, including ones that may not be implemented +yet. If your method uses integrals, also include a ``hint_Integral`` hint. +If there is more than one way to solve ODEs with your method, include a hint +for each one, as well as a ``_best`` hint. Your ``ode__best()`` +function should choose the best using min with ``ode_sol_simplicity`` as the +key argument. See +:obj:`~sympy.solvers.ode.single.HomogeneousCoeffBest`, for example. +The function that uses your method will be called ``ode_()``, so the +hint must only use characters that are allowed in a Python function name +(alphanumeric characters and the underscore '``_``' character). Include a +function for every hint, except for ``_Integral`` hints +(:py:meth:`~sympy.solvers.ode.dsolve` takes care of those automatically). +Hint names should be all lowercase, unless a word is commonly capitalized +(such as Integral or Bernoulli). If you have a hint that you do not want to +run with ``all_Integral`` that does not have an ``_Integral`` counterpart (such +as a best hint that would defeat the purpose of ``all_Integral``), you will +need to remove it manually in the :py:meth:`~sympy.solvers.ode.dsolve` code. +See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for +guidelines on writing a hint name. + +Determine *in general* how the solutions returned by your method compare with +other methods that can potentially solve the same ODEs. Then, put your hints +in the :py:data:`~sympy.solvers.ode.allhints` tuple in the order that they +should be called. The ordering of this tuple determines which hints are +default. Note that exceptions are ok, because it is easy for the user to +choose individual hints with :py:meth:`~sympy.solvers.ode.dsolve`. In +general, ``_Integral`` variants should go at the end of the list, and +``_best`` variants should go before the various hints they apply to. For +example, the ``undetermined_coefficients`` hint comes before the +``variation_of_parameters`` hint because, even though variation of parameters +is more general than undetermined coefficients, undetermined coefficients +generally returns cleaner results for the ODEs that it can solve than +variation of parameters does, and it does not require integration, so it is +much faster. + +Next, you need to have a match expression or a function that matches the type +of the ODE, which you should put in :py:meth:`~sympy.solvers.ode.classify_ode` +(if the match function is more than just a few lines. It should match the +ODE without solving for it as much as possible, so that +:py:meth:`~sympy.solvers.ode.classify_ode` remains fast and is not hindered by +bugs in solving code. Be sure to consider corner cases. For example, if your +solution method involves dividing by something, make sure you exclude the case +where that division will be 0. + +In most cases, the matching of the ODE will also give you the various parts +that you need to solve it. You should put that in a dictionary (``.match()`` +will do this for you), and add that as ``matching_hints['hint'] = matchdict`` +in the relevant part of :py:meth:`~sympy.solvers.ode.classify_ode`. +:py:meth:`~sympy.solvers.ode.classify_ode` will then send this to +:py:meth:`~sympy.solvers.ode.dsolve`, which will send it to your function as +the ``match`` argument. Your function should be named ``ode_(eq, func, +order, match)`. If you need to send more information, put it in the ``match`` +dictionary. For example, if you had to substitute in a dummy variable in +:py:meth:`~sympy.solvers.ode.classify_ode` to match the ODE, you will need to +pass it to your function using the `match` dict to access it. You can access +the independent variable using ``func.args[0]``, and the dependent variable +(the function you are trying to solve for) as ``func.func``. If, while trying +to solve the ODE, you find that you cannot, raise ``NotImplementedError``. +:py:meth:`~sympy.solvers.ode.dsolve` will catch this error with the ``all`` +meta-hint, rather than causing the whole routine to fail. + +Add a docstring to your function that describes the method employed. Like +with anything else in SymPy, you will need to add a doctest to the docstring, +in addition to real tests in ``test_ode.py``. Try to maintain consistency +with the other hint functions' docstrings. Add your method to the list at the +top of this docstring. Also, add your method to ``ode.rst`` in the +``docs/src`` directory, so that the Sphinx docs will pull its docstring into +the main SymPy documentation. Be sure to make the Sphinx documentation by +running ``make html`` from within the doc directory to verify that the +docstring formats correctly. + +If your solution method involves integrating, use :py:obj:`~.Integral` instead of +:py:meth:`~sympy.core.expr.Expr.integrate`. This allows the user to bypass +hard/slow integration by using the ``_Integral`` variant of your hint. In +most cases, calling :py:meth:`sympy.core.basic.Basic.doit` will integrate your +solution. If this is not the case, you will need to write special code in +:py:meth:`~sympy.solvers.ode.ode._handle_Integral`. Arbitrary constants should be +symbols named ``C1``, ``C2``, and so on. All solution methods should return +an equality instance. If you need an arbitrary number of arbitrary constants, +you can use ``constants = numbered_symbols(prefix='C', cls=Symbol, start=1)``. +If it is possible to solve for the dependent function in a general way, do so. +Otherwise, do as best as you can, but do not call solve in your +``ode_()`` function. :py:meth:`~sympy.solvers.ode.ode.odesimp` will attempt +to solve the solution for you, so you do not need to do that. Lastly, if your +ODE has a common simplification that can be applied to your solutions, you can +add a special case in :py:meth:`~sympy.solvers.ode.ode.odesimp` for it. For +example, solutions returned from the ``1st_homogeneous_coeff`` hints often +have many :obj:`~sympy.functions.elementary.exponential.log` terms, so +:py:meth:`~sympy.solvers.ode.ode.odesimp` calls +:py:meth:`~sympy.simplify.simplify.logcombine` on them (it also helps to write +the arbitrary constant as ``log(C1)`` instead of ``C1`` in this case). Also +consider common ways that you can rearrange your solution to have +:py:meth:`~sympy.solvers.ode.constantsimp` take better advantage of it. It is +better to put simplification in :py:meth:`~sympy.solvers.ode.ode.odesimp` than in +your method, because it can then be turned off with the simplify flag in +:py:meth:`~sympy.solvers.ode.dsolve`. If you have any extraneous +simplification in your function, be sure to only run it using ``if +match.get('simplify', True):``, especially if it can be slow or if it can +reduce the domain of the solution. + +Finally, as with every contribution to SymPy, your method will need to be +tested. Add a test for each method in ``test_ode.py``. Follow the +conventions there, i.e., test the solver using ``dsolve(eq, f(x), +hint=your_hint)``, and also test the solution using +:py:meth:`~sympy.solvers.ode.checkodesol` (you can put these in a separate +tests and skip/XFAIL if it runs too slow/does not work). Be sure to call your +hint specifically in :py:meth:`~sympy.solvers.ode.dsolve`, that way the test +will not be broken simply by the introduction of another matching hint. If your +method works for higher order (>1) ODEs, you will need to run ``sol = +constant_renumber(sol, 'C', 1, order)`` for each solution, where ``order`` is +the order of the ODE. This is because ``constant_renumber`` renumbers the +arbitrary constants by printing order, which is platform dependent. Try to +test every corner case of your solver, including a range of orders if it is a +`n`\th order solver, but if your solver is slow, such as if it involves hard +integration, try to keep the test run time down. + +Feel free to refactor existing hints to avoid duplicating code or creating +inconsistencies. If you can show that your method exactly duplicates an +existing method, including in the simplicity and speed of obtaining the +solutions, then you can remove the old, less general method. The existing +code is tested extensively in ``test_ode.py``, so if anything is broken, one +of those tests will surely fail. + +""" + +from sympy.core import Add, S, Mul, Pow, oo +from sympy.core.containers import Tuple +from sympy.core.expr import AtomicExpr, Expr +from sympy.core.function import (Function, Derivative, AppliedUndef, diff, + expand, expand_mul, Subs) +from sympy.core.multidimensional import vectorize +from sympy.core.numbers import nan, zoo, Number +from sympy.core.relational import Equality, Eq +from sympy.core.sorting import default_sort_key, ordered +from sympy.core.symbol import Symbol, Wild, Dummy, symbols +from sympy.core.sympify import sympify +from sympy.core.traversal import preorder_traversal + +from sympy.logic.boolalg import (BooleanAtom, BooleanTrue, + BooleanFalse) +from sympy.functions import exp, log, sqrt +from sympy.functions.combinatorial.factorials import factorial +from sympy.integrals.integrals import Integral +from sympy.polys import (Poly, terms_gcd, PolynomialError, lcm) +from sympy.polys.polytools import cancel +from sympy.series import Order +from sympy.series.series import series +from sympy.simplify import (collect, logcombine, powsimp, # type: ignore + separatevars, simplify, cse) +from sympy.simplify.radsimp import collect_const +from sympy.solvers import checksol, solve + +from sympy.utilities import numbered_symbols +from sympy.utilities.iterables import uniq, sift, iterable +from sympy.solvers.deutils import _preprocess, ode_order, _desolve + + +#: This is a list of hints in the order that they should be preferred by +#: :py:meth:`~sympy.solvers.ode.classify_ode`. In general, hints earlier in the +#: list should produce simpler solutions than those later in the list (for +#: ODEs that fit both). For now, the order of this list is based on empirical +#: observations by the developers of SymPy. +#: +#: The hint used by :py:meth:`~sympy.solvers.ode.dsolve` for a specific ODE +#: can be overridden (see the docstring). +#: +#: In general, ``_Integral`` hints are grouped at the end of the list, unless +#: there is a method that returns an unevaluable integral most of the time +#: (which go near the end of the list anyway). ``default``, ``all``, +#: ``best``, and ``all_Integral`` meta-hints should not be included in this +#: list, but ``_best`` and ``_Integral`` hints should be included. +allhints = ( + "factorable", + "nth_algebraic", + "separable", + "1st_exact", + "1st_linear", + "Bernoulli", + "1st_rational_riccati", + "Riccati_special_minus2", + "1st_homogeneous_coeff_best", + "1st_homogeneous_coeff_subs_indep_div_dep", + "1st_homogeneous_coeff_subs_dep_div_indep", + "almost_linear", + "linear_coefficients", + "separable_reduced", + "1st_power_series", + "lie_group", + "nth_linear_constant_coeff_homogeneous", + "nth_linear_euler_eq_homogeneous", + "nth_linear_constant_coeff_undetermined_coefficients", + "nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients", + "nth_linear_constant_coeff_variation_of_parameters", + "nth_linear_euler_eq_nonhomogeneous_variation_of_parameters", + "Liouville", + "2nd_linear_airy", + "2nd_linear_bessel", + "2nd_hypergeometric", + "2nd_hypergeometric_Integral", + "nth_order_reducible", + "2nd_power_series_ordinary", + "2nd_power_series_regular", + "nth_algebraic_Integral", + "separable_Integral", + "1st_exact_Integral", + "1st_linear_Integral", + "Bernoulli_Integral", + "1st_homogeneous_coeff_subs_indep_div_dep_Integral", + "1st_homogeneous_coeff_subs_dep_div_indep_Integral", + "almost_linear_Integral", + "linear_coefficients_Integral", + "separable_reduced_Integral", + "nth_linear_constant_coeff_variation_of_parameters_Integral", + "nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral", + "Liouville_Integral", + "2nd_nonlinear_autonomous_conserved", + "2nd_nonlinear_autonomous_conserved_Integral", + ) + + + +def get_numbered_constants(eq, num=1, start=1, prefix='C'): + """ + Returns a list of constants that do not occur + in eq already. + """ + + ncs = iter_numbered_constants(eq, start, prefix) + Cs = [next(ncs) for i in range(num)] + return (Cs[0] if num == 1 else tuple(Cs)) + + +def iter_numbered_constants(eq, start=1, prefix='C'): + """ + Returns an iterator of constants that do not occur + in eq already. + """ + + if isinstance(eq, (Expr, Eq)): + eq = [eq] + elif not iterable(eq): + raise ValueError("Expected Expr or iterable but got %s" % eq) + + atom_set = set().union(*[i.free_symbols for i in eq]) + func_set = set().union(*[i.atoms(Function) for i in eq]) + if func_set: + atom_set |= {Symbol(str(f.func)) for f in func_set} + return numbered_symbols(start=start, prefix=prefix, exclude=atom_set) + + +def dsolve(eq, func=None, hint="default", simplify=True, + ics= None, xi=None, eta=None, x0=0, n=6, **kwargs): + r""" + Solves any (supported) kind of ordinary differential equation and + system of ordinary differential equations. + + For single ordinary differential equation + ========================================= + + It is classified under this when number of equation in ``eq`` is one. + **Usage** + + ``dsolve(eq, f(x), hint)`` -> Solve ordinary differential equation + ``eq`` for function ``f(x)``, using method ``hint``. + + **Details** + + ``eq`` can be any supported ordinary differential equation (see the + :py:mod:`~sympy.solvers.ode` docstring for supported methods). + This can either be an :py:class:`~sympy.core.relational.Equality`, + or an expression, which is assumed to be equal to ``0``. + + ``f(x)`` is a function of one variable whose derivatives in that + variable make up the ordinary differential equation ``eq``. In + many cases it is not necessary to provide this; it will be + autodetected (and an error raised if it could not be detected). + + ``hint`` is the solving method that you want dsolve to use. Use + ``classify_ode(eq, f(x))`` to get all of the possible hints for an + ODE. The default hint, ``default``, will use whatever hint is + returned first by :py:meth:`~sympy.solvers.ode.classify_ode`. See + Hints below for more options that you can use for hint. + + ``simplify`` enables simplification by + :py:meth:`~sympy.solvers.ode.ode.odesimp`. See its docstring for more + information. Turn this off, for example, to disable solving of + solutions for ``func`` or simplification of arbitrary constants. + It will still integrate with this hint. Note that the solution may + contain more arbitrary constants than the order of the ODE with + this option enabled. + + ``xi`` and ``eta`` are the infinitesimal functions of an ordinary + differential equation. They are the infinitesimals of the Lie group + of point transformations for which the differential equation is + invariant. The user can specify values for the infinitesimals. If + nothing is specified, ``xi`` and ``eta`` are calculated using + :py:meth:`~sympy.solvers.ode.infinitesimals` with the help of various + heuristics. + + ``ics`` is the set of initial/boundary conditions for the differential equation. + It should be given in the form of ``{f(x0): x1, f(x).diff(x).subs(x, x2): + x3}`` and so on. For power series solutions, if no initial + conditions are specified ``f(0)`` is assumed to be ``C0`` and the power + series solution is calculated about 0. + + ``x0`` is the point about which the power series solution of a differential + equation is to be evaluated. + + ``n`` gives the exponent of the dependent variable up to which the power series + solution of a differential equation is to be evaluated. + + **Hints** + + Aside from the various solving methods, there are also some meta-hints + that you can pass to :py:meth:`~sympy.solvers.ode.dsolve`: + + ``default``: + This uses whatever hint is returned first by + :py:meth:`~sympy.solvers.ode.classify_ode`. This is the + default argument to :py:meth:`~sympy.solvers.ode.dsolve`. + + ``all``: + To make :py:meth:`~sympy.solvers.ode.dsolve` apply all + relevant classification hints, use ``dsolve(ODE, func, + hint="all")``. This will return a dictionary of + ``hint:solution`` terms. If a hint causes dsolve to raise the + ``NotImplementedError``, value of that hint's key will be the + exception object raised. The dictionary will also include + some special keys: + + - ``order``: The order of the ODE. See also + :py:meth:`~sympy.solvers.deutils.ode_order` in + ``deutils.py``. + - ``best``: The simplest hint; what would be returned by + ``best`` below. + - ``best_hint``: The hint that would produce the solution + given by ``best``. If more than one hint produces the best + solution, the first one in the tuple returned by + :py:meth:`~sympy.solvers.ode.classify_ode` is chosen. + - ``default``: The solution that would be returned by default. + This is the one produced by the hint that appears first in + the tuple returned by + :py:meth:`~sympy.solvers.ode.classify_ode`. + + ``all_Integral``: + This is the same as ``all``, except if a hint also has a + corresponding ``_Integral`` hint, it only returns the + ``_Integral`` hint. This is useful if ``all`` causes + :py:meth:`~sympy.solvers.ode.dsolve` to hang because of a + difficult or impossible integral. This meta-hint will also be + much faster than ``all``, because + :py:meth:`~sympy.core.expr.Expr.integrate` is an expensive + routine. + + ``best``: + To have :py:meth:`~sympy.solvers.ode.dsolve` try all methods + and return the simplest one. This takes into account whether + the solution is solvable in the function, whether it contains + any Integral classes (i.e. unevaluatable integrals), and + which one is the shortest in size. + + See also the :py:meth:`~sympy.solvers.ode.classify_ode` docstring for + more info on hints, and the :py:mod:`~sympy.solvers.ode` docstring for + a list of all supported hints. + + **Tips** + + - You can declare the derivative of an unknown function this way: + + >>> from sympy import Function, Derivative + >>> from sympy.abc import x # x is the independent variable + >>> f = Function("f")(x) # f is a function of x + >>> # f_ will be the derivative of f with respect to x + >>> f_ = Derivative(f, x) + + - See ``test_ode.py`` for many tests, which serves also as a set of + examples for how to use :py:meth:`~sympy.solvers.ode.dsolve`. + - :py:meth:`~sympy.solvers.ode.dsolve` always returns an + :py:class:`~sympy.core.relational.Equality` class (except for the + case when the hint is ``all`` or ``all_Integral``). If possible, it + solves the solution explicitly for the function being solved for. + Otherwise, it returns an implicit solution. + - Arbitrary constants are symbols named ``C1``, ``C2``, and so on. + - Because all solutions should be mathematically equivalent, some + hints may return the exact same result for an ODE. Often, though, + two different hints will return the same solution formatted + differently. The two should be equivalent. Also note that sometimes + the values of the arbitrary constants in two different solutions may + not be the same, because one constant may have "absorbed" other + constants into it. + - Do ``help(ode.ode_)`` to get help more information on a + specific hint, where ```` is the name of a hint without + ``_Integral``. + + For system of ordinary differential equations + ============================================= + + **Usage** + ``dsolve(eq, func)`` -> Solve a system of ordinary differential + equations ``eq`` for ``func`` being list of functions including + `x(t)`, `y(t)`, `z(t)` where number of functions in the list depends + upon the number of equations provided in ``eq``. + + **Details** + + ``eq`` can be any supported system of ordinary differential equations + This can either be an :py:class:`~sympy.core.relational.Equality`, + or an expression, which is assumed to be equal to ``0``. + + ``func`` holds ``x(t)`` and ``y(t)`` being functions of one variable which + together with some of their derivatives make up the system of ordinary + differential equation ``eq``. It is not necessary to provide this; it + will be autodetected (and an error raised if it could not be detected). + + **Hints** + + The hints are formed by parameters returned by classify_sysode, combining + them give hints name used later for forming method name. + + Examples + ======== + + >>> from sympy import Function, dsolve, Eq, Derivative, sin, cos, symbols + >>> from sympy.abc import x + >>> f = Function('f') + >>> dsolve(Derivative(f(x), x, x) + 9*f(x), f(x)) + Eq(f(x), C1*sin(3*x) + C2*cos(3*x)) + + >>> eq = sin(x)*cos(f(x)) + cos(x)*sin(f(x))*f(x).diff(x) + >>> dsolve(eq, hint='1st_exact') + [Eq(f(x), -acos(C1/cos(x)) + 2*pi), Eq(f(x), acos(C1/cos(x)))] + >>> dsolve(eq, hint='almost_linear') + [Eq(f(x), -acos(C1/cos(x)) + 2*pi), Eq(f(x), acos(C1/cos(x)))] + >>> t = symbols('t') + >>> x, y = symbols('x, y', cls=Function) + >>> eq = (Eq(Derivative(x(t),t), 12*t*x(t) + 8*y(t)), Eq(Derivative(y(t),t), 21*x(t) + 7*t*y(t))) + >>> dsolve(eq) + [Eq(x(t), C1*x0(t) + C2*x0(t)*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0(t)**2, t)), + Eq(y(t), C1*y0(t) + C2*(y0(t)*Integral(8*exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0(t)**2, t) + + exp(Integral(7*t, t))*exp(Integral(12*t, t))/x0(t)))] + >>> eq = (Eq(Derivative(x(t),t),x(t)*y(t)*sin(t)), Eq(Derivative(y(t),t),y(t)**2*sin(t))) + >>> dsolve(eq) + {Eq(x(t), -exp(C1)/(C2*exp(C1) - cos(t))), Eq(y(t), -1/(C1 - cos(t)))} + """ + if iterable(eq): + from sympy.solvers.ode.systems import dsolve_system + + # This may have to be changed in future + # when we have weakly and strongly + # connected components. This have to + # changed to show the systems that haven't + # been solved. + try: + sol = dsolve_system(eq, funcs=func, ics=ics, doit=True) + return sol[0] if len(sol) == 1 else sol + except NotImplementedError: + pass + + match = classify_sysode(eq, func) + + eq = match['eq'] + order = match['order'] + func = match['func'] + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + + # keep highest order term coefficient positive + for i in range(len(eq)): + for func_ in func: + if isinstance(func_, list): + pass + else: + if eq[i].coeff(diff(func[i],t,ode_order(eq[i], func[i]))).is_negative: + eq[i] = -eq[i] + match['eq'] = eq + if len(set(order.values()))!=1: + raise ValueError("It solves only those systems of equations whose orders are equal") + match['order'] = list(order.values())[0] + def recur_len(l): + return sum(recur_len(item) if isinstance(item,list) else 1 for item in l) + if recur_len(func) != len(eq): + raise ValueError("dsolve() and classify_sysode() work with " + "number of functions being equal to number of equations") + if match['type_of_equation'] is None: + raise NotImplementedError + else: + if match['is_linear'] == True: + solvefunc = globals()['sysode_linear_%(no_of_equation)seq_order%(order)s' % match] + else: + solvefunc = globals()['sysode_nonlinear_%(no_of_equation)seq_order%(order)s' % match] + sols = solvefunc(match) + if ics: + constants = Tuple(*sols).free_symbols - Tuple(*eq).free_symbols + solved_constants = solve_ics(sols, func, constants, ics) + return [sol.subs(solved_constants) for sol in sols] + return sols + else: + given_hint = hint # hint given by the user + + # See the docstring of _desolve for more details. + hints = _desolve(eq, func=func, + hint=hint, simplify=True, xi=xi, eta=eta, type='ode', ics=ics, + x0=x0, n=n, **kwargs) + eq = hints.pop('eq', eq) + all_ = hints.pop('all', False) + if all_: + retdict = {} + failed_hints = {} + gethints = classify_ode(eq, dict=True, hint='all') + orderedhints = gethints['ordered_hints'] + for hint in hints: + try: + rv = _helper_simplify(eq, hint, hints[hint], simplify) + except NotImplementedError as detail: + failed_hints[hint] = detail + else: + retdict[hint] = rv + func = hints[hint]['func'] + + retdict['best'] = min(list(retdict.values()), key=lambda x: + ode_sol_simplicity(x, func, trysolving=not simplify)) + if given_hint == 'best': + return retdict['best'] + for i in orderedhints: + if retdict['best'] == retdict.get(i, None): + retdict['best_hint'] = i + break + retdict['default'] = gethints['default'] + retdict['order'] = gethints['order'] + retdict.update(failed_hints) + return retdict + + else: + # The key 'hint' stores the hint needed to be solved for. + hint = hints['hint'] + return _helper_simplify(eq, hint, hints, simplify, ics=ics) + + +def _helper_simplify(eq, hint, match, simplify=True, ics=None, **kwargs): + r""" + Helper function of dsolve that calls the respective + :py:mod:`~sympy.solvers.ode` functions to solve for the ordinary + differential equations. This minimizes the computation in calling + :py:meth:`~sympy.solvers.deutils._desolve` multiple times. + """ + r = match + func = r['func'] + order = r['order'] + match = r[hint] + + if isinstance(match, SingleODESolver): + solvefunc = match + elif hint.endswith('_Integral'): + solvefunc = globals()['ode_' + hint[:-len('_Integral')]] + else: + solvefunc = globals()['ode_' + hint] + + free = eq.free_symbols + cons = lambda s: s.free_symbols.difference(free) + + if simplify: + # odesimp() will attempt to integrate, if necessary, apply constantsimp(), + # attempt to solve for func, and apply any other hint specific + # simplifications + if isinstance(solvefunc, SingleODESolver): + sols = solvefunc.get_general_solution() + else: + sols = solvefunc(eq, func, order, match) + if iterable(sols): + rv = [] + for s in sols: + simp = odesimp(eq, s, func, hint) + if iterable(simp): + rv.extend(simp) + else: + rv.append(simp) + else: + rv = odesimp(eq, sols, func, hint) + else: + # We still want to integrate (you can disable it separately with the hint) + if isinstance(solvefunc, SingleODESolver): + exprs = solvefunc.get_general_solution(simplify=False) + else: + match['simplify'] = False # Some hints can take advantage of this option + exprs = solvefunc(eq, func, order, match) + if isinstance(exprs, list): + rv = [_handle_Integral(expr, func, hint) for expr in exprs] + else: + rv = _handle_Integral(exprs, func, hint) + + if isinstance(rv, list): + assert all(isinstance(i, Eq) for i in rv), rv # if not => internal error + if simplify: + rv = _remove_redundant_solutions(eq, rv, order, func.args[0]) + if len(rv) == 1: + rv = rv[0] + if ics and 'power_series' not in hint: + if isinstance(rv, (Expr, Eq)): + solved_constants = solve_ics([rv], [r['func']], cons(rv), ics) + rv = rv.subs(solved_constants) + else: + rv1 = [] + for s in rv: + try: + solved_constants = solve_ics([s], [r['func']], cons(s), ics) + except ValueError: + continue + rv1.append(s.subs(solved_constants)) + if len(rv1) == 1: + return rv1[0] + rv = rv1 + return rv + + +def solve_ics(sols, funcs, constants, ics): + """ + Solve for the constants given initial conditions + + ``sols`` is a list of solutions. + + ``funcs`` is a list of functions. + + ``constants`` is a list of constants. + + ``ics`` is the set of initial/boundary conditions for the differential + equation. It should be given in the form of ``{f(x0): x1, + f(x).diff(x).subs(x, x2): x3}`` and so on. + + Returns a dictionary mapping constants to values. + ``solution.subs(constants)`` will replace the constants in ``solution``. + + Example + ======= + >>> # From dsolve(f(x).diff(x) - f(x), f(x)) + >>> from sympy import symbols, Eq, exp, Function + >>> from sympy.solvers.ode.ode import solve_ics + >>> f = Function('f') + >>> x, C1 = symbols('x C1') + >>> sols = [Eq(f(x), C1*exp(x))] + >>> funcs = [f(x)] + >>> constants = [C1] + >>> ics = {f(0): 2} + >>> solved_constants = solve_ics(sols, funcs, constants, ics) + >>> solved_constants + {C1: 2} + >>> sols[0].subs(solved_constants) + Eq(f(x), 2*exp(x)) + + """ + # Assume ics are of the form f(x0): value or Subs(diff(f(x), x, n), (x, + # x0)): value (currently checked by classify_ode). To solve, replace x + # with x0, f(x0) with value, then solve for constants. For f^(n)(x0), + # differentiate the solution n times, so that f^(n)(x) appears. + x = funcs[0].args[0] + diff_sols = [] + subs_sols = [] + diff_variables = set() + for funcarg, value in ics.items(): + if isinstance(funcarg, AppliedUndef): + x0 = funcarg.args[0] + matching_func = [f for f in funcs if f.func == funcarg.func][0] + S = sols + elif isinstance(funcarg, (Subs, Derivative)): + if isinstance(funcarg, Subs): + # Make sure it stays a subs. Otherwise subs below will produce + # a different looking term. + funcarg = funcarg.doit() + if isinstance(funcarg, Subs): + deriv = funcarg.expr + x0 = funcarg.point[0] + variables = funcarg.expr.variables + matching_func = deriv + elif isinstance(funcarg, Derivative): + deriv = funcarg + x0 = funcarg.variables[0] + variables = (x,)*len(funcarg.variables) + matching_func = deriv.subs(x0, x) + for sol in sols: + if sol.has(deriv.expr.func): + diff_sols.append(Eq(sol.lhs.diff(*variables), sol.rhs.diff(*variables))) + diff_variables.add(variables) + S = diff_sols + else: + raise NotImplementedError("Unrecognized initial condition") + + for sol in S: + if sol.has(matching_func): + sol2 = sol + sol2 = sol2.subs(x, x0) + sol2 = sol2.subs(funcarg, value) + # This check is necessary because of issue #15724 + if not isinstance(sol2, BooleanAtom) or not subs_sols: + subs_sols = [s for s in subs_sols if not isinstance(s, BooleanAtom)] + subs_sols.append(sol2) + + # TODO: Use solveset here + try: + solved_constants = solve(subs_sols, constants, dict=True) + except NotImplementedError: + solved_constants = [] + + # XXX: We can't differentiate between the solution not existing because of + # invalid initial conditions, and not existing because solve is not smart + # enough. If we could use solveset, this might be improvable, but for now, + # we use NotImplementedError in this case. + if not solved_constants: + raise ValueError("Couldn't solve for initial conditions") + + if solved_constants == True: + raise ValueError("Initial conditions did not produce any solutions for constants. Perhaps they are degenerate.") + + if len(solved_constants) > 1: + raise NotImplementedError("Initial conditions produced too many solutions for constants") + + return solved_constants[0] + +def classify_ode(eq, func=None, dict=False, ics=None, *, prep=True, xi=None, eta=None, n=None, **kwargs): + r""" + Returns a tuple of possible :py:meth:`~sympy.solvers.ode.dsolve` + classifications for an ODE. + + The tuple is ordered so that first item is the classification that + :py:meth:`~sympy.solvers.ode.dsolve` uses to solve the ODE by default. In + general, classifications at the near the beginning of the list will + produce better solutions faster than those near the end, thought there are + always exceptions. To make :py:meth:`~sympy.solvers.ode.dsolve` use a + different classification, use ``dsolve(ODE, func, + hint=)``. See also the + :py:meth:`~sympy.solvers.ode.dsolve` docstring for different meta-hints + you can use. + + If ``dict`` is true, :py:meth:`~sympy.solvers.ode.classify_ode` will + return a dictionary of ``hint:match`` expression terms. This is intended + for internal use by :py:meth:`~sympy.solvers.ode.dsolve`. Note that + because dictionaries are ordered arbitrarily, this will most likely not be + in the same order as the tuple. + + You can get help on different hints by executing + ``help(ode.ode_hintname)``, where ``hintname`` is the name of the hint + without ``_Integral``. + + See :py:data:`~sympy.solvers.ode.allhints` or the + :py:mod:`~sympy.solvers.ode` docstring for a list of all supported hints + that can be returned from :py:meth:`~sympy.solvers.ode.classify_ode`. + + Notes + ===== + + These are remarks on hint names. + + ``_Integral`` + + If a classification has ``_Integral`` at the end, it will return the + expression with an unevaluated :py:class:`~.Integral` + class in it. Note that a hint may do this anyway if + :py:meth:`~sympy.core.expr.Expr.integrate` cannot do the integral, + though just using an ``_Integral`` will do so much faster. Indeed, an + ``_Integral`` hint will always be faster than its corresponding hint + without ``_Integral`` because + :py:meth:`~sympy.core.expr.Expr.integrate` is an expensive routine. + If :py:meth:`~sympy.solvers.ode.dsolve` hangs, it is probably because + :py:meth:`~sympy.core.expr.Expr.integrate` is hanging on a tough or + impossible integral. Try using an ``_Integral`` hint or + ``all_Integral`` to get it return something. + + Note that some hints do not have ``_Integral`` counterparts. This is + because :py:func:`~sympy.integrals.integrals.integrate` is not used in + solving the ODE for those method. For example, `n`\th order linear + homogeneous ODEs with constant coefficients do not require integration + to solve, so there is no + ``nth_linear_homogeneous_constant_coeff_Integrate`` hint. You can + easily evaluate any unevaluated + :py:class:`~sympy.integrals.integrals.Integral`\s in an expression by + doing ``expr.doit()``. + + Ordinals + + Some hints contain an ordinal such as ``1st_linear``. This is to help + differentiate them from other hints, as well as from other methods + that may not be implemented yet. If a hint has ``nth`` in it, such as + the ``nth_linear`` hints, this means that the method used to applies + to ODEs of any order. + + ``indep`` and ``dep`` + + Some hints contain the words ``indep`` or ``dep``. These reference + the independent variable and the dependent function, respectively. For + example, if an ODE is in terms of `f(x)`, then ``indep`` will refer to + `x` and ``dep`` will refer to `f`. + + ``subs`` + + If a hints has the word ``subs`` in it, it means that the ODE is solved + by substituting the expression given after the word ``subs`` for a + single dummy variable. This is usually in terms of ``indep`` and + ``dep`` as above. The substituted expression will be written only in + characters allowed for names of Python objects, meaning operators will + be spelled out. For example, ``indep``/``dep`` will be written as + ``indep_div_dep``. + + ``coeff`` + + The word ``coeff`` in a hint refers to the coefficients of something + in the ODE, usually of the derivative terms. See the docstring for + the individual methods for more info (``help(ode)``). This is + contrast to ``coefficients``, as in ``undetermined_coefficients``, + which refers to the common name of a method. + + ``_best`` + + Methods that have more than one fundamental way to solve will have a + hint for each sub-method and a ``_best`` meta-classification. This + will evaluate all hints and return the best, using the same + considerations as the normal ``best`` meta-hint. + + + Examples + ======== + + >>> from sympy import Function, classify_ode, Eq + >>> from sympy.abc import x + >>> f = Function('f') + >>> classify_ode(Eq(f(x).diff(x), 0), f(x)) + ('nth_algebraic', + 'separable', + '1st_exact', + '1st_linear', + 'Bernoulli', + '1st_homogeneous_coeff_best', + '1st_homogeneous_coeff_subs_indep_div_dep', + '1st_homogeneous_coeff_subs_dep_div_indep', + '1st_power_series', 'lie_group', 'nth_linear_constant_coeff_homogeneous', + 'nth_linear_euler_eq_homogeneous', + 'nth_algebraic_Integral', 'separable_Integral', '1st_exact_Integral', + '1st_linear_Integral', 'Bernoulli_Integral', + '1st_homogeneous_coeff_subs_indep_div_dep_Integral', + '1st_homogeneous_coeff_subs_dep_div_indep_Integral') + >>> classify_ode(f(x).diff(x, 2) + 3*f(x).diff(x) + 2*f(x) - 4) + ('factorable', 'nth_linear_constant_coeff_undetermined_coefficients', + 'nth_linear_constant_coeff_variation_of_parameters', + 'nth_linear_constant_coeff_variation_of_parameters_Integral') + + """ + ics = sympify(ics) + + if func and len(func.args) != 1: + raise ValueError("dsolve() and classify_ode() only " + "work with functions of one variable, not %s" % func) + + if isinstance(eq, Equality): + eq = eq.lhs - eq.rhs + + # Some methods want the unprocessed equation + eq_orig = eq + + if prep or func is None: + eq, func_ = _preprocess(eq, func) + if func is None: + func = func_ + x = func.args[0] + f = func.func + y = Dummy('y') + terms = 5 if n is None else n + + order = ode_order(eq, f(x)) + # hint:matchdict or hint:(tuple of matchdicts) + # Also will contain "default": and "order":order items. + matching_hints = {"order": order} + + df = f(x).diff(x) + a = Wild('a', exclude=[f(x)]) + d = Wild('d', exclude=[df, f(x).diff(x, 2)]) + e = Wild('e', exclude=[df]) + n = Wild('n', exclude=[x, f(x), df]) + c1 = Wild('c1', exclude=[x]) + a3 = Wild('a3', exclude=[f(x), df, f(x).diff(x, 2)]) + b3 = Wild('b3', exclude=[f(x), df, f(x).diff(x, 2)]) + c3 = Wild('c3', exclude=[f(x), df, f(x).diff(x, 2)]) + boundary = {} # Used to extract initial conditions + C1 = Symbol("C1") + + # Preprocessing to get the initial conditions out + if ics is not None: + for funcarg in ics: + # Separating derivatives + if isinstance(funcarg, (Subs, Derivative)): + # f(x).diff(x).subs(x, 0) is a Subs, but f(x).diff(x).subs(x, + # y) is a Derivative + if isinstance(funcarg, Subs): + deriv = funcarg.expr + old = funcarg.variables[0] + new = funcarg.point[0] + elif isinstance(funcarg, Derivative): + deriv = funcarg + # No information on this. Just assume it was x + old = x + new = funcarg.variables[0] + + if (isinstance(deriv, Derivative) and isinstance(deriv.args[0], + AppliedUndef) and deriv.args[0].func == f and + len(deriv.args[0].args) == 1 and old == x and not + new.has(x) and all(i == deriv.variables[0] for i in + deriv.variables) and x not in ics[funcarg].free_symbols): + + dorder = ode_order(deriv, x) + temp = 'f' + str(dorder) + boundary.update({temp: new, temp + 'val': ics[funcarg]}) + else: + raise ValueError("Invalid boundary conditions for Derivatives") + + + # Separating functions + elif isinstance(funcarg, AppliedUndef): + if (funcarg.func == f and len(funcarg.args) == 1 and + not funcarg.args[0].has(x) and x not in ics[funcarg].free_symbols): + boundary.update({'f0': funcarg.args[0], 'f0val': ics[funcarg]}) + else: + raise ValueError("Invalid boundary conditions for Function") + + else: + raise ValueError("Enter boundary conditions of the form ics={f(point): value, f(x).diff(x, order).subs(x, point): value}") + + ode = SingleODEProblem(eq_orig, func, x, prep=prep, xi=xi, eta=eta) + user_hint = kwargs.get('hint', 'default') + # Used when dsolve is called without an explicit hint. + # We exit early to return the first valid match + early_exit = (user_hint=='default') + if user_hint.endswith('_Integral'): + user_hint = user_hint[:-len('_Integral')] + user_map = solver_map + # An explicit hint has been given to dsolve + # Skip matching code for other hints + if user_hint not in ['default', 'all', 'all_Integral', 'best'] and user_hint in solver_map: + user_map = {user_hint: solver_map[user_hint]} + + for hint in user_map: + solver = user_map[hint](ode) + if solver.matches(): + matching_hints[hint] = solver + if user_map[hint].has_integral: + matching_hints[hint + "_Integral"] = solver + if dict and early_exit: + matching_hints["default"] = hint + return matching_hints + + eq = expand(eq) + # Precondition to try remove f(x) from highest order derivative + reduced_eq = None + if eq.is_Add: + deriv_coef = eq.coeff(f(x).diff(x, order)) + if deriv_coef not in (1, 0): + r = deriv_coef.match(a*f(x)**c1) + if r and r[c1]: + den = f(x)**r[c1] + reduced_eq = Add(*[arg/den for arg in eq.args]) + if not reduced_eq: + reduced_eq = eq + + if order == 1: + + # NON-REDUCED FORM OF EQUATION matches + r = collect(eq, df, exact=True).match(d + e * df) + if r: + r['d'] = d + r['e'] = e + r['y'] = y + r[d] = r[d].subs(f(x), y) + r[e] = r[e].subs(f(x), y) + + # FIRST ORDER POWER SERIES WHICH NEEDS INITIAL CONDITIONS + # TODO: Hint first order series should match only if d/e is analytic. + # For now, only d/e and (d/e).diff(arg) is checked for existence at + # at a given point. + # This is currently done internally in ode_1st_power_series. + point = boundary.get('f0', 0) + value = boundary.get('f0val', C1) + check = cancel(r[d]/r[e]) + check1 = check.subs({x: point, y: value}) + if not check1.has(oo) and not check1.has(zoo) and \ + not check1.has(nan) and not check1.has(-oo): + check2 = (check1.diff(x)).subs({x: point, y: value}) + if not check2.has(oo) and not check2.has(zoo) and \ + not check2.has(nan) and not check2.has(-oo): + rseries = r.copy() + rseries.update({'terms': terms, 'f0': point, 'f0val': value}) + matching_hints["1st_power_series"] = rseries + + elif order == 2: + # Homogeneous second order differential equation of the form + # a3*f(x).diff(x, 2) + b3*f(x).diff(x) + c3 + # It has a definite power series solution at point x0 if, b3/a3 and c3/a3 + # are analytic at x0. + deq = a3*(f(x).diff(x, 2)) + b3*df + c3*f(x) + r = collect(reduced_eq, + [f(x).diff(x, 2), f(x).diff(x), f(x)]).match(deq) + ordinary = False + if r: + if not all(r[key].is_polynomial() for key in r): + n, d = reduced_eq.as_numer_denom() + reduced_eq = expand(n) + r = collect(reduced_eq, + [f(x).diff(x, 2), f(x).diff(x), f(x)]).match(deq) + if r and r[a3] != 0: + p = cancel(r[b3]/r[a3]) # Used below + q = cancel(r[c3]/r[a3]) # Used below + point = kwargs.get('x0', 0) + check = p.subs(x, point) + if not check.has(oo, nan, zoo, -oo): + check = q.subs(x, point) + if not check.has(oo, nan, zoo, -oo): + ordinary = True + r.update({'a3': a3, 'b3': b3, 'c3': c3, 'x0': point, 'terms': terms}) + matching_hints["2nd_power_series_ordinary"] = r + + # Checking if the differential equation has a regular singular point + # at x0. It has a regular singular point at x0, if (b3/a3)*(x - x0) + # and (c3/a3)*((x - x0)**2) are analytic at x0. + if not ordinary: + p = cancel((x - point)*p) + check = p.subs(x, point) + if not check.has(oo, nan, zoo, -oo): + q = cancel(((x - point)**2)*q) + check = q.subs(x, point) + if not check.has(oo, nan, zoo, -oo): + coeff_dict = {'p': p, 'q': q, 'x0': point, 'terms': terms} + matching_hints["2nd_power_series_regular"] = coeff_dict + + + # Order keys based on allhints. + retlist = [i for i in allhints if i in matching_hints] + if dict: + # Dictionaries are ordered arbitrarily, so make note of which + # hint would come first for dsolve(). Use an ordered dict in Py 3. + matching_hints["default"] = retlist[0] if retlist else None + matching_hints["ordered_hints"] = tuple(retlist) + return matching_hints + else: + return tuple(retlist) + + +def classify_sysode(eq, funcs=None, **kwargs): + r""" + Returns a dictionary of parameter names and values that define the system + of ordinary differential equations in ``eq``. + The parameters are further used in + :py:meth:`~sympy.solvers.ode.dsolve` for solving that system. + + Some parameter names and values are: + + 'is_linear' (boolean), which tells whether the given system is linear. + Note that "linear" here refers to the operator: terms such as ``x*diff(x,t)`` are + nonlinear, whereas terms like ``sin(t)*diff(x,t)`` are still linear operators. + + 'func' (list) contains the :py:class:`~sympy.core.function.Function`s that + appear with a derivative in the ODE, i.e. those that we are trying to solve + the ODE for. + + 'order' (dict) with the maximum derivative for each element of the 'func' + parameter. + + 'func_coeff' (dict or Matrix) with the coefficient for each triple ``(equation number, + function, order)```. The coefficients are those subexpressions that do not + appear in 'func', and hence can be considered constant for purposes of ODE + solving. The value of this parameter can also be a Matrix if the system of ODEs are + linear first order of the form X' = AX where X is the vector of dependent variables. + Here, this function returns the coefficient matrix A. + + 'eq' (list) with the equations from ``eq``, sympified and transformed into + expressions (we are solving for these expressions to be zero). + + 'no_of_equations' (int) is the number of equations (same as ``len(eq)``). + + 'type_of_equation' (string) is an internal classification of the type of + ODE. + + 'is_constant' (boolean), which tells if the system of ODEs is constant coefficient + or not. This key is temporary addition for now and is in the match dict only when + the system of ODEs is linear first order constant coefficient homogeneous. So, this + key's value is True for now if it is available else it does not exist. + + 'is_homogeneous' (boolean), which tells if the system of ODEs is homogeneous. Like the + key 'is_constant', this key is a temporary addition and it is True since this key value + is available only when the system is linear first order constant coefficient homogeneous. + + References + ========== + -https://eqworld.ipmnet.ru/en/solutions/sysode/sode-toc1.htm + -A. D. Polyanin and A. V. Manzhirov, Handbook of Mathematics for Engineers and Scientists + + Examples + ======== + + >>> from sympy import Function, Eq, symbols, diff + >>> from sympy.solvers.ode.ode import classify_sysode + >>> from sympy.abc import t + >>> f, x, y = symbols('f, x, y', cls=Function) + >>> k, l, m, n = symbols('k, l, m, n', Integer=True) + >>> x1 = diff(x(t), t) ; y1 = diff(y(t), t) + >>> x2 = diff(x(t), t, t) ; y2 = diff(y(t), t, t) + >>> eq = (Eq(x1, 12*x(t) - 6*y(t)), Eq(y1, 11*x(t) + 3*y(t))) + >>> classify_sysode(eq) + {'eq': [-12*x(t) + 6*y(t) + Derivative(x(t), t), -11*x(t) - 3*y(t) + Derivative(y(t), t)], 'func': [x(t), y(t)], + 'func_coeff': {(0, x(t), 0): -12, (0, x(t), 1): 1, (0, y(t), 0): 6, (0, y(t), 1): 0, (1, x(t), 0): -11, (1, x(t), 1): 0, (1, y(t), 0): -3, (1, y(t), 1): 1}, 'is_linear': True, 'no_of_equation': 2, 'order': {x(t): 1, y(t): 1}, 'type_of_equation': None} + >>> eq = (Eq(diff(x(t),t), 5*t*x(t) + t**2*y(t) + 2), Eq(diff(y(t),t), -t**2*x(t) + 5*t*y(t))) + >>> classify_sysode(eq) + {'eq': [-t**2*y(t) - 5*t*x(t) + Derivative(x(t), t) - 2, t**2*x(t) - 5*t*y(t) + Derivative(y(t), t)], + 'func': [x(t), y(t)], 'func_coeff': {(0, x(t), 0): -5*t, (0, x(t), 1): 1, (0, y(t), 0): -t**2, (0, y(t), 1): 0, + (1, x(t), 0): t**2, (1, x(t), 1): 0, (1, y(t), 0): -5*t, (1, y(t), 1): 1}, 'is_linear': True, 'no_of_equation': 2, + 'order': {x(t): 1, y(t): 1}, 'type_of_equation': None} + + """ + + # Sympify equations and convert iterables of equations into + # a list of equations + def _sympify(eq): + return list(map(sympify, eq if iterable(eq) else [eq])) + + eq, funcs = (_sympify(w) for w in [eq, funcs]) + for i, fi in enumerate(eq): + if isinstance(fi, Equality): + eq[i] = fi.lhs - fi.rhs + + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + matching_hints = {"no_of_equation":i+1} + matching_hints['eq'] = eq + if i==0: + raise ValueError("classify_sysode() works for systems of ODEs. " + "For scalar ODEs, classify_ode should be used") + + # find all the functions if not given + order = {} + if funcs==[None]: + funcs = _extract_funcs(eq) + + funcs = list(set(funcs)) + if len(funcs) != len(eq): + raise ValueError("Number of functions given is not equal to the number of equations %s" % funcs) + + # This logic of list of lists in funcs to + # be replaced later. + func_dict = {} + for func in funcs: + if not order.get(func, False): + max_order = 0 + for i, eqs_ in enumerate(eq): + order_ = ode_order(eqs_,func) + if max_order < order_: + max_order = order_ + eq_no = i + if eq_no in func_dict: + func_dict[eq_no] = [func_dict[eq_no], func] + else: + func_dict[eq_no] = func + order[func] = max_order + + funcs = [func_dict[i] for i in range(len(func_dict))] + matching_hints['func'] = funcs + for func in funcs: + if isinstance(func, list): + for func_elem in func: + if len(func_elem.args) != 1: + raise ValueError("dsolve() and classify_sysode() work with " + "functions of one variable only, not %s" % func) + else: + if func and len(func.args) != 1: + raise ValueError("dsolve() and classify_sysode() work with " + "functions of one variable only, not %s" % func) + + # find the order of all equation in system of odes + matching_hints["order"] = order + + # find coefficients of terms f(t), diff(f(t),t) and higher derivatives + # and similarly for other functions g(t), diff(g(t),t) in all equations. + # Here j denotes the equation number, funcs[l] denotes the function about + # which we are talking about and k denotes the order of function funcs[l] + # whose coefficient we are calculating. + def linearity_check(eqs, j, func, is_linear_): + for k in range(order[func] + 1): + func_coef[j, func, k] = collect(eqs.expand(), [diff(func, t, k)]).coeff(diff(func, t, k)) + if is_linear_ == True: + if func_coef[j, func, k] == 0: + if k == 0: + coef = eqs.as_independent(func, as_Add=True)[1] + for xr in range(1, ode_order(eqs,func) + 1): + coef -= eqs.as_independent(diff(func, t, xr), as_Add=True)[1] + if coef != 0: + is_linear_ = False + else: + if eqs.as_independent(diff(func, t, k), as_Add=True)[1]: + is_linear_ = False + else: + for func_ in funcs: + if isinstance(func_, list): + for elem_func_ in func_: + dep = func_coef[j, func, k].as_independent(elem_func_, as_Add=True)[1] + if dep != 0: + is_linear_ = False + else: + dep = func_coef[j, func, k].as_independent(func_, as_Add=True)[1] + if dep != 0: + is_linear_ = False + return is_linear_ + + func_coef = {} + is_linear = True + for j, eqs in enumerate(eq): + for func in funcs: + if isinstance(func, list): + for func_elem in func: + is_linear = linearity_check(eqs, j, func_elem, is_linear) + else: + is_linear = linearity_check(eqs, j, func, is_linear) + matching_hints['func_coeff'] = func_coef + matching_hints['is_linear'] = is_linear + + + if len(set(order.values())) == 1: + order_eq = list(matching_hints['order'].values())[0] + if matching_hints['is_linear'] == True: + if matching_hints['no_of_equation'] == 2: + if order_eq == 1: + type_of_equation = check_linear_2eq_order1(eq, funcs, func_coef) + else: + type_of_equation = None + # If the equation does not match up with any of the + # general case solvers in systems.py and the number + # of equations is greater than 2, then NotImplementedError + # should be raised. + else: + type_of_equation = None + + else: + if matching_hints['no_of_equation'] == 2: + if order_eq == 1: + type_of_equation = check_nonlinear_2eq_order1(eq, funcs, func_coef) + else: + type_of_equation = None + elif matching_hints['no_of_equation'] == 3: + if order_eq == 1: + type_of_equation = check_nonlinear_3eq_order1(eq, funcs, func_coef) + else: + type_of_equation = None + else: + type_of_equation = None + else: + type_of_equation = None + + matching_hints['type_of_equation'] = type_of_equation + + return matching_hints + + +def check_linear_2eq_order1(eq, func, func_coef): + x = func[0].func + y = func[1].func + fc = func_coef + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + r = {} + # for equations Eq(a1*diff(x(t),t), b1*x(t) + c1*y(t) + d1) + # and Eq(a2*diff(y(t),t), b2*x(t) + c2*y(t) + d2) + r['a1'] = fc[0,x(t),1] ; r['a2'] = fc[1,y(t),1] + r['b1'] = -fc[0,x(t),0]/fc[0,x(t),1] ; r['b2'] = -fc[1,x(t),0]/fc[1,y(t),1] + r['c1'] = -fc[0,y(t),0]/fc[0,x(t),1] ; r['c2'] = -fc[1,y(t),0]/fc[1,y(t),1] + forcing = [S.Zero,S.Zero] + for i in range(2): + for j in Add.make_args(eq[i]): + if not j.has(x(t), y(t)): + forcing[i] += j + if not (forcing[0].has(t) or forcing[1].has(t)): + # We can handle homogeneous case and simple constant forcings + r['d1'] = forcing[0] + r['d2'] = forcing[1] + else: + # Issue #9244: nonhomogeneous linear systems are not supported + return None + + # Conditions to check for type 6 whose equations are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and + # Eq(diff(y(t),t), a*[f(t) + a*h(t)]x(t) + a*[g(t) - h(t)]*y(t)) + p = 0 + q = 0 + p1 = cancel(r['b2']/(cancel(r['b2']/r['c2']).as_numer_denom()[0])) + p2 = cancel(r['b1']/(cancel(r['b1']/r['c1']).as_numer_denom()[0])) + for n, i in enumerate([p1, p2]): + for j in Mul.make_args(collect_const(i)): + if not j.has(t): + q = j + if q and n==0: + if ((r['b2']/j - r['b1'])/(r['c1'] - r['c2']/j)) == j: + p = 1 + elif q and n==1: + if ((r['b1']/j - r['b2'])/(r['c2'] - r['c1']/j)) == j: + p = 2 + # End of condition for type 6 + + if r['d1']!=0 or r['d2']!=0: + return None + else: + if not any(r[k].has(t) for k in 'a1 a2 b1 b2 c1 c2'.split()): + return None + else: + r['b1'] = r['b1']/r['a1'] ; r['b2'] = r['b2']/r['a2'] + r['c1'] = r['c1']/r['a1'] ; r['c2'] = r['c2']/r['a2'] + if p: + return "type6" + else: + # Equations for type 7 are Eq(diff(x(t),t), f(t)*x(t) + g(t)*y(t)) and Eq(diff(y(t),t), h(t)*x(t) + p(t)*y(t)) + return "type7" +def check_nonlinear_2eq_order1(eq, func, func_coef): + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + f = Wild('f') + g = Wild('g') + u, v = symbols('u, v', cls=Dummy) + def check_type(x, y): + r1 = eq[0].match(t*diff(x(t),t) - x(t) + f) + r2 = eq[1].match(t*diff(y(t),t) - y(t) + g) + if not (r1 and r2): + r1 = eq[0].match(diff(x(t),t) - x(t)/t + f/t) + r2 = eq[1].match(diff(y(t),t) - y(t)/t + g/t) + if not (r1 and r2): + r1 = (-eq[0]).match(t*diff(x(t),t) - x(t) + f) + r2 = (-eq[1]).match(t*diff(y(t),t) - y(t) + g) + if not (r1 and r2): + r1 = (-eq[0]).match(diff(x(t),t) - x(t)/t + f/t) + r2 = (-eq[1]).match(diff(y(t),t) - y(t)/t + g/t) + if r1 and r2 and not (r1[f].subs(diff(x(t),t),u).subs(diff(y(t),t),v).has(t) \ + or r2[g].subs(diff(x(t),t),u).subs(diff(y(t),t),v).has(t)): + return 'type5' + else: + return None + for func_ in func: + if isinstance(func_, list): + x = func[0][0].func + y = func[0][1].func + eq_type = check_type(x, y) + if not eq_type: + eq_type = check_type(y, x) + return eq_type + x = func[0].func + y = func[1].func + fc = func_coef + n = Wild('n', exclude=[x(t),y(t)]) + f1 = Wild('f1', exclude=[v,t]) + f2 = Wild('f2', exclude=[v,t]) + g1 = Wild('g1', exclude=[u,t]) + g2 = Wild('g2', exclude=[u,t]) + for i in range(2): + eqs = 0 + for terms in Add.make_args(eq[i]): + eqs += terms/fc[i,func[i],1] + eq[i] = eqs + r = eq[0].match(diff(x(t),t) - x(t)**n*f) + if r: + g = (diff(y(t),t) - eq[1])/r[f] + if r and not (g.has(x(t)) or g.subs(y(t),v).has(t) or r[f].subs(x(t),u).subs(y(t),v).has(t)): + return 'type1' + r = eq[0].match(diff(x(t),t) - exp(n*x(t))*f) + if r: + g = (diff(y(t),t) - eq[1])/r[f] + if r and not (g.has(x(t)) or g.subs(y(t),v).has(t) or r[f].subs(x(t),u).subs(y(t),v).has(t)): + return 'type2' + g = Wild('g') + r1 = eq[0].match(diff(x(t),t) - f) + r2 = eq[1].match(diff(y(t),t) - g) + if r1 and r2 and not (r1[f].subs(x(t),u).subs(y(t),v).has(t) or \ + r2[g].subs(x(t),u).subs(y(t),v).has(t)): + return 'type3' + r1 = eq[0].match(diff(x(t),t) - f) + r2 = eq[1].match(diff(y(t),t) - g) + num, den = ( + (r1[f].subs(x(t),u).subs(y(t),v))/ + (r2[g].subs(x(t),u).subs(y(t),v))).as_numer_denom() + R1 = num.match(f1*g1) + R2 = den.match(f2*g2) + # phi = (r1[f].subs(x(t),u).subs(y(t),v))/num + if R1 and R2: + return 'type4' + return None + + +def check_nonlinear_2eq_order2(eq, func, func_coef): + return None + +def check_nonlinear_3eq_order1(eq, func, func_coef): + x = func[0].func + y = func[1].func + z = func[2].func + fc = func_coef + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + u, v, w = symbols('u, v, w', cls=Dummy) + a = Wild('a', exclude=[x(t), y(t), z(t), t]) + b = Wild('b', exclude=[x(t), y(t), z(t), t]) + c = Wild('c', exclude=[x(t), y(t), z(t), t]) + f = Wild('f') + F1 = Wild('F1') + F2 = Wild('F2') + F3 = Wild('F3') + for i in range(3): + eqs = 0 + for terms in Add.make_args(eq[i]): + eqs += terms/fc[i,func[i],1] + eq[i] = eqs + r1 = eq[0].match(diff(x(t),t) - a*y(t)*z(t)) + r2 = eq[1].match(diff(y(t),t) - b*z(t)*x(t)) + r3 = eq[2].match(diff(z(t),t) - c*x(t)*y(t)) + if r1 and r2 and r3: + num1, den1 = r1[a].as_numer_denom() + num2, den2 = r2[b].as_numer_denom() + num3, den3 = r3[c].as_numer_denom() + if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]): + return 'type1' + r = eq[0].match(diff(x(t),t) - y(t)*z(t)*f) + if r: + r1 = collect_const(r[f]).match(a*f) + r2 = ((diff(y(t),t) - eq[1])/r1[f]).match(b*z(t)*x(t)) + r3 = ((diff(z(t),t) - eq[2])/r1[f]).match(c*x(t)*y(t)) + if r1 and r2 and r3: + num1, den1 = r1[a].as_numer_denom() + num2, den2 = r2[b].as_numer_denom() + num3, den3 = r3[c].as_numer_denom() + if solve([num1*u-den1*(v-w), num2*v-den2*(w-u), num3*w-den3*(u-v)],[u, v]): + return 'type2' + r = eq[0].match(diff(x(t),t) - (F2-F3)) + if r: + r1 = collect_const(r[F2]).match(c*F2) + r1.update(collect_const(r[F3]).match(b*F3)) + if r1: + if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): + r1[F2], r1[F3] = r1[F3], r1[F2] + r1[c], r1[b] = -r1[b], -r1[c] + r2 = eq[1].match(diff(y(t),t) - a*r1[F3] + r1[c]*F1) + if r2: + r3 = (eq[2] == diff(z(t),t) - r1[b]*r2[F1] + r2[a]*r1[F2]) + if r1 and r2 and r3: + return 'type3' + r = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3) + if r: + r1 = collect_const(r[F2]).match(c*F2) + r1.update(collect_const(r[F3]).match(b*F3)) + if r1: + if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): + r1[F2], r1[F3] = r1[F3], r1[F2] + r1[c], r1[b] = -r1[b], -r1[c] + r2 = (diff(y(t),t) - eq[1]).match(a*x(t)*r1[F3] - r1[c]*z(t)*F1) + if r2: + r3 = (diff(z(t),t) - eq[2] == r1[b]*y(t)*r2[F1] - r2[a]*x(t)*r1[F2]) + if r1 and r2 and r3: + return 'type4' + r = (diff(x(t),t) - eq[0]).match(x(t)*(F2 - F3)) + if r: + r1 = collect_const(r[F2]).match(c*F2) + r1.update(collect_const(r[F3]).match(b*F3)) + if r1: + if eq[1].has(r1[F2]) and not eq[1].has(r1[F3]): + r1[F2], r1[F3] = r1[F3], r1[F2] + r1[c], r1[b] = -r1[b], -r1[c] + r2 = (diff(y(t),t) - eq[1]).match(y(t)*(a*r1[F3] - r1[c]*F1)) + if r2: + r3 = (diff(z(t),t) - eq[2] == z(t)*(r1[b]*r2[F1] - r2[a]*r1[F2])) + if r1 and r2 and r3: + return 'type5' + return None + + +def check_nonlinear_3eq_order2(eq, func, func_coef): + return None + + +@vectorize(0) +def odesimp(ode, eq, func, hint): + r""" + Simplifies solutions of ODEs, including trying to solve for ``func`` and + running :py:meth:`~sympy.solvers.ode.constantsimp`. + + It may use knowledge of the type of solution that the hint returns to + apply additional simplifications. + + It also attempts to integrate any :py:class:`~sympy.integrals.integrals.Integral`\s + in the expression, if the hint is not an ``_Integral`` hint. + + This function should have no effect on expressions returned by + :py:meth:`~sympy.solvers.ode.dsolve`, as + :py:meth:`~sympy.solvers.ode.dsolve` already calls + :py:meth:`~sympy.solvers.ode.ode.odesimp`, but the individual hint functions + do not call :py:meth:`~sympy.solvers.ode.ode.odesimp` (because the + :py:meth:`~sympy.solvers.ode.dsolve` wrapper does). Therefore, this + function is designed for mainly internal use. + + Examples + ======== + + >>> from sympy import sin, symbols, dsolve, pprint, Function + >>> from sympy.solvers.ode.ode import odesimp + >>> x, u2, C1= symbols('x,u2,C1') + >>> f = Function('f') + + >>> eq = dsolve(x*f(x).diff(x) - f(x) - x*sin(f(x)/x), f(x), + ... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral', + ... simplify=False) + >>> pprint(eq, wrap_line=False) + x + ---- + f(x) + / + | + | / 1 \ + | -|u1 + -------| + | | /1 \| + | | sin|--|| + | \ \u1// + log(f(x)) = log(C1) + | ---------------- d(u1) + | 2 + | u1 + | + / + + >>> pprint(odesimp(eq, f(x), 1, {C1}, + ... hint='1st_homogeneous_coeff_subs_indep_div_dep' + ... )) #doctest: +SKIP + x + --------- = C1 + /f(x)\ + tan|----| + \2*x / + + """ + x = func.args[0] + f = func.func + C1 = get_numbered_constants(eq, num=1) + constants = eq.free_symbols - ode.free_symbols + + # First, integrate if the hint allows it. + eq = _handle_Integral(eq, func, hint) + if hint.startswith("nth_linear_euler_eq_nonhomogeneous"): + eq = simplify(eq) + if not isinstance(eq, Equality): + raise TypeError("eq should be an instance of Equality") + + # allow simplifications under assumption that symbols are nonzero + eq = eq.xreplace((_:={i: Dummy(nonzero=True) for i in constants})).xreplace({_[i]: i for i in _}) + + # Second, clean up the arbitrary constants. + # Right now, nth linear hints can put as many as 2*order constants in an + # expression. If that number grows with another hint, the third argument + # here should be raised accordingly, or constantsimp() rewritten to handle + # an arbitrary number of constants. + eq = constantsimp(eq, constants) + + # Lastly, now that we have cleaned up the expression, try solving for func. + # When CRootOf is implemented in solve(), we will want to return a CRootOf + # every time instead of an Equality. + + # Get the f(x) on the left if possible. + if eq.rhs == func and not eq.lhs.has(func): + eq = [Eq(eq.rhs, eq.lhs)] + + # make sure we are working with lists of solutions in simplified form. + if eq.lhs == func and not eq.rhs.has(func): + # The solution is already solved + eq = [eq] + + else: + # The solution is not solved, so try to solve it + try: + floats = any(i.is_Float for i in eq.atoms(Number)) + eqsol = solve(eq, func, force=True, rational=False if floats else None) + if not eqsol: + raise NotImplementedError + except (NotImplementedError, PolynomialError): + eq = [eq] + else: + def _expand(expr): + numer, denom = expr.as_numer_denom() + + if denom.is_Add: + return expr + else: + return powsimp(expr.expand(), combine='exp', deep=True) + + # XXX: the rest of odesimp() expects each ``t`` to be in a + # specific normal form: rational expression with numerator + # expanded, but with combined exponential functions (at + # least in this setup all tests pass). + eq = [Eq(f(x), _expand(t)) for t in eqsol] + + # special simplification of the lhs. + if hint.startswith("1st_homogeneous_coeff"): + for j, eqi in enumerate(eq): + newi = logcombine(eqi, force=True) + if isinstance(newi.lhs, log) and newi.rhs == 0: + newi = Eq(newi.lhs.args[0]/C1, C1) + eq[j] = newi + + # We cleaned up the constants before solving to help the solve engine with + # a simpler expression, but the solved expression could have introduced + # things like -C1, so rerun constantsimp() one last time before returning. + for i, eqi in enumerate(eq): + eq[i] = constantsimp(eqi, constants) + eq[i] = constant_renumber(eq[i], ode.free_symbols) + + # If there is only 1 solution, return it; + # otherwise return the list of solutions. + if len(eq) == 1: + eq = eq[0] + return eq + + +def ode_sol_simplicity(sol, func, trysolving=True): + r""" + Returns an extended integer representing how simple a solution to an ODE + is. + + The following things are considered, in order from most simple to least: + + - ``sol`` is solved for ``func``. + - ``sol`` is not solved for ``func``, but can be if passed to solve (e.g., + a solution returned by ``dsolve(ode, func, simplify=False``). + - If ``sol`` is not solved for ``func``, then base the result on the + length of ``sol``, as computed by ``len(str(sol))``. + - If ``sol`` has any unevaluated :py:class:`~sympy.integrals.integrals.Integral`\s, + this will automatically be considered less simple than any of the above. + + This function returns an integer such that if solution A is simpler than + solution B by above metric, then ``ode_sol_simplicity(sola, func) < + ode_sol_simplicity(solb, func)``. + + Currently, the following are the numbers returned, but if the heuristic is + ever improved, this may change. Only the ordering is guaranteed. + + +----------------------------------------------+-------------------+ + | Simplicity | Return | + +==============================================+===================+ + | ``sol`` solved for ``func`` | ``-2`` | + +----------------------------------------------+-------------------+ + | ``sol`` not solved for ``func`` but can be | ``-1`` | + +----------------------------------------------+-------------------+ + | ``sol`` is not solved nor solvable for | ``len(str(sol))`` | + | ``func`` | | + +----------------------------------------------+-------------------+ + | ``sol`` contains an | ``oo`` | + | :obj:`~sympy.integrals.integrals.Integral` | | + +----------------------------------------------+-------------------+ + + ``oo`` here means the SymPy infinity, which should compare greater than + any integer. + + If you already know :py:meth:`~sympy.solvers.solvers.solve` cannot solve + ``sol``, you can use ``trysolving=False`` to skip that step, which is the + only potentially slow step. For example, + :py:meth:`~sympy.solvers.ode.dsolve` with the ``simplify=False`` flag + should do this. + + If ``sol`` is a list of solutions, if the worst solution in the list + returns ``oo`` it returns that, otherwise it returns ``len(str(sol))``, + that is, the length of the string representation of the whole list. + + Examples + ======== + + This function is designed to be passed to ``min`` as the key argument, + such as ``min(listofsolutions, key=lambda i: ode_sol_simplicity(i, + f(x)))``. + + >>> from sympy import symbols, Function, Eq, tan, Integral + >>> from sympy.solvers.ode.ode import ode_sol_simplicity + >>> x, C1, C2 = symbols('x, C1, C2') + >>> f = Function('f') + + >>> ode_sol_simplicity(Eq(f(x), C1*x**2), f(x)) + -2 + >>> ode_sol_simplicity(Eq(x**2 + f(x), C1), f(x)) + -1 + >>> ode_sol_simplicity(Eq(f(x), C1*Integral(2*x, x)), f(x)) + oo + >>> eq1 = Eq(f(x)/tan(f(x)/(2*x)), C1) + >>> eq2 = Eq(f(x)/tan(f(x)/(2*x) + f(x)), C2) + >>> [ode_sol_simplicity(eq, f(x)) for eq in [eq1, eq2]] + [28, 35] + >>> min([eq1, eq2], key=lambda i: ode_sol_simplicity(i, f(x))) + Eq(f(x)/tan(f(x)/(2*x)), C1) + + """ + # TODO: if two solutions are solved for f(x), we still want to be + # able to get the simpler of the two + + # See the docstring for the coercion rules. We check easier (faster) + # things here first, to save time. + + if iterable(sol): + # See if there are Integrals + for i in sol: + if ode_sol_simplicity(i, func, trysolving=trysolving) == oo: + return oo + + return len(str(sol)) + + if sol.has(Integral): + return oo + + # Next, try to solve for func. This code will change slightly when CRootOf + # is implemented in solve(). Probably a CRootOf solution should fall + # somewhere between a normal solution and an unsolvable expression. + + # First, see if they are already solved + if sol.lhs == func and not sol.rhs.has(func) or \ + sol.rhs == func and not sol.lhs.has(func): + return -2 + # We are not so lucky, try solving manually + if trysolving: + try: + sols = solve(sol, func) + if not sols: + raise NotImplementedError + except NotImplementedError: + pass + else: + return -1 + + # Finally, a naive computation based on the length of the string version + # of the expression. This may favor combined fractions because they + # will not have duplicate denominators, and may slightly favor expressions + # with fewer additions and subtractions, as those are separated by spaces + # by the printer. + + # Additional ideas for simplicity heuristics are welcome, like maybe + # checking if a equation has a larger domain, or if constantsimp has + # introduced arbitrary constants numbered higher than the order of a + # given ODE that sol is a solution of. + return len(str(sol)) + + +def _extract_funcs(eqs): + funcs = [] + for eq in eqs: + derivs = [node for node in preorder_traversal(eq) if isinstance(node, Derivative)] + func = [] + for d in derivs: + func += list(d.atoms(AppliedUndef)) + for func_ in func: + funcs.append(func_) + funcs = list(uniq(funcs)) + + return funcs + + +def _get_constant_subexpressions(expr, Cs): + Cs = set(Cs) + Ces = [] + def _recursive_walk(expr): + expr_syms = expr.free_symbols + if expr_syms and expr_syms.issubset(Cs): + Ces.append(expr) + else: + if expr.func == exp: + expr = expr.expand(mul=True) + if expr.func in (Add, Mul): + d = sift(expr.args, lambda i : i.free_symbols.issubset(Cs)) + if len(d[True]) > 1: + x = expr.func(*d[True]) + if not x.is_number: + Ces.append(x) + elif isinstance(expr, Integral): + if expr.free_symbols.issubset(Cs) and \ + all(len(x) == 3 for x in expr.limits): + Ces.append(expr) + for i in expr.args: + _recursive_walk(i) + return + _recursive_walk(expr) + return Ces + +def __remove_linear_redundancies(expr, Cs): + cnts = {i: expr.count(i) for i in Cs} + Cs = [i for i in Cs if cnts[i] > 0] + + def _linear(expr): + if isinstance(expr, Add): + xs = [i for i in Cs if expr.count(i)==cnts[i] \ + and 0 == expr.diff(i, 2)] + d = {} + for x in xs: + y = expr.diff(x) + if y not in d: + d[y]=[] + d[y].append(x) + for y in d: + if len(d[y]) > 1: + d[y].sort(key=str) + for x in d[y][1:]: + expr = expr.subs(x, 0) + return expr + + def _recursive_walk(expr): + if len(expr.args) != 0: + expr = expr.func(*[_recursive_walk(i) for i in expr.args]) + expr = _linear(expr) + return expr + + if isinstance(expr, Equality): + lhs, rhs = [_recursive_walk(i) for i in expr.args] + f = lambda i: isinstance(i, Number) or i in Cs + if isinstance(lhs, Symbol) and lhs in Cs: + rhs, lhs = lhs, rhs + if lhs.func in (Add, Symbol) and rhs.func in (Add, Symbol): + dlhs = sift([lhs] if isinstance(lhs, AtomicExpr) else lhs.args, f) + drhs = sift([rhs] if isinstance(rhs, AtomicExpr) else rhs.args, f) + for i in [True, False]: + for hs in [dlhs, drhs]: + if i not in hs: + hs[i] = [0] + # this calculation can be simplified + lhs = Add(*dlhs[False]) - Add(*drhs[False]) + rhs = Add(*drhs[True]) - Add(*dlhs[True]) + elif lhs.func in (Mul, Symbol) and rhs.func in (Mul, Symbol): + dlhs = sift([lhs] if isinstance(lhs, AtomicExpr) else lhs.args, f) + if True in dlhs: + if False not in dlhs: + dlhs[False] = [1] + lhs = Mul(*dlhs[False]) + rhs = rhs/Mul(*dlhs[True]) + return Eq(lhs, rhs) + else: + return _recursive_walk(expr) + +@vectorize(0) +def constantsimp(expr, constants): + r""" + Simplifies an expression with arbitrary constants in it. + + This function is written specifically to work with + :py:meth:`~sympy.solvers.ode.dsolve`, and is not intended for general use. + + Simplification is done by "absorbing" the arbitrary constants into other + arbitrary constants, numbers, and symbols that they are not independent + of. + + The symbols must all have the same name with numbers after it, for + example, ``C1``, ``C2``, ``C3``. The ``symbolname`` here would be + '``C``', the ``startnumber`` would be 1, and the ``endnumber`` would be 3. + If the arbitrary constants are independent of the variable ``x``, then the + independent symbol would be ``x``. There is no need to specify the + dependent function, such as ``f(x)``, because it already has the + independent symbol, ``x``, in it. + + Because terms are "absorbed" into arbitrary constants and because + constants are renumbered after simplifying, the arbitrary constants in + expr are not necessarily equal to the ones of the same name in the + returned result. + + If two or more arbitrary constants are added, multiplied, or raised to the + power of each other, they are first absorbed together into a single + arbitrary constant. Then the new constant is combined into other terms if + necessary. + + Absorption of constants is done with limited assistance: + + 1. terms of :py:class:`~sympy.core.add.Add`\s are collected to try join + constants so `e^x (C_1 \cos(x) + C_2 \cos(x))` will simplify to `e^x + C_1 \cos(x)`; + + 2. powers with exponents that are :py:class:`~sympy.core.add.Add`\s are + expanded so `e^{C_1 + x}` will be simplified to `C_1 e^x`. + + Use :py:meth:`~sympy.solvers.ode.ode.constant_renumber` to renumber constants + after simplification or else arbitrary numbers on constants may appear, + e.g. `C_1 + C_3 x`. + + In rare cases, a single constant can be "simplified" into two constants. + Every differential equation solution should have as many arbitrary + constants as the order of the differential equation. The result here will + be technically correct, but it may, for example, have `C_1` and `C_2` in + an expression, when `C_1` is actually equal to `C_2`. Use your discretion + in such situations, and also take advantage of the ability to use hints in + :py:meth:`~sympy.solvers.ode.dsolve`. + + Examples + ======== + + >>> from sympy import symbols + >>> from sympy.solvers.ode.ode import constantsimp + >>> C1, C2, C3, x, y = symbols('C1, C2, C3, x, y') + >>> constantsimp(2*C1*x, {C1, C2, C3}) + C1*x + >>> constantsimp(C1 + 2 + x, {C1, C2, C3}) + C1 + x + >>> constantsimp(C1*C2 + 2 + C2 + C3*x, {C1, C2, C3}) + C1 + C3*x + + """ + # This function works recursively. The idea is that, for Mul, + # Add, Pow, and Function, if the class has a constant in it, then + # we can simplify it, which we do by recursing down and + # simplifying up. Otherwise, we can skip that part of the + # expression. + + Cs = constants + + orig_expr = expr + + constant_subexprs = _get_constant_subexpressions(expr, Cs) + for xe in constant_subexprs: + xes = list(xe.free_symbols) + if not xes: + continue + if all(expr.count(c) == xe.count(c) for c in xes): + xes.sort(key=str) + expr = expr.subs(xe, xes[0]) + + # try to perform common sub-expression elimination of constant terms + try: + commons, rexpr = cse(expr) + commons.reverse() + rexpr = rexpr[0] + for s in commons: + cs = list(s[1].atoms(Symbol)) + if len(cs) == 1 and cs[0] in Cs and \ + cs[0] not in rexpr.atoms(Symbol) and \ + not any(cs[0] in ex for ex in commons if ex != s): + rexpr = rexpr.subs(s[0], cs[0]) + else: + rexpr = rexpr.subs(*s) + expr = rexpr + except IndexError: + pass + expr = __remove_linear_redundancies(expr, Cs) + + def _conditional_term_factoring(expr): + new_expr = terms_gcd(expr, clear=False, deep=True, expand=False) + + # we do not want to factor exponentials, so handle this separately + if new_expr.is_Mul: + infac = False + asfac = False + for m in new_expr.args: + if isinstance(m, exp): + asfac = True + elif m.is_Add: + infac = any(isinstance(fi, exp) for t in m.args + for fi in Mul.make_args(t)) + if asfac and infac: + new_expr = expr + break + return new_expr + + expr = _conditional_term_factoring(expr) + + # call recursively if more simplification is possible + if orig_expr != expr: + return constantsimp(expr, Cs) + return expr + + +def constant_renumber(expr, variables=None, newconstants=None): + r""" + Renumber arbitrary constants in ``expr`` to use the symbol names as given + in ``newconstants``. In the process, this reorders expression terms in a + standard way. + + If ``newconstants`` is not provided then the new constant names will be + ``C1``, ``C2`` etc. Otherwise ``newconstants`` should be an iterable + giving the new symbols to use for the constants in order. + + The ``variables`` argument is a list of non-constant symbols. All other + free symbols found in ``expr`` are assumed to be constants and will be + renumbered. If ``variables`` is not given then any numbered symbol + beginning with ``C`` (e.g. ``C1``) is assumed to be a constant. + + Symbols are renumbered based on ``.sort_key()``, so they should be + numbered roughly in the order that they appear in the final, printed + expression. Note that this ordering is based in part on hashes, so it can + produce different results on different machines. + + The structure of this function is very similar to that of + :py:meth:`~sympy.solvers.ode.constantsimp`. + + Examples + ======== + + >>> from sympy import symbols + >>> from sympy.solvers.ode.ode import constant_renumber + >>> x, C1, C2, C3 = symbols('x,C1:4') + >>> expr = C3 + C2*x + C1*x**2 + >>> expr + C1*x**2 + C2*x + C3 + >>> constant_renumber(expr) + C1 + C2*x + C3*x**2 + + The ``variables`` argument specifies which are constants so that the + other symbols will not be renumbered: + + >>> constant_renumber(expr, [C1, x]) + C1*x**2 + C2 + C3*x + + The ``newconstants`` argument is used to specify what symbols to use when + replacing the constants: + + >>> constant_renumber(expr, [x], newconstants=symbols('E1:4')) + E1 + E2*x + E3*x**2 + + """ + + # System of expressions + if isinstance(expr, (set, list, tuple)): + return type(expr)(constant_renumber(Tuple(*expr), + variables=variables, newconstants=newconstants)) + + # Symbols in solution but not ODE are constants + if variables is not None: + variables = set(variables) + free_symbols = expr.free_symbols + constantsymbols = list(free_symbols - variables) + # Any Cn is a constant... + else: + variables = set() + isconstant = lambda s: s.startswith('C') and s[1:].isdigit() + constantsymbols = [sym for sym in expr.free_symbols if isconstant(sym.name)] + + # Find new constants checking that they aren't already in the ODE + if newconstants is None: + iter_constants = numbered_symbols(start=1, prefix='C', exclude=variables) + else: + iter_constants = (sym for sym in newconstants if sym not in variables) + + constants_found = [] + + # make a mapping to send all constantsymbols to S.One and use + # that to make sure that term ordering is not dependent on + # the indexed value of C + C_1 = [(ci, S.One) for ci in constantsymbols] + sort_key=lambda arg: default_sort_key(arg.subs(C_1)) + + def _constant_renumber(expr): + r""" + We need to have an internal recursive function + """ + + # For system of expressions + if isinstance(expr, Tuple): + renumbered = [_constant_renumber(e) for e in expr] + return Tuple(*renumbered) + + if isinstance(expr, Equality): + return Eq( + _constant_renumber(expr.lhs), + _constant_renumber(expr.rhs)) + + if type(expr) not in (Mul, Add, Pow) and not expr.is_Function and \ + not expr.has(*constantsymbols): + # Base case, as above. Hope there aren't constants inside + # of some other class, because they won't be renumbered. + return expr + elif expr.is_Piecewise: + return expr + elif expr in constantsymbols: + if expr not in constants_found: + constants_found.append(expr) + return expr + elif expr.is_Function or expr.is_Pow: + return expr.func( + *[_constant_renumber(x) for x in expr.args]) + else: + sortedargs = list(expr.args) + sortedargs.sort(key=sort_key) + return expr.func(*[_constant_renumber(x) for x in sortedargs]) + expr = _constant_renumber(expr) + + # Don't renumber symbols present in the ODE. + constants_found = [c for c in constants_found if c not in variables] + + # Renumbering happens here + subs_dict = dict(zip(constants_found, iter_constants)) + expr = expr.subs(subs_dict, simultaneous=True) + + return expr + + +def _handle_Integral(expr, func, hint): + r""" + Converts a solution with Integrals in it into an actual solution. + + For most hints, this simply runs ``expr.doit()``. + + """ + if hint == "nth_linear_constant_coeff_homogeneous": + sol = expr + elif not hint.endswith("_Integral"): + sol = expr.doit() + else: + sol = expr + return sol + + +# XXX: Should this function maybe go somewhere else? + + +def homogeneous_order(eq, *symbols): + r""" + Returns the order `n` if `g` is homogeneous and ``None`` if it is not + homogeneous. + + Determines if a function is homogeneous and if so of what order. A + function `f(x, y, \cdots)` is homogeneous of order `n` if `f(t x, t y, + \cdots) = t^n f(x, y, \cdots)`. + + If the function is of two variables, `F(x, y)`, then `f` being homogeneous + of any order is equivalent to being able to rewrite `F(x, y)` as `G(x/y)` + or `H(y/x)`. This fact is used to solve 1st order ordinary differential + equations whose coefficients are homogeneous of the same order (see the + docstrings of + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffSubsDepDivIndep` and + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffSubsIndepDivDep`). + + Symbols can be functions, but every argument of the function must be a + symbol, and the arguments of the function that appear in the expression + must match those given in the list of symbols. If a declared function + appears with different arguments than given in the list of symbols, + ``None`` is returned. + + Examples + ======== + + >>> from sympy import Function, homogeneous_order, sqrt + >>> from sympy.abc import x, y + >>> f = Function('f') + >>> homogeneous_order(f(x), f(x)) is None + True + >>> homogeneous_order(f(x,y), f(y, x), x, y) is None + True + >>> homogeneous_order(f(x), f(x), x) + 1 + >>> homogeneous_order(x**2*f(x)/sqrt(x**2+f(x)**2), x, f(x)) + 2 + >>> homogeneous_order(x**2+f(x), x, f(x)) is None + True + + """ + + if not symbols: + raise ValueError("homogeneous_order: no symbols were given.") + symset = set(symbols) + eq = sympify(eq) + + # The following are not supported + if eq.has(Order, Derivative): + return None + + # These are all constants + if (eq.is_Number or + eq.is_NumberSymbol or + eq.is_number + ): + return S.Zero + + # Replace all functions with dummy variables + dum = numbered_symbols(prefix='d', cls=Dummy) + newsyms = set() + for i in [j for j in symset if getattr(j, 'is_Function')]: + iargs = set(i.args) + if iargs.difference(symset): + return None + else: + dummyvar = next(dum) + eq = eq.subs(i, dummyvar) + symset.remove(i) + newsyms.add(dummyvar) + symset.update(newsyms) + + if not eq.free_symbols & symset: + return None + + # assuming order of a nested function can only be equal to zero + if isinstance(eq, Function): + return None if homogeneous_order( + eq.args[0], *tuple(symset)) != 0 else S.Zero + + # make the replacement of x with x*t and see if t can be factored out + t = Dummy('t', positive=True) # It is sufficient that t > 0 + eqs = separatevars(eq.subs([(i, t*i) for i in symset]), [t], dict=True)[t] + if eqs is S.One: + return S.Zero # there was no term with only t + i, d = eqs.as_independent(t, as_Add=False) + b, e = d.as_base_exp() + if b == t: + return e + + +def ode_2nd_power_series_ordinary(eq, func, order, match): + r""" + Gives a power series solution to a second order homogeneous differential + equation with polynomial coefficients at an ordinary point. A homogeneous + differential equation is of the form + + .. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) y(x) = 0 + + For simplicity it is assumed that `P(x)`, `Q(x)` and `R(x)` are polynomials, + it is sufficient that `\frac{Q(x)}{P(x)}` and `\frac{R(x)}{P(x)}` exists at + `x_{0}`. A recurrence relation is obtained by substituting `y` as `\sum_{n=0}^\infty a_{n}x^{n}`, + in the differential equation, and equating the nth term. Using this relation + various terms can be generated. + + + Examples + ======== + + >>> from sympy import dsolve, Function, pprint + >>> from sympy.abc import x + >>> f = Function("f") + >>> eq = f(x).diff(x, 2) + f(x) + >>> pprint(dsolve(eq, hint='2nd_power_series_ordinary')) + / 4 2 \ / 2\ + |x x | | x | / 6\ + f(x) = C2*|-- - -- + 1| + C1*x*|1 - --| + O\x / + \24 2 / \ 6 / + + + References + ========== + - https://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx + - George E. Simmons, "Differential Equations with Applications and + Historical Notes", p.p 176 - 184 + + """ + x = func.args[0] + f = func.func + C0, C1 = get_numbered_constants(eq, num=2) + n = Dummy("n", integer=True) + s = Wild("s") + k = Wild("k", exclude=[x]) + x0 = match['x0'] + terms = match['terms'] + p = match[match['a3']] + q = match[match['b3']] + r = match[match['c3']] + seriesdict = {} + recurr = Function("r") + + # Generating the recurrence relation which works this way: + # for the second order term the summation begins at n = 2. The coefficients + # p is multiplied with an*(n - 1)*(n - 2)*x**n-2 and a substitution is made such that + # the exponent of x becomes n. + # For example, if p is x, then the second degree recurrence term is + # an*(n - 1)*(n - 2)*x**n-1, substituting (n - 1) as n, it transforms to + # an+1*n*(n - 1)*x**n. + # A similar process is done with the first order and zeroth order term. + + coefflist = [(recurr(n), r), (n*recurr(n), q), (n*(n - 1)*recurr(n), p)] + for index, coeff in enumerate(coefflist): + if coeff[1]: + f2 = powsimp(expand((coeff[1]*(x - x0)**(n - index)).subs(x, x + x0))) + if f2.is_Add: + addargs = f2.args + else: + addargs = [f2] + for arg in addargs: + powm = arg.match(s*x**k) + term = coeff[0]*powm[s] + if not powm[k].is_Symbol: + term = term.subs(n, n - powm[k].as_independent(n)[0]) + startind = powm[k].subs(n, index) + # Seeing if the startterm can be reduced further. + # If it vanishes for n lesser than startind, it is + # equal to summation from n. + if startind: + for i in reversed(range(startind)): + if not term.subs(n, i): + seriesdict[term] = i + else: + seriesdict[term] = i + 1 + break + else: + seriesdict[term] = S.Zero + + # Stripping of terms so that the sum starts with the same number. + teq = S.Zero + suminit = seriesdict.values() + rkeys = seriesdict.keys() + req = Add(*rkeys) + if any(suminit): + maxval = max(suminit) + for term in seriesdict: + val = seriesdict[term] + if val != maxval: + for i in range(val, maxval): + teq += term.subs(n, val) + + finaldict = {} + if teq: + fargs = teq.atoms(AppliedUndef) + if len(fargs) == 1: + finaldict[fargs.pop()] = 0 + else: + maxf = max(fargs, key = lambda x: x.args[0]) + sol = solve(teq, maxf) + if isinstance(sol, list): + sol = sol[0] + finaldict[maxf] = sol + + # Finding the recurrence relation in terms of the largest term. + fargs = req.atoms(AppliedUndef) + maxf = max(fargs, key = lambda x: x.args[0]) + minf = min(fargs, key = lambda x: x.args[0]) + if minf.args[0].is_Symbol: + startiter = 0 + else: + startiter = -minf.args[0].as_independent(n)[0] + lhs = maxf + rhs = solve(req, maxf) + if isinstance(rhs, list): + rhs = rhs[0] + + # Checking how many values are already present + tcounter = len([t for t in finaldict.values() if t]) + + for _ in range(tcounter, terms - 3): # Assuming c0 and c1 to be arbitrary + check = rhs.subs(n, startiter) + nlhs = lhs.subs(n, startiter) + nrhs = check.subs(finaldict) + finaldict[nlhs] = nrhs + startiter += 1 + + # Post processing + series = C0 + C1*(x - x0) + for term in finaldict: + if finaldict[term]: + fact = term.args[0] + series += (finaldict[term].subs([(recurr(0), C0), (recurr(1), C1)])*( + x - x0)**fact) + series = collect(expand_mul(series), [C0, C1]) + Order(x**terms) + return Eq(f(x), series) + + +def ode_2nd_power_series_regular(eq, func, order, match): + r""" + Gives a power series solution to a second order homogeneous differential + equation with polynomial coefficients at a regular point. A second order + homogeneous differential equation is of the form + + .. math :: P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x) y(x) = 0 + + A point is said to regular singular at `x0` if `x - x0\frac{Q(x)}{P(x)}` + and `(x - x0)^{2}\frac{R(x)}{P(x)}` are analytic at `x0`. For simplicity + `P(x)`, `Q(x)` and `R(x)` are assumed to be polynomials. The algorithm for + finding the power series solutions is: + + 1. Try expressing `(x - x0)P(x)` and `((x - x0)^{2})Q(x)` as power series + solutions about x0. Find `p0` and `q0` which are the constants of the + power series expansions. + 2. Solve the indicial equation `f(m) = m(m - 1) + m*p0 + q0`, to obtain the + roots `m1` and `m2` of the indicial equation. + 3. If `m1 - m2` is a non integer there exists two series solutions. If + `m1 = m2`, there exists only one solution. If `m1 - m2` is an integer, + then the existence of one solution is confirmed. The other solution may + or may not exist. + + The power series solution is of the form `x^{m}\sum_{n=0}^\infty a_{n}x^{n}`. The + coefficients are determined by the following recurrence relation. + `a_{n} = -\frac{\sum_{k=0}^{n-1} q_{n-k} + (m + k)p_{n-k}}{f(m + n)}`. For the case + in which `m1 - m2` is an integer, it can be seen from the recurrence relation + that for the lower root `m`, when `n` equals the difference of both the + roots, the denominator becomes zero. So if the numerator is not equal to zero, + a second series solution exists. + + + Examples + ======== + + >>> from sympy import dsolve, Function, pprint + >>> from sympy.abc import x + >>> f = Function("f") + >>> eq = x*(f(x).diff(x, 2)) + 2*(f(x).diff(x)) + x*f(x) + >>> pprint(dsolve(eq, hint='2nd_power_series_regular')) + / 6 4 2 \ + | x x x | + / 4 2 \ C1*|- --- + -- - -- + 1| + |x x | \ 720 24 2 / / 6\ + f(x) = C2*|--- - -- + 1| + ------------------------ + O\x / + \120 6 / x + + + References + ========== + - George E. Simmons, "Differential Equations with Applications and + Historical Notes", p.p 176 - 184 + + """ + x = func.args[0] + f = func.func + C0, C1 = get_numbered_constants(eq, num=2) + m = Dummy("m") # for solving the indicial equation + x0 = match['x0'] + terms = match['terms'] + p = match['p'] + q = match['q'] + + # Generating the indicial equation + indicial = [] + for term in [p, q]: + if not term.has(x): + indicial.append(term) + else: + term = series(term, x=x, n=1, x0=x0) + if isinstance(term, Order): + indicial.append(S.Zero) + else: + for arg in term.args: + if not arg.has(x): + indicial.append(arg) + break + + p0, q0 = indicial + sollist = solve(m*(m - 1) + m*p0 + q0, m) + if sollist and isinstance(sollist, list) and all( + sol.is_real for sol in sollist): + serdict1 = {} + serdict2 = {} + if len(sollist) == 1: + # Only one series solution exists in this case. + m1 = m2 = sollist.pop() + if terms-m1-1 <= 0: + return Eq(f(x), Order(terms)) + serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0) + + else: + m1 = sollist[0] + m2 = sollist[1] + if m1 < m2: + m1, m2 = m2, m1 + # Irrespective of whether m1 - m2 is an integer or not, one + # Frobenius series solution exists. + serdict1 = _frobenius(terms-m1-1, m1, p0, q0, p, q, x0, x, C0) + if not (m1 - m2).is_integer: + # Second frobenius series solution exists. + serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1) + else: + # Check if second frobenius series solution exists. + serdict2 = _frobenius(terms-m2-1, m2, p0, q0, p, q, x0, x, C1, check=m1) + + if serdict1: + finalseries1 = C0 + for key in serdict1: + power = int(key.name[1:]) + finalseries1 += serdict1[key]*(x - x0)**power + finalseries1 = (x - x0)**m1*finalseries1 + finalseries2 = S.Zero + if serdict2: + for key in serdict2: + power = int(key.name[1:]) + finalseries2 += serdict2[key]*(x - x0)**power + finalseries2 += C1 + finalseries2 = (x - x0)**m2*finalseries2 + return Eq(f(x), collect(finalseries1 + finalseries2, + [C0, C1]) + Order(x**terms)) + + +def _frobenius(n, m, p0, q0, p, q, x0, x, c, check=None): + r""" + Returns a dict with keys as coefficients and values as their values in terms of C0 + """ + n = int(n) + # In cases where m1 - m2 is not an integer + m2 = check + + d = Dummy("d") + numsyms = numbered_symbols("C", start=0) + numsyms = [next(numsyms) for i in range(n + 1)] + serlist = [] + for ser in [p, q]: + # Order term not present + if ser.is_polynomial(x) and Poly(ser, x).degree() <= n: + if x0: + ser = ser.subs(x, x + x0) + dict_ = Poly(ser, x).as_dict() + # Order term present + else: + tseries = series(ser, x=x0, n=n+1) + # Removing order + dict_ = Poly(list(ordered(tseries.args))[: -1], x).as_dict() + # Fill in with zeros, if coefficients are zero. + for i in range(n + 1): + if (i,) not in dict_: + dict_[(i,)] = S.Zero + serlist.append(dict_) + + pseries = serlist[0] + qseries = serlist[1] + indicial = d*(d - 1) + d*p0 + q0 + frobdict = {} + for i in range(1, n + 1): + num = c*(m*pseries[(i,)] + qseries[(i,)]) + for j in range(1, i): + sym = Symbol("C" + str(j)) + num += frobdict[sym]*((m + j)*pseries[(i - j,)] + qseries[(i - j,)]) + + # Checking for cases when m1 - m2 is an integer. If num equals zero + # then a second Frobenius series solution cannot be found. If num is not zero + # then set constant as zero and proceed. + if m2 is not None and i == m2 - m: + if num: + return False + else: + frobdict[numsyms[i]] = S.Zero + else: + frobdict[numsyms[i]] = -num/(indicial.subs(d, m+i)) + + return frobdict + +def _remove_redundant_solutions(eq, solns, order, var): + r""" + Remove redundant solutions from the set of solutions. + + This function is needed because otherwise dsolve can return + redundant solutions. As an example consider: + + eq = Eq((f(x).diff(x, 2))*f(x).diff(x), 0) + + There are two ways to find solutions to eq. The first is to solve f(x).diff(x, 2) = 0 + leading to solution f(x)=C1 + C2*x. The second is to solve the equation f(x).diff(x) = 0 + leading to the solution f(x) = C1. In this particular case we then see + that the second solution is a special case of the first and we do not + want to return it. + + This does not always happen. If we have + + eq = Eq((f(x)**2-4)*(f(x).diff(x)-4), 0) + + then we get the algebraic solution f(x) = [-2, 2] and the integral solution + f(x) = x + C1 and in this case the two solutions are not equivalent wrt + initial conditions so both should be returned. + """ + def is_special_case_of(soln1, soln2): + return _is_special_case_of(soln1, soln2, eq, order, var) + + unique_solns = [] + for soln1 in solns: + for soln2 in unique_solns[:]: + if is_special_case_of(soln1, soln2): + break + elif is_special_case_of(soln2, soln1): + unique_solns.remove(soln2) + else: + unique_solns.append(soln1) + + return unique_solns + +def _is_special_case_of(soln1, soln2, eq, order, var): + r""" + True if soln1 is found to be a special case of soln2 wrt some value of the + constants that appear in soln2. False otherwise. + """ + # The solutions returned by dsolve may be given explicitly or implicitly. + # We will equate the sol1=(soln1.rhs - soln1.lhs), sol2=(soln2.rhs - soln2.lhs) + # of the two solutions. + # + # Since this is supposed to hold for all x it also holds for derivatives. + # For an order n ode we should be able to differentiate + # each solution n times to get n+1 equations. + # + # We then try to solve those n+1 equations for the integrations constants + # in sol2. If we can find a solution that does not depend on x then it + # means that some value of the constants in sol1 is a special case of + # sol2 corresponding to a particular choice of the integration constants. + + # In case the solution is in implicit form we subtract the sides + soln1 = soln1.rhs - soln1.lhs + soln2 = soln2.rhs - soln2.lhs + + # Work for the series solution + if soln1.has(Order) and soln2.has(Order): + if soln1.getO() == soln2.getO(): + soln1 = soln1.removeO() + soln2 = soln2.removeO() + else: + return False + elif soln1.has(Order) or soln2.has(Order): + return False + + constants1 = soln1.free_symbols.difference(eq.free_symbols) + constants2 = soln2.free_symbols.difference(eq.free_symbols) + + constants1_new = get_numbered_constants(Tuple(soln1, soln2), len(constants1)) + if len(constants1) == 1: + constants1_new = {constants1_new} + for c_old, c_new in zip(constants1, constants1_new): + soln1 = soln1.subs(c_old, c_new) + + # n equations for sol1 = sol2, sol1'=sol2', ... + lhs = soln1 + rhs = soln2 + eqns = [Eq(lhs, rhs)] + for n in range(1, order): + lhs = lhs.diff(var) + rhs = rhs.diff(var) + eq = Eq(lhs, rhs) + eqns.append(eq) + + # BooleanTrue/False awkwardly show up for trivial equations + if any(isinstance(eq, BooleanFalse) for eq in eqns): + return False + eqns = [eq for eq in eqns if not isinstance(eq, BooleanTrue)] + + try: + constant_solns = solve(eqns, constants2) + except NotImplementedError: + return False + + # Sometimes returns a dict and sometimes a list of dicts + if isinstance(constant_solns, dict): + constant_solns = [constant_solns] + + # after solving the issue 17418, maybe we don't need the following checksol code. + for constant_soln in constant_solns: + for eq in eqns: + eq=eq.rhs-eq.lhs + if checksol(eq, constant_soln) is not True: + return False + + # If any solution gives all constants as expressions that don't depend on + # x then there exists constants for soln2 that give soln1 + for constant_soln in constant_solns: + if not any(c.has(var) for c in constant_soln.values()): + return True + + return False + + +def ode_1st_power_series(eq, func, order, match): + r""" + The power series solution is a method which gives the Taylor series expansion + to the solution of a differential equation. + + For a first order differential equation `\frac{dy}{dx} = h(x, y)`, a power + series solution exists at a point `x = x_{0}` if `h(x, y)` is analytic at `x_{0}`. + The solution is given by + + .. math:: y(x) = y(x_{0}) + \sum_{n = 1}^{\infty} \frac{F_{n}(x_{0},b)(x - x_{0})^n}{n!}, + + where `y(x_{0}) = b` is the value of y at the initial value of `x_{0}`. + To compute the values of the `F_{n}(x_{0},b)` the following algorithm is + followed, until the required number of terms are generated. + + 1. `F_1 = h(x_{0}, b)` + 2. `F_{n+1} = \frac{\partial F_{n}}{\partial x} + \frac{\partial F_{n}}{\partial y}F_{1}` + + Examples + ======== + + >>> from sympy import Function, pprint, exp, dsolve + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = exp(x)*(f(x).diff(x)) - f(x) + >>> pprint(dsolve(eq, hint='1st_power_series')) + 3 4 5 + C1*x C1*x C1*x / 6\ + f(x) = C1 + C1*x - ----- + ----- + ----- + O\x / + 6 24 60 + + + References + ========== + + - Travis W. Walker, Analytic power series technique for solving first-order + differential equations, p.p 17, 18 + + """ + x = func.args[0] + y = match['y'] + f = func.func + h = -match[match['d']]/match[match['e']] + point = match['f0'] + value = match['f0val'] + terms = match['terms'] + + # First term + F = h + if not h: + return Eq(f(x), value) + + # Initialization + series = value + if terms > 1: + hc = h.subs({x: point, y: value}) + if hc.has(oo) or hc.has(nan) or hc.has(zoo): + # Derivative does not exist, not analytic + return Eq(f(x), oo) + elif hc: + series += hc*(x - point) + + for factcount in range(2, terms): + Fnew = F.diff(x) + F.diff(y)*h + Fnewc = Fnew.subs({x: point, y: value}) + # Same logic as above + if Fnewc.has(oo) or Fnewc.has(nan) or Fnewc.has(-oo) or Fnewc.has(zoo): + return Eq(f(x), oo) + series += Fnewc*((x - point)**factcount)/factorial(factcount) + F = Fnew + series += Order(x**terms) + return Eq(f(x), series) + + +def checkinfsol(eq, infinitesimals, func=None, order=None): + r""" + This function is used to check if the given infinitesimals are the + actual infinitesimals of the given first order differential equation. + This method is specific to the Lie Group Solver of ODEs. + + As of now, it simply checks, by substituting the infinitesimals in the + partial differential equation. + + + .. math:: \frac{\partial \eta}{\partial x} + \left(\frac{\partial \eta}{\partial y} + - \frac{\partial \xi}{\partial x}\right)*h + - \frac{\partial \xi}{\partial y}*h^{2} + - \xi\frac{\partial h}{\partial x} - \eta\frac{\partial h}{\partial y} = 0 + + + where `\eta`, and `\xi` are the infinitesimals and `h(x,y) = \frac{dy}{dx}` + + The infinitesimals should be given in the form of a list of dicts + ``[{xi(x, y): inf, eta(x, y): inf}]``, corresponding to the + output of the function infinitesimals. It returns a list + of values of the form ``[(True/False, sol)]`` where ``sol`` is the value + obtained after substituting the infinitesimals in the PDE. If it + is ``True``, then ``sol`` would be 0. + + """ + if isinstance(eq, Equality): + eq = eq.lhs - eq.rhs + if not func: + eq, func = _preprocess(eq) + variables = func.args + if len(variables) != 1: + raise ValueError("ODE's have only one independent variable") + else: + x = variables[0] + if not order: + order = ode_order(eq, func) + if order != 1: + raise NotImplementedError("Lie groups solver has been implemented " + "only for first order differential equations") + else: + df = func.diff(x) + a = Wild('a', exclude = [df]) + b = Wild('b', exclude = [df]) + match = collect(expand(eq), df).match(a*df + b) + + if match: + h = -simplify(match[b]/match[a]) + else: + try: + sol = solve(eq, df) + except NotImplementedError: + raise NotImplementedError("Infinitesimals for the " + "first order ODE could not be found") + else: + h = sol[0] # Find infinitesimals for one solution + + y = Dummy('y') + h = h.subs(func, y) + xi = Function('xi')(x, y) + eta = Function('eta')(x, y) + dxi = Function('xi')(x, func) + deta = Function('eta')(x, func) + pde = (eta.diff(x) + (eta.diff(y) - xi.diff(x))*h - + (xi.diff(y))*h**2 - xi*(h.diff(x)) - eta*(h.diff(y))) + soltup = [] + for sol in infinitesimals: + tsol = {xi: S(sol[dxi]).subs(func, y), + eta: S(sol[deta]).subs(func, y)} + sol = simplify(pde.subs(tsol).doit()) + if sol: + soltup.append((False, sol.subs(y, func))) + else: + soltup.append((True, 0)) + return soltup + + +def sysode_linear_2eq_order1(match_): + x = match_['func'][0].func + y = match_['func'][1].func + func = match_['func'] + fc = match_['func_coeff'] + eq = match_['eq'] + r = {} + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + for i in range(2): + eq[i] = Add(*[terms/fc[i,func[i],1] for terms in Add.make_args(eq[i])]) + + # for equations Eq(a1*diff(x(t),t), a*x(t) + b*y(t) + k1) + # and Eq(a2*diff(x(t),t), c*x(t) + d*y(t) + k2) + r['a'] = -fc[0,x(t),0]/fc[0,x(t),1] + r['c'] = -fc[1,x(t),0]/fc[1,y(t),1] + r['b'] = -fc[0,y(t),0]/fc[0,x(t),1] + r['d'] = -fc[1,y(t),0]/fc[1,y(t),1] + forcing = [S.Zero,S.Zero] + for i in range(2): + for j in Add.make_args(eq[i]): + if not j.has(x(t), y(t)): + forcing[i] += j + if not (forcing[0].has(t) or forcing[1].has(t)): + r['k1'] = forcing[0] + r['k2'] = forcing[1] + else: + raise NotImplementedError("Only homogeneous problems are supported" + + " (and constant inhomogeneity)") + + if match_['type_of_equation'] == 'type6': + sol = _linear_2eq_order1_type6(x, y, t, r, eq) + if match_['type_of_equation'] == 'type7': + sol = _linear_2eq_order1_type7(x, y, t, r, eq) + return sol + +def _linear_2eq_order1_type6(x, y, t, r, eq): + r""" + The equations of this type of ode are . + + .. math:: x' = f(t) x + g(t) y + + .. math:: y' = a [f(t) + a h(t)] x + a [g(t) - h(t)] y + + This is solved by first multiplying the first equation by `-a` and adding + it to the second equation to obtain + + .. math:: y' - a x' = -a h(t) (y - a x) + + Setting `U = y - ax` and integrating the equation we arrive at + + .. math:: y - ax = C_1 e^{-a \int h(t) \,dt} + + and on substituting the value of y in first equation give rise to first order ODEs. After solving for + `x`, we can obtain `y` by substituting the value of `x` in second equation. + + """ + C1, C2, C3, C4 = get_numbered_constants(eq, num=4) + p = 0 + q = 0 + p1 = cancel(r['c']/cancel(r['c']/r['d']).as_numer_denom()[0]) + p2 = cancel(r['a']/cancel(r['a']/r['b']).as_numer_denom()[0]) + for n, i in enumerate([p1, p2]): + for j in Mul.make_args(collect_const(i)): + if not j.has(t): + q = j + if q!=0 and n==0: + if ((r['c']/j - r['a'])/(r['b'] - r['d']/j)) == j: + p = 1 + s = j + break + if q!=0 and n==1: + if ((r['a']/j - r['c'])/(r['d'] - r['b']/j)) == j: + p = 2 + s = j + break + + if p == 1: + equ = diff(x(t),t) - r['a']*x(t) - r['b']*(s*x(t) + C1*exp(-s*Integral(r['b'] - r['d']/s, t))) + hint1 = classify_ode(equ)[1] + sol1 = dsolve(equ, hint=hint1+'_Integral').rhs + sol2 = s*sol1 + C1*exp(-s*Integral(r['b'] - r['d']/s, t)) + elif p ==2: + equ = diff(y(t),t) - r['c']*y(t) - r['d']*s*y(t) + C1*exp(-s*Integral(r['d'] - r['b']/s, t)) + hint1 = classify_ode(equ)[1] + sol2 = dsolve(equ, hint=hint1+'_Integral').rhs + sol1 = s*sol2 + C1*exp(-s*Integral(r['d'] - r['b']/s, t)) + return [Eq(x(t), sol1), Eq(y(t), sol2)] + +def _linear_2eq_order1_type7(x, y, t, r, eq): + r""" + The equations of this type of ode are . + + .. math:: x' = f(t) x + g(t) y + + .. math:: y' = h(t) x + p(t) y + + Differentiating the first equation and substituting the value of `y` + from second equation will give a second-order linear equation + + .. math:: g x'' - (fg + gp + g') x' + (fgp - g^{2} h + f g' - f' g) x = 0 + + This above equation can be easily integrated if following conditions are satisfied. + + 1. `fgp - g^{2} h + f g' - f' g = 0` + + 2. `fgp - g^{2} h + f g' - f' g = ag, fg + gp + g' = bg` + + If first condition is satisfied then it is solved by current dsolve solver and in second case it becomes + a constant coefficient differential equation which is also solved by current solver. + + Otherwise if the above condition fails then, + a particular solution is assumed as `x = x_0(t)` and `y = y_0(t)` + Then the general solution is expressed as + + .. math:: x = C_1 x_0(t) + C_2 x_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt + + .. math:: y = C_1 y_0(t) + C_2 [\frac{F(t) P(t)}{x_0(t)} + y_0(t) \int \frac{g(t) F(t) P(t)}{x_0^{2}(t)} \,dt] + + where C1 and C2 are arbitrary constants and + + .. math:: F(t) = e^{\int f(t) \,dt}, P(t) = e^{\int p(t) \,dt} + + """ + C1, C2, C3, C4 = get_numbered_constants(eq, num=4) + e1 = r['a']*r['b']*r['c'] - r['b']**2*r['c'] + r['a']*diff(r['b'],t) - diff(r['a'],t)*r['b'] + e2 = r['a']*r['c']*r['d'] - r['b']*r['c']**2 + diff(r['c'],t)*r['d'] - r['c']*diff(r['d'],t) + m1 = r['a']*r['b'] + r['b']*r['d'] + diff(r['b'],t) + m2 = r['a']*r['c'] + r['c']*r['d'] + diff(r['c'],t) + if e1 == 0: + sol1 = dsolve(r['b']*diff(x(t),t,t) - m1*diff(x(t),t)).rhs + sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs + elif e2 == 0: + sol2 = dsolve(r['c']*diff(y(t),t,t) - m2*diff(y(t),t)).rhs + sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs + elif not (e1/r['b']).has(t) and not (m1/r['b']).has(t): + sol1 = dsolve(diff(x(t),t,t) - (m1/r['b'])*diff(x(t),t) - (e1/r['b'])*x(t)).rhs + sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs + elif not (e2/r['c']).has(t) and not (m2/r['c']).has(t): + sol2 = dsolve(diff(y(t),t,t) - (m2/r['c'])*diff(y(t),t) - (e2/r['c'])*y(t)).rhs + sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs + else: + x0 = Function('x0')(t) # x0 and y0 being particular solutions + y0 = Function('y0')(t) + F = exp(Integral(r['a'],t)) + P = exp(Integral(r['d'],t)) + sol1 = C1*x0 + C2*x0*Integral(r['b']*F*P/x0**2, t) + sol2 = C1*y0 + C2*(F*P/x0 + y0*Integral(r['b']*F*P/x0**2, t)) + return [Eq(x(t), sol1), Eq(y(t), sol2)] + + +def sysode_nonlinear_2eq_order1(match_): + func = match_['func'] + eq = match_['eq'] + fc = match_['func_coeff'] + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + if match_['type_of_equation'] == 'type5': + sol = _nonlinear_2eq_order1_type5(func, t, eq) + return sol + x = func[0].func + y = func[1].func + for i in range(2): + eqs = 0 + for terms in Add.make_args(eq[i]): + eqs += terms/fc[i,func[i],1] + eq[i] = eqs + if match_['type_of_equation'] == 'type1': + sol = _nonlinear_2eq_order1_type1(x, y, t, eq) + elif match_['type_of_equation'] == 'type2': + sol = _nonlinear_2eq_order1_type2(x, y, t, eq) + elif match_['type_of_equation'] == 'type3': + sol = _nonlinear_2eq_order1_type3(x, y, t, eq) + elif match_['type_of_equation'] == 'type4': + sol = _nonlinear_2eq_order1_type4(x, y, t, eq) + return sol + + +def _nonlinear_2eq_order1_type1(x, y, t, eq): + r""" + Equations: + + .. math:: x' = x^n F(x,y) + + .. math:: y' = g(y) F(x,y) + + Solution: + + .. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2 + + where + + if `n \neq 1` + + .. math:: \varphi = [C_1 + (1-n) \int \frac{1}{g(y)} \,dy]^{\frac{1}{1-n}} + + if `n = 1` + + .. math:: \varphi = C_1 e^{\int \frac{1}{g(y)} \,dy} + + where `C_1` and `C_2` are arbitrary constants. + + """ + C1, C2 = get_numbered_constants(eq, num=2) + n = Wild('n', exclude=[x(t),y(t)]) + f = Wild('f') + u, v = symbols('u, v') + r = eq[0].match(diff(x(t),t) - x(t)**n*f) + g = ((diff(y(t),t) - eq[1])/r[f]).subs(y(t),v) + F = r[f].subs(x(t),u).subs(y(t),v) + n = r[n] + if n!=1: + phi = (C1 + (1-n)*Integral(1/g, v))**(1/(1-n)) + else: + phi = C1*exp(Integral(1/g, v)) + phi = phi.doit() + sol2 = solve(Integral(1/(g*F.subs(u,phi)), v).doit() - t - C2, v) + sol = [] + for sols in sol2: + sol.append(Eq(x(t),phi.subs(v, sols))) + sol.append(Eq(y(t), sols)) + return sol + +def _nonlinear_2eq_order1_type2(x, y, t, eq): + r""" + Equations: + + .. math:: x' = e^{\lambda x} F(x,y) + + .. math:: y' = g(y) F(x,y) + + Solution: + + .. math:: x = \varphi(y), \int \frac{1}{g(y) F(\varphi(y),y)} \,dy = t + C_2 + + where + + if `\lambda \neq 0` + + .. math:: \varphi = -\frac{1}{\lambda} log(C_1 - \lambda \int \frac{1}{g(y)} \,dy) + + if `\lambda = 0` + + .. math:: \varphi = C_1 + \int \frac{1}{g(y)} \,dy + + where `C_1` and `C_2` are arbitrary constants. + + """ + C1, C2 = get_numbered_constants(eq, num=2) + n = Wild('n', exclude=[x(t),y(t)]) + f = Wild('f') + u, v = symbols('u, v') + r = eq[0].match(diff(x(t),t) - exp(n*x(t))*f) + g = ((diff(y(t),t) - eq[1])/r[f]).subs(y(t),v) + F = r[f].subs(x(t),u).subs(y(t),v) + n = r[n] + if n: + phi = -1/n*log(C1 - n*Integral(1/g, v)) + else: + phi = C1 + Integral(1/g, v) + phi = phi.doit() + sol2 = solve(Integral(1/(g*F.subs(u,phi)), v).doit() - t - C2, v) + sol = [] + for sols in sol2: + sol.append(Eq(x(t),phi.subs(v, sols))) + sol.append(Eq(y(t), sols)) + return sol + +def _nonlinear_2eq_order1_type3(x, y, t, eq): + r""" + Autonomous system of general form + + .. math:: x' = F(x,y) + + .. math:: y' = G(x,y) + + Assuming `y = y(x, C_1)` where `C_1` is an arbitrary constant is the general + solution of the first-order equation + + .. math:: F(x,y) y'_x = G(x,y) + + Then the general solution of the original system of equations has the form + + .. math:: \int \frac{1}{F(x,y(x,C_1))} \,dx = t + C_1 + + """ + C1, C2, C3, C4 = get_numbered_constants(eq, num=4) + v = Function('v') + u = Symbol('u') + f = Wild('f') + g = Wild('g') + r1 = eq[0].match(diff(x(t),t) - f) + r2 = eq[1].match(diff(y(t),t) - g) + F = r1[f].subs(x(t), u).subs(y(t), v(u)) + G = r2[g].subs(x(t), u).subs(y(t), v(u)) + sol2r = dsolve(Eq(diff(v(u), u), G/F)) + if isinstance(sol2r, Equality): + sol2r = [sol2r] + for sol2s in sol2r: + sol1 = solve(Integral(1/F.subs(v(u), sol2s.rhs), u).doit() - t - C2, u) + sol = [] + for sols in sol1: + sol.append(Eq(x(t), sols)) + sol.append(Eq(y(t), (sol2s.rhs).subs(u, sols))) + return sol + +def _nonlinear_2eq_order1_type4(x, y, t, eq): + r""" + Equation: + + .. math:: x' = f_1(x) g_1(y) \phi(x,y,t) + + .. math:: y' = f_2(x) g_2(y) \phi(x,y,t) + + First integral: + + .. math:: \int \frac{f_2(x)}{f_1(x)} \,dx - \int \frac{g_1(y)}{g_2(y)} \,dy = C + + where `C` is an arbitrary constant. + + On solving the first integral for `x` (resp., `y` ) and on substituting the + resulting expression into either equation of the original solution, one + arrives at a first-order equation for determining `y` (resp., `x` ). + + """ + C1, C2 = get_numbered_constants(eq, num=2) + u, v = symbols('u, v') + U, V = symbols('U, V', cls=Function) + f = Wild('f') + g = Wild('g') + f1 = Wild('f1', exclude=[v,t]) + f2 = Wild('f2', exclude=[v,t]) + g1 = Wild('g1', exclude=[u,t]) + g2 = Wild('g2', exclude=[u,t]) + r1 = eq[0].match(diff(x(t),t) - f) + r2 = eq[1].match(diff(y(t),t) - g) + num, den = ( + (r1[f].subs(x(t),u).subs(y(t),v))/ + (r2[g].subs(x(t),u).subs(y(t),v))).as_numer_denom() + R1 = num.match(f1*g1) + R2 = den.match(f2*g2) + phi = (r1[f].subs(x(t),u).subs(y(t),v))/num + F1 = R1[f1]; F2 = R2[f2] + G1 = R1[g1]; G2 = R2[g2] + sol1r = solve(Integral(F2/F1, u).doit() - Integral(G1/G2,v).doit() - C1, u) + sol2r = solve(Integral(F2/F1, u).doit() - Integral(G1/G2,v).doit() - C1, v) + sol = [] + for sols in sol1r: + sol.append(Eq(y(t), dsolve(diff(V(t),t) - F2.subs(u,sols).subs(v,V(t))*G2.subs(v,V(t))*phi.subs(u,sols).subs(v,V(t))).rhs)) + for sols in sol2r: + sol.append(Eq(x(t), dsolve(diff(U(t),t) - F1.subs(u,U(t))*G1.subs(v,sols).subs(u,U(t))*phi.subs(v,sols).subs(u,U(t))).rhs)) + return set(sol) + +def _nonlinear_2eq_order1_type5(func, t, eq): + r""" + Clairaut system of ODEs + + .. math:: x = t x' + F(x',y') + + .. math:: y = t y' + G(x',y') + + The following are solutions of the system + + `(i)` straight lines: + + .. math:: x = C_1 t + F(C_1, C_2), y = C_2 t + G(C_1, C_2) + + where `C_1` and `C_2` are arbitrary constants; + + `(ii)` envelopes of the above lines; + + `(iii)` continuously differentiable lines made up from segments of the lines + `(i)` and `(ii)`. + + """ + C1, C2 = get_numbered_constants(eq, num=2) + f = Wild('f') + g = Wild('g') + def check_type(x, y): + r1 = eq[0].match(t*diff(x(t),t) - x(t) + f) + r2 = eq[1].match(t*diff(y(t),t) - y(t) + g) + if not (r1 and r2): + r1 = eq[0].match(diff(x(t),t) - x(t)/t + f/t) + r2 = eq[1].match(diff(y(t),t) - y(t)/t + g/t) + if not (r1 and r2): + r1 = (-eq[0]).match(t*diff(x(t),t) - x(t) + f) + r2 = (-eq[1]).match(t*diff(y(t),t) - y(t) + g) + if not (r1 and r2): + r1 = (-eq[0]).match(diff(x(t),t) - x(t)/t + f/t) + r2 = (-eq[1]).match(diff(y(t),t) - y(t)/t + g/t) + return [r1, r2] + for func_ in func: + if isinstance(func_, list): + x = func[0][0].func + y = func[0][1].func + [r1, r2] = check_type(x, y) + if not (r1 and r2): + [r1, r2] = check_type(y, x) + x, y = y, x + x1 = diff(x(t),t); y1 = diff(y(t),t) + return {Eq(x(t), C1*t + r1[f].subs(x1,C1).subs(y1,C2)), Eq(y(t), C2*t + r2[g].subs(x1,C1).subs(y1,C2))} + +def sysode_nonlinear_3eq_order1(match_): + x = match_['func'][0].func + y = match_['func'][1].func + z = match_['func'][2].func + eq = match_['eq'] + t = list(list(eq[0].atoms(Derivative))[0].atoms(Symbol))[0] + if match_['type_of_equation'] == 'type1': + sol = _nonlinear_3eq_order1_type1(x, y, z, t, eq) + if match_['type_of_equation'] == 'type2': + sol = _nonlinear_3eq_order1_type2(x, y, z, t, eq) + if match_['type_of_equation'] == 'type3': + sol = _nonlinear_3eq_order1_type3(x, y, z, t, eq) + if match_['type_of_equation'] == 'type4': + sol = _nonlinear_3eq_order1_type4(x, y, z, t, eq) + if match_['type_of_equation'] == 'type5': + sol = _nonlinear_3eq_order1_type5(x, y, z, t, eq) + return sol + +def _nonlinear_3eq_order1_type1(x, y, z, t, eq): + r""" + Equations: + + .. math:: a x' = (b - c) y z, \enspace b y' = (c - a) z x, \enspace c z' = (a - b) x y + + First Integrals: + + .. math:: a x^{2} + b y^{2} + c z^{2} = C_1 + + .. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2 + + where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and + `z` and on substituting the resulting expressions into the first equation of the + system, we arrives at a separable first-order equation on `x`. Similarly doing that + for other two equations, we will arrive at first order equation on `y` and `z` too. + + References + ========== + -https://eqworld.ipmnet.ru/en/solutions/sysode/sode0401.pdf + + """ + C1, C2 = get_numbered_constants(eq, num=2) + u, v, w = symbols('u, v, w') + p = Wild('p', exclude=[x(t), y(t), z(t), t]) + q = Wild('q', exclude=[x(t), y(t), z(t), t]) + s = Wild('s', exclude=[x(t), y(t), z(t), t]) + r = (diff(x(t),t) - eq[0]).match(p*y(t)*z(t)) + r.update((diff(y(t),t) - eq[1]).match(q*z(t)*x(t))) + r.update((diff(z(t),t) - eq[2]).match(s*x(t)*y(t))) + n1, d1 = r[p].as_numer_denom() + n2, d2 = r[q].as_numer_denom() + n3, d3 = r[s].as_numer_denom() + val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, d3*u-d3*v-n3*w],[u,v]) + vals = [val[v], val[u]] + c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1]) + b = vals[0].subs(w, c) + a = vals[1].subs(w, c) + y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b))) + z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c))) + z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c))) + x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a))) + x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a))) + y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b))) + sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x) + sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y) + sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z) + return [sol1, sol2, sol3] + + +def _nonlinear_3eq_order1_type2(x, y, z, t, eq): + r""" + Equations: + + .. math:: a x' = (b - c) y z f(x, y, z, t) + + .. math:: b y' = (c - a) z x f(x, y, z, t) + + .. math:: c z' = (a - b) x y f(x, y, z, t) + + First Integrals: + + .. math:: a x^{2} + b y^{2} + c z^{2} = C_1 + + .. math:: a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} = C_2 + + where `C_1` and `C_2` are arbitrary constants. On solving the integrals for `y` and + `z` and on substituting the resulting expressions into the first equation of the + system, we arrives at a first-order differential equations on `x`. Similarly doing + that for other two equations we will arrive at first order equation on `y` and `z`. + + References + ========== + -https://eqworld.ipmnet.ru/en/solutions/sysode/sode0402.pdf + + """ + C1, C2 = get_numbered_constants(eq, num=2) + u, v, w = symbols('u, v, w') + p = Wild('p', exclude=[x(t), y(t), z(t), t]) + q = Wild('q', exclude=[x(t), y(t), z(t), t]) + s = Wild('s', exclude=[x(t), y(t), z(t), t]) + f = Wild('f') + r1 = (diff(x(t),t) - eq[0]).match(y(t)*z(t)*f) + r = collect_const(r1[f]).match(p*f) + r.update(((diff(y(t),t) - eq[1])/r[f]).match(q*z(t)*x(t))) + r.update(((diff(z(t),t) - eq[2])/r[f]).match(s*x(t)*y(t))) + n1, d1 = r[p].as_numer_denom() + n2, d2 = r[q].as_numer_denom() + n3, d3 = r[s].as_numer_denom() + val = solve([n1*u-d1*v+d1*w, d2*u+n2*v-d2*w, -d3*u+d3*v+n3*w],[u,v]) + vals = [val[v], val[u]] + c = lcm(vals[0].as_numer_denom()[1], vals[1].as_numer_denom()[1]) + a = vals[0].subs(w, c) + b = vals[1].subs(w, c) + y_x = sqrt(((c*C1-C2) - a*(c-a)*x(t)**2)/(b*(c-b))) + z_x = sqrt(((b*C1-C2) - a*(b-a)*x(t)**2)/(c*(b-c))) + z_y = sqrt(((a*C1-C2) - b*(a-b)*y(t)**2)/(c*(a-c))) + x_y = sqrt(((c*C1-C2) - b*(c-b)*y(t)**2)/(a*(c-a))) + x_z = sqrt(((b*C1-C2) - c*(b-c)*z(t)**2)/(a*(b-a))) + y_z = sqrt(((a*C1-C2) - c*(a-c)*z(t)**2)/(b*(a-b))) + sol1 = dsolve(a*diff(x(t),t) - (b-c)*y_x*z_x*r[f]) + sol2 = dsolve(b*diff(y(t),t) - (c-a)*z_y*x_y*r[f]) + sol3 = dsolve(c*diff(z(t),t) - (a-b)*x_z*y_z*r[f]) + return [sol1, sol2, sol3] + +def _nonlinear_3eq_order1_type3(x, y, z, t, eq): + r""" + Equations: + + .. math:: x' = c F_2 - b F_3, \enspace y' = a F_3 - c F_1, \enspace z' = b F_1 - a F_2 + + where `F_n = F_n(x, y, z, t)`. + + 1. First Integral: + + .. math:: a x + b y + c z = C_1, + + where C is an arbitrary constant. + + 2. If we assume function `F_n` to be independent of `t`,i.e, `F_n` = `F_n (x, y, z)` + Then, on eliminating `t` and `z` from the first two equation of the system, one + arrives at the first-order equation + + .. math:: \frac{dy}{dx} = \frac{a F_3 (x, y, z) - c F_1 (x, y, z)}{c F_2 (x, y, z) - + b F_3 (x, y, z)} + + where `z = \frac{1}{c} (C_1 - a x - b y)` + + References + ========== + -https://eqworld.ipmnet.ru/en/solutions/sysode/sode0404.pdf + + """ + C1 = get_numbered_constants(eq, num=1) + u, v, w = symbols('u, v, w') + fu, fv, fw = symbols('u, v, w', cls=Function) + p = Wild('p', exclude=[x(t), y(t), z(t), t]) + q = Wild('q', exclude=[x(t), y(t), z(t), t]) + s = Wild('s', exclude=[x(t), y(t), z(t), t]) + F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) + r1 = (diff(x(t), t) - eq[0]).match(F2-F3) + r = collect_const(r1[F2]).match(s*F2) + r.update(collect_const(r1[F3]).match(q*F3)) + if eq[1].has(r[F2]) and not eq[1].has(r[F3]): + r[F2], r[F3] = r[F3], r[F2] + r[s], r[q] = -r[q], -r[s] + r.update((diff(y(t), t) - eq[1]).match(p*r[F3] - r[s]*F1)) + a = r[p]; b = r[q]; c = r[s] + F1 = r[F1].subs(x(t), u).subs(y(t),v).subs(z(t), w) + F2 = r[F2].subs(x(t), u).subs(y(t),v).subs(z(t), w) + F3 = r[F3].subs(x(t), u).subs(y(t),v).subs(z(t), w) + z_xy = (C1-a*u-b*v)/c + y_zx = (C1-a*u-c*w)/b + x_yz = (C1-b*v-c*w)/a + y_x = dsolve(diff(fv(u),u) - ((a*F3-c*F1)/(c*F2-b*F3)).subs(w,z_xy).subs(v,fv(u))).rhs + z_x = dsolve(diff(fw(u),u) - ((b*F1-a*F2)/(c*F2-b*F3)).subs(v,y_zx).subs(w,fw(u))).rhs + z_y = dsolve(diff(fw(v),v) - ((b*F1-a*F2)/(a*F3-c*F1)).subs(u,x_yz).subs(w,fw(v))).rhs + x_y = dsolve(diff(fu(v),v) - ((c*F2-b*F3)/(a*F3-c*F1)).subs(w,z_xy).subs(u,fu(v))).rhs + y_z = dsolve(diff(fv(w),w) - ((a*F3-c*F1)/(b*F1-a*F2)).subs(u,x_yz).subs(v,fv(w))).rhs + x_z = dsolve(diff(fu(w),w) - ((c*F2-b*F3)/(b*F1-a*F2)).subs(v,y_zx).subs(u,fu(w))).rhs + sol1 = dsolve(diff(fu(t),t) - (c*F2 - b*F3).subs(v,y_x).subs(w,z_x).subs(u,fu(t))).rhs + sol2 = dsolve(diff(fv(t),t) - (a*F3 - c*F1).subs(u,x_y).subs(w,z_y).subs(v,fv(t))).rhs + sol3 = dsolve(diff(fw(t),t) - (b*F1 - a*F2).subs(u,x_z).subs(v,y_z).subs(w,fw(t))).rhs + return [sol1, sol2, sol3] + +def _nonlinear_3eq_order1_type4(x, y, z, t, eq): + r""" + Equations: + + .. math:: x' = c z F_2 - b y F_3, \enspace y' = a x F_3 - c z F_1, \enspace z' = b y F_1 - a x F_2 + + where `F_n = F_n (x, y, z, t)` + + 1. First integral: + + .. math:: a x^{2} + b y^{2} + c z^{2} = C_1 + + where `C` is an arbitrary constant. + + 2. Assuming the function `F_n` is independent of `t`: `F_n = F_n (x, y, z)`. Then on + eliminating `t` and `z` from the first two equations of the system, one arrives at + the first-order equation + + .. math:: \frac{dy}{dx} = \frac{a x F_3 (x, y, z) - c z F_1 (x, y, z)} + {c z F_2 (x, y, z) - b y F_3 (x, y, z)} + + where `z = \pm \sqrt{\frac{1}{c} (C_1 - a x^{2} - b y^{2})}` + + References + ========== + -https://eqworld.ipmnet.ru/en/solutions/sysode/sode0405.pdf + + """ + C1 = get_numbered_constants(eq, num=1) + u, v, w = symbols('u, v, w') + p = Wild('p', exclude=[x(t), y(t), z(t), t]) + q = Wild('q', exclude=[x(t), y(t), z(t), t]) + s = Wild('s', exclude=[x(t), y(t), z(t), t]) + F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) + r1 = eq[0].match(diff(x(t),t) - z(t)*F2 + y(t)*F3) + r = collect_const(r1[F2]).match(s*F2) + r.update(collect_const(r1[F3]).match(q*F3)) + if eq[1].has(r[F2]) and not eq[1].has(r[F3]): + r[F2], r[F3] = r[F3], r[F2] + r[s], r[q] = -r[q], -r[s] + r.update((diff(y(t),t) - eq[1]).match(p*x(t)*r[F3] - r[s]*z(t)*F1)) + a = r[p]; b = r[q]; c = r[s] + F1 = r[F1].subs(x(t),u).subs(y(t),v).subs(z(t),w) + F2 = r[F2].subs(x(t),u).subs(y(t),v).subs(z(t),w) + F3 = r[F3].subs(x(t),u).subs(y(t),v).subs(z(t),w) + x_yz = sqrt((C1 - b*v**2 - c*w**2)/a) + y_zx = sqrt((C1 - c*w**2 - a*u**2)/b) + z_xy = sqrt((C1 - a*u**2 - b*v**2)/c) + y_x = dsolve(diff(v(u),u) - ((a*u*F3-c*w*F1)/(c*w*F2-b*v*F3)).subs(w,z_xy).subs(v,v(u))).rhs + z_x = dsolve(diff(w(u),u) - ((b*v*F1-a*u*F2)/(c*w*F2-b*v*F3)).subs(v,y_zx).subs(w,w(u))).rhs + z_y = dsolve(diff(w(v),v) - ((b*v*F1-a*u*F2)/(a*u*F3-c*w*F1)).subs(u,x_yz).subs(w,w(v))).rhs + x_y = dsolve(diff(u(v),v) - ((c*w*F2-b*v*F3)/(a*u*F3-c*w*F1)).subs(w,z_xy).subs(u,u(v))).rhs + y_z = dsolve(diff(v(w),w) - ((a*u*F3-c*w*F1)/(b*v*F1-a*u*F2)).subs(u,x_yz).subs(v,v(w))).rhs + x_z = dsolve(diff(u(w),w) - ((c*w*F2-b*v*F3)/(b*v*F1-a*u*F2)).subs(v,y_zx).subs(u,u(w))).rhs + sol1 = dsolve(diff(u(t),t) - (c*w*F2 - b*v*F3).subs(v,y_x).subs(w,z_x).subs(u,u(t))).rhs + sol2 = dsolve(diff(v(t),t) - (a*u*F3 - c*w*F1).subs(u,x_y).subs(w,z_y).subs(v,v(t))).rhs + sol3 = dsolve(diff(w(t),t) - (b*v*F1 - a*u*F2).subs(u,x_z).subs(v,y_z).subs(w,w(t))).rhs + return [sol1, sol2, sol3] + +def _nonlinear_3eq_order1_type5(x, y, z, t, eq): + r""" + .. math:: x' = x (c F_2 - b F_3), \enspace y' = y (a F_3 - c F_1), \enspace z' = z (b F_1 - a F_2) + + where `F_n = F_n (x, y, z, t)` and are arbitrary functions. + + First Integral: + + .. math:: \left|x\right|^{a} \left|y\right|^{b} \left|z\right|^{c} = C_1 + + where `C` is an arbitrary constant. If the function `F_n` is independent of `t`, + then, by eliminating `t` and `z` from the first two equations of the system, one + arrives at a first-order equation. + + References + ========== + -https://eqworld.ipmnet.ru/en/solutions/sysode/sode0406.pdf + + """ + C1 = get_numbered_constants(eq, num=1) + u, v, w = symbols('u, v, w') + fu, fv, fw = symbols('u, v, w', cls=Function) + p = Wild('p', exclude=[x(t), y(t), z(t), t]) + q = Wild('q', exclude=[x(t), y(t), z(t), t]) + s = Wild('s', exclude=[x(t), y(t), z(t), t]) + F1, F2, F3 = symbols('F1, F2, F3', cls=Wild) + r1 = eq[0].match(diff(x(t), t) - x(t)*F2 + x(t)*F3) + r = collect_const(r1[F2]).match(s*F2) + r.update(collect_const(r1[F3]).match(q*F3)) + if eq[1].has(r[F2]) and not eq[1].has(r[F3]): + r[F2], r[F3] = r[F3], r[F2] + r[s], r[q] = -r[q], -r[s] + r.update((diff(y(t), t) - eq[1]).match(y(t)*(p*r[F3] - r[s]*F1))) + a = r[p]; b = r[q]; c = r[s] + F1 = r[F1].subs(x(t), u).subs(y(t), v).subs(z(t), w) + F2 = r[F2].subs(x(t), u).subs(y(t), v).subs(z(t), w) + F3 = r[F3].subs(x(t), u).subs(y(t), v).subs(z(t), w) + x_yz = (C1*v**-b*w**-c)**-a + y_zx = (C1*w**-c*u**-a)**-b + z_xy = (C1*u**-a*v**-b)**-c + y_x = dsolve(diff(fv(u), u) - ((v*(a*F3 - c*F1))/(u*(c*F2 - b*F3))).subs(w, z_xy).subs(v, fv(u))).rhs + z_x = dsolve(diff(fw(u), u) - ((w*(b*F1 - a*F2))/(u*(c*F2 - b*F3))).subs(v, y_zx).subs(w, fw(u))).rhs + z_y = dsolve(diff(fw(v), v) - ((w*(b*F1 - a*F2))/(v*(a*F3 - c*F1))).subs(u, x_yz).subs(w, fw(v))).rhs + x_y = dsolve(diff(fu(v), v) - ((u*(c*F2 - b*F3))/(v*(a*F3 - c*F1))).subs(w, z_xy).subs(u, fu(v))).rhs + y_z = dsolve(diff(fv(w), w) - ((v*(a*F3 - c*F1))/(w*(b*F1 - a*F2))).subs(u, x_yz).subs(v, fv(w))).rhs + x_z = dsolve(diff(fu(w), w) - ((u*(c*F2 - b*F3))/(w*(b*F1 - a*F2))).subs(v, y_zx).subs(u, fu(w))).rhs + sol1 = dsolve(diff(fu(t), t) - (u*(c*F2 - b*F3)).subs(v, y_x).subs(w, z_x).subs(u, fu(t))).rhs + sol2 = dsolve(diff(fv(t), t) - (v*(a*F3 - c*F1)).subs(u, x_y).subs(w, z_y).subs(v, fv(t))).rhs + sol3 = dsolve(diff(fw(t), t) - (w*(b*F1 - a*F2)).subs(u, x_z).subs(v, y_z).subs(w, fw(t))).rhs + return [sol1, sol2, sol3] + + +#This import is written at the bottom to avoid circular imports. +from .single import SingleODEProblem, SingleODESolver, solver_map diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/riccati.py b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/riccati.py new file mode 100644 index 0000000000000000000000000000000000000000..2ef66ed0896d39bee8fba1b74a0c93734742fc1f --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/riccati.py @@ -0,0 +1,893 @@ +r""" +This module contains :py:meth:`~sympy.solvers.ode.riccati.solve_riccati`, +a function which gives all rational particular solutions to first order +Riccati ODEs. A general first order Riccati ODE is given by - + +.. math:: y' = b_0(x) + b_1(x)w + b_2(x)w^2 + +where `b_0, b_1` and `b_2` can be arbitrary rational functions of `x` +with `b_2 \ne 0`. When `b_2 = 0`, the equation is not a Riccati ODE +anymore and becomes a Linear ODE. Similarly, when `b_0 = 0`, the equation +is a Bernoulli ODE. The algorithm presented below can find rational +solution(s) to all ODEs with `b_2 \ne 0` that have a rational solution, +or prove that no rational solution exists for the equation. + +Background +========== + +A Riccati equation can be transformed to its normal form + +.. math:: y' + y^2 = a(x) + +using the transformation + +.. math:: y = -b_2(x) - \frac{b'_2(x)}{2 b_2(x)} - \frac{b_1(x)}{2} + +where `a(x)` is given by + +.. math:: a(x) = \frac{1}{4}\left(\frac{b_2'}{b_2} + b_1\right)^2 - \frac{1}{2}\left(\frac{b_2'}{b_2} + b_1\right)' - b_0 b_2 + +Thus, we can develop an algorithm to solve for the Riccati equation +in its normal form, which would in turn give us the solution for +the original Riccati equation. + +Algorithm +========= + +The algorithm implemented here is presented in the Ph.D thesis +"Rational and Algebraic Solutions of First-Order Algebraic ODEs" +by N. Thieu Vo. The entire thesis can be found here - +https://www3.risc.jku.at/publications/download/risc_5387/PhDThesisThieu.pdf + +We have only implemented the Rational Riccati solver (Algorithm 11, +Pg 78-82 in Thesis). Before we proceed towards the implementation +of the algorithm, a few definitions to understand are - + +1. Valuation of a Rational Function at `\infty`: + The valuation of a rational function `p(x)` at `\infty` is equal + to the difference between the degree of the denominator and the + numerator of `p(x)`. + + NOTE: A general definition of valuation of a rational function + at any value of `x` can be found in Pg 63 of the thesis, but + is not of any interest for this algorithm. + +2. Zeros and Poles of a Rational Function: + Let `a(x) = \frac{S(x)}{T(x)}, T \ne 0` be a rational function + of `x`. Then - + + a. The Zeros of `a(x)` are the roots of `S(x)`. + b. The Poles of `a(x)` are the roots of `T(x)`. However, `\infty` + can also be a pole of a(x). We say that `a(x)` has a pole at + `\infty` if `a(\frac{1}{x})` has a pole at 0. + +Every pole is associated with an order that is equal to the multiplicity +of its appearance as a root of `T(x)`. A pole is called a simple pole if +it has an order 1. Similarly, a pole is called a multiple pole if it has +an order `\ge` 2. + +Necessary Conditions +==================== + +For a Riccati equation in its normal form, + +.. math:: y' + y^2 = a(x) + +we can define + +a. A pole is called a movable pole if it is a pole of `y(x)` and is not +a pole of `a(x)`. +b. Similarly, a pole is called a non-movable pole if it is a pole of both +`y(x)` and `a(x)`. + +Then, the algorithm states that a rational solution exists only if - + +a. Every pole of `a(x)` must be either a simple pole or a multiple pole +of even order. +b. The valuation of `a(x)` at `\infty` must be even or be `\ge` 2. + +This algorithm finds all possible rational solutions for the Riccati ODE. +If no rational solutions are found, it means that no rational solutions +exist. + +The algorithm works for Riccati ODEs where the coefficients are rational +functions in the independent variable `x` with rational number coefficients +i.e. in `Q(x)`. The coefficients in the rational function cannot be floats, +irrational numbers, symbols or any other kind of expression. The reasons +for this are - + +1. When using symbols, different symbols could take the same value and this +would affect the multiplicity of poles if symbols are present here. + +2. An integer degree bound is required to calculate a polynomial solution +to an auxiliary differential equation, which in turn gives the particular +solution for the original ODE. If symbols/floats/irrational numbers are +present, we cannot determine if the expression for the degree bound is an +integer or not. + +Solution +======== + +With these definitions, we can state a general form for the solution of +the equation. `y(x)` must have the form - + +.. math:: y(x) = \sum_{i=1}^{n} \sum_{j=1}^{r_i} \frac{c_{ij}}{(x - x_i)^j} + \sum_{i=1}^{m} \frac{1}{x - \chi_i} + \sum_{i=0}^{N} d_i x^i + +where `x_1, x_2, \dots, x_n` are non-movable poles of `a(x)`, +`\chi_1, \chi_2, \dots, \chi_m` are movable poles of `a(x)`, and the values +of `N, n, r_1, r_2, \dots, r_n` can be determined from `a(x)`. The +coefficient vectors `(d_0, d_1, \dots, d_N)` and `(c_{i1}, c_{i2}, \dots, c_{i r_i})` +can be determined from `a(x)`. We will have 2 choices each of these vectors +and part of the procedure is figuring out which of the 2 should be used +to get the solution correctly. + +Implementation +============== + +In this implementation, we use ``Poly`` to represent a rational function +rather than using ``Expr`` since ``Poly`` is much faster. Since we cannot +represent rational functions directly using ``Poly``, we instead represent +a rational function with 2 ``Poly`` objects - one for its numerator and +the other for its denominator. + +The code is written to match the steps given in the thesis (Pg 82) + +Step 0 : Match the equation - +Find `b_0, b_1` and `b_2`. If `b_2 = 0` or no such functions exist, raise +an error + +Step 1 : Transform the equation to its normal form as explained in the +theory section. + +Step 2 : Initialize an empty set of solutions, ``sol``. + +Step 3 : If `a(x) = 0`, append `\frac{1}/{(x - C1)}` to ``sol``. + +Step 4 : If `a(x)` is a rational non-zero number, append `\pm \sqrt{a}` +to ``sol``. + +Step 5 : Find the poles and their multiplicities of `a(x)`. Let +the number of poles be `n`. Also find the valuation of `a(x)` at +`\infty` using ``val_at_inf``. + +NOTE: Although the algorithm considers `\infty` as a pole, it is +not mentioned if it a part of the set of finite poles. `\infty` +is NOT a part of the set of finite poles. If a pole exists at +`\infty`, we use its multiplicity to find the laurent series of +`a(x)` about `\infty`. + +Step 6 : Find `n` c-vectors (one for each pole) and 1 d-vector using +``construct_c`` and ``construct_d``. Now, determine all the ``2**(n + 1)`` +combinations of choosing between 2 choices for each of the `n` c-vectors +and 1 d-vector. + +NOTE: The equation for `d_{-1}` in Case 4 (Pg 80) has a printinig +mistake. The term `- d_N` must be replaced with `-N d_N`. The same +has been explained in the code as well. + +For each of these above combinations, do + +Step 8 : Compute `m` in ``compute_m_ybar``. `m` is the degree bound of +the polynomial solution we must find for the auxiliary equation. + +Step 9 : In ``compute_m_ybar``, compute ybar as well where ``ybar`` is +one part of y(x) - + +.. math:: \overline{y}(x) = \sum_{i=1}^{n} \sum_{j=1}^{r_i} \frac{c_{ij}}{(x - x_i)^j} + \sum_{i=0}^{N} d_i x^i + +Step 10 : If `m` is a non-negative integer - + +Step 11: Find a polynomial solution of degree `m` for the auxiliary equation. + +There are 2 cases possible - + + a. `m` is a non-negative integer: We can solve for the coefficients + in `p(x)` using Undetermined Coefficients. + + b. `m` is not a non-negative integer: In this case, we cannot find + a polynomial solution to the auxiliary equation, and hence, we ignore + this value of `m`. + +Step 12 : For each `p(x)` that exists, append `ybar + \frac{p'(x)}{p(x)}` +to ``sol``. + +Step 13 : For each solution in ``sol``, apply an inverse transformation, +so that the solutions of the original equation are found using the +solutions of the equation in its normal form. +""" + + +from itertools import product +from sympy.core import S +from sympy.core.add import Add +from sympy.core.numbers import oo, Float +from sympy.core.function import count_ops +from sympy.core.relational import Eq +from sympy.core.symbol import symbols, Symbol, Dummy +from sympy.functions import sqrt, exp +from sympy.functions.elementary.complexes import sign +from sympy.integrals.integrals import Integral +from sympy.polys.domains import ZZ +from sympy.polys.polytools import Poly +from sympy.polys.polyroots import roots +from sympy.solvers.solveset import linsolve + + +def riccati_normal(w, x, b1, b2): + """ + Given a solution `w(x)` to the equation + + .. math:: w'(x) = b_0(x) + b_1(x)*w(x) + b_2(x)*w(x)^2 + + and rational function coefficients `b_1(x)` and + `b_2(x)`, this function transforms the solution to + give a solution `y(x)` for its corresponding normal + Riccati ODE + + .. math:: y'(x) + y(x)^2 = a(x) + + using the transformation + + .. math:: y(x) = -b_2(x)*w(x) - b'_2(x)/(2*b_2(x)) - b_1(x)/2 + """ + return -b2*w - b2.diff(x)/(2*b2) - b1/2 + + +def riccati_inverse_normal(y, x, b1, b2, bp=None): + """ + Inverse transforming the solution to the normal + Riccati ODE to get the solution to the Riccati ODE. + """ + # bp is the expression which is independent of the solution + # and hence, it need not be computed again + if bp is None: + bp = -b2.diff(x)/(2*b2**2) - b1/(2*b2) + # w(x) = -y(x)/b2(x) - b2'(x)/(2*b2(x)^2) - b1(x)/(2*b2(x)) + return -y/b2 + bp + + +def riccati_reduced(eq, f, x): + """ + Convert a Riccati ODE into its corresponding + normal Riccati ODE. + """ + match, funcs = match_riccati(eq, f, x) + # If equation is not a Riccati ODE, exit + if not match: + return False + # Using the rational functions, find the expression for a(x) + b0, b1, b2 = funcs + a = -b0*b2 + b1**2/4 - b1.diff(x)/2 + 3*b2.diff(x)**2/(4*b2**2) + b1*b2.diff(x)/(2*b2) - \ + b2.diff(x, 2)/(2*b2) + # Normal form of Riccati ODE is f'(x) + f(x)^2 = a(x) + return f(x).diff(x) + f(x)**2 - a + +def linsolve_dict(eq, syms): + """ + Get the output of linsolve as a dict + """ + # Convert tuple type return value of linsolve + # to a dictionary for ease of use + sol = linsolve(eq, syms) + if not sol: + return {} + return dict(zip(syms, list(sol)[0])) + + +def match_riccati(eq, f, x): + """ + A function that matches and returns the coefficients + if an equation is a Riccati ODE + + Parameters + ========== + + eq: Equation to be matched + f: Dependent variable + x: Independent variable + + Returns + ======= + + match: True if equation is a Riccati ODE, False otherwise + funcs: [b0, b1, b2] if match is True, [] otherwise. Here, + b0, b1 and b2 are rational functions which match the equation. + """ + # Group terms based on f(x) + if isinstance(eq, Eq): + eq = eq.lhs - eq.rhs + eq = eq.expand().collect(f(x)) + cf = eq.coeff(f(x).diff(x)) + + # There must be an f(x).diff(x) term. + # eq must be an Add object since we are using the expanded + # equation and it must have atleast 2 terms (b2 != 0) + if cf != 0 and isinstance(eq, Add): + + # Divide all coefficients by the coefficient of f(x).diff(x) + # and add the terms again to get the same equation + eq = Add(*((x/cf).cancel() for x in eq.args)).collect(f(x)) + + # Match the equation with the pattern + b1 = -eq.coeff(f(x)) + b2 = -eq.coeff(f(x)**2) + b0 = (f(x).diff(x) - b1*f(x) - b2*f(x)**2 - eq).expand() + funcs = [b0, b1, b2] + + # Check if coefficients are not symbols and floats + if any(len(x.atoms(Symbol)) > 1 or len(x.atoms(Float)) for x in funcs): + return False, [] + + # If b_0(x) contains f(x), it is not a Riccati ODE + if len(b0.atoms(f)) or not all((b2 != 0, b0.is_rational_function(x), + b1.is_rational_function(x), b2.is_rational_function(x))): + return False, [] + return True, funcs + return False, [] + + +def val_at_inf(num, den, x): + # Valuation of a rational function at oo = deg(denom) - deg(numer) + return den.degree(x) - num.degree(x) + + +def check_necessary_conds(val_inf, muls): + """ + The necessary conditions for a rational solution + to exist are as follows - + + i) Every pole of a(x) must be either a simple pole + or a multiple pole of even order. + + ii) The valuation of a(x) at infinity must be even + or be greater than or equal to 2. + + Here, a simple pole is a pole with multiplicity 1 + and a multiple pole is a pole with multiplicity + greater than 1. + """ + return (val_inf >= 2 or (val_inf <= 0 and val_inf%2 == 0)) and \ + all(mul == 1 or (mul%2 == 0 and mul >= 2) for mul in muls) + + +def inverse_transform_poly(num, den, x): + """ + A function to make the substitution + x -> 1/x in a rational function that + is represented using Poly objects for + numerator and denominator. + """ + # Declare for reuse + one = Poly(1, x) + xpoly = Poly(x, x) + + # Check if degree of numerator is same as denominator + pwr = val_at_inf(num, den, x) + if pwr >= 0: + # Denominator has greater degree. Substituting x with + # 1/x would make the extra power go to the numerator + if num.expr != 0: + num = num.transform(one, xpoly) * x**pwr + den = den.transform(one, xpoly) + else: + # Numerator has greater degree. Substituting x with + # 1/x would make the extra power go to the denominator + num = num.transform(one, xpoly) + den = den.transform(one, xpoly) * x**(-pwr) + return num.cancel(den, include=True) + + +def limit_at_inf(num, den, x): + """ + Find the limit of a rational function + at oo + """ + # pwr = degree(num) - degree(den) + pwr = -val_at_inf(num, den, x) + # Numerator has a greater degree than denominator + # Limit at infinity would depend on the sign of the + # leading coefficients of numerator and denominator + if pwr > 0: + return oo*sign(num.LC()/den.LC()) + # Degree of numerator is equal to that of denominator + # Limit at infinity is just the ratio of leading coeffs + elif pwr == 0: + return num.LC()/den.LC() + # Degree of numerator is less than that of denominator + # Limit at infinity is just 0 + else: + return 0 + + +def construct_c_case_1(num, den, x, pole): + # Find the coefficient of 1/(x - pole)**2 in the + # Laurent series expansion of a(x) about pole. + num1, den1 = (num*Poly((x - pole)**2, x, extension=True)).cancel(den, include=True) + r = (num1.subs(x, pole))/(den1.subs(x, pole)) + + # If multiplicity is 2, the coefficient to be added + # in the c-vector is c = (1 +- sqrt(1 + 4*r))/2 + if r != -S(1)/4: + return [[(1 + sqrt(1 + 4*r))/2], [(1 - sqrt(1 + 4*r))/2]] + return [[S.Half]] + + +def construct_c_case_2(num, den, x, pole, mul): + # Generate the coefficients using the recurrence + # relation mentioned in (5.14) in the thesis (Pg 80) + + # r_i = mul/2 + ri = mul//2 + + # Find the Laurent series coefficients about the pole + ser = rational_laurent_series(num, den, x, pole, mul, 6) + + # Start with an empty memo to store the coefficients + # This is for the plus case + cplus = [0 for i in range(ri)] + + # Base Case + cplus[ri-1] = sqrt(ser[2*ri]) + + # Iterate backwards to find all coefficients + s = ri - 1 + sm = 0 + for s in range(ri-1, 0, -1): + sm = 0 + for j in range(s+1, ri): + sm += cplus[j-1]*cplus[ri+s-j-1] + if s!= 1: + cplus[s-1] = (ser[ri+s] - sm)/(2*cplus[ri-1]) + + # Memo for the minus case + cminus = [-x for x in cplus] + + # Find the 0th coefficient in the recurrence + cplus[0] = (ser[ri+s] - sm - ri*cplus[ri-1])/(2*cplus[ri-1]) + cminus[0] = (ser[ri+s] - sm - ri*cminus[ri-1])/(2*cminus[ri-1]) + + # Add both the plus and minus cases' coefficients + if cplus != cminus: + return [cplus, cminus] + return cplus + + +def construct_c_case_3(): + # If multiplicity is 1, the coefficient to be added + # in the c-vector is 1 (no choice) + return [[1]] + + +def construct_c(num, den, x, poles, muls): + """ + Helper function to calculate the coefficients + in the c-vector for each pole. + """ + c = [] + for pole, mul in zip(poles, muls): + c.append([]) + + # Case 3 + if mul == 1: + # Add the coefficients from Case 3 + c[-1].extend(construct_c_case_3()) + + # Case 1 + elif mul == 2: + # Add the coefficients from Case 1 + c[-1].extend(construct_c_case_1(num, den, x, pole)) + + # Case 2 + else: + # Add the coefficients from Case 2 + c[-1].extend(construct_c_case_2(num, den, x, pole, mul)) + + return c + + +def construct_d_case_4(ser, N): + # Initialize an empty vector + dplus = [0 for i in range(N+2)] + # d_N = sqrt(a_{2*N}) + dplus[N] = sqrt(ser[2*N]) + + # Use the recurrence relations to find + # the value of d_s + for s in range(N-1, -2, -1): + sm = 0 + for j in range(s+1, N): + sm += dplus[j]*dplus[N+s-j] + if s != -1: + dplus[s] = (ser[N+s] - sm)/(2*dplus[N]) + + # Coefficients for the case of d_N = -sqrt(a_{2*N}) + dminus = [-x for x in dplus] + + # The third equation in Eq 5.15 of the thesis is WRONG! + # d_N must be replaced with N*d_N in that equation. + dplus[-1] = (ser[N+s] - N*dplus[N] - sm)/(2*dplus[N]) + dminus[-1] = (ser[N+s] - N*dminus[N] - sm)/(2*dminus[N]) + + if dplus != dminus: + return [dplus, dminus] + return dplus + + +def construct_d_case_5(ser): + # List to store coefficients for plus case + dplus = [0, 0] + + # d_0 = sqrt(a_0) + dplus[0] = sqrt(ser[0]) + + # d_(-1) = a_(-1)/(2*d_0) + dplus[-1] = ser[-1]/(2*dplus[0]) + + # Coefficients for the minus case are just the negative + # of the coefficients for the positive case. + dminus = [-x for x in dplus] + + if dplus != dminus: + return [dplus, dminus] + return dplus + + +def construct_d_case_6(num, den, x): + # s_oo = lim x->0 1/x**2 * a(1/x) which is equivalent to + # s_oo = lim x->oo x**2 * a(x) + s_inf = limit_at_inf(Poly(x**2, x)*num, den, x) + + # d_(-1) = (1 +- sqrt(1 + 4*s_oo))/2 + if s_inf != -S(1)/4: + return [[(1 + sqrt(1 + 4*s_inf))/2], [(1 - sqrt(1 + 4*s_inf))/2]] + return [[S.Half]] + + +def construct_d(num, den, x, val_inf): + """ + Helper function to calculate the coefficients + in the d-vector based on the valuation of the + function at oo. + """ + N = -val_inf//2 + # Multiplicity of oo as a pole + mul = -val_inf if val_inf < 0 else 0 + ser = rational_laurent_series(num, den, x, oo, mul, 1) + + # Case 4 + if val_inf < 0: + d = construct_d_case_4(ser, N) + + # Case 5 + elif val_inf == 0: + d = construct_d_case_5(ser) + + # Case 6 + else: + d = construct_d_case_6(num, den, x) + + return d + + +def rational_laurent_series(num, den, x, r, m, n): + r""" + The function computes the Laurent series coefficients + of a rational function. + + Parameters + ========== + + num: A Poly object that is the numerator of `f(x)`. + den: A Poly object that is the denominator of `f(x)`. + x: The variable of expansion of the series. + r: The point of expansion of the series. + m: Multiplicity of r if r is a pole of `f(x)`. Should + be zero otherwise. + n: Order of the term upto which the series is expanded. + + Returns + ======= + + series: A dictionary that has power of the term as key + and coefficient of that term as value. + + Below is a basic outline of how the Laurent series of a + rational function `f(x)` about `x_0` is being calculated - + + 1. Substitute `x + x_0` in place of `x`. If `x_0` + is a pole of `f(x)`, multiply the expression by `x^m` + where `m` is the multiplicity of `x_0`. Denote the + the resulting expression as g(x). We do this substitution + so that we can now find the Laurent series of g(x) about + `x = 0`. + + 2. We can then assume that the Laurent series of `g(x)` + takes the following form - + + .. math:: g(x) = \frac{num(x)}{den(x)} = \sum_{m = 0}^{\infty} a_m x^m + + where `a_m` denotes the Laurent series coefficients. + + 3. Multiply the denominator to the RHS of the equation + and form a recurrence relation for the coefficients `a_m`. + """ + one = Poly(1, x, extension=True) + + if r == oo: + # Series at x = oo is equal to first transforming + # the function from x -> 1/x and finding the + # series at x = 0 + num, den = inverse_transform_poly(num, den, x) + r = S(0) + + if r: + # For an expansion about a non-zero point, a + # transformation from x -> x + r must be made + num = num.transform(Poly(x + r, x, extension=True), one) + den = den.transform(Poly(x + r, x, extension=True), one) + + # Remove the pole from the denominator if the series + # expansion is about one of the poles + num, den = (num*x**m).cancel(den, include=True) + + # Equate coefficients for the first terms (base case) + maxdegree = 1 + max(num.degree(), den.degree()) + syms = symbols(f'a:{maxdegree}', cls=Dummy) + diff = num - den * Poly(syms[::-1], x) + coeff_diffs = diff.all_coeffs()[::-1][:maxdegree] + (coeffs, ) = linsolve(coeff_diffs, syms) + + # Use the recursion relation for the rest + recursion = den.all_coeffs()[::-1] + div, rec_rhs = recursion[0], recursion[1:] + series = list(coeffs) + while len(series) < n: + next_coeff = Add(*(c*series[-1-n] for n, c in enumerate(rec_rhs))) / div + series.append(-next_coeff) + series = {m - i: val for i, val in enumerate(series)} + return series + +def compute_m_ybar(x, poles, choice, N): + """ + Helper function to calculate - + + 1. m - The degree bound for the polynomial + solution that must be found for the auxiliary + differential equation. + + 2. ybar - Part of the solution which can be + computed using the poles, c and d vectors. + """ + ybar = 0 + m = Poly(choice[-1][-1], x, extension=True) + + # Calculate the first (nested) summation for ybar + # as given in Step 9 of the Thesis (Pg 82) + dybar = [] + for i, polei in enumerate(poles): + for j, cij in enumerate(choice[i]): + dybar.append(cij/(x - polei)**(j + 1)) + m -=Poly(choice[i][0], x, extension=True) # can't accumulate Poly and use with Add + ybar += Add(*dybar) + + # Calculate the second summation for ybar + for i in range(N+1): + ybar += choice[-1][i]*x**i + return (m.expr, ybar) + + +def solve_aux_eq(numa, dena, numy, deny, x, m): + """ + Helper function to find a polynomial solution + of degree m for the auxiliary differential + equation. + """ + # Assume that the solution is of the type + # p(x) = C_0 + C_1*x + ... + C_{m-1}*x**(m-1) + x**m + psyms = symbols(f'C0:{m}', cls=Dummy) + K = ZZ[psyms] + psol = Poly(K.gens, x, domain=K) + Poly(x**m, x, domain=K) + + # Eq (5.16) in Thesis - Pg 81 + auxeq = (dena*(numy.diff(x)*deny - numy*deny.diff(x) + numy**2) - numa*deny**2)*psol + if m >= 1: + px = psol.diff(x) + auxeq += px*(2*numy*deny*dena) + if m >= 2: + auxeq += px.diff(x)*(deny**2*dena) + if m != 0: + # m is a non-zero integer. Find the constant terms using undetermined coefficients + return psol, linsolve_dict(auxeq.all_coeffs(), psyms), True + else: + # m == 0 . Check if 1 (x**0) is a solution to the auxiliary equation + return S.One, auxeq, auxeq == 0 + + +def remove_redundant_sols(sol1, sol2, x): + """ + Helper function to remove redundant + solutions to the differential equation. + """ + # If y1 and y2 are redundant solutions, there is + # some value of the arbitrary constant for which + # they will be equal + + syms1 = sol1.atoms(Symbol, Dummy) + syms2 = sol2.atoms(Symbol, Dummy) + num1, den1 = [Poly(e, x, extension=True) for e in sol1.together().as_numer_denom()] + num2, den2 = [Poly(e, x, extension=True) for e in sol2.together().as_numer_denom()] + # Cross multiply + e = num1*den2 - den1*num2 + # Check if there are any constants + syms = list(e.atoms(Symbol, Dummy)) + if len(syms): + # Find values of constants for which solutions are equal + redn = linsolve(e.all_coeffs(), syms) + if len(redn): + # Return the general solution over a particular solution + if len(syms1) > len(syms2): + return sol2 + # If both have constants, return the lesser complex solution + elif len(syms1) == len(syms2): + return sol1 if count_ops(syms1) >= count_ops(syms2) else sol2 + else: + return sol1 + + +def get_gen_sol_from_part_sol(part_sols, a, x): + """" + Helper function which computes the general + solution for a Riccati ODE from its particular + solutions. + + There are 3 cases to find the general solution + from the particular solutions for a Riccati ODE + depending on the number of particular solution(s) + we have - 1, 2 or 3. + + For more information, see Section 6 of + "Methods of Solution of the Riccati Differential Equation" + by D. R. Haaheim and F. M. Stein + """ + + # If no particular solutions are found, a general + # solution cannot be found + if len(part_sols) == 0: + return [] + + # In case of a single particular solution, the general + # solution can be found by using the substitution + # y = y1 + 1/z and solving a Bernoulli ODE to find z. + elif len(part_sols) == 1: + y1 = part_sols[0] + i = exp(Integral(2*y1, x)) + z = i * Integral(a/i, x) + z = z.doit() + if a == 0 or z == 0: + return y1 + return y1 + 1/z + + # In case of 2 particular solutions, the general solution + # can be found by solving a separable equation. This is + # the most common case, i.e. most Riccati ODEs have 2 + # rational particular solutions. + elif len(part_sols) == 2: + y1, y2 = part_sols + # One of them already has a constant + if len(y1.atoms(Dummy)) + len(y2.atoms(Dummy)) > 0: + u = exp(Integral(y2 - y1, x)).doit() + # Introduce a constant + else: + C1 = Dummy('C1') + u = C1*exp(Integral(y2 - y1, x)).doit() + if u == 1: + return y2 + return (y2*u - y1)/(u - 1) + + # In case of 3 particular solutions, a closed form + # of the general solution can be obtained directly + else: + y1, y2, y3 = part_sols[:3] + C1 = Dummy('C1') + return (C1 + 1)*y2*(y1 - y3)/(C1*y1 + y2 - (C1 + 1)*y3) + + +def solve_riccati(fx, x, b0, b1, b2, gensol=False): + """ + The main function that gives particular/general + solutions to Riccati ODEs that have atleast 1 + rational particular solution. + """ + # Step 1 : Convert to Normal Form + a = -b0*b2 + b1**2/4 - b1.diff(x)/2 + 3*b2.diff(x)**2/(4*b2**2) + b1*b2.diff(x)/(2*b2) - \ + b2.diff(x, 2)/(2*b2) + a_t = a.together() + num, den = [Poly(e, x, extension=True) for e in a_t.as_numer_denom()] + num, den = num.cancel(den, include=True) + + # Step 2 + presol = [] + + # Step 3 : a(x) is 0 + if num == 0: + presol.append(1/(x + Dummy('C1'))) + + # Step 4 : a(x) is a non-zero constant + elif x not in num.free_symbols.union(den.free_symbols): + presol.extend([sqrt(a), -sqrt(a)]) + + # Step 5 : Find poles and valuation at infinity + poles = roots(den, x) + poles, muls = list(poles.keys()), list(poles.values()) + val_inf = val_at_inf(num, den, x) + + if len(poles): + # Check necessary conditions (outlined in the module docstring) + if not check_necessary_conds(val_inf, muls): + raise ValueError("Rational Solution doesn't exist") + + # Step 6 + # Construct c-vectors for each singular point + c = construct_c(num, den, x, poles, muls) + + # Construct d vectors for each singular point + d = construct_d(num, den, x, val_inf) + + # Step 7 : Iterate over all possible combinations and return solutions + # For each possible combination, generate an array of 0's and 1's + # where 0 means pick 1st choice and 1 means pick the second choice. + + # NOTE: We could exit from the loop if we find 3 particular solutions, + # but it is not implemented here as - + # a. Finding 3 particular solutions is very rare. Most of the time, + # only 2 particular solutions are found. + # b. In case we exit after finding 3 particular solutions, it might + # happen that 1 or 2 of them are redundant solutions. So, instead of + # spending some more time in computing the particular solutions, + # we will end up computing the general solution from a single + # particular solution which is usually slower than computing the + # general solution from 2 or 3 particular solutions. + c.append(d) + choices = product(*c) + for choice in choices: + m, ybar = compute_m_ybar(x, poles, choice, -val_inf//2) + numy, deny = [Poly(e, x, extension=True) for e in ybar.together().as_numer_denom()] + # Step 10 : Check if a valid solution exists. If yes, also check + # if m is a non-negative integer + if m.is_nonnegative == True and m.is_integer == True: + + # Step 11 : Find polynomial solutions of degree m for the auxiliary equation + psol, coeffs, exists = solve_aux_eq(num, den, numy, deny, x, m) + + # Step 12 : If valid polynomial solution exists, append solution. + if exists: + # m == 0 case + if psol == 1 and coeffs == 0: + # p(x) = 1, so p'(x)/p(x) term need not be added + presol.append(ybar) + # m is a positive integer and there are valid coefficients + elif len(coeffs): + # Substitute the valid coefficients to get p(x) + psol = psol.xreplace(coeffs) + # y(x) = ybar(x) + p'(x)/p(x) + presol.append(ybar + psol.diff(x)/psol) + + # Remove redundant solutions from the list of existing solutions + remove = set() + for i in range(len(presol)): + for j in range(i+1, len(presol)): + rem = remove_redundant_sols(presol[i], presol[j], x) + if rem is not None: + remove.add(rem) + sols = [x for x in presol if x not in remove] + + # Step 15 : Inverse transform the solutions of the equation in normal form + bp = -b2.diff(x)/(2*b2**2) - b1/(2*b2) + + # If general solution is required, compute it from the particular solutions + if gensol: + sols = [get_gen_sol_from_part_sol(sols, a, x)] + + # Inverse transform the particular solutions + presol = [Eq(fx, riccati_inverse_normal(y, x, b1, b2, bp).cancel(extension=True)) for y in sols] + return presol diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/single.py b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/single.py new file mode 100644 index 0000000000000000000000000000000000000000..7d46931122f11a1592097e6a7117192d39bae10e --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/single.py @@ -0,0 +1,2979 @@ +# +# This is the module for ODE solver classes for single ODEs. +# + +from __future__ import annotations +from typing import ClassVar, Iterator + +from .riccati import match_riccati, solve_riccati +from sympy.core import Add, S, Pow, Rational +from sympy.core.cache import cached_property +from sympy.core.exprtools import factor_terms +from sympy.core.expr import Expr +from sympy.core.function import AppliedUndef, Derivative, diff, Function, expand, Subs, _mexpand +from sympy.core.numbers import zoo +from sympy.core.relational import Equality, Eq +from sympy.core.symbol import Symbol, Dummy, Wild +from sympy.core.mul import Mul +from sympy.functions import exp, tan, log, sqrt, besselj, bessely, cbrt, airyai, airybi +from sympy.integrals import Integral +from sympy.polys import Poly +from sympy.polys.polytools import cancel, factor, degree +from sympy.simplify import collect, simplify, separatevars, logcombine, posify # type: ignore +from sympy.simplify.radsimp import fraction +from sympy.utilities import numbered_symbols +from sympy.solvers.solvers import solve +from sympy.solvers.deutils import ode_order, _preprocess +from sympy.polys.matrices.linsolve import _lin_eq2dict +from sympy.polys.solvers import PolyNonlinearError +from .hypergeometric import equivalence_hypergeometric, match_2nd_2F1_hypergeometric, \ + get_sol_2F1_hypergeometric, match_2nd_hypergeometric +from .nonhomogeneous import _get_euler_characteristic_eq_sols, _get_const_characteristic_eq_sols, \ + _solve_undetermined_coefficients, _solve_variation_of_parameters, _test_term, _undetermined_coefficients_match, \ + _get_simplified_sol +from .lie_group import _ode_lie_group + + +class ODEMatchError(NotImplementedError): + """Raised if a SingleODESolver is asked to solve an ODE it does not match""" + pass + + +class SingleODEProblem: + """Represents an ordinary differential equation (ODE) + + This class is used internally in the by dsolve and related + functions/classes so that properties of an ODE can be computed + efficiently. + + Examples + ======== + + This class is used internally by dsolve. To instantiate an instance + directly first define an ODE problem: + + >>> from sympy import Function, Symbol + >>> x = Symbol('x') + >>> f = Function('f') + >>> eq = f(x).diff(x, 2) + + Now you can create a SingleODEProblem instance and query its properties: + + >>> from sympy.solvers.ode.single import SingleODEProblem + >>> problem = SingleODEProblem(f(x).diff(x), f(x), x) + >>> problem.eq + Derivative(f(x), x) + >>> problem.func + f(x) + >>> problem.sym + x + """ + + # Instance attributes: + eq = None # type: Expr + func = None # type: AppliedUndef + sym = None # type: Symbol + _order = None # type: int + _eq_expanded = None # type: Expr + _eq_preprocessed = None # type: Expr + _eq_high_order_free = None + + def __init__(self, eq, func, sym, prep=True, **kwargs): + assert isinstance(eq, Expr) + assert isinstance(func, AppliedUndef) + assert isinstance(sym, Symbol) + assert isinstance(prep, bool) + self.eq = eq + self.func = func + self.sym = sym + self.prep = prep + self.params = kwargs + + @cached_property + def order(self) -> int: + return ode_order(self.eq, self.func) + + @cached_property + def eq_preprocessed(self) -> Expr: + return self._get_eq_preprocessed() + + @cached_property + def eq_high_order_free(self) -> Expr: + a = Wild('a', exclude=[self.func]) + c1 = Wild('c1', exclude=[self.sym]) + # Precondition to try remove f(x) from highest order derivative + reduced_eq = None + if self.eq.is_Add: + deriv_coef = self.eq.coeff(self.func.diff(self.sym, self.order)) + if deriv_coef not in (1, 0): + r = deriv_coef.match(a*self.func**c1) + if r and r[c1]: + den = self.func**r[c1] + reduced_eq = Add(*[arg/den for arg in self.eq.args]) + if not reduced_eq: + reduced_eq = expand(self.eq) + return reduced_eq + + @cached_property + def eq_expanded(self) -> Expr: + return expand(self.eq_preprocessed) + + def _get_eq_preprocessed(self) -> Expr: + if self.prep: + process_eq, process_func = _preprocess(self.eq, self.func) + if process_func != self.func: + raise ValueError + else: + process_eq = self.eq + return process_eq + + def get_numbered_constants(self, num=1, start=1, prefix='C') -> list[Symbol]: + """ + Returns a list of constants that do not occur + in eq already. + """ + ncs = self.iter_numbered_constants(start, prefix) + Cs = [next(ncs) for i in range(num)] + return Cs + + def iter_numbered_constants(self, start=1, prefix='C') -> Iterator[Symbol]: + """ + Returns an iterator of constants that do not occur + in eq already. + """ + atom_set = self.eq.free_symbols + func_set = self.eq.atoms(Function) + if func_set: + atom_set |= {Symbol(str(f.func)) for f in func_set} + return numbered_symbols(start=start, prefix=prefix, exclude=atom_set) + + @cached_property + def is_autonomous(self): + u = Dummy('u') + x = self.sym + syms = self.eq.subs(self.func, u).free_symbols + return x not in syms + + def get_linear_coefficients(self, eq, func, order): + r""" + Matches a differential equation to the linear form: + + .. math:: a_n(x) y^{(n)} + \cdots + a_1(x)y' + a_0(x) y + B(x) = 0 + + Returns a dict of order:coeff terms, where order is the order of the + derivative on each term, and coeff is the coefficient of that derivative. + The key ``-1`` holds the function `B(x)`. Returns ``None`` if the ODE is + not linear. This function assumes that ``func`` has already been checked + to be good. + + Examples + ======== + + >>> from sympy import Function, cos, sin + >>> from sympy.abc import x + >>> from sympy.solvers.ode.single import SingleODEProblem + >>> f = Function('f') + >>> eq = f(x).diff(x, 3) + 2*f(x).diff(x) + \ + ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) - \ + ... sin(x) + >>> obj = SingleODEProblem(eq, f(x), x) + >>> obj.get_linear_coefficients(eq, f(x), 3) + {-1: x - sin(x), 0: -1, 1: cos(x) + 2, 2: x, 3: 1} + >>> eq = f(x).diff(x, 3) + 2*f(x).diff(x) + \ + ... x*f(x).diff(x, 2) + cos(x)*f(x).diff(x) + x - f(x) - \ + ... sin(f(x)) + >>> obj = SingleODEProblem(eq, f(x), x) + >>> obj.get_linear_coefficients(eq, f(x), 3) == None + True + + """ + f = func.func + x = func.args[0] + symset = {Derivative(f(x), x, i) for i in range(order+1)} + try: + rhs, lhs_terms = _lin_eq2dict(eq, symset) + except PolyNonlinearError: + return None + + if rhs.has(func) or any(c.has(func) for c in lhs_terms.values()): + return None + terms = {i: lhs_terms.get(f(x).diff(x, i), S.Zero) for i in range(order+1)} + terms[-1] = rhs + return terms + + # TODO: Add methods that can be used by many ODE solvers: + # order + # is_linear() + # get_linear_coefficients() + # eq_prepared (the ODE in prepared form) + + +class SingleODESolver: + """ + Base class for Single ODE solvers. + + Subclasses should implement the _matches and _get_general_solution + methods. This class is not intended to be instantiated directly but its + subclasses are as part of dsolve. + + Examples + ======== + + You can use a subclass of SingleODEProblem to solve a particular type of + ODE. We first define a particular ODE problem: + + >>> from sympy import Function, Symbol + >>> x = Symbol('x') + >>> f = Function('f') + >>> eq = f(x).diff(x, 2) + + Now we solve this problem using the NthAlgebraic solver which is a + subclass of SingleODESolver: + + >>> from sympy.solvers.ode.single import NthAlgebraic, SingleODEProblem + >>> problem = SingleODEProblem(eq, f(x), x) + >>> solver = NthAlgebraic(problem) + >>> solver.get_general_solution() + [Eq(f(x), _C*x + _C)] + + The normal way to solve an ODE is to use dsolve (which would use + NthAlgebraic and other solvers internally). When using dsolve a number of + other things are done such as evaluating integrals, simplifying the + solution and renumbering the constants: + + >>> from sympy import dsolve + >>> dsolve(eq, hint='nth_algebraic') + Eq(f(x), C1 + C2*x) + """ + + # Subclasses should store the hint name (the argument to dsolve) in this + # attribute + hint: ClassVar[str] + + # Subclasses should define this to indicate if they support an _Integral + # hint. + has_integral: ClassVar[bool] + + # The ODE to be solved + ode_problem = None # type: SingleODEProblem + + # Cache whether or not the equation has matched the method + _matched: bool | None = None + + # Subclasses should store in this attribute the list of order(s) of ODE + # that subclass can solve or leave it to None if not specific to any order + order: list | None = None + + def __init__(self, ode_problem): + self.ode_problem = ode_problem + + def matches(self) -> bool: + if self.order is not None and self.ode_problem.order not in self.order: + self._matched = False + return self._matched + + if self._matched is None: + self._matched = self._matches() + return self._matched + + def get_general_solution(self, *, simplify: bool = True) -> list[Equality]: + if not self.matches(): + msg = "%s solver cannot solve:\n%s" + raise ODEMatchError(msg % (self.hint, self.ode_problem.eq)) + return self._get_general_solution(simplify_flag=simplify) + + def _matches(self) -> bool: + msg = "Subclasses of SingleODESolver should implement matches." + raise NotImplementedError(msg) + + def _get_general_solution(self, *, simplify_flag: bool = True) -> list[Equality]: + msg = "Subclasses of SingleODESolver should implement get_general_solution." + raise NotImplementedError(msg) + + +class SinglePatternODESolver(SingleODESolver): + '''Superclass for ODE solvers based on pattern matching''' + + def wilds(self): + prob = self.ode_problem + f = prob.func.func + x = prob.sym + order = prob.order + return self._wilds(f, x, order) + + def wilds_match(self): + match = self._wilds_match + return [match.get(w, S.Zero) for w in self.wilds()] + + def _matches(self): + eq = self.ode_problem.eq_expanded + f = self.ode_problem.func.func + x = self.ode_problem.sym + order = self.ode_problem.order + df = f(x).diff(x, order) + + if order not in [1, 2]: + return False + + pattern = self._equation(f(x), x, order) + + if not pattern.coeff(df).has(Wild): + eq = expand(eq / eq.coeff(df)) + eq = eq.collect([f(x).diff(x), f(x)], func = cancel) + + self._wilds_match = match = eq.match(pattern) + if match is not None: + return self._verify(f(x)) + return False + + def _verify(self, fx) -> bool: + return True + + def _wilds(self, f, x, order): + msg = "Subclasses of SingleODESolver should implement _wilds" + raise NotImplementedError(msg) + + def _equation(self, fx, x, order): + msg = "Subclasses of SingleODESolver should implement _equation" + raise NotImplementedError(msg) + + +class NthAlgebraic(SingleODESolver): + r""" + Solves an `n`\th order ordinary differential equation using algebra and + integrals. + + There is no general form for the kind of equation that this can solve. The + the equation is solved algebraically treating differentiation as an + invertible algebraic function. + + Examples + ======== + + >>> from sympy import Function, dsolve, Eq + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = Eq(f(x) * (f(x).diff(x)**2 - 1), 0) + >>> dsolve(eq, f(x), hint='nth_algebraic') + [Eq(f(x), 0), Eq(f(x), C1 - x), Eq(f(x), C1 + x)] + + Note that this solver can return algebraic solutions that do not have any + integration constants (f(x) = 0 in the above example). + """ + + hint = 'nth_algebraic' + has_integral = True # nth_algebraic_Integral hint + + def _matches(self): + r""" + Matches any differential equation that nth_algebraic can solve. Uses + `sympy.solve` but teaches it how to integrate derivatives. + + This involves calling `sympy.solve` and does most of the work of finding a + solution (apart from evaluating the integrals). + """ + eq = self.ode_problem.eq + func = self.ode_problem.func + var = self.ode_problem.sym + + # Derivative that solve can handle: + diffx = self._get_diffx(var) + + # Replace derivatives wrt the independent variable with diffx + def replace(eq, var): + def expand_diffx(*args): + differand, diffs = args[0], args[1:] + toreplace = differand + for v, n in diffs: + for _ in range(n): + if v == var: + toreplace = diffx(toreplace) + else: + toreplace = Derivative(toreplace, v) + return toreplace + return eq.replace(Derivative, expand_diffx) + + # Restore derivatives in solution afterwards + def unreplace(eq, var): + return eq.replace(diffx, lambda e: Derivative(e, var)) + + subs_eqn = replace(eq, var) + try: + # turn off simplification to protect Integrals that have + # _t instead of fx in them and would otherwise factor + # as t_*Integral(1, x) + solns = solve(subs_eqn, func, simplify=False) + except NotImplementedError: + solns = [] + + solns = [simplify(unreplace(soln, var)) for soln in solns] + solns = [Equality(func, soln) for soln in solns] + + self.solutions = solns + return len(solns) != 0 + + def _get_general_solution(self, *, simplify_flag: bool = True): + return self.solutions + + # This needs to produce an invertible function but the inverse depends + # which variable we are integrating with respect to. Since the class can + # be stored in cached results we need to ensure that we always get the + # same class back for each particular integration variable so we store these + # classes in a global dict: + _diffx_stored: dict[Symbol, type[Function]] = {} + + @staticmethod + def _get_diffx(var): + diffcls = NthAlgebraic._diffx_stored.get(var, None) + + if diffcls is None: + # A class that behaves like Derivative wrt var but is "invertible". + class diffx(Function): + def inverse(self): + # don't use integrate here because fx has been replaced by _t + # in the equation; integrals will not be correct while solve + # is at work. + return lambda expr: Integral(expr, var) + Dummy('C') + + diffcls = NthAlgebraic._diffx_stored.setdefault(var, diffx) + + return diffcls + + +class FirstExact(SinglePatternODESolver): + r""" + Solves 1st order exact ordinary differential equations. + + A 1st order differential equation is called exact if it is the total + differential of a function. That is, the differential equation + + .. math:: P(x, y) \,\partial{}x + Q(x, y) \,\partial{}y = 0 + + is exact if there is some function `F(x, y)` such that `P(x, y) = + \partial{}F/\partial{}x` and `Q(x, y) = \partial{}F/\partial{}y`. It can + be shown that a necessary and sufficient condition for a first order ODE + to be exact is that `\partial{}P/\partial{}y = \partial{}Q/\partial{}x`. + Then, the solution will be as given below:: + + >>> from sympy import Function, Eq, Integral, symbols, pprint + >>> x, y, t, x0, y0, C1= symbols('x,y,t,x0,y0,C1') + >>> P, Q, F= map(Function, ['P', 'Q', 'F']) + >>> pprint(Eq(Eq(F(x, y), Integral(P(t, y), (t, x0, x)) + + ... Integral(Q(x0, t), (t, y0, y))), C1)) + x y + / / + | | + F(x, y) = | P(t, y) dt + | Q(x0, t) dt = C1 + | | + / / + x0 y0 + + Where the first partials of `P` and `Q` exist and are continuous in a + simply connected region. + + A note: SymPy currently has no way to represent inert substitution on an + expression, so the hint ``1st_exact_Integral`` will return an integral + with `dy`. This is supposed to represent the function that you are + solving for. + + Examples + ======== + + >>> from sympy import Function, dsolve, cos, sin + >>> from sympy.abc import x + >>> f = Function('f') + >>> dsolve(cos(f(x)) - (x*sin(f(x)) - f(x)**2)*f(x).diff(x), + ... f(x), hint='1st_exact') + Eq(x*cos(f(x)) + f(x)**3/3, C1) + + References + ========== + + - https://en.wikipedia.org/wiki/Exact_differential_equation + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 73 + + # indirect doctest + + """ + hint = "1st_exact" + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + P = Wild('P', exclude=[f(x).diff(x)]) + Q = Wild('Q', exclude=[f(x).diff(x)]) + return P, Q + + def _equation(self, fx, x, order): + P, Q = self.wilds() + return P + Q*fx.diff(x) + + def _verify(self, fx) -> bool: + P, Q = self.wilds() + x = self.ode_problem.sym + y = Dummy('y') + + m, n = self.wilds_match() + + m = m.subs(fx, y) + n = n.subs(fx, y) + numerator = cancel(m.diff(y) - n.diff(x)) + + if numerator.is_zero: + # Is exact + return True + else: + # The following few conditions try to convert a non-exact + # differential equation into an exact one. + # References: + # 1. Differential equations with applications + # and historical notes - George E. Simmons + # 2. https://math.okstate.edu/people/binegar/2233-S99/2233-l12.pdf + + factor_n = cancel(numerator/n) + factor_m = cancel(-numerator/m) + if y not in factor_n.free_symbols: + # If (dP/dy - dQ/dx) / Q = f(x) + # then exp(integral(f(x))*equation becomes exact + factor = factor_n + integration_variable = x + elif x not in factor_m.free_symbols: + # If (dP/dy - dQ/dx) / -P = f(y) + # then exp(integral(f(y))*equation becomes exact + factor = factor_m + integration_variable = y + else: + # Couldn't convert to exact + return False + + factor = exp(Integral(factor, integration_variable)) + m *= factor + n *= factor + self._wilds_match[P] = m.subs(y, fx) + self._wilds_match[Q] = n.subs(y, fx) + return True + + def _get_general_solution(self, *, simplify_flag: bool = True): + m, n = self.wilds_match() + fx = self.ode_problem.func + x = self.ode_problem.sym + (C1,) = self.ode_problem.get_numbered_constants(num=1) + y = Dummy('y') + + m = m.subs(fx, y) + n = n.subs(fx, y) + + gen_sol = Eq(Subs(Integral(m, x) + + Integral(n - Integral(m, x).diff(y), y), y, fx), C1) + return [gen_sol] + + +class FirstLinear(SinglePatternODESolver): + r""" + Solves 1st order linear differential equations. + + These are differential equations of the form + + .. math:: dy/dx + P(x) y = Q(x)\text{.} + + These kinds of differential equations can be solved in a general way. The + integrating factor `e^{\int P(x) \,dx}` will turn the equation into a + separable equation. The general solution is:: + + >>> from sympy import Function, dsolve, Eq, pprint, diff, sin + >>> from sympy.abc import x + >>> f, P, Q = map(Function, ['f', 'P', 'Q']) + >>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x)) + >>> pprint(genform) + d + P(x)*f(x) + --(f(x)) = Q(x) + dx + >>> pprint(dsolve(genform, f(x), hint='1st_linear_Integral')) + / / \ + | | | + | | / | / + | | | | | + | | | P(x) dx | - | P(x) dx + | | | | | + | | / | / + f(x) = |C1 + | Q(x)*e dx|*e + | | | + \ / / + + + Examples + ======== + + >>> f = Function('f') + >>> pprint(dsolve(Eq(x*diff(f(x), x) - f(x), x**2*sin(x)), + ... f(x), '1st_linear')) + f(x) = x*(C1 - cos(x)) + + References + ========== + + - https://en.wikipedia.org/wiki/Linear_differential_equation#First-order_equation_with_variable_coefficients + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 92 + + # indirect doctest + + """ + hint = '1st_linear' + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + P = Wild('P', exclude=[f(x)]) + Q = Wild('Q', exclude=[f(x), f(x).diff(x)]) + return P, Q + + def _equation(self, fx, x, order): + P, Q = self.wilds() + return fx.diff(x) + P*fx - Q + + def _get_general_solution(self, *, simplify_flag: bool = True): + P, Q = self.wilds_match() + fx = self.ode_problem.func + x = self.ode_problem.sym + (C1,) = self.ode_problem.get_numbered_constants(num=1) + gensol = Eq(fx, ((C1 + Integral(Q*exp(Integral(P, x)), x)) + * exp(-Integral(P, x)))) + return [gensol] + + +class AlmostLinear(SinglePatternODESolver): + r""" + Solves an almost-linear differential equation. + + The general form of an almost linear differential equation is + + .. math:: a(x) g'(f(x)) f'(x) + b(x) g(f(x)) + c(x) + + Here `f(x)` is the function to be solved for (the dependent variable). + The substitution `g(f(x)) = u(x)` leads to a linear differential equation + for `u(x)` of the form `a(x) u' + b(x) u + c(x) = 0`. This can be solved + for `u(x)` by the `first_linear` hint and then `f(x)` is found by solving + `g(f(x)) = u(x)`. + + See Also + ======== + :obj:`sympy.solvers.ode.single.FirstLinear` + + Examples + ======== + + >>> from sympy import dsolve, Function, pprint, sin, cos + >>> from sympy.abc import x + >>> f = Function('f') + >>> d = f(x).diff(x) + >>> eq = x*d + x*f(x) + 1 + >>> dsolve(eq, f(x), hint='almost_linear') + Eq(f(x), (C1 - Ei(x))*exp(-x)) + >>> pprint(dsolve(eq, f(x), hint='almost_linear')) + -x + f(x) = (C1 - Ei(x))*e + >>> example = cos(f(x))*f(x).diff(x) + sin(f(x)) + 1 + >>> pprint(example) + d + sin(f(x)) + cos(f(x))*--(f(x)) + 1 + dx + >>> pprint(dsolve(example, f(x), hint='almost_linear')) + / -x \ / -x \ + [f(x) = pi - asin\C1*e - 1/, f(x) = asin\C1*e - 1/] + + + References + ========== + + - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications + of the ACM, Volume 14, Number 8, August 1971, pp. 558 + """ + hint = "almost_linear" + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + P = Wild('P', exclude=[f(x).diff(x)]) + Q = Wild('Q', exclude=[f(x).diff(x)]) + return P, Q + + def _equation(self, fx, x, order): + P, Q = self.wilds() + return P*fx.diff(x) + Q + + def _verify(self, fx): + a, b = self.wilds_match() + c, b = b.as_independent(fx) if b.is_Add else (S.Zero, b) + # a, b and c are the function a(x), b(x) and c(x) respectively. + # c(x) is obtained by separating out b as terms with and without fx i.e, l(y) + # The following conditions checks if the given equation is an almost-linear differential equation using the fact that + # a(x)*(l(y))' / l(y)' is independent of l(y) + + if b.diff(fx) != 0 and not simplify(b.diff(fx)/a).has(fx): + self.ly = factor_terms(b).as_independent(fx, as_Add=False)[1] # Gives the term containing fx i.e., l(y) + self.ax = a / self.ly.diff(fx) + self.cx = -c # cx is taken as -c(x) to simplify expression in the solution integral + self.bx = factor_terms(b) / self.ly + return True + + return False + + def _get_general_solution(self, *, simplify_flag: bool = True): + x = self.ode_problem.sym + (C1,) = self.ode_problem.get_numbered_constants(num=1) + gensol = Eq(self.ly, ((C1 + Integral((self.cx/self.ax)*exp(Integral(self.bx/self.ax, x)), x)) + * exp(-Integral(self.bx/self.ax, x)))) + + return [gensol] + + +class Bernoulli(SinglePatternODESolver): + r""" + Solves Bernoulli differential equations. + + These are equations of the form + + .. math:: dy/dx + P(x) y = Q(x) y^n\text{, }n \ne 1`\text{.} + + The substitution `w = 1/y^{1-n}` will transform an equation of this form + into one that is linear (see the docstring of + :obj:`~sympy.solvers.ode.single.FirstLinear`). The general solution is:: + + >>> from sympy import Function, dsolve, Eq, pprint + >>> from sympy.abc import x, n + >>> f, P, Q = map(Function, ['f', 'P', 'Q']) + >>> genform = Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)**n) + >>> pprint(genform) + d n + P(x)*f(x) + --(f(x)) = Q(x)*f (x) + dx + >>> pprint(dsolve(genform, f(x), hint='Bernoulli_Integral'), num_columns=110) + -1 + ----- + n - 1 + // / / \ \ + || | | | | + || | / | / | / | + || | | | | | | | + || | -(n - 1)* | P(x) dx | -(n - 1)* | P(x) dx | (n - 1)* | P(x) dx| + || | | | | | | | + || | / | / | / | + f(x) = ||C1 - n* | Q(x)*e dx + | Q(x)*e dx|*e | + || | | | | + \\ / / / / + + + Note that the equation is separable when `n = 1` (see the docstring of + :obj:`~sympy.solvers.ode.single.Separable`). + + >>> pprint(dsolve(Eq(f(x).diff(x) + P(x)*f(x), Q(x)*f(x)), f(x), + ... hint='separable_Integral')) + f(x) + / + | / + | 1 | + | - dy = C1 + | (-P(x) + Q(x)) dx + | y | + | / + / + + + Examples + ======== + + >>> from sympy import Function, dsolve, Eq, pprint, log + >>> from sympy.abc import x + >>> f = Function('f') + + >>> pprint(dsolve(Eq(x*f(x).diff(x) + f(x), log(x)*f(x)**2), + ... f(x), hint='Bernoulli')) + 1 + f(x) = ----------------- + C1*x + log(x) + 1 + + References + ========== + + - https://en.wikipedia.org/wiki/Bernoulli_differential_equation + + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 95 + + # indirect doctest + + """ + hint = "Bernoulli" + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + P = Wild('P', exclude=[f(x)]) + Q = Wild('Q', exclude=[f(x)]) + n = Wild('n', exclude=[x, f(x), f(x).diff(x)]) + return P, Q, n + + def _equation(self, fx, x, order): + P, Q, n = self.wilds() + return fx.diff(x) + P*fx - Q*fx**n + + def _get_general_solution(self, *, simplify_flag: bool = True): + P, Q, n = self.wilds_match() + fx = self.ode_problem.func + x = self.ode_problem.sym + (C1,) = self.ode_problem.get_numbered_constants(num=1) + if n==1: + gensol = Eq(log(fx), ( + C1 + Integral((-P + Q), x) + )) + else: + gensol = Eq(fx**(1-n), ( + (C1 - (n - 1) * Integral(Q*exp(-n*Integral(P, x)) + * exp(Integral(P, x)), x) + ) * exp(-(1 - n)*Integral(P, x))) + ) + return [gensol] + + +class Factorable(SingleODESolver): + r""" + Solves equations having a solvable factor. + + This function is used to solve the equation having factors. Factors may be of type algebraic or ode. It + will try to solve each factor independently. Factors will be solved by calling dsolve. We will return the + list of solutions. + + Examples + ======== + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = (f(x)**2-4)*(f(x).diff(x)+f(x)) + >>> pprint(dsolve(eq, f(x))) + -x + [f(x) = 2, f(x) = -2, f(x) = C1*e ] + + + """ + hint = "factorable" + has_integral = False + + def _matches(self): + eq_orig = self.ode_problem.eq + f = self.ode_problem.func.func + x = self.ode_problem.sym + df = f(x).diff(x) + self.eqs = [] + eq = eq_orig.collect(f(x), func = cancel) + eq = fraction(factor(eq))[0] + factors = Mul.make_args(factor(eq)) + roots = [fac.as_base_exp() for fac in factors if len(fac.args)!=0] + if len(roots)>1 or roots[0][1]>1: + for base, expo in roots: + if base.has(f(x)): + self.eqs.append(base) + if len(self.eqs)>0: + return True + roots = solve(eq, df) + if len(roots)>0: + self.eqs = [(df - root) for root in roots] + # Avoid infinite recursion + matches = self.eqs != [eq_orig] + return matches + for i in factors: + if i.has(f(x)): + self.eqs.append(i) + return len(self.eqs)>0 and len(factors)>1 + + def _get_general_solution(self, *, simplify_flag: bool = True): + func = self.ode_problem.func.func + x = self.ode_problem.sym + eqns = self.eqs + sols = [] + for eq in eqns: + try: + sol = dsolve(eq, func(x)) + except NotImplementedError: + continue + else: + if isinstance(sol, list): + sols.extend(sol) + else: + sols.append(sol) + + if sols == []: + raise NotImplementedError("The given ODE " + str(eq) + " cannot be solved by" + + " the factorable group method") + return sols + + +class RiccatiSpecial(SinglePatternODESolver): + r""" + The general Riccati equation has the form + + .. math:: dy/dx = f(x) y^2 + g(x) y + h(x)\text{.} + + While it does not have a general solution [1], the "special" form, `dy/dx + = a y^2 - b x^c`, does have solutions in many cases [2]. This routine + returns a solution for `a(dy/dx) = b y^2 + c y/x + d/x^2` that is obtained + by using a suitable change of variables to reduce it to the special form + and is valid when neither `a` nor `b` are zero and either `c` or `d` is + zero. + + >>> from sympy.abc import x, a, b, c, d + >>> from sympy import dsolve, checkodesol, pprint, Function + >>> f = Function('f') + >>> y = f(x) + >>> genform = a*y.diff(x) - (b*y**2 + c*y/x + d/x**2) + >>> sol = dsolve(genform, y, hint="Riccati_special_minus2") + >>> pprint(sol, wrap_line=False) + / / __________________ \\ + | __________________ | / 2 || + | / 2 | \/ 4*b*d - (a + c) *log(x)|| + -|a + c - \/ 4*b*d - (a + c) *tan|C1 + ----------------------------|| + \ \ 2*a // + f(x) = ------------------------------------------------------------------------ + 2*b*x + + >>> checkodesol(genform, sol, order=1)[0] + True + + References + ========== + + - https://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Riccati + - https://eqworld.ipmnet.ru/en/solutions/ode/ode0106.pdf - + https://eqworld.ipmnet.ru/en/solutions/ode/ode0123.pdf + """ + hint = "Riccati_special_minus2" + has_integral = False + order = [1] + + def _wilds(self, f, x, order): + a = Wild('a', exclude=[x, f(x), f(x).diff(x), 0]) + b = Wild('b', exclude=[x, f(x), f(x).diff(x), 0]) + c = Wild('c', exclude=[x, f(x), f(x).diff(x)]) + d = Wild('d', exclude=[x, f(x), f(x).diff(x)]) + return a, b, c, d + + def _equation(self, fx, x, order): + a, b, c, d = self.wilds() + return a*fx.diff(x) + b*fx**2 + c*fx/x + d/x**2 + + def _get_general_solution(self, *, simplify_flag: bool = True): + a, b, c, d = self.wilds_match() + fx = self.ode_problem.func + x = self.ode_problem.sym + (C1,) = self.ode_problem.get_numbered_constants(num=1) + mu = sqrt(4*d*b - (a - c)**2) + + gensol = Eq(fx, (a - c - mu*tan(mu/(2*a)*log(x) + C1))/(2*b*x)) + return [gensol] + + +class RationalRiccati(SinglePatternODESolver): + r""" + Gives general solutions to the first order Riccati differential + equations that have atleast one rational particular solution. + + .. math :: y' = b_0(x) + b_1(x) y + b_2(x) y^2 + + where `b_0`, `b_1` and `b_2` are rational functions of `x` + with `b_2 \ne 0` (`b_2 = 0` would make it a Bernoulli equation). + + Examples + ======== + + >>> from sympy import Symbol, Function, dsolve, checkodesol + >>> f = Function('f') + >>> x = Symbol('x') + + >>> eq = -x**4*f(x)**2 + x**3*f(x).diff(x) + x**2*f(x) + 20 + >>> sol = dsolve(eq, hint="1st_rational_riccati") + >>> sol + Eq(f(x), (4*C1 - 5*x**9 - 4)/(x**2*(C1 + x**9 - 1))) + >>> checkodesol(eq, sol) + (True, 0) + + References + ========== + + - Riccati ODE: https://en.wikipedia.org/wiki/Riccati_equation + - N. Thieu Vo - Rational and Algebraic Solutions of First-Order Algebraic ODEs: + Algorithm 11, pp. 78 - https://www3.risc.jku.at/publications/download/risc_5387/PhDThesisThieu.pdf + """ + has_integral = False + hint = "1st_rational_riccati" + order = [1] + + def _wilds(self, f, x, order): + b0 = Wild('b0', exclude=[f(x), f(x).diff(x)]) + b1 = Wild('b1', exclude=[f(x), f(x).diff(x)]) + b2 = Wild('b2', exclude=[f(x), f(x).diff(x)]) + return (b0, b1, b2) + + def _equation(self, fx, x, order): + b0, b1, b2 = self.wilds() + return fx.diff(x) - b0 - b1*fx - b2*fx**2 + + def _matches(self): + eq = self.ode_problem.eq_expanded + f = self.ode_problem.func.func + x = self.ode_problem.sym + order = self.ode_problem.order + + if order != 1: + return False + + match, funcs = match_riccati(eq, f, x) + if not match: + return False + _b0, _b1, _b2 = funcs + b0, b1, b2 = self.wilds() + self._wilds_match = match = {b0: _b0, b1: _b1, b2: _b2} + return True + + def _get_general_solution(self, *, simplify_flag: bool = True): + # Match the equation + b0, b1, b2 = self.wilds_match() + fx = self.ode_problem.func + x = self.ode_problem.sym + return solve_riccati(fx, x, b0, b1, b2, gensol=True) + + +class SecondNonlinearAutonomousConserved(SinglePatternODESolver): + r""" + Gives solution for the autonomous second order nonlinear + differential equation of the form + + .. math :: f''(x) = g(f(x)) + + The solution for this differential equation can be computed + by multiplying by `f'(x)` and integrating on both sides, + converting it into a first order differential equation. + + Examples + ======== + + >>> from sympy import Function, symbols, dsolve + >>> f, g = symbols('f g', cls=Function) + >>> x = symbols('x') + + >>> eq = f(x).diff(x, 2) - g(f(x)) + >>> dsolve(eq, simplify=False) + [Eq(Integral(1/sqrt(C1 + 2*Integral(g(_u), _u)), (_u, f(x))), C2 + x), + Eq(Integral(1/sqrt(C1 + 2*Integral(g(_u), _u)), (_u, f(x))), C2 - x)] + + >>> from sympy import exp, log + >>> eq = f(x).diff(x, 2) - exp(f(x)) + log(f(x)) + >>> dsolve(eq, simplify=False) + [Eq(Integral(1/sqrt(-2*_u*log(_u) + 2*_u + C1 + 2*exp(_u)), (_u, f(x))), C2 + x), + Eq(Integral(1/sqrt(-2*_u*log(_u) + 2*_u + C1 + 2*exp(_u)), (_u, f(x))), C2 - x)] + + References + ========== + + - https://eqworld.ipmnet.ru/en/solutions/ode/ode0301.pdf + """ + hint = "2nd_nonlinear_autonomous_conserved" + has_integral = True + order = [2] + + def _wilds(self, f, x, order): + fy = Wild('fy', exclude=[0, f(x).diff(x), f(x).diff(x, 2)]) + return (fy, ) + + def _equation(self, fx, x, order): + fy = self.wilds()[0] + return fx.diff(x, 2) + fy + + def _verify(self, fx): + return self.ode_problem.is_autonomous + + def _get_general_solution(self, *, simplify_flag: bool = True): + g = self.wilds_match()[0] + fx = self.ode_problem.func + x = self.ode_problem.sym + u = Dummy('u') + g = g.subs(fx, u) + C1, C2 = self.ode_problem.get_numbered_constants(num=2) + inside = -2*Integral(g, u) + C1 + lhs = Integral(1/sqrt(inside), (u, fx)) + return [Eq(lhs, C2 + x), Eq(lhs, C2 - x)] + + +class Liouville(SinglePatternODESolver): + r""" + Solves 2nd order Liouville differential equations. + + The general form of a Liouville ODE is + + .. math:: \frac{d^2 y}{dx^2} + g(y) \left(\! + \frac{dy}{dx}\!\right)^2 + h(x) + \frac{dy}{dx}\text{.} + + The general solution is: + + >>> from sympy import Function, dsolve, Eq, pprint, diff + >>> from sympy.abc import x + >>> f, g, h = map(Function, ['f', 'g', 'h']) + >>> genform = Eq(diff(f(x),x,x) + g(f(x))*diff(f(x),x)**2 + + ... h(x)*diff(f(x),x), 0) + >>> pprint(genform) + 2 2 + /d \ d d + g(f(x))*|--(f(x))| + h(x)*--(f(x)) + ---(f(x)) = 0 + \dx / dx 2 + dx + >>> pprint(dsolve(genform, f(x), hint='Liouville_Integral')) + f(x) + / / + | | + | / | / + | | | | + | - | h(x) dx | | g(y) dy + | | | | + | / | / + C1 + C2* | e dx + | e dy = 0 + | | + / / + + Examples + ======== + + >>> from sympy import Function, dsolve, Eq, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(diff(f(x), x, x) + diff(f(x), x)**2/f(x) + + ... diff(f(x), x)/x, f(x), hint='Liouville')) + ________________ ________________ + [f(x) = -\/ C1 + C2*log(x) , f(x) = \/ C1 + C2*log(x) ] + + References + ========== + + - Goldstein and Braun, "Advanced Methods for the Solution of Differential + Equations", pp. 98 + - https://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Liouville + + # indirect doctest + + """ + hint = "Liouville" + has_integral = True + order = [2] + + def _wilds(self, f, x, order): + d = Wild('d', exclude=[f(x).diff(x), f(x).diff(x, 2)]) + e = Wild('e', exclude=[f(x).diff(x)]) + k = Wild('k', exclude=[f(x).diff(x)]) + return d, e, k + + def _equation(self, fx, x, order): + # Liouville ODE in the form + # f(x).diff(x, 2) + g(f(x))*(f(x).diff(x))**2 + h(x)*f(x).diff(x) + # See Goldstein and Braun, "Advanced Methods for the Solution of + # Differential Equations", pg. 98 + d, e, k = self.wilds() + return d*fx.diff(x, 2) + e*fx.diff(x)**2 + k*fx.diff(x) + + def _verify(self, fx): + d, e, k = self.wilds_match() + self.y = Dummy('y') + x = self.ode_problem.sym + self.g = simplify(e/d).subs(fx, self.y) + self.h = simplify(k/d).subs(fx, self.y) + if self.y in self.h.free_symbols or x in self.g.free_symbols: + return False + return True + + def _get_general_solution(self, *, simplify_flag: bool = True): + d, e, k = self.wilds_match() + fx = self.ode_problem.func + x = self.ode_problem.sym + C1, C2 = self.ode_problem.get_numbered_constants(num=2) + int = Integral(exp(Integral(self.g, self.y)), (self.y, None, fx)) + gen_sol = Eq(int + C1*Integral(exp(-Integral(self.h, x)), x) + C2, 0) + + return [gen_sol] + + +class Separable(SinglePatternODESolver): + r""" + Solves separable 1st order differential equations. + + This is any differential equation that can be written as `P(y) + \tfrac{dy}{dx} = Q(x)`. The solution can then just be found by + rearranging terms and integrating: `\int P(y) \,dy = \int Q(x) \,dx`. + This hint uses :py:meth:`sympy.simplify.simplify.separatevars` as its back + end, so if a separable equation is not caught by this solver, it is most + likely the fault of that function. + :py:meth:`~sympy.simplify.simplify.separatevars` is + smart enough to do most expansion and factoring necessary to convert a + separable equation `F(x, y)` into the proper form `P(x)\cdot{}Q(y)`. The + general solution is:: + + >>> from sympy import Function, dsolve, Eq, pprint + >>> from sympy.abc import x + >>> a, b, c, d, f = map(Function, ['a', 'b', 'c', 'd', 'f']) + >>> genform = Eq(a(x)*b(f(x))*f(x).diff(x), c(x)*d(f(x))) + >>> pprint(genform) + d + a(x)*b(f(x))*--(f(x)) = c(x)*d(f(x)) + dx + >>> pprint(dsolve(genform, f(x), hint='separable_Integral')) + f(x) + / / + | | + | b(y) | c(x) + | ---- dy = C1 + | ---- dx + | d(y) | a(x) + | | + / / + + Examples + ======== + + >>> from sympy import Function, dsolve, Eq + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(Eq(f(x)*f(x).diff(x) + x, 3*x*f(x)**2), f(x), + ... hint='separable', simplify=False)) + / 2 \ 2 + log\3*f (x) - 1/ x + ---------------- = C1 + -- + 6 2 + + References + ========== + + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 52 + + # indirect doctest + + """ + hint = "separable" + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + d = Wild('d', exclude=[f(x).diff(x), f(x).diff(x, 2)]) + e = Wild('e', exclude=[f(x).diff(x)]) + return d, e + + def _equation(self, fx, x, order): + d, e = self.wilds() + return d + e*fx.diff(x) + + def _verify(self, fx): + d, e = self.wilds_match() + self.y = Dummy('y') + x = self.ode_problem.sym + d = separatevars(d.subs(fx, self.y)) + e = separatevars(e.subs(fx, self.y)) + # m1[coeff]*m1[x]*m1[y] + m2[coeff]*m2[x]*m2[y]*y' + self.m1 = separatevars(d, dict=True, symbols=(x, self.y)) + self.m2 = separatevars(e, dict=True, symbols=(x, self.y)) + if self.m1 and self.m2: + return True + return False + + def _get_match_object(self): + fx = self.ode_problem.func + x = self.ode_problem.sym + return self.m1, self.m2, x, fx + + def _get_general_solution(self, *, simplify_flag: bool = True): + m1, m2, x, fx = self._get_match_object() + (C1,) = self.ode_problem.get_numbered_constants(num=1) + int = Integral(m2['coeff']*m2[self.y]/m1[self.y], + (self.y, None, fx)) + gen_sol = Eq(int, Integral(-m1['coeff']*m1[x]/ + m2[x], x) + C1) + return [gen_sol] + + +class SeparableReduced(Separable): + r""" + Solves a differential equation that can be reduced to the separable form. + + The general form of this equation is + + .. math:: y' + (y/x) H(x^n y) = 0\text{}. + + This can be solved by substituting `u(y) = x^n y`. The equation then + reduces to the separable form `\frac{u'}{u (\mathrm{power} - H(u))} - + \frac{1}{x} = 0`. + + The general solution is: + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x, n + >>> f, g = map(Function, ['f', 'g']) + >>> genform = f(x).diff(x) + (f(x)/x)*g(x**n*f(x)) + >>> pprint(genform) + / n \ + d f(x)*g\x *f(x)/ + --(f(x)) + --------------- + dx x + >>> pprint(dsolve(genform, hint='separable_reduced')) + n + x *f(x) + / + | + | 1 + | ------------ dy = C1 + log(x) + | y*(n - g(y)) + | + / + + See Also + ======== + :obj:`sympy.solvers.ode.single.Separable` + + Examples + ======== + + >>> from sympy import dsolve, Function, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> d = f(x).diff(x) + >>> eq = (x - x**2*f(x))*d - f(x) + >>> dsolve(eq, hint='separable_reduced') + [Eq(f(x), (1 - sqrt(C1*x**2 + 1))/x), Eq(f(x), (sqrt(C1*x**2 + 1) + 1)/x)] + >>> pprint(dsolve(eq, hint='separable_reduced')) + ___________ ___________ + / 2 / 2 + 1 - \/ C1*x + 1 \/ C1*x + 1 + 1 + [f(x) = ------------------, f(x) = ------------------] + x x + + References + ========== + + - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications + of the ACM, Volume 14, Number 8, August 1971, pp. 558 + """ + hint = "separable_reduced" + has_integral = True + order = [1] + + def _degree(self, expr, x): + # Made this function to calculate the degree of + # x in an expression. If expr will be of form + # x**p*y, (wheare p can be variables/rationals) then it + # will return p. + for val in expr: + if val.has(x): + if isinstance(val, Pow) and val.as_base_exp()[0] == x: + return (val.as_base_exp()[1]) + elif val == x: + return (val.as_base_exp()[1]) + else: + return self._degree(val.args, x) + return 0 + + def _powers(self, expr): + # this function will return all the different relative power of x w.r.t f(x). + # expr = x**p * f(x)**q then it will return {p/q}. + pows = set() + fx = self.ode_problem.func + x = self.ode_problem.sym + self.y = Dummy('y') + if isinstance(expr, Add): + exprs = expr.atoms(Add) + elif isinstance(expr, Mul): + exprs = expr.atoms(Mul) + elif isinstance(expr, Pow): + exprs = expr.atoms(Pow) + else: + exprs = {expr} + + for arg in exprs: + if arg.has(x): + _, u = arg.as_independent(x, fx) + pow = self._degree((u.subs(fx, self.y), ), x)/self._degree((u.subs(fx, self.y), ), self.y) + pows.add(pow) + return pows + + def _verify(self, fx): + num, den = self.wilds_match() + x = self.ode_problem.sym + factor = simplify(x/fx*num/den) + # Try representing factor in terms of x^n*y + # where n is lowest power of x in factor; + # first remove terms like sqrt(2)*3 from factor.atoms(Mul) + num, dem = factor.as_numer_denom() + num = expand(num) + dem = expand(dem) + pows = self._powers(num) + pows.update(self._powers(dem)) + pows = list(pows) + if(len(pows)==1) and pows[0]!=zoo: + self.t = Dummy('t') + self.r2 = {'t': self.t} + num = num.subs(x**pows[0]*fx, self.t) + dem = dem.subs(x**pows[0]*fx, self.t) + test = num/dem + free = test.free_symbols + if len(free) == 1 and free.pop() == self.t: + self.r2.update({'power' : pows[0], 'u' : test}) + return True + return False + return False + + def _get_match_object(self): + fx = self.ode_problem.func + x = self.ode_problem.sym + u = self.r2['u'].subs(self.r2['t'], self.y) + ycoeff = 1/(self.y*(self.r2['power'] - u)) + m1 = {self.y: 1, x: -1/x, 'coeff': 1} + m2 = {self.y: ycoeff, x: 1, 'coeff': 1} + return m1, m2, x, x**self.r2['power']*fx + + +class HomogeneousCoeffSubsDepDivIndep(SinglePatternODESolver): + r""" + Solves a 1st order differential equation with homogeneous coefficients + using the substitution `u_1 = \frac{\text{}}{\text{}}`. + + This is a differential equation + + .. math:: P(x, y) + Q(x, y) dy/dx = 0 + + such that `P` and `Q` are homogeneous and of the same order. A function + `F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`. + Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`. See + also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`. + + If the coefficients `P` and `Q` in the differential equation above are + homogeneous functions of the same order, then it can be shown that the + substitution `y = u_1 x` (i.e. `u_1 = y/x`) will turn the differential + equation into an equation separable in the variables `x` and `u`. If + `h(u_1)` is the function that results from making the substitution `u_1 = + f(x)/x` on `P(x, f(x))` and `g(u_2)` is the function that results from the + substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) + + Q(x, f(x)) f'(x) = 0`, then the general solution is:: + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x + >>> f, g, h = map(Function, ['f', 'g', 'h']) + >>> genform = g(f(x)/x) + h(f(x)/x)*f(x).diff(x) + >>> pprint(genform) + /f(x)\ /f(x)\ d + g|----| + h|----|*--(f(x)) + \ x / \ x / dx + >>> pprint(dsolve(genform, f(x), + ... hint='1st_homogeneous_coeff_subs_dep_div_indep_Integral')) + f(x) + ---- + x + / + | + | -h(u1) + log(x) = C1 + | ---------------- d(u1) + | u1*h(u1) + g(u1) + | + / + + Where `u_1 h(u_1) + g(u_1) \ne 0` and `x \ne 0`. + + See also the docstrings of + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffBest` and + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffSubsIndepDivDep`. + + Examples + ======== + + >>> from sympy import Function, dsolve + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), + ... hint='1st_homogeneous_coeff_subs_dep_div_indep', simplify=False)) + / 3 \ + |3*f(x) f (x)| + log|------ + -----| + | x 3 | + \ x / + log(x) = log(C1) - ------------------- + 3 + + References + ========== + + - https://en.wikipedia.org/wiki/Homogeneous_differential_equation + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 59 + + # indirect doctest + + """ + hint = "1st_homogeneous_coeff_subs_dep_div_indep" + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + d = Wild('d', exclude=[f(x).diff(x), f(x).diff(x, 2)]) + e = Wild('e', exclude=[f(x).diff(x)]) + return d, e + + def _equation(self, fx, x, order): + d, e = self.wilds() + return d + e*fx.diff(x) + + def _verify(self, fx): + self.d, self.e = self.wilds_match() + self.y = Dummy('y') + x = self.ode_problem.sym + self.d = separatevars(self.d.subs(fx, self.y)) + self.e = separatevars(self.e.subs(fx, self.y)) + ordera = homogeneous_order(self.d, x, self.y) + orderb = homogeneous_order(self.e, x, self.y) + if ordera == orderb and ordera is not None: + self.u = Dummy('u') + if simplify((self.d + self.u*self.e).subs({x: 1, self.y: self.u})) != 0: + return True + return False + return False + + def _get_match_object(self): + fx = self.ode_problem.func + x = self.ode_problem.sym + self.u1 = Dummy('u1') + xarg = 0 + yarg = 0 + return [self.d, self.e, fx, x, self.u, self.u1, self.y, xarg, yarg] + + def _get_general_solution(self, *, simplify_flag: bool = True): + d, e, fx, x, u, u1, y, xarg, yarg = self._get_match_object() + (C1,) = self.ode_problem.get_numbered_constants(num=1) + int = Integral( + (-e/(d + u1*e)).subs({x: 1, y: u1}), + (u1, None, fx/x)) + sol = logcombine(Eq(log(x), int + log(C1)), force=True) + gen_sol = sol.subs(fx, u).subs(((u, u - yarg), (x, x - xarg), (u, fx))) + return [gen_sol] + + +class HomogeneousCoeffSubsIndepDivDep(SinglePatternODESolver): + r""" + Solves a 1st order differential equation with homogeneous coefficients + using the substitution `u_2 = \frac{\text{}}{\text{}}`. + + This is a differential equation + + .. math:: P(x, y) + Q(x, y) dy/dx = 0 + + such that `P` and `Q` are homogeneous and of the same order. A function + `F(x, y)` is homogeneous of order `n` if `F(x t, y t) = t^n F(x, y)`. + Equivalently, `F(x, y)` can be rewritten as `G(y/x)` or `H(x/y)`. See + also the docstring of :py:meth:`~sympy.solvers.ode.homogeneous_order`. + + If the coefficients `P` and `Q` in the differential equation above are + homogeneous functions of the same order, then it can be shown that the + substitution `x = u_2 y` (i.e. `u_2 = x/y`) will turn the differential + equation into an equation separable in the variables `y` and `u_2`. If + `h(u_2)` is the function that results from making the substitution `u_2 = + x/f(x)` on `P(x, f(x))` and `g(u_2)` is the function that results from the + substitution on `Q(x, f(x))` in the differential equation `P(x, f(x)) + + Q(x, f(x)) f'(x) = 0`, then the general solution is: + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x + >>> f, g, h = map(Function, ['f', 'g', 'h']) + >>> genform = g(x/f(x)) + h(x/f(x))*f(x).diff(x) + >>> pprint(genform) + / x \ / x \ d + g|----| + h|----|*--(f(x)) + \f(x)/ \f(x)/ dx + >>> pprint(dsolve(genform, f(x), + ... hint='1st_homogeneous_coeff_subs_indep_div_dep_Integral')) + x + ---- + f(x) + / + | + | -g(u1) + | ---------------- d(u1) + | u1*g(u1) + h(u1) + | + / + + f(x) = C1*e + + Where `u_1 g(u_1) + h(u_1) \ne 0` and `f(x) \ne 0`. + + See also the docstrings of + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffBest` and + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffSubsDepDivIndep`. + + Examples + ======== + + >>> from sympy import Function, pprint, dsolve + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), + ... hint='1st_homogeneous_coeff_subs_indep_div_dep', + ... simplify=False)) + / 2 \ + |3*x | + log|----- + 1| + | 2 | + \f (x) / + log(f(x)) = log(C1) - -------------- + 3 + + References + ========== + + - https://en.wikipedia.org/wiki/Homogeneous_differential_equation + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 59 + + # indirect doctest + + """ + hint = "1st_homogeneous_coeff_subs_indep_div_dep" + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + d = Wild('d', exclude=[f(x).diff(x), f(x).diff(x, 2)]) + e = Wild('e', exclude=[f(x).diff(x)]) + return d, e + + def _equation(self, fx, x, order): + d, e = self.wilds() + return d + e*fx.diff(x) + + def _verify(self, fx): + self.d, self.e = self.wilds_match() + self.y = Dummy('y') + x = self.ode_problem.sym + self.d = separatevars(self.d.subs(fx, self.y)) + self.e = separatevars(self.e.subs(fx, self.y)) + ordera = homogeneous_order(self.d, x, self.y) + orderb = homogeneous_order(self.e, x, self.y) + if ordera == orderb and ordera is not None: + self.u = Dummy('u') + if simplify((self.e + self.u*self.d).subs({x: self.u, self.y: 1})) != 0: + return True + return False + return False + + def _get_match_object(self): + fx = self.ode_problem.func + x = self.ode_problem.sym + self.u1 = Dummy('u1') + xarg = 0 + yarg = 0 + return [self.d, self.e, fx, x, self.u, self.u1, self.y, xarg, yarg] + + def _get_general_solution(self, *, simplify_flag: bool = True): + d, e, fx, x, u, u1, y, xarg, yarg = self._get_match_object() + (C1,) = self.ode_problem.get_numbered_constants(num=1) + int = Integral(simplify((-d/(e + u1*d)).subs({x: u1, y: 1})), (u1, None, x/fx)) # type: ignore + sol = logcombine(Eq(log(fx), int + log(C1)), force=True) + gen_sol = sol.subs(fx, u).subs(((u, u - yarg), (x, x - xarg), (u, fx))) + return [gen_sol] + + +class HomogeneousCoeffBest(HomogeneousCoeffSubsIndepDivDep, HomogeneousCoeffSubsDepDivIndep): + r""" + Returns the best solution to an ODE from the two hints + ``1st_homogeneous_coeff_subs_dep_div_indep`` and + ``1st_homogeneous_coeff_subs_indep_div_dep``. + + This is as determined by :py:meth:`~sympy.solvers.ode.ode.ode_sol_simplicity`. + + See the + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffSubsIndepDivDep` + and + :obj:`~sympy.solvers.ode.single.HomogeneousCoeffSubsDepDivIndep` + docstrings for more information on these hints. Note that there is no + ``ode_1st_homogeneous_coeff_best_Integral`` hint. + + Examples + ======== + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(2*x*f(x) + (x**2 + f(x)**2)*f(x).diff(x), f(x), + ... hint='1st_homogeneous_coeff_best', simplify=False)) + / 2 \ + |3*x | + log|----- + 1| + | 2 | + \f (x) / + log(f(x)) = log(C1) - -------------- + 3 + + References + ========== + + - https://en.wikipedia.org/wiki/Homogeneous_differential_equation + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 59 + + # indirect doctest + + """ + hint = "1st_homogeneous_coeff_best" + has_integral = False + order = [1] + + def _verify(self, fx): + if HomogeneousCoeffSubsIndepDivDep._verify(self, fx) and HomogeneousCoeffSubsDepDivIndep._verify(self, fx): + return True + return False + + def _get_general_solution(self, *, simplify_flag: bool = True): + # There are two substitutions that solve the equation, u1=y/x and u2=x/y + # # They produce different integrals, so try them both and see which + # # one is easier + sol1 = HomogeneousCoeffSubsIndepDivDep._get_general_solution(self) + sol2 = HomogeneousCoeffSubsDepDivIndep._get_general_solution(self) + fx = self.ode_problem.func + if simplify_flag: + sol1 = odesimp(self.ode_problem.eq, *sol1, fx, "1st_homogeneous_coeff_subs_indep_div_dep") + sol2 = odesimp(self.ode_problem.eq, *sol2, fx, "1st_homogeneous_coeff_subs_dep_div_indep") + return min([sol1, sol2], key=lambda x: ode_sol_simplicity(x, fx, trysolving=not simplify)) + + +class LinearCoefficients(HomogeneousCoeffBest): + r""" + Solves a differential equation with linear coefficients. + + The general form of a differential equation with linear coefficients is + + .. math:: y' + F\left(\!\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + + c_2}\!\right) = 0\text{,} + + where `a_1`, `b_1`, `c_1`, `a_2`, `b_2`, `c_2` are constants and `a_1 b_2 + - a_2 b_1 \ne 0`. + + This can be solved by substituting: + + .. math:: x = x' + \frac{b_2 c_1 - b_1 c_2}{a_2 b_1 - a_1 b_2} + + y = y' + \frac{a_1 c_2 - a_2 c_1}{a_2 b_1 - a_1 + b_2}\text{.} + + This substitution reduces the equation to a homogeneous differential + equation. + + See Also + ======== + :obj:`sympy.solvers.ode.single.HomogeneousCoeffBest` + :obj:`sympy.solvers.ode.single.HomogeneousCoeffSubsIndepDivDep` + :obj:`sympy.solvers.ode.single.HomogeneousCoeffSubsDepDivIndep` + + Examples + ======== + + >>> from sympy import dsolve, Function, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> df = f(x).diff(x) + >>> eq = (x + f(x) + 1)*df + (f(x) - 6*x + 1) + >>> dsolve(eq, hint='linear_coefficients') + [Eq(f(x), -x - sqrt(C1 + 7*x**2) - 1), Eq(f(x), -x + sqrt(C1 + 7*x**2) - 1)] + >>> pprint(dsolve(eq, hint='linear_coefficients')) + ___________ ___________ + / 2 / 2 + [f(x) = -x - \/ C1 + 7*x - 1, f(x) = -x + \/ C1 + 7*x - 1] + + + References + ========== + + - Joel Moses, "Symbolic Integration - The Stormy Decade", Communications + of the ACM, Volume 14, Number 8, August 1971, pp. 558 + """ + hint = "linear_coefficients" + has_integral = True + order = [1] + + def _wilds(self, f, x, order): + d = Wild('d', exclude=[f(x).diff(x), f(x).diff(x, 2)]) + e = Wild('e', exclude=[f(x).diff(x)]) + return d, e + + def _equation(self, fx, x, order): + d, e = self.wilds() + return d + e*fx.diff(x) + + def _verify(self, fx): + self.d, self.e = self.wilds_match() + a, b = self.wilds() + F = self.d/self.e + x = self.ode_problem.sym + params = self._linear_coeff_match(F, fx) + if params: + self.xarg, self.yarg = params + u = Dummy('u') + t = Dummy('t') + self.y = Dummy('y') + # Dummy substitution for df and f(x). + dummy_eq = self.ode_problem.eq.subs(((fx.diff(x), t), (fx, u))) + reps = ((x, x + self.xarg), (u, u + self.yarg), (t, fx.diff(x)), (u, fx)) + dummy_eq = simplify(dummy_eq.subs(reps)) + # get the re-cast values for e and d + r2 = collect(expand(dummy_eq), [fx.diff(x), fx]).match(a*fx.diff(x) + b) + if r2: + self.d, self.e = r2[b], r2[a] + orderd = homogeneous_order(self.d, x, fx) + ordere = homogeneous_order(self.e, x, fx) + if orderd == ordere and orderd is not None: + self.d = self.d.subs(fx, self.y) + self.e = self.e.subs(fx, self.y) + return True + return False + return False + + def _linear_coeff_match(self, expr, func): + r""" + Helper function to match hint ``linear_coefficients``. + + Matches the expression to the form `(a_1 x + b_1 f(x) + c_1)/(a_2 x + b_2 + f(x) + c_2)` where the following conditions hold: + + 1. `a_1`, `b_1`, `c_1`, `a_2`, `b_2`, `c_2` are Rationals; + 2. `c_1` or `c_2` are not equal to zero; + 3. `a_2 b_1 - a_1 b_2` is not equal to zero. + + Return ``xarg``, ``yarg`` where + + 1. ``xarg`` = `(b_2 c_1 - b_1 c_2)/(a_2 b_1 - a_1 b_2)` + 2. ``yarg`` = `(a_1 c_2 - a_2 c_1)/(a_2 b_1 - a_1 b_2)` + + + Examples + ======== + + >>> from sympy import Function, sin + >>> from sympy.abc import x + >>> from sympy.solvers.ode.single import LinearCoefficients + >>> f = Function('f') + >>> eq = (-25*f(x) - 8*x + 62)/(4*f(x) + 11*x - 11) + >>> obj = LinearCoefficients(eq) + >>> obj._linear_coeff_match(eq, f(x)) + (1/9, 22/9) + >>> eq = sin((-5*f(x) - 8*x + 6)/(4*f(x) + x - 1)) + >>> obj = LinearCoefficients(eq) + >>> obj._linear_coeff_match(eq, f(x)) + (19/27, 2/27) + >>> eq = sin(f(x)/x) + >>> obj = LinearCoefficients(eq) + >>> obj._linear_coeff_match(eq, f(x)) + + """ + f = func.func + x = func.args[0] + def abc(eq): + r''' + Internal function of _linear_coeff_match + that returns Rationals a, b, c + if eq is a*x + b*f(x) + c, else None. + ''' + eq = _mexpand(eq) + c = eq.as_independent(x, f(x), as_Add=True)[0] + if not c.is_Rational: + return + a = eq.coeff(x) + if not a.is_Rational: + return + b = eq.coeff(f(x)) + if not b.is_Rational: + return + if eq == a*x + b*f(x) + c: + return a, b, c + + def match(arg): + r''' + Internal function of _linear_coeff_match that returns Rationals a1, + b1, c1, a2, b2, c2 and a2*b1 - a1*b2 of the expression (a1*x + b1*f(x) + + c1)/(a2*x + b2*f(x) + c2) if one of c1 or c2 and a2*b1 - a1*b2 is + non-zero, else None. + ''' + n, d = arg.together().as_numer_denom() + m = abc(n) + if m is not None: + a1, b1, c1 = m + m = abc(d) + if m is not None: + a2, b2, c2 = m + d = a2*b1 - a1*b2 + if (c1 or c2) and d: + return a1, b1, c1, a2, b2, c2, d + + m = [fi.args[0] for fi in expr.atoms(Function) if fi.func != f and + len(fi.args) == 1 and not fi.args[0].is_Function] or {expr} + m1 = match(m.pop()) + if m1 and all(match(mi) == m1 for mi in m): + a1, b1, c1, a2, b2, c2, denom = m1 + return (b2*c1 - b1*c2)/denom, (a1*c2 - a2*c1)/denom + + def _get_match_object(self): + fx = self.ode_problem.func + x = self.ode_problem.sym + self.u1 = Dummy('u1') + u = Dummy('u') + return [self.d, self.e, fx, x, u, self.u1, self.y, self.xarg, self.yarg] + + +class NthOrderReducible(SingleODESolver): + r""" + Solves ODEs that only involve derivatives of the dependent variable using + a substitution of the form `f^n(x) = g(x)`. + + For example any second order ODE of the form `f''(x) = h(f'(x), x)` can be + transformed into a pair of 1st order ODEs `g'(x) = h(g(x), x)` and + `f'(x) = g(x)`. Usually the 1st order ODE for `g` is easier to solve. If + that gives an explicit solution for `g` then `f` is found simply by + integration. + + + Examples + ======== + + >>> from sympy import Function, dsolve, Eq + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = Eq(x*f(x).diff(x)**2 + f(x).diff(x, 2), 0) + >>> dsolve(eq, f(x), hint='nth_order_reducible') + ... # doctest: +NORMALIZE_WHITESPACE + Eq(f(x), C1 - sqrt(-1/C2)*log(-C2*sqrt(-1/C2) + x) + sqrt(-1/C2)*log(C2*sqrt(-1/C2) + x)) + + """ + hint = "nth_order_reducible" + has_integral = False + + def _matches(self): + # Any ODE that can be solved with a substitution and + # repeated integration e.g.: + # `d^2/dx^2(y) + x*d/dx(y) = constant + #f'(x) must be finite for this to work + eq = self.ode_problem.eq_preprocessed + func = self.ode_problem.func + x = self.ode_problem.sym + r""" + Matches any differential equation that can be rewritten with a smaller + order. Only derivatives of ``func`` alone, wrt a single variable, + are considered, and only in them should ``func`` appear. + """ + # ODE only handles functions of 1 variable so this affirms that state + assert len(func.args) == 1 + vc = [d.variable_count[0] for d in eq.atoms(Derivative) + if d.expr == func and len(d.variable_count) == 1] + ords = [c for v, c in vc if v == x] + if len(ords) < 2: + return False + self.smallest = min(ords) + # make sure func does not appear outside of derivatives + D = Dummy() + if eq.subs(func.diff(x, self.smallest), D).has(func): + return False + return True + + def _get_general_solution(self, *, simplify_flag: bool = True): + eq = self.ode_problem.eq + f = self.ode_problem.func.func + x = self.ode_problem.sym + n = self.smallest + # get a unique function name for g + names = [a.name for a in eq.atoms(AppliedUndef)] + while True: + name = Dummy().name + if name not in names: + g = Function(name) + break + w = f(x).diff(x, n) + geq = eq.subs(w, g(x)) + gsol = dsolve(geq, g(x)) + + if not isinstance(gsol, list): + gsol = [gsol] + + # Might be multiple solutions to the reduced ODE: + fsol = [] + for gsoli in gsol: + fsoli = dsolve(gsoli.subs(g(x), w), f(x)) # or do integration n times + fsol.append(fsoli) + + return fsol + + +class SecondHypergeometric(SingleODESolver): + r""" + Solves 2nd order linear differential equations. + + It computes special function solutions which can be expressed using the + 2F1, 1F1 or 0F1 hypergeometric functions. + + .. math:: y'' + A(x) y' + B(x) y = 0\text{,} + + where `A` and `B` are rational functions. + + These kinds of differential equations have solution of non-Liouvillian form. + + Given linear ODE can be obtained from 2F1 given by + + .. math:: (x^2 - x) y'' + ((a + b + 1) x - c) y' + b a y = 0\text{,} + + where {a, b, c} are arbitrary constants. + + Notes + ===== + + The algorithm should find any solution of the form + + .. math:: y = P(x) _pF_q(..; ..;\frac{\alpha x^k + \beta}{\gamma x^k + \delta})\text{,} + + where pFq is any of 2F1, 1F1 or 0F1 and `P` is an "arbitrary function". + Currently only the 2F1 case is implemented in SymPy but the other cases are + described in the paper and could be implemented in future (contributions + welcome!). + + + Examples + ======== + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = (x*x - x)*f(x).diff(x,2) + (5*x - 1)*f(x).diff(x) + 4*f(x) + >>> pprint(dsolve(eq, f(x), '2nd_hypergeometric')) + _ + / / 4 \\ |_ /-1, -1 | \ + |C1 + C2*|log(x) + -----||* | | | x| + \ \ x + 1// 2 1 \ 1 | / + f(x) = -------------------------------------------- + 3 + (x - 1) + + + References + ========== + + - "Non-Liouvillian solutions for second order linear ODEs" by L. Chan, E.S. Cheb-Terrab + + """ + hint = "2nd_hypergeometric" + has_integral = True + + def _matches(self): + eq = self.ode_problem.eq_preprocessed + func = self.ode_problem.func + r = match_2nd_hypergeometric(eq, func) + self.match_object = None + if r: + A, B = r + d = equivalence_hypergeometric(A, B, func) + if d: + if d['type'] == "2F1": + self.match_object = match_2nd_2F1_hypergeometric(d['I0'], d['k'], d['sing_point'], func) + if self.match_object is not None: + self.match_object.update({'A':A, 'B':B}) + # We can extend it for 1F1 and 0F1 type also. + return self.match_object is not None + + def _get_general_solution(self, *, simplify_flag: bool = True): + eq = self.ode_problem.eq + func = self.ode_problem.func + if self.match_object['type'] == "2F1": + sol = get_sol_2F1_hypergeometric(eq, func, self.match_object) + if sol is None: + raise NotImplementedError("The given ODE " + str(eq) + " cannot be solved by" + + " the hypergeometric method") + + return [sol] + + +class NthLinearConstantCoeffHomogeneous(SingleODESolver): + r""" + Solves an `n`\th order linear homogeneous differential equation with + constant coefficients. + + This is an equation of the form + + .. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + + a_0 f(x) = 0\text{.} + + These equations can be solved in a general manner, by taking the roots of + the characteristic equation `a_n m^n + a_{n-1} m^{n-1} + \cdots + a_1 m + + a_0 = 0`. The solution will then be the sum of `C_n x^i e^{r x}` terms, + for each where `C_n` is an arbitrary constant, `r` is a root of the + characteristic equation and `i` is one of each from 0 to the multiplicity + of the root - 1 (for example, a root 3 of multiplicity 2 would create the + terms `C_1 e^{3 x} + C_2 x e^{3 x}`). The exponential is usually expanded + for complex roots using Euler's equation `e^{I x} = \cos(x) + I \sin(x)`. + Complex roots always come in conjugate pairs in polynomials with real + coefficients, so the two roots will be represented (after simplifying the + constants) as `e^{a x} \left(C_1 \cos(b x) + C_2 \sin(b x)\right)`. + + If SymPy cannot find exact roots to the characteristic equation, a + :py:class:`~sympy.polys.rootoftools.ComplexRootOf` instance will be return + instead. + + >>> from sympy import Function, dsolve + >>> from sympy.abc import x + >>> f = Function('f') + >>> dsolve(f(x).diff(x, 5) + 10*f(x).diff(x) - 2*f(x), f(x), + ... hint='nth_linear_constant_coeff_homogeneous') + ... # doctest: +NORMALIZE_WHITESPACE + Eq(f(x), C5*exp(x*CRootOf(_x**5 + 10*_x - 2, 0)) + + (C1*sin(x*im(CRootOf(_x**5 + 10*_x - 2, 1))) + + C2*cos(x*im(CRootOf(_x**5 + 10*_x - 2, 1))))*exp(x*re(CRootOf(_x**5 + 10*_x - 2, 1))) + + (C3*sin(x*im(CRootOf(_x**5 + 10*_x - 2, 3))) + + C4*cos(x*im(CRootOf(_x**5 + 10*_x - 2, 3))))*exp(x*re(CRootOf(_x**5 + 10*_x - 2, 3)))) + + Note that because this method does not involve integration, there is no + ``nth_linear_constant_coeff_homogeneous_Integral`` hint. + + Examples + ======== + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(f(x).diff(x, 4) + 2*f(x).diff(x, 3) - + ... 2*f(x).diff(x, 2) - 6*f(x).diff(x) + 5*f(x), f(x), + ... hint='nth_linear_constant_coeff_homogeneous')) + x -2*x + f(x) = (C1 + C2*x)*e + (C3*sin(x) + C4*cos(x))*e + + References + ========== + + - https://en.wikipedia.org/wiki/Linear_differential_equation section: + Nonhomogeneous_equation_with_constant_coefficients + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 211 + + # indirect doctest + + """ + hint = "nth_linear_constant_coeff_homogeneous" + has_integral = False + + def _matches(self): + eq = self.ode_problem.eq_high_order_free + func = self.ode_problem.func + order = self.ode_problem.order + x = self.ode_problem.sym + self.r = self.ode_problem.get_linear_coefficients(eq, func, order) + if order and self.r and not any(self.r[i].has(x) for i in self.r if i >= 0): + if not self.r[-1]: + return True + else: + return False + return False + + def _get_general_solution(self, *, simplify_flag: bool = True): + fx = self.ode_problem.func + order = self.ode_problem.order + roots, collectterms = _get_const_characteristic_eq_sols(self.r, fx, order) + # A generator of constants + constants = self.ode_problem.get_numbered_constants(num=len(roots)) + gsol = Add(*[i*j for (i, j) in zip(constants, roots)]) + gsol = Eq(fx, gsol) + if simplify_flag: + gsol = _get_simplified_sol([gsol], fx, collectterms) + + return [gsol] + + +class NthLinearConstantCoeffVariationOfParameters(SingleODESolver): + r""" + Solves an `n`\th order linear differential equation with constant + coefficients using the method of variation of parameters. + + This method works on any differential equations of the form + + .. math:: f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + a_0 + f(x) = P(x)\text{.} + + This method works by assuming that the particular solution takes the form + + .. math:: \sum_{x=1}^{n} c_i(x) y_i(x)\text{,} + + where `y_i` is the `i`\th solution to the homogeneous equation. The + solution is then solved using Wronskian's and Cramer's Rule. The + particular solution is given by + + .. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \,dx + \right) y_i(x) \text{,} + + where `W(x)` is the Wronskian of the fundamental system (the system of `n` + linearly independent solutions to the homogeneous equation), and `W_i(x)` + is the Wronskian of the fundamental system with the `i`\th column replaced + with `[0, 0, \cdots, 0, P(x)]`. + + This method is general enough to solve any `n`\th order inhomogeneous + linear differential equation with constant coefficients, but sometimes + SymPy cannot simplify the Wronskian well enough to integrate it. If this + method hangs, try using the + ``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and + simplifying the integrals manually. Also, prefer using + ``nth_linear_constant_coeff_undetermined_coefficients`` when it + applies, because it does not use integration, making it faster and more + reliable. + + Warning, using simplify=False with + 'nth_linear_constant_coeff_variation_of_parameters' in + :py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will + not attempt to simplify the Wronskian before integrating. It is + recommended that you only use simplify=False with + 'nth_linear_constant_coeff_variation_of_parameters_Integral' for this + method, especially if the solution to the homogeneous equation has + trigonometric functions in it. + + Examples + ======== + + >>> from sympy import Function, dsolve, pprint, exp, log + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(f(x).diff(x, 3) - 3*f(x).diff(x, 2) + + ... 3*f(x).diff(x) - f(x) - exp(x)*log(x), f(x), + ... hint='nth_linear_constant_coeff_variation_of_parameters')) + / / / x*log(x) 11*x\\\ x + f(x) = |C1 + x*|C2 + x*|C3 + -------- - ----|||*e + \ \ \ 6 36 /// + + References + ========== + + - https://en.wikipedia.org/wiki/Variation_of_parameters + - https://planetmath.org/VariationOfParameters + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 233 + + # indirect doctest + + """ + hint = "nth_linear_constant_coeff_variation_of_parameters" + has_integral = True + + def _matches(self): + eq = self.ode_problem.eq_high_order_free + func = self.ode_problem.func + order = self.ode_problem.order + x = self.ode_problem.sym + self.r = self.ode_problem.get_linear_coefficients(eq, func, order) + + if order and self.r and not any(self.r[i].has(x) for i in self.r if i >= 0): + if self.r[-1]: + return True + else: + return False + return False + + def _get_general_solution(self, *, simplify_flag: bool = True): + eq = self.ode_problem.eq_high_order_free + f = self.ode_problem.func.func + x = self.ode_problem.sym + order = self.ode_problem.order + roots, collectterms = _get_const_characteristic_eq_sols(self.r, f(x), order) + # A generator of constants + constants = self.ode_problem.get_numbered_constants(num=len(roots)) + homogen_sol = Add(*[i*j for (i, j) in zip(constants, roots)]) + homogen_sol = Eq(f(x), homogen_sol) + homogen_sol = _solve_variation_of_parameters(eq, f(x), roots, homogen_sol, order, self.r, simplify_flag) + if simplify_flag: + homogen_sol = _get_simplified_sol([homogen_sol], f(x), collectterms) + return [homogen_sol] + + +class NthLinearConstantCoeffUndeterminedCoefficients(SingleODESolver): + r""" + Solves an `n`\th order linear differential equation with constant + coefficients using the method of undetermined coefficients. + + This method works on differential equations of the form + + .. math:: a_n f^{(n)}(x) + a_{n-1} f^{(n-1)}(x) + \cdots + a_1 f'(x) + + a_0 f(x) = P(x)\text{,} + + where `P(x)` is a function that has a finite number of linearly + independent derivatives. + + Functions that fit this requirement are finite sums functions of the form + `a x^i e^{b x} \sin(c x + d)` or `a x^i e^{b x} \cos(c x + d)`, where `i` + is a non-negative integer and `a`, `b`, `c`, and `d` are constants. For + example any polynomial in `x`, functions like `x^2 e^{2 x}`, `x \sin(x)`, + and `e^x \cos(x)` can all be used. Products of `\sin`'s and `\cos`'s have + a finite number of derivatives, because they can be expanded into `\sin(a + x)` and `\cos(b x)` terms. However, SymPy currently cannot do that + expansion, so you will need to manually rewrite the expression in terms of + the above to use this method. So, for example, you will need to manually + convert `\sin^2(x)` into `(1 + \cos(2 x))/2` to properly apply the method + of undetermined coefficients on it. + + This method works by creating a trial function from the expression and all + of its linear independent derivatives and substituting them into the + original ODE. The coefficients for each term will be a system of linear + equations, which are be solved for and substituted, giving the solution. + If any of the trial functions are linearly dependent on the solution to + the homogeneous equation, they are multiplied by sufficient `x` to make + them linearly independent. + + Examples + ======== + + >>> from sympy import Function, dsolve, pprint, exp, cos + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(f(x).diff(x, 2) + 2*f(x).diff(x) + f(x) - + ... 4*exp(-x)*x**2 + cos(2*x), f(x), + ... hint='nth_linear_constant_coeff_undetermined_coefficients')) + / / 3\\ + | | x || -x 4*sin(2*x) 3*cos(2*x) + f(x) = |C1 + x*|C2 + --||*e - ---------- + ---------- + \ \ 3 // 25 25 + + References + ========== + + - https://en.wikipedia.org/wiki/Method_of_undetermined_coefficients + - M. Tenenbaum & H. Pollard, "Ordinary Differential Equations", + Dover 1963, pp. 221 + + # indirect doctest + + """ + hint = "nth_linear_constant_coeff_undetermined_coefficients" + has_integral = False + + def _matches(self): + eq = self.ode_problem.eq_high_order_free + func = self.ode_problem.func + order = self.ode_problem.order + x = self.ode_problem.sym + self.r = self.ode_problem.get_linear_coefficients(eq, func, order) + does_match = False + if order and self.r and not any(self.r[i].has(x) for i in self.r if i >= 0): + if self.r[-1]: + eq_homogeneous = Add(eq, -self.r[-1]) + undetcoeff = _undetermined_coefficients_match(self.r[-1], x, func, eq_homogeneous) + if undetcoeff['test']: + self.trialset = undetcoeff['trialset'] + does_match = True + return does_match + + def _get_general_solution(self, *, simplify_flag: bool = True): + eq = self.ode_problem.eq + f = self.ode_problem.func.func + x = self.ode_problem.sym + order = self.ode_problem.order + roots, collectterms = _get_const_characteristic_eq_sols(self.r, f(x), order) + # A generator of constants + constants = self.ode_problem.get_numbered_constants(num=len(roots)) + homogen_sol = Add(*[i*j for (i, j) in zip(constants, roots)]) + homogen_sol = Eq(f(x), homogen_sol) + self.r.update({'list': roots, 'sol': homogen_sol, 'simpliy_flag': simplify_flag}) + gsol = _solve_undetermined_coefficients(eq, f(x), order, self.r, self.trialset) + if simplify_flag: + gsol = _get_simplified_sol([gsol], f(x), collectterms) + return [gsol] + + +class NthLinearEulerEqHomogeneous(SingleODESolver): + r""" + Solves an `n`\th order linear homogeneous variable-coefficient + Cauchy-Euler equidimensional ordinary differential equation. + + This is an equation with form `0 = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) + \cdots`. + + These equations can be solved in a general manner, by substituting + solutions of the form `f(x) = x^r`, and deriving a characteristic equation + for `r`. When there are repeated roots, we include extra terms of the + form `C_{r k} \ln^k(x) x^r`, where `C_{r k}` is an arbitrary integration + constant, `r` is a root of the characteristic equation, and `k` ranges + over the multiplicity of `r`. In the cases where the roots are complex, + solutions of the form `C_1 x^a \sin(b \log(x)) + C_2 x^a \cos(b \log(x))` + are returned, based on expansions with Euler's formula. The general + solution is the sum of the terms found. If SymPy cannot find exact roots + to the characteristic equation, a + :py:obj:`~.ComplexRootOf` instance will be returned + instead. + + >>> from sympy import Function, dsolve + >>> from sympy.abc import x + >>> f = Function('f') + >>> dsolve(4*x**2*f(x).diff(x, 2) + f(x), f(x), + ... hint='nth_linear_euler_eq_homogeneous') + ... # doctest: +NORMALIZE_WHITESPACE + Eq(f(x), sqrt(x)*(C1 + C2*log(x))) + + Note that because this method does not involve integration, there is no + ``nth_linear_euler_eq_homogeneous_Integral`` hint. + + The following is for internal use: + + - ``returns = 'sol'`` returns the solution to the ODE. + - ``returns = 'list'`` returns a list of linearly independent solutions, + corresponding to the fundamental solution set, for use with non + homogeneous solution methods like variation of parameters and + undetermined coefficients. Note that, though the solutions should be + linearly independent, this function does not explicitly check that. You + can do ``assert simplify(wronskian(sollist)) != 0`` to check for linear + independence. Also, ``assert len(sollist) == order`` will need to pass. + - ``returns = 'both'``, return a dictionary ``{'sol': , + 'list': }``. + + Examples + ======== + + >>> from sympy import Function, dsolve, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = f(x).diff(x, 2)*x**2 - 4*f(x).diff(x)*x + 6*f(x) + >>> pprint(dsolve(eq, f(x), + ... hint='nth_linear_euler_eq_homogeneous')) + 2 + f(x) = x *(C1 + C2*x) + + References + ========== + + - https://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation + - C. Bender & S. Orszag, "Advanced Mathematical Methods for Scientists and + Engineers", Springer 1999, pp. 12 + + # indirect doctest + + """ + hint = "nth_linear_euler_eq_homogeneous" + has_integral = False + + def _matches(self): + eq = self.ode_problem.eq_preprocessed + f = self.ode_problem.func.func + order = self.ode_problem.order + x = self.ode_problem.sym + match = self.ode_problem.get_linear_coefficients(eq, f(x), order) + self.r = None + does_match = False + + if order and match: + coeff = match[order] + factor = x**order / coeff + self.r = {i: factor*match[i] for i in match} + if self.r and all(_test_term(self.r[i], f(x), i) for i in + self.r if i >= 0): + if not self.r[-1]: + does_match = True + return does_match + + def _get_general_solution(self, *, simplify_flag: bool = True): + fx = self.ode_problem.func + eq = self.ode_problem.eq + homogen_sol = _get_euler_characteristic_eq_sols(eq, fx, self.r)[0] + return [homogen_sol] + + +class NthLinearEulerEqNonhomogeneousVariationOfParameters(SingleODESolver): + r""" + Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional + ordinary differential equation using variation of parameters. + + This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) + \cdots`. + + This method works by assuming that the particular solution takes the form + + .. math:: \sum_{x=1}^{n} c_i(x) y_i(x) {a_n} {x^n} \text{, } + + where `y_i` is the `i`\th solution to the homogeneous equation. The + solution is then solved using Wronskian's and Cramer's Rule. The + particular solution is given by multiplying eq given below with `a_n x^{n}` + + .. math:: \sum_{x=1}^n \left( \int \frac{W_i(x)}{W(x)} \, dx + \right) y_i(x) \text{, } + + where `W(x)` is the Wronskian of the fundamental system (the system of `n` + linearly independent solutions to the homogeneous equation), and `W_i(x)` + is the Wronskian of the fundamental system with the `i`\th column replaced + with `[0, 0, \cdots, 0, \frac{x^{- n}}{a_n} g{\left(x \right)}]`. + + This method is general enough to solve any `n`\th order inhomogeneous + linear differential equation, but sometimes SymPy cannot simplify the + Wronskian well enough to integrate it. If this method hangs, try using the + ``nth_linear_constant_coeff_variation_of_parameters_Integral`` hint and + simplifying the integrals manually. Also, prefer using + ``nth_linear_constant_coeff_undetermined_coefficients`` when it + applies, because it does not use integration, making it faster and more + reliable. + + Warning, using simplify=False with + 'nth_linear_constant_coeff_variation_of_parameters' in + :py:meth:`~sympy.solvers.ode.dsolve` may cause it to hang, because it will + not attempt to simplify the Wronskian before integrating. It is + recommended that you only use simplify=False with + 'nth_linear_constant_coeff_variation_of_parameters_Integral' for this + method, especially if the solution to the homogeneous equation has + trigonometric functions in it. + + Examples + ======== + + >>> from sympy import Function, dsolve, Derivative + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - x**4 + >>> dsolve(eq, f(x), + ... hint='nth_linear_euler_eq_nonhomogeneous_variation_of_parameters').expand() + Eq(f(x), C1*x + C2*x**2 + x**4/6) + + """ + hint = "nth_linear_euler_eq_nonhomogeneous_variation_of_parameters" + has_integral = True + + def _matches(self): + eq = self.ode_problem.eq_preprocessed + f = self.ode_problem.func.func + order = self.ode_problem.order + x = self.ode_problem.sym + match = self.ode_problem.get_linear_coefficients(eq, f(x), order) + self.r = None + does_match = False + + if order and match: + coeff = match[order] + factor = x**order / coeff + self.r = {i: factor*match[i] for i in match} + if self.r and all(_test_term(self.r[i], f(x), i) for i in + self.r if i >= 0): + if self.r[-1]: + does_match = True + + return does_match + + def _get_general_solution(self, *, simplify_flag: bool = True): + eq = self.ode_problem.eq + f = self.ode_problem.func.func + x = self.ode_problem.sym + order = self.ode_problem.order + homogen_sol, roots = _get_euler_characteristic_eq_sols(eq, f(x), self.r) + self.r[-1] = self.r[-1]/self.r[order] + sol = _solve_variation_of_parameters(eq, f(x), roots, homogen_sol, order, self.r, simplify_flag) + + return [Eq(f(x), homogen_sol.rhs + (sol.rhs - homogen_sol.rhs)*self.r[order])] + + +class NthLinearEulerEqNonhomogeneousUndeterminedCoefficients(SingleODESolver): + r""" + Solves an `n`\th order linear non homogeneous Cauchy-Euler equidimensional + ordinary differential equation using undetermined coefficients. + + This is an equation with form `g(x) = a_0 f(x) + a_1 x f'(x) + a_2 x^2 f''(x) + \cdots`. + + These equations can be solved in a general manner, by substituting + solutions of the form `x = exp(t)`, and deriving a characteristic equation + of form `g(exp(t)) = b_0 f(t) + b_1 f'(t) + b_2 f''(t) \cdots` which can + be then solved by nth_linear_constant_coeff_undetermined_coefficients if + g(exp(t)) has finite number of linearly independent derivatives. + + Functions that fit this requirement are finite sums functions of the form + `a x^i e^{b x} \sin(c x + d)` or `a x^i e^{b x} \cos(c x + d)`, where `i` + is a non-negative integer and `a`, `b`, `c`, and `d` are constants. For + example any polynomial in `x`, functions like `x^2 e^{2 x}`, `x \sin(x)`, + and `e^x \cos(x)` can all be used. Products of `\sin`'s and `\cos`'s have + a finite number of derivatives, because they can be expanded into `\sin(a + x)` and `\cos(b x)` terms. However, SymPy currently cannot do that + expansion, so you will need to manually rewrite the expression in terms of + the above to use this method. So, for example, you will need to manually + convert `\sin^2(x)` into `(1 + \cos(2 x))/2` to properly apply the method + of undetermined coefficients on it. + + After replacement of x by exp(t), this method works by creating a trial function + from the expression and all of its linear independent derivatives and + substituting them into the original ODE. The coefficients for each term + will be a system of linear equations, which are be solved for and + substituted, giving the solution. If any of the trial functions are linearly + dependent on the solution to the homogeneous equation, they are multiplied + by sufficient `x` to make them linearly independent. + + Examples + ======== + + >>> from sympy import dsolve, Function, Derivative, log + >>> from sympy.abc import x + >>> f = Function('f') + >>> eq = x**2*Derivative(f(x), x, x) - 2*x*Derivative(f(x), x) + 2*f(x) - log(x) + >>> dsolve(eq, f(x), + ... hint='nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients').expand() + Eq(f(x), C1*x + C2*x**2 + log(x)/2 + 3/4) + + """ + hint = "nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients" + has_integral = False + + def _matches(self): + eq = self.ode_problem.eq_high_order_free + f = self.ode_problem.func.func + order = self.ode_problem.order + x = self.ode_problem.sym + match = self.ode_problem.get_linear_coefficients(eq, f(x), order) + self.r = None + does_match = False + + if order and match: + coeff = match[order] + factor = x**order / coeff + self.r = {i: factor*match[i] for i in match} + if self.r and all(_test_term(self.r[i], f(x), i) for i in + self.r if i >= 0): + if self.r[-1]: + e, re = posify(self.r[-1].subs(x, exp(x))) + undetcoeff = _undetermined_coefficients_match(e.subs(re), x) + if undetcoeff['test']: + does_match = True + return does_match + + def _get_general_solution(self, *, simplify_flag: bool = True): + f = self.ode_problem.func.func + x = self.ode_problem.sym + chareq, eq, symbol = S.Zero, S.Zero, Dummy('x') + for i in self.r.keys(): + if i >= 0: + chareq += (self.r[i]*diff(x**symbol, x, i)*x**-symbol).expand() + + for i in range(1, degree(Poly(chareq, symbol))+1): + eq += chareq.coeff(symbol**i)*diff(f(x), x, i) + + if chareq.as_coeff_add(symbol)[0]: + eq += chareq.as_coeff_add(symbol)[0]*f(x) + e, re = posify(self.r[-1].subs(x, exp(x))) + eq += e.subs(re) + + self.const_undet_instance = NthLinearConstantCoeffUndeterminedCoefficients(SingleODEProblem(eq, f(x), x)) + sol = self.const_undet_instance.get_general_solution(simplify = simplify_flag)[0] + sol = sol.subs(x, log(x)) + sol = sol.subs(f(log(x)), f(x)).expand() + + return [sol] + + +class SecondLinearBessel(SingleODESolver): + r""" + Gives solution of the Bessel differential equation + + .. math :: x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} y(x) + (x^2-n^2) y(x) + + if `n` is integer then the solution is of the form ``Eq(f(x), C0 besselj(n,x) + + C1 bessely(n,x))`` as both the solutions are linearly independent else if + `n` is a fraction then the solution is of the form ``Eq(f(x), C0 besselj(n,x) + + C1 besselj(-n,x))`` which can also transform into ``Eq(f(x), C0 besselj(n,x) + + C1 bessely(n,x))``. + + Examples + ======== + + >>> from sympy.abc import x + >>> from sympy import Symbol + >>> v = Symbol('v', positive=True) + >>> from sympy import dsolve, Function + >>> f = Function('f') + >>> y = f(x) + >>> genform = x**2*y.diff(x, 2) + x*y.diff(x) + (x**2 - v**2)*y + >>> dsolve(genform) + Eq(f(x), C1*besselj(v, x) + C2*bessely(v, x)) + + References + ========== + + https://math24.net/bessel-differential-equation.html + + """ + hint = "2nd_linear_bessel" + has_integral = False + + def _matches(self): + eq = self.ode_problem.eq_high_order_free + f = self.ode_problem.func + order = self.ode_problem.order + x = self.ode_problem.sym + df = f.diff(x) + a = Wild('a', exclude=[f,df]) + b = Wild('b', exclude=[x, f,df]) + a4 = Wild('a4', exclude=[x,f,df]) + b4 = Wild('b4', exclude=[x,f,df]) + c4 = Wild('c4', exclude=[x,f,df]) + d4 = Wild('d4', exclude=[x,f,df]) + a3 = Wild('a3', exclude=[f, df, f.diff(x, 2)]) + b3 = Wild('b3', exclude=[f, df, f.diff(x, 2)]) + c3 = Wild('c3', exclude=[f, df, f.diff(x, 2)]) + deq = a3*(f.diff(x, 2)) + b3*df + c3*f + r = collect(eq, + [f.diff(x, 2), df, f]).match(deq) + if order == 2 and r: + if not all(r[key].is_polynomial() for key in r): + n, d = eq.as_numer_denom() + eq = expand(n) + r = collect(eq, + [f.diff(x, 2), df, f]).match(deq) + + if r and r[a3] != 0: + # leading coeff of f(x).diff(x, 2) + coeff = factor(r[a3]).match(a4*(x-b)**b4) + + if coeff: + # if coeff[b4] = 0 means constant coefficient + if coeff[b4] == 0: + return False + point = coeff[b] + else: + return False + + if point: + r[a3] = simplify(r[a3].subs(x, x+point)) + r[b3] = simplify(r[b3].subs(x, x+point)) + r[c3] = simplify(r[c3].subs(x, x+point)) + + # making a3 in the form of x**2 + r[a3] = cancel(r[a3]/(coeff[a4]*(x)**(-2+coeff[b4]))) + r[b3] = cancel(r[b3]/(coeff[a4]*(x)**(-2+coeff[b4]))) + r[c3] = cancel(r[c3]/(coeff[a4]*(x)**(-2+coeff[b4]))) + # checking if b3 is of form c*(x-b) + coeff1 = factor(r[b3]).match(a4*(x)) + if coeff1 is None: + return False + # c3 maybe of very complex form so I am simply checking (a - b) form + # if yes later I will match with the standerd form of bessel in a and b + # a, b are wild variable defined above. + _coeff2 = expand(r[c3]).match(a - b) + if _coeff2 is None: + return False + # matching with standerd form for c3 + coeff2 = factor(_coeff2[a]).match(c4**2*(x)**(2*a4)) + if coeff2 is None: + return False + + if _coeff2[b] == 0: + coeff2[d4] = 0 + else: + coeff2[d4] = factor(_coeff2[b]).match(d4**2)[d4] + + self.rn = {'n':coeff2[d4], 'a4':coeff2[c4], 'd4':coeff2[a4]} + self.rn['c4'] = coeff1[a4] + self.rn['b4'] = point + return True + return False + + def _get_general_solution(self, *, simplify_flag: bool = True): + f = self.ode_problem.func.func + x = self.ode_problem.sym + n = self.rn['n'] + a4 = self.rn['a4'] + c4 = self.rn['c4'] + d4 = self.rn['d4'] + b4 = self.rn['b4'] + n = sqrt(n**2 + Rational(1, 4)*(c4 - 1)**2) + (C1, C2) = self.ode_problem.get_numbered_constants(num=2) + return [Eq(f(x), ((x**(Rational(1-c4,2)))*(C1*besselj(n/d4,a4*x**d4/d4) + + C2*bessely(n/d4,a4*x**d4/d4))).subs(x, x-b4))] + + +class SecondLinearAiry(SingleODESolver): + r""" + Gives solution of the Airy differential equation + + .. math :: \frac{d^2y}{dx^2} + (a + b x) y(x) = 0 + + in terms of Airy special functions airyai and airybi. + + Examples + ======== + + >>> from sympy import dsolve, Function + >>> from sympy.abc import x + >>> f = Function("f") + >>> eq = f(x).diff(x, 2) - x*f(x) + >>> dsolve(eq) + Eq(f(x), C1*airyai(x) + C2*airybi(x)) + """ + hint = "2nd_linear_airy" + has_integral = False + + def _matches(self): + eq = self.ode_problem.eq_high_order_free + f = self.ode_problem.func + order = self.ode_problem.order + x = self.ode_problem.sym + df = f.diff(x) + a4 = Wild('a4', exclude=[x,f,df]) + b4 = Wild('b4', exclude=[x,f,df]) + match = self.ode_problem.get_linear_coefficients(eq, f, order) + does_match = False + if order == 2 and match and match[2] != 0: + if match[1].is_zero: + self.rn = cancel(match[0]/match[2]).match(a4+b4*x) + if self.rn and self.rn[b4] != 0: + self.rn = {'b':self.rn[a4],'m':self.rn[b4]} + does_match = True + return does_match + + def _get_general_solution(self, *, simplify_flag: bool = True): + f = self.ode_problem.func.func + x = self.ode_problem.sym + (C1, C2) = self.ode_problem.get_numbered_constants(num=2) + b = self.rn['b'] + m = self.rn['m'] + if m.is_positive: + arg = - b/cbrt(m)**2 - cbrt(m)*x + elif m.is_negative: + arg = - b/cbrt(-m)**2 + cbrt(-m)*x + else: + arg = - b/cbrt(-m)**2 + cbrt(-m)*x + + return [Eq(f(x), C1*airyai(arg) + C2*airybi(arg))] + + +class LieGroup(SingleODESolver): + r""" + This hint implements the Lie group method of solving first order differential + equations. The aim is to convert the given differential equation from the + given coordinate system into another coordinate system where it becomes + invariant under the one-parameter Lie group of translations. The converted + ODE can be easily solved by quadrature. It makes use of the + :py:meth:`sympy.solvers.ode.infinitesimals` function which returns the + infinitesimals of the transformation. + + The coordinates `r` and `s` can be found by solving the following Partial + Differential Equations. + + .. math :: \xi\frac{\partial r}{\partial x} + \eta\frac{\partial r}{\partial y} + = 0 + + .. math :: \xi\frac{\partial s}{\partial x} + \eta\frac{\partial s}{\partial y} + = 1 + + The differential equation becomes separable in the new coordinate system + + .. math :: \frac{ds}{dr} = \frac{\frac{\partial s}{\partial x} + + h(x, y)\frac{\partial s}{\partial y}}{ + \frac{\partial r}{\partial x} + h(x, y)\frac{\partial r}{\partial y}} + + After finding the solution by integration, it is then converted back to the original + coordinate system by substituting `r` and `s` in terms of `x` and `y` again. + + Examples + ======== + + >>> from sympy import Function, dsolve, exp, pprint + >>> from sympy.abc import x + >>> f = Function('f') + >>> pprint(dsolve(f(x).diff(x) + 2*x*f(x) - x*exp(-x**2), f(x), + ... hint='lie_group')) + / 2\ 2 + | x | -x + f(x) = |C1 + --|*e + \ 2 / + + + References + ========== + + - Solving differential equations by Symmetry Groups, + John Starrett, pp. 1 - pp. 14 + + """ + hint = "lie_group" + has_integral = False + + def _has_additional_params(self): + return 'xi' in self.ode_problem.params and 'eta' in self.ode_problem.params + + def _matches(self): + eq = self.ode_problem.eq + f = self.ode_problem.func.func + order = self.ode_problem.order + x = self.ode_problem.sym + df = f(x).diff(x) + y = Dummy('y') + d = Wild('d', exclude=[df, f(x).diff(x, 2)]) + e = Wild('e', exclude=[df]) + does_match = False + if self._has_additional_params() and order == 1: + xi = self.ode_problem.params['xi'] + eta = self.ode_problem.params['eta'] + self.r3 = {'xi': xi, 'eta': eta} + r = collect(eq, df, exact=True).match(d + e * df) + if r: + r['d'] = d + r['e'] = e + r['y'] = y + r[d] = r[d].subs(f(x), y) + r[e] = r[e].subs(f(x), y) + self.r3.update(r) + does_match = True + return does_match + + def _get_general_solution(self, *, simplify_flag: bool = True): + eq = self.ode_problem.eq + x = self.ode_problem.sym + func = self.ode_problem.func + order = self.ode_problem.order + df = func.diff(x) + + try: + eqsol = solve(eq, df) + except NotImplementedError: + eqsol = [] + + desols = [] + for s in eqsol: + sol = _ode_lie_group(s, func, order, match=self.r3) + if sol: + desols.extend(sol) + + if desols == []: + raise NotImplementedError("The given ODE " + str(eq) + " cannot be solved by" + + " the lie group method") + return desols + + +solver_map = { + 'factorable': Factorable, + 'nth_linear_constant_coeff_homogeneous': NthLinearConstantCoeffHomogeneous, + 'nth_linear_euler_eq_homogeneous': NthLinearEulerEqHomogeneous, + 'nth_linear_constant_coeff_undetermined_coefficients': NthLinearConstantCoeffUndeterminedCoefficients, + 'nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients': NthLinearEulerEqNonhomogeneousUndeterminedCoefficients, + 'separable': Separable, + '1st_exact': FirstExact, + '1st_linear': FirstLinear, + 'Bernoulli': Bernoulli, + 'Riccati_special_minus2': RiccatiSpecial, + '1st_rational_riccati': RationalRiccati, + '1st_homogeneous_coeff_best': HomogeneousCoeffBest, + '1st_homogeneous_coeff_subs_indep_div_dep': HomogeneousCoeffSubsIndepDivDep, + '1st_homogeneous_coeff_subs_dep_div_indep': HomogeneousCoeffSubsDepDivIndep, + 'almost_linear': AlmostLinear, + 'linear_coefficients': LinearCoefficients, + 'separable_reduced': SeparableReduced, + 'nth_linear_constant_coeff_variation_of_parameters': NthLinearConstantCoeffVariationOfParameters, + 'nth_linear_euler_eq_nonhomogeneous_variation_of_parameters': NthLinearEulerEqNonhomogeneousVariationOfParameters, + 'Liouville': Liouville, + '2nd_linear_airy': SecondLinearAiry, + '2nd_linear_bessel': SecondLinearBessel, + '2nd_hypergeometric': SecondHypergeometric, + 'nth_order_reducible': NthOrderReducible, + '2nd_nonlinear_autonomous_conserved': SecondNonlinearAutonomousConserved, + 'nth_algebraic': NthAlgebraic, + 'lie_group': LieGroup, + } + +# Avoid circular import: +from .ode import dsolve, ode_sol_simplicity, odesimp, homogeneous_order diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/subscheck.py b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/subscheck.py new file mode 100644 index 0000000000000000000000000000000000000000..6ac7fba7d364bf599e928ccf591b5bef096576d0 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/subscheck.py @@ -0,0 +1,392 @@ +from sympy.core import S, Pow +from sympy.core.function import (Derivative, AppliedUndef, diff) +from sympy.core.relational import Equality, Eq +from sympy.core.symbol import Dummy +from sympy.core.sympify import sympify + +from sympy.logic.boolalg import BooleanAtom +from sympy.functions import exp +from sympy.series import Order +from sympy.simplify.simplify import simplify, posify, besselsimp +from sympy.simplify.trigsimp import trigsimp +from sympy.simplify.sqrtdenest import sqrtdenest +from sympy.solvers import solve +from sympy.solvers.deutils import _preprocess, ode_order +from sympy.utilities.iterables import iterable, is_sequence + + +def sub_func_doit(eq, func, new): + r""" + When replacing the func with something else, we usually want the + derivative evaluated, so this function helps in making that happen. + + Examples + ======== + + >>> from sympy import Derivative, symbols, Function + >>> from sympy.solvers.ode.subscheck import sub_func_doit + >>> x, z = symbols('x, z') + >>> y = Function('y') + + >>> sub_func_doit(3*Derivative(y(x), x) - 1, y(x), x) + 2 + + >>> sub_func_doit(x*Derivative(y(x), x) - y(x)**2 + y(x), y(x), + ... 1/(x*(z + 1/x))) + x*(-1/(x**2*(z + 1/x)) + 1/(x**3*(z + 1/x)**2)) + 1/(x*(z + 1/x)) + ...- 1/(x**2*(z + 1/x)**2) + """ + reps= {func: new} + for d in eq.atoms(Derivative): + if d.expr == func: + reps[d] = new.diff(*d.variable_count) + else: + reps[d] = d.xreplace({func: new}).doit(deep=False) + return eq.xreplace(reps) + + +def checkodesol(ode, sol, func=None, order='auto', solve_for_func=True): + r""" + Substitutes ``sol`` into ``ode`` and checks that the result is ``0``. + + This works when ``func`` is one function, like `f(x)` or a list of + functions like `[f(x), g(x)]` when `ode` is a system of ODEs. ``sol`` can + be a single solution or a list of solutions. Each solution may be an + :py:class:`~sympy.core.relational.Equality` that the solution satisfies, + e.g. ``Eq(f(x), C1), Eq(f(x) + C1, 0)``; or simply an + :py:class:`~sympy.core.expr.Expr`, e.g. ``f(x) - C1``. In most cases it + will not be necessary to explicitly identify the function, but if the + function cannot be inferred from the original equation it can be supplied + through the ``func`` argument. + + If a sequence of solutions is passed, the same sort of container will be + used to return the result for each solution. + + It tries the following methods, in order, until it finds zero equivalence: + + 1. Substitute the solution for `f` in the original equation. This only + works if ``ode`` is solved for `f`. It will attempt to solve it first + unless ``solve_for_func == False``. + 2. Take `n` derivatives of the solution, where `n` is the order of + ``ode``, and check to see if that is equal to the solution. This only + works on exact ODEs. + 3. Take the 1st, 2nd, ..., `n`\th derivatives of the solution, each time + solving for the derivative of `f` of that order (this will always be + possible because `f` is a linear operator). Then back substitute each + derivative into ``ode`` in reverse order. + + This function returns a tuple. The first item in the tuple is ``True`` if + the substitution results in ``0``, and ``False`` otherwise. The second + item in the tuple is what the substitution results in. It should always + be ``0`` if the first item is ``True``. Sometimes this function will + return ``False`` even when an expression is identically equal to ``0``. + This happens when :py:meth:`~sympy.simplify.simplify.simplify` does not + reduce the expression to ``0``. If an expression returned by this + function vanishes identically, then ``sol`` really is a solution to + the ``ode``. + + If this function seems to hang, it is probably because of a hard + simplification. + + To use this function to test, test the first item of the tuple. + + Examples + ======== + + >>> from sympy import (Eq, Function, checkodesol, symbols, + ... Derivative, exp) + >>> x, C1, C2 = symbols('x,C1,C2') + >>> f, g = symbols('f g', cls=Function) + >>> checkodesol(f(x).diff(x), Eq(f(x), C1)) + (True, 0) + >>> assert checkodesol(f(x).diff(x), C1)[0] + >>> assert not checkodesol(f(x).diff(x), x)[0] + >>> checkodesol(f(x).diff(x, 2), x**2) + (False, 2) + + >>> eqs = [Eq(Derivative(f(x), x), f(x)), Eq(Derivative(g(x), x), g(x))] + >>> sol = [Eq(f(x), C1*exp(x)), Eq(g(x), C2*exp(x))] + >>> checkodesol(eqs, sol) + (True, [0, 0]) + + """ + if iterable(ode): + return checksysodesol(ode, sol, func=func) + + if not isinstance(ode, Equality): + ode = Eq(ode, 0) + if func is None: + try: + _, func = _preprocess(ode.lhs) + except ValueError: + funcs = [s.atoms(AppliedUndef) for s in ( + sol if is_sequence(sol, set) else [sol])] + funcs = set().union(*funcs) + if len(funcs) != 1: + raise ValueError( + 'must pass func arg to checkodesol for this case.') + func = funcs.pop() + if not isinstance(func, AppliedUndef) or len(func.args) != 1: + raise ValueError( + "func must be a function of one variable, not %s" % func) + if is_sequence(sol, set): + return type(sol)([checkodesol(ode, i, order=order, solve_for_func=solve_for_func) for i in sol]) + + if not isinstance(sol, Equality): + sol = Eq(func, sol) + elif sol.rhs == func: + sol = sol.reversed + + if order == 'auto': + order = ode_order(ode, func) + solved = sol.lhs == func and not sol.rhs.has(func) + if solve_for_func and not solved: + rhs = solve(sol, func) + if rhs: + eqs = [Eq(func, t) for t in rhs] + if len(rhs) == 1: + eqs = eqs[0] + return checkodesol(ode, eqs, order=order, + solve_for_func=False) + + x = func.args[0] + + # Handle series solutions here + if sol.has(Order): + assert sol.lhs == func + Oterm = sol.rhs.getO() + solrhs = sol.rhs.removeO() + + Oexpr = Oterm.expr + assert isinstance(Oexpr, Pow) + sorder = Oexpr.exp + assert Oterm == Order(x**sorder) + + odesubs = (ode.lhs-ode.rhs).subs(func, solrhs).doit().expand() + + neworder = Order(x**(sorder - order)) + odesubs = odesubs + neworder + assert odesubs.getO() == neworder + residual = odesubs.removeO() + + return (residual == 0, residual) + + s = True + testnum = 0 + while s: + if testnum == 0: + # First pass, try substituting a solved solution directly into the + # ODE. This has the highest chance of succeeding. + ode_diff = ode.lhs - ode.rhs + + if sol.lhs == func: + s = sub_func_doit(ode_diff, func, sol.rhs) + s = besselsimp(s) + else: + testnum += 1 + continue + ss = simplify(s.rewrite(exp)) + if ss: + # with the new numer_denom in power.py, if we do a simple + # expansion then testnum == 0 verifies all solutions. + s = ss.expand(force=True) + else: + s = 0 + testnum += 1 + elif testnum == 1: + # Second pass. If we cannot substitute f, try seeing if the nth + # derivative is equal, this will only work for odes that are exact, + # by definition. + s = simplify( + trigsimp(diff(sol.lhs, x, order) - diff(sol.rhs, x, order)) - + trigsimp(ode.lhs) + trigsimp(ode.rhs)) + # s2 = simplify( + # diff(sol.lhs, x, order) - diff(sol.rhs, x, order) - \ + # ode.lhs + ode.rhs) + testnum += 1 + elif testnum == 2: + # Third pass. Try solving for df/dx and substituting that into the + # ODE. Thanks to Chris Smith for suggesting this method. Many of + # the comments below are his, too. + # The method: + # - Take each of 1..n derivatives of the solution. + # - Solve each nth derivative for d^(n)f/dx^(n) + # (the differential of that order) + # - Back substitute into the ODE in decreasing order + # (i.e., n, n-1, ...) + # - Check the result for zero equivalence + if sol.lhs == func and not sol.rhs.has(func): + diffsols = {0: sol.rhs} + elif sol.rhs == func and not sol.lhs.has(func): + diffsols = {0: sol.lhs} + else: + diffsols = {} + sol = sol.lhs - sol.rhs + for i in range(1, order + 1): + # Differentiation is a linear operator, so there should always + # be 1 solution. Nonetheless, we test just to make sure. + # We only need to solve once. After that, we automatically + # have the solution to the differential in the order we want. + if i == 1: + ds = sol.diff(x) + try: + sdf = solve(ds, func.diff(x, i)) + if not sdf: + raise NotImplementedError + except NotImplementedError: + testnum += 1 + break + else: + diffsols[i] = sdf[0] + else: + # This is what the solution says df/dx should be. + diffsols[i] = diffsols[i - 1].diff(x) + + # Make sure the above didn't fail. + if testnum > 2: + continue + else: + # Substitute it into ODE to check for self consistency. + lhs, rhs = ode.lhs, ode.rhs + for i in range(order, -1, -1): + if i == 0 and 0 not in diffsols: + # We can only substitute f(x) if the solution was + # solved for f(x). + break + lhs = sub_func_doit(lhs, func.diff(x, i), diffsols[i]) + rhs = sub_func_doit(rhs, func.diff(x, i), diffsols[i]) + ode_or_bool = Eq(lhs, rhs) + ode_or_bool = simplify(ode_or_bool) + + if isinstance(ode_or_bool, (bool, BooleanAtom)): + if ode_or_bool: + lhs = rhs = S.Zero + else: + lhs = ode_or_bool.lhs + rhs = ode_or_bool.rhs + # No sense in overworking simplify -- just prove that the + # numerator goes to zero + num = trigsimp((lhs - rhs).as_numer_denom()[0]) + # since solutions are obtained using force=True we test + # using the same level of assumptions + ## replace function with dummy so assumptions will work + _func = Dummy('func') + num = num.subs(func, _func) + ## posify the expression + num, reps = posify(num) + s = simplify(num).xreplace(reps).xreplace({_func: func}) + testnum += 1 + else: + break + + if not s: + return (True, s) + elif s is True: # The code above never was able to change s + raise NotImplementedError("Unable to test if " + str(sol) + + " is a solution to " + str(ode) + ".") + else: + return (False, s) + + +def checksysodesol(eqs, sols, func=None): + r""" + Substitutes corresponding ``sols`` for each functions into each ``eqs`` and + checks that the result of substitutions for each equation is ``0``. The + equations and solutions passed can be any iterable. + + This only works when each ``sols`` have one function only, like `x(t)` or `y(t)`. + For each function, ``sols`` can have a single solution or a list of solutions. + In most cases it will not be necessary to explicitly identify the function, + but if the function cannot be inferred from the original equation it + can be supplied through the ``func`` argument. + + When a sequence of equations is passed, the same sequence is used to return + the result for each equation with each function substituted with corresponding + solutions. + + It tries the following method to find zero equivalence for each equation: + + Substitute the solutions for functions, like `x(t)` and `y(t)` into the + original equations containing those functions. + This function returns a tuple. The first item in the tuple is ``True`` if + the substitution results for each equation is ``0``, and ``False`` otherwise. + The second item in the tuple is what the substitution results in. Each element + of the ``list`` should always be ``0`` corresponding to each equation if the + first item is ``True``. Note that sometimes this function may return ``False``, + but with an expression that is identically equal to ``0``, instead of returning + ``True``. This is because :py:meth:`~sympy.simplify.simplify.simplify` cannot + reduce the expression to ``0``. If an expression returned by each function + vanishes identically, then ``sols`` really is a solution to ``eqs``. + + If this function seems to hang, it is probably because of a difficult simplification. + + Examples + ======== + + >>> from sympy import Eq, diff, symbols, sin, cos, exp, sqrt, S, Function + >>> from sympy.solvers.ode.subscheck import checksysodesol + >>> C1, C2 = symbols('C1:3') + >>> t = symbols('t') + >>> x, y = symbols('x, y', cls=Function) + >>> eq = (Eq(diff(x(t),t), x(t) + y(t) + 17), Eq(diff(y(t),t), -2*x(t) + y(t) + 12)) + >>> sol = [Eq(x(t), (C1*sin(sqrt(2)*t) + C2*cos(sqrt(2)*t))*exp(t) - S(5)/3), + ... Eq(y(t), (sqrt(2)*C1*cos(sqrt(2)*t) - sqrt(2)*C2*sin(sqrt(2)*t))*exp(t) - S(46)/3)] + >>> checksysodesol(eq, sol) + (True, [0, 0]) + >>> eq = (Eq(diff(x(t),t),x(t)*y(t)**4), Eq(diff(y(t),t),y(t)**3)) + >>> sol = [Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), -sqrt(2)*sqrt(-1/(C2 + t))/2), + ... Eq(x(t), C1*exp(-1/(4*(C2 + t)))), Eq(y(t), sqrt(2)*sqrt(-1/(C2 + t))/2)] + >>> checksysodesol(eq, sol) + (True, [0, 0]) + + """ + def _sympify(eq): + return list(map(sympify, eq if iterable(eq) else [eq])) + eqs = _sympify(eqs) + for i in range(len(eqs)): + if isinstance(eqs[i], Equality): + eqs[i] = eqs[i].lhs - eqs[i].rhs + if func is None: + funcs = [] + for eq in eqs: + derivs = eq.atoms(Derivative) + func = set().union(*[d.atoms(AppliedUndef) for d in derivs]) + funcs.extend(func) + funcs = list(set(funcs)) + if not all(isinstance(func, AppliedUndef) and len(func.args) == 1 for func in funcs)\ + and len({func.args for func in funcs})!=1: + raise ValueError("func must be a function of one variable, not %s" % func) + for sol in sols: + if len(sol.atoms(AppliedUndef)) != 1: + raise ValueError("solutions should have one function only") + if len(funcs) != len({sol.lhs for sol in sols}): + raise ValueError("number of solutions provided does not match the number of equations") + dictsol = {} + for sol in sols: + func = list(sol.atoms(AppliedUndef))[0] + if sol.rhs == func: + sol = sol.reversed + solved = sol.lhs == func and not sol.rhs.has(func) + if not solved: + rhs = solve(sol, func) + if not rhs: + raise NotImplementedError + else: + rhs = sol.rhs + dictsol[func] = rhs + checkeq = [] + for eq in eqs: + for func in funcs: + eq = sub_func_doit(eq, func, dictsol[func]) + ss = simplify(eq) + if ss != 0: + eq = ss.expand(force=True) + if eq != 0: + eq = sqrtdenest(eq).simplify() + else: + eq = 0 + checkeq.append(eq) + if len(set(checkeq)) == 1 and list(set(checkeq))[0] == 0: + return (True, checkeq) + else: + return (False, checkeq) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/systems.py b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/systems.py new file mode 100644 index 0000000000000000000000000000000000000000..2d2c9b57a969c7fb5c67c06ce952fa398e22a48d --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/systems.py @@ -0,0 +1,2135 @@ +from sympy.core import Add, Mul, S +from sympy.core.containers import Tuple +from sympy.core.exprtools import factor_terms +from sympy.core.numbers import I +from sympy.core.relational import Eq, Equality +from sympy.core.sorting import default_sort_key, ordered +from sympy.core.symbol import Dummy, Symbol +from sympy.core.function import (expand_mul, expand, Derivative, + AppliedUndef, Function, Subs) +from sympy.functions import (exp, im, cos, sin, re, Piecewise, + piecewise_fold, sqrt, log) +from sympy.functions.combinatorial.factorials import factorial +from sympy.matrices import zeros, Matrix, NonSquareMatrixError, MatrixBase, eye +from sympy.polys import Poly, together +from sympy.simplify import collect, radsimp, signsimp # type: ignore +from sympy.simplify.powsimp import powdenest, powsimp +from sympy.simplify.ratsimp import ratsimp +from sympy.simplify.simplify import simplify +from sympy.sets.sets import FiniteSet +from sympy.solvers.deutils import ode_order +from sympy.solvers.solveset import NonlinearError, solveset +from sympy.utilities.iterables import (connected_components, iterable, + strongly_connected_components) +from sympy.utilities.misc import filldedent +from sympy.integrals.integrals import Integral, integrate + + +def _get_func_order(eqs, funcs): + return {func: max(ode_order(eq, func) for eq in eqs) for func in funcs} + + +class ODEOrderError(ValueError): + """Raised by linear_ode_to_matrix if the system has the wrong order""" + pass + + +class ODENonlinearError(NonlinearError): + """Raised by linear_ode_to_matrix if the system is nonlinear""" + pass + + +def _simpsol(soleq): + lhs = soleq.lhs + sol = soleq.rhs + sol = powsimp(sol) + gens = list(sol.atoms(exp)) + p = Poly(sol, *gens, expand=False) + gens = [factor_terms(g) for g in gens] + if not gens: + gens = p.gens + syms = [Symbol('C1'), Symbol('C2')] + terms = [] + for coeff, monom in zip(p.coeffs(), p.monoms()): + coeff = piecewise_fold(coeff) + if isinstance(coeff, Piecewise): + coeff = Piecewise(*((ratsimp(coef).collect(syms), cond) for coef, cond in coeff.args)) + else: + coeff = ratsimp(coeff).collect(syms) + monom = Mul(*(g ** i for g, i in zip(gens, monom))) + terms.append(coeff * monom) + return Eq(lhs, Add(*terms)) + + +def _solsimp(e, t): + no_t, has_t = powsimp(expand_mul(e)).as_independent(t) + + no_t = ratsimp(no_t) + has_t = has_t.replace(exp, lambda a: exp(factor_terms(a))) + + return no_t + has_t + + +def simpsol(sol, wrt1, wrt2, doit=True): + """Simplify solutions from dsolve_system.""" + + # The parameter sol is the solution as returned by dsolve (list of Eq). + # + # The parameters wrt1 and wrt2 are lists of symbols to be collected for + # with those in wrt1 being collected for first. This allows for collecting + # on any factors involving the independent variable before collecting on + # the integration constants or vice versa using e.g.: + # + # sol = simpsol(sol, [t], [C1, C2]) # t first, constants after + # sol = simpsol(sol, [C1, C2], [t]) # constants first, t after + # + # If doit=True (default) then simpsol will begin by evaluating any + # unevaluated integrals. Since many integrals will appear multiple times + # in the solutions this is done intelligently by computing each integral + # only once. + # + # The strategy is to first perform simple cancellation with factor_terms + # and then multiply out all brackets with expand_mul. This gives an Add + # with many terms. + # + # We split each term into two multiplicative factors dep and coeff where + # all factors that involve wrt1 are in dep and any constant factors are in + # coeff e.g. + # sqrt(2)*C1*exp(t) -> ( exp(t), sqrt(2)*C1 ) + # + # The dep factors are simplified using powsimp to combine expanded + # exponential factors e.g. + # exp(a*t)*exp(b*t) -> exp(t*(a+b)) + # + # We then collect coefficients for all terms having the same (simplified) + # dep. The coefficients are then simplified using together and ratsimp and + # lastly by recursively applying the same transformation to the + # coefficients to collect on wrt2. + # + # Finally the result is recombined into an Add and signsimp is used to + # normalise any minus signs. + + def simprhs(rhs, rep, wrt1, wrt2): + """Simplify the rhs of an ODE solution""" + if rep: + rhs = rhs.subs(rep) + rhs = factor_terms(rhs) + rhs = simp_coeff_dep(rhs, wrt1, wrt2) + rhs = signsimp(rhs) + return rhs + + def simp_coeff_dep(expr, wrt1, wrt2=None): + """Split rhs into terms, split terms into dep and coeff and collect on dep""" + add_dep_terms = lambda e: e.is_Add and e.has(*wrt1) + expandable = lambda e: e.is_Mul and any(map(add_dep_terms, e.args)) + expand_func = lambda e: expand_mul(e, deep=False) + expand_mul_mod = lambda e: e.replace(expandable, expand_func) + terms = Add.make_args(expand_mul_mod(expr)) + dc = {} + for term in terms: + coeff, dep = term.as_independent(*wrt1, as_Add=False) + # Collect together the coefficients for terms that have the same + # dependence on wrt1 (after dep is normalised using simpdep). + dep = simpdep(dep, wrt1) + + # See if the dependence on t cancels out... + if dep is not S.One: + dep2 = factor_terms(dep) + if not dep2.has(*wrt1): + coeff *= dep2 + dep = S.One + + if dep not in dc: + dc[dep] = coeff + else: + dc[dep] += coeff + # Apply the method recursively to the coefficients but this time + # collecting on wrt2 rather than wrt2. + termpairs = ((simpcoeff(c, wrt2), d) for d, c in dc.items()) + if wrt2 is not None: + termpairs = ((simp_coeff_dep(c, wrt2), d) for c, d in termpairs) + return Add(*(c * d for c, d in termpairs)) + + def simpdep(term, wrt1): + """Normalise factors involving t with powsimp and recombine exp""" + def canonicalise(a): + # Using factor_terms here isn't quite right because it leads to things + # like exp(t*(1+t)) that we don't want. We do want to cancel factors + # and pull out a common denominator but ideally the numerator would be + # expressed as a standard form polynomial in t so we expand_mul + # and collect afterwards. + a = factor_terms(a) + num, den = a.as_numer_denom() + num = expand_mul(num) + num = collect(num, wrt1) + return num / den + + term = powsimp(term) + rep = {e: exp(canonicalise(e.args[0])) for e in term.atoms(exp)} + term = term.subs(rep) + return term + + def simpcoeff(coeff, wrt2): + """Bring to a common fraction and cancel with ratsimp""" + coeff = together(coeff) + if coeff.is_polynomial(): + # Calling ratsimp can be expensive. The main reason is to simplify + # sums of terms with irrational denominators so we limit ourselves + # to the case where the expression is polynomial in any symbols. + # Maybe there's a better approach... + coeff = ratsimp(radsimp(coeff)) + # collect on secondary variables first and any remaining symbols after + if wrt2 is not None: + syms = list(wrt2) + list(ordered(coeff.free_symbols - set(wrt2))) + else: + syms = list(ordered(coeff.free_symbols)) + coeff = collect(coeff, syms) + coeff = together(coeff) + return coeff + + # There are often repeated integrals. Collect unique integrals and + # evaluate each once and then substitute into the final result to replace + # all occurrences in each of the solution equations. + if doit: + integrals = set().union(*(s.atoms(Integral) for s in sol)) + rep = {i: factor_terms(i).doit() for i in integrals} + else: + rep = {} + + sol = [Eq(s.lhs, simprhs(s.rhs, rep, wrt1, wrt2)) for s in sol] + return sol + + +def linodesolve_type(A, t, b=None): + r""" + Helper function that determines the type of the system of ODEs for solving with :obj:`sympy.solvers.ode.systems.linodesolve()` + + Explanation + =========== + + This function takes in the coefficient matrix and/or the non-homogeneous term + and returns the type of the equation that can be solved by :obj:`sympy.solvers.ode.systems.linodesolve()`. + + If the system is constant coefficient homogeneous, then "type1" is returned + + If the system is constant coefficient non-homogeneous, then "type2" is returned + + If the system is non-constant coefficient homogeneous, then "type3" is returned + + If the system is non-constant coefficient non-homogeneous, then "type4" is returned + + If the system has a non-constant coefficient matrix which can be factorized into constant + coefficient matrix, then "type5" or "type6" is returned for when the system is homogeneous or + non-homogeneous respectively. + + Note that, if the system of ODEs is of "type3" or "type4", then along with the type, + the commutative antiderivative of the coefficient matrix is also returned. + + If the system cannot be solved by :obj:`sympy.solvers.ode.systems.linodesolve()`, then + NotImplementedError is raised. + + Parameters + ========== + + A : Matrix + Coefficient matrix of the system of ODEs + b : Matrix or None + Non-homogeneous term of the system. The default value is None. + If this argument is None, then the system is assumed to be homogeneous. + + Examples + ======== + + >>> from sympy import symbols, Matrix + >>> from sympy.solvers.ode.systems import linodesolve_type + >>> t = symbols("t") + >>> A = Matrix([[1, 1], [2, 3]]) + >>> b = Matrix([t, 1]) + + >>> linodesolve_type(A, t) + {'antiderivative': None, 'type_of_equation': 'type1'} + + >>> linodesolve_type(A, t, b=b) + {'antiderivative': None, 'type_of_equation': 'type2'} + + >>> A_t = Matrix([[1, t], [-t, 1]]) + + >>> linodesolve_type(A_t, t) + {'antiderivative': Matrix([ + [ t, t**2/2], + [-t**2/2, t]]), 'type_of_equation': 'type3'} + + >>> linodesolve_type(A_t, t, b=b) + {'antiderivative': Matrix([ + [ t, t**2/2], + [-t**2/2, t]]), 'type_of_equation': 'type4'} + + >>> A_non_commutative = Matrix([[1, t], [t, -1]]) + >>> linodesolve_type(A_non_commutative, t) + Traceback (most recent call last): + ... + NotImplementedError: + The system does not have a commutative antiderivative, it cannot be + solved by linodesolve. + + Returns + ======= + + Dict + + Raises + ====== + + NotImplementedError + When the coefficient matrix does not have a commutative antiderivative + + See Also + ======== + + linodesolve: Function for which linodesolve_type gets the information + + """ + + match = {} + is_non_constant = not _matrix_is_constant(A, t) + is_non_homogeneous = not (b is None or b.is_zero_matrix) + type = "type{}".format(int("{}{}".format(int(is_non_constant), int(is_non_homogeneous)), 2) + 1) + + B = None + match.update({"type_of_equation": type, "antiderivative": B}) + + if is_non_constant: + B, is_commuting = _is_commutative_anti_derivative(A, t) + if not is_commuting: + raise NotImplementedError(filldedent(''' + The system does not have a commutative antiderivative, it cannot be solved + by linodesolve. + ''')) + + match['antiderivative'] = B + match.update(_first_order_type5_6_subs(A, t, b=b)) + + return match + + +def _first_order_type5_6_subs(A, t, b=None): + match = {} + + factor_terms = _factor_matrix(A, t) + is_homogeneous = b is None or b.is_zero_matrix + + if factor_terms is not None: + t_ = Symbol("{}_".format(t)) + F_t = integrate(factor_terms[0], t) + inverse = solveset(Eq(t_, F_t), t) + + # Note: A simple way to check if a function is invertible + # or not. + if isinstance(inverse, FiniteSet) and not inverse.has(Piecewise)\ + and len(inverse) == 1: + + A = factor_terms[1] + if not is_homogeneous: + b = b / factor_terms[0] + b = b.subs(t, list(inverse)[0]) + type = "type{}".format(5 + (not is_homogeneous)) + match.update({'func_coeff': A, 'tau': F_t, + 't_': t_, 'type_of_equation': type, 'rhs': b}) + + return match + + +def linear_ode_to_matrix(eqs, funcs, t, order): + r""" + Convert a linear system of ODEs to matrix form + + Explanation + =========== + + Express a system of linear ordinary differential equations as a single + matrix differential equation [1]. For example the system $x' = x + y + 1$ + and $y' = x - y$ can be represented as + + .. math:: A_1 X' = A_0 X + b + + where $A_1$ and $A_0$ are $2 \times 2$ matrices and $b$, $X$ and $X'$ are + $2 \times 1$ matrices with $X = [x, y]^T$. + + Higher-order systems are represented with additional matrices e.g. a + second-order system would look like + + .. math:: A_2 X'' = A_1 X' + A_0 X + b + + Examples + ======== + + >>> from sympy import Function, Symbol, Matrix, Eq + >>> from sympy.solvers.ode.systems import linear_ode_to_matrix + >>> t = Symbol('t') + >>> x = Function('x') + >>> y = Function('y') + + We can create a system of linear ODEs like + + >>> eqs = [ + ... Eq(x(t).diff(t), x(t) + y(t) + 1), + ... Eq(y(t).diff(t), x(t) - y(t)), + ... ] + >>> funcs = [x(t), y(t)] + >>> order = 1 # 1st order system + + Now ``linear_ode_to_matrix`` can represent this as a matrix + differential equation. + + >>> (A1, A0), b = linear_ode_to_matrix(eqs, funcs, t, order) + >>> A1 + Matrix([ + [1, 0], + [0, 1]]) + >>> A0 + Matrix([ + [1, 1], + [1, -1]]) + >>> b + Matrix([ + [1], + [0]]) + + The original equations can be recovered from these matrices: + + >>> eqs_mat = Matrix([eq.lhs - eq.rhs for eq in eqs]) + >>> X = Matrix(funcs) + >>> A1 * X.diff(t) - A0 * X - b == eqs_mat + True + + If the system of equations has a maximum order greater than the + order of the system specified, a ODEOrderError exception is raised. + + >>> eqs = [Eq(x(t).diff(t, 2), x(t).diff(t) + x(t)), Eq(y(t).diff(t), y(t) + x(t))] + >>> linear_ode_to_matrix(eqs, funcs, t, 1) + Traceback (most recent call last): + ... + ODEOrderError: Cannot represent system in 1-order form + + If the system of equations is nonlinear, then ODENonlinearError is + raised. + + >>> eqs = [Eq(x(t).diff(t), x(t) + y(t)), Eq(y(t).diff(t), y(t)**2 + x(t))] + >>> linear_ode_to_matrix(eqs, funcs, t, 1) + Traceback (most recent call last): + ... + ODENonlinearError: The system of ODEs is nonlinear. + + Parameters + ========== + + eqs : list of SymPy expressions or equalities + The equations as expressions (assumed equal to zero). + funcs : list of applied functions + The dependent variables of the system of ODEs. + t : symbol + The independent variable. + order : int + The order of the system of ODEs. + + Returns + ======= + + The tuple ``(As, b)`` where ``As`` is a tuple of matrices and ``b`` is the + the matrix representing the rhs of the matrix equation. + + Raises + ====== + + ODEOrderError + When the system of ODEs have an order greater than what was specified + ODENonlinearError + When the system of ODEs is nonlinear + + See Also + ======== + + linear_eq_to_matrix: for systems of linear algebraic equations. + + References + ========== + + .. [1] https://en.wikipedia.org/wiki/Matrix_differential_equation + + """ + from sympy.solvers.solveset import linear_eq_to_matrix + + if any(ode_order(eq, func) > order for eq in eqs for func in funcs): + msg = "Cannot represent system in {}-order form" + raise ODEOrderError(msg.format(order)) + + As = [] + + for o in range(order, -1, -1): + # Work from the highest derivative down + syms = [func.diff(t, o) for func in funcs] + + # Ai is the matrix for X(t).diff(t, o) + # eqs is minus the remainder of the equations. + try: + Ai, b = linear_eq_to_matrix(eqs, syms) + except NonlinearError: + raise ODENonlinearError("The system of ODEs is nonlinear.") + + Ai = Ai.applyfunc(expand_mul) + + As.append(Ai if o == order else -Ai) + + if o: + eqs = [-eq for eq in b] + else: + rhs = b + + return As, rhs + + +def matrix_exp(A, t): + r""" + Matrix exponential $\exp(A*t)$ for the matrix ``A`` and scalar ``t``. + + Explanation + =========== + + This functions returns the $\exp(A*t)$ by doing a simple + matrix multiplication: + + .. math:: \exp(A*t) = P * expJ * P^{-1} + + where $expJ$ is $\exp(J*t)$. $J$ is the Jordan normal + form of $A$ and $P$ is matrix such that: + + .. math:: A = P * J * P^{-1} + + The matrix exponential $\exp(A*t)$ appears in the solution of linear + differential equations. For example if $x$ is a vector and $A$ is a matrix + then the initial value problem + + .. math:: \frac{dx(t)}{dt} = A \times x(t), x(0) = x0 + + has the unique solution + + .. math:: x(t) = \exp(A t) x0 + + Examples + ======== + + >>> from sympy import Symbol, Matrix, pprint + >>> from sympy.solvers.ode.systems import matrix_exp + >>> t = Symbol('t') + + We will consider a 2x2 matrix for comupting the exponential + + >>> A = Matrix([[2, -5], [2, -4]]) + >>> pprint(A) + [2 -5] + [ ] + [2 -4] + + Now, exp(A*t) is given as follows: + + >>> pprint(matrix_exp(A, t)) + [ -t -t -t ] + [3*e *sin(t) + e *cos(t) -5*e *sin(t) ] + [ ] + [ -t -t -t ] + [ 2*e *sin(t) - 3*e *sin(t) + e *cos(t)] + + Parameters + ========== + + A : Matrix + The matrix $A$ in the expression $\exp(A*t)$ + t : Symbol + The independent variable + + See Also + ======== + + matrix_exp_jordan_form: For exponential of Jordan normal form + + References + ========== + + .. [1] https://en.wikipedia.org/wiki/Jordan_normal_form + .. [2] https://en.wikipedia.org/wiki/Matrix_exponential + + """ + P, expJ = matrix_exp_jordan_form(A, t) + return P * expJ * P.inv() + + +def matrix_exp_jordan_form(A, t): + r""" + Matrix exponential $\exp(A*t)$ for the matrix *A* and scalar *t*. + + Explanation + =========== + + Returns the Jordan form of the $\exp(A*t)$ along with the matrix $P$ such that: + + .. math:: + \exp(A*t) = P * expJ * P^{-1} + + Examples + ======== + + >>> from sympy import Matrix, Symbol + >>> from sympy.solvers.ode.systems import matrix_exp, matrix_exp_jordan_form + >>> t = Symbol('t') + + We will consider a 2x2 defective matrix. This shows that our method + works even for defective matrices. + + >>> A = Matrix([[1, 1], [0, 1]]) + + It can be observed that this function gives us the Jordan normal form + and the required invertible matrix P. + + >>> P, expJ = matrix_exp_jordan_form(A, t) + + Here, it is shown that P and expJ returned by this function is correct + as they satisfy the formula: P * expJ * P_inverse = exp(A*t). + + >>> P * expJ * P.inv() == matrix_exp(A, t) + True + + Parameters + ========== + + A : Matrix + The matrix $A$ in the expression $\exp(A*t)$ + t : Symbol + The independent variable + + References + ========== + + .. [1] https://en.wikipedia.org/wiki/Defective_matrix + .. [2] https://en.wikipedia.org/wiki/Jordan_matrix + .. [3] https://en.wikipedia.org/wiki/Jordan_normal_form + + """ + + N, M = A.shape + if N != M: + raise ValueError('Needed square matrix but got shape (%s, %s)' % (N, M)) + elif A.has(t): + raise ValueError('Matrix A should not depend on t') + + def jordan_chains(A): + '''Chains from Jordan normal form analogous to M.eigenvects(). + Returns a dict with eignevalues as keys like: + {e1: [[v111,v112,...], [v121, v122,...]], e2:...} + where vijk is the kth vector in the jth chain for eigenvalue i. + ''' + P, blocks = A.jordan_cells() + basis = [P[:,i] for i in range(P.shape[1])] + n = 0 + chains = {} + for b in blocks: + eigval = b[0, 0] + size = b.shape[0] + if eigval not in chains: + chains[eigval] = [] + chains[eigval].append(basis[n:n+size]) + n += size + return chains + + eigenchains = jordan_chains(A) + + # Needed for consistency across Python versions + eigenchains_iter = sorted(eigenchains.items(), key=default_sort_key) + isreal = not A.has(I) + + blocks = [] + vectors = [] + seen_conjugate = set() + for e, chains in eigenchains_iter: + for chain in chains: + n = len(chain) + if isreal and e != e.conjugate() and e.conjugate() in eigenchains: + if e in seen_conjugate: + continue + seen_conjugate.add(e.conjugate()) + exprt = exp(re(e) * t) + imrt = im(e) * t + imblock = Matrix([[cos(imrt), sin(imrt)], + [-sin(imrt), cos(imrt)]]) + expJblock2 = Matrix(n, n, lambda i,j: + imblock * t**(j-i) / factorial(j-i) if j >= i + else zeros(2, 2)) + expJblock = Matrix(2*n, 2*n, lambda i,j: expJblock2[i//2,j//2][i%2,j%2]) + + blocks.append(exprt * expJblock) + for i in range(n): + vectors.append(re(chain[i])) + vectors.append(im(chain[i])) + else: + vectors.extend(chain) + fun = lambda i,j: t**(j-i)/factorial(j-i) if j >= i else 0 + expJblock = Matrix(n, n, fun) + blocks.append(exp(e * t) * expJblock) + + expJ = Matrix.diag(*blocks) + P = Matrix(N, N, lambda i,j: vectors[j][i]) + + return P, expJ + + +# Note: To add a docstring example with tau +def linodesolve(A, t, b=None, B=None, type="auto", doit=False, + tau=None): + r""" + System of n equations linear first-order differential equations + + Explanation + =========== + + This solver solves the system of ODEs of the following form: + + .. math:: + X'(t) = A(t) X(t) + b(t) + + Here, $A(t)$ is the coefficient matrix, $X(t)$ is the vector of n independent variables, + $b(t)$ is the non-homogeneous term and $X'(t)$ is the derivative of $X(t)$ + + Depending on the properties of $A(t)$ and $b(t)$, this solver evaluates the solution + differently. + + When $A(t)$ is constant coefficient matrix and $b(t)$ is zero vector i.e. system is homogeneous, + the system is "type1". The solution is: + + .. math:: + X(t) = \exp(A t) C + + Here, $C$ is a vector of constants and $A$ is the constant coefficient matrix. + + When $A(t)$ is constant coefficient matrix and $b(t)$ is non-zero i.e. system is non-homogeneous, + the system is "type2". The solution is: + + .. math:: + X(t) = e^{A t} ( \int e^{- A t} b \,dt + C) + + When $A(t)$ is coefficient matrix such that its commutative with its antiderivative $B(t)$ and + $b(t)$ is a zero vector i.e. system is homogeneous, the system is "type3". The solution is: + + .. math:: + X(t) = \exp(B(t)) C + + When $A(t)$ is commutative with its antiderivative $B(t)$ and $b(t)$ is non-zero i.e. system is + non-homogeneous, the system is "type4". The solution is: + + .. math:: + X(t) = e^{B(t)} ( \int e^{-B(t)} b(t) \,dt + C) + + When $A(t)$ is a coefficient matrix such that it can be factorized into a scalar and a constant + coefficient matrix: + + .. math:: + A(t) = f(t) * A + + Where $f(t)$ is a scalar expression in the independent variable $t$ and $A$ is a constant matrix, + then we can do the following substitutions: + + .. math:: + tau = \int f(t) dt, X(t) = Y(tau), b(t) = b(f^{-1}(tau)) + + Here, the substitution for the non-homogeneous term is done only when its non-zero. + Using these substitutions, our original system becomes: + + .. math:: + Y'(tau) = A * Y(tau) + b(tau)/f(tau) + + The above system can be easily solved using the solution for "type1" or "type2" depending + on the homogeneity of the system. After we get the solution for $Y(tau)$, we substitute the + solution for $tau$ as $t$ to get back $X(t)$ + + .. math:: + X(t) = Y(tau) + + Systems of "type5" and "type6" have a commutative antiderivative but we use this solution + because its faster to compute. + + The final solution is the general solution for all the four equations since a constant coefficient + matrix is always commutative with its antidervative. + + An additional feature of this function is, if someone wants to substitute for value of the independent + variable, they can pass the substitution `tau` and the solution will have the independent variable + substituted with the passed expression(`tau`). + + Parameters + ========== + + A : Matrix + Coefficient matrix of the system of linear first order ODEs. + t : Symbol + Independent variable in the system of ODEs. + b : Matrix or None + Non-homogeneous term in the system of ODEs. If None is passed, + a homogeneous system of ODEs is assumed. + B : Matrix or None + Antiderivative of the coefficient matrix. If the antiderivative + is not passed and the solution requires the term, then the solver + would compute it internally. + type : String + Type of the system of ODEs passed. Depending on the type, the + solution is evaluated. The type values allowed and the corresponding + system it solves are: "type1" for constant coefficient homogeneous + "type2" for constant coefficient non-homogeneous, "type3" for non-constant + coefficient homogeneous, "type4" for non-constant coefficient non-homogeneous, + "type5" and "type6" for non-constant coefficient homogeneous and non-homogeneous + systems respectively where the coefficient matrix can be factorized to a constant + coefficient matrix. + The default value is "auto" which will let the solver decide the correct type of + the system passed. + doit : Boolean + Evaluate the solution if True, default value is False + tau: Expression + Used to substitute for the value of `t` after we get the solution of the system. + + Examples + ======== + + To solve the system of ODEs using this function directly, several things must be + done in the right order. Wrong inputs to the function will lead to incorrect results. + + >>> from sympy import symbols, Function, Eq + >>> from sympy.solvers.ode.systems import canonical_odes, linear_ode_to_matrix, linodesolve, linodesolve_type + >>> from sympy.solvers.ode.subscheck import checkodesol + >>> f, g = symbols("f, g", cls=Function) + >>> x, a = symbols("x, a") + >>> funcs = [f(x), g(x)] + >>> eqs = [Eq(f(x).diff(x) - f(x), a*g(x) + 1), Eq(g(x).diff(x) + g(x), a*f(x))] + + Here, it is important to note that before we derive the coefficient matrix, it is + important to get the system of ODEs into the desired form. For that we will use + :obj:`sympy.solvers.ode.systems.canonical_odes()`. + + >>> eqs = canonical_odes(eqs, funcs, x) + >>> eqs + [[Eq(Derivative(f(x), x), a*g(x) + f(x) + 1), Eq(Derivative(g(x), x), a*f(x) - g(x))]] + + Now, we will use :obj:`sympy.solvers.ode.systems.linear_ode_to_matrix()` to get the coefficient matrix and the + non-homogeneous term if it is there. + + >>> eqs = eqs[0] + >>> (A1, A0), b = linear_ode_to_matrix(eqs, funcs, x, 1) + >>> A = A0 + + We have the coefficient matrices and the non-homogeneous term ready. Now, we can use + :obj:`sympy.solvers.ode.systems.linodesolve_type()` to get the information for the system of ODEs + to finally pass it to the solver. + + >>> system_info = linodesolve_type(A, x, b=b) + >>> sol_vector = linodesolve(A, x, b=b, B=system_info['antiderivative'], type=system_info['type_of_equation']) + + Now, we can prove if the solution is correct or not by using :obj:`sympy.solvers.ode.checkodesol()` + + >>> sol = [Eq(f, s) for f, s in zip(funcs, sol_vector)] + >>> checkodesol(eqs, sol) + (True, [0, 0]) + + We can also use the doit method to evaluate the solutions passed by the function. + + >>> sol_vector_evaluated = linodesolve(A, x, b=b, type="type2", doit=True) + + Now, we will look at a system of ODEs which is non-constant. + + >>> eqs = [Eq(f(x).diff(x), f(x) + x*g(x)), Eq(g(x).diff(x), -x*f(x) + g(x))] + + The system defined above is already in the desired form, so we do not have to convert it. + + >>> (A1, A0), b = linear_ode_to_matrix(eqs, funcs, x, 1) + >>> A = A0 + + A user can also pass the commutative antiderivative required for type3 and type4 system of ODEs. + Passing an incorrect one will lead to incorrect results. If the coefficient matrix is not commutative + with its antiderivative, then :obj:`sympy.solvers.ode.systems.linodesolve_type()` raises a NotImplementedError. + If it does have a commutative antiderivative, then the function just returns the information about the system. + + >>> system_info = linodesolve_type(A, x, b=b) + + Now, we can pass the antiderivative as an argument to get the solution. If the system information is not + passed, then the solver will compute the required arguments internally. + + >>> sol_vector = linodesolve(A, x, b=b) + + Once again, we can verify the solution obtained. + + >>> sol = [Eq(f, s) for f, s in zip(funcs, sol_vector)] + >>> checkodesol(eqs, sol) + (True, [0, 0]) + + Returns + ======= + + List + + Raises + ====== + + ValueError + This error is raised when the coefficient matrix, non-homogeneous term + or the antiderivative, if passed, are not a matrix or + do not have correct dimensions + NonSquareMatrixError + When the coefficient matrix or its antiderivative, if passed is not a + square matrix + NotImplementedError + If the coefficient matrix does not have a commutative antiderivative + + See Also + ======== + + linear_ode_to_matrix: Coefficient matrix computation function + canonical_odes: System of ODEs representation change + linodesolve_type: Getting information about systems of ODEs to pass in this solver + + """ + + if not isinstance(A, MatrixBase): + raise ValueError(filldedent('''\ + The coefficients of the system of ODEs should be of type Matrix + ''')) + + if not A.is_square: + raise NonSquareMatrixError(filldedent('''\ + The coefficient matrix must be a square + ''')) + + if b is not None: + if not isinstance(b, MatrixBase): + raise ValueError(filldedent('''\ + The non-homogeneous terms of the system of ODEs should be of type Matrix + ''')) + + if A.rows != b.rows: + raise ValueError(filldedent('''\ + The system of ODEs should have the same number of non-homogeneous terms and the number of + equations + ''')) + + if B is not None: + if not isinstance(B, MatrixBase): + raise ValueError(filldedent('''\ + The antiderivative of coefficients of the system of ODEs should be of type Matrix + ''')) + + if not B.is_square: + raise NonSquareMatrixError(filldedent('''\ + The antiderivative of the coefficient matrix must be a square + ''')) + + if A.rows != B.rows: + raise ValueError(filldedent('''\ + The coefficient matrix and its antiderivative should have same dimensions + ''')) + + if not any(type == "type{}".format(i) for i in range(1, 7)) and not type == "auto": + raise ValueError(filldedent('''\ + The input type should be a valid one + ''')) + + n = A.rows + + # constants = numbered_symbols(prefix='C', cls=Dummy, start=const_idx+1) + Cvect = Matrix([Dummy() for _ in range(n)]) + + if b is None and any(type == typ for typ in ["type2", "type4", "type6"]): + b = zeros(n, 1) + + is_transformed = tau is not None + passed_type = type + + if type == "auto": + system_info = linodesolve_type(A, t, b=b) + type = system_info["type_of_equation"] + B = system_info["antiderivative"] + + if type in ("type5", "type6"): + is_transformed = True + if passed_type != "auto": + if tau is None: + system_info = _first_order_type5_6_subs(A, t, b=b) + if not system_info: + raise ValueError(filldedent(''' + The system passed isn't {}. + '''.format(type))) + + tau = system_info['tau'] + t = system_info['t_'] + A = system_info['A'] + b = system_info['b'] + + intx_wrtt = lambda x: Integral(x, t) if x else 0 + if type in ("type1", "type2", "type5", "type6"): + P, J = matrix_exp_jordan_form(A, t) + P = simplify(P) + + if type in ("type1", "type5"): + sol_vector = P * (J * Cvect) + else: + Jinv = J.subs(t, -t) + sol_vector = P * J * ((Jinv * P.inv() * b).applyfunc(intx_wrtt) + Cvect) + else: + if B is None: + B, _ = _is_commutative_anti_derivative(A, t) + + if type == "type3": + sol_vector = B.exp() * Cvect + else: + sol_vector = B.exp() * (((-B).exp() * b).applyfunc(intx_wrtt) + Cvect) + + if is_transformed: + sol_vector = sol_vector.subs(t, tau) + + gens = sol_vector.atoms(exp) + + if type != "type1": + sol_vector = [expand_mul(s) for s in sol_vector] + + sol_vector = [collect(s, ordered(gens), exact=True) for s in sol_vector] + + if doit: + sol_vector = [s.doit() for s in sol_vector] + + return sol_vector + + +def _matrix_is_constant(M, t): + """Checks if the matrix M is independent of t or not.""" + return all(coef.as_independent(t, as_Add=True)[1] == 0 for coef in M) + + +def canonical_odes(eqs, funcs, t): + r""" + Function that solves for highest order derivatives in a system + + Explanation + =========== + + This function inputs a system of ODEs and based on the system, + the dependent variables and their highest order, returns the system + in the following form: + + .. math:: + X'(t) = A(t) X(t) + b(t) + + Here, $X(t)$ is the vector of dependent variables of lower order, $A(t)$ is + the coefficient matrix, $b(t)$ is the non-homogeneous term and $X'(t)$ is the + vector of dependent variables in their respective highest order. We use the term + canonical form to imply the system of ODEs which is of the above form. + + If the system passed has a non-linear term with multiple solutions, then a list of + systems is returned in its canonical form. + + Parameters + ========== + + eqs : List + List of the ODEs + funcs : List + List of dependent variables + t : Symbol + Independent variable + + Examples + ======== + + >>> from sympy import symbols, Function, Eq, Derivative + >>> from sympy.solvers.ode.systems import canonical_odes + >>> f, g = symbols("f g", cls=Function) + >>> x, y = symbols("x y") + >>> funcs = [f(x), g(x)] + >>> eqs = [Eq(f(x).diff(x) - 7*f(x), 12*g(x)), Eq(g(x).diff(x) + g(x), 20*f(x))] + + >>> canonical_eqs = canonical_odes(eqs, funcs, x) + >>> canonical_eqs + [[Eq(Derivative(f(x), x), 7*f(x) + 12*g(x)), Eq(Derivative(g(x), x), 20*f(x) - g(x))]] + + >>> system = [Eq(Derivative(f(x), x)**2 - 2*Derivative(f(x), x) + 1, 4), Eq(-y*f(x) + Derivative(g(x), x), 0)] + + >>> canonical_system = canonical_odes(system, funcs, x) + >>> canonical_system + [[Eq(Derivative(f(x), x), -1), Eq(Derivative(g(x), x), y*f(x))], [Eq(Derivative(f(x), x), 3), Eq(Derivative(g(x), x), y*f(x))]] + + Returns + ======= + + List + + """ + from sympy.solvers.solvers import solve + + order = _get_func_order(eqs, funcs) + + canon_eqs = solve(eqs, *[func.diff(t, order[func]) for func in funcs], dict=True) + + systems = [] + for eq in canon_eqs: + system = [Eq(func.diff(t, order[func]), eq[func.diff(t, order[func])]) for func in funcs] + systems.append(system) + + return systems + + +def _is_commutative_anti_derivative(A, t): + r""" + Helper function for determining if the Matrix passed is commutative with its antiderivative + + Explanation + =========== + + This function checks if the Matrix $A$ passed is commutative with its antiderivative with respect + to the independent variable $t$. + + .. math:: + B(t) = \int A(t) dt + + The function outputs two values, first one being the antiderivative $B(t)$, second one being a + boolean value, if True, then the matrix $A(t)$ passed is commutative with $B(t)$, else the matrix + passed isn't commutative with $B(t)$. + + Parameters + ========== + + A : Matrix + The matrix which has to be checked + t : Symbol + Independent variable + + Examples + ======== + + >>> from sympy import symbols, Matrix + >>> from sympy.solvers.ode.systems import _is_commutative_anti_derivative + >>> t = symbols("t") + >>> A = Matrix([[1, t], [-t, 1]]) + + >>> B, is_commuting = _is_commutative_anti_derivative(A, t) + >>> is_commuting + True + + Returns + ======= + + Matrix, Boolean + + """ + B = integrate(A, t) + is_commuting = (B*A - A*B).applyfunc(expand).applyfunc(factor_terms).is_zero_matrix + + is_commuting = False if is_commuting is None else is_commuting + + return B, is_commuting + + +def _factor_matrix(A, t): + term = None + for element in A: + temp_term = element.as_independent(t)[1] + if temp_term.has(t): + term = temp_term + break + + if term is not None: + A_factored = (A/term).applyfunc(ratsimp) + can_factor = _matrix_is_constant(A_factored, t) + term = (term, A_factored) if can_factor else None + + return term + + +def _is_second_order_type2(A, t): + term = _factor_matrix(A, t) + is_type2 = False + + if term is not None: + term = 1/term[0] + is_type2 = term.is_polynomial() + + if is_type2: + poly = Poly(term.expand(), t) + monoms = poly.monoms() + + if monoms[0][0] in (2, 4): + cs = _get_poly_coeffs(poly, 4) + a, b, c, d, e = cs + + a1 = powdenest(sqrt(a), force=True) + c1 = powdenest(sqrt(e), force=True) + b1 = powdenest(sqrt(c - 2*a1*c1), force=True) + + is_type2 = (b == 2*a1*b1) and (d == 2*b1*c1) + term = a1*t**2 + b1*t + c1 + + else: + is_type2 = False + + return is_type2, term + + +def _get_poly_coeffs(poly, order): + cs = [0 for _ in range(order+1)] + for c, m in zip(poly.coeffs(), poly.monoms()): + cs[-1-m[0]] = c + return cs + + +def _match_second_order_type(A1, A0, t, b=None): + r""" + Works only for second order system in its canonical form. + + Type 0: Constant coefficient matrix, can be simply solved by + introducing dummy variables. + Type 1: When the substitution: $U = t*X' - X$ works for reducing + the second order system to first order system. + Type 2: When the system is of the form: $poly * X'' = A*X$ where + $poly$ is square of a quadratic polynomial with respect to + *t* and $A$ is a constant coefficient matrix. + + """ + match = {"type_of_equation": "type0"} + n = A1.shape[0] + + if _matrix_is_constant(A1, t) and _matrix_is_constant(A0, t): + return match + + if (A1 + A0*t).applyfunc(expand_mul).is_zero_matrix: + match.update({"type_of_equation": "type1", "A1": A1}) + + elif A1.is_zero_matrix and (b is None or b.is_zero_matrix): + is_type2, term = _is_second_order_type2(A0, t) + if is_type2: + a, b, c = _get_poly_coeffs(Poly(term, t), 2) + A = (A0*(term**2).expand()).applyfunc(ratsimp) + (b**2/4 - a*c)*eye(n, n) + tau = integrate(1/term, t) + t_ = Symbol("{}_".format(t)) + match.update({"type_of_equation": "type2", "A0": A, + "g(t)": sqrt(term), "tau": tau, "is_transformed": True, + "t_": t_}) + + return match + + +def _second_order_subs_type1(A, b, funcs, t): + r""" + For a linear, second order system of ODEs, a particular substitution. + + A system of the below form can be reduced to a linear first order system of + ODEs: + .. math:: + X'' = A(t) * (t*X' - X) + b(t) + + By substituting: + .. math:: U = t*X' - X + + To get the system: + .. math:: U' = t*(A(t)*U + b(t)) + + Where $U$ is the vector of dependent variables, $X$ is the vector of dependent + variables in `funcs` and $X'$ is the first order derivative of $X$ with respect to + $t$. It may or may not reduce the system into linear first order system of ODEs. + + Then a check is made to determine if the system passed can be reduced or not, if + this substitution works, then the system is reduced and its solved for the new + substitution. After we get the solution for $U$: + + .. math:: U = a(t) + + We substitute and return the reduced system: + + .. math:: + a(t) = t*X' - X + + Parameters + ========== + + A: Matrix + Coefficient matrix($A(t)*t$) of the second order system of this form. + b: Matrix + Non-homogeneous term($b(t)$) of the system of ODEs. + funcs: List + List of dependent variables + t: Symbol + Independent variable of the system of ODEs. + + Returns + ======= + + List + + """ + + U = Matrix([t*func.diff(t) - func for func in funcs]) + + sol = linodesolve(A, t, t*b) + reduced_eqs = [Eq(u, s) for s, u in zip(sol, U)] + reduced_eqs = canonical_odes(reduced_eqs, funcs, t)[0] + + return reduced_eqs + + +def _second_order_subs_type2(A, funcs, t_): + r""" + Returns a second order system based on the coefficient matrix passed. + + Explanation + =========== + + This function returns a system of second order ODE of the following form: + + .. math:: + X'' = A * X + + Here, $X$ is the vector of dependent variables, but a bit modified, $A$ is the + coefficient matrix passed. + + Along with returning the second order system, this function also returns the new + dependent variables with the new independent variable `t_` passed. + + Parameters + ========== + + A: Matrix + Coefficient matrix of the system + funcs: List + List of old dependent variables + t_: Symbol + New independent variable + + Returns + ======= + + List, List + + """ + func_names = [func.func.__name__ for func in funcs] + new_funcs = [Function(Dummy("{}_".format(name)))(t_) for name in func_names] + rhss = A * Matrix(new_funcs) + new_eqs = [Eq(func.diff(t_, 2), rhs) for func, rhs in zip(new_funcs, rhss)] + + return new_eqs, new_funcs + + +def _is_euler_system(As, t): + return all(_matrix_is_constant((A*t**i).applyfunc(ratsimp), t) for i, A in enumerate(As)) + + +def _classify_linear_system(eqs, funcs, t, is_canon=False): + r""" + Returns a dictionary with details of the eqs if the system passed is linear + and can be classified by this function else returns None + + Explanation + =========== + + This function takes the eqs, converts it into a form Ax = b where x is a vector of terms + containing dependent variables and their derivatives till their maximum order. If it is + possible to convert eqs into Ax = b, then all the equations in eqs are linear otherwise + they are non-linear. + + To check if the equations are constant coefficient, we need to check if all the terms in + A obtained above are constant or not. + + To check if the equations are homogeneous or not, we need to check if b is a zero matrix + or not. + + Parameters + ========== + + eqs: List + List of ODEs + funcs: List + List of dependent variables + t: Symbol + Independent variable of the equations in eqs + is_canon: Boolean + If True, then this function will not try to get the + system in canonical form. Default value is False + + Returns + ======= + + match = { + 'no_of_equation': len(eqs), + 'eq': eqs, + 'func': funcs, + 'order': order, + 'is_linear': is_linear, + 'is_constant': is_constant, + 'is_homogeneous': is_homogeneous, + } + + Dict or list of Dicts or None + Dict with values for keys: + 1. no_of_equation: Number of equations + 2. eq: The set of equations + 3. func: List of dependent variables + 4. order: A dictionary that gives the order of the + dependent variable in eqs + 5. is_linear: Boolean value indicating if the set of + equations are linear or not. + 6. is_constant: Boolean value indicating if the set of + equations have constant coefficients or not. + 7. is_homogeneous: Boolean value indicating if the set of + equations are homogeneous or not. + 8. commutative_antiderivative: Antiderivative of the coefficient + matrix if the coefficient matrix is non-constant + and commutative with its antiderivative. This key + may or may not exist. + 9. is_general: Boolean value indicating if the system of ODEs is + solvable using one of the general case solvers or not. + 10. rhs: rhs of the non-homogeneous system of ODEs in Matrix form. This + key may or may not exist. + 11. is_higher_order: True if the system passed has an order greater than 1. + This key may or may not exist. + 12. is_second_order: True if the system passed is a second order ODE. This + key may or may not exist. + This Dict is the answer returned if the eqs are linear and constant + coefficient. Otherwise, None is returned. + + """ + + # Error for i == 0 can be added but isn't for now + + # Check for len(funcs) == len(eqs) + if len(funcs) != len(eqs): + raise ValueError("Number of functions given is not equal to the number of equations %s" % funcs) + + # ValueError when functions have more than one arguments + for func in funcs: + if len(func.args) != 1: + raise ValueError("dsolve() and classify_sysode() work with " + "functions of one variable only, not %s" % func) + + # Getting the func_dict and order using the helper + # function + order = _get_func_order(eqs, funcs) + system_order = max(order[func] for func in funcs) + is_higher_order = system_order > 1 + is_second_order = system_order == 2 and all(order[func] == 2 for func in funcs) + + # Not adding the check if the len(func.args) for + # every func in funcs is 1 + + # Linearity check + try: + + canon_eqs = canonical_odes(eqs, funcs, t) if not is_canon else [eqs] + if len(canon_eqs) == 1: + As, b = linear_ode_to_matrix(canon_eqs[0], funcs, t, system_order) + else: + + match = { + 'is_implicit': True, + 'canon_eqs': canon_eqs + } + + return match + + # When the system of ODEs is non-linear, an ODENonlinearError is raised. + # This function catches the error and None is returned. + except ODENonlinearError: + return None + + is_linear = True + + # Homogeneous check + is_homogeneous = True if b.is_zero_matrix else False + + # Is general key is used to identify if the system of ODEs can be solved by + # one of the general case solvers or not. + match = { + 'no_of_equation': len(eqs), + 'eq': eqs, + 'func': funcs, + 'order': order, + 'is_linear': is_linear, + 'is_homogeneous': is_homogeneous, + 'is_general': True + } + + if not is_homogeneous: + match['rhs'] = b + + is_constant = all(_matrix_is_constant(A_, t) for A_ in As) + + # The match['is_linear'] check will be added in the future when this + # function becomes ready to deal with non-linear systems of ODEs + + if not is_higher_order: + A = As[1] + match['func_coeff'] = A + + # Constant coefficient check + is_constant = _matrix_is_constant(A, t) + match['is_constant'] = is_constant + + try: + system_info = linodesolve_type(A, t, b=b) + except NotImplementedError: + return None + + match.update(system_info) + antiderivative = match.pop("antiderivative") + + if not is_constant: + match['commutative_antiderivative'] = antiderivative + + return match + else: + match['type_of_equation'] = "type0" + + if is_second_order: + A1, A0 = As[1:] + + match_second_order = _match_second_order_type(A1, A0, t) + match.update(match_second_order) + + match['is_second_order'] = True + + # If system is constant, then no need to check if its in euler + # form or not. It will be easier and faster to directly proceed + # to solve it. + if match['type_of_equation'] == "type0" and not is_constant: + is_euler = _is_euler_system(As, t) + if is_euler: + t_ = Symbol('{}_'.format(t)) + match.update({'is_transformed': True, 'type_of_equation': 'type1', + 't_': t_}) + else: + is_jordan = lambda M: M == Matrix.jordan_block(M.shape[0], M[0, 0]) + terms = _factor_matrix(As[-1], t) + if all(A.is_zero_matrix for A in As[1:-1]) and terms is not None and not is_jordan(terms[1]): + P, J = terms[1].jordan_form() + match.update({'type_of_equation': 'type2', 'J': J, + 'f(t)': terms[0], 'P': P, 'is_transformed': True}) + + if match['type_of_equation'] != 'type0' and is_second_order: + match.pop('is_second_order', None) + + match['is_higher_order'] = is_higher_order + + return match + +def _preprocess_eqs(eqs): + processed_eqs = [] + for eq in eqs: + processed_eqs.append(eq if isinstance(eq, Equality) else Eq(eq, 0)) + + return processed_eqs + + +def _eqs2dict(eqs, funcs): + eqsorig = {} + eqsmap = {} + funcset = set(funcs) + for eq in eqs: + f1, = eq.lhs.atoms(AppliedUndef) + f2s = (eq.rhs.atoms(AppliedUndef) - {f1}) & funcset + eqsmap[f1] = f2s + eqsorig[f1] = eq + return eqsmap, eqsorig + + +def _dict2graph(d): + nodes = list(d) + edges = [(f1, f2) for f1, f2s in d.items() for f2 in f2s] + G = (nodes, edges) + return G + + +def _is_type1(scc, t): + eqs, funcs = scc + + try: + (A1, A0), b = linear_ode_to_matrix(eqs, funcs, t, 1) + except (ODENonlinearError, ODEOrderError): + return False + + if _matrix_is_constant(A0, t) and b.is_zero_matrix: + return True + + return False + + +def _combine_type1_subsystems(subsystem, funcs, t): + indices = [i for i, sys in enumerate(zip(subsystem, funcs)) if _is_type1(sys, t)] + remove = set() + for ip, i in enumerate(indices): + for j in indices[ip+1:]: + if any(eq2.has(funcs[i]) for eq2 in subsystem[j]): + subsystem[j] = subsystem[i] + subsystem[j] + remove.add(i) + subsystem = [sys for i, sys in enumerate(subsystem) if i not in remove] + return subsystem + + +def _component_division(eqs, funcs, t): + + # Assuming that each eq in eqs is in canonical form, + # that is, [f(x).diff(x) = .., g(x).diff(x) = .., etc] + # and that the system passed is in its first order + eqsmap, eqsorig = _eqs2dict(eqs, funcs) + + subsystems = [] + for cc in connected_components(_dict2graph(eqsmap)): + eqsmap_c = {f: eqsmap[f] for f in cc} + sccs = strongly_connected_components(_dict2graph(eqsmap_c)) + subsystem = [[eqsorig[f] for f in scc] for scc in sccs] + subsystem = _combine_type1_subsystems(subsystem, sccs, t) + subsystems.append(subsystem) + + return subsystems + + +# Returns: List of equations +def _linear_ode_solver(match): + t = match['t'] + funcs = match['func'] + + rhs = match.get('rhs', None) + tau = match.get('tau', None) + t = match['t_'] if 't_' in match else t + A = match['func_coeff'] + + # Note: To make B None when the matrix has constant + # coefficient + B = match.get('commutative_antiderivative', None) + type = match['type_of_equation'] + + sol_vector = linodesolve(A, t, b=rhs, B=B, + type=type, tau=tau) + + sol = [Eq(f, s) for f, s in zip(funcs, sol_vector)] + + return sol + + +def _select_equations(eqs, funcs, key=lambda x: x): + eq_dict = {e.lhs: e.rhs for e in eqs} + return [Eq(f, eq_dict[key(f)]) for f in funcs] + + +def _higher_order_ode_solver(match): + eqs = match["eq"] + funcs = match["func"] + t = match["t"] + sysorder = match['order'] + type = match.get('type_of_equation', "type0") + + is_second_order = match.get('is_second_order', False) + is_transformed = match.get('is_transformed', False) + is_euler = is_transformed and type == "type1" + is_higher_order_type2 = is_transformed and type == "type2" and 'P' in match + + if is_second_order: + new_eqs, new_funcs = _second_order_to_first_order(eqs, funcs, t, + A1=match.get("A1", None), A0=match.get("A0", None), + b=match.get("rhs", None), type=type, + t_=match.get("t_", None)) + else: + new_eqs, new_funcs = _higher_order_to_first_order(eqs, sysorder, t, funcs=funcs, + type=type, J=match.get('J', None), + f_t=match.get('f(t)', None), + P=match.get('P', None), b=match.get('rhs', None)) + + if is_transformed: + t = match.get('t_', t) + + if not is_higher_order_type2: + new_eqs = _select_equations(new_eqs, [f.diff(t) for f in new_funcs]) + + sol = None + + # NotImplementedError may be raised when the system may be actually + # solvable if it can be just divided into sub-systems + try: + if not is_higher_order_type2: + sol = _strong_component_solver(new_eqs, new_funcs, t) + except NotImplementedError: + sol = None + + # Dividing the system only when it becomes essential + if sol is None: + try: + sol = _component_solver(new_eqs, new_funcs, t) + except NotImplementedError: + sol = None + + if sol is None: + return sol + + is_second_order_type2 = is_second_order and type == "type2" + + underscores = '__' if is_transformed else '_' + + sol = _select_equations(sol, funcs, + key=lambda x: Function(Dummy('{}{}0'.format(x.func.__name__, underscores)))(t)) + + if match.get("is_transformed", False): + if is_second_order_type2: + g_t = match["g(t)"] + tau = match["tau"] + sol = [Eq(s.lhs, s.rhs.subs(t, tau) * g_t) for s in sol] + elif is_euler: + t = match['t'] + tau = match['t_'] + sol = [s.subs(tau, log(t)) for s in sol] + elif is_higher_order_type2: + P = match['P'] + sol_vector = P * Matrix([s.rhs for s in sol]) + sol = [Eq(f, s) for f, s in zip(funcs, sol_vector)] + + return sol + + +# Returns: List of equations or None +# If None is returned by this solver, then the system +# of ODEs cannot be solved directly by dsolve_system. +def _strong_component_solver(eqs, funcs, t): + from sympy.solvers.ode.ode import dsolve, constant_renumber + + match = _classify_linear_system(eqs, funcs, t, is_canon=True) + sol = None + + # Assuming that we can't get an implicit system + # since we are already canonical equations from + # dsolve_system + if match: + match['t'] = t + + if match.get('is_higher_order', False): + sol = _higher_order_ode_solver(match) + + elif match.get('is_linear', False): + sol = _linear_ode_solver(match) + + # Note: For now, only linear systems are handled by this function + # hence, the match condition is added. This can be removed later. + if sol is None and len(eqs) == 1: + sol = dsolve(eqs[0], func=funcs[0]) + variables = Tuple(eqs[0]).free_symbols + new_constants = [Dummy() for _ in range(ode_order(eqs[0], funcs[0]))] + sol = constant_renumber(sol, variables=variables, newconstants=new_constants) + sol = [sol] + + # To add non-linear case here in future + + return sol + + +def _get_funcs_from_canon(eqs): + return [eq.lhs.args[0] for eq in eqs] + + +# Returns: List of Equations(a solution) +def _weak_component_solver(wcc, t): + + # We will divide the systems into sccs + # only when the wcc cannot be solved as + # a whole + eqs = [] + for scc in wcc: + eqs += scc + funcs = _get_funcs_from_canon(eqs) + + sol = _strong_component_solver(eqs, funcs, t) + if sol: + return sol + + sol = [] + + for scc in wcc: + eqs = scc + funcs = _get_funcs_from_canon(eqs) + + # Substituting solutions for the dependent + # variables solved in previous SCC, if any solved. + comp_eqs = [eq.subs({s.lhs: s.rhs for s in sol}) for eq in eqs] + scc_sol = _strong_component_solver(comp_eqs, funcs, t) + + if scc_sol is None: + raise NotImplementedError(filldedent(''' + The system of ODEs passed cannot be solved by dsolve_system. + ''')) + + # scc_sol: List of equations + # scc_sol is a solution + sol += scc_sol + + return sol + + +# Returns: List of Equations(a solution) +def _component_solver(eqs, funcs, t): + components = _component_division(eqs, funcs, t) + sol = [] + + for wcc in components: + + # wcc_sol: List of Equations + sol += _weak_component_solver(wcc, t) + + # sol: List of Equations + return sol + + +def _second_order_to_first_order(eqs, funcs, t, type="auto", A1=None, + A0=None, b=None, t_=None): + r""" + Expects the system to be in second order and in canonical form + + Explanation + =========== + + Reduces a second order system into a first order one depending on the type of second + order system. + 1. "type0": If this is passed, then the system will be reduced to first order by + introducing dummy variables. + 2. "type1": If this is passed, then a particular substitution will be used to reduce the + the system into first order. + 3. "type2": If this is passed, then the system will be transformed with new dependent + variables and independent variables. This transformation is a part of solving + the corresponding system of ODEs. + + `A1` and `A0` are the coefficient matrices from the system and it is assumed that the + second order system has the form given below: + + .. math:: + A2 * X'' = A1 * X' + A0 * X + b + + Here, $A2$ is the coefficient matrix for the vector $X''$ and $b$ is the non-homogeneous + term. + + Default value for `b` is None but if `A1` and `A0` are passed and `b` is not passed, then the + system will be assumed homogeneous. + + """ + is_a1 = A1 is None + is_a0 = A0 is None + + if (type == "type1" and is_a1) or (type == "type2" and is_a0)\ + or (type == "auto" and (is_a1 or is_a0)): + (A2, A1, A0), b = linear_ode_to_matrix(eqs, funcs, t, 2) + + if not A2.is_Identity: + raise ValueError(filldedent(''' + The system must be in its canonical form. + ''')) + + if type == "auto": + match = _match_second_order_type(A1, A0, t) + type = match["type_of_equation"] + A1 = match.get("A1", None) + A0 = match.get("A0", None) + + sys_order = dict.fromkeys(funcs, 2) + + if type == "type1": + if b is None: + b = zeros(len(eqs)) + eqs = _second_order_subs_type1(A1, b, funcs, t) + sys_order = dict.fromkeys(funcs, 1) + + if type == "type2": + if t_ is None: + t_ = Symbol("{}_".format(t)) + t = t_ + eqs, funcs = _second_order_subs_type2(A0, funcs, t_) + sys_order = dict.fromkeys(funcs, 2) + + return _higher_order_to_first_order(eqs, sys_order, t, funcs=funcs) + + +def _higher_order_type2_to_sub_systems(J, f_t, funcs, t, max_order, b=None, P=None): + + # Note: To add a test for this ValueError + if J is None or f_t is None or not _matrix_is_constant(J, t): + raise ValueError(filldedent(''' + Correctly input for args 'A' and 'f_t' for Linear, Higher Order, + Type 2 + ''')) + + if P is None and b is not None and not b.is_zero_matrix: + raise ValueError(filldedent(''' + Provide the keyword 'P' for matrix P in A = P * J * P-1. + ''')) + + new_funcs = Matrix([Function(Dummy('{}__0'.format(f.func.__name__)))(t) for f in funcs]) + new_eqs = new_funcs.diff(t, max_order) - f_t * J * new_funcs + + if b is not None and not b.is_zero_matrix: + new_eqs -= P.inv() * b + + new_eqs = canonical_odes(new_eqs, new_funcs, t)[0] + + return new_eqs, new_funcs + + +def _higher_order_to_first_order(eqs, sys_order, t, funcs=None, type="type0", **kwargs): + if funcs is None: + funcs = sys_order.keys() + + # Standard Cauchy Euler system + if type == "type1": + t_ = Symbol('{}_'.format(t)) + new_funcs = [Function(Dummy('{}_'.format(f.func.__name__)))(t_) for f in funcs] + max_order = max(sys_order[func] for func in funcs) + subs_dict = dict(zip(funcs, new_funcs)) + subs_dict[t] = exp(t_) + + free_function = Function(Dummy()) + + def _get_coeffs_from_subs_expression(expr): + if isinstance(expr, Subs): + free_symbol = expr.args[1][0] + term = expr.args[0] + return {ode_order(term, free_symbol): 1} + + if isinstance(expr, Mul): + coeff = expr.args[0] + order = list(_get_coeffs_from_subs_expression(expr.args[1]).keys())[0] + return {order: coeff} + + if isinstance(expr, Add): + coeffs = {} + for arg in expr.args: + + if isinstance(arg, Mul): + coeffs.update(_get_coeffs_from_subs_expression(arg)) + + else: + order = list(_get_coeffs_from_subs_expression(arg).keys())[0] + coeffs[order] = 1 + + return coeffs + + for o in range(1, max_order + 1): + expr = free_function(log(t_)).diff(t_, o)*t_**o + coeff_dict = _get_coeffs_from_subs_expression(expr) + coeffs = [coeff_dict[order] if order in coeff_dict else 0 for order in range(o + 1)] + expr_to_subs = sum(free_function(t_).diff(t_, i) * c for i, c in + enumerate(coeffs)) / t**o + subs_dict.update({f.diff(t, o): expr_to_subs.subs(free_function(t_), nf) + for f, nf in zip(funcs, new_funcs)}) + + new_eqs = [eq.subs(subs_dict) for eq in eqs] + new_sys_order = {nf: sys_order[f] for f, nf in zip(funcs, new_funcs)} + + new_eqs = canonical_odes(new_eqs, new_funcs, t_)[0] + + return _higher_order_to_first_order(new_eqs, new_sys_order, t_, funcs=new_funcs) + + # Systems of the form: X(n)(t) = f(t)*A*X + b + # where X(n)(t) is the nth derivative of the vector of dependent variables + # with respect to the independent variable and A is a constant matrix. + if type == "type2": + J = kwargs.get('J', None) + f_t = kwargs.get('f_t', None) + b = kwargs.get('b', None) + P = kwargs.get('P', None) + max_order = max(sys_order[func] for func in funcs) + + return _higher_order_type2_to_sub_systems(J, f_t, funcs, t, max_order, P=P, b=b) + + # Note: To be changed to this after doit option is disabled for default cases + # new_sysorder = _get_func_order(new_eqs, new_funcs) + # + # return _higher_order_to_first_order(new_eqs, new_sysorder, t, funcs=new_funcs) + + new_funcs = [] + + for prev_func in funcs: + func_name = prev_func.func.__name__ + func = Function(Dummy('{}_0'.format(func_name)))(t) + new_funcs.append(func) + subs_dict = {prev_func: func} + new_eqs = [] + + for i in range(1, sys_order[prev_func]): + new_func = Function(Dummy('{}_{}'.format(func_name, i)))(t) + subs_dict[prev_func.diff(t, i)] = new_func + new_funcs.append(new_func) + + prev_f = subs_dict[prev_func.diff(t, i-1)] + new_eq = Eq(prev_f.diff(t), new_func) + new_eqs.append(new_eq) + + eqs = [eq.subs(subs_dict) for eq in eqs] + new_eqs + + return eqs, new_funcs + + +def dsolve_system(eqs, funcs=None, t=None, ics=None, doit=False, simplify=True): + r""" + Solves any(supported) system of Ordinary Differential Equations + + Explanation + =========== + + This function takes a system of ODEs as an input, determines if the + it is solvable by this function, and returns the solution if found any. + + This function can handle: + 1. Linear, First Order, Constant coefficient homogeneous system of ODEs + 2. Linear, First Order, Constant coefficient non-homogeneous system of ODEs + 3. Linear, First Order, non-constant coefficient homogeneous system of ODEs + 4. Linear, First Order, non-constant coefficient non-homogeneous system of ODEs + 5. Any implicit system which can be divided into system of ODEs which is of the above 4 forms + 6. Any higher order linear system of ODEs that can be reduced to one of the 5 forms of systems described above. + + The types of systems described above are not limited by the number of equations, i.e. this + function can solve the above types irrespective of the number of equations in the system passed. + But, the bigger the system, the more time it will take to solve the system. + + This function returns a list of solutions. Each solution is a list of equations where LHS is + the dependent variable and RHS is an expression in terms of the independent variable. + + Among the non constant coefficient types, not all the systems are solvable by this function. Only + those which have either a coefficient matrix with a commutative antiderivative or those systems which + may be divided further so that the divided systems may have coefficient matrix with commutative antiderivative. + + Parameters + ========== + + eqs : List + system of ODEs to be solved + funcs : List or None + List of dependent variables that make up the system of ODEs + t : Symbol or None + Independent variable in the system of ODEs + ics : Dict or None + Set of initial boundary/conditions for the system of ODEs + doit : Boolean + Evaluate the solutions if True. Default value is True. Can be + set to false if the integral evaluation takes too much time and/or + is not required. + simplify: Boolean + Simplify the solutions for the systems. Default value is True. + Can be set to false if simplification takes too much time and/or + is not required. + + Examples + ======== + + >>> from sympy import symbols, Eq, Function + >>> from sympy.solvers.ode.systems import dsolve_system + >>> f, g = symbols("f g", cls=Function) + >>> x = symbols("x") + + >>> eqs = [Eq(f(x).diff(x), g(x)), Eq(g(x).diff(x), f(x))] + >>> dsolve_system(eqs) + [[Eq(f(x), -C1*exp(-x) + C2*exp(x)), Eq(g(x), C1*exp(-x) + C2*exp(x))]] + + You can also pass the initial conditions for the system of ODEs: + + >>> dsolve_system(eqs, ics={f(0): 1, g(0): 0}) + [[Eq(f(x), exp(x)/2 + exp(-x)/2), Eq(g(x), exp(x)/2 - exp(-x)/2)]] + + Optionally, you can pass the dependent variables and the independent + variable for which the system is to be solved: + + >>> funcs = [f(x), g(x)] + >>> dsolve_system(eqs, funcs=funcs, t=x) + [[Eq(f(x), -C1*exp(-x) + C2*exp(x)), Eq(g(x), C1*exp(-x) + C2*exp(x))]] + + Lets look at an implicit system of ODEs: + + >>> eqs = [Eq(f(x).diff(x)**2, g(x)**2), Eq(g(x).diff(x), g(x))] + >>> dsolve_system(eqs) + [[Eq(f(x), C1 - C2*exp(x)), Eq(g(x), C2*exp(x))], [Eq(f(x), C1 + C2*exp(x)), Eq(g(x), C2*exp(x))]] + + Returns + ======= + + List of List of Equations + + Raises + ====== + + NotImplementedError + When the system of ODEs is not solvable by this function. + ValueError + When the parameters passed are not in the required form. + + """ + from sympy.solvers.ode.ode import solve_ics, _extract_funcs, constant_renumber + + if not iterable(eqs): + raise ValueError(filldedent(''' + List of equations should be passed. The input is not valid. + ''')) + + eqs = _preprocess_eqs(eqs) + + if funcs is not None and not isinstance(funcs, list): + raise ValueError(filldedent(''' + Input to the funcs should be a list of functions. + ''')) + + if funcs is None: + funcs = _extract_funcs(eqs) + + if any(len(func.args) != 1 for func in funcs): + raise ValueError(filldedent(''' + dsolve_system can solve a system of ODEs with only one independent + variable. + ''')) + + if len(eqs) != len(funcs): + raise ValueError(filldedent(''' + Number of equations and number of functions do not match + ''')) + + if t is not None and not isinstance(t, Symbol): + raise ValueError(filldedent(''' + The independent variable must be of type Symbol + ''')) + + if t is None: + t = list(list(eqs[0].atoms(Derivative))[0].atoms(Symbol))[0] + + sols = [] + canon_eqs = canonical_odes(eqs, funcs, t) + + for canon_eq in canon_eqs: + try: + sol = _strong_component_solver(canon_eq, funcs, t) + except NotImplementedError: + sol = None + + if sol is None: + sol = _component_solver(canon_eq, funcs, t) + + sols.append(sol) + + if sols: + final_sols = [] + variables = Tuple(*eqs).free_symbols + + for sol in sols: + + sol = _select_equations(sol, funcs) + sol = constant_renumber(sol, variables=variables) + + if ics: + constants = Tuple(*sol).free_symbols - variables + solved_constants = solve_ics(sol, funcs, constants, ics) + sol = [s.subs(solved_constants) for s in sol] + + if simplify: + constants = Tuple(*sol).free_symbols - variables + sol = simpsol(sol, [t], constants, doit=doit) + + final_sols.append(sol) + + sols = final_sols + + return sols diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__init__.py b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__init__.py new file mode 100644 index 0000000000000000000000000000000000000000..e69de29bb2d1d6434b8b29ae775ad8c2e48c5391 diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_lie_group.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_lie_group.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..fa3805b8707a71b06ce6c6bfda54a1ba5c2ff298 Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_lie_group.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_ode.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_ode.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..3fe1a5508a1ac3fa5300ae27aecc9f237c0e9dba Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_ode.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_single.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_single.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..ca103f5b82f4fe8cbc5d49316d1477118189634d Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_single.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_subscheck.cpython-310.pyc b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_subscheck.cpython-310.pyc new file mode 100644 index 0000000000000000000000000000000000000000..6258b0fe316f7875c0572b7de9e3739ba3154317 Binary files /dev/null and b/pllava/lib/python3.10/site-packages/sympy/solvers/ode/tests/__pycache__/test_subscheck.cpython-310.pyc differ diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/pde.py b/pllava/lib/python3.10/site-packages/sympy/solvers/pde.py new file mode 100644 index 0000000000000000000000000000000000000000..75e5503145af859e0f9bf9e95f3fd9d88bbfc5dc --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/pde.py @@ -0,0 +1,971 @@ +""" +This module contains pdsolve() and different helper functions that it +uses. It is heavily inspired by the ode module and hence the basic +infrastructure remains the same. + +**Functions in this module** + + These are the user functions in this module: + + - pdsolve() - Solves PDE's + - classify_pde() - Classifies PDEs into possible hints for dsolve(). + - pde_separate() - Separate variables in partial differential equation either by + additive or multiplicative separation approach. + + These are the helper functions in this module: + + - pde_separate_add() - Helper function for searching additive separable solutions. + - pde_separate_mul() - Helper function for searching multiplicative + separable solutions. + +**Currently implemented solver methods** + +The following methods are implemented for solving partial differential +equations. See the docstrings of the various pde_hint() functions for +more information on each (run help(pde)): + + - 1st order linear homogeneous partial differential equations + with constant coefficients. + - 1st order linear general partial differential equations + with constant coefficients. + - 1st order linear partial differential equations with + variable coefficients. + +""" +from functools import reduce + +from itertools import combinations_with_replacement +from sympy.simplify import simplify # type: ignore +from sympy.core import Add, S +from sympy.core.function import Function, expand, AppliedUndef, Subs +from sympy.core.relational import Equality, Eq +from sympy.core.symbol import Symbol, Wild, symbols +from sympy.functions import exp +from sympy.integrals.integrals import Integral, integrate +from sympy.utilities.iterables import has_dups, is_sequence +from sympy.utilities.misc import filldedent + +from sympy.solvers.deutils import _preprocess, ode_order, _desolve +from sympy.solvers.solvers import solve +from sympy.simplify.radsimp import collect + +import operator + + +allhints = ( + "1st_linear_constant_coeff_homogeneous", + "1st_linear_constant_coeff", + "1st_linear_constant_coeff_Integral", + "1st_linear_variable_coeff" + ) + + +def pdsolve(eq, func=None, hint='default', dict=False, solvefun=None, **kwargs): + """ + Solves any (supported) kind of partial differential equation. + + **Usage** + + pdsolve(eq, f(x,y), hint) -> Solve partial differential equation + eq for function f(x,y), using method hint. + + **Details** + + ``eq`` can be any supported partial differential equation (see + the pde docstring for supported methods). This can either + be an Equality, or an expression, which is assumed to be + equal to 0. + + ``f(x,y)`` is a function of two variables whose derivatives in that + variable make up the partial differential equation. In many + cases it is not necessary to provide this; it will be autodetected + (and an error raised if it could not be detected). + + ``hint`` is the solving method that you want pdsolve to use. Use + classify_pde(eq, f(x,y)) to get all of the possible hints for + a PDE. The default hint, 'default', will use whatever hint + is returned first by classify_pde(). See Hints below for + more options that you can use for hint. + + ``solvefun`` is the convention used for arbitrary functions returned + by the PDE solver. If not set by the user, it is set by default + to be F. + + **Hints** + + Aside from the various solving methods, there are also some + meta-hints that you can pass to pdsolve(): + + "default": + This uses whatever hint is returned first by + classify_pde(). This is the default argument to + pdsolve(). + + "all": + To make pdsolve apply all relevant classification hints, + use pdsolve(PDE, func, hint="all"). This will return a + dictionary of hint:solution terms. If a hint causes + pdsolve to raise the NotImplementedError, value of that + hint's key will be the exception object raised. The + dictionary will also include some special keys: + + - order: The order of the PDE. See also ode_order() in + deutils.py + - default: The solution that would be returned by + default. This is the one produced by the hint that + appears first in the tuple returned by classify_pde(). + + "all_Integral": + This is the same as "all", except if a hint also has a + corresponding "_Integral" hint, it only returns the + "_Integral" hint. This is useful if "all" causes + pdsolve() to hang because of a difficult or impossible + integral. This meta-hint will also be much faster than + "all", because integrate() is an expensive routine. + + See also the classify_pde() docstring for more info on hints, + and the pde docstring for a list of all supported hints. + + **Tips** + - You can declare the derivative of an unknown function this way: + + >>> from sympy import Function, Derivative + >>> from sympy.abc import x, y # x and y are the independent variables + >>> f = Function("f")(x, y) # f is a function of x and y + >>> # fx will be the partial derivative of f with respect to x + >>> fx = Derivative(f, x) + >>> # fy will be the partial derivative of f with respect to y + >>> fy = Derivative(f, y) + + - See test_pde.py for many tests, which serves also as a set of + examples for how to use pdsolve(). + - pdsolve always returns an Equality class (except for the case + when the hint is "all" or "all_Integral"). Note that it is not possible + to get an explicit solution for f(x, y) as in the case of ODE's + - Do help(pde.pde_hintname) to get help more information on a + specific hint + + + Examples + ======== + + >>> from sympy.solvers.pde import pdsolve + >>> from sympy import Function, Eq + >>> from sympy.abc import x, y + >>> f = Function('f') + >>> u = f(x, y) + >>> ux = u.diff(x) + >>> uy = u.diff(y) + >>> eq = Eq(1 + (2*(ux/u)) + (3*(uy/u)), 0) + >>> pdsolve(eq) + Eq(f(x, y), F(3*x - 2*y)*exp(-2*x/13 - 3*y/13)) + + """ + + if not solvefun: + solvefun = Function('F') + + # See the docstring of _desolve for more details. + hints = _desolve(eq, func=func, hint=hint, simplify=True, + type='pde', **kwargs) + eq = hints.pop('eq', False) + all_ = hints.pop('all', False) + + if all_: + # TODO : 'best' hint should be implemented when adequate + # number of hints are added. + pdedict = {} + failed_hints = {} + gethints = classify_pde(eq, dict=True) + pdedict.update({'order': gethints['order'], + 'default': gethints['default']}) + for hint in hints: + try: + rv = _helper_simplify(eq, hint, hints[hint]['func'], + hints[hint]['order'], hints[hint][hint], solvefun) + except NotImplementedError as detail: + failed_hints[hint] = detail + else: + pdedict[hint] = rv + pdedict.update(failed_hints) + return pdedict + + else: + return _helper_simplify(eq, hints['hint'], hints['func'], + hints['order'], hints[hints['hint']], solvefun) + + +def _helper_simplify(eq, hint, func, order, match, solvefun): + """Helper function of pdsolve that calls the respective + pde functions to solve for the partial differential + equations. This minimizes the computation in + calling _desolve multiple times. + """ + + if hint.endswith("_Integral"): + solvefunc = globals()[ + "pde_" + hint[:-len("_Integral")]] + else: + solvefunc = globals()["pde_" + hint] + return _handle_Integral(solvefunc(eq, func, order, + match, solvefun), func, order, hint) + + +def _handle_Integral(expr, func, order, hint): + r""" + Converts a solution with integrals in it into an actual solution. + + Simplifies the integral mainly using doit() + """ + if hint.endswith("_Integral"): + return expr + + elif hint == "1st_linear_constant_coeff": + return simplify(expr.doit()) + + else: + return expr + + +def classify_pde(eq, func=None, dict=False, *, prep=True, **kwargs): + """ + Returns a tuple of possible pdsolve() classifications for a PDE. + + The tuple is ordered so that first item is the classification that + pdsolve() uses to solve the PDE by default. In general, + classifications near the beginning of the list will produce + better solutions faster than those near the end, though there are + always exceptions. To make pdsolve use a different classification, + use pdsolve(PDE, func, hint=). See also the pdsolve() + docstring for different meta-hints you can use. + + If ``dict`` is true, classify_pde() will return a dictionary of + hint:match expression terms. This is intended for internal use by + pdsolve(). Note that because dictionaries are ordered arbitrarily, + this will most likely not be in the same order as the tuple. + + You can get help on different hints by doing help(pde.pde_hintname), + where hintname is the name of the hint without "_Integral". + + See sympy.pde.allhints or the sympy.pde docstring for a list of all + supported hints that can be returned from classify_pde. + + + Examples + ======== + + >>> from sympy.solvers.pde import classify_pde + >>> from sympy import Function, Eq + >>> from sympy.abc import x, y + >>> f = Function('f') + >>> u = f(x, y) + >>> ux = u.diff(x) + >>> uy = u.diff(y) + >>> eq = Eq(1 + (2*(ux/u)) + (3*(uy/u)), 0) + >>> classify_pde(eq) + ('1st_linear_constant_coeff_homogeneous',) + """ + + if func and len(func.args) != 2: + raise NotImplementedError("Right now only partial " + "differential equations of two variables are supported") + + if prep or func is None: + prep, func_ = _preprocess(eq, func) + if func is None: + func = func_ + + if isinstance(eq, Equality): + if eq.rhs != 0: + return classify_pde(eq.lhs - eq.rhs, func) + eq = eq.lhs + + f = func.func + x = func.args[0] + y = func.args[1] + fx = f(x,y).diff(x) + fy = f(x,y).diff(y) + + # TODO : For now pde.py uses support offered by the ode_order function + # to find the order with respect to a multi-variable function. An + # improvement could be to classify the order of the PDE on the basis of + # individual variables. + order = ode_order(eq, f(x,y)) + + # hint:matchdict or hint:(tuple of matchdicts) + # Also will contain "default": and "order":order items. + matching_hints = {'order': order} + + if not order: + if dict: + matching_hints["default"] = None + return matching_hints + return () + + eq = expand(eq) + + a = Wild('a', exclude = [f(x,y)]) + b = Wild('b', exclude = [f(x,y), fx, fy, x, y]) + c = Wild('c', exclude = [f(x,y), fx, fy, x, y]) + d = Wild('d', exclude = [f(x,y), fx, fy, x, y]) + e = Wild('e', exclude = [f(x,y), fx, fy]) + n = Wild('n', exclude = [x, y]) + # Try removing the smallest power of f(x,y) + # from the highest partial derivatives of f(x,y) + reduced_eq = eq + if eq.is_Add: + power = None + for i in set(combinations_with_replacement((x,y), order)): + coeff = eq.coeff(f(x,y).diff(*i)) + if coeff == 1: + continue + match = coeff.match(a*f(x,y)**n) + if match and match[a]: + if power is None or match[n] < power: + power = match[n] + if power: + den = f(x,y)**power + reduced_eq = Add(*[arg/den for arg in eq.args]) + + if order == 1: + reduced_eq = collect(reduced_eq, f(x, y)) + r = reduced_eq.match(b*fx + c*fy + d*f(x,y) + e) + if r: + if not r[e]: + ## Linear first-order homogeneous partial-differential + ## equation with constant coefficients + r.update({'b': b, 'c': c, 'd': d}) + matching_hints["1st_linear_constant_coeff_homogeneous"] = r + elif r[b]**2 + r[c]**2 != 0: + ## Linear first-order general partial-differential + ## equation with constant coefficients + r.update({'b': b, 'c': c, 'd': d, 'e': e}) + matching_hints["1st_linear_constant_coeff"] = r + matching_hints["1st_linear_constant_coeff_Integral"] = r + + else: + b = Wild('b', exclude=[f(x, y), fx, fy]) + c = Wild('c', exclude=[f(x, y), fx, fy]) + d = Wild('d', exclude=[f(x, y), fx, fy]) + r = reduced_eq.match(b*fx + c*fy + d*f(x,y) + e) + if r: + r.update({'b': b, 'c': c, 'd': d, 'e': e}) + matching_hints["1st_linear_variable_coeff"] = r + + # Order keys based on allhints. + rettuple = tuple(i for i in allhints if i in matching_hints) + + if dict: + # Dictionaries are ordered arbitrarily, so make note of which + # hint would come first for pdsolve(). Use an ordered dict in Py 3. + matching_hints["default"] = None + matching_hints["ordered_hints"] = rettuple + for i in allhints: + if i in matching_hints: + matching_hints["default"] = i + break + return matching_hints + return rettuple + + +def checkpdesol(pde, sol, func=None, solve_for_func=True): + """ + Checks if the given solution satisfies the partial differential + equation. + + pde is the partial differential equation which can be given in the + form of an equation or an expression. sol is the solution for which + the pde is to be checked. This can also be given in an equation or + an expression form. If the function is not provided, the helper + function _preprocess from deutils is used to identify the function. + + If a sequence of solutions is passed, the same sort of container will be + used to return the result for each solution. + + The following methods are currently being implemented to check if the + solution satisfies the PDE: + + 1. Directly substitute the solution in the PDE and check. If the + solution has not been solved for f, then it will solve for f + provided solve_for_func has not been set to False. + + If the solution satisfies the PDE, then a tuple (True, 0) is returned. + Otherwise a tuple (False, expr) where expr is the value obtained + after substituting the solution in the PDE. However if a known solution + returns False, it may be due to the inability of doit() to simplify it to zero. + + Examples + ======== + + >>> from sympy import Function, symbols + >>> from sympy.solvers.pde import checkpdesol, pdsolve + >>> x, y = symbols('x y') + >>> f = Function('f') + >>> eq = 2*f(x,y) + 3*f(x,y).diff(x) + 4*f(x,y).diff(y) + >>> sol = pdsolve(eq) + >>> assert checkpdesol(eq, sol)[0] + >>> eq = x*f(x,y) + f(x,y).diff(x) + >>> checkpdesol(eq, sol) + (False, (x*F(4*x - 3*y) - 6*F(4*x - 3*y)/25 + 4*Subs(Derivative(F(_xi_1), _xi_1), _xi_1, 4*x - 3*y))*exp(-6*x/25 - 8*y/25)) + """ + + # Converting the pde into an equation + if not isinstance(pde, Equality): + pde = Eq(pde, 0) + + # If no function is given, try finding the function present. + if func is None: + try: + _, func = _preprocess(pde.lhs) + except ValueError: + funcs = [s.atoms(AppliedUndef) for s in ( + sol if is_sequence(sol, set) else [sol])] + funcs = set().union(funcs) + if len(funcs) != 1: + raise ValueError( + 'must pass func arg to checkpdesol for this case.') + func = funcs.pop() + + # If the given solution is in the form of a list or a set + # then return a list or set of tuples. + if is_sequence(sol, set): + return type(sol)([checkpdesol( + pde, i, func=func, + solve_for_func=solve_for_func) for i in sol]) + + # Convert solution into an equation + if not isinstance(sol, Equality): + sol = Eq(func, sol) + elif sol.rhs == func: + sol = sol.reversed + + # Try solving for the function + solved = sol.lhs == func and not sol.rhs.has(func) + if solve_for_func and not solved: + solved = solve(sol, func) + if solved: + if len(solved) == 1: + return checkpdesol(pde, Eq(func, solved[0]), + func=func, solve_for_func=False) + else: + return checkpdesol(pde, [Eq(func, t) for t in solved], + func=func, solve_for_func=False) + + # try direct substitution of the solution into the PDE and simplify + if sol.lhs == func: + pde = pde.lhs - pde.rhs + s = simplify(pde.subs(func, sol.rhs).doit()) + return s is S.Zero, s + + raise NotImplementedError(filldedent(''' + Unable to test if %s is a solution to %s.''' % (sol, pde))) + + + +def pde_1st_linear_constant_coeff_homogeneous(eq, func, order, match, solvefun): + r""" + Solves a first order linear homogeneous + partial differential equation with constant coefficients. + + The general form of this partial differential equation is + + .. math:: a \frac{\partial f(x,y)}{\partial x} + + b \frac{\partial f(x,y)}{\partial y} + c f(x,y) = 0 + + where `a`, `b` and `c` are constants. + + The general solution is of the form: + + .. math:: + f(x, y) = F(- a y + b x ) e^{- \frac{c (a x + b y)}{a^2 + b^2}} + + and can be found in SymPy with ``pdsolve``:: + + >>> from sympy.solvers import pdsolve + >>> from sympy.abc import x, y, a, b, c + >>> from sympy import Function, pprint + >>> f = Function('f') + >>> u = f(x,y) + >>> ux = u.diff(x) + >>> uy = u.diff(y) + >>> genform = a*ux + b*uy + c*u + >>> pprint(genform) + d d + a*--(f(x, y)) + b*--(f(x, y)) + c*f(x, y) + dx dy + + >>> pprint(pdsolve(genform)) + -c*(a*x + b*y) + --------------- + 2 2 + a + b + f(x, y) = F(-a*y + b*x)*e + + Examples + ======== + + >>> from sympy import pdsolve + >>> from sympy import Function, pprint + >>> from sympy.abc import x,y + >>> f = Function('f') + >>> pdsolve(f(x,y) + f(x,y).diff(x) + f(x,y).diff(y)) + Eq(f(x, y), F(x - y)*exp(-x/2 - y/2)) + >>> pprint(pdsolve(f(x,y) + f(x,y).diff(x) + f(x,y).diff(y))) + x y + - - - - + 2 2 + f(x, y) = F(x - y)*e + + References + ========== + + - Viktor Grigoryan, "Partial Differential Equations" + Math 124A - Fall 2010, pp.7 + + """ + # TODO : For now homogeneous first order linear PDE's having + # two variables are implemented. Once there is support for + # solving systems of ODE's, this can be extended to n variables. + + f = func.func + x = func.args[0] + y = func.args[1] + b = match[match['b']] + c = match[match['c']] + d = match[match['d']] + return Eq(f(x,y), exp(-S(d)/(b**2 + c**2)*(b*x + c*y))*solvefun(c*x - b*y)) + + +def pde_1st_linear_constant_coeff(eq, func, order, match, solvefun): + r""" + Solves a first order linear partial differential equation + with constant coefficients. + + The general form of this partial differential equation is + + .. math:: a \frac{\partial f(x,y)}{\partial x} + + b \frac{\partial f(x,y)}{\partial y} + + c f(x,y) = G(x,y) + + where `a`, `b` and `c` are constants and `G(x, y)` can be an arbitrary + function in `x` and `y`. + + The general solution of the PDE is: + + .. math:: + f(x, y) = \left. \left[F(\eta) + \frac{1}{a^2 + b^2} + \int\limits^{a x + b y} G\left(\frac{a \xi + b \eta}{a^2 + b^2}, + \frac{- a \eta + b \xi}{a^2 + b^2} \right) + e^{\frac{c \xi}{a^2 + b^2}}\, d\xi\right] + e^{- \frac{c \xi}{a^2 + b^2}} + \right|_{\substack{\eta=- a y + b x\\ \xi=a x + b y }}\, , + + where `F(\eta)` is an arbitrary single-valued function. The solution + can be found in SymPy with ``pdsolve``:: + + >>> from sympy.solvers import pdsolve + >>> from sympy.abc import x, y, a, b, c + >>> from sympy import Function, pprint + >>> f = Function('f') + >>> G = Function('G') + >>> u = f(x, y) + >>> ux = u.diff(x) + >>> uy = u.diff(y) + >>> genform = a*ux + b*uy + c*u - G(x,y) + >>> pprint(genform) + d d + a*--(f(x, y)) + b*--(f(x, y)) + c*f(x, y) - G(x, y) + dx dy + >>> pprint(pdsolve(genform, hint='1st_linear_constant_coeff_Integral')) + // a*x + b*y \ \| + || / | || + || | | || + || | c*xi | || + || | ------- | || + || | 2 2 | || + || | /a*xi + b*eta -a*eta + b*xi\ a + b | || + || | G|------------, -------------|*e d(xi)| || + || | | 2 2 2 2 | | || + || | \ a + b a + b / | -c*xi || + || | | -------|| + || / | 2 2|| + || | a + b || + f(x, y) = ||F(eta) + -------------------------------------------------------|*e || + || 2 2 | || + \\ a + b / /|eta=-a*y + b*x, xi=a*x + b*y + + Examples + ======== + + >>> from sympy.solvers.pde import pdsolve + >>> from sympy import Function, pprint, exp + >>> from sympy.abc import x,y + >>> f = Function('f') + >>> eq = -2*f(x,y).diff(x) + 4*f(x,y).diff(y) + 5*f(x,y) - exp(x + 3*y) + >>> pdsolve(eq) + Eq(f(x, y), (F(4*x + 2*y)*exp(x/2) + exp(x + 4*y)/15)*exp(-y)) + + References + ========== + + - Viktor Grigoryan, "Partial Differential Equations" + Math 124A - Fall 2010, pp.7 + + """ + + # TODO : For now homogeneous first order linear PDE's having + # two variables are implemented. Once there is support for + # solving systems of ODE's, this can be extended to n variables. + xi, eta = symbols("xi eta") + f = func.func + x = func.args[0] + y = func.args[1] + b = match[match['b']] + c = match[match['c']] + d = match[match['d']] + e = -match[match['e']] + expterm = exp(-S(d)/(b**2 + c**2)*xi) + functerm = solvefun(eta) + solvedict = solve((b*x + c*y - xi, c*x - b*y - eta), x, y) + # Integral should remain as it is in terms of xi, + # doit() should be done in _handle_Integral. + genterm = (1/S(b**2 + c**2))*Integral( + (1/expterm*e).subs(solvedict), (xi, b*x + c*y)) + return Eq(f(x,y), Subs(expterm*(functerm + genterm), + (eta, xi), (c*x - b*y, b*x + c*y))) + + +def pde_1st_linear_variable_coeff(eq, func, order, match, solvefun): + r""" + Solves a first order linear partial differential equation + with variable coefficients. The general form of this partial + differential equation is + + .. math:: a(x, y) \frac{\partial f(x, y)}{\partial x} + + b(x, y) \frac{\partial f(x, y)}{\partial y} + + c(x, y) f(x, y) = G(x, y) + + where `a(x, y)`, `b(x, y)`, `c(x, y)` and `G(x, y)` are arbitrary + functions in `x` and `y`. This PDE is converted into an ODE by + making the following transformation: + + 1. `\xi` as `x` + + 2. `\eta` as the constant in the solution to the differential + equation `\frac{dy}{dx} = -\frac{b}{a}` + + Making the previous substitutions reduces it to the linear ODE + + .. math:: a(\xi, \eta)\frac{du}{d\xi} + c(\xi, \eta)u - G(\xi, \eta) = 0 + + which can be solved using ``dsolve``. + + >>> from sympy.abc import x, y + >>> from sympy import Function, pprint + >>> a, b, c, G, f= [Function(i) for i in ['a', 'b', 'c', 'G', 'f']] + >>> u = f(x,y) + >>> ux = u.diff(x) + >>> uy = u.diff(y) + >>> genform = a(x, y)*u + b(x, y)*ux + c(x, y)*uy - G(x,y) + >>> pprint(genform) + d d + -G(x, y) + a(x, y)*f(x, y) + b(x, y)*--(f(x, y)) + c(x, y)*--(f(x, y)) + dx dy + + + Examples + ======== + + >>> from sympy.solvers.pde import pdsolve + >>> from sympy import Function, pprint + >>> from sympy.abc import x,y + >>> f = Function('f') + >>> eq = x*(u.diff(x)) - y*(u.diff(y)) + y**2*u - y**2 + >>> pdsolve(eq) + Eq(f(x, y), F(x*y)*exp(y**2/2) + 1) + + References + ========== + + - Viktor Grigoryan, "Partial Differential Equations" + Math 124A - Fall 2010, pp.7 + + """ + from sympy.solvers.ode import dsolve + + xi, eta = symbols("xi eta") + f = func.func + x = func.args[0] + y = func.args[1] + b = match[match['b']] + c = match[match['c']] + d = match[match['d']] + e = -match[match['e']] + + + if not d: + # To deal with cases like b*ux = e or c*uy = e + if not (b and c): + if c: + try: + tsol = integrate(e/c, y) + except NotImplementedError: + raise NotImplementedError("Unable to find a solution" + " due to inability of integrate") + else: + return Eq(f(x,y), solvefun(x) + tsol) + if b: + try: + tsol = integrate(e/b, x) + except NotImplementedError: + raise NotImplementedError("Unable to find a solution" + " due to inability of integrate") + else: + return Eq(f(x,y), solvefun(y) + tsol) + + if not c: + # To deal with cases when c is 0, a simpler method is used. + # The PDE reduces to b*(u.diff(x)) + d*u = e, which is a linear ODE in x + plode = f(x).diff(x)*b + d*f(x) - e + sol = dsolve(plode, f(x)) + syms = sol.free_symbols - plode.free_symbols - {x, y} + rhs = _simplify_variable_coeff(sol.rhs, syms, solvefun, y) + return Eq(f(x, y), rhs) + + if not b: + # To deal with cases when b is 0, a simpler method is used. + # The PDE reduces to c*(u.diff(y)) + d*u = e, which is a linear ODE in y + plode = f(y).diff(y)*c + d*f(y) - e + sol = dsolve(plode, f(y)) + syms = sol.free_symbols - plode.free_symbols - {x, y} + rhs = _simplify_variable_coeff(sol.rhs, syms, solvefun, x) + return Eq(f(x, y), rhs) + + dummy = Function('d') + h = (c/b).subs(y, dummy(x)) + sol = dsolve(dummy(x).diff(x) - h, dummy(x)) + if isinstance(sol, list): + sol = sol[0] + solsym = sol.free_symbols - h.free_symbols - {x, y} + if len(solsym) == 1: + solsym = solsym.pop() + etat = (solve(sol, solsym)[0]).subs(dummy(x), y) + ysub = solve(eta - etat, y)[0] + deq = (b*(f(x).diff(x)) + d*f(x) - e).subs(y, ysub) + final = (dsolve(deq, f(x), hint='1st_linear')).rhs + if isinstance(final, list): + final = final[0] + finsyms = final.free_symbols - deq.free_symbols - {x, y} + rhs = _simplify_variable_coeff(final, finsyms, solvefun, etat) + return Eq(f(x, y), rhs) + + else: + raise NotImplementedError("Cannot solve the partial differential equation due" + " to inability of constantsimp") + + +def _simplify_variable_coeff(sol, syms, func, funcarg): + r""" + Helper function to replace constants by functions in 1st_linear_variable_coeff + """ + eta = Symbol("eta") + if len(syms) == 1: + sym = syms.pop() + final = sol.subs(sym, func(funcarg)) + + else: + for sym in syms: + final = sol.subs(sym, func(funcarg)) + + return simplify(final.subs(eta, funcarg)) + + +def pde_separate(eq, fun, sep, strategy='mul'): + """Separate variables in partial differential equation either by additive + or multiplicative separation approach. It tries to rewrite an equation so + that one of the specified variables occurs on a different side of the + equation than the others. + + :param eq: Partial differential equation + + :param fun: Original function F(x, y, z) + + :param sep: List of separated functions [X(x), u(y, z)] + + :param strategy: Separation strategy. You can choose between additive + separation ('add') and multiplicative separation ('mul') which is + default. + + Examples + ======== + + >>> from sympy import E, Eq, Function, pde_separate, Derivative as D + >>> from sympy.abc import x, t + >>> u, X, T = map(Function, 'uXT') + + >>> eq = Eq(D(u(x, t), x), E**(u(x, t))*D(u(x, t), t)) + >>> pde_separate(eq, u(x, t), [X(x), T(t)], strategy='add') + [exp(-X(x))*Derivative(X(x), x), exp(T(t))*Derivative(T(t), t)] + + >>> eq = Eq(D(u(x, t), x, 2), D(u(x, t), t, 2)) + >>> pde_separate(eq, u(x, t), [X(x), T(t)], strategy='mul') + [Derivative(X(x), (x, 2))/X(x), Derivative(T(t), (t, 2))/T(t)] + + See Also + ======== + pde_separate_add, pde_separate_mul + """ + + do_add = False + if strategy == 'add': + do_add = True + elif strategy == 'mul': + do_add = False + else: + raise ValueError('Unknown strategy: %s' % strategy) + + if isinstance(eq, Equality): + if eq.rhs != 0: + return pde_separate(Eq(eq.lhs - eq.rhs, 0), fun, sep, strategy) + else: + return pde_separate(Eq(eq, 0), fun, sep, strategy) + + if eq.rhs != 0: + raise ValueError("Value should be 0") + + # Handle arguments + orig_args = list(fun.args) + subs_args = [arg for s in sep for arg in s.args] + + if do_add: + functions = reduce(operator.add, sep) + else: + functions = reduce(operator.mul, sep) + + # Check whether variables match + if len(subs_args) != len(orig_args): + raise ValueError("Variable counts do not match") + # Check for duplicate arguments like [X(x), u(x, y)] + if has_dups(subs_args): + raise ValueError("Duplicate substitution arguments detected") + # Check whether the variables match + if set(orig_args) != set(subs_args): + raise ValueError("Arguments do not match") + + # Substitute original function with separated... + result = eq.lhs.subs(fun, functions).doit() + + # Divide by terms when doing multiplicative separation + if not do_add: + eq = 0 + for i in result.args: + eq += i/functions + result = eq + + svar = subs_args[0] + dvar = subs_args[1:] + return _separate(result, svar, dvar) + + +def pde_separate_add(eq, fun, sep): + """ + Helper function for searching additive separable solutions. + + Consider an equation of two independent variables x, y and a dependent + variable w, we look for the product of two functions depending on different + arguments: + + `w(x, y, z) = X(x) + y(y, z)` + + Examples + ======== + + >>> from sympy import E, Eq, Function, pde_separate_add, Derivative as D + >>> from sympy.abc import x, t + >>> u, X, T = map(Function, 'uXT') + + >>> eq = Eq(D(u(x, t), x), E**(u(x, t))*D(u(x, t), t)) + >>> pde_separate_add(eq, u(x, t), [X(x), T(t)]) + [exp(-X(x))*Derivative(X(x), x), exp(T(t))*Derivative(T(t), t)] + + """ + return pde_separate(eq, fun, sep, strategy='add') + + +def pde_separate_mul(eq, fun, sep): + """ + Helper function for searching multiplicative separable solutions. + + Consider an equation of two independent variables x, y and a dependent + variable w, we look for the product of two functions depending on different + arguments: + + `w(x, y, z) = X(x)*u(y, z)` + + Examples + ======== + + >>> from sympy import Function, Eq, pde_separate_mul, Derivative as D + >>> from sympy.abc import x, y + >>> u, X, Y = map(Function, 'uXY') + + >>> eq = Eq(D(u(x, y), x, 2), D(u(x, y), y, 2)) + >>> pde_separate_mul(eq, u(x, y), [X(x), Y(y)]) + [Derivative(X(x), (x, 2))/X(x), Derivative(Y(y), (y, 2))/Y(y)] + + """ + return pde_separate(eq, fun, sep, strategy='mul') + + +def _separate(eq, dep, others): + """Separate expression into two parts based on dependencies of variables.""" + + # FIRST PASS + # Extract derivatives depending our separable variable... + terms = set() + for term in eq.args: + if term.is_Mul: + for i in term.args: + if i.is_Derivative and not i.has(*others): + terms.add(term) + continue + elif term.is_Derivative and not term.has(*others): + terms.add(term) + # Find the factor that we need to divide by + div = set() + for term in terms: + ext, sep = term.expand().as_independent(dep) + # Failed? + if sep.has(*others): + return None + div.add(ext) + # FIXME: Find lcm() of all the divisors and divide with it, instead of + # current hack :( + # https://github.com/sympy/sympy/issues/4597 + if len(div) > 0: + # double sum required or some tests will fail + eq = Add(*[simplify(Add(*[term/i for i in div])) for term in eq.args]) + # SECOND PASS - separate the derivatives + div = set() + lhs = rhs = 0 + for term in eq.args: + # Check, whether we have already term with independent variable... + if not term.has(*others): + lhs += term + continue + # ...otherwise, try to separate + temp, sep = term.expand().as_independent(dep) + # Failed? + if sep.has(*others): + return None + # Extract the divisors + div.add(sep) + rhs -= term.expand() + # Do the division + fulldiv = reduce(operator.add, div) + lhs = simplify(lhs/fulldiv).expand() + rhs = simplify(rhs/fulldiv).expand() + # ...and check whether we were successful :) + if lhs.has(*others) or rhs.has(dep): + return None + return [lhs, rhs] diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/polysys.py b/pllava/lib/python3.10/site-packages/sympy/solvers/polysys.py new file mode 100644 index 0000000000000000000000000000000000000000..eb4787e2fc32c3128bbf7de4770fadc0d211f9e8 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/polysys.py @@ -0,0 +1,437 @@ +"""Solvers of systems of polynomial equations. """ +import itertools + +from sympy.core import S +from sympy.core.sorting import default_sort_key +from sympy.polys import Poly, groebner, roots +from sympy.polys.polytools import parallel_poly_from_expr +from sympy.polys.polyerrors import (ComputationFailed, + PolificationFailed, CoercionFailed) +from sympy.simplify import rcollect +from sympy.utilities import postfixes +from sympy.utilities.misc import filldedent + + +class SolveFailed(Exception): + """Raised when solver's conditions were not met. """ + + +def solve_poly_system(seq, *gens, strict=False, **args): + """ + Return a list of solutions for the system of polynomial equations + or else None. + + Parameters + ========== + + seq: a list/tuple/set + Listing all the equations that are needed to be solved + gens: generators + generators of the equations in seq for which we want the + solutions + strict: a boolean (default is False) + if strict is True, NotImplementedError will be raised if + the solution is known to be incomplete (which can occur if + not all solutions are expressible in radicals) + args: Keyword arguments + Special options for solving the equations. + + + Returns + ======= + + List[Tuple] + a list of tuples with elements being solutions for the + symbols in the order they were passed as gens + None + None is returned when the computed basis contains only the ground. + + Examples + ======== + + >>> from sympy import solve_poly_system + >>> from sympy.abc import x, y + + >>> solve_poly_system([x*y - 2*y, 2*y**2 - x**2], x, y) + [(0, 0), (2, -sqrt(2)), (2, sqrt(2))] + + >>> solve_poly_system([x**5 - x + y**3, y**2 - 1], x, y, strict=True) + Traceback (most recent call last): + ... + UnsolvableFactorError + + """ + try: + polys, opt = parallel_poly_from_expr(seq, *gens, **args) + except PolificationFailed as exc: + raise ComputationFailed('solve_poly_system', len(seq), exc) + + if len(polys) == len(opt.gens) == 2: + f, g = polys + + if all(i <= 2 for i in f.degree_list() + g.degree_list()): + try: + return solve_biquadratic(f, g, opt) + except SolveFailed: + pass + + return solve_generic(polys, opt, strict=strict) + + +def solve_biquadratic(f, g, opt): + """Solve a system of two bivariate quadratic polynomial equations. + + Parameters + ========== + + f: a single Expr or Poly + First equation + g: a single Expr or Poly + Second Equation + opt: an Options object + For specifying keyword arguments and generators + + Returns + ======= + + List[Tuple] + a list of tuples with elements being solutions for the + symbols in the order they were passed as gens + None + None is returned when the computed basis contains only the ground. + + Examples + ======== + + >>> from sympy import Options, Poly + >>> from sympy.abc import x, y + >>> from sympy.solvers.polysys import solve_biquadratic + >>> NewOption = Options((x, y), {'domain': 'ZZ'}) + + >>> a = Poly(y**2 - 4 + x, y, x, domain='ZZ') + >>> b = Poly(y*2 + 3*x - 7, y, x, domain='ZZ') + >>> solve_biquadratic(a, b, NewOption) + [(1/3, 3), (41/27, 11/9)] + + >>> a = Poly(y + x**2 - 3, y, x, domain='ZZ') + >>> b = Poly(-y + x - 4, y, x, domain='ZZ') + >>> solve_biquadratic(a, b, NewOption) + [(7/2 - sqrt(29)/2, -sqrt(29)/2 - 1/2), (sqrt(29)/2 + 7/2, -1/2 + \ + sqrt(29)/2)] + """ + G = groebner([f, g]) + + if len(G) == 1 and G[0].is_ground: + return None + + if len(G) != 2: + raise SolveFailed + + x, y = opt.gens + p, q = G + if not p.gcd(q).is_ground: + # not 0-dimensional + raise SolveFailed + + p = Poly(p, x, expand=False) + p_roots = [rcollect(expr, y) for expr in roots(p).keys()] + + q = q.ltrim(-1) + q_roots = list(roots(q).keys()) + + solutions = [(p_root.subs(y, q_root), q_root) for q_root, p_root in + itertools.product(q_roots, p_roots)] + + return sorted(solutions, key=default_sort_key) + + +def solve_generic(polys, opt, strict=False): + """ + Solve a generic system of polynomial equations. + + Returns all possible solutions over C[x_1, x_2, ..., x_m] of a + set F = { f_1, f_2, ..., f_n } of polynomial equations, using + Groebner basis approach. For now only zero-dimensional systems + are supported, which means F can have at most a finite number + of solutions. If the basis contains only the ground, None is + returned. + + The algorithm works by the fact that, supposing G is the basis + of F with respect to an elimination order (here lexicographic + order is used), G and F generate the same ideal, they have the + same set of solutions. By the elimination property, if G is a + reduced, zero-dimensional Groebner basis, then there exists an + univariate polynomial in G (in its last variable). This can be + solved by computing its roots. Substituting all computed roots + for the last (eliminated) variable in other elements of G, new + polynomial system is generated. Applying the above procedure + recursively, a finite number of solutions can be found. + + The ability of finding all solutions by this procedure depends + on the root finding algorithms. If no solutions were found, it + means only that roots() failed, but the system is solvable. To + overcome this difficulty use numerical algorithms instead. + + Parameters + ========== + + polys: a list/tuple/set + Listing all the polynomial equations that are needed to be solved + opt: an Options object + For specifying keyword arguments and generators + strict: a boolean + If strict is True, NotImplementedError will be raised if the solution + is known to be incomplete + + Returns + ======= + + List[Tuple] + a list of tuples with elements being solutions for the + symbols in the order they were passed as gens + None + None is returned when the computed basis contains only the ground. + + References + ========== + + .. [Buchberger01] B. Buchberger, Groebner Bases: A Short + Introduction for Systems Theorists, In: R. Moreno-Diaz, + B. Buchberger, J.L. Freire, Proceedings of EUROCAST'01, + February, 2001 + + .. [Cox97] D. Cox, J. Little, D. O'Shea, Ideals, Varieties + and Algorithms, Springer, Second Edition, 1997, pp. 112 + + Raises + ======== + + NotImplementedError + If the system is not zero-dimensional (does not have a finite + number of solutions) + + UnsolvableFactorError + If ``strict`` is True and not all solution components are + expressible in radicals + + Examples + ======== + + >>> from sympy import Poly, Options + >>> from sympy.solvers.polysys import solve_generic + >>> from sympy.abc import x, y + >>> NewOption = Options((x, y), {'domain': 'ZZ'}) + + >>> a = Poly(x - y + 5, x, y, domain='ZZ') + >>> b = Poly(x + y - 3, x, y, domain='ZZ') + >>> solve_generic([a, b], NewOption) + [(-1, 4)] + + >>> a = Poly(x - 2*y + 5, x, y, domain='ZZ') + >>> b = Poly(2*x - y - 3, x, y, domain='ZZ') + >>> solve_generic([a, b], NewOption) + [(11/3, 13/3)] + + >>> a = Poly(x**2 + y, x, y, domain='ZZ') + >>> b = Poly(x + y*4, x, y, domain='ZZ') + >>> solve_generic([a, b], NewOption) + [(0, 0), (1/4, -1/16)] + + >>> a = Poly(x**5 - x + y**3, x, y, domain='ZZ') + >>> b = Poly(y**2 - 1, x, y, domain='ZZ') + >>> solve_generic([a, b], NewOption, strict=True) + Traceback (most recent call last): + ... + UnsolvableFactorError + + """ + def _is_univariate(f): + """Returns True if 'f' is univariate in its last variable. """ + for monom in f.monoms(): + if any(monom[:-1]): + return False + + return True + + def _subs_root(f, gen, zero): + """Replace generator with a root so that the result is nice. """ + p = f.as_expr({gen: zero}) + + if f.degree(gen) >= 2: + p = p.expand(deep=False) + + return p + + def _solve_reduced_system(system, gens, entry=False): + """Recursively solves reduced polynomial systems. """ + if len(system) == len(gens) == 1: + # the below line will produce UnsolvableFactorError if + # strict=True and the solution from `roots` is incomplete + zeros = list(roots(system[0], gens[-1], strict=strict).keys()) + return [(zero,) for zero in zeros] + + basis = groebner(system, gens, polys=True) + + if len(basis) == 1 and basis[0].is_ground: + if not entry: + return [] + else: + return None + + univariate = list(filter(_is_univariate, basis)) + + if len(basis) < len(gens): + raise NotImplementedError(filldedent(''' + only zero-dimensional systems supported + (finite number of solutions) + ''')) + + if len(univariate) == 1: + f = univariate.pop() + else: + raise NotImplementedError(filldedent(''' + only zero-dimensional systems supported + (finite number of solutions) + ''')) + + gens = f.gens + gen = gens[-1] + + # the below line will produce UnsolvableFactorError if + # strict=True and the solution from `roots` is incomplete + zeros = list(roots(f.ltrim(gen), strict=strict).keys()) + + if not zeros: + return [] + + if len(basis) == 1: + return [(zero,) for zero in zeros] + + solutions = [] + + for zero in zeros: + new_system = [] + new_gens = gens[:-1] + + for b in basis[:-1]: + eq = _subs_root(b, gen, zero) + + if eq is not S.Zero: + new_system.append(eq) + + for solution in _solve_reduced_system(new_system, new_gens): + solutions.append(solution + (zero,)) + + if solutions and len(solutions[0]) != len(gens): + raise NotImplementedError(filldedent(''' + only zero-dimensional systems supported + (finite number of solutions) + ''')) + return solutions + + try: + result = _solve_reduced_system(polys, opt.gens, entry=True) + except CoercionFailed: + raise NotImplementedError + + if result is not None: + return sorted(result, key=default_sort_key) + + +def solve_triangulated(polys, *gens, **args): + """ + Solve a polynomial system using Gianni-Kalkbrenner algorithm. + + The algorithm proceeds by computing one Groebner basis in the ground + domain and then by iteratively computing polynomial factorizations in + appropriately constructed algebraic extensions of the ground domain. + + Parameters + ========== + + polys: a list/tuple/set + Listing all the equations that are needed to be solved + gens: generators + generators of the equations in polys for which we want the + solutions + args: Keyword arguments + Special options for solving the equations + + Returns + ======= + + List[Tuple] + A List of tuples. Solutions for symbols that satisfy the + equations listed in polys + + Examples + ======== + + >>> from sympy import solve_triangulated + >>> from sympy.abc import x, y, z + + >>> F = [x**2 + y + z - 1, x + y**2 + z - 1, x + y + z**2 - 1] + + >>> solve_triangulated(F, x, y, z) + [(0, 0, 1), (0, 1, 0), (1, 0, 0)] + + References + ========== + + 1. Patrizia Gianni, Teo Mora, Algebraic Solution of System of + Polynomial Equations using Groebner Bases, AAECC-5 on Applied Algebra, + Algebraic Algorithms and Error-Correcting Codes, LNCS 356 247--257, 1989 + + """ + G = groebner(polys, gens, polys=True) + G = list(reversed(G)) + + domain = args.get('domain') + + if domain is not None: + for i, g in enumerate(G): + G[i] = g.set_domain(domain) + + f, G = G[0].ltrim(-1), G[1:] + dom = f.get_domain() + + zeros = f.ground_roots() + solutions = {((zero,), dom) for zero in zeros} + + var_seq = reversed(gens[:-1]) + vars_seq = postfixes(gens[1:]) + + for var, vars in zip(var_seq, vars_seq): + _solutions = set() + + for values, dom in solutions: + H, mapping = [], list(zip(vars, values)) + + for g in G: + _vars = (var,) + vars + + if g.has_only_gens(*_vars) and g.degree(var) != 0: + h = g.ltrim(var).eval(dict(mapping)) + + if g.degree(var) == h.degree(): + H.append(h) + + p = min(H, key=lambda h: h.degree()) + zeros = p.ground_roots() + + for zero in zeros: + if not zero.is_Rational: + dom_zero = dom.algebraic_field(zero) + else: + dom_zero = dom + + _solutions.add(((zero,) + values, dom_zero)) + + solutions = _solutions + + solutions = list(solutions) + + for i, (solution, _) in enumerate(solutions): + solutions[i] = solution + + return sorted(solutions, key=default_sort_key) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/simplex.py b/pllava/lib/python3.10/site-packages/sympy/solvers/simplex.py new file mode 100644 index 0000000000000000000000000000000000000000..f3940844c8e5eaa5aa2cfb38be88c0cbb170c5e5 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/simplex.py @@ -0,0 +1,1141 @@ +"""Tools for optimizing a linear function for a given simplex. + +For the linear objective function ``f`` with linear constraints +expressed using `Le`, `Ge` or `Eq` can be found with ``lpmin`` or +``lpmax``. The symbols are **unbounded** unless specifically +constrained. + +As an alternative, the matrices describing the objective and the +constraints, and an optional list of bounds can be passed to +``linprog`` which will solve for the minimization of ``C*x`` +under constraints ``A*x <= b`` and/or ``Aeq*x = beq``, and +individual bounds for variables given as ``(lo, hi)``. The values +returned are **nonnegative** unless bounds are provided that +indicate otherwise. + +Errors that might be raised are UnboundedLPError when there is no +finite solution for the system or InfeasibleLPError when the +constraints represent impossible conditions (i.e. a non-existant + simplex). + +Here is a simple 1-D system: minimize `x` given that ``x >= 1``. + + >>> from sympy.solvers.simplex import lpmin, linprog + >>> from sympy.abc import x + + The function and a list with the constraint is passed directly + to `lpmin`: + + >>> lpmin(x, [x >= 1]) + (1, {x: 1}) + + For `linprog` the matrix for the objective is `[1]` and the + uivariate constraint can be passed as a bound with None acting + as infinity: + + >>> linprog([1], bounds=(1, None)) + (1, [1]) + + Or the matrices, corresponding to ``x >= 1`` expressed as + ``-x <= -1`` as required by the routine, can be passed: + + >>> linprog([1], [-1], [-1]) + (1, [1]) + + If there is no limit for the objective, an error is raised. + In this case there is a valid region of interest (simplex) + but no limit to how small ``x`` can be: + + >>> lpmin(x, []) + Traceback (most recent call last): + ... + sympy.solvers.simplex.UnboundedLPError: + Objective function can assume arbitrarily large values! + + An error is raised if there is no possible solution: + + >>> lpmin(x,[x<=1,x>=2]) + Traceback (most recent call last): + ... + sympy.solvers.simplex.InfeasibleLPError: + Inconsistent/False constraint +""" + +from sympy.core import sympify +from sympy.core.exprtools import factor_terms +from sympy.core.relational import Le, Ge, Eq +from sympy.core.singleton import S +from sympy.core.symbol import Dummy +from sympy.core.sorting import ordered +from sympy.functions.elementary.complexes import sign +from sympy.matrices.dense import Matrix, zeros +from sympy.solvers.solveset import linear_eq_to_matrix +from sympy.utilities.iterables import numbered_symbols +from sympy.utilities.misc import filldedent + + +class UnboundedLPError(Exception): + """ + A linear programing problem is said to be unbounded if its objective + function can assume arbitrarily large values. + + Example + ======= + + Suppose you want to maximize + 2x + subject to + x >= 0 + + There's no upper limit that 2x can take. + """ + + pass + + +class InfeasibleLPError(Exception): + """ + A linear programing problem is considered infeasible if its + constraint set is empty. That is, if the set of all vectors + satisfying the contraints is empty, then the problem is infeasible. + + Example + ======= + + Suppose you want to maximize + x + subject to + x >= 10 + x <= 9 + + No x can satisfy those constraints. + """ + + pass + + +def _pivot(M, i, j): + """ + The pivot element `M[i, j]` is inverted and the rest of the matrix + modified and returned as a new matrix; original is left unmodified. + + Example + ======= + + >>> from sympy.matrices.dense import Matrix + >>> from sympy.solvers.simplex import _pivot + >>> from sympy import var + >>> Matrix(3, 3, var('a:i')) + Matrix([ + [a, b, c], + [d, e, f], + [g, h, i]]) + >>> _pivot(_, 1, 0) + Matrix([ + [-a/d, -a*e/d + b, -a*f/d + c], + [ 1/d, e/d, f/d], + [-g/d, h - e*g/d, i - f*g/d]]) + """ + Mi, Mj, Mij = M[i, :], M[:, j], M[i, j] + if Mij == 0: + raise ZeroDivisionError( + "Tried to pivot about zero-valued entry.") + A = M - Mj * (Mi / Mij) + A[i, :] = Mi / Mij + A[:, j] = -Mj / Mij + A[i, j] = 1 / Mij + return A + + +def _choose_pivot_row(A, B, candidate_rows, pivot_col, Y): + # Choose row with smallest ratio + # If there are ties, pick using Bland's rule + return min(candidate_rows, key=lambda i: (B[i] / A[i, pivot_col], Y[i])) + + +def _simplex(A, B, C, D=None, dual=False): + """Return ``(o, x, y)`` obtained from the two-phase simplex method + using Bland's rule: ``o`` is the minimum value of primal, + ``Cx - D``, under constraints ``Ax <= B`` (with ``x >= 0``) and + the maximum of the dual, ``y^{T}B - D``, under constraints + ``A^{T}*y >= C^{T}`` (with ``y >= 0``). To compute the dual of + the system, pass `dual=True` and ``(o, y, x)`` will be returned. + + Note: the nonnegative constraints for ``x`` and ``y`` supercede + any values of ``A`` and ``B`` that are inconsistent with that + assumption, so if a constraint of ``x >= -1`` is represented + in ``A`` and ``B``, no value will be obtained that is negative; if + a constraint of ``x <= -1`` is represented, an error will be + raised since no solution is possible. + + This routine relies on the ability of determining whether an + expression is 0 or not. This is guaranteed if the input contains + only Float or Rational entries. It will raise a TypeError if + a relationship does not evaluate to True or False. + + Examples + ======== + + >>> from sympy.solvers.simplex import _simplex + >>> from sympy import Matrix + + Consider the simple minimization of ``f = x + y + 1`` under the + constraint that ``y + 2*x >= 4``. This is the "standard form" of + a minimization. + + In the nonnegative quadrant, this inequality describes a area above + a triangle with vertices at (0, 4), (0, 0) and (2, 0). The minimum + of ``f`` occurs at (2, 0). Define A, B, C, D for the standard + minimization: + + >>> A = Matrix([[2, 1]]) + >>> B = Matrix([4]) + >>> C = Matrix([[1, 1]]) + >>> D = Matrix([-1]) + + Confirm that this is the system of interest: + + >>> from sympy.abc import x, y + >>> X = Matrix([x, y]) + >>> (C*X - D)[0] + x + y + 1 + >>> [i >= j for i, j in zip(A*X, B)] + [2*x + y >= 4] + + Since `_simplex` will do a minimization for constraints given as + ``A*x <= B``, the signs of ``A`` and ``B`` must be negated since + the currently correspond to a greater-than inequality: + + >>> _simplex(-A, -B, C, D) + (3, [2, 0], [1/2]) + + The dual of minimizing ``f`` is maximizing ``F = c*y - d`` for + ``a*y <= b`` where ``a``, ``b``, ``c``, ``d`` are derived from the + transpose of the matrix representation of the standard minimization: + + >>> tr = lambda a, b, c, d: [i.T for i in (a, c, b, d)] + >>> a, b, c, d = tr(A, B, C, D) + + This time ``a*x <= b`` is the expected inequality for the `_simplex` + method, but to maximize ``F``, the sign of ``c`` and ``d`` must be + changed (so that minimizing the negative will give the negative of + the maximum of ``F``): + + >>> _simplex(a, b, -c, -d) + (-3, [1/2], [2, 0]) + + The negative of ``F`` and the min of ``f`` are the same. The dual + point `[1/2]` is the value of ``y`` that minimized ``F = c*y - d`` + under constraints a*x <= b``: + + >>> y = Matrix(['y']) + >>> (c*y - d)[0] + 4*y + 1 + >>> [i <= j for i, j in zip(a*y,b)] + [2*y <= 1, y <= 1] + + In this 1-dimensional dual system, the more restrictive contraint is + the first which limits ``y`` between 0 and 1/2 and the maximum of + ``F`` is attained at the nonzero value, hence is ``4*(1/2) + 1 = 3``. + + In this case the values for ``x`` and ``y`` were the same when the + dual representation was solved. This is not always the case (though + the value of the function will be the same). + + >>> l = [[1, 1], [-1, 1], [0, 1], [-1, 0]], [5, 1, 2, -1], [[1, 1]], [-1] + >>> A, B, C, D = [Matrix(i) for i in l] + >>> _simplex(A, B, -C, -D) + (-6, [3, 2], [1, 0, 0, 0]) + >>> _simplex(A, B, -C, -D, dual=True) # [5, 0] != [3, 2] + (-6, [1, 0, 0, 0], [5, 0]) + + In both cases the function has the same value: + + >>> Matrix(C)*Matrix([3, 2]) == Matrix(C)*Matrix([5, 0]) + True + + See Also + ======== + _lp - poses min/max problem in form compatible with _simplex + lpmin - minimization which calls _lp + lpmax - maximimzation which calls _lp + + References + ========== + + .. [1] Thomas S. Ferguson, LINEAR PROGRAMMING: A Concise Introduction + web.tecnico.ulisboa.pt/mcasquilho/acad/or/ftp/FergusonUCLA_lp.pdf + + """ + A, B, C, D = [Matrix(i) for i in (A, B, C, D or [0])] + if dual: + _o, d, p = _simplex(-A.T, C.T, B.T, -D) + return -_o, d, p + + if A and B: + M = Matrix([[A, B], [C, D]]) + else: + if A or B: + raise ValueError("must give A and B") + # no constraints given + M = Matrix([[C, D]]) + n = M.cols - 1 + m = M.rows - 1 + + if not all(i.is_Float or i.is_Rational for i in M): + # with literal Float and Rational we are guaranteed the + # ability of determining whether an expression is 0 or not + raise TypeError(filldedent(""" + Only rationals and floats are allowed. + """ + ) + ) + + # x variables have priority over y variables during Bland's rule + # since False < True + X = [(False, j) for j in range(n)] + Y = [(True, i) for i in range(m)] + + # Phase 1: find a feasible solution or determine none exist + + ## keep track of last pivot row and column + last = None + + while True: + B = M[:-1, -1] + A = M[:-1, :-1] + if all(B[i] >= 0 for i in range(B.rows)): + # We have found a feasible solution + break + + # Find k: first row with a negative rightmost entry + for k in range(B.rows): + if B[k] < 0: + break # use current value of k below + else: + pass # error will raise below + + # Choose pivot column, c + piv_cols = [_ for _ in range(A.cols) if A[k, _] < 0] + if not piv_cols: + raise InfeasibleLPError(filldedent(""" + The constraint set is empty!""")) + _, c = min((X[i], i) for i in piv_cols) # Bland's rule + + # Choose pivot row, r + piv_rows = [_ for _ in range(A.rows) if A[_, c] > 0 and B[_] > 0] + piv_rows.append(k) + r = _choose_pivot_row(A, B, piv_rows, c, Y) + + # check for oscillation + if (r, c) == last: + # Not sure what to do here; it looks like there will be + # oscillations; see o1 test added at this commit to + # see a system with no solution and the o2 for one + # with a solution. In the case of o2, the solution + # from linprog is the same as the one from lpmin, but + # the matrices created in the lpmin case are different + # than those created without replacements in linprog and + # the matrices in the linprog case lead to oscillations. + # If the matrices could be re-written in linprog like + # lpmin does, this behavior could be avoided and then + # perhaps the oscillating case would only occur when + # there is no solution. For now, the output is checked + # before exit if oscillations were detected and an + # error is raised there if the solution was invalid. + # + # cf section 6 of Ferguson for a non-cycling modification + last = True + break + last = r, c + + M = _pivot(M, r, c) + X[c], Y[r] = Y[r], X[c] + + # Phase 2: from a feasible solution, pivot to optimal + while True: + B = M[:-1, -1] + A = M[:-1, :-1] + C = M[-1, :-1] + + # Choose a pivot column, c + piv_cols = [] + piv_cols = [_ for _ in range(n) if C[_] < 0] + if not piv_cols: + break + _, c = min((X[i], i) for i in piv_cols) # Bland's rule + + # Choose a pivot row, r + piv_rows = [_ for _ in range(m) if A[_, c] > 0] + if not piv_rows: + raise UnboundedLPError(filldedent(""" + Objective function can assume + arbitrarily large values!""")) + r = _choose_pivot_row(A, B, piv_rows, c, Y) + + M = _pivot(M, r, c) + X[c], Y[r] = Y[r], X[c] + + argmax = [None] * n + argmin_dual = [None] * m + + for i, (v, n) in enumerate(X): + if v == False: + argmax[n] = 0 + else: + argmin_dual[n] = M[-1, i] + + for i, (v, n) in enumerate(Y): + if v == True: + argmin_dual[n] = 0 + else: + argmax[n] = M[i, -1] + + if last and not all(i >= 0 for i in argmax + argmin_dual): + raise InfeasibleLPError(filldedent(""" + Oscillating system led to invalid solution. + If you believe there was a valid solution, please + report this as a bug.""")) + return -M[-1, -1], argmax, argmin_dual + + +## routines that use _simplex or support those that do + + +def _abcd(M, list=False): + """return parts of M as matrices or lists + + Examples + ======== + + >>> from sympy import Matrix + >>> from sympy.solvers.simplex import _abcd + + >>> m = Matrix(3, 3, range(9)); m + Matrix([ + [0, 1, 2], + [3, 4, 5], + [6, 7, 8]]) + >>> a, b, c, d = _abcd(m) + >>> a + Matrix([ + [0, 1], + [3, 4]]) + >>> b + Matrix([ + [2], + [5]]) + >>> c + Matrix([[6, 7]]) + >>> d + Matrix([[8]]) + + The matrices can be returned as compact lists, too: + + >>> L = a, b, c, d = _abcd(m, list=True); L + ([[0, 1], [3, 4]], [2, 5], [[6, 7]], [8]) + """ + + def aslist(i): + l = i.tolist() + if len(l[0]) == 1: # col vector + return [i[0] for i in l] + return l + + m = M[:-1, :-1], M[:-1, -1], M[-1, :-1], M[-1:, -1:] + if not list: + return m + return tuple([aslist(i) for i in m]) + + +def _m(a, b, c, d=None): + """return Matrix([[a, b], [c, d]]) from matrices + in Matrix or list form. + + Examples + ======== + + >>> from sympy import Matrix + >>> from sympy.solvers.simplex import _abcd, _m + >>> m = Matrix(3, 3, range(9)) + >>> L = _abcd(m, list=True); L + ([[0, 1], [3, 4]], [2, 5], [[6, 7]], [8]) + >>> _abcd(m) + (Matrix([ + [0, 1], + [3, 4]]), Matrix([ + [2], + [5]]), Matrix([[6, 7]]), Matrix([[8]])) + >>> assert m == _m(*L) == _m(*_) + """ + a, b, c, d = [Matrix(i) for i in (a, b, c, d or [0])] + return Matrix([[a, b], [c, d]]) + + +def _primal_dual(M, factor=True): + """return primal and dual function and constraints + assuming that ``M = Matrix([[A, b], [c, d]])`` and the + function ``c*x - d`` is being minimized with ``Ax >= b`` + for nonnegative values of ``x``. The dual and its + constraints will be for maximizing `b.T*y - d` subject + to ``A.T*y <= c.T``. + + Examples + ======== + + >>> from sympy.solvers.simplex import _primal_dual, lpmin, lpmax + >>> from sympy import Matrix + + The following matrix represents the primal task of + minimizing x + y + 7 for y >= x + 1 and y >= -2*x + 3. + The dual task seeks to maximize x + 3*y + 7 with + 2*y - x <= 1 and and x + y <= 1: + + >>> M = Matrix([ + ... [-1, 1, 1], + ... [ 2, 1, 3], + ... [ 1, 1, -7]]) + >>> p, d = _primal_dual(M) + + The minimum of the primal and maximum of the dual are the same + (though they occur at different points): + + >>> lpmin(*p) + (28/3, {x1: 2/3, x2: 5/3}) + >>> lpmax(*d) + (28/3, {y1: 1/3, y2: 2/3}) + + If the equivalent (but canonical) inequalities are + desired, leave `factor=True`, otherwise the unmodified + inequalities for M will be returned. + + >>> m = Matrix([ + ... [-3, -2, 4, -2], + ... [ 2, 0, 0, -2], + ... [ 0, 1, -3, 0]]) + + >>> _primal_dual(m, False) # last condition is 2*x1 >= -2 + ((x2 - 3*x3, + [-3*x1 - 2*x2 + 4*x3 >= -2, 2*x1 >= -2]), + (-2*y1 - 2*y2, + [-3*y1 + 2*y2 <= 0, -2*y1 <= 1, 4*y1 <= -3])) + + >>> _primal_dual(m) # condition now x1 >= -1 + ((x2 - 3*x3, + [-3*x1 - 2*x2 + 4*x3 >= -2, x1 >= -1]), + (-2*y1 - 2*y2, + [-3*y1 + 2*y2 <= 0, -2*y1 <= 1, 4*y1 <= -3])) + + If you pass the transpose of the matrix, the primal will be + identified as the standard minimization problem and the + dual as the standard maximization: + + >>> _primal_dual(m.T) + ((-2*x1 - 2*x2, + [-3*x1 + 2*x2 >= 0, -2*x1 >= 1, 4*x1 >= -3]), + (y2 - 3*y3, + [-3*y1 - 2*y2 + 4*y3 <= -2, y1 <= -1])) + + A matrix must have some size or else None will be returned for + the functions: + + >>> _primal_dual(Matrix([[1, 2]])) + ((x1 - 2, []), (-2, [])) + + >>> _primal_dual(Matrix([])) + ((None, []), (None, [])) + + References + ========== + + .. [1] David Galvin, Relations between Primal and Dual + www3.nd.edu/~dgalvin1/30210/30210_F07/presentations/dual_opt.pdf + """ + if not M: + return (None, []), (None, []) + if not hasattr(M, "shape"): + if len(M) not in (3, 4): + raise ValueError("expecting Matrix or 3 or 4 lists") + M = _m(*M) + m, n = [i - 1 for i in M.shape] + A, b, c, d = _abcd(M) + d = d[0] + _ = lambda x: numbered_symbols(x, start=1) + x = Matrix([i for i, j in zip(_("x"), range(n))]) + yT = Matrix([i for i, j in zip(_("y"), range(m))]).T + + def ineq(L, r, op): + rv = [] + for r in (op(i, j) for i, j in zip(L, r)): + if r == True: + continue + elif r == False: + return [False] + if factor: + f = factor_terms(r) + if f.lhs.is_Mul and f.rhs % f.lhs.args[0] == 0: + assert len(f.lhs.args) == 2, f.lhs + k = f.lhs.args[0] + r = r.func(sign(k) * f.lhs.args[1], f.rhs // abs(k)) + rv.append(r) + return rv + + eq = lambda x, d: x[0] - d if x else -d + F = eq(c * x, d) + f = eq(yT * b, d) + return (F, ineq(A * x, b, Ge)), (f, ineq(yT * A, c, Le)) + + +def _rel_as_nonpos(constr, syms): + """return `(np, d, aux)` where `np` is a list of nonpositive + expressions that represent the given constraints (possibly + rewritten in terms of auxilliary variables) expressible with + nonnegative symbols, and `d` is a dictionary mapping a given + symbols to an expression with an auxilliary variable. In some + cases a symbol will be used as part of the change of variables, + e.g. x: x - z1 instead of x: z1 - z2. + + If any constraint is False/empty, return None. All variables in + ``constr`` are assumed to be unbounded unless explicitly indicated + otherwise with a univariate constraint, e.g. ``x >= 0`` will + restrict ``x`` to nonnegative values. + + The ``syms`` must be included so all symbols can be given an + unbounded assumption if they are not otherwise bound with + univariate conditions like ``x <= 3``. + + Examples + ======== + + >>> from sympy.solvers.simplex import _rel_as_nonpos + >>> from sympy.abc import x, y + >>> _rel_as_nonpos([x >= y, x >= 0, y >= 0], (x, y)) + ([-x + y], {}, []) + >>> _rel_as_nonpos([x >= 3, x <= 5], [x]) + ([_z1 - 2], {x: _z1 + 3}, [_z1]) + >>> _rel_as_nonpos([x <= 5], [x]) + ([], {x: 5 - _z1}, [_z1]) + >>> _rel_as_nonpos([x >= 1], [x]) + ([], {x: _z1 + 1}, [_z1]) + """ + r = {} # replacements to handle change of variables + np = [] # nonpositive expressions + aux = [] # auxilliary symbols added + ui = numbered_symbols("z", start=1, cls=Dummy) # auxilliary symbols + univariate = {} # {x: interval} for univariate constraints + unbound = [] # symbols designated as unbound + syms = set(syms) # the expected syms of the system + + # separate out univariates + for i in constr: + if i == True: + continue # ignore + if i == False: + return # no solution + if i.has(S.Infinity, S.NegativeInfinity): + raise ValueError("only finite bounds are permitted") + if isinstance(i, (Le, Ge)): + i = i.lts - i.gts + freei = i.free_symbols + if freei - syms: + raise ValueError( + "unexpected symbol(s) in constraint: %s" % (freei - syms) + ) + if len(freei) > 1: + np.append(i) + elif freei: + x = freei.pop() + if x in unbound: + continue # will handle later + ivl = Le(i, 0, evaluate=False).as_set() + if x not in univariate: + univariate[x] = ivl + else: + univariate[x] &= ivl + elif i: + return False + else: + raise TypeError(filldedent(""" + only equalities like Eq(x, y) or non-strict + inequalities like x >= y are allowed in lp, not %s""" % i)) + + # introduce auxilliary variables as needed for univariate + # inequalities + for x in syms: + i = univariate.get(x, True) + if not i: + return None # no solution possible + if i == True: + unbound.append(x) + continue + a, b = i.inf, i.sup + if a.is_infinite: + u = next(ui) + r[x] = b - u + aux.append(u) + elif b.is_infinite: + if a: + u = next(ui) + r[x] = a + u + aux.append(u) + else: + # standard nonnegative relationship + pass + else: + u = next(ui) + aux.append(u) + # shift so u = x - a => x = u + a + r[x] = u + a + # add constraint for u <= b - a + # since when u = b-a then x = u + a = b - a + a = b: + # the upper limit for x + np.append(u - (b - a)) + + # make change of variables for unbound variables + for x in unbound: + u = next(ui) + r[x] = u - x # reusing x + aux.append(u) + + return np, r, aux + + +def _lp_matrices(objective, constraints): + """return A, B, C, D, r, x+X, X for maximizing + objective = Cx - D with constraints Ax <= B, introducing + introducing auxilliary variables, X, as necessary to make + replacements of symbols as given in r, {xi: expression with Xj}, + so all variables in x+X will take on nonnegative values. + + Every univariate condition creates a semi-infinite + condition, e.g. a single ``x <= 3`` creates the + interval ``[-oo, 3]`` while ``x <= 3`` and ``x >= 2`` + create an interval ``[2, 3]``. Variables not in a univariate + expression will take on nonnegative values. + """ + + # sympify input and collect free symbols + F = sympify(objective) + np = [sympify(i) for i in constraints] + syms = set.union(*[i.free_symbols for i in [F] + np], set()) + + # change Eq(x, y) to x - y <= 0 and y - x <= 0 + for i in range(len(np)): + if isinstance(np[i], Eq): + np[i] = np[i].lhs - np[i].rhs <= 0 + np.append(-np[i].lhs <= 0) + + # convert constraints to nonpositive expressions + _ = _rel_as_nonpos(np, syms) + if _ is None: + raise InfeasibleLPError(filldedent(""" + Inconsistent/False constraint""")) + np, r, aux = _ + + # do change of variables + F = F.xreplace(r) + np = [i.xreplace(r) for i in np] + + # convert to matrices + xx = list(ordered(syms)) + aux + A, B = linear_eq_to_matrix(np, xx) + C, D = linear_eq_to_matrix([F], xx) + return A, B, C, D, r, xx, aux + + +def _lp(min_max, f, constr): + """Return the optimization (min or max) of ``f`` with the given + constraints. All variables are unbounded unless constrained. + + If `min_max` is 'max' then the results corresponding to the + maximization of ``f`` will be returned, else the minimization. + The constraints can be given as Le, Ge or Eq expressions. + + Examples + ======== + + >>> from sympy.solvers.simplex import _lp as lp + >>> from sympy import Eq + >>> from sympy.abc import x, y, z + >>> f = x + y - 2*z + >>> c = [7*x + 4*y - 7*z <= 3, 3*x - y + 10*z <= 6] + >>> c += [i >= 0 for i in (x, y, z)] + >>> lp(min, f, c) + (-6/5, {x: 0, y: 0, z: 3/5}) + + By passing max, the maximum value for f under the constraints + is returned (if possible): + + >>> lp(max, f, c) + (3/4, {x: 0, y: 3/4, z: 0}) + + Constraints that are equalities will require that the solution + also satisfy them: + + >>> lp(max, f, c + [Eq(y - 9*x, 1)]) + (5/7, {x: 0, y: 1, z: 1/7}) + + All symbols are reported, even if they are not in the objective + function: + + >>> lp(min, x, [y + x >= 3, x >= 0]) + (0, {x: 0, y: 3}) + """ + # get the matrix components for the system expressed + # in terms of only nonnegative variables + A, B, C, D, r, xx, aux = _lp_matrices(f, constr) + + how = str(min_max).lower() + if "max" in how: + # _simplex minimizes for Ax <= B so we + # have to change the sign of the function + # and negate the optimal value returned + _o, p, d = _simplex(A, B, -C, -D) + o = -_o + elif "min" in how: + o, p, d = _simplex(A, B, C, D) + else: + raise ValueError("expecting min or max") + + # restore original variables and remove aux from p + p = dict(zip(xx, p)) + if r: # p has original symbols and auxilliary symbols + # if r has x: x - z1 use values from p to update + r = {k: v.xreplace(p) for k, v in r.items()} + # then use the actual value of x (= x - z1) in p + p.update(r) + # don't show aux + p = {k: p[k] for k in ordered(p) if k not in aux} + + # not returning dual since there may be extra constraints + # when a variable has finite bounds + return o, p + + +def lpmin(f, constr): + """return minimum of linear equation ``f`` under + linear constraints expressed using Ge, Le or Eq. + + All variables are unbounded unless constrained. + + Examples + ======== + + >>> from sympy.solvers.simplex import lpmin + >>> from sympy import Eq + >>> from sympy.abc import x, y + >>> lpmin(x, [2*x - 3*y >= -1, Eq(x + 3*y, 2), x <= 2*y]) + (1/3, {x: 1/3, y: 5/9}) + + Negative values for variables are permitted unless explicitly + exluding, so minimizing ``x`` for ``x <= 3`` is an + unbounded problem while the following has a bounded solution: + + >>> lpmin(x, [x >= 0, x <= 3]) + (0, {x: 0}) + + Without indicating that ``x`` is nonnegative, there + is no minimum for this objective: + + >>> lpmin(x, [x <= 3]) + Traceback (most recent call last): + ... + sympy.solvers.simplex.UnboundedLPError: + Objective function can assume arbitrarily large values! + + See Also + ======== + linprog, lpmax + """ + return _lp(min, f, constr) + + +def lpmax(f, constr): + """return maximum of linear equation ``f`` under + linear constraints expressed using Ge, Le or Eq. + + All variables are unbounded unless constrained. + + Examples + ======== + + >>> from sympy.solvers.simplex import lpmax + >>> from sympy import Eq + >>> from sympy.abc import x, y + >>> lpmax(x, [2*x - 3*y >= -1, Eq(x+ 3*y,2), x <= 2*y]) + (4/5, {x: 4/5, y: 2/5}) + + Negative values for variables are permitted unless explicitly + exluding: + + >>> lpmax(x, [x <= -1]) + (-1, {x: -1}) + + If a non-negative constraint is added for x, there is no + possible solution: + + >>> lpmax(x, [x <= -1, x >= 0]) + Traceback (most recent call last): + ... + sympy.solvers.simplex.InfeasibleLPError: inconsistent/False constraint + + See Also + ======== + linprog, lpmin + """ + return _lp(max, f, constr) + + +def _handle_bounds(bounds): + # introduce auxilliary variables as needed for univariate + # inequalities + + unbound = [] + R = [0] * len(bounds) # a (growing) row of zeros + + def n(): + return len(R) - 1 + + def Arow(inc=1): + R.extend([0] * inc) + return R[:] + + row = [] + for x, (a, b) in enumerate(bounds): + if a is None and b is None: + unbound.append(x) + elif a is None: + # r[x] = b - u + A = Arow() + A[x] = 1 + A[n()] = 1 + B = [b] + row.append((A, B)) + A = [0] * len(A) + A[x] = -1 + A[n()] = -1 + B = [-b] + row.append((A, B)) + elif b is None: + if a: + # r[x] = a + u + A = Arow() + A[x] = 1 + A[n()] = -1 + B = [a] + row.append((A, B)) + A = [0] * len(A) + A[x] = -1 + A[n()] = 1 + B = [-a] + row.append((A, B)) + else: + # standard nonnegative relationship + pass + else: + # r[x] = u + a + A = Arow() + A[x] = 1 + A[n()] = -1 + B = [a] + row.append((A, B)) + A = [0] * len(A) + A[x] = -1 + A[n()] = 1 + B = [-a] + row.append((A, B)) + # u <= b - a + A = [0] * len(A) + A[x] = 0 + A[n()] = 1 + B = [b - a] + row.append((A, B)) + + # make change of variables for unbound variables + for x in unbound: + # r[x] = u - v + A = Arow(2) + B = [0] + A[x] = 1 + A[n()] = 1 + A[n() - 1] = -1 + row.append((A, B)) + A = [0] * len(A) + A[x] = -1 + A[n()] = -1 + A[n() - 1] = 1 + row.append((A, B)) + + return Matrix([r+[0]*(len(R) - len(r)) for r,_ in row] + ), Matrix([i[1] for i in row]) + + +def linprog(c, A=None, b=None, A_eq=None, b_eq=None, bounds=None): + """Return the minimization of ``c*x`` with the given + constraints ``A*x <= b`` and ``A_eq*x = b_eq``. Unless bounds + are given, variables will have nonnegative values in the solution. + + If ``A`` is not given, then the dimension of the system will + be determined by the length of ``C``. + + By default, all variables will be nonnegative. If ``bounds`` + is given as a single tuple, ``(lo, hi)``, then all variables + will be constrained to be between ``lo`` and ``hi``. Use + None for a ``lo`` or ``hi`` if it is unconstrained in the + negative or positive direction, respectively, e.g. + ``(None, 0)`` indicates nonpositive values. To set + individual ranges, pass a list with length equal to the + number of columns in ``A``, each element being a tuple; if + only a few variables take on non-default values they can be + passed as a dictionary with keys giving the corresponding + column to which the variable is assigned, e.g. ``bounds={2: + (1, 4)}`` would limit the 3rd variable to have a value in + range ``[1, 4]``. + + Examples + ======== + + >>> from sympy.solvers.simplex import linprog + >>> from sympy import symbols, Eq, linear_eq_to_matrix as M, Matrix + >>> x = x1, x2, x3, x4 = symbols('x1:5') + >>> X = Matrix(x) + >>> c, d = M(5*x2 + x3 + 4*x4 - x1, x) + >>> a, b = M([5*x2 + 2*x3 + 5*x4 - (x1 + 5)], x) + >>> aeq, beq = M([Eq(3*x2 + x4, 2), Eq(-x1 + x3 + 2*x4, 1)], x) + >>> constr = [i <= j for i,j in zip(a*X, b)] + >>> constr += [Eq(i, j) for i,j in zip(aeq*X, beq)] + >>> linprog(c, a, b, aeq, beq) + (9/2, [0, 1/2, 0, 1/2]) + >>> assert all(i.subs(dict(zip(x, _[1]))) for i in constr) + + See Also + ======== + lpmin, lpmax + """ + + ## the objective + C = Matrix(c) + if C.rows != 1 and C.cols == 1: + C = C.T + if C.rows != 1: + raise ValueError("C must be a single row.") + + ## the inequalities + if not A: + if b: + raise ValueError("A and b must both be given") + # the governing equations will be simple constraints + # on variables + A, b = zeros(0, C.cols), zeros(C.cols, 1) + else: + A, b = [Matrix(i) for i in (A, b)] + + if A.cols != C.cols: + raise ValueError("number of columns in A and C must match") + + ## the equalities + if A_eq is None: + if not b_eq is None: + raise ValueError("A_eq and b_eq must both be given") + else: + A_eq, b_eq = [Matrix(i) for i in (A_eq, b_eq)] + # if x == y then x <= y and x >= y (-x <= -y) + A = A.col_join(A_eq) + A = A.col_join(-A_eq) + b = b.col_join(b_eq) + b = b.col_join(-b_eq) + + if not (bounds is None or bounds == {} or bounds == (0, None)): + ## the bounds are interpreted + if type(bounds) is tuple and len(bounds) == 2: + bounds = [bounds] * A.cols + elif len(bounds) == A.cols and all( + type(i) is tuple and len(i) == 2 for i in bounds): + pass # individual bounds + elif type(bounds) is dict and all( + type(i) is tuple and len(i) == 2 + for i in bounds.values()): + # sparse bounds + db = bounds + bounds = [(0, None)] * A.cols + while db: + i, j = db.popitem() + bounds[i] = j # IndexError if out-of-bounds indices + else: + raise ValueError("unexpected bounds %s" % bounds) + A_, b_ = _handle_bounds(bounds) + aux = A_.cols - A.cols + if A: + A = Matrix([[A, zeros(A.rows, aux)], [A_]]) + b = b.col_join(b_) + else: + A = A_ + b = b_ + C = C.row_join(zeros(1, aux)) + else: + aux = -A.cols # set so -aux will give all cols below + + o, p, d = _simplex(A, b, C) + return o, p[:-aux] # don't include aux values + +def show_linprog(c, A=None, b=None, A_eq=None, b_eq=None, bounds=None): + from sympy import symbols + ## the objective + C = Matrix(c) + if C.rows != 1 and C.cols == 1: + C = C.T + if C.rows != 1: + raise ValueError("C must be a single row.") + + ## the inequalities + if not A: + if b: + raise ValueError("A and b must both be given") + # the governing equations will be simple constraints + # on variables + A, b = zeros(0, C.cols), zeros(C.cols, 1) + else: + A, b = [Matrix(i) for i in (A, b)] + + if A.cols != C.cols: + raise ValueError("number of columns in A and C must match") + + ## the equalities + if A_eq is None: + if not b_eq is None: + raise ValueError("A_eq and b_eq must both be given") + else: + A_eq, b_eq = [Matrix(i) for i in (A_eq, b_eq)] + + if not (bounds is None or bounds == {} or bounds == (0, None)): + ## the bounds are interpreted + if type(bounds) is tuple and len(bounds) == 2: + bounds = [bounds] * A.cols + elif len(bounds) == A.cols and all( + type(i) is tuple and len(i) == 2 for i in bounds): + pass # individual bounds + elif type(bounds) is dict and all( + type(i) is tuple and len(i) == 2 + for i in bounds.values()): + # sparse bounds + db = bounds + bounds = [(0, None)] * A.cols + while db: + i, j = db.popitem() + bounds[i] = j # IndexError if out-of-bounds indices + else: + raise ValueError("unexpected bounds %s" % bounds) + + x = Matrix(symbols('x1:%s' % (A.cols+1))) + f,c = (C*x)[0], [i<=j for i,j in zip(A*x, b)] + [Eq(i,j) for i,j in zip(A_eq*x,b_eq)] + for i, (lo, hi) in enumerate(bounds): + if lo is None and hi is None: + continue + if lo is None: + c.append(x[i]<=hi) + elif hi is None: + c.append(x[i]>=lo) + else: + c.append(x[i]>=lo) + c.append(x[i]<=hi) + return f,c diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/solvers.py b/pllava/lib/python3.10/site-packages/sympy/solvers/solvers.py new file mode 100644 index 0000000000000000000000000000000000000000..68b196cf305d48d790a79cfb907f4e05bd8817b1 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/solvers.py @@ -0,0 +1,3678 @@ +""" +This module contain solvers for all kinds of equations: + + - algebraic or transcendental, use solve() + + - recurrence, use rsolve() + + - differential, use dsolve() + + - nonlinear (numerically), use nsolve() + (you will need a good starting point) + +""" +from __future__ import annotations + +from sympy.core import (S, Add, Symbol, Dummy, Expr, Mul) +from sympy.core.assumptions import check_assumptions +from sympy.core.exprtools import factor_terms +from sympy.core.function import (expand_mul, expand_log, Derivative, + AppliedUndef, UndefinedFunction, nfloat, + Function, expand_power_exp, _mexpand, expand, + expand_func) +from sympy.core.logic import fuzzy_not +from sympy.core.numbers import Float, Rational, _illegal +from sympy.core.intfunc import integer_log, ilcm +from sympy.core.power import Pow +from sympy.core.relational import Eq, Ne +from sympy.core.sorting import ordered, default_sort_key +from sympy.core.sympify import sympify, _sympify +from sympy.core.traversal import preorder_traversal +from sympy.logic.boolalg import And, BooleanAtom + +from sympy.functions import (log, exp, LambertW, cos, sin, tan, acos, asin, atan, + Abs, re, im, arg, sqrt, atan2) +from sympy.functions.combinatorial.factorials import binomial +from sympy.functions.elementary.hyperbolic import HyperbolicFunction +from sympy.functions.elementary.piecewise import piecewise_fold, Piecewise +from sympy.functions.elementary.trigonometric import TrigonometricFunction +from sympy.integrals.integrals import Integral +from sympy.ntheory.factor_ import divisors +from sympy.simplify import (simplify, collect, powsimp, posify, # type: ignore + powdenest, nsimplify, denom, logcombine, sqrtdenest, fraction, + separatevars) +from sympy.simplify.sqrtdenest import sqrt_depth +from sympy.simplify.fu import TR1, TR2i, TR10, TR11 +from sympy.strategies.rl import rebuild +from sympy.matrices.exceptions import NonInvertibleMatrixError +from sympy.matrices import Matrix, zeros +from sympy.polys import roots, cancel, factor, Poly +from sympy.polys.solvers import sympy_eqs_to_ring, solve_lin_sys +from sympy.polys.polyerrors import GeneratorsNeeded, PolynomialError +from sympy.polys.polytools import gcd +from sympy.utilities.lambdify import lambdify +from sympy.utilities.misc import filldedent, debugf +from sympy.utilities.iterables import (connected_components, + generate_bell, uniq, iterable, is_sequence, subsets, flatten, sift) +from sympy.utilities.decorator import conserve_mpmath_dps + +from mpmath import findroot + +from sympy.solvers.polysys import solve_poly_system + +from types import GeneratorType +from collections import defaultdict +from itertools import combinations, product + +import warnings + + +def recast_to_symbols(eqs, symbols): + """ + Return (e, s, d) where e and s are versions of *eqs* and + *symbols* in which any non-Symbol objects in *symbols* have + been replaced with generic Dummy symbols and d is a dictionary + that can be used to restore the original expressions. + + Examples + ======== + + >>> from sympy.solvers.solvers import recast_to_symbols + >>> from sympy import symbols, Function + >>> x, y = symbols('x y') + >>> fx = Function('f')(x) + >>> eqs, syms = [fx + 1, x, y], [fx, y] + >>> e, s, d = recast_to_symbols(eqs, syms); (e, s, d) + ([_X0 + 1, x, y], [_X0, y], {_X0: f(x)}) + + The original equations and symbols can be restored using d: + + >>> assert [i.xreplace(d) for i in eqs] == eqs + >>> assert [d.get(i, i) for i in s] == syms + + """ + if not iterable(eqs) and iterable(symbols): + raise ValueError('Both eqs and symbols must be iterable') + orig = list(symbols) + symbols = list(ordered(symbols)) + swap_sym = {} + i = 0 + for s in symbols: + if not isinstance(s, Symbol) and s not in swap_sym: + swap_sym[s] = Dummy('X%d' % i) + i += 1 + new_f = [] + for i in eqs: + isubs = getattr(i, 'subs', None) + if isubs is not None: + new_f.append(isubs(swap_sym)) + else: + new_f.append(i) + restore = {v: k for k, v in swap_sym.items()} + return new_f, [swap_sym.get(i, i) for i in orig], restore + + +def _ispow(e): + """Return True if e is a Pow or is exp.""" + return isinstance(e, Expr) and (e.is_Pow or isinstance(e, exp)) + + +def _simple_dens(f, symbols): + # when checking if a denominator is zero, we can just check the + # base of powers with nonzero exponents since if the base is zero + # the power will be zero, too. To keep it simple and fast, we + # limit simplification to exponents that are Numbers + dens = set() + for d in denoms(f, symbols): + if d.is_Pow and d.exp.is_Number: + if d.exp.is_zero: + continue # foo**0 is never 0 + d = d.base + dens.add(d) + return dens + + +def denoms(eq, *symbols): + """ + Return (recursively) set of all denominators that appear in *eq* + that contain any symbol in *symbols*; if *symbols* are not + provided then all denominators will be returned. + + Examples + ======== + + >>> from sympy.solvers.solvers import denoms + >>> from sympy.abc import x, y, z + + >>> denoms(x/y) + {y} + + >>> denoms(x/(y*z)) + {y, z} + + >>> denoms(3/x + y/z) + {x, z} + + >>> denoms(x/2 + y/z) + {2, z} + + If *symbols* are provided then only denominators containing + those symbols will be returned: + + >>> denoms(1/x + 1/y + 1/z, y, z) + {y, z} + + """ + + pot = preorder_traversal(eq) + dens = set() + for p in pot: + # Here p might be Tuple or Relational + # Expr subtrees (e.g. lhs and rhs) will be traversed after by pot + if not isinstance(p, Expr): + continue + den = denom(p) + if den is S.One: + continue + dens.update(Mul.make_args(den)) + if not symbols: + return dens + elif len(symbols) == 1: + if iterable(symbols[0]): + symbols = symbols[0] + return {d for d in dens if any(s in d.free_symbols for s in symbols)} + + +def checksol(f, symbol, sol=None, **flags): + """ + Checks whether sol is a solution of equation f == 0. + + Explanation + =========== + + Input can be either a single symbol and corresponding value + or a dictionary of symbols and values. When given as a dictionary + and flag ``simplify=True``, the values in the dictionary will be + simplified. *f* can be a single equation or an iterable of equations. + A solution must satisfy all equations in *f* to be considered valid; + if a solution does not satisfy any equation, False is returned; if one or + more checks are inconclusive (and none are False) then None is returned. + + Examples + ======== + + >>> from sympy import checksol, symbols + >>> 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 + + To check if an expression is zero using ``checksol()``, pass it + as *f* and send an empty dictionary for *symbol*: + + >>> checksol(x**2 + x - x*(x + 1), {}) + 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. + 'warn=True (default is False)' + show a warning if checksol() could not conclude. + 'simplify=True (default)' + simplify solution before substituting into function and + simplify the function before trying specific simplifications + 'force=True (default is False)' + make positive all symbols without assumptions regarding sign. + + """ + from sympy.physics.units import Unit + + minimal = flags.get('minimal', False) + + 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 iterable(f): + if not f: + raise ValueError('no functions to check') + rv = True + for fi in f: + check = checksol(fi, sol, **flags) + if check: + continue + if check is False: + return False + rv = None # don't return, wait to see if there's a False + return rv + + f = _sympify(f) + + if f.is_number: + return f.is_zero + + if isinstance(f, Poly): + f = f.as_expr() + elif isinstance(f, (Eq, Ne)): + if f.rhs in (S.true, S.false): + f = f.reversed + B, E = f.args + if isinstance(B, BooleanAtom): + f = f.subs(sol) + if not f.is_Boolean: + return + elif isinstance(f, Eq): + f = Add(f.lhs, -f.rhs, evaluate=False) + + if isinstance(f, BooleanAtom): + return bool(f) + elif not f.is_Relational and not f: + return True + + illegal = set(_illegal) + if any(sympify(v).atoms() & illegal for k, v in sol.items()): + return False + + attempt = -1 + numerical = flags.get('numerical', True) + while 1: + attempt += 1 + if attempt == 0: + val = f.subs(sol) + if isinstance(val, Mul): + val = val.as_independent(Unit)[0] + if val.atoms() & illegal: + return False + elif attempt == 1: + if not val.is_number: + if not val.is_constant(*list(sol.keys()), simplify=not minimal): + return False + # there are free symbols -- simple expansion might work + _, val = val.as_content_primitive() + val = _mexpand(val.as_numer_denom()[0], recursive=True) + elif attempt == 2: + if minimal: + return + if flags.get('simplify', True): + for k in sol: + sol[k] = simplify(sol[k]) + # start over without the failed expanded form, possibly + # with a simplified solution + val = simplify(f.subs(sol)) + if flags.get('force', True): + val, reps = posify(val) + # expansion may work now, so try again and check + exval = _mexpand(val, recursive=True) + if exval.is_number: + # we can decide now + val = exval + else: + # if there are no radicals and no functions then this can't be + # zero anymore -- can it? + pot = preorder_traversal(expand_mul(val)) + seen = set() + saw_pow_func = False + for p in pot: + if p in seen: + continue + seen.add(p) + if p.is_Pow and not p.exp.is_Integer: + saw_pow_func = True + elif p.is_Function: + saw_pow_func = True + elif isinstance(p, UndefinedFunction): + saw_pow_func = True + if saw_pow_func: + break + if saw_pow_func is False: + return False + if flags.get('force', True): + # don't do a zero check with the positive assumptions in place + val = val.subs(reps) + nz = fuzzy_not(val.is_zero) + if nz is not None: + # issue 5673: nz may be True even when False + # so these are just hacks to keep a false positive + # from being returned + + # HACK 1: LambertW (issue 5673) + if val.is_number and val.has(LambertW): + # don't eval this to verify solution since if we got here, + # numerical must be False + return None + + # add other HACKs here if necessary, otherwise we assume + # the nz value is correct + return not nz + break + if val.is_Rational: + return val == 0 + if numerical and val.is_number: + return (abs(val.n(18).n(12, chop=True)) < 1e-9) is S.true + + if flags.get('warn', False): + warnings.warn("\n\tWarning: could not verify solution %s." % sol) + # returns None if it can't conclude + # TODO: improve solution testing + + +def solve(f, *symbols, **flags): + r""" + Algebraically solves equations and systems of equations. + + Explanation + =========== + + Currently supported: + - polynomial + - transcendental + - piecewise combinations of the above + - systems of linear and polynomial equations + - systems containing relational expressions + - systems implied by undetermined coefficients + + Examples + ======== + + The default output varies according to the input and might + be a list (possibly empty), a dictionary, a list of + dictionaries or tuples, or an expression involving relationals. + For specifics regarding different forms of output that may appear, see :ref:`solve_output`. + Let it suffice here to say that to obtain a uniform output from + `solve` use ``dict=True`` or ``set=True`` (see below). + + >>> from sympy import solve, Poly, Eq, Matrix, Symbol + >>> from sympy.abc import x, y, z, a, b + + The expressions that are passed can be Expr, Equality, or Poly + classes (or lists of the same); a Matrix is considered to be a + list of all the elements of the matrix: + + >>> solve(x - 3, x) + [3] + >>> solve(Eq(x, 3), x) + [3] + >>> solve(Poly(x - 3), x) + [3] + >>> solve(Matrix([[x, x + y]]), x, y) == solve([x, x + y], x, y) + True + + If no symbols are indicated to be of interest and the equation is + univariate, a list of values is returned; otherwise, the keys in + a dictionary will indicate which (of all the variables used in + the expression(s)) variables and solutions were found: + + >>> solve(x**2 - 4) + [-2, 2] + >>> solve((x - a)*(y - b)) + [{a: x}, {b: y}] + >>> solve([x - 3, y - 1]) + {x: 3, y: 1} + >>> solve([x - 3, y**2 - 1]) + [{x: 3, y: -1}, {x: 3, y: 1}] + + If you pass symbols for which solutions are sought, the output will vary + depending on the number of symbols you passed, whether you are passing + a list of expressions or not, and whether a linear system was solved. + Uniform output is attained by using ``dict=True`` or ``set=True``. + + >>> #### *** feel free to skip to the stars below *** #### + >>> from sympy import TableForm + >>> h = [None, ';|;'.join(['e', 's', 'solve(e, s)', 'solve(e, s, dict=True)', + ... 'solve(e, s, set=True)']).split(';')] + >>> t = [] + >>> for e, s in [ + ... (x - y, y), + ... (x - y, [x, y]), + ... (x**2 - y, [x, y]), + ... ([x - 3, y -1], [x, y]), + ... ]: + ... how = [{}, dict(dict=True), dict(set=True)] + ... res = [solve(e, s, **f) for f in how] + ... t.append([e, '|', s, '|'] + [res[0], '|', res[1], '|', res[2]]) + ... + >>> # ******************************************************* # + >>> TableForm(t, headings=h, alignments="<") + e | s | solve(e, s) | solve(e, s, dict=True) | solve(e, s, set=True) + --------------------------------------------------------------------------------------- + x - y | y | [x] | [{y: x}] | ([y], {(x,)}) + x - y | [x, y] | [(y, y)] | [{x: y}] | ([x, y], {(y, y)}) + x**2 - y | [x, y] | [(x, x**2)] | [{y: x**2}] | ([x, y], {(x, x**2)}) + [x - 3, y - 1] | [x, y] | {x: 3, y: 1} | [{x: 3, y: 1}] | ([x, y], {(3, 1)}) + + * If any equation does not depend on the symbol(s) given, it will be + eliminated from the equation set and an answer may be given + implicitly in terms of variables that were not of interest: + + >>> solve([x - y, y - 3], x) + {x: y} + + When you pass all but one of the free symbols, an attempt + is made to find a single solution based on the method of + undetermined coefficients. If it succeeds, a dictionary of values + is returned. If you want an algebraic solutions for one + or more of the symbols, pass the expression to be solved in a list: + + >>> e = a*x + b - 2*x - 3 + >>> solve(e, [a, b]) + {a: 2, b: 3} + >>> solve([e], [a, b]) + {a: -b/x + (2*x + 3)/x} + + When there is no solution for any given symbol which will make all + expressions zero, the empty list is returned (or an empty set in + the tuple when ``set=True``): + + >>> from sympy import sqrt + >>> solve(3, x) + [] + >>> solve(x - 3, y) + [] + >>> solve(sqrt(x) + 1, x, set=True) + ([x], set()) + + When an object other than a Symbol is given as a symbol, it is + isolated algebraically and an implicit solution may be obtained. + This is mostly provided as a convenience to save you from replacing + the object with a Symbol and solving for that Symbol. It will only + work if the specified object can be replaced with a Symbol using the + subs method: + + >>> from sympy import exp, Function + >>> f = Function('f') + + >>> solve(f(x) - x, f(x)) + [x] + >>> solve(f(x).diff(x) - f(x) - x, f(x).diff(x)) + [x + f(x)] + >>> solve(f(x).diff(x) - f(x) - x, f(x)) + [-x + Derivative(f(x), x)] + >>> solve(x + exp(x)**2, exp(x), set=True) + ([exp(x)], {(-sqrt(-x),), (sqrt(-x),)}) + + >>> from sympy import Indexed, IndexedBase, Tuple + >>> A = IndexedBase('A') + >>> eqs = Tuple(A[1] + A[2] - 3, A[1] - A[2] + 1) + >>> solve(eqs, eqs.atoms(Indexed)) + {A[1]: 1, A[2]: 2} + + * To solve for a function within a derivative, use :func:`~.dsolve`. + + To solve for a symbol implicitly, use implicit=True: + + >>> solve(x + exp(x), x) + [-LambertW(1)] + >>> solve(x + exp(x), x, implicit=True) + [-exp(x)] + + It is possible to solve for anything in an expression that can be + replaced with a symbol using :obj:`~sympy.core.basic.Basic.subs`: + + >>> solve(x + 2 + sqrt(3), x + 2) + [-sqrt(3)] + >>> solve((x + 2 + sqrt(3), x + 4 + y), y, x + 2) + {y: -2 + sqrt(3), x + 2: -sqrt(3)} + + * Nothing heroic is done in this implicit solving so you may end up + with a symbol still in the solution: + + >>> eqs = (x*y + 3*y + sqrt(3), x + 4 + y) + >>> solve(eqs, y, x + 2) + {y: -sqrt(3)/(x + 3), x + 2: -2*x/(x + 3) - 6/(x + 3) + sqrt(3)/(x + 3)} + >>> solve(eqs, y*x, x) + {x: -y - 4, x*y: -3*y - sqrt(3)} + + * If you attempt to solve for a number, remember that the number + you have obtained does not necessarily mean that the value is + equivalent to the expression obtained: + + >>> solve(sqrt(2) - 1, 1) + [sqrt(2)] + >>> solve(x - y + 1, 1) # /!\ -1 is targeted, too + [x/(y - 1)] + >>> [_.subs(z, -1) for _ in solve((x - y + 1).subs(-1, z), 1)] + [-x + y] + + **Additional Examples** + + ``solve()`` with check=True (default) will run through the symbol tags to + eliminate unwanted solutions. If no assumptions are included, all possible + solutions will be returned: + + >>> x = Symbol("x") + >>> solve(x**2 - 1) + [-1, 1] + + By setting the ``positive`` flag, only one solution will be returned: + + >>> pos = Symbol("pos", positive=True) + >>> solve(pos**2 - 1) + [1] + + When the solutions are checked, those that make any denominator zero + are automatically excluded. If you do not want to exclude such solutions, + then use the check=False option: + + >>> from sympy import sin, limit + >>> solve(sin(x)/x) # 0 is excluded + [pi] + + If ``check=False``, then a solution to the numerator being zero is found + but the value of $x = 0$ is a spurious solution since $\sin(x)/x$ has the well + known limit (without discontinuity) of 1 at $x = 0$: + + >>> solve(sin(x)/x, check=False) + [0, pi] + + In the following case, however, the limit exists and is equal to the + value of $x = 0$ that is excluded when check=True: + + >>> eq = x**2*(1/x - z**2/x) + >>> solve(eq, x) + [] + >>> solve(eq, x, check=False) + [0] + >>> limit(eq, x, 0, '-') + 0 + >>> limit(eq, x, 0, '+') + 0 + + **Solving Relationships** + + When one or more expressions passed to ``solve`` is a relational, + a relational result is returned (and the ``dict`` and ``set`` flags + are ignored): + + >>> solve(x < 3) + (-oo < x) & (x < 3) + >>> solve([x < 3, x**2 > 4], x) + ((-oo < x) & (x < -2)) | ((2 < x) & (x < 3)) + >>> solve([x + y - 3, x > 3], x) + (3 < x) & (x < oo) & Eq(x, 3 - y) + + Although checking of assumptions on symbols in relationals + is not done, setting assumptions will affect how certain + relationals might automatically simplify: + + >>> solve(x**2 > 4) + ((-oo < x) & (x < -2)) | ((2 < x) & (x < oo)) + + >>> r = Symbol('r', real=True) + >>> solve(r**2 > 4) + (2 < r) | (r < -2) + + There is currently no algorithm in SymPy that allows you to use + relationships to resolve more than one variable. So the following + does not determine that ``q < 0`` (and trying to solve for ``r`` + and ``q`` will raise an error): + + >>> from sympy import symbols + >>> r, q = symbols('r, q', real=True) + >>> solve([r + q - 3, r > 3], r) + (3 < r) & Eq(r, 3 - q) + + You can directly call the routine that ``solve`` calls + when it encounters a relational: :func:`~.reduce_inequalities`. + It treats Expr like Equality. + + >>> from sympy import reduce_inequalities + >>> reduce_inequalities([x**2 - 4]) + Eq(x, -2) | Eq(x, 2) + + If each relationship contains only one symbol of interest, + the expressions can be processed for multiple symbols: + + >>> reduce_inequalities([0 <= x - 1, y < 3], [x, y]) + (-oo < y) & (1 <= x) & (x < oo) & (y < 3) + + But an error is raised if any relationship has more than one + symbol of interest: + + >>> reduce_inequalities([0 <= x*y - 1, y < 3], [x, y]) + Traceback (most recent call last): + ... + NotImplementedError: + inequality has more than one symbol of interest. + + **Disabling High-Order Explicit Solutions** + + When solving polynomial expressions, you might not want explicit solutions + (which can be quite long). If the expression is univariate, ``CRootOf`` + instances will be returned instead: + + >>> solve(x**3 - x + 1) + [-1/((-1/2 - sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)) - + (-1/2 - sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)/3, + -(-1/2 + sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)/3 - + 1/((-1/2 + sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)), + -(3*sqrt(69)/2 + 27/2)**(1/3)/3 - + 1/(3*sqrt(69)/2 + 27/2)**(1/3)] + >>> solve(x**3 - x + 1, cubics=False) + [CRootOf(x**3 - x + 1, 0), + CRootOf(x**3 - x + 1, 1), + CRootOf(x**3 - x + 1, 2)] + + If the expression is multivariate, no solution might be returned: + + >>> solve(x**3 - x + a, x, cubics=False) + [] + + Sometimes solutions will be obtained even when a flag is False because the + expression could be factored. In the following example, the equation can + be factored as the product of a linear and a quadratic factor so explicit + solutions (which did not require solving a cubic expression) are obtained: + + >>> eq = x**3 + 3*x**2 + x - 1 + >>> solve(eq, cubics=False) + [-1, -1 + sqrt(2), -sqrt(2) - 1] + + **Solving Equations Involving Radicals** + + Because of SymPy's use of the principle root, some solutions + to radical equations will be missed unless check=False: + + >>> from sympy import root + >>> eq = root(x**3 - 3*x**2, 3) + 1 - x + >>> solve(eq) + [] + >>> solve(eq, check=False) + [1/3] + + In the above example, there is only a single solution to the + equation. Other expressions will yield spurious roots which + must be checked manually; roots which give a negative argument + to odd-powered radicals will also need special checking: + + >>> from sympy import real_root, S + >>> eq = root(x, 3) - root(x, 5) + S(1)/7 + >>> solve(eq) # this gives 2 solutions but misses a 3rd + [CRootOf(7*x**5 - 7*x**3 + 1, 1)**15, + CRootOf(7*x**5 - 7*x**3 + 1, 2)**15] + >>> sol = solve(eq, check=False) + >>> [abs(eq.subs(x,i).n(2)) for i in sol] + [0.48, 0.e-110, 0.e-110, 0.052, 0.052] + + The first solution is negative so ``real_root`` must be used to see that it + satisfies the expression: + + >>> abs(real_root(eq.subs(x, sol[0])).n(2)) + 0.e-110 + + If the roots of the equation are not real then more care will be + necessary to find the roots, especially for higher order equations. + Consider the following expression: + + >>> expr = root(x, 3) - root(x, 5) + + We will construct a known value for this expression at x = 3 by selecting + the 1-th root for each radical: + + >>> expr1 = root(x, 3, 1) - root(x, 5, 1) + >>> v = expr1.subs(x, -3) + + The ``solve`` function is unable to find any exact roots to this equation: + + >>> eq = Eq(expr, v); eq1 = Eq(expr1, v) + >>> solve(eq, check=False), solve(eq1, check=False) + ([], []) + + The function ``unrad``, however, can be used to get a form of the equation + for which numerical roots can be found: + + >>> from sympy.solvers.solvers import unrad + >>> from sympy import nroots + >>> e, (p, cov) = unrad(eq) + >>> pvals = nroots(e) + >>> inversion = solve(cov, x)[0] + >>> xvals = [inversion.subs(p, i) for i in pvals] + + Although ``eq`` or ``eq1`` could have been used to find ``xvals``, the + solution can only be verified with ``expr1``: + + >>> z = expr - v + >>> [xi.n(chop=1e-9) for xi in xvals if abs(z.subs(x, xi).n()) < 1e-9] + [] + >>> z1 = expr1 - v + >>> [xi.n(chop=1e-9) for xi in xvals if abs(z1.subs(x, xi).n()) < 1e-9] + [-3.0] + + Parameters + ========== + + f : + - a single Expr or Poly that must be zero + - an Equality + - a Relational expression + - a Boolean + - iterable of one or more of the above + + symbols : (object(s) to solve for) specified as + - none given (other non-numeric objects will be used) + - single symbol + - denested list of symbols + (e.g., ``solve(f, x, y)``) + - ordered iterable of symbols + (e.g., ``solve(f, [x, y])``) + + flags : + dict=True (default is False) + Return list (perhaps empty) of solution mappings. + set=True (default is False) + Return list of symbols and set of tuple(s) of solution(s). + exclude=[] (default) + Do not try to solve for any of the free symbols in exclude; + if expressions are given, the free symbols in them will + be extracted automatically. + check=True (default) + If False, do not do any testing of solutions. This can be + useful if you want to include solutions that make any + denominator zero. + numerical=True (default) + Do a fast numerical check if *f* has only one symbol. + minimal=True (default is False) + A very fast, minimal testing. + warn=True (default is False) + Show a warning if ``checksol()`` could not conclude. + simplify=True (default) + Simplify all but polynomials of order 3 or greater before + returning them and (if check is not False) use the + general simplify function on the solutions and the + expression obtained when they are substituted into the + function which should be zero. + force=True (default is False) + Make positive all symbols without assumptions regarding sign. + rational=True (default) + Recast Floats as Rational; if this option is not used, the + system containing Floats may fail to solve because of issues + with polys. If rational=None, Floats will be recast as + rationals but the answer will be recast as Floats. If the + flag is False then nothing will be done to the Floats. + manual=True (default is False) + Do not use the polys/matrix method to solve a system of + equations, solve them one at a time as you might "manually." + implicit=True (default is False) + Allows ``solve`` to return a solution for a pattern in terms of + other functions that contain that pattern; this is only + needed if the pattern is inside of some invertible function + like cos, exp, ect. + particular=True (default is False) + Instructs ``solve`` to try to find a particular solution to + a linear system with as many zeros as possible; this is very + expensive. + quick=True (default is False; ``particular`` must be True) + Selects a fast heuristic to find a solution with many zeros + whereas a value of False uses the very slow method guaranteed + to find the largest number of zeros possible. + cubics=True (default) + Return explicit solutions when cubic expressions are encountered. + When False, quartics and quintics are disabled, too. + quartics=True (default) + Return explicit solutions when quartic expressions are encountered. + When False, quintics are disabled, too. + quintics=True (default) + Return explicit solutions (if possible) when quintic expressions + are encountered. + + See Also + ======== + + rsolve: For solving recurrence relationships + dsolve: For solving differential equations + + """ + from .inequalities import reduce_inequalities + + # checking/recording flags + ########################################################################### + + # set solver types explicitly; as soon as one is False + # all the rest will be False + hints = ('cubics', 'quartics', 'quintics') + default = True + for k in hints: + default = flags.setdefault(k, bool(flags.get(k, default))) + + # allow solution to contain symbol if True: + implicit = flags.get('implicit', False) + + # record desire to see warnings + warn = flags.get('warn', False) + + # this flag will be needed for quick exits below, so record + # now -- but don't record `dict` yet since it might change + as_set = flags.get('set', False) + + # keeping track of how f was passed + bare_f = not iterable(f) + + # check flag usage for particular/quick which should only be used + # with systems of equations + if flags.get('quick', None) is not None: + if not flags.get('particular', None): + raise ValueError('when using `quick`, `particular` should be True') + if flags.get('particular', False) and bare_f: + raise ValueError(filldedent(""" + The 'particular/quick' flag is usually used with systems of + equations. Either pass your equation in a list or + consider using a solver like `diophantine` if you are + looking for a solution in integers.""")) + + # sympify everything, creating list of expressions and list of symbols + ########################################################################### + + def _sympified_list(w): + return list(map(sympify, w if iterable(w) else [w])) + f, symbols = (_sympified_list(w) for w in [f, symbols]) + + # preprocess symbol(s) + ########################################################################### + + ordered_symbols = None # were the symbols in a well defined order? + if not symbols: + # get symbols from equations + symbols = set().union(*[fi.free_symbols for fi in f]) + if len(symbols) < len(f): + for fi in f: + pot = preorder_traversal(fi) + for p in pot: + if isinstance(p, AppliedUndef): + if not as_set: + flags['dict'] = True # better show symbols + symbols.add(p) + pot.skip() # don't go any deeper + ordered_symbols = False + symbols = list(ordered(symbols)) # to make it canonical + else: + if len(symbols) == 1 and iterable(symbols[0]): + symbols = symbols[0] + ordered_symbols = symbols and is_sequence(symbols, + include=GeneratorType) + _symbols = list(uniq(symbols)) + if len(_symbols) != len(symbols): + ordered_symbols = False + symbols = list(ordered(symbols)) + else: + symbols = _symbols + + # check for duplicates + if len(symbols) != len(set(symbols)): + raise ValueError('duplicate symbols given') + # remove those not of interest + exclude = flags.pop('exclude', set()) + if exclude: + if isinstance(exclude, Expr): + exclude = [exclude] + exclude = set().union(*[e.free_symbols for e in sympify(exclude)]) + symbols = [s for s in symbols if s not in exclude] + + # preprocess equation(s) + ########################################################################### + + # automatically ignore True values + if isinstance(f, list): + f = [s for s in f if s is not S.true] + + # handle canonicalization of equation types + for i, fi in enumerate(f): + if isinstance(fi, (Eq, Ne)): + if 'ImmutableDenseMatrix' in [type(a).__name__ for a in fi.args]: + fi = fi.lhs - fi.rhs + else: + L, R = fi.args + if isinstance(R, BooleanAtom): + L, R = R, L + if isinstance(L, BooleanAtom): + if isinstance(fi, Ne): + L = ~L + if R.is_Relational: + fi = ~R if L is S.false else R + elif R.is_Symbol: + return L + elif R.is_Boolean and (~R).is_Symbol: + return ~L + else: + raise NotImplementedError(filldedent(''' + Unanticipated argument of Eq when other arg + is True or False. + ''')) + elif isinstance(fi, Eq): + fi = Add(fi.lhs, -fi.rhs, evaluate=False) + f[i] = fi + + # *** dispatch and handle as a system of relationals + # ************************************************** + if fi.is_Relational: + if len(symbols) != 1: + raise ValueError("can only solve for one symbol at a time") + if warn and symbols[0].assumptions0: + warnings.warn(filldedent(""" + \tWarning: assumptions about variable '%s' are + not handled currently.""" % symbols[0])) + return reduce_inequalities(f, symbols=symbols) + + # convert Poly to expression + if isinstance(fi, Poly): + f[i] = fi.as_expr() + + # rewrite hyperbolics in terms of exp if they have symbols of + # interest + f[i] = f[i].replace(lambda w: isinstance(w, HyperbolicFunction) and \ + w.has_free(*symbols), lambda w: w.rewrite(exp)) + + # if we have a Matrix, we need to iterate over its elements again + if f[i].is_Matrix: + bare_f = False + f.extend(list(f[i])) + f[i] = S.Zero + + # if we can split it into real and imaginary parts then do so + freei = f[i].free_symbols + if freei and all(s.is_extended_real or s.is_imaginary for s in freei): + fr, fi = f[i].as_real_imag() + # accept as long as new re, im, arg or atan2 are not introduced + had = f[i].atoms(re, im, arg, atan2) + if fr and fi and fr != fi and not any( + i.atoms(re, im, arg, atan2) - had for i in (fr, fi)): + if bare_f: + bare_f = False + f[i: i + 1] = [fr, fi] + + # real/imag handling ----------------------------- + if any(isinstance(fi, (bool, BooleanAtom)) for fi in f): + if as_set: + return [], set() + return [] + + for i, fi in enumerate(f): + # Abs + while True: + was = fi + fi = fi.replace(Abs, lambda arg: + separatevars(Abs(arg)).rewrite(Piecewise) if arg.has(*symbols) + else Abs(arg)) + if was == fi: + break + + for e in fi.find(Abs): + if e.has(*symbols): + raise NotImplementedError('solving %s when the argument ' + 'is not real or imaginary.' % e) + + # arg + fi = fi.replace(arg, lambda a: arg(a).rewrite(atan2).rewrite(atan)) + + # save changes + f[i] = fi + + # see if re(s) or im(s) appear + freim = [fi for fi in f if fi.has(re, im)] + if freim: + irf = [] + for s in symbols: + if s.is_real or s.is_imaginary: + continue # neither re(x) nor im(x) will appear + # if re(s) or im(s) appear, the auxiliary equation must be present + if any(fi.has(re(s), im(s)) for fi in freim): + irf.append((s, re(s) + S.ImaginaryUnit*im(s))) + if irf: + for s, rhs in irf: + f = [fi.xreplace({s: rhs}) for fi in f] + [s - rhs] + symbols.extend([re(s), im(s)]) + if bare_f: + bare_f = False + flags['dict'] = True + # end of real/imag handling ----------------------------- + + # we can solve for non-symbol entities by replacing them with Dummy symbols + f, symbols, swap_sym = recast_to_symbols(f, symbols) + # this set of symbols (perhaps recast) is needed below + symset = set(symbols) + + # get rid of equations that have no symbols of interest; we don't + # try to solve them because the user didn't ask and they might be + # hard to solve; this means that solutions may be given in terms + # of the eliminated equations e.g. solve((x-y, y-3), x) -> {x: y} + newf = [] + for fi in f: + # let the solver handle equations that.. + # - have no symbols but are expressions + # - have symbols of interest + # - have no symbols of interest but are constant + # but when an expression is not constant and has no symbols of + # interest, it can't change what we obtain for a solution from + # the remaining equations so we don't include it; and if it's + # zero it can be removed and if it's not zero, there is no + # solution for the equation set as a whole + # + # The reason for doing this filtering is to allow an answer + # to be obtained to queries like solve((x - y, y), x); without + # this mod the return value is [] + ok = False + if fi.free_symbols & symset: + ok = True + else: + if fi.is_number: + if fi.is_Number: + if fi.is_zero: + continue + return [] + ok = True + else: + if fi.is_constant(): + ok = True + if ok: + newf.append(fi) + if not newf: + if as_set: + return symbols, set() + return [] + f = newf + del newf + + # mask off any Object that we aren't going to invert: Derivative, + # Integral, etc... so that solving for anything that they contain will + # give an implicit solution + seen = set() + non_inverts = set() + for fi in f: + pot = preorder_traversal(fi) + for p in pot: + if not isinstance(p, Expr) or isinstance(p, Piecewise): + pass + elif (isinstance(p, bool) or + not p.args or + p in symset or + p.is_Add or p.is_Mul or + p.is_Pow and not implicit or + p.is_Function and not implicit) and p.func not in (re, im): + continue + elif p not in seen: + seen.add(p) + if p.free_symbols & symset: + non_inverts.add(p) + else: + continue + pot.skip() + del seen + non_inverts = dict(list(zip(non_inverts, [Dummy() for _ in non_inverts]))) + f = [fi.subs(non_inverts) for fi in f] + + # Both xreplace and subs are needed below: xreplace to force substitution + # inside Derivative, subs to handle non-straightforward substitutions + non_inverts = [(v, k.xreplace(swap_sym).subs(swap_sym)) for k, v in non_inverts.items()] + + # rationalize Floats + floats = False + if flags.get('rational', True) is not False: + for i, fi in enumerate(f): + if fi.has(Float): + floats = True + f[i] = nsimplify(fi, rational=True) + + # capture any denominators before rewriting since + # they may disappear after the rewrite, e.g. issue 14779 + flags['_denominators'] = _simple_dens(f[0], symbols) + + # Any embedded piecewise functions need to be brought out to the + # top level so that the appropriate strategy gets selected. + # However, this is necessary only if one of the piecewise + # functions depends on one of the symbols we are solving for. + def _has_piecewise(e): + if e.is_Piecewise: + return e.has(*symbols) + return any(_has_piecewise(a) for a in e.args) + for i, fi in enumerate(f): + if _has_piecewise(fi): + f[i] = piecewise_fold(fi) + + # expand angles of sums; in general, expand_trig will allow + # more roots to be found but this is not a great solultion + # to not returning a parametric solution, otherwise + # many values can be returned that have a simple + # relationship between values + targs = {t for fi in f for t in fi.atoms(TrigonometricFunction)} + if len(targs) > 1: + add, other = sift(targs, lambda x: x.args[0].is_Add, binary=True) + add, other = [[i for i in l if i.has_free(*symbols)] for l in (add, other)] + trep = {} + for t in add: + a = t.args[0] + ind, dep = a.as_independent(*symbols) + if dep in symbols or -dep in symbols: + # don't let expansion expand wrt anything in ind + n = Dummy() if not ind.is_Number else ind + trep[t] = TR10(t.func(dep + n)).xreplace({n: ind}) + if other and len(other) <= 2: + base = gcd(*[i.args[0] for i in other]) if len(other) > 1 else other[0].args[0] + for i in other: + trep[i] = TR11(i, base) + f = [fi.xreplace(trep) for fi in f] + + # + # try to get a solution + ########################################################################### + if bare_f: + solution = None + if len(symbols) != 1: + solution = _solve_undetermined(f[0], symbols, flags) + if not solution: + solution = _solve(f[0], *symbols, **flags) + else: + linear, solution = _solve_system(f, symbols, **flags) + assert type(solution) is list + assert not solution or type(solution[0]) is dict, solution + # + # postprocessing + ########################################################################### + # capture as_dict flag now (as_set already captured) + as_dict = flags.get('dict', False) + + # define how solution will get unpacked + tuple_format = lambda s: [tuple([i.get(x, x) for x in symbols]) for i in s] + if as_dict or as_set: + unpack = None + elif bare_f: + if len(symbols) == 1: + unpack = lambda s: [i[symbols[0]] for i in s] + elif len(solution) == 1 and len(solution[0]) == len(symbols): + # undetermined linear coeffs solution + unpack = lambda s: s[0] + elif ordered_symbols: + unpack = tuple_format + else: + unpack = lambda s: s + else: + if solution: + if linear and len(solution) == 1: + # if you want the tuple solution for the linear + # case, use `set=True` + unpack = lambda s: s[0] + elif ordered_symbols: + unpack = tuple_format + else: + unpack = lambda s: s + else: + unpack = None + + # Restore masked-off objects + if non_inverts and type(solution) is list: + solution = [{k: v.subs(non_inverts) for k, v in s.items()} + for s in solution] + + # Restore original "symbols" if a dictionary is returned. + # This is not necessary for + # - the single univariate equation case + # since the symbol will have been removed from the solution; + # - the nonlinear poly_system since that only supports zero-dimensional + # systems and those results come back as a list + # + # ** unless there were Derivatives with the symbols, but those were handled + # above. + if swap_sym: + symbols = [swap_sym.get(k, k) for k in symbols] + for i, sol in enumerate(solution): + solution[i] = {swap_sym.get(k, k): v.subs(swap_sym) + for k, v in sol.items()} + + # Get assumptions about symbols, to filter solutions. + # Note that if assumptions about a solution can't be verified, it is still + # returned. + check = flags.get('check', True) + + # restore floats + if floats and solution and flags.get('rational', None) is None: + solution = nfloat(solution, exponent=False) + # nfloat might reveal more duplicates + solution = _remove_duplicate_solutions(solution) + + if check and solution: # assumption checking + warn = flags.get('warn', False) + got_None = [] # solutions for which one or more symbols gave None + no_False = [] # solutions for which no symbols gave False + for sol in solution: + a_None = False + for symb, val in sol.items(): + test = check_assumptions(val, **symb.assumptions0) + if test: + continue + if test is False: + break + a_None = True + else: + no_False.append(sol) + if a_None: + got_None.append(sol) + + solution = no_False + if warn and got_None: + warnings.warn(filldedent(""" + \tWarning: assumptions concerning following solution(s) + cannot be checked:""" + '\n\t' + + ', '.join(str(s) for s in got_None))) + + # + # done + ########################################################################### + + if not solution: + if as_set: + return symbols, set() + return [] + + # make orderings canonical for list of dictionaries + if not as_set: # for set, no point in ordering + solution = [{k: s[k] for k in ordered(s)} for s in solution] + solution.sort(key=default_sort_key) + + if not (as_set or as_dict): + return unpack(solution) + + if as_dict: + return solution + + # set output: (symbols, {t1, t2, ...}) from list of dictionaries; + # include all symbols for those that like a verbose solution + # and to resolve any differences in dictionary keys. + # + # The set results can easily be used to make a verbose dict as + # k, v = solve(eqs, syms, set=True) + # sol = [dict(zip(k,i)) for i in v] + # + if ordered_symbols: + k = symbols # keep preferred order + else: + # just unify the symbols for which solutions were found + k = list(ordered(set(flatten(tuple(i.keys()) for i in solution)))) + return k, {tuple([s.get(ki, ki) for ki in k]) for s in solution} + + +def _solve_undetermined(g, symbols, flags): + """solve helper to return a list with one dict (solution) else None + + A direct call to solve_undetermined_coeffs is more flexible and + can return both multiple solutions and handle more than one independent + variable. Here, we have to be more cautious to keep from solving + something that does not look like an undetermined coeffs system -- + to minimize the surprise factor since singularities that cancel are not + prohibited in solve_undetermined_coeffs. + """ + if g.free_symbols - set(symbols): + sol = solve_undetermined_coeffs(g, symbols, **dict(flags, dict=True, set=None)) + if len(sol) == 1: + return sol + + +def _solve(f, *symbols, **flags): + """Return a checked solution for *f* in terms of one or more of the + symbols in the form of a list of dictionaries. + + If no method is implemented to solve the equation, a NotImplementedError + will be raised. In the case that conversion of an expression to a Poly + gives None a ValueError will be raised. + """ + + not_impl_msg = "No algorithms are implemented to solve equation %s" + + if len(symbols) != 1: + # look for solutions for desired symbols that are independent + # of symbols already solved for, e.g. if we solve for x = y + # then no symbol having x in its solution will be returned. + + # First solve for linear symbols (since that is easier and limits + # solution size) and then proceed with symbols appearing + # in a non-linear fashion. Ideally, if one is solving a single + # expression for several symbols, they would have to be + # appear in factors of an expression, but we do not here + # attempt factorization. XXX perhaps handling a Mul + # should come first in this routine whether there is + # one or several symbols. + nonlin_s = [] + got_s = set() + rhs_s = set() + result = [] + for s in symbols: + xi, v = solve_linear(f, symbols=[s]) + if xi == s: + # no need to check but we should simplify if desired + if flags.get('simplify', True): + v = simplify(v) + vfree = v.free_symbols + if vfree & got_s: + # was linear, but has redundant relationship + # e.g. x - y = 0 has y == x is redundant for x == y + # so ignore + continue + rhs_s |= vfree + got_s.add(xi) + result.append({xi: v}) + elif xi: # there might be a non-linear solution if xi is not 0 + nonlin_s.append(s) + if not nonlin_s: + return result + for s in nonlin_s: + try: + soln = _solve(f, s, **flags) + for sol in soln: + if sol[s].free_symbols & got_s: + # depends on previously solved symbols: ignore + continue + got_s.add(s) + result.append(sol) + except NotImplementedError: + continue + if got_s: + return result + else: + raise NotImplementedError(not_impl_msg % f) + + # solve f for a single variable + + symbol = symbols[0] + + # expand binomials only if it has the unknown symbol + f = f.replace(lambda e: isinstance(e, binomial) and e.has(symbol), + lambda e: expand_func(e)) + + # checking will be done unless it is turned off before making a + # recursive call; the variables `checkdens` and `check` are + # captured here (for reference below) in case flag value changes + flags['check'] = checkdens = check = flags.pop('check', True) + + # build up solutions if f is a Mul + if f.is_Mul: + result = set() + for m in f.args: + if m in {S.NegativeInfinity, S.ComplexInfinity, S.Infinity}: + result = set() + break + soln = _vsolve(m, symbol, **flags) + result.update(set(soln)) + result = [{symbol: v} for v in result] + if check: + # all solutions have been checked but now we must + # check that the solutions do not set denominators + # in any factor to zero + dens = flags.get('_denominators', _simple_dens(f, symbols)) + result = [s for s in result if + not any(checksol(den, s, **flags) for den in + dens)] + # set flags for quick exit at end; solutions for each + # factor were already checked and simplified + check = False + flags['simplify'] = False + + elif f.is_Piecewise: + result = set() + if any(e.is_zero for e, c in f.args): + f = f.simplify() # failure imminent w/o help + + cond = neg = True + for expr, cnd in f.args: + # the explicit condition for this expr is the current cond + # and none of the previous conditions + cond = And(neg, cnd) + neg = And(neg, ~cond) + + if expr.is_zero and cond.simplify() != False: + raise NotImplementedError(filldedent(''' + An expression is already zero when %s. + This means that in this *region* the solution + is zero but solve can only represent discrete, + not interval, solutions. If this is a spurious + interval it might be resolved with simplification + of the Piecewise conditions.''' % cond)) + candidates = _vsolve(expr, symbol, **flags) + + for candidate in candidates: + if candidate in result: + # an unconditional value was already there + continue + try: + v = cond.subs(symbol, candidate) + _eval_simplify = getattr(v, '_eval_simplify', None) + if _eval_simplify is not None: + # unconditionally take the simplification of v + v = _eval_simplify(ratio=2, measure=lambda x: 1) + except TypeError: + # incompatible type with condition(s) + continue + if v == False: + continue + if v == True: + result.add(candidate) + else: + result.add(Piecewise( + (candidate, v), + (S.NaN, True))) + # solutions already checked and simplified + # **************************************** + return [{symbol: r} for r in result] + else: + # first see if it really depends on symbol and whether there + # is only a linear solution + f_num, sol = solve_linear(f, symbols=symbols) + if f_num.is_zero or sol is S.NaN: + return [] + elif f_num.is_Symbol: + # no need to check but simplify if desired + if flags.get('simplify', True): + sol = simplify(sol) + return [{f_num: sol}] + + poly = None + # check for a single Add generator + if not f_num.is_Add: + add_args = [i for i in f_num.atoms(Add) + if symbol in i.free_symbols] + if len(add_args) == 1: + gen = add_args[0] + spart = gen.as_independent(symbol)[1].as_base_exp()[0] + if spart == symbol: + try: + poly = Poly(f_num, spart) + except PolynomialError: + pass + + result = False # no solution was obtained + msg = '' # there is no failure message + + # Poly is generally robust enough to convert anything to + # a polynomial and tell us the different generators that it + # contains, so we will inspect the generators identified by + # polys to figure out what to do. + + # try to identify a single generator that will allow us to solve this + # as a polynomial, followed (perhaps) by a change of variables if the + # generator is not a symbol + + try: + if poly is None: + poly = Poly(f_num) + if poly is None: + raise ValueError('could not convert %s to Poly' % f_num) + except GeneratorsNeeded: + simplified_f = simplify(f_num) + if simplified_f != f_num: + return _solve(simplified_f, symbol, **flags) + raise ValueError('expression appears to be a constant') + + gens = [g for g in poly.gens if g.has(symbol)] + + def _as_base_q(x): + """Return (b**e, q) for x = b**(p*e/q) where p/q is the leading + Rational of the exponent of x, e.g. exp(-2*x/3) -> (exp(x), 3) + """ + b, e = x.as_base_exp() + if e.is_Rational: + return b, e.q + if not e.is_Mul: + return x, 1 + c, ee = e.as_coeff_Mul() + if c.is_Rational and c is not S.One: # c could be a Float + return b**ee, c.q + return x, 1 + + if len(gens) > 1: + # If there is more than one generator, it could be that the + # generators have the same base but different powers, e.g. + # >>> Poly(exp(x) + 1/exp(x)) + # Poly(exp(-x) + exp(x), exp(-x), exp(x), domain='ZZ') + # + # If unrad was not disabled then there should be no rational + # exponents appearing as in + # >>> Poly(sqrt(x) + sqrt(sqrt(x))) + # Poly(sqrt(x) + x**(1/4), sqrt(x), x**(1/4), domain='ZZ') + + bases, qs = list(zip(*[_as_base_q(g) for g in gens])) + bases = set(bases) + + if len(bases) > 1 or not all(q == 1 for q in qs): + funcs = {b for b in bases if b.is_Function} + + trig = {_ for _ in funcs if + isinstance(_, TrigonometricFunction)} + other = funcs - trig + if not other and len(funcs.intersection(trig)) > 1: + newf = None + if f_num.is_Add and len(f_num.args) == 2: + # check for sin(x)**p = cos(x)**p + _args = f_num.args + t = a, b = [i.atoms(Function).intersection( + trig) for i in _args] + if all(len(i) == 1 for i in t): + a, b = [i.pop() for i in t] + if isinstance(a, cos): + a, b = b, a + _args = _args[::-1] + if isinstance(a, sin) and isinstance(b, cos + ) and a.args[0] == b.args[0]: + # sin(x) + cos(x) = 0 -> tan(x) + 1 = 0 + newf, _d = (TR2i(_args[0]/_args[1]) + 1 + ).as_numer_denom() + if not _d.is_Number: + newf = None + if newf is None: + newf = TR1(f_num).rewrite(tan) + if newf != f_num: + # don't check the rewritten form --check + # solutions in the un-rewritten form below + flags['check'] = False + result = _solve(newf, symbol, **flags) + flags['check'] = check + + # just a simple case - see if replacement of single function + # clears all symbol-dependent functions, e.g. + # log(x) - log(log(x) - 1) - 3 can be solved even though it has + # two generators. + + if result is False and funcs: + funcs = list(ordered(funcs)) # put shallowest function first + f1 = funcs[0] + t = Dummy('t') + # perform the substitution + ftry = f_num.subs(f1, t) + + # if no Functions left, we can proceed with usual solve + if not ftry.has(symbol): + cv_sols = _solve(ftry, t, **flags) + cv_inv = list(ordered(_vsolve(t - f1, symbol, **flags)))[0] + result = [{symbol: cv_inv.subs(sol)} for sol in cv_sols] + + if result is False: + msg = 'multiple generators %s' % gens + + else: + # e.g. case where gens are exp(x), exp(-x) + u = bases.pop() + t = Dummy('t') + inv = _vsolve(u - t, symbol, **flags) + if isinstance(u, (Pow, exp)): + # this will be resolved by factor in _tsolve but we might + # as well try a simple expansion here to get things in + # order so something like the following will work now without + # having to factor: + # + # >>> eq = (exp(I*(-x-2))+exp(I*(x+2))) + # >>> eq.subs(exp(x),y) # fails + # exp(I*(-x - 2)) + exp(I*(x + 2)) + # >>> eq.expand().subs(exp(x),y) # works + # y**I*exp(2*I) + y**(-I)*exp(-2*I) + def _expand(p): + b, e = p.as_base_exp() + e = expand_mul(e) + return expand_power_exp(b**e) + ftry = f_num.replace( + lambda w: w.is_Pow or isinstance(w, exp), + _expand).subs(u, t) + if not ftry.has(symbol): + soln = _solve(ftry, t, **flags) + result = [{symbol: i.subs(s)} for i in inv for s in soln] + + elif len(gens) == 1: + + # There is only one generator that we are interested in, but + # there may have been more than one generator identified by + # polys (e.g. for symbols other than the one we are interested + # in) so recast the poly in terms of our generator of interest. + # Also use composite=True with f_num since Poly won't update + # poly as documented in issue 8810. + + poly = Poly(f_num, gens[0], composite=True) + + # if we aren't on the tsolve-pass, use roots + if not flags.pop('tsolve', False): + soln = None + deg = poly.degree() + flags['tsolve'] = True + hints = ('cubics', 'quartics', 'quintics') + solvers = {h: flags.get(h) for h in hints} + soln = roots(poly, **solvers) + if sum(soln.values()) < deg: + # e.g. roots(32*x**5 + 400*x**4 + 2032*x**3 + + # 5000*x**2 + 6250*x + 3189) -> {} + # so all_roots is used and RootOf instances are + # returned *unless* the system is multivariate + # or high-order EX domain. + try: + soln = poly.all_roots() + except NotImplementedError: + if not flags.get('incomplete', True): + raise NotImplementedError( + filldedent(''' + Neither high-order multivariate polynomials + nor sorting of EX-domain polynomials is supported. + If you want to see any results, pass keyword incomplete=True to + solve; to see numerical values of roots + for univariate expressions, use nroots. + ''')) + else: + pass + else: + soln = list(soln.keys()) + + if soln is not None: + u = poly.gen + if u != symbol: + try: + t = Dummy('t') + inv = _vsolve(u - t, symbol, **flags) + soln = {i.subs(t, s) for i in inv for s in soln} + except NotImplementedError: + # perhaps _tsolve can handle f_num + soln = None + else: + check = False # only dens need to be checked + if soln is not None: + if len(soln) > 2: + # if the flag wasn't set then unset it since high-order + # results are quite long. Perhaps one could base this + # decision on a certain critical length of the + # roots. In addition, wester test M2 has an expression + # whose roots can be shown to be real with the + # unsimplified form of the solution whereas only one of + # the simplified forms appears to be real. + flags['simplify'] = flags.get('simplify', False) + if soln is not None: + result = [{symbol: v} for v in soln] + + # fallback if above fails + # ----------------------- + if result is False: + # try unrad + if flags.pop('_unrad', True): + try: + u = unrad(f_num, symbol) + except (ValueError, NotImplementedError): + u = False + if u: + eq, cov = u + if cov: + isym, ieq = cov + inv = _vsolve(ieq, symbol, **flags)[0] + rv = {inv.subs(xi) for xi in _solve(eq, isym, **flags)} + else: + try: + rv = set(_vsolve(eq, symbol, **flags)) + except NotImplementedError: + rv = None + if rv is not None: + result = [{symbol: v} for v in rv] + # if the flag wasn't set then unset it since unrad results + # can be quite long or of very high order + flags['simplify'] = flags.get('simplify', False) + else: + pass # for coverage + + # try _tsolve + if result is False: + flags.pop('tsolve', None) # allow tsolve to be used on next pass + try: + soln = _tsolve(f_num, symbol, **flags) + if soln is not None: + result = [{symbol: v} for v in soln] + except PolynomialError: + pass + # ----------- end of fallback ---------------------------- + + if result is False: + raise NotImplementedError('\n'.join([msg, not_impl_msg % f])) + + result = _remove_duplicate_solutions(result) + + if flags.get('simplify', True): + result = [{k: d[k].simplify() for k in d} for d in result] + # Simplification might reveal more duplicates + result = _remove_duplicate_solutions(result) + # we just simplified the solution so we now set the flag to + # False so the simplification doesn't happen again in checksol() + flags['simplify'] = False + + if checkdens: + # reject any result that makes any denom. affirmatively 0; + # if in doubt, keep it + dens = _simple_dens(f, symbols) + result = [r for r in result if + not any(checksol(d, r, **flags) + for d in dens)] + if check: + # keep only results if the check is not False + result = [r for r in result if + checksol(f_num, r, **flags) is not False] + return result + + +def _remove_duplicate_solutions(solutions: list[dict[Expr, Expr]] + ) -> list[dict[Expr, Expr]]: + """Remove duplicates from a list of dicts""" + solutions_set = set() + solutions_new = [] + + for sol in solutions: + solset = frozenset(sol.items()) + if solset not in solutions_set: + solutions_new.append(sol) + solutions_set.add(solset) + + return solutions_new + + +def _solve_system(exprs, symbols, **flags): + """return ``(linear, solution)`` where ``linear`` is True + if the system was linear, else False; ``solution`` + is a list of dictionaries giving solutions for the symbols + """ + if not exprs: + return False, [] + + if flags.pop('_split', True): + # Split the system into connected components + V = exprs + symsset = set(symbols) + exprsyms = {e: e.free_symbols & symsset for e in exprs} + E = [] + sym_indices = {sym: i for i, sym in enumerate(symbols)} + for n, e1 in enumerate(exprs): + for e2 in exprs[:n]: + # Equations are connected if they share a symbol + if exprsyms[e1] & exprsyms[e2]: + E.append((e1, e2)) + G = V, E + subexprs = connected_components(G) + if len(subexprs) > 1: + subsols = [] + linear = True + for subexpr in subexprs: + subsyms = set() + for e in subexpr: + subsyms |= exprsyms[e] + subsyms = sorted(subsyms, key = lambda x: sym_indices[x]) + flags['_split'] = False # skip split step + _linear, subsol = _solve_system(subexpr, subsyms, **flags) + if linear: + linear = linear and _linear + if not isinstance(subsol, list): + subsol = [subsol] + subsols.append(subsol) + # Full solution is cartesion product of subsystems + sols = [] + for soldicts in product(*subsols): + sols.append(dict(item for sd in soldicts + for item in sd.items())) + return linear, sols + + polys = [] + dens = set() + failed = [] + result = [] + solved_syms = [] + linear = True + manual = flags.get('manual', False) + checkdens = check = flags.get('check', True) + + for j, g in enumerate(exprs): + dens.update(_simple_dens(g, symbols)) + i, d = _invert(g, *symbols) + if d in symbols: + if linear: + linear = solve_linear(g, 0, [d])[0] == d + g = d - i + g = g.as_numer_denom()[0] + if manual: + failed.append(g) + continue + + poly = g.as_poly(*symbols, extension=True) + + if poly is not None: + polys.append(poly) + else: + failed.append(g) + + if polys: + if all(p.is_linear for p in polys): + n, m = len(polys), len(symbols) + matrix = zeros(n, m + 1) + + for i, poly in enumerate(polys): + for monom, coeff in poly.terms(): + try: + j = monom.index(1) + matrix[i, j] = coeff + except ValueError: + matrix[i, m] = -coeff + + # returns a dictionary ({symbols: values}) or None + if flags.pop('particular', False): + result = minsolve_linear_system(matrix, *symbols, **flags) + else: + result = solve_linear_system(matrix, *symbols, **flags) + result = [result] if result else [] + if failed: + if result: + solved_syms = list(result[0].keys()) # there is only one result dict + else: + solved_syms = [] + # linear doesn't change + else: + linear = False + if len(symbols) > len(polys): + + free = set().union(*[p.free_symbols for p in polys]) + free = list(ordered(free.intersection(symbols))) + got_s = set() + result = [] + for syms in subsets(free, len(polys)): + try: + # returns [], None or list of tuples + res = solve_poly_system(polys, *syms) + if res: + for r in set(res): + skip = False + for r1 in r: + if got_s and any(ss in r1.free_symbols + for ss in got_s): + # sol depends on previously + # solved symbols: discard it + skip = True + if not skip: + got_s.update(syms) + result.append(dict(list(zip(syms, r)))) + except NotImplementedError: + pass + if got_s: + solved_syms = list(got_s) + else: + raise NotImplementedError('no valid subset found') + else: + try: + result = solve_poly_system(polys, *symbols) + if result: + solved_syms = symbols + result = [dict(list(zip(solved_syms, r))) for r in set(result)] + except NotImplementedError: + failed.extend([g.as_expr() for g in polys]) + solved_syms = [] + + # convert None or [] to [{}] + result = result or [{}] + + if failed: + linear = False + # For each failed equation, see if we can solve for one of the + # remaining symbols from that equation. If so, we update the + # solution set and continue with the next failed equation, + # repeating until we are done or we get an equation that can't + # be solved. + def _ok_syms(e, sort=False): + rv = e.free_symbols & legal + + # Solve first for symbols that have lower degree in the equation. + # Ideally we want to solve firstly for symbols that appear linearly + # with rational coefficients e.g. if e = x*y + z then we should + # solve for z first. + def key(sym): + ep = e.as_poly(sym) + if ep is None: + complexity = (S.Infinity, S.Infinity, S.Infinity) + else: + coeff_syms = ep.LC().free_symbols + complexity = (ep.degree(), len(coeff_syms & rv), len(coeff_syms)) + return complexity + (default_sort_key(sym),) + + if sort: + rv = sorted(rv, key=key) + return rv + + legal = set(symbols) # what we are interested in + # sort so equation with the fewest potential symbols is first + u = Dummy() # used in solution checking + for eq in ordered(failed, lambda _: len(_ok_syms(_))): + newresult = [] + bad_results = [] + hit = False + for r in result: + got_s = set() + # update eq with everything that is known so far + eq2 = eq.subs(r) + # if check is True then we see if it satisfies this + # equation, otherwise we just accept it + if check and r: + b = checksol(u, u, eq2, minimal=True) + if b is not None: + # this solution is sufficient to know whether + # it is valid or not so we either accept or + # reject it, then continue + if b: + newresult.append(r) + else: + bad_results.append(r) + continue + # search for a symbol amongst those available that + # can be solved for + ok_syms = _ok_syms(eq2, sort=True) + if not ok_syms: + if r: + newresult.append(r) + break # skip as it's independent of desired symbols + for s in ok_syms: + try: + soln = _vsolve(eq2, s, **flags) + except NotImplementedError: + continue + # put each solution in r and append the now-expanded + # result in the new result list; use copy since the + # solution for s is being added in-place + for sol in soln: + if got_s and any(ss in sol.free_symbols for ss in got_s): + # sol depends on previously solved symbols: discard it + continue + rnew = r.copy() + for k, v in r.items(): + rnew[k] = v.subs(s, sol) + # and add this new solution + rnew[s] = sol + # check that it is independent of previous solutions + iset = set(rnew.items()) + for i in newresult: + if len(i) < len(iset) and not set(i.items()) - iset: + # this is a superset of a known solution that + # is smaller + break + else: + # keep it + newresult.append(rnew) + hit = True + got_s.add(s) + if not hit: + raise NotImplementedError('could not solve %s' % eq2) + else: + result = newresult + for b in bad_results: + if b in result: + result.remove(b) + + if not result: + return False, [] + + # rely on linear/polynomial system solvers to simplify + # XXX the following tests show that the expressions + # returned are not the same as they would be if simplify + # were applied to this: + # sympy/solvers/ode/tests/test_systems/test__classify_linear_system + # sympy/solvers/tests/test_solvers/test_issue_4886 + # so the docs should be updated to reflect that or else + # the following should be `bool(failed) or not linear` + default_simplify = bool(failed) + if flags.get('simplify', default_simplify): + for r in result: + for k in r: + r[k] = simplify(r[k]) + flags['simplify'] = False # don't need to do so in checksol now + + if checkdens: + result = [r for r in result + if not any(checksol(d, r, **flags) for d in dens)] + + if check and not linear: + result = [r for r in result + if not any(checksol(e, r, **flags) is False for e in exprs)] + + result = [r for r in result if r] + return linear, result + + +def solve_linear(lhs, rhs=0, symbols=[], exclude=[]): + r""" + Return a tuple derived from ``f = lhs - rhs`` that is one of + the following: ``(0, 1)``, ``(0, 0)``, ``(symbol, solution)``, ``(n, d)``. + + Explanation + =========== + + ``(0, 1)`` meaning that ``f`` is independent of the symbols in *symbols* + that are not in *exclude*. + + ``(0, 0)`` meaning that there is no solution to the equation amongst the + symbols given. If the first element of the tuple is not zero, then the + function is guaranteed to be dependent on a symbol in *symbols*. + + ``(symbol, solution)`` where symbol appears linearly in the numerator of + ``f``, is in *symbols* (if given), and is not in *exclude* (if given). No + simplification is done to ``f`` other than a ``mul=True`` expansion, so the + solution will correspond strictly to a unique solution. + + ``(n, d)`` where ``n`` and ``d`` are the numerator and denominator of ``f`` + when the numerator was not linear in any symbol of interest; ``n`` will + never be a symbol unless a solution for that symbol was found (in which case + the second element is the solution, not the denominator). + + Examples + ======== + + >>> from sympy import cancel, Pow + + ``f`` is independent of the symbols in *symbols* that are not in + *exclude*: + + >>> from sympy import cos, sin, solve_linear + >>> from sympy.abc import x, y, z + >>> eq = y*cos(x)**2 + y*sin(x)**2 - y # = y*(1 - 1) = 0 + >>> solve_linear(eq) + (0, 1) + >>> eq = cos(x)**2 + sin(x)**2 # = 1 + >>> solve_linear(eq) + (0, 1) + >>> solve_linear(x, exclude=[x]) + (0, 1) + + The variable ``x`` appears as a linear variable in each of the + following: + + >>> 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 the numerator of the expression is a symbol, then ``(0, 0)`` is + returned if the solution for that symbol would have set any + denominator to 0: + + >>> eq = 1/(1/x - 2) + >>> eq.as_numer_denom() + (x, 1 - 2*x) + >>> solve_linear(eq) + (0, 0) + + But automatic rewriting may cause a symbol in the denominator to + appear in the numerator so a solution will be returned: + + >>> (1/x)**-1 + x + >>> solve_linear((1/x)**-1) + (x, 0) + + Use an unevaluated expression to avoid this: + + >>> solve_linear(Pow(1/x, -1, evaluate=False)) + (0, 0) + + If ``x`` is allowed to cancel in the following expression, then it + appears to be linear in ``x``, but this sort of cancellation is not + done by ``solve_linear`` so the solution will always satisfy the + original expression without causing a division by zero error. + + >>> eq = x**2*(1/x - z**2/x) + >>> solve_linear(cancel(eq)) + (x, 0) + >>> solve_linear(eq) + (x**2*(1 - z**2), x) + + A list of symbols for which a solution is desired may be given: + + >>> solve_linear(x + y + z, symbols=[y]) + (y, -x - z) + + A list of symbols to ignore may also be given: + + >>> solve_linear(x + y + z, exclude=[x]) + (y, -x - z) + + (A solution for ``y`` is obtained because it is the first variable + from the canonically sorted list of symbols that had a linear + solution.) + + """ + if isinstance(lhs, Eq): + if rhs: + raise ValueError(filldedent(''' + If lhs is an Equality, rhs must be 0 but was %s''' % rhs)) + rhs = lhs.rhs + lhs = lhs.lhs + dens = None + eq = lhs - rhs + n, d = eq.as_numer_denom() + if not n: + return S.Zero, S.One + + free = n.free_symbols + if not symbols: + symbols = free + else: + bad = [s for s in symbols if not s.is_Symbol] + if bad: + if len(bad) == 1: + bad = bad[0] + if len(symbols) == 1: + eg = 'solve(%s, %s)' % (eq, symbols[0]) + else: + eg = 'solve(%s, *%s)' % (eq, list(symbols)) + raise ValueError(filldedent(''' + solve_linear only handles symbols, not %s. To isolate + non-symbols use solve, e.g. >>> %s <<<. + ''' % (bad, eg))) + symbols = free.intersection(symbols) + symbols = symbols.difference(exclude) + if not symbols: + return S.Zero, S.One + + # derivatives are easy to do but tricky to analyze to see if they + # are going to disallow a linear solution, so for simplicity we + # just evaluate the ones that have the symbols of interest + derivs = defaultdict(list) + for der in n.atoms(Derivative): + csym = der.free_symbols & symbols + for c in csym: + derivs[c].append(der) + + all_zero = True + for xi in sorted(symbols, key=default_sort_key): # canonical order + # if there are derivatives in this var, calculate them now + if isinstance(derivs[xi], list): + derivs[xi] = {der: der.doit() for der in derivs[xi]} + newn = n.subs(derivs[xi]) + dnewn_dxi = newn.diff(xi) + # dnewn_dxi can be nonzero if it survives differentation by any + # of its free symbols + free = dnewn_dxi.free_symbols + if dnewn_dxi and (not free or any(dnewn_dxi.diff(s) for s in free) or free == symbols): + all_zero = False + if dnewn_dxi is S.NaN: + break + if xi not in dnewn_dxi.free_symbols: + vi = -1/dnewn_dxi*(newn.subs(xi, 0)) + if dens is None: + dens = _simple_dens(eq, symbols) + if not any(checksol(di, {xi: vi}, minimal=True) is True + for di in dens): + # simplify any trivial integral + irep = [(i, i.doit()) for i in vi.atoms(Integral) if + i.function.is_number] + # do a slight bit of simplification + vi = expand_mul(vi.subs(irep)) + return xi, vi + if all_zero: + return S.Zero, S.One + if n.is_Symbol: # no solution for this symbol was found + return S.Zero, S.Zero + return n, d + + +def minsolve_linear_system(system, *symbols, **flags): + r""" + Find a particular solution to a linear system. + + Explanation + =========== + + In particular, try to find a solution with the minimal possible number + of non-zero variables using a naive algorithm with exponential complexity. + If ``quick=True``, a heuristic is used. + + """ + quick = flags.get('quick', False) + # Check if there are any non-zero solutions at all + s0 = solve_linear_system(system, *symbols, **flags) + if not s0 or all(v == 0 for v in s0.values()): + return s0 + if quick: + # We just solve the system and try to heuristically find a nice + # solution. + s = solve_linear_system(system, *symbols) + def update(determined, solution): + delete = [] + for k, v in solution.items(): + solution[k] = v.subs(determined) + if not solution[k].free_symbols: + delete.append(k) + determined[k] = solution[k] + for k in delete: + del solution[k] + determined = {} + update(determined, s) + while s: + # NOTE sort by default_sort_key to get deterministic result + k = max((k for k in s.values()), + key=lambda x: (len(x.free_symbols), default_sort_key(x))) + kfree = k.free_symbols + x = next(reversed(list(ordered(kfree)))) + if len(kfree) != 1: + determined[x] = S.Zero + else: + val = _vsolve(k, x, check=False)[0] + if not val and not any(v.subs(x, val) for v in s.values()): + determined[x] = S.One + else: + determined[x] = val + update(determined, s) + return determined + else: + # We try to select n variables which we want to be non-zero. + # All others will be assumed zero. We try to solve the modified system. + # If there is a non-trivial solution, just set the free variables to + # one. If we do this for increasing n, trying all combinations of + # variables, we will find an optimal solution. + # We speed up slightly by starting at one less than the number of + # variables the quick method manages. + N = len(symbols) + bestsol = minsolve_linear_system(system, *symbols, quick=True) + n0 = len([x for x in bestsol.values() if x != 0]) + for n in range(n0 - 1, 1, -1): + debugf('minsolve: %s', n) + thissol = None + for nonzeros in combinations(range(N), n): + subm = Matrix([system.col(i).T for i in nonzeros] + [system.col(-1).T]).T + s = solve_linear_system(subm, *[symbols[i] for i in nonzeros]) + if s and not all(v == 0 for v in s.values()): + subs = [(symbols[v], S.One) for v in nonzeros] + for k, v in s.items(): + s[k] = v.subs(subs) + for sym in symbols: + if sym not in s: + if symbols.index(sym) in nonzeros: + s[sym] = S.One + else: + s[sym] = S.Zero + thissol = s + break + if thissol is None: + break + bestsol = thissol + return bestsol + + +def solve_linear_system(system, *symbols, **flags): + r""" + Solve system of $N$ linear equations with $M$ variables, which means + both under- and overdetermined systems are supported. + + Explanation + =========== + + The possible number of solutions is zero, one, or infinite. Respectively, + this procedure will return None or a dictionary with solutions. In the + case of underdetermined systems, all arbitrary parameters are skipped. + This may cause a situation in which an empty dictionary is returned. + In that case, all symbols can be assigned arbitrary values. + + Input to this function is a $N\times M + 1$ 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. + + Examples + ======== + + >>> 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} + + A degenerate system returns an empty dictionary: + + >>> system = Matrix(( (0,0,0), (0,0,0) )) + >>> solve_linear_system(system, x, y) + {} + + """ + assert system.shape[1] == len(symbols) + 1 + + # This is just a wrapper for solve_lin_sys + eqs = list(system * Matrix(symbols + (-1,))) + eqs, ring = sympy_eqs_to_ring(eqs, symbols) + sol = solve_lin_sys(eqs, ring, _raw=False) + if sol is not None: + sol = {sym:val for sym, val in sol.items() if sym != val} + return sol + + +def solve_undetermined_coeffs(equ, coeffs, *syms, **flags): + r""" + Solve a system of equations in $k$ parameters that is formed by + matching coefficients in variables ``coeffs`` that are on + factors dependent on the remaining variables (or those given + explicitly by ``syms``. + + Explanation + =========== + + The result of this function is a dictionary with symbolic values of those + parameters with respect to coefficients in $q$ -- empty if there + is no solution or coefficients do not appear in the equation -- else + None (if the system was not recognized). If there is more than one + solution, the solutions are passed as a list. The output can be modified using + the same semantics as for `solve` since the flags that are passed are sent + directly to `solve` so, for example the flag ``dict=True`` will always return a list + of solutions as dictionaries. + + This function accepts both Equality and Expr class instances. + The solving process is most efficient when symbols are specified + in addition to parameters to be determined, but an attempt to + determine them (if absent) will be made. If an expected solution is not + obtained (and symbols were not specified) try specifying them. + + Examples + ======== + + >>> from sympy import Eq, solve_undetermined_coeffs + >>> from sympy.abc import a, b, c, h, p, k, x, y + + >>> solve_undetermined_coeffs(Eq(a*x + a + b, x/2), [a, b], x) + {a: 1/2, b: -1/2} + >>> solve_undetermined_coeffs(a - 2, [a]) + {a: 2} + + The equation can be nonlinear in the symbols: + + >>> X, Y, Z = y, x**y, y*x**y + >>> eq = a*X + b*Y + c*Z - X - 2*Y - 3*Z + >>> coeffs = a, b, c + >>> syms = x, y + >>> solve_undetermined_coeffs(eq, coeffs, syms) + {a: 1, b: 2, c: 3} + + And the system can be nonlinear in coefficients, too, but if + there is only a single solution, it will be returned as a + dictionary: + + >>> eq = a*x**2 + b*x + c - ((x - h)**2 + 4*p*k)/4/p + >>> solve_undetermined_coeffs(eq, (h, p, k), x) + {h: -b/(2*a), k: (4*a*c - b**2)/(4*a), p: 1/(4*a)} + + Multiple solutions are always returned in a list: + + >>> solve_undetermined_coeffs(a**2*x + b - x, [a, b], x) + [{a: -1, b: 0}, {a: 1, b: 0}] + + Using flag ``dict=True`` (in keeping with semantics in :func:`~.solve`) + will force the result to always be a list with any solutions + as elements in that list. + + >>> solve_undetermined_coeffs(a*x - 2*x, [a], dict=True) + [{a: 2}] + """ + if not (coeffs and all(i.is_Symbol for i in coeffs)): + raise ValueError('must provide symbols for coeffs') + + if isinstance(equ, Eq): + eq = equ.lhs - equ.rhs + else: + eq = equ + + ceq = cancel(eq) + xeq = _mexpand(ceq.as_numer_denom()[0], recursive=True) + + free = xeq.free_symbols + coeffs = free & set(coeffs) + if not coeffs: + return ([], {}) if flags.get('set', None) else [] # solve(0, x) -> [] + + if not syms: + # e.g. A*exp(x) + B - (exp(x) + y) separated into parts that + # don't/do depend on coeffs gives + # -(exp(x) + y), A*exp(x) + B + # then see what symbols are common to both + # {x} = {x, A, B} - {x, y} + ind, dep = xeq.as_independent(*coeffs, as_Add=True) + dfree = dep.free_symbols + syms = dfree & ind.free_symbols + if not syms: + # but if the system looks like (a + b)*x + b - c + # then {} = {a, b, x} - c + # so calculate {x} = {a, b, x} - {a, b} + syms = dfree - set(coeffs) + if not syms: + syms = [Dummy()] + else: + if len(syms) == 1 and iterable(syms[0]): + syms = syms[0] + e, s, _ = recast_to_symbols([xeq], syms) + xeq = e[0] + syms = s + + # find the functional forms in which symbols appear + + gens = set(xeq.as_coefficients_dict(*syms).keys()) - {1} + cset = set(coeffs) + if any(g.has_xfree(cset) for g in gens): + return # a generator contained a coefficient symbol + + # make sure we are working with symbols for generators + + e, gens, _ = recast_to_symbols([xeq], list(gens)) + xeq = e[0] + + # collect coefficients in front of generators + + system = list(collect(xeq, gens, evaluate=False).values()) + + # get a solution + + soln = solve(system, coeffs, **flags) + + # unpack unless told otherwise if length is 1 + + settings = flags.get('dict', None) or flags.get('set', None) + if type(soln) is dict or settings or len(soln) != 1: + return soln + return soln[0] + + +def solve_linear_system_LU(matrix, syms): + """ + Solves the augmented matrix system using ``LUsolve`` and returns a + dictionary in which solutions are keyed to the symbols of *syms* as ordered. + + Explanation + =========== + + The matrix must be invertible. + + Examples + ======== + + >>> from sympy import Matrix, solve_linear_system_LU + >>> from sympy.abc import x, y, z + + >>> solve_linear_system_LU(Matrix([ + ... [1, 2, 0, 1], + ... [3, 2, 2, 1], + ... [2, 0, 0, 1]]), [x, y, z]) + {x: 1/2, y: 1/4, z: -1/2} + + See Also + ======== + + LUsolve + + """ + if matrix.rows != matrix.cols - 1: + raise ValueError("Rows should be equal to columns - 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 + + +def det_perm(M): + """ + Return the determinant of *M* by using permutations to select factors. + + Explanation + =========== + + For sizes larger than 8 the number of permutations becomes prohibitively + large, or if there are no symbols in the matrix, it is better to use the + standard determinant routines (e.g., ``M.det()``.) + + See Also + ======== + + det_minor + det_quick + + """ + args = [] + s = True + n = M.rows + list_ = M.flat() + for perm in generate_bell(n): + fac = [] + idx = 0 + for j in perm: + fac.append(list_[idx + j]) + idx += n + term = Mul(*fac) # disaster with unevaluated Mul -- takes forever for n=7 + args.append(term if s else -term) + s = not s + return Add(*args) + + +def det_minor(M): + """ + Return the ``det(M)`` computed from minors without + introducing new nesting in products. + + See Also + ======== + + det_perm + det_quick + + """ + n = M.rows + if n == 2: + return M[0, 0]*M[1, 1] - M[1, 0]*M[0, 1] + else: + return sum((1, -1)[i % 2]*Add(*[M[0, i]*d for d in + Add.make_args(det_minor(M.minor_submatrix(0, i)))]) + if M[0, i] else S.Zero for i in range(n)) + + +def det_quick(M, method=None): + """ + Return ``det(M)`` assuming that either + there are lots of zeros or the size of the matrix + is small. If this assumption is not met, then the normal + Matrix.det function will be used with method = ``method``. + + See Also + ======== + + det_minor + det_perm + + """ + if any(i.has(Symbol) for i in M): + if M.rows < 8 and all(i.has(Symbol) for i in M): + return det_perm(M) + return det_minor(M) + else: + return M.det(method=method) if method else M.det() + + +def inv_quick(M): + """Return the inverse of ``M``, assuming that either + there are lots of zeros or the size of the matrix + is small. + """ + if not all(i.is_Number for i in M): + if not any(i.is_Number for i in M): + det = lambda _: det_perm(_) + else: + det = lambda _: det_minor(_) + else: + return M.inv() + n = M.rows + d = det(M) + if d == S.Zero: + raise NonInvertibleMatrixError("Matrix det == 0; not invertible") + ret = zeros(n) + s1 = -1 + for i in range(n): + s = s1 = -s1 + for j in range(n): + di = det(M.minor_submatrix(i, j)) + ret[j, i] = s*di/d + s = -s + return ret + + +# these are functions that have multiple inverse values per period +multi_inverses = { + sin: lambda x: (asin(x), S.Pi - asin(x)), + cos: lambda x: (acos(x), 2*S.Pi - acos(x)), +} + + +def _vsolve(e, s, **flags): + """return list of scalar values for the solution of e for symbol s""" + return [i[s] for i in _solve(e, s, **flags)] + + +def _tsolve(eq, sym, **flags): + """ + Helper for ``_solve`` that solves a transcendental equation with respect + to the given symbol. Various equations containing powers and logarithms, + can be solved. + + There is currently no guarantee that all solutions will be returned or + that a real solution will be favored over a complex one. + + Either a list of potential solutions will be returned or None will be + returned (in the case that no method was known to get a solution + for the equation). All other errors (like the inability to cast an + expression as a Poly) are unhandled. + + Examples + ======== + + >>> from sympy import log, ordered + >>> from sympy.solvers.solvers import _tsolve as tsolve + >>> from sympy.abc import x + + >>> list(ordered(tsolve(3**(2*x + 5) - 4, x))) + [-5/2 + log(2)/log(3), (-5*log(3)/2 + log(2) + I*pi)/log(3)] + + >>> tsolve(log(x) + 2*x, x) + [LambertW(2)/2] + + """ + if 'tsolve_saw' not in flags: + flags['tsolve_saw'] = [] + if eq in flags['tsolve_saw']: + return None + else: + flags['tsolve_saw'].append(eq) + + rhs, lhs = _invert(eq, sym) + + if lhs == sym: + return [rhs] + try: + if lhs.is_Add: + # it's time to try factoring; powdenest is used + # to try get powers in standard form for better factoring + f = factor(powdenest(lhs - rhs)) + if f.is_Mul: + return _vsolve(f, sym, **flags) + if rhs: + f = logcombine(lhs, force=flags.get('force', True)) + if f.count(log) != lhs.count(log): + if isinstance(f, log): + return _vsolve(f.args[0] - exp(rhs), sym, **flags) + return _tsolve(f - rhs, sym, **flags) + + elif lhs.is_Pow: + if lhs.exp.is_Integer: + if lhs - rhs != eq: + return _vsolve(lhs - rhs, sym, **flags) + + if sym not in lhs.exp.free_symbols: + return _vsolve(lhs.base - rhs**(1/lhs.exp), sym, **flags) + + # _tsolve calls this with Dummy before passing the actual number in. + if any(t.is_Dummy for t in rhs.free_symbols): + raise NotImplementedError # _tsolve will call here again... + + # a ** g(x) == 0 + if not rhs: + # f(x)**g(x) only has solutions where f(x) == 0 and g(x) != 0 at + # the same place + sol_base = _vsolve(lhs.base, sym, **flags) + return [s for s in sol_base if lhs.exp.subs(sym, s) != 0] # XXX use checksol here? + + # a ** g(x) == b + if not lhs.base.has(sym): + if lhs.base == 0: + return _vsolve(lhs.exp, sym, **flags) if rhs != 0 else [] + + # Gets most solutions... + if lhs.base == rhs.as_base_exp()[0]: + # handles case when bases are equal + sol = _vsolve(lhs.exp - rhs.as_base_exp()[1], sym, **flags) + else: + # handles cases when bases are not equal and exp + # may or may not be equal + f = exp(log(lhs.base)*lhs.exp) - exp(log(rhs)) + sol = _vsolve(f, sym, **flags) + + # Check for duplicate solutions + def equal(expr1, expr2): + _ = Dummy() + eq = checksol(expr1 - _, _, expr2) + if eq is None: + if nsimplify(expr1) != nsimplify(expr2): + return False + # they might be coincidentally the same + # so check more rigorously + eq = expr1.equals(expr2) # XXX expensive but necessary? + return eq + + # Guess a rational exponent + e_rat = nsimplify(log(abs(rhs))/log(abs(lhs.base))) + e_rat = simplify(posify(e_rat)[0]) + n, d = fraction(e_rat) + if expand(lhs.base**n - rhs**d) == 0: + sol = [s for s in sol if not equal(lhs.exp.subs(sym, s), e_rat)] + sol.extend(_vsolve(lhs.exp - e_rat, sym, **flags)) + + return list(set(sol)) + + # f(x) ** g(x) == c + else: + sol = [] + logform = lhs.exp*log(lhs.base) - log(rhs) + if logform != lhs - rhs: + try: + sol.extend(_vsolve(logform, sym, **flags)) + except NotImplementedError: + pass + + # Collect possible solutions and check with substitution later. + check = [] + if rhs == 1: + # f(x) ** g(x) = 1 -- g(x)=0 or f(x)=+-1 + check.extend(_vsolve(lhs.exp, sym, **flags)) + check.extend(_vsolve(lhs.base - 1, sym, **flags)) + check.extend(_vsolve(lhs.base + 1, sym, **flags)) + elif rhs.is_Rational: + for d in (i for i in divisors(abs(rhs.p)) if i != 1): + e, t = integer_log(rhs.p, d) + if not t: + continue # rhs.p != d**b + for s in divisors(abs(rhs.q)): + if s**e== rhs.q: + r = Rational(d, s) + check.extend(_vsolve(lhs.base - r, sym, **flags)) + check.extend(_vsolve(lhs.base + r, sym, **flags)) + check.extend(_vsolve(lhs.exp - e, sym, **flags)) + elif rhs.is_irrational: + b_l, e_l = lhs.base.as_base_exp() + n, d = (e_l*lhs.exp).as_numer_denom() + b, e = sqrtdenest(rhs).as_base_exp() + check = [sqrtdenest(i) for i in (_vsolve(lhs.base - b, sym, **flags))] + check.extend([sqrtdenest(i) for i in (_vsolve(lhs.exp - e, sym, **flags))]) + if e_l*d != 1: + check.extend(_vsolve(b_l**n - rhs**(e_l*d), sym, **flags)) + for s in check: + ok = checksol(eq, sym, s) + if ok is None: + ok = eq.subs(sym, s).equals(0) + if ok: + sol.append(s) + return list(set(sol)) + + elif lhs.is_Function and len(lhs.args) == 1: + if lhs.func in multi_inverses: + # sin(x) = 1/3 -> x - asin(1/3) & x - (pi - asin(1/3)) + soln = [] + for i in multi_inverses[type(lhs)](rhs): + soln.extend(_vsolve(lhs.args[0] - i, sym, **flags)) + return list(set(soln)) + elif lhs.func == LambertW: + return _vsolve(lhs.args[0] - rhs*exp(rhs), sym, **flags) + + rewrite = lhs.rewrite(exp) + rewrite = rebuild(rewrite) # avoid rewrites involving evaluate=False + if rewrite != lhs: + return _vsolve(rewrite - rhs, sym, **flags) + except NotImplementedError: + pass + + # maybe it is a lambert pattern + if flags.pop('bivariate', True): + # lambert forms may need some help being recognized, e.g. changing + # 2**(3*x) + x**3*log(2)**3 + 3*x**2*log(2)**2 + 3*x*log(2) + 1 + # to 2**(3*x) + (x*log(2) + 1)**3 + + # make generator in log have exponent of 1 + logs = eq.atoms(log) + spow = min( + {i.exp for j in logs for i in j.atoms(Pow) + if i.base == sym} or {1}) + if spow != 1: + p = sym**spow + u = Dummy('bivariate-cov') + ueq = eq.subs(p, u) + if not ueq.has_free(sym): + sol = _vsolve(ueq, u, **flags) + inv = _vsolve(p - u, sym) + return [i.subs(u, s) for i in inv for s in sol] + + g = _filtered_gens(eq.as_poly(), sym) + up_or_log = set() + for gi in g: + if isinstance(gi, (exp, log)) or (gi.is_Pow and gi.base == S.Exp1): + up_or_log.add(gi) + elif gi.is_Pow: + gisimp = powdenest(expand_power_exp(gi)) + if gisimp.is_Pow and sym in gisimp.exp.free_symbols: + up_or_log.add(gi) + eq_down = expand_log(expand_power_exp(eq)).subs( + dict(list(zip(up_or_log, [0]*len(up_or_log))))) + eq = expand_power_exp(factor(eq_down, deep=True) + (eq - eq_down)) + rhs, lhs = _invert(eq, sym) + if lhs.has(sym): + try: + poly = lhs.as_poly() + g = _filtered_gens(poly, sym) + _eq = lhs - rhs + sols = _solve_lambert(_eq, sym, g) + # use a simplified form if it satisfies eq + # and has fewer operations + for n, s in enumerate(sols): + ns = nsimplify(s) + if ns != s and ns.count_ops() <= s.count_ops(): + ok = checksol(_eq, sym, ns) + if ok is None: + ok = _eq.subs(sym, ns).equals(0) + if ok: + sols[n] = ns + return sols + except NotImplementedError: + # maybe it's a convoluted function + if len(g) == 2: + try: + gpu = bivariate_type(lhs - rhs, *g) + if gpu is None: + raise NotImplementedError + g, p, u = gpu + flags['bivariate'] = False + inversion = _tsolve(g - u, sym, **flags) + if inversion: + sol = _vsolve(p, u, **flags) + return list({i.subs(u, s) + for i in inversion for s in sol}) + except NotImplementedError: + pass + else: + pass + + if flags.pop('force', True): + flags['force'] = False + pos, reps = posify(lhs - rhs) + if rhs == S.ComplexInfinity: + return [] + for u, s in reps.items(): + if s == sym: + break + else: + u = sym + if pos.has(u): + try: + soln = _vsolve(pos, u, **flags) + return [s.subs(reps) for s in soln] + except NotImplementedError: + pass + else: + pass # here for coverage + + return # here for coverage + + +# TODO: option for calculating J numerically + +@conserve_mpmath_dps +def nsolve(*args, dict=False, **kwargs): + r""" + Solve a nonlinear equation system numerically: ``nsolve(f, [args,] x0, + modules=['mpmath'], **kwargs)``. + + Explanation + =========== + + ``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``. + + If the keyword arguments contain ``dict=True`` (default is False) ``nsolve`` + will return a list (perhaps empty) of solution mappings. This might be + especially useful if you want to use ``nsolve`` as a fallback to solve since + using the dict argument for both methods produces return values of + consistent type structure. Please note: to keep this consistent with + ``solve``, the solution will be returned in a list even though ``nsolve`` + (currently at least) only finds one solution at a time. + + Overdetermined systems are supported. + + Examples + ======== + + >>> from sympy import Symbol, nsolve + >>> import mpmath + >>> 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))) + Matrix([[-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 + + To solve with higher precision than the default, use the prec argument: + + >>> from sympy import cos + >>> nsolve(cos(x) - x, 1) + 0.739085133215161 + >>> nsolve(cos(x) - x, 1, prec=50) + 0.73908513321516064165531208767387340401341175890076 + >>> cos(_) + 0.73908513321516064165531208767387340401341175890076 + + To solve for complex roots of real functions, a nonreal initial point + must be specified: + + >>> from sympy import I + >>> nsolve(x**2 + 2, I) + 1.4142135623731*I + + ``mpmath.findroot`` is used and you can find their more extensive + documentation, especially concerning keyword parameters and + available solvers. Note, however, that functions which are very + steep near the root, the verification of the solution may fail. In + this case you should use the flag ``verify=False`` and + independently verify the solution. + + >>> from sympy import cos, cosh + >>> f = cos(x)*cosh(x) - 1 + >>> nsolve(f, 3.14*100) + Traceback (most recent call last): + ... + ValueError: Could not find root within given tolerance. (1.39267e+230 > 2.1684e-19) + >>> ans = nsolve(f, 3.14*100, verify=False); ans + 312.588469032184 + >>> f.subs(x, ans).n(2) + 2.1e+121 + >>> (f/f.diff(x)).subs(x, ans).n(2) + 7.4e-15 + + One might safely skip the verification if bounds of the root are known + and a bisection method is used: + + >>> bounds = lambda i: (3.14*i, 3.14*(i + 1)) + >>> nsolve(f, bounds(100), solver='bisect', verify=False) + 315.730061685774 + + Alternatively, a function may be better behaved when the + denominator is ignored. Since this is not always the case, however, + the decision of what function to use is left to the discretion of + the user. + + >>> eq = x**2/(1 - x)/(1 - 2*x)**2 - 100 + >>> nsolve(eq, 0.46) + Traceback (most recent call last): + ... + ValueError: Could not find root within given tolerance. (10000 > 2.1684e-19) + Try another starting point or tweak arguments. + >>> nsolve(eq.as_numer_denom()[0], 0.46) + 0.46792545969349058 + + """ + # there are several other SymPy functions that use method= so + # guard against that here + if 'method' in kwargs: + raise ValueError(filldedent(''' + Keyword "method" should not be used in this context. When using + some mpmath solvers directly, the keyword "method" is + used, but when using nsolve (and findroot) the keyword to use is + "solver".''')) + + if 'prec' in kwargs: + import mpmath + mpmath.mp.dps = kwargs.pop('prec') + + # keyword argument to return result as a dictionary + as_dict = dict + from builtins import dict # to unhide the builtin + + # interpret arguments + if len(args) == 3: + f = args[0] + fargs = args[1] + x0 = args[2] + if iterable(fargs) and iterable(x0): + if len(x0) != len(fargs): + raise TypeError('nsolve expected exactly %i guess vectors, got %i' + % (len(fargs), len(x0))) + elif len(args) == 2: + f = args[0] + fargs = None + x0 = args[1] + if iterable(f): + raise TypeError('nsolve expected 3 arguments, got 2') + 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 iterable(f): + f = list(f) + for i, fi in enumerate(f): + if isinstance(fi, Eq): + f[i] = fi.lhs - fi.rhs + f = Matrix(f).T + if iterable(x0): + x0 = list(x0) + if not isinstance(f, Matrix): + # assume it's a SymPy expression + if isinstance(f, Eq): + f = f.lhs - f.rhs + elif f.is_Relational: + raise TypeError('nsolve cannot accept inequalities') + syms = f.free_symbols + if fargs is None: + fargs = syms.copy().pop() + if not (len(syms) == 1 and (fargs in syms or fargs[0] in syms)): + raise ValueError(filldedent(''' + expected a one-dimensional and numerical function''')) + + # the function is much better behaved if there is no denominator + # but sending the numerator is left to the user since sometimes + # the function is better behaved when the denominator is present + # e.g., issue 11768 + + f = lambdify(fargs, f, modules) + x = sympify(findroot(f, x0, **kwargs)) + if as_dict: + return [{fargs: x}] + return x + + if len(fargs) > f.cols: + raise NotImplementedError(filldedent(''' + 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) + if as_dict: + return [dict(zip(fargs, [sympify(xi) for xi in x]))] + return Matrix(x) + + +def _invert(eq, *symbols, **kwargs): + """ + Return tuple (i, d) where ``i`` is independent of *symbols* and ``d`` + contains symbols. + + Explanation + =========== + + ``i`` and ``d`` are obtained after recursively using algebraic inversion + until an uninvertible ``d`` remains. If there are no free symbols then + ``d`` will be zero. Some (but not necessarily all) solutions to the + expression ``i - d`` will be related to the solutions of the original + expression. + + Examples + ======== + + >>> from sympy.solvers.solvers import _invert as invert + >>> from sympy import sqrt, cos + >>> from sympy.abc import x, y + >>> invert(x - 3) + (3, x) + >>> invert(3) + (3, 0) + >>> invert(2*cos(x) - 1) + (1/2, cos(x)) + >>> invert(sqrt(x) - 3) + (3, sqrt(x)) + >>> invert(sqrt(x) + y, x) + (-y, sqrt(x)) + >>> invert(sqrt(x) + y, y) + (-sqrt(x), y) + >>> invert(sqrt(x) + y, x, y) + (0, sqrt(x) + y) + + If there is more than one symbol in a power's base and the exponent + is not an Integer, then the principal root will be used for the + inversion: + + >>> invert(sqrt(x + y) - 2) + (4, x + y) + >>> invert(sqrt(x + y) + 2) # note +2 instead of -2 + (4, x + y) + + If the exponent is an Integer, setting ``integer_power`` to True + will force the principal root to be selected: + + >>> invert(x**2 - 4, integer_power=True) + (2, x) + + """ + eq = sympify(eq) + if eq.args: + # make sure we are working with flat eq + eq = eq.func(*eq.args) + free = eq.free_symbols + if not symbols: + symbols = free + if not free & set(symbols): + return eq, S.Zero + + dointpow = bool(kwargs.get('integer_power', False)) + + lhs = eq + rhs = S.Zero + while True: + was = lhs + while True: + indep, dep = lhs.as_independent(*symbols) + + # dep + indep == rhs + if lhs.is_Add: + # this indicates we have done it all + if indep.is_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 + + # collect like-terms in symbols + if lhs.is_Add: + terms = {} + for a in lhs.args: + i, d = a.as_independent(*symbols) + terms.setdefault(d, []).append(i) + if any(len(v) > 1 for v in terms.values()): + args = [] + for d, i in terms.items(): + if len(i) > 1: + args.append(Add(*i)*d) + else: + args.append(i[0]*d) + lhs = Add(*args) + + # if it's a two-term Add with rhs = 0 and two powers we can get the + # dependent terms together, e.g. 3*f(x) + 2*g(x) -> f(x)/g(x) = -2/3 + if lhs.is_Add and not rhs and len(lhs.args) == 2 and \ + not lhs.is_polynomial(*symbols): + a, b = ordered(lhs.args) + ai, ad = a.as_independent(*symbols) + bi, bd = b.as_independent(*symbols) + if any(_ispow(i) for i in (ad, bd)): + a_base, a_exp = ad.as_base_exp() + b_base, b_exp = bd.as_base_exp() + if a_base == b_base and a_exp.extract_additively(b_exp) is None: + # a = -b and exponents do not have canceling terms/factors + # e.g. if exponents were 3*x and x then the ratio would have + # an exponent of 2*x: one of the roots would be lost + rat = powsimp(powdenest(ad/bd)) + lhs = rat + rhs = -bi/ai + else: + rat = ad/bd + _lhs = powsimp(ad/bd) + if _lhs != rat: + lhs = _lhs + rhs = -bi/ai + elif ai == -bi: + if isinstance(ad, Function) and ad.func == bd.func: + if len(ad.args) == len(bd.args) == 1: + lhs = ad.args[0] - bd.args[0] + elif len(ad.args) == len(bd.args): + # should be able to solve + # f(x, y) - f(2 - x, 0) == 0 -> x == 1 + raise NotImplementedError( + 'equal function with more than 1 argument') + else: + raise ValueError( + 'function with different numbers of args') + + elif lhs.is_Mul and any(_ispow(a) for a in lhs.args): + lhs = powsimp(powdenest(lhs)) + + if lhs.is_Function: + if hasattr(lhs, 'inverse') and lhs.inverse() is not None and len(lhs.args) == 1: + # -1 + # f(x) = g -> x = f (g) + # + # /!\ inverse should not be defined if there are multiple values + # for the function -- these are handled in _tsolve + # + rhs = lhs.inverse()(rhs) + lhs = lhs.args[0] + elif isinstance(lhs, atan2): + y, x = lhs.args + lhs = 2*atan(y/(sqrt(x**2 + y**2) + x)) + elif lhs.func == rhs.func: + if len(lhs.args) == len(rhs.args) == 1: + lhs = lhs.args[0] + rhs = rhs.args[0] + elif len(lhs.args) == len(rhs.args): + # should be able to solve + # f(x, y) == f(2, 3) -> x == 2 + # f(x, x + y) == f(2, 3) -> x == 2 + raise NotImplementedError( + 'equal function with more than 1 argument') + else: + raise ValueError( + 'function with different numbers of args') + + + if rhs and lhs.is_Pow and lhs.exp.is_Integer and lhs.exp < 0: + lhs = 1/lhs + rhs = 1/rhs + + # base**a = b -> base = b**(1/a) if + # a is an Integer and dointpow=True (this gives real branch of root) + # a is not an Integer and the equation is multivariate and the + # base has more than 1 symbol in it + # The rationale for this is that right now the multi-system solvers + # doesn't try to resolve generators to see, for example, if the whole + # system is written in terms of sqrt(x + y) so it will just fail, so we + # do that step here. + if lhs.is_Pow and ( + lhs.exp.is_Integer and dointpow or not lhs.exp.is_Integer and + len(symbols) > 1 and len(lhs.base.free_symbols & set(symbols)) > 1): + rhs = rhs**(1/lhs.exp) + lhs = lhs.base + + if lhs == was: + break + return rhs, lhs + + +def unrad(eq, *syms, **flags): + """ + Remove radicals with symbolic arguments and return (eq, cov), + None, or raise an error. + + Explanation + =========== + + None is returned if there are no radicals to remove. + + NotImplementedError is raised if there are radicals and they cannot be + removed or if the relationship between the original symbols and the + change of variable needed to rewrite the system as a polynomial cannot + be solved. + + Otherwise the tuple, ``(eq, cov)``, is returned where: + + *eq*, ``cov`` + *eq* is an equation without radicals (in the symbol(s) of + interest) whose solutions are a superset of the solutions to the + original expression. *eq* might be rewritten in terms of a new + variable; the relationship to the original variables is given by + ``cov`` which is a list containing ``v`` and ``v**p - b`` where + ``p`` is the power needed to clear the radical and ``b`` is the + radical now expressed as a polynomial in the symbols of interest. + For example, for sqrt(2 - x) the tuple would be + ``(c, c**2 - 2 + x)``. The solutions of *eq* will contain + solutions to the original equation (if there are any). + + *syms* + An iterable of symbols which, if provided, will limit the focus of + radical removal: only radicals with one or more of the symbols of + interest will be cleared. All free symbols are used if *syms* is not + set. + + *flags* are used internally for communication during recursive calls. + Two options are also recognized: + + ``take``, when defined, is interpreted as a single-argument function + that returns True if a given Pow should be handled. + + Radicals can be removed from an expression if: + + * All bases of the radicals are the same; a change of variables is + done in this case. + * If all radicals appear in one term of the expression. + * There are only four terms with sqrt() factors or there are less than + four terms having sqrt() factors. + * There are only two terms with radicals. + + Examples + ======== + + >>> from sympy.solvers.solvers import unrad + >>> from sympy.abc import x + >>> from sympy import sqrt, Rational, root + + >>> unrad(sqrt(x)*x**Rational(1, 3) + 2) + (x**5 - 64, []) + >>> unrad(sqrt(x) + root(x + 1, 3)) + (-x**3 + x**2 + 2*x + 1, []) + >>> eq = sqrt(x) + root(x, 3) - 2 + >>> unrad(eq) + (_p**3 + _p**2 - 2, [_p, _p**6 - x]) + + """ + + uflags = {"check": False, "simplify": False} + + def _cov(p, e): + if cov: + # XXX - uncovered + oldp, olde = cov + if Poly(e, p).degree(p) in (1, 2): + cov[:] = [p, olde.subs(oldp, _vsolve(e, p, **uflags)[0])] + else: + raise NotImplementedError + else: + cov[:] = [p, e] + + def _canonical(eq, cov): + if cov: + # change symbol to vanilla so no solutions are eliminated + p, e = cov + rep = {p: Dummy(p.name)} + eq = eq.xreplace(rep) + cov = [p.xreplace(rep), e.xreplace(rep)] + + # remove constants and powers of factors since these don't change + # the location of the root; XXX should factor or factor_terms be used? + eq = factor_terms(_mexpand(eq.as_numer_denom()[0], recursive=True), clear=True) + if eq.is_Mul: + args = [] + for f in eq.args: + if f.is_number: + continue + if f.is_Pow: + args.append(f.base) + else: + args.append(f) + eq = Mul(*args) # leave as Mul for more efficient solving + + # make the sign canonical + margs = list(Mul.make_args(eq)) + changed = False + for i, m in enumerate(margs): + if m.could_extract_minus_sign(): + margs[i] = -m + changed = True + if changed: + eq = Mul(*margs, evaluate=False) + + return eq, cov + + def _Q(pow): + # return leading Rational of denominator of Pow's exponent + c = pow.as_base_exp()[1].as_coeff_Mul()[0] + if not c.is_Rational: + return S.One + return c.q + + # define the _take method that will determine whether a term is of interest + def _take(d): + # return True if coefficient of any factor's exponent's den is not 1 + for pow in Mul.make_args(d): + if not pow.is_Pow: + continue + if _Q(pow) == 1: + continue + if pow.free_symbols & syms: + return True + return False + _take = flags.setdefault('_take', _take) + + if isinstance(eq, Eq): + eq = eq.lhs - eq.rhs # XXX legacy Eq as Eqn support + elif not isinstance(eq, Expr): + return + + cov, nwas, rpt = [flags.setdefault(k, v) for k, v in + sorted({"cov": [], "n": None, "rpt": 0}.items())] + + # preconditioning + eq = powdenest(factor_terms(eq, radical=True, clear=True)) + eq = eq.as_numer_denom()[0] + eq = _mexpand(eq, recursive=True) + if eq.is_number: + return + + # see if there are radicals in symbols of interest + syms = set(syms) or eq.free_symbols # _take uses this + poly = eq.as_poly() + gens = [g for g in poly.gens if _take(g)] + if not gens: + return + + # recast poly in terms of eigen-gens + poly = eq.as_poly(*gens) + + # not a polynomial e.g. 1 + sqrt(x)*exp(sqrt(x)) with gen sqrt(x) + if poly is None: + return + + # - an exponent has a symbol of interest (don't handle) + if any(g.exp.has(*syms) for g in gens): + return + + def _rads_bases_lcm(poly): + # if all the bases are the same or all the radicals are in one + # term, `lcm` will be the lcm of the denominators of the + # exponents of the radicals + lcm = 1 + rads = set() + bases = set() + for g in poly.gens: + q = _Q(g) + if q != 1: + rads.add(g) + lcm = ilcm(lcm, q) + bases.add(g.base) + return rads, bases, lcm + rads, bases, lcm = _rads_bases_lcm(poly) + + covsym = Dummy('p', nonnegative=True) + + # only keep in syms symbols that actually appear in radicals; + # and update gens + newsyms = set() + for r in rads: + newsyms.update(syms & r.free_symbols) + if newsyms != syms: + syms = newsyms + # get terms together that have common generators + drad = dict(zip(rads, range(len(rads)))) + rterms = {(): []} + args = Add.make_args(poly.as_expr()) + for t in args: + if _take(t): + common = set(t.as_poly().gens).intersection(rads) + key = tuple(sorted([drad[i] for i in common])) + else: + key = () + rterms.setdefault(key, []).append(t) + others = Add(*rterms.pop(())) + rterms = [Add(*rterms[k]) for k in rterms.keys()] + + # the output will depend on the order terms are processed, so + # make it canonical quickly + rterms = list(reversed(list(ordered(rterms)))) + + ok = False # we don't have a solution yet + depth = sqrt_depth(eq) + + if len(rterms) == 1 and not (rterms[0].is_Add and lcm > 2): + eq = rterms[0]**lcm - ((-others)**lcm) + ok = True + else: + if len(rterms) == 1 and rterms[0].is_Add: + rterms = list(rterms[0].args) + if len(bases) == 1: + b = bases.pop() + if len(syms) > 1: + x = b.free_symbols + else: + x = syms + x = list(ordered(x))[0] + try: + inv = _vsolve(covsym**lcm - b, x, **uflags) + if not inv: + raise NotImplementedError + eq = poly.as_expr().subs(b, covsym**lcm).subs(x, inv[0]) + _cov(covsym, covsym**lcm - b) + return _canonical(eq, cov) + except NotImplementedError: + pass + + if len(rterms) == 2: + if not others: + eq = rterms[0]**lcm - (-rterms[1])**lcm + ok = True + elif not log(lcm, 2).is_Integer: + # the lcm-is-power-of-two case is handled below + r0, r1 = rterms + if flags.get('_reverse', False): + r1, r0 = r0, r1 + i0 = _rads0, _bases0, lcm0 = _rads_bases_lcm(r0.as_poly()) + i1 = _rads1, _bases1, lcm1 = _rads_bases_lcm(r1.as_poly()) + for reverse in range(2): + if reverse: + i0, i1 = i1, i0 + r0, r1 = r1, r0 + _rads1, _, lcm1 = i1 + _rads1 = Mul(*_rads1) + t1 = _rads1**lcm1 + c = covsym**lcm1 - t1 + for x in syms: + try: + sol = _vsolve(c, x, **uflags) + if not sol: + raise NotImplementedError + neweq = r0.subs(x, sol[0]) + covsym*r1/_rads1 + \ + others + tmp = unrad(neweq, covsym) + if tmp: + eq, newcov = tmp + if newcov: + newp, newc = newcov + _cov(newp, c.subs(covsym, + _vsolve(newc, covsym, **uflags)[0])) + else: + _cov(covsym, c) + else: + eq = neweq + _cov(covsym, c) + ok = True + break + except NotImplementedError: + if reverse: + raise NotImplementedError( + 'no successful change of variable found') + else: + pass + if ok: + break + elif len(rterms) == 3: + # two cube roots and another with order less than 5 + # (so an analytical solution can be found) or a base + # that matches one of the cube root bases + info = [_rads_bases_lcm(i.as_poly()) for i in rterms] + RAD = 0 + BASES = 1 + LCM = 2 + if info[0][LCM] != 3: + info.append(info.pop(0)) + rterms.append(rterms.pop(0)) + elif info[1][LCM] != 3: + info.append(info.pop(1)) + rterms.append(rterms.pop(1)) + if info[0][LCM] == info[1][LCM] == 3: + if info[1][BASES] != info[2][BASES]: + info[0], info[1] = info[1], info[0] + rterms[0], rterms[1] = rterms[1], rterms[0] + if info[1][BASES] == info[2][BASES]: + eq = rterms[0]**3 + (rterms[1] + rterms[2] + others)**3 + ok = True + elif info[2][LCM] < 5: + # a*root(A, 3) + b*root(B, 3) + others = c + a, b, c, d, A, B = [Dummy(i) for i in 'abcdAB'] + # zz represents the unraded expression into which the + # specifics for this case are substituted + zz = (c - d)*(A**3*a**9 + 3*A**2*B*a**6*b**3 - + 3*A**2*a**6*c**3 + 9*A**2*a**6*c**2*d - 9*A**2*a**6*c*d**2 + + 3*A**2*a**6*d**3 + 3*A*B**2*a**3*b**6 + 21*A*B*a**3*b**3*c**3 - + 63*A*B*a**3*b**3*c**2*d + 63*A*B*a**3*b**3*c*d**2 - + 21*A*B*a**3*b**3*d**3 + 3*A*a**3*c**6 - 18*A*a**3*c**5*d + + 45*A*a**3*c**4*d**2 - 60*A*a**3*c**3*d**3 + 45*A*a**3*c**2*d**4 - + 18*A*a**3*c*d**5 + 3*A*a**3*d**6 + B**3*b**9 - 3*B**2*b**6*c**3 + + 9*B**2*b**6*c**2*d - 9*B**2*b**6*c*d**2 + 3*B**2*b**6*d**3 + + 3*B*b**3*c**6 - 18*B*b**3*c**5*d + 45*B*b**3*c**4*d**2 - + 60*B*b**3*c**3*d**3 + 45*B*b**3*c**2*d**4 - 18*B*b**3*c*d**5 + + 3*B*b**3*d**6 - c**9 + 9*c**8*d - 36*c**7*d**2 + 84*c**6*d**3 - + 126*c**5*d**4 + 126*c**4*d**5 - 84*c**3*d**6 + 36*c**2*d**7 - + 9*c*d**8 + d**9) + def _t(i): + b = Mul(*info[i][RAD]) + return cancel(rterms[i]/b), Mul(*info[i][BASES]) + aa, AA = _t(0) + bb, BB = _t(1) + cc = -rterms[2] + dd = others + eq = zz.xreplace(dict(zip( + (a, A, b, B, c, d), + (aa, AA, bb, BB, cc, dd)))) + ok = True + # handle power-of-2 cases + if not ok: + if log(lcm, 2).is_Integer and (not others and + len(rterms) == 4 or len(rterms) < 4): + def _norm2(a, b): + return a**2 + b**2 + 2*a*b + + if len(rterms) == 4: + # (r0+r1)**2 - (r2+r3)**2 + r0, r1, r2, r3 = rterms + eq = _norm2(r0, r1) - _norm2(r2, r3) + ok = True + elif len(rterms) == 3: + # (r1+r2)**2 - (r0+others)**2 + r0, r1, r2 = rterms + eq = _norm2(r1, r2) - _norm2(r0, others) + ok = True + elif len(rterms) == 2: + # r0**2 - (r1+others)**2 + r0, r1 = rterms + eq = r0**2 - _norm2(r1, others) + ok = True + + new_depth = sqrt_depth(eq) if ok else depth + rpt += 1 # XXX how many repeats with others unchanging is enough? + if not ok or ( + nwas is not None and len(rterms) == nwas and + new_depth is not None and new_depth == depth and + rpt > 3): + raise NotImplementedError('Cannot remove all radicals') + + flags.update({"cov": cov, "n": len(rterms), "rpt": rpt}) + neq = unrad(eq, *syms, **flags) + if neq: + eq, cov = neq + eq, cov = _canonical(eq, cov) + return eq, cov + + +# delayed imports +from sympy.solvers.bivariate import ( + bivariate_type, _solve_lambert, _filtered_gens) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/solveset.py b/pllava/lib/python3.10/site-packages/sympy/solvers/solveset.py new file mode 100644 index 0000000000000000000000000000000000000000..4d1bd62439f816f304fb1c812ab54555144a2b2a --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/solveset.py @@ -0,0 +1,4124 @@ +""" +This module contains functions to: + + - solve a single equation for a single variable, in any domain either real or complex. + + - solve a single transcendental equation for a single variable in any domain either real or complex. + (currently supports solving in real domain only) + + - solve a system of linear equations with N variables and M equations. + + - solve a system of Non Linear Equations with N variables and M equations +""" +from sympy.core.sympify import sympify +from sympy.core import (S, Pow, Dummy, pi, Expr, Wild, Mul, + Add, Basic) +from sympy.core.containers import Tuple +from sympy.core.function import (Lambda, expand_complex, AppliedUndef, + expand_log, _mexpand, expand_trig, nfloat) +from sympy.core.mod import Mod +from sympy.core.numbers import I, Number, Rational, oo +from sympy.core.intfunc import integer_log +from sympy.core.relational import Eq, Ne, Relational +from sympy.core.sorting import default_sort_key, ordered +from sympy.core.symbol import Symbol, _uniquely_named_symbol +from sympy.core.sympify import _sympify +from sympy.core.traversal import preorder_traversal +from sympy.external.gmpy import gcd as number_gcd, lcm as number_lcm +from sympy.polys.matrices.linsolve import _linear_eq_to_dict +from sympy.polys.polyroots import UnsolvableFactorError +from sympy.simplify.simplify import simplify, fraction, trigsimp, nsimplify +from sympy.simplify import powdenest, logcombine +from sympy.functions import (log, tan, cot, sin, cos, sec, csc, exp, + acos, asin, atan, acot, acsc, asec, + piecewise_fold, Piecewise) +from sympy.functions.combinatorial.numbers import totient +from sympy.functions.elementary.complexes import Abs, arg, re, im +from sympy.functions.elementary.hyperbolic import (HyperbolicFunction, + sinh, cosh, tanh, coth, sech, csch, + asinh, acosh, atanh, acoth, asech, acsch) +from sympy.functions.elementary.miscellaneous import real_root +from sympy.functions.elementary.trigonometric import TrigonometricFunction +from sympy.logic.boolalg import And, BooleanTrue +from sympy.sets import (FiniteSet, imageset, Interval, Intersection, + Union, ConditionSet, ImageSet, Complement, Contains) +from sympy.sets.sets import Set, ProductSet +from sympy.matrices import zeros, Matrix, MatrixBase +from sympy.ntheory.factor_ import divisors +from sympy.ntheory.residue_ntheory import discrete_log, nthroot_mod +from sympy.polys import (roots, Poly, degree, together, PolynomialError, + RootOf, factor, lcm, gcd) +from sympy.polys.polyerrors import CoercionFailed +from sympy.polys.polytools import invert, groebner, poly +from sympy.polys.solvers import (sympy_eqs_to_ring, solve_lin_sys, + PolyNonlinearError) +from sympy.polys.matrices.linsolve import _linsolve +from sympy.solvers.solvers import (checksol, denoms, unrad, + _simple_dens, recast_to_symbols) +from sympy.solvers.polysys import solve_poly_system +from sympy.utilities import filldedent +from sympy.utilities.iterables import (numbered_symbols, has_dups, + is_sequence, iterable) +from sympy.calculus.util import periodicity, continuous_domain, function_range + + +from types import GeneratorType + + +class NonlinearError(ValueError): + """Raised when unexpectedly encountering nonlinear equations""" + pass + + +def _masked(f, *atoms): + """Return ``f``, with all objects given by ``atoms`` replaced with + Dummy symbols, ``d``, and the list of replacements, ``(d, e)``, + where ``e`` is an object of type given by ``atoms`` in which + any other instances of atoms have been recursively replaced with + Dummy symbols, too. The tuples are ordered so that if they are + applied in sequence, the origin ``f`` will be restored. + + Examples + ======== + + >>> from sympy import cos + >>> from sympy.abc import x + >>> from sympy.solvers.solveset import _masked + + >>> f = cos(cos(x) + 1) + >>> f, reps = _masked(cos(1 + cos(x)), cos) + >>> f + _a1 + >>> reps + [(_a1, cos(_a0 + 1)), (_a0, cos(x))] + >>> for d, e in reps: + ... f = f.xreplace({d: e}) + >>> f + cos(cos(x) + 1) + """ + sym = numbered_symbols('a', cls=Dummy, real=True) + mask = [] + for a in ordered(f.atoms(*atoms)): + for i in mask: + a = a.replace(*i) + mask.append((a, next(sym))) + for i, (o, n) in enumerate(mask): + f = f.replace(o, n) + mask[i] = (n, o) + mask = list(reversed(mask)) + return f, mask + + +def _invert(f_x, y, x, domain=S.Complexes): + r""" + Reduce the complex valued equation $f(x) = y$ to a set of equations + + $$\left\{g(x) = h_1(y),\ g(x) = h_2(y),\ \dots,\ g(x) = h_n(y) \right\}$$ + + where $g(x)$ is a simpler function than $f(x)$. The return value is a tuple + $(g(x), \mathrm{set}_h)$, where $g(x)$ is a function of $x$ and $\mathrm{set}_h$ is + the set of function $\left\{h_1(y), h_2(y), \dots, h_n(y)\right\}$. + Here, $y$ is not necessarily a symbol. + + $\mathrm{set}_h$ contains the functions, along with the information + about the domain in which they are valid, through set + operations. For instance, if :math:`y = |x| - n` is inverted + in the real domain, then $\mathrm{set}_h$ is not simply + $\{-n, n\}$ as the nature of `n` is unknown; rather, it is: + + $$ \left(\left[0, \infty\right) \cap \left\{n\right\}\right) \cup + \left(\left(-\infty, 0\right] \cap \left\{- n\right\}\right)$$ + + By default, the complex domain is used which means that inverting even + seemingly simple functions like $\exp(x)$ will give very different + results from those obtained in the real domain. + (In the case of $\exp(x)$, the inversion via $\log$ is multi-valued + in the complex domain, having infinitely many branches.) + + If you are working with real values only (or you are not sure which + function to use) you should probably set the domain to + ``S.Reals`` (or use ``invert_real`` which does that automatically). + + + Examples + ======== + + >>> from sympy.solvers.solveset import invert_complex, invert_real + >>> from sympy.abc import x, y + >>> from sympy import exp + + When does exp(x) == y? + + >>> invert_complex(exp(x), y, x) + (x, ImageSet(Lambda(_n, I*(2*_n*pi + arg(y)) + log(Abs(y))), Integers)) + >>> invert_real(exp(x), y, x) + (x, Intersection({log(y)}, Reals)) + + When does exp(x) == 1? + + >>> invert_complex(exp(x), 1, x) + (x, ImageSet(Lambda(_n, 2*_n*I*pi), Integers)) + >>> invert_real(exp(x), 1, x) + (x, {0}) + + See Also + ======== + invert_real, invert_complex + """ + x = sympify(x) + if not x.is_Symbol: + raise ValueError("x must be a symbol") + f_x = sympify(f_x) + if x not in f_x.free_symbols: + raise ValueError("Inverse of constant function doesn't exist") + y = sympify(y) + if x in y.free_symbols: + raise ValueError("y should be independent of x ") + + if domain.is_subset(S.Reals): + x1, s = _invert_real(f_x, FiniteSet(y), x) + else: + x1, s = _invert_complex(f_x, FiniteSet(y), x) + + # f couldn't be inverted completely; return unmodified. + if x1 != x: + return x1, s + + # Avoid adding gratuitous intersections with S.Complexes. Actual + # conditions should be handled by the respective inverters. + if domain is S.Complexes: + return x1, s + + if isinstance(s, FiniteSet): + return x1, s.intersect(domain) + + # "Fancier" solution sets like those obtained by inversion of trigonometric + # functions already include general validity conditions (i.e. conditions on + # the domain of the respective inverse functions), so we should avoid adding + # blanket intesections with S.Reals. But subsets of R (or C) must still be + # accounted for. + if domain is S.Reals: + return x1, s + else: + return x1, s.intersect(domain) + + +invert_complex = _invert + + +def invert_real(f_x, y, x): + """ + Inverts a real-valued function. Same as :func:`invert_complex`, but sets + the domain to ``S.Reals`` before inverting. + """ + return _invert(f_x, y, x, S.Reals) + + +def _invert_real(f, g_ys, symbol): + """Helper function for _invert.""" + + if f == symbol or g_ys is S.EmptySet: + return (symbol, g_ys) + + n = Dummy('n', real=True) + + if isinstance(f, exp) or (f.is_Pow and f.base == S.Exp1): + return _invert_real(f.exp, + imageset(Lambda(n, log(n)), g_ys), + symbol) + + if hasattr(f, 'inverse') and f.inverse() is not None and not isinstance(f, ( + TrigonometricFunction, + HyperbolicFunction, + )): + if len(f.args) > 1: + raise ValueError("Only functions with one argument are supported.") + return _invert_real(f.args[0], + imageset(Lambda(n, f.inverse()(n)), g_ys), + symbol) + + if isinstance(f, Abs): + return _invert_abs(f.args[0], g_ys, symbol) + + if f.is_Add: + # f = g + h + g, h = f.as_independent(symbol) + if g is not S.Zero: + return _invert_real(h, imageset(Lambda(n, n - g), g_ys), symbol) + + if f.is_Mul: + # f = g*h + g, h = f.as_independent(symbol) + + if g is not S.One: + return _invert_real(h, imageset(Lambda(n, n/g), g_ys), symbol) + + if f.is_Pow: + base, expo = f.args + base_has_sym = base.has(symbol) + expo_has_sym = expo.has(symbol) + + if not expo_has_sym: + + if expo.is_rational: + num, den = expo.as_numer_denom() + + if den % 2 == 0 and num % 2 == 1 and den.is_zero is False: + # Here we have f(x)**(num/den) = y + # where den is nonzero and even and y is an element + # of the set g_ys. + # den is even, so we are only interested in the cases + # where both f(x) and y are positive. + # Restricting y to be positive (using the set g_ys_pos) + # means that y**(den/num) is always positive. + # Therefore it isn't necessary to also constrain f(x) + # to be positive because we are only going to + # find solutions of f(x) = y**(d/n) + # where the rhs is already required to be positive. + root = Lambda(n, real_root(n, expo)) + g_ys_pos = g_ys & Interval(0, oo) + res = imageset(root, g_ys_pos) + _inv, _set = _invert_real(base, res, symbol) + return (_inv, _set) + + if den % 2 == 1: + root = Lambda(n, real_root(n, expo)) + res = imageset(root, g_ys) + if num % 2 == 0: + neg_res = imageset(Lambda(n, -n), res) + return _invert_real(base, res + neg_res, symbol) + if num % 2 == 1: + return _invert_real(base, res, symbol) + + elif expo.is_irrational: + root = Lambda(n, real_root(n, expo)) + g_ys_pos = g_ys & Interval(0, oo) + res = imageset(root, g_ys_pos) + return _invert_real(base, res, symbol) + + else: + # indeterminate exponent, e.g. Float or parity of + # num, den of rational could not be determined + pass # use default return + + if not base_has_sym: + rhs = g_ys.args[0] + if base.is_positive: + return _invert_real(expo, + imageset(Lambda(n, log(n, base, evaluate=False)), g_ys), symbol) + elif base.is_negative: + s, b = integer_log(rhs, base) + if b: + return _invert_real(expo, FiniteSet(s), symbol) + else: + return (expo, S.EmptySet) + elif base.is_zero: + one = Eq(rhs, 1) + if one == S.true: + # special case: 0**x - 1 + return _invert_real(expo, FiniteSet(0), symbol) + elif one == S.false: + return (expo, S.EmptySet) + + if isinstance(f, (TrigonometricFunction, HyperbolicFunction)): + return _invert_trig_hyp_real(f, g_ys, symbol) + + return (f, g_ys) + + +# Dictionaries of inverses will be cached after first use. +_trig_inverses = None +_hyp_inverses = None + +def _invert_trig_hyp_real(f, g_ys, symbol): + """Helper function for inverting trigonometric and hyperbolic functions. + + This helper only handles inversion over the reals. + + For trigonometric functions only finite `g_ys` sets are implemented. + + For hyperbolic functions the set `g_ys` is checked against the domain of the + respective inverse functions. Infinite `g_ys` sets are also supported. + """ + + if isinstance(f, HyperbolicFunction): + n = Dummy('n', real=True) + + if isinstance(f, sinh): + # asinh is defined over R. + return _invert_real(f.args[0], imageset(n, asinh(n), g_ys), symbol) + + if isinstance(f, cosh): + g_ys_dom = g_ys.intersect(Interval(1, oo)) + if isinstance(g_ys_dom, Intersection): + # could not properly resolve domain check + if isinstance(g_ys, FiniteSet): + # If g_ys is a `FiniteSet`` it should be sufficient to just + # let the calling `_invert_real()` add an intersection with + # `S.Reals` (or a subset `domain`) to ensure that only valid + # (real) solutions are returned. + # This avoids adding "too many" Intersections or + # ConditionSets in the returned set. + g_ys_dom = g_ys + else: + return (f, g_ys) + return _invert_real(f.args[0], Union( + imageset(n, acosh(n), g_ys_dom), + imageset(n, -acosh(n), g_ys_dom)), symbol) + + if isinstance(f, sech): + g_ys_dom = g_ys.intersect(Interval.Lopen(0, 1)) + if isinstance(g_ys_dom, Intersection): + if isinstance(g_ys, FiniteSet): + g_ys_dom = g_ys + else: + return (f, g_ys) + return _invert_real(f.args[0], Union( + imageset(n, asech(n), g_ys_dom), + imageset(n, -asech(n), g_ys_dom)), symbol) + + if isinstance(f, tanh): + g_ys_dom = g_ys.intersect(Interval.open(-1, 1)) + if isinstance(g_ys_dom, Intersection): + if isinstance(g_ys, FiniteSet): + g_ys_dom = g_ys + else: + return (f, g_ys) + return _invert_real(f.args[0], + imageset(n, atanh(n), g_ys_dom), symbol) + + if isinstance(f, coth): + g_ys_dom = g_ys - Interval(-1, 1) + if isinstance(g_ys_dom, Complement): + if isinstance(g_ys, FiniteSet): + g_ys_dom = g_ys + else: + return (f, g_ys) + return _invert_real(f.args[0], + imageset(n, acoth(n), g_ys_dom), symbol) + + if isinstance(f, csch): + g_ys_dom = g_ys - FiniteSet(0) + if isinstance(g_ys_dom, Complement): + if isinstance(g_ys, FiniteSet): + g_ys_dom = g_ys + else: + return (f, g_ys) + return _invert_real(f.args[0], + imageset(n, acsch(n), g_ys_dom), symbol) + + elif isinstance(f, TrigonometricFunction) and isinstance(g_ys, FiniteSet): + def _get_trig_inverses(func): + global _trig_inverses + if _trig_inverses is None: + _trig_inverses = { + sin : ((asin, lambda y: pi-asin(y)), 2*pi, Interval(-1, 1)), + cos : ((acos, lambda y: -acos(y)), 2*pi, Interval(-1, 1)), + tan : ((atan,), pi, S.Reals), + cot : ((acot,), pi, S.Reals), + sec : ((asec, lambda y: -asec(y)), 2*pi, + Union(Interval(-oo, -1), Interval(1, oo))), + csc : ((acsc, lambda y: pi-acsc(y)), 2*pi, + Union(Interval(-oo, -1), Interval(1, oo)))} + return _trig_inverses[func] + + invs, period, rng = _get_trig_inverses(f.func) + n = Dummy('n', integer=True) + def create_return_set(g): + # returns ConditionSet that will be part of the final (x, set) tuple + invsimg = Union(*[ + imageset(n, period*n + inv(g), S.Integers) for inv in invs]) + inv_f, inv_g_ys = _invert_real(f.args[0], invsimg, symbol) + if inv_f == symbol: # inversion successful + conds = rng.contains(g) + return ConditionSet(symbol, conds, inv_g_ys) + else: + return ConditionSet(symbol, Eq(f, g), S.Reals) + + retset = Union(*[create_return_set(g) for g in g_ys]) + return (symbol, retset) + + else: + return (f, g_ys) + + +def _invert_trig_hyp_complex(f, g_ys, symbol): + """Helper function for inverting trigonometric and hyperbolic functions. + + This helper only handles inversion over the complex numbers. + Only finite `g_ys` sets are implemented. + + Handling of singularities is only implemented for hyperbolic equations. + In case of a symbolic element g in g_ys a ConditionSet may be returned. + """ + + if isinstance(f, TrigonometricFunction) and isinstance(g_ys, FiniteSet): + def inv(trig): + if isinstance(trig, (sin, csc)): + F = asin if isinstance(trig, sin) else acsc + return ( + lambda a: 2*n*pi + F(a), + lambda a: 2*n*pi + pi - F(a)) + if isinstance(trig, (cos, sec)): + F = acos if isinstance(trig, cos) else asec + return ( + lambda a: 2*n*pi + F(a), + lambda a: 2*n*pi - F(a)) + if isinstance(trig, (tan, cot)): + return (lambda a: n*pi + trig.inverse()(a),) + + n = Dummy('n', integer=True) + invs = S.EmptySet + for L in inv(f): + invs += Union(*[imageset(Lambda(n, L(g)), S.Integers) for g in g_ys]) + return _invert_complex(f.args[0], invs, symbol) + + elif isinstance(f, HyperbolicFunction) and isinstance(g_ys, FiniteSet): + # There are two main options regarding singularities / domain checking + # for symbolic elements in g_ys: + # 1. Add a "catch-all" intersection with S.Complexes. + # 2. ConditionSets. + # At present ConditionSets seem to work better and have the additional + # benefit of representing the precise conditions that must be satisfied. + # The conditions are also rather straightforward. (At most two isolated + # points.) + def _get_hyp_inverses(func): + global _hyp_inverses + if _hyp_inverses is None: + _hyp_inverses = { + sinh : ((asinh, lambda y: I*pi-asinh(y)), 2*I*pi, ()), + cosh : ((acosh, lambda y: -acosh(y)), 2*I*pi, ()), + tanh : ((atanh,), I*pi, (-1, 1)), + coth : ((acoth,), I*pi, (-1, 1)), + sech : ((asech, lambda y: -asech(y)), 2*I*pi, (0, )), + csch : ((acsch, lambda y: I*pi-acsch(y)), 2*I*pi, (0, ))} + return _hyp_inverses[func] + + # invs: iterable of main inverses, e.g. (acosh, -acosh). + # excl: iterable of singularities to be checked for. + invs, period, excl = _get_hyp_inverses(f.func) + n = Dummy('n', integer=True) + def create_return_set(g): + # returns ConditionSet that will be part of the final (x, set) tuple + invsimg = Union(*[ + imageset(n, period*n + inv(g), S.Integers) for inv in invs]) + inv_f, inv_g_ys = _invert_complex(f.args[0], invsimg, symbol) + if inv_f == symbol: # inversion successful + conds = And(*[Ne(g, e) for e in excl]) + return ConditionSet(symbol, conds, inv_g_ys) + else: + return ConditionSet(symbol, Eq(f, g), S.Complexes) + + retset = Union(*[create_return_set(g) for g in g_ys]) + return (symbol, retset) + + else: + return (f, g_ys) + + +def _invert_complex(f, g_ys, symbol): + """Helper function for _invert.""" + + if f == symbol or g_ys is S.EmptySet: + return (symbol, g_ys) + + n = Dummy('n') + + if f.is_Add: + # f = g + h + g, h = f.as_independent(symbol) + if g is not S.Zero: + return _invert_complex(h, imageset(Lambda(n, n - g), g_ys), symbol) + + if f.is_Mul: + # f = g*h + g, h = f.as_independent(symbol) + + if g is not S.One: + if g in {S.NegativeInfinity, S.ComplexInfinity, S.Infinity}: + return (h, S.EmptySet) + return _invert_complex(h, imageset(Lambda(n, n/g), g_ys), symbol) + + if f.is_Pow: + base, expo = f.args + # special case: g**r = 0 + # Could be improved like `_invert_real` to handle more general cases. + if expo.is_Rational and g_ys == FiniteSet(0): + if expo.is_positive: + return _invert_complex(base, g_ys, symbol) + + if hasattr(f, 'inverse') and f.inverse() is not None and \ + not isinstance(f, TrigonometricFunction) and \ + not isinstance(f, HyperbolicFunction) and \ + not isinstance(f, exp): + if len(f.args) > 1: + raise ValueError("Only functions with one argument are supported.") + return _invert_complex(f.args[0], + imageset(Lambda(n, f.inverse()(n)), g_ys), symbol) + + if isinstance(f, exp) or (f.is_Pow and f.base == S.Exp1): + if isinstance(g_ys, ImageSet): + # can solve up to `(d*exp(exp(...(exp(a*x + b))...) + c)` format. + # Further can be improved to `(d*exp(exp(...(exp(a*x**n + b*x**(n-1) + ... + f))...) + c)`. + g_ys_expr = g_ys.lamda.expr + g_ys_vars = g_ys.lamda.variables + k = Dummy('k{}'.format(len(g_ys_vars))) + g_ys_vars_1 = (k,) + g_ys_vars + exp_invs = Union(*[imageset(Lambda((g_ys_vars_1,), (I*(2*k*pi + arg(g_ys_expr)) + + log(Abs(g_ys_expr)))), S.Integers**(len(g_ys_vars_1)))]) + return _invert_complex(f.exp, exp_invs, symbol) + + elif isinstance(g_ys, FiniteSet): + exp_invs = Union(*[imageset(Lambda(n, I*(2*n*pi + arg(g_y)) + + log(Abs(g_y))), S.Integers) + for g_y in g_ys if g_y != 0]) + return _invert_complex(f.exp, exp_invs, symbol) + + if isinstance(f, (TrigonometricFunction, HyperbolicFunction)): + return _invert_trig_hyp_complex(f, g_ys, symbol) + + return (f, g_ys) + + +def _invert_abs(f, g_ys, symbol): + """Helper function for inverting absolute value functions. + + Returns the complete result of inverting an absolute value + function along with the conditions which must also be satisfied. + + If it is certain that all these conditions are met, a :class:`~.FiniteSet` + of all possible solutions is returned. If any condition cannot be + satisfied, an :class:`~.EmptySet` is returned. Otherwise, a + :class:`~.ConditionSet` of the solutions, with all the required conditions + specified, is returned. + + """ + if not g_ys.is_FiniteSet: + # this could be used for FiniteSet, but the + # results are more compact if they aren't, e.g. + # ConditionSet(x, Contains(n, Interval(0, oo)), {-n, n}) vs + # Union(Intersection(Interval(0, oo), {n}), Intersection(Interval(-oo, 0), {-n})) + # for the solution of abs(x) - n + pos = Intersection(g_ys, Interval(0, S.Infinity)) + parg = _invert_real(f, pos, symbol) + narg = _invert_real(-f, pos, symbol) + if parg[0] != narg[0]: + raise NotImplementedError + return parg[0], Union(narg[1], parg[1]) + + # check conditions: all these must be true. If any are unknown + # then return them as conditions which must be satisfied + unknown = [] + for a in g_ys.args: + ok = a.is_nonnegative if a.is_Number else a.is_positive + if ok is None: + unknown.append(a) + elif not ok: + return symbol, S.EmptySet + if unknown: + conditions = And(*[Contains(i, Interval(0, oo)) + for i in unknown]) + else: + conditions = True + n = Dummy('n', real=True) + # this is slightly different than above: instead of solving + # +/-f on positive values, here we solve for f on +/- g_ys + g_x, values = _invert_real(f, Union( + imageset(Lambda(n, n), g_ys), + imageset(Lambda(n, -n), g_ys)), symbol) + return g_x, ConditionSet(g_x, conditions, values) + + +def domain_check(f, symbol, p): + """Returns False if point p is infinite or any subexpression of f + is infinite or becomes so after replacing symbol with p. If none of + these conditions is met then True will be returned. + + Examples + ======== + + >>> from sympy import Mul, oo + >>> from sympy.abc import x + >>> from sympy.solvers.solveset import domain_check + >>> g = 1/(1 + (1/(x + 1))**2) + >>> domain_check(g, x, -1) + False + >>> domain_check(x**2, x, 0) + True + >>> domain_check(1/x, x, oo) + False + + * The function relies on the assumption that the original form + of the equation has not been changed by automatic simplification. + + >>> domain_check(x/x, x, 0) # x/x is automatically simplified to 1 + True + + * To deal with automatic evaluations use evaluate=False: + + >>> domain_check(Mul(x, 1/x, evaluate=False), x, 0) + False + """ + f, p = sympify(f), sympify(p) + if p.is_infinite: + return False + return _domain_check(f, symbol, p) + + +def _domain_check(f, symbol, p): + # helper for domain check + if f.is_Atom and f.is_finite: + return True + elif f.subs(symbol, p).is_infinite: + return False + elif isinstance(f, Piecewise): + # Check the cases of the Piecewise in turn. There might be invalid + # expressions in later cases that don't apply e.g. + # solveset(Piecewise((0, Eq(x, 0)), (1/x, True)), x) + for expr, cond in f.args: + condsubs = cond.subs(symbol, p) + if condsubs is S.false: + continue + elif condsubs is S.true: + return _domain_check(expr, symbol, p) + else: + # We don't know which case of the Piecewise holds. On this + # basis we cannot decide whether any solution is in or out of + # the domain. Ideally this function would allow returning a + # symbolic condition for the validity of the solution that + # could be handled in the calling code. In the mean time we'll + # give this particular solution the benefit of the doubt and + # let it pass. + return True + else: + # TODO : We should not blindly recurse through all args of arbitrary expressions like this + return all(_domain_check(g, symbol, p) + for g in f.args) + + +def _is_finite_with_finite_vars(f, domain=S.Complexes): + """ + Return True if the given expression is finite. For symbols that + do not assign a value for `complex` and/or `real`, the domain will + be used to assign a value; symbols that do not assign a value + for `finite` will be made finite. All other assumptions are + left unmodified. + """ + def assumptions(s): + A = s.assumptions0 + A.setdefault('finite', A.get('finite', True)) + if domain.is_subset(S.Reals): + # if this gets set it will make complex=True, too + A.setdefault('real', True) + else: + # don't change 'real' because being complex implies + # nothing about being real + A.setdefault('complex', True) + return A + + reps = {s: Dummy(**assumptions(s)) for s in f.free_symbols} + return f.xreplace(reps).is_finite + + +def _is_function_class_equation(func_class, f, symbol): + """ Tests whether the equation is an equation of the given function class. + + The given equation belongs to the given function class if it is + comprised of functions of the function class which are multiplied by + or added to expressions independent of the symbol. In addition, the + arguments of all such functions must be linear in the symbol as well. + + Examples + ======== + + >>> from sympy.solvers.solveset import _is_function_class_equation + >>> from sympy import tan, sin, tanh, sinh, exp + >>> from sympy.abc import x + >>> from sympy.functions.elementary.trigonometric import TrigonometricFunction + >>> from sympy.functions.elementary.hyperbolic import HyperbolicFunction + >>> _is_function_class_equation(TrigonometricFunction, exp(x) + tan(x), x) + False + >>> _is_function_class_equation(TrigonometricFunction, tan(x) + sin(x), x) + True + >>> _is_function_class_equation(TrigonometricFunction, tan(x**2), x) + False + >>> _is_function_class_equation(TrigonometricFunction, tan(x + 2), x) + True + >>> _is_function_class_equation(HyperbolicFunction, tanh(x) + sinh(x), x) + True + """ + if f.is_Mul or f.is_Add: + return all(_is_function_class_equation(func_class, arg, symbol) + for arg in f.args) + + if f.is_Pow: + if not f.exp.has(symbol): + return _is_function_class_equation(func_class, f.base, symbol) + else: + return False + + if not f.has(symbol): + return True + + if isinstance(f, func_class): + try: + g = Poly(f.args[0], symbol) + return g.degree() <= 1 + except PolynomialError: + return False + else: + return False + + +def _solve_as_rational(f, symbol, domain): + """ solve rational functions""" + f = together(_mexpand(f, recursive=True), deep=True) + g, h = fraction(f) + if not h.has(symbol): + try: + return _solve_as_poly(g, symbol, domain) + except NotImplementedError: + # The polynomial formed from g could end up having + # coefficients in a ring over which finding roots + # isn't implemented yet, e.g. ZZ[a] for some symbol a + return ConditionSet(symbol, Eq(f, 0), domain) + except CoercionFailed: + # contained oo, zoo or nan + return S.EmptySet + else: + valid_solns = _solveset(g, symbol, domain) + invalid_solns = _solveset(h, symbol, domain) + return valid_solns - invalid_solns + + +class _SolveTrig1Error(Exception): + """Raised when _solve_trig1 heuristics do not apply""" + +def _solve_trig(f, symbol, domain): + """Function to call other helpers to solve trigonometric equations """ + # If f is composed of a single trig function (potentially appearing multiple + # times) we should solve by either inverting directly or inverting after a + # suitable change of variable. + # + # _solve_trig is currently only called by _solveset for trig/hyperbolic + # functions of an argument linear in x. Inverting a symbolic argument should + # include a guard against division by zero in order to have a result that is + # consistent with similar processing done by _solve_trig1. + # (Ideally _invert should add these conditions by itself.) + trig_expr, count = None, 0 + for expr in preorder_traversal(f): + if isinstance(expr, (TrigonometricFunction, + HyperbolicFunction)) and expr.has(symbol): + if not trig_expr: + trig_expr, count = expr, 1 + elif expr == trig_expr: + count += 1 + else: + trig_expr, count = False, 0 + break + if count == 1: + # direct inversion + x, sol = _invert(f, 0, symbol, domain) + if x == symbol: + cond = True + if trig_expr.free_symbols - {symbol}: + a, h = trig_expr.args[0].as_independent(symbol, as_Add=True) + m, h = h.as_independent(symbol, as_Add=False) + num, den = m.as_numer_denom() + cond = Ne(num, 0) & Ne(den, 0) + return ConditionSet(symbol, cond, sol) + else: + return ConditionSet(symbol, Eq(f, 0), domain) + elif count: + # solve by change of variable + y = Dummy('y') + f_cov = f.subs(trig_expr, y) + sol_cov = solveset(f_cov, y, domain) + if isinstance(sol_cov, FiniteSet): + return Union( + *[_solve_trig(trig_expr-s, symbol, domain) for s in sol_cov]) + + sol = None + try: + # multiple trig/hyp functions; solve by rewriting to exp + sol = _solve_trig1(f, symbol, domain) + except _SolveTrig1Error: + try: + # multiple trig/hyp functions; solve by rewriting to tan(x/2) + sol = _solve_trig2(f, symbol, domain) + except ValueError: + raise NotImplementedError(filldedent(''' + Solution to this kind of trigonometric equations + is yet to be implemented''')) + return sol + + +def _solve_trig1(f, symbol, domain): + """Primary solver for trigonometric and hyperbolic equations + + Returns either the solution set as a ConditionSet (auto-evaluated to a + union of ImageSets if no variables besides 'symbol' are involved) or + raises _SolveTrig1Error if f == 0 cannot be solved. + + Notes + ===== + Algorithm: + 1. Do a change of variable x -> mu*x in arguments to trigonometric and + hyperbolic functions, in order to reduce them to small integers. (This + step is crucial to keep the degrees of the polynomials of step 4 low.) + 2. Rewrite trigonometric/hyperbolic functions as exponentials. + 3. Proceed to a 2nd change of variable, replacing exp(I*x) or exp(x) by y. + 4. Solve the resulting rational equation. + 5. Use invert_complex or invert_real to return to the original variable. + 6. If the coefficients of 'symbol' were symbolic in nature, add the + necessary consistency conditions in a ConditionSet. + + """ + # Prepare change of variable + x = Dummy('x') + if _is_function_class_equation(HyperbolicFunction, f, symbol): + cov = exp(x) + inverter = invert_real if domain.is_subset(S.Reals) else invert_complex + else: + cov = exp(I*x) + inverter = invert_complex + + f = trigsimp(f) + f_original = f + trig_functions = f.atoms(TrigonometricFunction, HyperbolicFunction) + trig_arguments = [e.args[0] for e in trig_functions] + # trigsimp may have reduced the equation to an expression + # that is independent of 'symbol' (e.g. cos**2+sin**2) + if not any(a.has(symbol) for a in trig_arguments): + return solveset(f_original, symbol, domain) + + denominators = [] + numerators = [] + for ar in trig_arguments: + try: + poly_ar = Poly(ar, symbol) + except PolynomialError: + raise _SolveTrig1Error("trig argument is not a polynomial") + if poly_ar.degree() > 1: # degree >1 still bad + raise _SolveTrig1Error("degree of variable must not exceed one") + if poly_ar.degree() == 0: # degree 0, don't care + continue + c = poly_ar.all_coeffs()[0] # got the coefficient of 'symbol' + numerators.append(fraction(c)[0]) + denominators.append(fraction(c)[1]) + + mu = lcm(denominators)/gcd(numerators) + f = f.subs(symbol, mu*x) + f = f.rewrite(exp) + f = together(f) + g, h = fraction(f) + y = Dummy('y') + g, h = g.expand(), h.expand() + g, h = g.subs(cov, y), h.subs(cov, y) + if g.has(x) or h.has(x): + raise _SolveTrig1Error("change of variable not possible") + + solns = solveset_complex(g, y) - solveset_complex(h, y) + if isinstance(solns, ConditionSet): + raise _SolveTrig1Error("polynomial has ConditionSet solution") + + if isinstance(solns, FiniteSet): + if any(isinstance(s, RootOf) for s in solns): + raise _SolveTrig1Error("polynomial results in RootOf object") + # revert the change of variable + cov = cov.subs(x, symbol/mu) + result = Union(*[inverter(cov, s, symbol)[1] for s in solns]) + # In case of symbolic coefficients, the solution set is only valid + # if numerator and denominator of mu are non-zero. + if mu.has(Symbol): + syms = (mu).atoms(Symbol) + munum, muden = fraction(mu) + condnum = munum.as_independent(*syms, as_Add=False)[1] + condden = muden.as_independent(*syms, as_Add=False)[1] + cond = And(Ne(condnum, 0), Ne(condden, 0)) + else: + cond = True + # Actual conditions are returned as part of the ConditionSet. Adding an + # intersection with C would only complicate some solution sets due to + # current limitations of intersection code. (e.g. #19154) + if domain is S.Complexes: + # This is a slight abuse of ConditionSet. Ideally this should + # be some kind of "PiecewiseSet". (See #19507 discussion) + return ConditionSet(symbol, cond, result) + else: + return ConditionSet(symbol, cond, Intersection(result, domain)) + elif solns is S.EmptySet: + return S.EmptySet + else: + raise _SolveTrig1Error("polynomial solutions must form FiniteSet") + + +def _solve_trig2(f, symbol, domain): + """Secondary helper to solve trigonometric equations, + called when first helper fails """ + f = trigsimp(f) + f_original = f + trig_functions = f.atoms(sin, cos, tan, sec, cot, csc) + trig_arguments = [e.args[0] for e in trig_functions] + denominators = [] + numerators = [] + + # todo: This solver can be extended to hyperbolics if the + # analogous change of variable to tanh (instead of tan) + # is used. + if not trig_functions: + return ConditionSet(symbol, Eq(f_original, 0), domain) + + # todo: The pre-processing below (extraction of numerators, denominators, + # gcd, lcm, mu, etc.) should be updated to the enhanced version in + # _solve_trig1. (See #19507) + for ar in trig_arguments: + try: + poly_ar = Poly(ar, symbol) + except PolynomialError: + raise ValueError("give up, we cannot solve if this is not a polynomial in x") + if poly_ar.degree() > 1: # degree >1 still bad + raise ValueError("degree of variable inside polynomial should not exceed one") + if poly_ar.degree() == 0: # degree 0, don't care + continue + c = poly_ar.all_coeffs()[0] # got the coefficient of 'symbol' + try: + numerators.append(Rational(c).p) + denominators.append(Rational(c).q) + except TypeError: + return ConditionSet(symbol, Eq(f_original, 0), domain) + + x = Dummy('x') + + mu = Rational(2)*number_lcm(*denominators)/number_gcd(*numerators) + f = f.subs(symbol, mu*x) + f = f.rewrite(tan) + f = expand_trig(f) + f = together(f) + + g, h = fraction(f) + y = Dummy('y') + g, h = g.expand(), h.expand() + g, h = g.subs(tan(x), y), h.subs(tan(x), y) + + if g.has(x) or h.has(x): + return ConditionSet(symbol, Eq(f_original, 0), domain) + solns = solveset(g, y, S.Reals) - solveset(h, y, S.Reals) + + if isinstance(solns, FiniteSet): + result = Union(*[invert_real(tan(symbol/mu), s, symbol)[1] + for s in solns]) + dsol = invert_real(tan(symbol/mu), oo, symbol)[1] + if degree(h) > degree(g): # If degree(denom)>degree(num) then there + result = Union(result, dsol) # would be another sol at Lim(denom-->oo) + return Intersection(result, domain) + elif solns is S.EmptySet: + return S.EmptySet + else: + return ConditionSet(symbol, Eq(f_original, 0), S.Reals) + + +def _solve_as_poly(f, symbol, domain=S.Complexes): + """ + Solve the equation using polynomial techniques if it already is a + polynomial equation or, with a change of variables, can be made so. + """ + result = None + if f.is_polynomial(symbol): + solns = roots(f, symbol, cubics=True, quartics=True, + quintics=True, domain='EX') + num_roots = sum(solns.values()) + if degree(f, symbol) <= num_roots: + result = FiniteSet(*solns.keys()) + else: + poly = Poly(f, symbol) + solns = poly.all_roots() + if poly.degree() <= len(solns): + result = FiniteSet(*solns) + else: + result = ConditionSet(symbol, Eq(f, 0), domain) + else: + poly = Poly(f) + if poly is None: + result = ConditionSet(symbol, Eq(f, 0), domain) + gens = [g for g in poly.gens if g.has(symbol)] + + if len(gens) == 1: + poly = Poly(poly, gens[0]) + gen = poly.gen + deg = poly.degree() + poly = Poly(poly.as_expr(), poly.gen, composite=True) + poly_solns = FiniteSet(*roots(poly, cubics=True, quartics=True, + quintics=True).keys()) + + if len(poly_solns) < deg: + result = ConditionSet(symbol, Eq(f, 0), domain) + + if gen != symbol: + y = Dummy('y') + inverter = invert_real if domain.is_subset(S.Reals) else invert_complex + lhs, rhs_s = inverter(gen, y, symbol) + if lhs == symbol: + result = Union(*[rhs_s.subs(y, s) for s in poly_solns]) + if isinstance(result, FiniteSet) and isinstance(gen, Pow + ) and gen.base.is_Rational: + result = FiniteSet(*[expand_log(i) for i in result]) + else: + result = ConditionSet(symbol, Eq(f, 0), domain) + else: + result = ConditionSet(symbol, Eq(f, 0), domain) + + if result is not None: + if isinstance(result, FiniteSet): + # this is to simplify solutions like -sqrt(-I) to sqrt(2)/2 + # - sqrt(2)*I/2. We are not expanding for solution with symbols + # or undefined functions because that makes the solution more complicated. + # For example, expand_complex(a) returns re(a) + I*im(a) + if all(s.atoms(Symbol, AppliedUndef) == set() and not isinstance(s, RootOf) + for s in result): + s = Dummy('s') + result = imageset(Lambda(s, expand_complex(s)), result) + if isinstance(result, FiniteSet) and domain != S.Complexes: + # Avoid adding gratuitous intersections with S.Complexes. Actual + # conditions should be handled elsewhere. + result = result.intersection(domain) + return result + else: + return ConditionSet(symbol, Eq(f, 0), domain) + + +def _solve_radical(f, unradf, symbol, solveset_solver): + """ Helper function to solve equations with radicals """ + res = unradf + eq, cov = res if res else (f, []) + if not cov: + result = solveset_solver(eq, symbol) - \ + Union(*[solveset_solver(g, symbol) for g in denoms(f, symbol)]) + else: + y, yeq = cov + if not solveset_solver(y - I, y): + yreal = Dummy('yreal', real=True) + yeq = yeq.xreplace({y: yreal}) + eq = eq.xreplace({y: yreal}) + y = yreal + g_y_s = solveset_solver(yeq, symbol) + f_y_sols = solveset_solver(eq, y) + result = Union(*[imageset(Lambda(y, g_y), f_y_sols) + for g_y in g_y_s]) + + def check_finiteset(solutions): + f_set = [] # solutions for FiniteSet + c_set = [] # solutions for ConditionSet + for s in solutions: + if checksol(f, symbol, s): + f_set.append(s) + else: + c_set.append(s) + return FiniteSet(*f_set) + ConditionSet(symbol, Eq(f, 0), FiniteSet(*c_set)) + + def check_set(solutions): + if solutions is S.EmptySet: + return solutions + elif isinstance(solutions, ConditionSet): + # XXX: Maybe the base set should be checked? + return solutions + elif isinstance(solutions, FiniteSet): + return check_finiteset(solutions) + elif isinstance(solutions, Complement): + A, B = solutions.args + return Complement(check_set(A), B) + elif isinstance(solutions, Union): + return Union(*[check_set(s) for s in solutions.args]) + else: + # XXX: There should be more cases checked here. The cases above + # are all those that come up in the test suite for now. + return solutions + + solution_set = check_set(result) + + return solution_set + + +def _solve_abs(f, symbol, domain): + """ Helper function to solve equation involving absolute value function """ + if not domain.is_subset(S.Reals): + raise ValueError(filldedent(''' + Absolute values cannot be inverted in the + complex domain.''')) + p, q, r = Wild('p'), Wild('q'), Wild('r') + pattern_match = f.match(p*Abs(q) + r) or {} + f_p, f_q, f_r = [pattern_match.get(i, S.Zero) for i in (p, q, r)] + + if not (f_p.is_zero or f_q.is_zero): + domain = continuous_domain(f_q, symbol, domain) + from .inequalities import solve_univariate_inequality + q_pos_cond = solve_univariate_inequality(f_q >= 0, symbol, + relational=False, domain=domain, continuous=True) + q_neg_cond = q_pos_cond.complement(domain) + + sols_q_pos = solveset_real(f_p*f_q + f_r, + symbol).intersect(q_pos_cond) + sols_q_neg = solveset_real(f_p*(-f_q) + f_r, + symbol).intersect(q_neg_cond) + return Union(sols_q_pos, sols_q_neg) + else: + return ConditionSet(symbol, Eq(f, 0), domain) + + +def solve_decomposition(f, symbol, domain): + """ + Function to solve equations via the principle of "Decomposition + and Rewriting". + + Examples + ======== + >>> from sympy import exp, sin, Symbol, pprint, S + >>> from sympy.solvers.solveset import solve_decomposition as sd + >>> x = Symbol('x') + >>> f1 = exp(2*x) - 3*exp(x) + 2 + >>> sd(f1, x, S.Reals) + {0, log(2)} + >>> f2 = sin(x)**2 + 2*sin(x) + 1 + >>> pprint(sd(f2, x, S.Reals), use_unicode=False) + 3*pi + {2*n*pi + ---- | n in Integers} + 2 + >>> f3 = sin(x + 2) + >>> pprint(sd(f3, x, S.Reals), use_unicode=False) + {2*n*pi - 2 | n in Integers} U {2*n*pi - 2 + pi | n in Integers} + + """ + from sympy.solvers.decompogen import decompogen + # decompose the given function + g_s = decompogen(f, symbol) + # `y_s` represents the set of values for which the function `g` is to be + # solved. + # `solutions` represent the solutions of the equations `g = y_s` or + # `g = 0` depending on the type of `y_s`. + # As we are interested in solving the equation: f = 0 + y_s = FiniteSet(0) + for g in g_s: + frange = function_range(g, symbol, domain) + y_s = Intersection(frange, y_s) + result = S.EmptySet + if isinstance(y_s, FiniteSet): + for y in y_s: + solutions = solveset(Eq(g, y), symbol, domain) + if not isinstance(solutions, ConditionSet): + result += solutions + + else: + if isinstance(y_s, ImageSet): + iter_iset = (y_s,) + + elif isinstance(y_s, Union): + iter_iset = y_s.args + + elif y_s is S.EmptySet: + # y_s is not in the range of g in g_s, so no solution exists + #in the given domain + return S.EmptySet + + for iset in iter_iset: + new_solutions = solveset(Eq(iset.lamda.expr, g), symbol, domain) + dummy_var = tuple(iset.lamda.expr.free_symbols)[0] + (base_set,) = iset.base_sets + if isinstance(new_solutions, FiniteSet): + new_exprs = new_solutions + + elif isinstance(new_solutions, Intersection): + if isinstance(new_solutions.args[1], FiniteSet): + new_exprs = new_solutions.args[1] + + for new_expr in new_exprs: + result += ImageSet(Lambda(dummy_var, new_expr), base_set) + + if result is S.EmptySet: + return ConditionSet(symbol, Eq(f, 0), domain) + + y_s = result + + return y_s + + +def _solveset(f, symbol, domain, _check=False): + """Helper for solveset to return a result from an expression + that has already been sympify'ed and is known to contain the + given symbol.""" + # _check controls whether the answer is checked or not + from sympy.simplify.simplify import signsimp + + if isinstance(f, BooleanTrue): + return domain + + orig_f = f + if f.is_Mul: + coeff, f = f.as_independent(symbol, as_Add=False) + if coeff in {S.ComplexInfinity, S.NegativeInfinity, S.Infinity}: + f = together(orig_f) + elif f.is_Add: + a, h = f.as_independent(symbol) + m, h = h.as_independent(symbol, as_Add=False) + if m not in {S.ComplexInfinity, S.Zero, S.Infinity, + S.NegativeInfinity}: + f = a/m + h # XXX condition `m != 0` should be added to soln + + # assign the solvers to use + solver = lambda f, x, domain=domain: _solveset(f, x, domain) + inverter = lambda f, rhs, symbol: _invert(f, rhs, symbol, domain) + + result = S.EmptySet + + if f.expand().is_zero: + return domain + elif not f.has(symbol): + return S.EmptySet + elif f.is_Mul and all(_is_finite_with_finite_vars(m, domain) + for m in f.args): + # if f(x) and g(x) are both finite we can say that the solution of + # f(x)*g(x) == 0 is same as Union(f(x) == 0, g(x) == 0) is not true in + # general. g(x) can grow to infinitely large for the values where + # f(x) == 0. To be sure that we are not silently allowing any + # wrong solutions we are using this technique only if both f and g are + # finite for a finite input. + result = Union(*[solver(m, symbol) for m in f.args]) + elif (_is_function_class_equation(TrigonometricFunction, f, symbol) or \ + _is_function_class_equation(HyperbolicFunction, f, symbol)): + result = _solve_trig(f, symbol, domain) + elif isinstance(f, arg): + a = f.args[0] + result = Intersection(_solveset(re(a) > 0, symbol, domain), + _solveset(im(a), symbol, domain)) + elif f.is_Piecewise: + expr_set_pairs = f.as_expr_set_pairs(domain) + for (expr, in_set) in expr_set_pairs: + if in_set.is_Relational: + in_set = in_set.as_set() + solns = solver(expr, symbol, in_set) + result += solns + elif isinstance(f, Eq): + result = solver(Add(f.lhs, -f.rhs, evaluate=False), symbol, domain) + + elif f.is_Relational: + from .inequalities import solve_univariate_inequality + try: + result = solve_univariate_inequality( + f, symbol, domain=domain, relational=False) + except NotImplementedError: + result = ConditionSet(symbol, f, domain) + return result + elif _is_modular(f, symbol): + result = _solve_modular(f, symbol, domain) + else: + lhs, rhs_s = inverter(f, 0, symbol) + if lhs == symbol: + # do some very minimal simplification since + # repeated inversion may have left the result + # in a state that other solvers (e.g. poly) + # would have simplified; this is done here + # rather than in the inverter since here it + # is only done once whereas there it would + # be repeated for each step of the inversion + if isinstance(rhs_s, FiniteSet): + rhs_s = FiniteSet(*[Mul(* + signsimp(i).as_content_primitive()) + for i in rhs_s]) + result = rhs_s + + elif isinstance(rhs_s, FiniteSet): + for equation in [lhs - rhs for rhs in rhs_s]: + if equation == f: + u = unrad(f, symbol) + if u: + result += _solve_radical(equation, u, + symbol, + solver) + elif equation.has(Abs): + result += _solve_abs(f, symbol, domain) + else: + result_rational = _solve_as_rational(equation, symbol, domain) + if not isinstance(result_rational, ConditionSet): + result += result_rational + else: + # may be a transcendental type equation + t_result = _transolve(equation, symbol, domain) + if isinstance(t_result, ConditionSet): + # might need factoring; this is expensive so we + # have delayed until now. To avoid recursion + # errors look for a non-trivial factoring into + # a product of symbol dependent terms; I think + # that something that factors as a Pow would + # have already been recognized by now. + factored = equation.factor() + if factored.is_Mul and equation != factored: + _, dep = factored.as_independent(symbol) + if not dep.is_Add: + # non-trivial factoring of equation + # but use form with constants + # in case they need special handling + t_results = [] + for fac in Mul.make_args(factored): + if fac.has(symbol): + t_results.append(solver(fac, symbol)) + t_result = Union(*t_results) + result += t_result + else: + result += solver(equation, symbol) + + elif rhs_s is not S.EmptySet: + result = ConditionSet(symbol, Eq(f, 0), domain) + + if isinstance(result, ConditionSet): + if isinstance(f, Expr): + num, den = f.as_numer_denom() + if den.has(symbol): + _result = _solveset(num, symbol, domain) + if not isinstance(_result, ConditionSet): + singularities = _solveset(den, symbol, domain) + result = _result - singularities + + if _check: + if isinstance(result, ConditionSet): + # it wasn't solved or has enumerated all conditions + # -- leave it alone + return result + + # whittle away all but the symbol-containing core + # to use this for testing + if isinstance(orig_f, Expr): + fx = orig_f.as_independent(symbol, as_Add=True)[1] + fx = fx.as_independent(symbol, as_Add=False)[1] + else: + fx = orig_f + + if isinstance(result, FiniteSet): + # check the result for invalid solutions + result = FiniteSet(*[s for s in result + if isinstance(s, RootOf) + or domain_check(fx, symbol, s)]) + + return result + + +def _is_modular(f, symbol): + """ + Helper function to check below mentioned types of modular equations. + ``A - Mod(B, C) = 0`` + + A -> This can or cannot be a function of symbol. + B -> This is surely a function of symbol. + C -> It is an integer. + + Parameters + ========== + + f : Expr + The equation to be checked. + + symbol : Symbol + The concerned variable for which the equation is to be checked. + + Examples + ======== + + >>> from sympy import symbols, exp, Mod + >>> from sympy.solvers.solveset import _is_modular as check + >>> x, y = symbols('x y') + >>> check(Mod(x, 3) - 1, x) + True + >>> check(Mod(x, 3) - 1, y) + False + >>> check(Mod(x, 3)**2 - 5, x) + False + >>> check(Mod(x, 3)**2 - y, x) + False + >>> check(exp(Mod(x, 3)) - 1, x) + False + >>> check(Mod(3, y) - 1, y) + False + """ + + if not f.has(Mod): + return False + + # extract modterms from f. + modterms = list(f.atoms(Mod)) + + return (len(modterms) == 1 and # only one Mod should be present + modterms[0].args[0].has(symbol) and # B-> function of symbol + modterms[0].args[1].is_integer and # C-> to be an integer. + any(isinstance(term, Mod) + for term in list(_term_factors(f))) # free from other funcs + ) + + +def _invert_modular(modterm, rhs, n, symbol): + """ + Helper function to invert modular equation. + ``Mod(a, m) - rhs = 0`` + + Generally it is inverted as (a, ImageSet(Lambda(n, m*n + rhs), S.Integers)). + More simplified form will be returned if possible. + + If it is not invertible then (modterm, rhs) is returned. + + The following cases arise while inverting equation ``Mod(a, m) - rhs = 0``: + + 1. If a is symbol then m*n + rhs is the required solution. + + 2. If a is an instance of ``Add`` then we try to find two symbol independent + parts of a and the symbol independent part gets transferred to the other + side and again the ``_invert_modular`` is called on the symbol + dependent part. + + 3. If a is an instance of ``Mul`` then same as we done in ``Add`` we separate + out the symbol dependent and symbol independent parts and transfer the + symbol independent part to the rhs with the help of invert and again the + ``_invert_modular`` is called on the symbol dependent part. + + 4. If a is an instance of ``Pow`` then two cases arise as following: + + - If a is of type (symbol_indep)**(symbol_dep) then the remainder is + evaluated with the help of discrete_log function and then the least + period is being found out with the help of totient function. + period*n + remainder is the required solution in this case. + For reference: (https://en.wikipedia.org/wiki/Euler's_theorem) + + - If a is of type (symbol_dep)**(symbol_indep) then we try to find all + primitive solutions list with the help of nthroot_mod function. + m*n + rem is the general solution where rem belongs to solutions list + from nthroot_mod function. + + Parameters + ========== + + modterm, rhs : Expr + The modular equation to be inverted, ``modterm - rhs = 0`` + + symbol : Symbol + The variable in the equation to be inverted. + + n : Dummy + Dummy variable for output g_n. + + Returns + ======= + + A tuple (f_x, g_n) is being returned where f_x is modular independent function + of symbol and g_n being set of values f_x can have. + + Examples + ======== + + >>> from sympy import symbols, exp, Mod, Dummy, S + >>> from sympy.solvers.solveset import _invert_modular as invert_modular + >>> x, y = symbols('x y') + >>> n = Dummy('n') + >>> invert_modular(Mod(exp(x), 7), S(5), n, x) + (Mod(exp(x), 7), 5) + >>> invert_modular(Mod(x, 7), S(5), n, x) + (x, ImageSet(Lambda(_n, 7*_n + 5), Integers)) + >>> invert_modular(Mod(3*x + 8, 7), S(5), n, x) + (x, ImageSet(Lambda(_n, 7*_n + 6), Integers)) + >>> invert_modular(Mod(x**4, 7), S(5), n, x) + (x, EmptySet) + >>> invert_modular(Mod(2**(x**2 + x + 1), 7), S(2), n, x) + (x**2 + x + 1, ImageSet(Lambda(_n, 3*_n + 1), Naturals0)) + + """ + a, m = modterm.args + + if rhs.is_integer is False: + return symbol, S.EmptySet + + if rhs.is_real is False or any(term.is_real is False + for term in list(_term_factors(a))): + # Check for complex arguments + return modterm, rhs + + if abs(rhs) >= abs(m): + # if rhs has value greater than value of m. + return symbol, S.EmptySet + + if a == symbol: + return symbol, ImageSet(Lambda(n, m*n + rhs), S.Integers) + + if a.is_Add: + # g + h = a + g, h = a.as_independent(symbol) + if g is not S.Zero: + x_indep_term = rhs - Mod(g, m) + return _invert_modular(Mod(h, m), Mod(x_indep_term, m), n, symbol) + + if a.is_Mul: + # g*h = a + g, h = a.as_independent(symbol) + if g is not S.One: + x_indep_term = rhs*invert(g, m) + return _invert_modular(Mod(h, m), Mod(x_indep_term, m), n, symbol) + + if a.is_Pow: + # base**expo = a + base, expo = a.args + if expo.has(symbol) and not base.has(symbol): + # remainder -> solution independent of n of equation. + # m, rhs are made coprime by dividing number_gcd(m, rhs) + if not m.is_Integer and rhs.is_Integer and a.base.is_Integer: + return modterm, rhs + + mdiv = m.p // number_gcd(m.p, rhs.p) + try: + remainder = discrete_log(mdiv, rhs.p, a.base.p) + except ValueError: # log does not exist + return modterm, rhs + # period -> coefficient of n in the solution and also referred as + # the least period of expo in which it is repeats itself. + # (a**(totient(m)) - 1) divides m. Here is link of theorem: + # (https://en.wikipedia.org/wiki/Euler's_theorem) + period = totient(m) + for p in divisors(period): + # there might a lesser period exist than totient(m). + if pow(a.base, p, m / number_gcd(m.p, a.base.p)) == 1: + period = p + break + # recursion is not applied here since _invert_modular is currently + # not smart enough to handle infinite rhs as here expo has infinite + # rhs = ImageSet(Lambda(n, period*n + remainder), S.Naturals0). + return expo, ImageSet(Lambda(n, period*n + remainder), S.Naturals0) + elif base.has(symbol) and not expo.has(symbol): + try: + remainder_list = nthroot_mod(rhs, expo, m, all_roots=True) + if remainder_list == []: + return symbol, S.EmptySet + except (ValueError, NotImplementedError): + return modterm, rhs + g_n = S.EmptySet + for rem in remainder_list: + g_n += ImageSet(Lambda(n, m*n + rem), S.Integers) + return base, g_n + + return modterm, rhs + + +def _solve_modular(f, symbol, domain): + r""" + Helper function for solving modular equations of type ``A - Mod(B, C) = 0``, + where A can or cannot be a function of symbol, B is surely a function of + symbol and C is an integer. + + Currently ``_solve_modular`` is only able to solve cases + where A is not a function of symbol. + + Parameters + ========== + + f : Expr + The modular equation to be solved, ``f = 0`` + + symbol : Symbol + The variable in the equation to be solved. + + domain : Set + A set over which the equation is solved. It has to be a subset of + Integers. + + Returns + ======= + + A set of integer solutions satisfying the given modular equation. + A ``ConditionSet`` if the equation is unsolvable. + + Examples + ======== + + >>> from sympy.solvers.solveset import _solve_modular as solve_modulo + >>> from sympy import S, Symbol, sin, Intersection, Interval, Mod + >>> x = Symbol('x') + >>> solve_modulo(Mod(5*x - 8, 7) - 3, x, S.Integers) + ImageSet(Lambda(_n, 7*_n + 5), Integers) + >>> solve_modulo(Mod(5*x - 8, 7) - 3, x, S.Reals) # domain should be subset of integers. + ConditionSet(x, Eq(Mod(5*x + 6, 7) - 3, 0), Reals) + >>> solve_modulo(-7 + Mod(x, 5), x, S.Integers) + EmptySet + >>> solve_modulo(Mod(12**x, 21) - 18, x, S.Integers) + ImageSet(Lambda(_n, 6*_n + 2), Naturals0) + >>> solve_modulo(Mod(sin(x), 7) - 3, x, S.Integers) # not solvable + ConditionSet(x, Eq(Mod(sin(x), 7) - 3, 0), Integers) + >>> solve_modulo(3 - Mod(x, 5), x, Intersection(S.Integers, Interval(0, 100))) + Intersection(ImageSet(Lambda(_n, 5*_n + 3), Integers), Range(0, 101, 1)) + """ + # extract modterm and g_y from f + unsolved_result = ConditionSet(symbol, Eq(f, 0), domain) + modterm = list(f.atoms(Mod))[0] + rhs = -S.One*(f.subs(modterm, S.Zero)) + if f.as_coefficients_dict()[modterm].is_negative: + # checks if coefficient of modterm is negative in main equation. + rhs *= -S.One + + if not domain.is_subset(S.Integers): + return unsolved_result + + if rhs.has(symbol): + # TODO Case: A-> function of symbol, can be extended here + # in future. + return unsolved_result + + n = Dummy('n', integer=True) + f_x, g_n = _invert_modular(modterm, rhs, n, symbol) + + if f_x == modterm and g_n == rhs: + return unsolved_result + + if f_x == symbol: + if domain is not S.Integers: + return domain.intersect(g_n) + return g_n + + if isinstance(g_n, ImageSet): + lamda_expr = g_n.lamda.expr + lamda_vars = g_n.lamda.variables + base_sets = g_n.base_sets + sol_set = _solveset(f_x - lamda_expr, symbol, S.Integers) + if isinstance(sol_set, FiniteSet): + tmp_sol = S.EmptySet + for sol in sol_set: + tmp_sol += ImageSet(Lambda(lamda_vars, sol), *base_sets) + sol_set = tmp_sol + else: + sol_set = ImageSet(Lambda(lamda_vars, sol_set), *base_sets) + return domain.intersect(sol_set) + + return unsolved_result + + +def _term_factors(f): + """ + Iterator to get the factors of all terms present + in the given equation. + + Parameters + ========== + f : Expr + Equation that needs to be addressed + + Returns + ======= + Factors of all terms present in the equation. + + Examples + ======== + + >>> from sympy import symbols + >>> from sympy.solvers.solveset import _term_factors + >>> x = symbols('x') + >>> list(_term_factors(-2 - x**2 + x*(x + 1))) + [-2, -1, x**2, x, x + 1] + """ + for add_arg in Add.make_args(f): + yield from Mul.make_args(add_arg) + + +def _solve_exponential(lhs, rhs, symbol, domain): + r""" + Helper function for solving (supported) exponential equations. + + Exponential equations are the sum of (currently) at most + two terms with one or both of them having a power with a + symbol-dependent exponent. + + For example + + .. math:: 5^{2x + 3} - 5^{3x - 1} + + .. math:: 4^{5 - 9x} - e^{2 - x} + + Parameters + ========== + + lhs, rhs : Expr + The exponential equation to be solved, `lhs = rhs` + + symbol : Symbol + The variable in which the equation is solved + + domain : Set + A set over which the equation is solved. + + Returns + ======= + + A set of solutions satisfying the given equation. + A ``ConditionSet`` if the equation is unsolvable or + if the assumptions are not properly defined, in that case + a different style of ``ConditionSet`` is returned having the + solution(s) of the equation with the desired assumptions. + + Examples + ======== + + >>> from sympy.solvers.solveset import _solve_exponential as solve_expo + >>> from sympy import symbols, S + >>> x = symbols('x', real=True) + >>> a, b = symbols('a b') + >>> solve_expo(2**x + 3**x - 5**x, 0, x, S.Reals) # not solvable + ConditionSet(x, Eq(2**x + 3**x - 5**x, 0), Reals) + >>> solve_expo(a**x - b**x, 0, x, S.Reals) # solvable but incorrect assumptions + ConditionSet(x, (a > 0) & (b > 0), {0}) + >>> solve_expo(3**(2*x) - 2**(x + 3), 0, x, S.Reals) + {-3*log(2)/(-2*log(3) + log(2))} + >>> solve_expo(2**x - 4**x, 0, x, S.Reals) + {0} + + * Proof of correctness of the method + + The logarithm function is the inverse of the exponential function. + The defining relation between exponentiation and logarithm is: + + .. math:: {\log_b x} = y \enspace if \enspace b^y = x + + Therefore if we are given an equation with exponent terms, we can + convert every term to its corresponding logarithmic form. This is + achieved by taking logarithms and expanding the equation using + logarithmic identities so that it can easily be handled by ``solveset``. + + For example: + + .. math:: 3^{2x} = 2^{x + 3} + + Taking log both sides will reduce the equation to + + .. math:: (2x)\log(3) = (x + 3)\log(2) + + This form can be easily handed by ``solveset``. + """ + unsolved_result = ConditionSet(symbol, Eq(lhs - rhs, 0), domain) + newlhs = powdenest(lhs) + if lhs != newlhs: + # it may also be advantageous to factor the new expr + neweq = factor(newlhs - rhs) + if neweq != (lhs - rhs): + return _solveset(neweq, symbol, domain) # try again with _solveset + + if not (isinstance(lhs, Add) and len(lhs.args) == 2): + # solving for the sum of more than two powers is possible + # but not yet implemented + return unsolved_result + + if rhs != 0: + return unsolved_result + + a, b = list(ordered(lhs.args)) + a_term = a.as_independent(symbol)[1] + b_term = b.as_independent(symbol)[1] + + a_base, a_exp = a_term.as_base_exp() + b_base, b_exp = b_term.as_base_exp() + + if domain.is_subset(S.Reals): + conditions = And( + a_base > 0, + b_base > 0, + Eq(im(a_exp), 0), + Eq(im(b_exp), 0)) + else: + conditions = And( + Ne(a_base, 0), + Ne(b_base, 0)) + + L, R = (expand_log(log(i), force=True) for i in (a, -b)) + solutions = _solveset(L - R, symbol, domain) + + return ConditionSet(symbol, conditions, solutions) + + +def _is_exponential(f, symbol): + r""" + Return ``True`` if one or more terms contain ``symbol`` only in + exponents, else ``False``. + + Parameters + ========== + + f : Expr + The equation to be checked + + symbol : Symbol + The variable in which the equation is checked + + Examples + ======== + + >>> from sympy import symbols, cos, exp + >>> from sympy.solvers.solveset import _is_exponential as check + >>> x, y = symbols('x y') + >>> check(y, y) + False + >>> check(x**y - 1, y) + True + >>> check(x**y*2**y - 1, y) + True + >>> check(exp(x + 3) + 3**x, x) + True + >>> check(cos(2**x), x) + False + + * Philosophy behind the helper + + The function extracts each term of the equation and checks if it is + of exponential form w.r.t ``symbol``. + """ + rv = False + for expr_arg in _term_factors(f): + if symbol not in expr_arg.free_symbols: + continue + if (isinstance(expr_arg, Pow) and + symbol not in expr_arg.base.free_symbols or + isinstance(expr_arg, exp)): + rv = True # symbol in exponent + else: + return False # dependent on symbol in non-exponential way + return rv + + +def _solve_logarithm(lhs, rhs, symbol, domain): + r""" + Helper to solve logarithmic equations which are reducible + to a single instance of `\log`. + + Logarithmic equations are (currently) the equations that contains + `\log` terms which can be reduced to a single `\log` term or + a constant using various logarithmic identities. + + For example: + + .. math:: \log(x) + \log(x - 4) + + can be reduced to: + + .. math:: \log(x(x - 4)) + + Parameters + ========== + + lhs, rhs : Expr + The logarithmic equation to be solved, `lhs = rhs` + + symbol : Symbol + The variable in which the equation is solved + + domain : Set + A set over which the equation is solved. + + Returns + ======= + + A set of solutions satisfying the given equation. + A ``ConditionSet`` if the equation is unsolvable. + + Examples + ======== + + >>> from sympy import symbols, log, S + >>> from sympy.solvers.solveset import _solve_logarithm as solve_log + >>> x = symbols('x') + >>> f = log(x - 3) + log(x + 3) + >>> solve_log(f, 0, x, S.Reals) + {-sqrt(10), sqrt(10)} + + * Proof of correctness + + A logarithm is another way to write exponent and is defined by + + .. math:: {\log_b x} = y \enspace if \enspace b^y = x + + When one side of the equation contains a single logarithm, the + equation can be solved by rewriting the equation as an equivalent + exponential equation as defined above. But if one side contains + more than one logarithm, we need to use the properties of logarithm + to condense it into a single logarithm. + + Take for example + + .. math:: \log(2x) - 15 = 0 + + contains single logarithm, therefore we can directly rewrite it to + exponential form as + + .. math:: x = \frac{e^{15}}{2} + + But if the equation has more than one logarithm as + + .. math:: \log(x - 3) + \log(x + 3) = 0 + + we use logarithmic identities to convert it into a reduced form + + Using, + + .. math:: \log(a) + \log(b) = \log(ab) + + the equation becomes, + + .. math:: \log((x - 3)(x + 3)) + + This equation contains one logarithm and can be solved by rewriting + to exponents. + """ + new_lhs = logcombine(lhs, force=True) + new_f = new_lhs - rhs + + return _solveset(new_f, symbol, domain) + + +def _is_logarithmic(f, symbol): + r""" + Return ``True`` if the equation is in the form + `a\log(f(x)) + b\log(g(x)) + ... + c` else ``False``. + + Parameters + ========== + + f : Expr + The equation to be checked + + symbol : Symbol + The variable in which the equation is checked + + Returns + ======= + + ``True`` if the equation is logarithmic otherwise ``False``. + + Examples + ======== + + >>> from sympy import symbols, tan, log + >>> from sympy.solvers.solveset import _is_logarithmic as check + >>> x, y = symbols('x y') + >>> check(log(x + 2) - log(x + 3), x) + True + >>> check(tan(log(2*x)), x) + False + >>> check(x*log(x), x) + False + >>> check(x + log(x), x) + False + >>> check(y + log(x), x) + True + + * Philosophy behind the helper + + The function extracts each term and checks whether it is + logarithmic w.r.t ``symbol``. + """ + rv = False + for term in Add.make_args(f): + saw_log = False + for term_arg in Mul.make_args(term): + if symbol not in term_arg.free_symbols: + continue + if isinstance(term_arg, log): + if saw_log: + return False # more than one log in term + saw_log = True + else: + return False # dependent on symbol in non-log way + if saw_log: + rv = True + return rv + + +def _is_lambert(f, symbol): + r""" + If this returns ``False`` then the Lambert solver (``_solve_lambert``) will not be called. + + Explanation + =========== + + Quick check for cases that the Lambert solver might be able to handle. + + 1. Equations containing more than two operands and `symbol`s involving any of + `Pow`, `exp`, `HyperbolicFunction`,`TrigonometricFunction`, `log` terms. + + 2. In `Pow`, `exp` the exponent should have `symbol` whereas for + `HyperbolicFunction`,`TrigonometricFunction`, `log` should contain `symbol`. + + 3. For `HyperbolicFunction`,`TrigonometricFunction` the number of trigonometric functions in + equation should be less than number of symbols. (since `A*cos(x) + B*sin(x) - c` + is not the Lambert type). + + Some forms of lambert equations are: + 1. X**X = C + 2. X*(B*log(X) + D)**A = C + 3. A*log(B*X + A) + d*X = C + 4. (B*X + A)*exp(d*X + g) = C + 5. g*exp(B*X + h) - B*X = C + 6. A*D**(E*X + g) - B*X = C + 7. A*cos(X) + B*sin(X) - D*X = C + 8. A*cosh(X) + B*sinh(X) - D*X = C + + Where X is any variable, + A, B, C, D, E are any constants, + g, h are linear functions or log terms. + + Parameters + ========== + + f : Expr + The equation to be checked + + symbol : Symbol + The variable in which the equation is checked + + Returns + ======= + + If this returns ``False`` then the Lambert solver (``_solve_lambert``) will not be called. + + Examples + ======== + + >>> from sympy.solvers.solveset import _is_lambert + >>> from sympy import symbols, cosh, sinh, log + >>> x = symbols('x') + + >>> _is_lambert(3*log(x) - x*log(3), x) + True + >>> _is_lambert(log(log(x - 3)) + log(x-3), x) + True + >>> _is_lambert(cosh(x) - sinh(x), x) + False + >>> _is_lambert((x**2 - 2*x + 1).subs(x, (log(x) + 3*x)**2 - 1), x) + True + + See Also + ======== + + _solve_lambert + + """ + term_factors = list(_term_factors(f.expand())) + + # total number of symbols in equation + no_of_symbols = len([arg for arg in term_factors if arg.has(symbol)]) + # total number of trigonometric terms in equation + no_of_trig = len([arg for arg in term_factors \ + if arg.has(HyperbolicFunction, TrigonometricFunction)]) + + if f.is_Add and no_of_symbols >= 2: + # `log`, `HyperbolicFunction`, `TrigonometricFunction` should have symbols + # and no_of_trig < no_of_symbols + lambert_funcs = (log, HyperbolicFunction, TrigonometricFunction) + if any(isinstance(arg, lambert_funcs)\ + for arg in term_factors if arg.has(symbol)): + if no_of_trig < no_of_symbols: + return True + # here, `Pow`, `exp` exponent should have symbols + elif any(isinstance(arg, (Pow, exp)) \ + for arg in term_factors if (arg.as_base_exp()[1]).has(symbol)): + return True + return False + + +def _transolve(f, symbol, domain): + r""" + Function to solve transcendental equations. It is a helper to + ``solveset`` and should be used internally. ``_transolve`` + currently supports the following class of equations: + + - Exponential equations + - Logarithmic equations + + Parameters + ========== + + f : Any transcendental equation that needs to be solved. + This needs to be an expression, which is assumed + to be equal to ``0``. + + symbol : The variable for which the equation is solved. + This needs to be of class ``Symbol``. + + domain : A set over which the equation is solved. + This needs to be of class ``Set``. + + Returns + ======= + + Set + A set of values for ``symbol`` for which ``f`` is equal to + zero. An ``EmptySet`` is returned if ``f`` does not have solutions + in respective domain. A ``ConditionSet`` is returned as unsolved + object if algorithms to evaluate complete solution are not + yet implemented. + + How to use ``_transolve`` + ========================= + + ``_transolve`` should not be used as an independent function, because + it assumes that the equation (``f``) and the ``symbol`` comes from + ``solveset`` and might have undergone a few modification(s). + To use ``_transolve`` as an independent function the equation (``f``) + and the ``symbol`` should be passed as they would have been by + ``solveset``. + + Examples + ======== + + >>> from sympy.solvers.solveset import _transolve as transolve + >>> from sympy.solvers.solvers import _tsolve as tsolve + >>> from sympy import symbols, S, pprint + >>> x = symbols('x', real=True) # assumption added + >>> transolve(5**(x - 3) - 3**(2*x + 1), x, S.Reals) + {-(log(3) + 3*log(5))/(-log(5) + 2*log(3))} + + How ``_transolve`` works + ======================== + + ``_transolve`` uses two types of helper functions to solve equations + of a particular class: + + Identifying helpers: To determine whether a given equation + belongs to a certain class of equation or not. Returns either + ``True`` or ``False``. + + Solving helpers: Once an equation is identified, a corresponding + helper either solves the equation or returns a form of the equation + that ``solveset`` might better be able to handle. + + * Philosophy behind the module + + The purpose of ``_transolve`` is to take equations which are not + already polynomial in their generator(s) and to either recast them + as such through a valid transformation or to solve them outright. + A pair of helper functions for each class of supported + transcendental functions are employed for this purpose. One + identifies the transcendental form of an equation and the other + either solves it or recasts it into a tractable form that can be + solved by ``solveset``. + For example, an equation in the form `ab^{f(x)} - cd^{g(x)} = 0` + can be transformed to + `\log(a) + f(x)\log(b) - \log(c) - g(x)\log(d) = 0` + (under certain assumptions) and this can be solved with ``solveset`` + if `f(x)` and `g(x)` are in polynomial form. + + How ``_transolve`` is better than ``_tsolve`` + ============================================= + + 1) Better output + + ``_transolve`` provides expressions in a more simplified form. + + Consider a simple exponential equation + + >>> f = 3**(2*x) - 2**(x + 3) + >>> pprint(transolve(f, x, S.Reals), use_unicode=False) + -3*log(2) + {------------------} + -2*log(3) + log(2) + >>> pprint(tsolve(f, x), use_unicode=False) + / 3 \ + | --------| + | log(2/9)| + [-log\2 /] + + 2) Extensible + + The API of ``_transolve`` is designed such that it is easily + extensible, i.e. the code that solves a given class of + equations is encapsulated in a helper and not mixed in with + the code of ``_transolve`` itself. + + 3) Modular + + ``_transolve`` is designed to be modular i.e, for every class of + equation a separate helper for identification and solving is + implemented. This makes it easy to change or modify any of the + method implemented directly in the helpers without interfering + with the actual structure of the API. + + 4) Faster Computation + + Solving equation via ``_transolve`` is much faster as compared to + ``_tsolve``. In ``solve``, attempts are made computing every possibility + to get the solutions. This series of attempts makes solving a bit + slow. In ``_transolve``, computation begins only after a particular + type of equation is identified. + + How to add new class of equations + ================================= + + Adding a new class of equation solver is a three-step procedure: + + - Identify the type of the equations + + Determine the type of the class of equations to which they belong: + it could be of ``Add``, ``Pow``, etc. types. Separate internal functions + are used for each type. Write identification and solving helpers + and use them from within the routine for the given type of equation + (after adding it, if necessary). Something like: + + .. code-block:: python + + def add_type(lhs, rhs, x): + .... + if _is_exponential(lhs, x): + new_eq = _solve_exponential(lhs, rhs, x) + .... + rhs, lhs = eq.as_independent(x) + if lhs.is_Add: + result = add_type(lhs, rhs, x) + + - Define the identification helper. + + - Define the solving helper. + + Apart from this, a few other things needs to be taken care while + adding an equation solver: + + - Naming conventions: + Name of the identification helper should be as + ``_is_class`` where class will be the name or abbreviation + of the class of equation. The solving helper will be named as + ``_solve_class``. + For example: for exponential equations it becomes + ``_is_exponential`` and ``_solve_expo``. + - The identifying helpers should take two input parameters, + the equation to be checked and the variable for which a solution + is being sought, while solving helpers would require an additional + domain parameter. + - Be sure to consider corner cases. + - Add tests for each helper. + - Add a docstring to your helper that describes the method + implemented. + The documentation of the helpers should identify: + + - the purpose of the helper, + - the method used to identify and solve the equation, + - a proof of correctness + - the return values of the helpers + """ + + def add_type(lhs, rhs, symbol, domain): + """ + Helper for ``_transolve`` to handle equations of + ``Add`` type, i.e. equations taking the form as + ``a*f(x) + b*g(x) + .... = c``. + For example: 4**x + 8**x = 0 + """ + result = ConditionSet(symbol, Eq(lhs - rhs, 0), domain) + + # check if it is exponential type equation + if _is_exponential(lhs, symbol): + result = _solve_exponential(lhs, rhs, symbol, domain) + # check if it is logarithmic type equation + elif _is_logarithmic(lhs, symbol): + result = _solve_logarithm(lhs, rhs, symbol, domain) + + return result + + result = ConditionSet(symbol, Eq(f, 0), domain) + + # invert_complex handles the call to the desired inverter based + # on the domain specified. + lhs, rhs_s = invert_complex(f, 0, symbol, domain) + + if isinstance(rhs_s, FiniteSet): + assert (len(rhs_s.args)) == 1 + rhs = rhs_s.args[0] + + if lhs.is_Add: + result = add_type(lhs, rhs, symbol, domain) + else: + result = rhs_s + + return result + + +def solveset(f, symbol=None, domain=S.Complexes): + r"""Solves a given inequality or equation with set as output + + Parameters + ========== + + f : Expr or a relational. + The target equation or inequality + symbol : Symbol + The variable for which the equation is solved + domain : Set + The domain over which the equation is solved + + Returns + ======= + + Set + A set of values for `symbol` for which `f` is True or is equal to + zero. An :class:`~.EmptySet` is returned if `f` is False or nonzero. + A :class:`~.ConditionSet` is returned as unsolved object if algorithms + to evaluate complete solution are not yet implemented. + + ``solveset`` claims to be complete in the solution set that it returns. + + Raises + ====== + + NotImplementedError + The algorithms to solve inequalities in complex domain are + not yet implemented. + ValueError + The input is not valid. + RuntimeError + It is a bug, please report to the github issue tracker. + + + Notes + ===== + + Python interprets 0 and 1 as False and True, respectively, but + in this function they refer to solutions of an expression. So 0 and 1 + return the domain and EmptySet, respectively, while True and False + return the opposite (as they are assumed to be solutions of relational + expressions). + + + See Also + ======== + + solveset_real: solver for real domain + solveset_complex: solver for complex domain + + Examples + ======== + + >>> from sympy import exp, sin, Symbol, pprint, S, Eq + >>> from sympy.solvers.solveset import solveset, solveset_real + + * The default domain is complex. Not specifying a domain will lead + to the solving of the equation in the complex domain (and this + is not affected by the assumptions on the symbol): + + >>> x = Symbol('x') + >>> pprint(solveset(exp(x) - 1, x), use_unicode=False) + {2*n*I*pi | n in Integers} + + >>> x = Symbol('x', real=True) + >>> pprint(solveset(exp(x) - 1, x), use_unicode=False) + {2*n*I*pi | n in Integers} + + * If you want to use ``solveset`` to solve the equation in the + real domain, provide a real domain. (Using ``solveset_real`` + does this automatically.) + + >>> R = S.Reals + >>> x = Symbol('x') + >>> solveset(exp(x) - 1, x, R) + {0} + >>> solveset_real(exp(x) - 1, x) + {0} + + The solution is unaffected by assumptions on the symbol: + + >>> p = Symbol('p', positive=True) + >>> pprint(solveset(p**2 - 4)) + {-2, 2} + + When a :class:`~.ConditionSet` is returned, symbols with assumptions that + would alter the set are replaced with more generic symbols: + + >>> i = Symbol('i', imaginary=True) + >>> solveset(Eq(i**2 + i*sin(i), 1), i, domain=S.Reals) + ConditionSet(_R, Eq(_R**2 + _R*sin(_R) - 1, 0), Reals) + + * Inequalities can be solved over the real domain only. Use of a complex + domain leads to a NotImplementedError. + + >>> solveset(exp(x) > 1, x, R) + Interval.open(0, oo) + + """ + f = sympify(f) + symbol = sympify(symbol) + + if f is S.true: + return domain + + if f is S.false: + return S.EmptySet + + if not isinstance(f, (Expr, Relational, Number)): + raise ValueError("%s is not a valid SymPy expression" % f) + + if not isinstance(symbol, (Expr, Relational)) and symbol is not None: + raise ValueError("%s is not a valid SymPy symbol" % (symbol,)) + + if not isinstance(domain, Set): + raise ValueError("%s is not a valid domain" %(domain)) + + free_symbols = f.free_symbols + + if f.has(Piecewise): + f = piecewise_fold(f) + + if symbol is None and not free_symbols: + b = Eq(f, 0) + if b is S.true: + return domain + elif b is S.false: + return S.EmptySet + else: + raise NotImplementedError(filldedent(''' + relationship between value and 0 is unknown: %s''' % b)) + + if symbol is None: + if len(free_symbols) == 1: + symbol = free_symbols.pop() + elif free_symbols: + raise ValueError(filldedent(''' + The independent variable must be specified for a + multivariate equation.''')) + elif not isinstance(symbol, Symbol): + f, s, swap = recast_to_symbols([f], [symbol]) + # the xreplace will be needed if a ConditionSet is returned + return solveset(f[0], s[0], domain).xreplace(swap) + + # solveset should ignore assumptions on symbols + newsym = None + if domain.is_subset(S.Reals): + if symbol._assumptions_orig != {'real': True}: + newsym = Dummy('R', real=True) + elif domain.is_subset(S.Complexes): + if symbol._assumptions_orig != {'complex': True}: + newsym = Dummy('C', complex=True) + + if newsym is not None: + rv = solveset(f.xreplace({symbol: newsym}), newsym, domain) + # try to use the original symbol if possible + try: + _rv = rv.xreplace({newsym: symbol}) + except TypeError: + _rv = rv + if rv.dummy_eq(_rv): + rv = _rv + return rv + + # Abs has its own handling method which avoids the + # rewriting property that the first piece of abs(x) + # is for x >= 0 and the 2nd piece for x < 0 -- solutions + # can look better if the 2nd condition is x <= 0. Since + # the solution is a set, duplication of results is not + # an issue, e.g. {y, -y} when y is 0 will be {0} + f, mask = _masked(f, Abs) + f = f.rewrite(Piecewise) # everything that's not an Abs + for d, e in mask: + # everything *in* an Abs + e = e.func(e.args[0].rewrite(Piecewise)) + f = f.xreplace({d: e}) + f = piecewise_fold(f) + + return _solveset(f, symbol, domain, _check=True) + + +def solveset_real(f, symbol): + return solveset(f, symbol, S.Reals) + + +def solveset_complex(f, symbol): + return solveset(f, symbol, S.Complexes) + + +def _solveset_multi(eqs, syms, domains): + '''Basic implementation of a multivariate solveset. + + For internal use (not ready for public consumption)''' + + rep = {} + for sym, dom in zip(syms, domains): + if dom is S.Reals: + rep[sym] = Symbol(sym.name, real=True) + eqs = [eq.subs(rep) for eq in eqs] + syms = [sym.subs(rep) for sym in syms] + + syms = tuple(syms) + + if len(eqs) == 0: + return ProductSet(*domains) + + if len(syms) == 1: + sym = syms[0] + domain = domains[0] + solsets = [solveset(eq, sym, domain) for eq in eqs] + solset = Intersection(*solsets) + return ImageSet(Lambda((sym,), (sym,)), solset).doit() + + eqs = sorted(eqs, key=lambda eq: len(eq.free_symbols & set(syms))) + + for n, eq in enumerate(eqs): + sols = [] + all_handled = True + for sym in syms: + if sym not in eq.free_symbols: + continue + sol = solveset(eq, sym, domains[syms.index(sym)]) + + if isinstance(sol, FiniteSet): + i = syms.index(sym) + symsp = syms[:i] + syms[i+1:] + domainsp = domains[:i] + domains[i+1:] + eqsp = eqs[:n] + eqs[n+1:] + for s in sol: + eqsp_sub = [eq.subs(sym, s) for eq in eqsp] + sol_others = _solveset_multi(eqsp_sub, symsp, domainsp) + fun = Lambda((symsp,), symsp[:i] + (s,) + symsp[i:]) + sols.append(ImageSet(fun, sol_others).doit()) + else: + all_handled = False + if all_handled: + return Union(*sols) + + +def solvify(f, symbol, domain): + """Solves an equation using solveset and returns the solution in accordance + with the `solve` output API. + + Returns + ======= + + We classify the output based on the type of solution returned by `solveset`. + + Solution | Output + ---------------------------------------- + FiniteSet | list + + ImageSet, | list (if `f` is periodic) + Union | + + Union | list (with FiniteSet) + + EmptySet | empty list + + Others | None + + + Raises + ====== + + NotImplementedError + A ConditionSet is the input. + + Examples + ======== + + >>> from sympy.solvers.solveset import solvify + >>> from sympy.abc import x + >>> from sympy import S, tan, sin, exp + >>> solvify(x**2 - 9, x, S.Reals) + [-3, 3] + >>> solvify(sin(x) - 1, x, S.Reals) + [pi/2] + >>> solvify(tan(x), x, S.Reals) + [0] + >>> solvify(exp(x) - 1, x, S.Complexes) + + >>> solvify(exp(x) - 1, x, S.Reals) + [0] + + """ + solution_set = solveset(f, symbol, domain) + result = None + if solution_set is S.EmptySet: + result = [] + + elif isinstance(solution_set, ConditionSet): + raise NotImplementedError('solveset is unable to solve this equation.') + + elif isinstance(solution_set, FiniteSet): + result = list(solution_set) + + else: + period = periodicity(f, symbol) + if period is not None: + solutions = S.EmptySet + iter_solutions = () + if isinstance(solution_set, ImageSet): + iter_solutions = (solution_set,) + elif isinstance(solution_set, Union): + if all(isinstance(i, ImageSet) for i in solution_set.args): + iter_solutions = solution_set.args + + for solution in iter_solutions: + solutions += solution.intersect(Interval(0, period, False, True)) + + if isinstance(solutions, FiniteSet): + result = list(solutions) + + else: + solution = solution_set.intersect(domain) + if isinstance(solution, Union): + # concerned about only FiniteSet with Union but not about ImageSet + # if required could be extend + if any(isinstance(i, FiniteSet) for i in solution.args): + result = [sol for soln in solution.args \ + for sol in soln.args if isinstance(soln,FiniteSet)] + else: + return None + + elif isinstance(solution, FiniteSet): + result += solution + + return result + + +############################################################################### +################################ LINSOLVE ##################################### +############################################################################### + + +def linear_coeffs(eq, *syms, dict=False): + """Return a list whose elements are the coefficients of the + corresponding symbols in the sum of terms in ``eq``. + The additive constant is returned as the last element of the + list. + + Raises + ====== + + NonlinearError + The equation contains a nonlinear term + ValueError + duplicate or unordered symbols are passed + + Parameters + ========== + + dict - (default False) when True, return coefficients as a + dictionary with coefficients keyed to syms that were present; + key 1 gives the constant term + + Examples + ======== + + >>> from sympy.solvers.solveset import linear_coeffs + >>> from sympy.abc import x, y, z + >>> linear_coeffs(3*x + 2*y - 1, x, y) + [3, 2, -1] + + It is not necessary to expand the expression: + + >>> linear_coeffs(x + y*(z*(x*3 + 2) + 3), x) + [3*y*z + 1, y*(2*z + 3)] + + When nonlinear is detected, an error will be raised: + + * even if they would cancel after expansion (so the + situation does not pass silently past the caller's + attention) + + >>> eq = 1/x*(x - 1) + 1/x + >>> linear_coeffs(eq.expand(), x) + [0, 1] + >>> linear_coeffs(eq, x) + Traceback (most recent call last): + ... + NonlinearError: + nonlinear in given generators + + * when there are cross terms + + >>> linear_coeffs(x*(y + 1), x, y) + Traceback (most recent call last): + ... + NonlinearError: + symbol-dependent cross-terms encountered + + * when there are terms that contain an expression + dependent on the symbols that is not linear + + >>> linear_coeffs(x**2, x) + Traceback (most recent call last): + ... + NonlinearError: + nonlinear in given generators + """ + eq = _sympify(eq) + if len(syms) == 1 and iterable(syms[0]) and not isinstance(syms[0], Basic): + raise ValueError('expecting unpacked symbols, *syms') + symset = set(syms) + if len(symset) != len(syms): + raise ValueError('duplicate symbols given') + try: + d, c = _linear_eq_to_dict([eq], symset) + d = d[0] + c = c[0] + except PolyNonlinearError as err: + raise NonlinearError(str(err)) + if dict: + if c: + d[S.One] = c + return d + rv = [S.Zero]*(len(syms) + 1) + rv[-1] = c + for i, k in enumerate(syms): + if k not in d: + continue + rv[i] = d[k] + return rv + + +def linear_eq_to_matrix(equations, *symbols): + r""" + Converts a given System of Equations into Matrix form. + Here `equations` must be a linear system of equations in + `symbols`. Element ``M[i, j]`` corresponds to the coefficient + of the jth symbol in the ith equation. + + The Matrix form corresponds to the augmented matrix form. + For example: + + .. math:: 4x + 2y + 3z = 1 + .. math:: 3x + y + z = -6 + .. math:: 2x + 4y + 9z = 2 + + This system will return $A$ and $b$ as: + + $$ A = \left[\begin{array}{ccc} + 4 & 2 & 3 \\ + 3 & 1 & 1 \\ + 2 & 4 & 9 + \end{array}\right] \ \ b = \left[\begin{array}{c} + 1 \\ -6 \\ 2 + \end{array}\right] $$ + + The only simplification performed is to convert + ``Eq(a, b)`` $\Rightarrow a - b$. + + Raises + ====== + + NonlinearError + The equations contain a nonlinear term. + ValueError + The symbols are not given or are not unique. + + Examples + ======== + + >>> from sympy import linear_eq_to_matrix, symbols + >>> c, x, y, z = symbols('c, x, y, z') + + The coefficients (numerical or symbolic) of the symbols will + be returned as matrices: + + >>> eqns = [c*x + z - 1 - c, y + z, x - y] + >>> A, b = linear_eq_to_matrix(eqns, [x, y, z]) + >>> A + Matrix([ + [c, 0, 1], + [0, 1, 1], + [1, -1, 0]]) + >>> b + Matrix([ + [c + 1], + [ 0], + [ 0]]) + + This routine does not simplify expressions and will raise an error + if nonlinearity is encountered: + + >>> eqns = [ + ... (x**2 - 3*x)/(x - 3) - 3, + ... y**2 - 3*y - y*(y - 4) + x - 4] + >>> linear_eq_to_matrix(eqns, [x, y]) + Traceback (most recent call last): + ... + NonlinearError: + symbol-dependent term can be ignored using `strict=False` + + Simplifying these equations will discard the removable singularity + in the first and reveal the linear structure of the second: + + >>> [e.simplify() for e in eqns] + [x - 3, x + y - 4] + + Any such simplification needed to eliminate nonlinear terms must + be done *before* calling this routine. + """ + if not symbols: + raise ValueError(filldedent(''' + Symbols must be given, for which coefficients + are to be found. + ''')) + + # Check if 'symbols' is a set and raise an error if it is + if isinstance(symbols[0], set): + raise TypeError( + "Unordered 'set' type is not supported as input for symbols.") + + if hasattr(symbols[0], '__iter__'): + symbols = symbols[0] + + if has_dups(symbols): + raise ValueError('Symbols must be unique') + + equations = sympify(equations) + if isinstance(equations, MatrixBase): + equations = list(equations) + elif isinstance(equations, (Expr, Eq)): + equations = [equations] + elif not is_sequence(equations): + raise ValueError(filldedent(''' + Equation(s) must be given as a sequence, Expr, + Eq or Matrix. + ''')) + + # construct the dictionaries + try: + eq, c = _linear_eq_to_dict(equations, symbols) + except PolyNonlinearError as err: + raise NonlinearError(str(err)) + # prepare output matrices + n, m = shape = len(eq), len(symbols) + ix = dict(zip(symbols, range(m))) + A = zeros(*shape) + for row, d in enumerate(eq): + for k in d: + col = ix[k] + A[row, col] = d[k] + b = Matrix(n, 1, [-i for i in c]) + return A, b + + +def linsolve(system, *symbols): + r""" + Solve system of $N$ linear equations with $M$ variables; both + underdetermined and overdetermined systems are supported. + The possible number of solutions is zero, one or infinite. + Zero solutions throws a ValueError, whereas infinite + solutions are represented parametrically in terms of the given + symbols. For unique solution a :class:`~.FiniteSet` of ordered tuples + is returned. + + All standard input formats are supported: + For the given set of equations, the respective input types + are given below: + + .. math:: 3x + 2y - z = 1 + .. math:: 2x - 2y + 4z = -2 + .. math:: 2x - y + 2z = 0 + + * Augmented matrix form, ``system`` given below: + + $$ \text{system} = \left[{array}{cccc} + 3 & 2 & -1 & 1\\ + 2 & -2 & 4 & -2\\ + 2 & -1 & 2 & 0 + \end{array}\right] $$ + + :: + + system = Matrix([[3, 2, -1, 1], [2, -2, 4, -2], [2, -1, 2, 0]]) + + * List of equations form + + :: + + system = [3x + 2y - z - 1, 2x - 2y + 4z + 2, 2x - y + 2z] + + * Input $A$ and $b$ in matrix form (from $Ax = b$) are given as: + + $$ A = \left[\begin{array}{ccc} + 3 & 2 & -1 \\ + 2 & -2 & 4 \\ + 2 & -1 & 2 + \end{array}\right] \ \ b = \left[\begin{array}{c} + 1 \\ -2 \\ 0 + \end{array}\right] $$ + + :: + + A = Matrix([[3, 2, -1], [2, -2, 4], [2, -1, 2]]) + b = Matrix([[1], [-2], [0]]) + system = (A, b) + + Symbols can always be passed but are actually only needed + when 1) a system of equations is being passed and 2) the + system is passed as an underdetermined matrix and one wants + to control the name of the free variables in the result. + An error is raised if no symbols are used for case 1, but if + no symbols are provided for case 2, internally generated symbols + will be provided. When providing symbols for case 2, there should + be at least as many symbols are there are columns in matrix A. + + The algorithm used here is Gauss-Jordan elimination, which + results, after elimination, in a row echelon form matrix. + + Returns + ======= + + A FiniteSet containing an ordered tuple of values for the + unknowns for which the `system` has a solution. (Wrapping + the tuple in FiniteSet is used to maintain a consistent + output format throughout solveset.) + + Returns EmptySet, if the linear system is inconsistent. + + Raises + ====== + + ValueError + The input is not valid. + The symbols are not given. + + Examples + ======== + + >>> from sympy import Matrix, linsolve, symbols + >>> x, y, z = symbols("x, y, z") + >>> A = Matrix([[1, 2, 3], [4, 5, 6], [7, 8, 10]]) + >>> b = Matrix([3, 6, 9]) + >>> A + Matrix([ + [1, 2, 3], + [4, 5, 6], + [7, 8, 10]]) + >>> b + Matrix([ + [3], + [6], + [9]]) + >>> linsolve((A, b), [x, y, z]) + {(-1, 2, 0)} + + * Parametric Solution: In case the system is underdetermined, the + function will return a parametric solution in terms of the given + symbols. Those that are free will be returned unchanged. e.g. in + the system below, `z` is returned as the solution for variable z; + it can take on any value. + + >>> A = Matrix([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) + >>> b = Matrix([3, 6, 9]) + >>> linsolve((A, b), x, y, z) + {(z - 1, 2 - 2*z, z)} + + If no symbols are given, internally generated symbols will be used. + The ``tau0`` in the third position indicates (as before) that the third + variable -- whatever it is named -- can take on any value: + + >>> linsolve((A, b)) + {(tau0 - 1, 2 - 2*tau0, tau0)} + + * List of equations as input + + >>> Eqns = [3*x + 2*y - z - 1, 2*x - 2*y + 4*z + 2, - x + y/2 - z] + >>> linsolve(Eqns, x, y, z) + {(1, -2, -2)} + + * Augmented matrix as input + + >>> aug = Matrix([[2, 1, 3, 1], [2, 6, 8, 3], [6, 8, 18, 5]]) + >>> aug + Matrix([ + [2, 1, 3, 1], + [2, 6, 8, 3], + [6, 8, 18, 5]]) + >>> linsolve(aug, x, y, z) + {(3/10, 2/5, 0)} + + * Solve for symbolic coefficients + + >>> a, b, c, d, e, f = symbols('a, b, c, d, e, f') + >>> eqns = [a*x + b*y - c, d*x + e*y - f] + >>> linsolve(eqns, x, y) + {((-b*f + c*e)/(a*e - b*d), (a*f - c*d)/(a*e - b*d))} + + * A degenerate system returns solution as set of given + symbols. + + >>> system = Matrix(([0, 0, 0], [0, 0, 0], [0, 0, 0])) + >>> linsolve(system, x, y) + {(x, y)} + + * For an empty system linsolve returns empty set + + >>> linsolve([], x) + EmptySet + + * An error is raised if any nonlinearity is detected, even + if it could be removed with expansion + + >>> linsolve([x*(1/x - 1)], x) + Traceback (most recent call last): + ... + NonlinearError: nonlinear term: 1/x + + >>> linsolve([x*(y + 1)], x, y) + Traceback (most recent call last): + ... + NonlinearError: nonlinear cross-term: x*(y + 1) + + >>> linsolve([x**2 - 1], x) + Traceback (most recent call last): + ... + NonlinearError: nonlinear term: x**2 + """ + if not system: + return S.EmptySet + + # If second argument is an iterable + if symbols and hasattr(symbols[0], '__iter__'): + symbols = symbols[0] + sym_gen = isinstance(symbols, GeneratorType) + dup_msg = 'duplicate symbols given' + + + b = None # if we don't get b the input was bad + # unpack system + + if hasattr(system, '__iter__'): + + # 1). (A, b) + if len(system) == 2 and isinstance(system[0], MatrixBase): + A, b = system + + # 2). (eq1, eq2, ...) + if not isinstance(system[0], MatrixBase): + if sym_gen or not symbols: + raise ValueError(filldedent(''' + When passing a system of equations, the explicit + symbols for which a solution is being sought must + be given as a sequence, too. + ''')) + if len(set(symbols)) != len(symbols): + raise ValueError(dup_msg) + + # + # Pass to the sparse solver implemented in polys. It is important + # that we do not attempt to convert the equations to a matrix + # because that would be very inefficient for large sparse systems + # of equations. + # + eqs = system + eqs = [sympify(eq) for eq in eqs] + try: + sol = _linsolve(eqs, symbols) + except PolyNonlinearError as exc: + # e.g. cos(x) contains an element of the set of generators + raise NonlinearError(str(exc)) + + if sol is None: + return S.EmptySet + + sol = FiniteSet(Tuple(*(sol.get(sym, sym) for sym in symbols))) + return sol + + elif isinstance(system, MatrixBase) and not ( + symbols and not isinstance(symbols, GeneratorType) and + isinstance(symbols[0], MatrixBase)): + # 3). A augmented with b + A, b = system[:, :-1], system[:, -1:] + + if b is None: + raise ValueError("Invalid arguments") + if sym_gen: + symbols = [next(symbols) for i in range(A.cols)] + symset = set(symbols) + if any(symset & (A.free_symbols | b.free_symbols)): + raise ValueError(filldedent(''' + At least one of the symbols provided + already appears in the system to be solved. + One way to avoid this is to use Dummy symbols in + the generator, e.g. numbered_symbols('%s', cls=Dummy) + ''' % symbols[0].name.rstrip('1234567890'))) + elif len(symset) != len(symbols): + raise ValueError(dup_msg) + + if not symbols: + symbols = [Dummy() for _ in range(A.cols)] + name = _uniquely_named_symbol('tau', (A, b), + compare=lambda i: str(i).rstrip('1234567890')).name + gen = numbered_symbols(name) + else: + gen = None + + # This is just a wrapper for solve_lin_sys + eqs = [] + rows = A.tolist() + for rowi, bi in zip(rows, b): + terms = [elem * sym for elem, sym in zip(rowi, symbols) if elem] + terms.append(-bi) + eqs.append(Add(*terms)) + + eqs, ring = sympy_eqs_to_ring(eqs, symbols) + sol = solve_lin_sys(eqs, ring, _raw=False) + if sol is None: + return S.EmptySet + #sol = {sym:val for sym, val in sol.items() if sym != val} + sol = FiniteSet(Tuple(*(sol.get(sym, sym) for sym in symbols))) + + if gen is not None: + solsym = sol.free_symbols + rep = {sym: next(gen) for sym in symbols if sym in solsym} + sol = sol.subs(rep) + + return sol + + +############################################################################## +# ------------------------------nonlinsolve ---------------------------------# +############################################################################## + + +def _return_conditionset(eqs, symbols): + # return conditionset + eqs = (Eq(lhs, 0) for lhs in eqs) + condition_set = ConditionSet( + Tuple(*symbols), And(*eqs), S.Complexes**len(symbols)) + return condition_set + + +def substitution(system, symbols, result=[{}], known_symbols=[], + exclude=[], all_symbols=None): + r""" + Solves the `system` using substitution method. It is used in + :func:`~.nonlinsolve`. This will be called from :func:`~.nonlinsolve` when any + equation(s) is non polynomial equation. + + Parameters + ========== + + system : list of equations + The target system of equations + symbols : list of symbols to be solved. + The variable(s) for which the system is solved + known_symbols : list of solved symbols + Values are known for these variable(s) + result : An empty list or list of dict + If No symbol values is known then empty list otherwise + symbol as keys and corresponding value in dict. + exclude : Set of expression. + Mostly denominator expression(s) of the equations of the system. + Final solution should not satisfy these expressions. + all_symbols : known_symbols + symbols(unsolved). + + Returns + ======= + + A FiniteSet of ordered tuple of values of `all_symbols` for which the + `system` has solution. Order of values in the tuple is same as symbols + present in the parameter `all_symbols`. If parameter `all_symbols` is None + then same as symbols present in the parameter `symbols`. + + Please note that general FiniteSet is unordered, the solution returned + here is not simply a FiniteSet of solutions, rather it is a FiniteSet of + ordered tuple, i.e. the first & only argument to FiniteSet is a tuple of + solutions, which is ordered, & hence the returned solution is ordered. + + Also note that solution could also have been returned as an ordered tuple, + FiniteSet is just a wrapper `{}` around the tuple. It has no other + significance except for the fact it is just used to maintain a consistent + output format throughout the solveset. + + Raises + ====== + + ValueError + The input is not valid. + The symbols are not given. + AttributeError + The input symbols are not :class:`~.Symbol` type. + + Examples + ======== + + >>> from sympy import symbols, substitution + >>> x, y = symbols('x, y', real=True) + >>> substitution([x + y], [x], [{y: 1}], [y], set([]), [x, y]) + {(-1, 1)} + + * When you want a soln not satisfying $x + 1 = 0$ + + >>> substitution([x + y], [x], [{y: 1}], [y], set([x + 1]), [y, x]) + EmptySet + >>> substitution([x + y], [x], [{y: 1}], [y], set([x - 1]), [y, x]) + {(1, -1)} + >>> substitution([x + y - 1, y - x**2 + 5], [x, y]) + {(-3, 4), (2, -1)} + + * Returns both real and complex solution + + >>> x, y, z = symbols('x, y, z') + >>> from sympy import exp, sin + >>> substitution([exp(x) - sin(y), y**2 - 4], [x, y]) + {(ImageSet(Lambda(_n, I*(2*_n*pi + pi) + log(sin(2))), Integers), -2), + (ImageSet(Lambda(_n, 2*_n*I*pi + log(sin(2))), Integers), 2)} + + >>> eqs = [z**2 + exp(2*x) - sin(y), -3 + exp(-y)] + >>> substitution(eqs, [y, z]) + {(-log(3), -sqrt(-exp(2*x) - sin(log(3)))), + (-log(3), sqrt(-exp(2*x) - sin(log(3)))), + (ImageSet(Lambda(_n, 2*_n*I*pi - log(3)), Integers), + ImageSet(Lambda(_n, -sqrt(-exp(2*x) + sin(2*_n*I*pi - log(3)))), Integers)), + (ImageSet(Lambda(_n, 2*_n*I*pi - log(3)), Integers), + ImageSet(Lambda(_n, sqrt(-exp(2*x) + sin(2*_n*I*pi - log(3)))), Integers))} + + """ + + if not system: + return S.EmptySet + + for i, e in enumerate(system): + if isinstance(e, Eq): + system[i] = e.lhs - e.rhs + + if not symbols: + msg = ('Symbols must be given, for which solution of the ' + 'system is to be found.') + raise ValueError(filldedent(msg)) + + if not is_sequence(symbols): + msg = ('symbols should be given as a sequence, e.g. a list.' + 'Not type %s: %s') + raise TypeError(filldedent(msg % (type(symbols), symbols))) + + if not getattr(symbols[0], 'is_Symbol', False): + msg = ('Iterable of symbols must be given as ' + 'second argument, not type %s: %s') + raise ValueError(filldedent(msg % (type(symbols[0]), symbols[0]))) + + # By default `all_symbols` will be same as `symbols` + if all_symbols is None: + all_symbols = symbols + + old_result = result + # storing complements and intersection for particular symbol + complements = {} + intersections = {} + + # when total_solveset_call equals total_conditionset + # it means that solveset failed to solve all eqs. + total_conditionset = -1 + total_solveset_call = -1 + + def _unsolved_syms(eq, sort=False): + """Returns the unsolved symbol present + in the equation `eq`. + """ + free = eq.free_symbols + unsolved = (free - set(known_symbols)) & set(all_symbols) + if sort: + unsolved = list(unsolved) + unsolved.sort(key=default_sort_key) + return unsolved + + # sort such that equation with the fewest potential symbols is first. + # means eq with less number of variable first in the list. + eqs_in_better_order = list( + ordered(system, lambda _: len(_unsolved_syms(_)))) + + def add_intersection_complement(result, intersection_dict, complement_dict): + # If solveset has returned some intersection/complement + # for any symbol, it will be added in the final solution. + final_result = [] + for res in result: + res_copy = res + for key_res, value_res in res.items(): + intersect_set, complement_set = None, None + for key_sym, value_sym in intersection_dict.items(): + if key_sym == key_res: + intersect_set = value_sym + for key_sym, value_sym in complement_dict.items(): + if key_sym == key_res: + complement_set = value_sym + if intersect_set or complement_set: + new_value = FiniteSet(value_res) + if intersect_set and intersect_set != S.Complexes: + new_value = Intersection(new_value, intersect_set) + if complement_set: + new_value = Complement(new_value, complement_set) + if new_value is S.EmptySet: + res_copy = None + break + elif new_value.is_FiniteSet and len(new_value) == 1: + res_copy[key_res] = set(new_value).pop() + else: + res_copy[key_res] = new_value + + if res_copy is not None: + final_result.append(res_copy) + return final_result + + def _extract_main_soln(sym, sol, soln_imageset): + """Separate the Complements, Intersections, ImageSet lambda expr and + its base_set. This function returns the unmasked sol from different classes + of sets and also returns the appended ImageSet elements in a + soln_imageset dict: `{unmasked element: ImageSet}`. + """ + # if there is union, then need to check + # Complement, Intersection, Imageset. + # Order should not be changed. + if isinstance(sol, ConditionSet): + # extracts any solution in ConditionSet + sol = sol.base_set + + if isinstance(sol, Complement): + # extract solution and complement + complements[sym] = sol.args[1] + sol = sol.args[0] + # complement will be added at the end + # using `add_intersection_complement` method + + # if there is union of Imageset or other in soln. + # no testcase is written for this if block + if isinstance(sol, Union): + sol_args = sol.args + sol = S.EmptySet + # We need in sequence so append finteset elements + # and then imageset or other. + for sol_arg2 in sol_args: + if isinstance(sol_arg2, FiniteSet): + sol += sol_arg2 + else: + # ImageSet, Intersection, complement then + # append them directly + sol += FiniteSet(sol_arg2) + + if isinstance(sol, Intersection): + # Interval/Set will be at 0th index always + if sol.args[0] not in (S.Reals, S.Complexes): + # Sometimes solveset returns soln with intersection + # S.Reals or S.Complexes. We don't consider that + # intersection. + intersections[sym] = sol.args[0] + sol = sol.args[1] + # after intersection and complement Imageset should + # be checked. + if isinstance(sol, ImageSet): + soln_imagest = sol + expr2 = sol.lamda.expr + sol = FiniteSet(expr2) + soln_imageset[expr2] = soln_imagest + + if not isinstance(sol, FiniteSet): + sol = FiniteSet(sol) + return sol, soln_imageset + + def _check_exclude(rnew, imgset_yes): + rnew_ = rnew + if imgset_yes: + # replace all dummy variables (Imageset lambda variables) + # with zero before `checksol`. Considering fundamental soln + # for `checksol`. + rnew_copy = rnew.copy() + dummy_n = imgset_yes[0] + for key_res, value_res in rnew_copy.items(): + rnew_copy[key_res] = value_res.subs(dummy_n, 0) + rnew_ = rnew_copy + # satisfy_exclude == true if it satisfies the expr of `exclude` list. + try: + # something like : `Mod(-log(3), 2*I*pi)` can't be + # simplified right now, so `checksol` returns `TypeError`. + # when this issue is fixed this try block should be + # removed. Mod(-log(3), 2*I*pi) == -log(3) + satisfy_exclude = any( + checksol(d, rnew_) for d in exclude) + except TypeError: + satisfy_exclude = None + return satisfy_exclude + + def _restore_imgset(rnew, original_imageset, newresult): + restore_sym = set(rnew.keys()) & \ + set(original_imageset.keys()) + for key_sym in restore_sym: + img = original_imageset[key_sym] + rnew[key_sym] = img + if rnew not in newresult: + newresult.append(rnew) + + def _append_eq(eq, result, res, delete_soln, n=None): + u = Dummy('u') + if n: + eq = eq.subs(n, 0) + satisfy = eq if eq in (True, False) else checksol(u, u, eq, minimal=True) + if satisfy is False: + delete_soln = True + res = {} + else: + result.append(res) + return result, res, delete_soln + + def _append_new_soln(rnew, sym, sol, imgset_yes, soln_imageset, + original_imageset, newresult, eq=None): + """If `rnew` (A dict ) contains valid soln + append it to `newresult` list. + `imgset_yes` is (base, dummy_var) if there was imageset in previously + calculated result(otherwise empty tuple). `original_imageset` is dict + of imageset expr and imageset from this result. + `soln_imageset` dict of imageset expr and imageset of new soln. + """ + satisfy_exclude = _check_exclude(rnew, imgset_yes) + delete_soln = False + # soln should not satisfy expr present in `exclude` list. + if not satisfy_exclude: + local_n = None + # if it is imageset + if imgset_yes: + local_n = imgset_yes[0] + base = imgset_yes[1] + if sym and sol: + # when `sym` and `sol` is `None` means no new + # soln. In that case we will append rnew directly after + # substituting original imagesets in rnew values if present + # (second last line of this function using _restore_imgset) + dummy_list = list(sol.atoms(Dummy)) + # use one dummy `n` which is in + # previous imageset + local_n_list = [ + local_n for i in range( + 0, len(dummy_list))] + + dummy_zip = zip(dummy_list, local_n_list) + lam = Lambda(local_n, sol.subs(dummy_zip)) + rnew[sym] = ImageSet(lam, base) + if eq is not None: + newresult, rnew, delete_soln = _append_eq( + eq, newresult, rnew, delete_soln, local_n) + elif eq is not None: + newresult, rnew, delete_soln = _append_eq( + eq, newresult, rnew, delete_soln) + elif sol in soln_imageset.keys(): + rnew[sym] = soln_imageset[sol] + # restore original imageset + _restore_imgset(rnew, original_imageset, newresult) + else: + newresult.append(rnew) + elif satisfy_exclude: + delete_soln = True + rnew = {} + _restore_imgset(rnew, original_imageset, newresult) + return newresult, delete_soln + + def _new_order_result(result, eq): + # separate first, second priority. `res` that makes `eq` value equals + # to zero, should be used first then other result(second priority). + # If it is not done then we may miss some soln. + first_priority = [] + second_priority = [] + for res in result: + if not any(isinstance(val, ImageSet) for val in res.values()): + if eq.subs(res) == 0: + first_priority.append(res) + else: + second_priority.append(res) + if first_priority or second_priority: + return first_priority + second_priority + return result + + def _solve_using_known_values(result, solver): + """Solves the system using already known solution + (result contains the dict ). + solver is :func:`~.solveset_complex` or :func:`~.solveset_real`. + """ + # stores imageset . + soln_imageset = {} + total_solvest_call = 0 + total_conditionst = 0 + + # sort equations so the one with the fewest potential + # symbols appears first + for index, eq in enumerate(eqs_in_better_order): + newresult = [] + # if imageset, expr is used to solve for other symbol + imgset_yes = False + for res in result: + original_imageset = {} + got_symbol = set() # symbols solved in one iteration + # find the imageset and use its expr. + for k, v in res.items(): + if isinstance(v, ImageSet): + res[k] = v.lamda.expr + original_imageset[k] = v + dummy_n = v.lamda.expr.atoms(Dummy).pop() + (base,) = v.base_sets + imgset_yes = (dummy_n, base) + assert not isinstance(v, FiniteSet) # if so, internal error + # update eq with everything that is known so far + eq2 = eq.subs(res).expand() + if imgset_yes and not eq2.has(imgset_yes[0]): + # The substituted equation simplified in such a way that + # it's no longer necessary to encapsulate a potential new + # solution in an ImageSet. (E.g. at the previous step some + # {n*2*pi} was found as partial solution for one of the + # unknowns, but its main solution expression n*2*pi has now + # been substituted in a trigonometric function.) + imgset_yes = False + + unsolved_syms = _unsolved_syms(eq2, sort=True) + if not unsolved_syms: + if res: + newresult, delete_res = _append_new_soln( + res, None, None, imgset_yes, soln_imageset, + original_imageset, newresult, eq2) + if delete_res: + # `delete_res` is true, means substituting `res` in + # eq2 doesn't return `zero` or deleting the `res` + # (a soln) since it satisfies expr of `exclude` + # list. + result.remove(res) + continue # skip as it's independent of desired symbols + depen1, depen2 = eq2.as_independent(*unsolved_syms) + if (depen1.has(Abs) or depen2.has(Abs)) and solver == solveset_complex: + # Absolute values cannot be inverted in the + # complex domain + continue + soln_imageset = {} + for sym in unsolved_syms: + not_solvable = False + try: + soln = solver(eq2, sym) + total_solvest_call += 1 + soln_new = S.EmptySet + if isinstance(soln, Complement): + # separate solution and complement + complements[sym] = soln.args[1] + soln = soln.args[0] + # complement will be added at the end + if isinstance(soln, Intersection): + # Interval will be at 0th index always + if soln.args[0] != Interval(-oo, oo): + # sometimes solveset returns soln + # with intersection S.Reals, to confirm that + # soln is in domain=S.Reals + intersections[sym] = soln.args[0] + soln_new += soln.args[1] + soln = soln_new if soln_new else soln + if index > 0 and solver == solveset_real: + # one symbol's real soln, another symbol may have + # corresponding complex soln. + if not isinstance(soln, (ImageSet, ConditionSet)): + soln += solveset_complex(eq2, sym) # might give ValueError with Abs + except (NotImplementedError, ValueError): + # If solveset is not able to solve equation `eq2`. Next + # time we may get soln using next equation `eq2` + continue + if isinstance(soln, ConditionSet): + if soln.base_set in (S.Reals, S.Complexes): + soln = S.EmptySet + # don't do `continue` we may get soln + # in terms of other symbol(s) + not_solvable = True + total_conditionst += 1 + else: + soln = soln.base_set + + if soln is not S.EmptySet: + soln, soln_imageset = _extract_main_soln( + sym, soln, soln_imageset) + + for sol in soln: + # sol is not a `Union` since we checked it + # before this loop + sol, soln_imageset = _extract_main_soln( + sym, sol, soln_imageset) + sol = set(sol).pop() # XXX what if there are more solutions? + free = sol.free_symbols + if got_symbol and any( + ss in free for ss in got_symbol + ): + # sol depends on previously solved symbols + # then continue + continue + rnew = res.copy() + # put each solution in res and append the new result + # in the new result list (solution for symbol `s`) + # along with old results. + for k, v in res.items(): + if isinstance(v, Expr) and isinstance(sol, Expr): + # if any unsolved symbol is present + # Then subs known value + rnew[k] = v.subs(sym, sol) + # and add this new solution + if sol in soln_imageset.keys(): + # replace all lambda variables with 0. + imgst = soln_imageset[sol] + rnew[sym] = imgst.lamda( + *[0 for i in range(0, len( + imgst.lamda.variables))]) + else: + rnew[sym] = sol + newresult, delete_res = _append_new_soln( + rnew, sym, sol, imgset_yes, soln_imageset, + original_imageset, newresult) + if delete_res: + # deleting the `res` (a soln) since it satisfies + # eq of `exclude` list + result.remove(res) + # solution got for sym + if not not_solvable: + got_symbol.add(sym) + # next time use this new soln + if newresult: + result = newresult + return result, total_solvest_call, total_conditionst + + new_result_real, solve_call1, cnd_call1 = _solve_using_known_values( + old_result, solveset_real) + new_result_complex, solve_call2, cnd_call2 = _solve_using_known_values( + old_result, solveset_complex) + + # If total_solveset_call is equal to total_conditionset + # then solveset failed to solve all of the equations. + # In this case we return a ConditionSet here. + total_conditionset += (cnd_call1 + cnd_call2) + total_solveset_call += (solve_call1 + solve_call2) + + if total_conditionset == total_solveset_call and total_solveset_call != -1: + return _return_conditionset(eqs_in_better_order, all_symbols) + + # don't keep duplicate solutions + filtered_complex = [] + for i in list(new_result_complex): + for j in list(new_result_real): + if i.keys() != j.keys(): + continue + if all(a.dummy_eq(b) for a, b in zip(i.values(), j.values()) \ + if not (isinstance(a, int) and isinstance(b, int))): + break + else: + filtered_complex.append(i) + # overall result + result = new_result_real + filtered_complex + + result_all_variables = [] + result_infinite = [] + for res in result: + if not res: + # means {None : None} + continue + # If length < len(all_symbols) means infinite soln. + # Some or all the soln is dependent on 1 symbol. + # eg. {x: y+2} then final soln {x: y+2, y: y} + if len(res) < len(all_symbols): + solved_symbols = res.keys() + unsolved = list(filter( + lambda x: x not in solved_symbols, all_symbols)) + for unsolved_sym in unsolved: + res[unsolved_sym] = unsolved_sym + result_infinite.append(res) + if res not in result_all_variables: + result_all_variables.append(res) + + if result_infinite: + # we have general soln + # eg : [{x: -1, y : 1}, {x : -y, y: y}] then + # return [{x : -y, y : y}] + result_all_variables = result_infinite + if intersections or complements: + result_all_variables = add_intersection_complement( + result_all_variables, intersections, complements) + + # convert to ordered tuple + result = S.EmptySet + for r in result_all_variables: + temp = [r[symb] for symb in all_symbols] + result += FiniteSet(tuple(temp)) + return result + + +def _solveset_work(system, symbols): + soln = solveset(system[0], symbols[0]) + if isinstance(soln, FiniteSet): + _soln = FiniteSet(*[(s,) for s in soln]) + return _soln + else: + return FiniteSet(tuple(FiniteSet(soln))) + + +def _handle_positive_dimensional(polys, symbols, denominators): + from sympy.polys.polytools import groebner + # substitution method where new system is groebner basis of the system + _symbols = list(symbols) + _symbols.sort(key=default_sort_key) + basis = groebner(polys, _symbols, polys=True) + new_system = [] + for poly_eq in basis: + new_system.append(poly_eq.as_expr()) + result = [{}] + result = substitution( + new_system, symbols, result, [], + denominators) + return result + + +def _handle_zero_dimensional(polys, symbols, system): + # solve 0 dimensional poly system using `solve_poly_system` + result = solve_poly_system(polys, *symbols) + # May be some extra soln is added because + # we used `unrad` in `_separate_poly_nonpoly`, so + # need to check and remove if it is not a soln. + result_update = S.EmptySet + for res in result: + dict_sym_value = dict(list(zip(symbols, res))) + if all(checksol(eq, dict_sym_value) for eq in system): + result_update += FiniteSet(res) + return result_update + + +def _separate_poly_nonpoly(system, symbols): + polys = [] + polys_expr = [] + nonpolys = [] + # unrad_changed stores a list of expressions containing + # radicals that were processed using unrad + # this is useful if solutions need to be checked later. + unrad_changed = [] + denominators = set() + poly = None + for eq in system: + # Store denom expressions that contain symbols + denominators.update(_simple_dens(eq, symbols)) + # Convert equality to expression + if isinstance(eq, Eq): + eq = eq.lhs - eq.rhs + # try to remove sqrt and rational power + without_radicals = unrad(simplify(eq), *symbols) + if without_radicals: + unrad_changed.append(eq) + eq_unrad, cov = without_radicals + if not cov: + eq = eq_unrad + if isinstance(eq, Expr): + eq = eq.as_numer_denom()[0] + poly = eq.as_poly(*symbols, extension=True) + elif simplify(eq).is_number: + continue + if poly is not None: + polys.append(poly) + polys_expr.append(poly.as_expr()) + else: + nonpolys.append(eq) + return polys, polys_expr, nonpolys, denominators, unrad_changed + + +def _handle_poly(polys, symbols): + # _handle_poly(polys, symbols) -> (poly_sol, poly_eqs) + # + # We will return possible solution information to nonlinsolve as well as a + # new system of polynomial equations to be solved if we cannot solve + # everything directly here. The new system of polynomial equations will be + # a lex-order Groebner basis for the original system. The lex basis + # hopefully separate some of the variables and equations and give something + # easier for substitution to work with. + + # The format for representing solution sets in nonlinsolve and substitution + # is a list of dicts. These are the special cases: + no_information = [{}] # No equations solved yet + no_solutions = [] # The system is inconsistent and has no solutions. + + # If there is no need to attempt further solution of these equations then + # we return no equations: + no_equations = [] + + inexact = any(not p.domain.is_Exact for p in polys) + if inexact: + # The use of Groebner over RR is likely to result incorrectly in an + # inconsistent Groebner basis. So, convert any float coefficients to + # Rational before computing the Groebner basis. + polys = [poly(nsimplify(p, rational=True)) for p in polys] + + # Compute a Groebner basis in grevlex order wrt the ordering given. We will + # try to convert this to lex order later. Usually it seems to be more + # efficient to compute a lex order basis by computing a grevlex basis and + # converting to lex with fglm. + basis = groebner(polys, symbols, order='grevlex', polys=False) + + # + # No solutions (inconsistent equations)? + # + if 1 in basis: + + # No solutions: + poly_sol = no_solutions + poly_eqs = no_equations + + # + # Finite number of solutions (zero-dimensional case) + # + elif basis.is_zero_dimensional: + + # Convert Groebner basis to lex ordering + basis = basis.fglm('lex') + + # Convert polynomial coefficients back to float before calling + # solve_poly_system + if inexact: + basis = [nfloat(p) for p in basis] + + # Solve the zero-dimensional case using solve_poly_system if possible. + # If some polynomials have factors that cannot be solved in radicals + # then this will fail. Using solve_poly_system(..., strict=True) + # ensures that we either get a complete solution set in radicals or + # UnsolvableFactorError will be raised. + try: + result = solve_poly_system(basis, *symbols, strict=True) + except UnsolvableFactorError: + # Failure... not fully solvable in radicals. Return the lex-order + # basis for substitution to handle. + poly_sol = no_information + poly_eqs = list(basis) + else: + # Success! We have a finite solution set and solve_poly_system has + # succeeded in finding all solutions. Return the solutions and also + # an empty list of remaining equations to be solved. + poly_sol = [dict(zip(symbols, res)) for res in result] + poly_eqs = no_equations + + # + # Infinite families of solutions (positive-dimensional case) + # + else: + # In this case the grevlex basis cannot be converted to lex using the + # fglm method and also solve_poly_system cannot solve the equations. We + # would like to return a lex basis but since we can't use fglm we + # compute the lex basis directly here. The time required to recompute + # the basis is generally significantly less than the time required by + # substitution to solve the new system. + poly_sol = no_information + poly_eqs = list(groebner(polys, symbols, order='lex', polys=False)) + + if inexact: + poly_eqs = [nfloat(p) for p in poly_eqs] + + return poly_sol, poly_eqs + + +def nonlinsolve(system, *symbols): + r""" + Solve system of $N$ nonlinear equations with $M$ variables, which means both + under and overdetermined systems are supported. Positive dimensional + system is also supported (A system with infinitely many solutions is said + to be positive-dimensional). In a positive dimensional system the solution will + be dependent on at least one symbol. Returns both real solution + and complex solution (if they exist). + + Parameters + ========== + + system : list of equations + The target system of equations + symbols : list of Symbols + symbols should be given as a sequence eg. list + + Returns + ======= + + A :class:`~.FiniteSet` of ordered tuple of values of `symbols` for which the `system` + has solution. Order of values in the tuple is same as symbols present in + the parameter `symbols`. + + Please note that general :class:`~.FiniteSet` is unordered, the solution + returned here is not simply a :class:`~.FiniteSet` of solutions, rather it + is a :class:`~.FiniteSet` of ordered tuple, i.e. the first and only + argument to :class:`~.FiniteSet` is a tuple of solutions, which is + ordered, and, hence ,the returned solution is ordered. + + Also note that solution could also have been returned as an ordered tuple, + FiniteSet is just a wrapper ``{}`` around the tuple. It has no other + significance except for the fact it is just used to maintain a consistent + output format throughout the solveset. + + For the given set of equations, the respective input types + are given below: + + .. math:: xy - 1 = 0 + .. math:: 4x^2 + y^2 - 5 = 0 + + :: + + system = [x*y - 1, 4*x**2 + y**2 - 5] + symbols = [x, y] + + Raises + ====== + + ValueError + The input is not valid. + The symbols are not given. + AttributeError + The input symbols are not `Symbol` type. + + Examples + ======== + + >>> from sympy import symbols, nonlinsolve + >>> x, y, z = symbols('x, y, z', real=True) + >>> nonlinsolve([x*y - 1, 4*x**2 + y**2 - 5], [x, y]) + {(-1, -1), (-1/2, -2), (1/2, 2), (1, 1)} + + 1. Positive dimensional system and complements: + + >>> from sympy import pprint + >>> from sympy.polys.polytools import is_zero_dimensional + >>> a, b, c, d = symbols('a, b, c, d', extended_real=True) + >>> eq1 = a + b + c + d + >>> eq2 = a*b + b*c + c*d + d*a + >>> eq3 = a*b*c + b*c*d + c*d*a + d*a*b + >>> eq4 = a*b*c*d - 1 + >>> system = [eq1, eq2, eq3, eq4] + >>> is_zero_dimensional(system) + False + >>> pprint(nonlinsolve(system, [a, b, c, d]), use_unicode=False) + -1 1 1 -1 + {(---, -d, -, {d} \ {0}), (-, -d, ---, {d} \ {0})} + d d d d + >>> nonlinsolve([(x+y)**2 - 4, x + y - 2], [x, y]) + {(2 - y, y)} + + 2. If some of the equations are non-polynomial then `nonlinsolve` + will call the ``substitution`` function and return real and complex solutions, + if present. + + >>> from sympy import exp, sin + >>> nonlinsolve([exp(x) - sin(y), y**2 - 4], [x, y]) + {(ImageSet(Lambda(_n, I*(2*_n*pi + pi) + log(sin(2))), Integers), -2), + (ImageSet(Lambda(_n, 2*_n*I*pi + log(sin(2))), Integers), 2)} + + 3. If system is non-linear polynomial and zero-dimensional then it + returns both solution (real and complex solutions, if present) using + :func:`~.solve_poly_system`: + + >>> from sympy import sqrt + >>> nonlinsolve([x**2 - 2*y**2 -2, x*y - 2], [x, y]) + {(-2, -1), (2, 1), (-sqrt(2)*I, sqrt(2)*I), (sqrt(2)*I, -sqrt(2)*I)} + + 4. ``nonlinsolve`` can solve some linear (zero or positive dimensional) + system (because it uses the :func:`sympy.polys.polytools.groebner` function to get the + groebner basis and then uses the ``substitution`` function basis as the + new `system`). But it is not recommended to solve linear system using + ``nonlinsolve``, because :func:`~.linsolve` is better for general linear systems. + + >>> nonlinsolve([x + 2*y -z - 3, x - y - 4*z + 9, y + z - 4], [x, y, z]) + {(3*z - 5, 4 - z, z)} + + 5. System having polynomial equations and only real solution is + solved using :func:`~.solve_poly_system`: + + >>> e1 = sqrt(x**2 + y**2) - 10 + >>> e2 = sqrt(y**2 + (-x + 10)**2) - 3 + >>> nonlinsolve((e1, e2), (x, y)) + {(191/20, -3*sqrt(391)/20), (191/20, 3*sqrt(391)/20)} + >>> nonlinsolve([x**2 + 2/y - 2, x + y - 3], [x, y]) + {(1, 2), (1 - sqrt(5), 2 + sqrt(5)), (1 + sqrt(5), 2 - sqrt(5))} + >>> nonlinsolve([x**2 + 2/y - 2, x + y - 3], [y, x]) + {(2, 1), (2 - sqrt(5), 1 + sqrt(5)), (2 + sqrt(5), 1 - sqrt(5))} + + 6. It is better to use symbols instead of trigonometric functions or + :class:`~.Function`. For example, replace $\sin(x)$ with a symbol, replace + $f(x)$ with a symbol and so on. Get a solution from ``nonlinsolve`` and then + use :func:`~.solveset` to get the value of $x$. + + How nonlinsolve is better than old solver ``_solve_system`` : + ============================================================= + + 1. A positive dimensional system solver: nonlinsolve can return + solution for positive dimensional system. It finds the + Groebner Basis of the positive dimensional system(calling it as + basis) then we can start solving equation(having least number of + variable first in the basis) using solveset and substituting that + solved solutions into other equation(of basis) to get solution in + terms of minimum variables. Here the important thing is how we + are substituting the known values and in which equations. + + 2. Real and complex solutions: nonlinsolve returns both real + and complex solution. If all the equations in the system are polynomial + then using :func:`~.solve_poly_system` both real and complex solution is returned. + If all the equations in the system are not polynomial equation then goes to + ``substitution`` method with this polynomial and non polynomial equation(s), + to solve for unsolved variables. Here to solve for particular variable + solveset_real and solveset_complex is used. For both real and complex + solution ``_solve_using_known_values`` is used inside ``substitution`` + (``substitution`` will be called when any non-polynomial equation is present). + If a solution is valid its general solution is added to the final result. + + 3. :class:`~.Complement` and :class:`~.Intersection` will be added: + nonlinsolve maintains dict for complements and intersections. If solveset + find complements or/and intersections with any interval or set during the + execution of ``substitution`` function, then complement or/and + intersection for that variable is added before returning final solution. + + """ + if not system: + return S.EmptySet + + if not symbols: + msg = ('Symbols must be given, for which solution of the ' + 'system is to be found.') + raise ValueError(filldedent(msg)) + + if hasattr(symbols[0], '__iter__'): + symbols = symbols[0] + + if not is_sequence(symbols) or not symbols: + msg = ('Symbols must be given, for which solution of the ' + 'system is to be found.') + raise IndexError(filldedent(msg)) + + symbols = list(map(_sympify, symbols)) + system, symbols, swap = recast_to_symbols(system, symbols) + if swap: + soln = nonlinsolve(system, symbols) + return FiniteSet(*[tuple(i.xreplace(swap) for i in s) for s in soln]) + + if len(system) == 1 and len(symbols) == 1: + return _solveset_work(system, symbols) + + # main code of def nonlinsolve() starts from here + + polys, polys_expr, nonpolys, denominators, unrad_changed = \ + _separate_poly_nonpoly(system, symbols) + + poly_eqs = [] + poly_sol = [{}] + + if polys: + poly_sol, poly_eqs = _handle_poly(polys, symbols) + if poly_sol and poly_sol[0]: + poly_syms = set().union(*(eq.free_symbols for eq in polys)) + unrad_syms = set().union(*(eq.free_symbols for eq in unrad_changed)) + if unrad_syms == poly_syms and unrad_changed: + # if all the symbols have been solved by _handle_poly + # and unrad has been used then check solutions + poly_sol = [sol for sol in poly_sol if checksol(unrad_changed, sol)] + + # Collect together the unsolved polynomials with the non-polynomial + # equations. + remaining = poly_eqs + nonpolys + + # to_tuple converts a solution dictionary to a tuple containing the + # value for each symbol + to_tuple = lambda sol: tuple(sol[s] for s in symbols) + + if not remaining: + # If there is nothing left to solve then return the solution from + # solve_poly_system directly. + return FiniteSet(*map(to_tuple, poly_sol)) + else: + # Here we handle: + # + # 1. The Groebner basis if solve_poly_system failed. + # 2. The Groebner basis in the positive-dimensional case. + # 3. Any non-polynomial equations + # + # If solve_poly_system did succeed then we pass those solutions in as + # preliminary results. + subs_res = substitution(remaining, symbols, result=poly_sol, exclude=denominators) + + if not isinstance(subs_res, FiniteSet): + return subs_res + + # check solutions produced by substitution. 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Function +from sympy.core.numbers import I +from sympy.core.power import Pow +from sympy.core.relational import Eq +from sympy.core.singleton import S +from sympy.core.symbol import Symbol +from sympy.functions.elementary.exponential import (exp, log) +from sympy.functions.elementary.hyperbolic import (cosh, sinh) +from sympy.functions.elementary.miscellaneous import sqrt +from sympy.functions.elementary.trigonometric import (acos, cos, sin) +from sympy.integrals.integrals import Integral +from sympy.solvers.ode.ode import constantsimp, constant_renumber +from sympy.testing.pytest import XFAIL + + +x = Symbol('x') +y = Symbol('y') +z = Symbol('z') +u2 = Symbol('u2') +_a = Symbol('_a') +C1 = Symbol('C1') +C2 = Symbol('C2') +C3 = Symbol('C3') +f = Function('f') + + +def test_constant_mul(): + # We want C1 (Constant) below to absorb the y's, but not the x's + assert constant_renumber(constantsimp(y*C1, [C1])) == C1*y + assert constant_renumber(constantsimp(C1*y, [C1])) == C1*y + assert constant_renumber(constantsimp(x*C1, [C1])) == x*C1 + assert constant_renumber(constantsimp(C1*x, [C1])) == x*C1 + assert constant_renumber(constantsimp(2*C1, [C1])) == C1 + assert constant_renumber(constantsimp(C1*2, [C1])) == C1 + assert constant_renumber(constantsimp(y*C1*x, [C1, y])) == C1*x + assert constant_renumber(constantsimp(x*y*C1, [C1, y])) == x*C1 + assert constant_renumber(constantsimp(y*x*C1, [C1, y])) == x*C1 + assert constant_renumber(constantsimp(C1*x*y, [C1, y])) == C1*x + assert constant_renumber(constantsimp(x*C1*y, [C1, y])) == x*C1 + assert constant_renumber(constantsimp(C1*y*(y + 1), [C1])) == C1*y*(y+1) + assert constant_renumber(constantsimp(y*C1*(y + 1), [C1])) == C1*y*(y+1) + assert constant_renumber(constantsimp(x*(y*C1), [C1])) == x*y*C1 + assert constant_renumber(constantsimp(x*(C1*y), [C1])) == x*y*C1 + assert constant_renumber(constantsimp(C1*(x*y), [C1, y])) == C1*x + assert constant_renumber(constantsimp((x*y)*C1, [C1, y])) == x*C1 + assert constant_renumber(constantsimp((y*x)*C1, [C1, y])) == x*C1 + assert constant_renumber(constantsimp(y*(y + 1)*C1, [C1, y])) == C1 + assert constant_renumber(constantsimp((C1*x)*y, [C1, y])) == C1*x + assert constant_renumber(constantsimp(y*(x*C1), [C1, y])) == x*C1 + assert constant_renumber(constantsimp((x*C1)*y, [C1, y])) == x*C1 + assert constant_renumber(constantsimp(C1*x*y*x*y*2, [C1, y])) == C1*x**2 + assert constant_renumber(constantsimp(C1*x*y*z, [C1, y, z])) == C1*x + assert constant_renumber(constantsimp(C1*x*y**2*sin(z), [C1, y, z])) == C1*x + assert constant_renumber(constantsimp(C1*C1, [C1])) == C1 + assert constant_renumber(constantsimp(C1*C2, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C2*C2, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C1*C1*C2, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C1*x*2**x, [C1])) == C1*x*2**x + +def test_constant_add(): + assert constant_renumber(constantsimp(C1 + C1, [C1])) == C1 + assert constant_renumber(constantsimp(C1 + 2, [C1])) == C1 + assert constant_renumber(constantsimp(2 + C1, [C1])) == C1 + assert constant_renumber(constantsimp(C1 + y, [C1, y])) == C1 + assert constant_renumber(constantsimp(C1 + x, [C1])) == C1 + x + assert constant_renumber(constantsimp(C1 + C1, [C1])) == C1 + assert constant_renumber(constantsimp(C1 + C2, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C2 + C1, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C1 + C2 + C1, [C1, C2])) == C1 + + +def test_constant_power_as_base(): + assert constant_renumber(constantsimp(C1**C1, [C1])) == C1 + assert constant_renumber(constantsimp(Pow(C1, C1), [C1])) == C1 + assert constant_renumber(constantsimp(C1**C1, [C1])) == C1 + assert constant_renumber(constantsimp(C1**C2, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C2**C1, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C2**C2, [C1, C2])) == C1 + assert constant_renumber(constantsimp(C1**y, [C1, y])) == C1 + assert constant_renumber(constantsimp(C1**x, [C1])) == C1**x + assert constant_renumber(constantsimp(C1**2, [C1])) == C1 + assert constant_renumber( + constantsimp(C1**(x*y), [C1])) == C1**(x*y) + + +def test_constant_power_as_exp(): + assert constant_renumber(constantsimp(x**C1, [C1])) == x**C1 + assert constant_renumber(constantsimp(y**C1, [C1, y])) == C1 + assert constant_renumber(constantsimp(x**y**C1, [C1, y])) == x**C1 + assert constant_renumber( + constantsimp((x**y)**C1, [C1])) == (x**y)**C1 + assert constant_renumber( + constantsimp(x**(y**C1), [C1, y])) == x**C1 + assert constant_renumber(constantsimp(x**C1**y, [C1, y])) == x**C1 + assert constant_renumber( + constantsimp(x**(C1**y), [C1, y])) == x**C1 + assert constant_renumber( + constantsimp((x**C1)**y, [C1])) == (x**C1)**y + assert constant_renumber(constantsimp(2**C1, [C1])) == C1 + assert constant_renumber(constantsimp(S(2)**C1, [C1])) == C1 + assert constant_renumber(constantsimp(exp(C1), [C1])) == C1 + assert constant_renumber( + constantsimp(exp(C1 + x), [C1])) == C1*exp(x) + assert constant_renumber(constantsimp(Pow(2, C1), [C1])) == C1 + + +def test_constant_function(): + assert constant_renumber(constantsimp(sin(C1), [C1])) == C1 + assert constant_renumber(constantsimp(f(C1), [C1])) == C1 + assert constant_renumber(constantsimp(f(C1, C1), [C1])) == C1 + assert constant_renumber(constantsimp(f(C1, C2), [C1, C2])) == C1 + assert constant_renumber(constantsimp(f(C2, C1), [C1, C2])) == C1 + assert constant_renumber(constantsimp(f(C2, C2), [C1, C2])) == C1 + assert constant_renumber( + constantsimp(f(C1, x), [C1])) == f(C1, x) + assert constant_renumber(constantsimp(f(C1, y), [C1, y])) == C1 + assert constant_renumber(constantsimp(f(y, C1), [C1, y])) == C1 + assert constant_renumber(constantsimp(f(C1, y, C2), [C1, C2, y])) == C1 + + +def test_constant_function_multiple(): + # The rules to not renumber in this case would be too complicated, and + # dsolve is not likely to ever encounter anything remotely like this. + assert constant_renumber( + constantsimp(f(C1, C1, x), [C1])) == f(C1, C1, x) + + +def test_constant_multiple(): + assert constant_renumber(constantsimp(C1*2 + 2, [C1])) == C1 + assert constant_renumber(constantsimp(x*2/C1, [C1])) == C1*x + assert constant_renumber(constantsimp(C1**2*2 + 2, [C1])) == C1 + assert constant_renumber( + constantsimp(sin(2*C1) + x + sqrt(2), [C1])) == C1 + x + assert constant_renumber(constantsimp(2*C1 + C2, [C1, C2])) == C1 + +def test_constant_repeated(): + assert C1 + C1*x == constant_renumber( C1 + C1*x) + +def test_ode_solutions(): + # only a few examples here, the rest will be tested in the actual dsolve tests + assert constant_renumber(constantsimp(C1*exp(2*x) + exp(x)*(C2 + C3), [C1, C2, C3])) == \ + constant_renumber(C1*exp(x) + C2*exp(2*x)) + assert constant_renumber( + constantsimp(Eq(f(x), I*C1*sinh(x/3) + C2*cosh(x/3)), [C1, C2]) + ) == constant_renumber(Eq(f(x), C1*sinh(x/3) + C2*cosh(x/3))) + assert constant_renumber(constantsimp(Eq(f(x), acos((-C1)/cos(x))), [C1])) == \ + Eq(f(x), acos(C1/cos(x))) + assert constant_renumber( + constantsimp(Eq(log(f(x)/C1) + 2*exp(x/f(x)), 0), [C1]) + ) == Eq(log(C1*f(x)) + 2*exp(x/f(x)), 0) + assert constant_renumber(constantsimp(Eq(log(x*sqrt(2)*sqrt(1/x)*sqrt(f(x)) + /C1) + x**2/(2*f(x)**2), 0), [C1])) == \ + Eq(log(C1*sqrt(x)*sqrt(f(x))) + x**2/(2*f(x)**2), 0) + assert constant_renumber(constantsimp(Eq(-exp(-f(x)/x)*sin(f(x)/x)/2 + log(x/C1) - + cos(f(x)/x)*exp(-f(x)/x)/2, 0), [C1])) == \ + Eq(-exp(-f(x)/x)*sin(f(x)/x)/2 + log(C1*x) - cos(f(x)/x)* + exp(-f(x)/x)/2, 0) + assert constant_renumber(constantsimp(Eq(-Integral(-1/(sqrt(1 - u2**2)*u2), + (u2, _a, x/f(x))) + log(f(x)/C1), 0), [C1])) == \ + Eq(-Integral(-1/(u2*sqrt(1 - u2**2)), (u2, _a, x/f(x))) + + log(C1*f(x)), 0) + assert [constantsimp(i, [C1]) for i in [Eq(f(x), sqrt(-C1*x + x**2)), Eq(f(x), -sqrt(-C1*x + x**2))]] == \ + [Eq(f(x), sqrt(x*(C1 + x))), Eq(f(x), -sqrt(x*(C1 + x)))] + + +@XFAIL +def test_nonlocal_simplification(): + assert constantsimp(C1 + C2+x*C2, [C1, C2]) == C1 + C2*x + + +def test_constant_Eq(): + # C1 on the rhs is well-tested, but the lhs is only tested here + assert constantsimp(Eq(C1, 3 + f(x)*x), [C1]) == Eq(x*f(x), C1) + assert constantsimp(Eq(C1, 3 * f(x)*x), [C1]) == Eq(f(x)*x, C1) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_decompogen.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_decompogen.py new file mode 100644 index 0000000000000000000000000000000000000000..1ba03f4b42558231b626b6ed169f8b0a81a72bf9 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_decompogen.py @@ -0,0 +1,59 @@ +from sympy.solvers.decompogen import decompogen, compogen +from sympy.core.symbol import symbols +from sympy.functions.elementary.complexes import Abs +from sympy.functions.elementary.exponential import exp +from sympy.functions.elementary.miscellaneous import sqrt, Max +from sympy.functions.elementary.trigonometric import (cos, sin) +from sympy.testing.pytest import XFAIL, raises + +x, y = symbols('x y') + + +def test_decompogen(): + assert decompogen(sin(cos(x)), x) == [sin(x), cos(x)] + assert decompogen(sin(x)**2 + sin(x) + 1, x) == [x**2 + x + 1, sin(x)] + assert decompogen(sqrt(6*x**2 - 5), x) == [sqrt(x), 6*x**2 - 5] + assert decompogen(sin(sqrt(cos(x**2 + 1))), x) == [sin(x), sqrt(x), cos(x), x**2 + 1] + assert decompogen(Abs(cos(x)**2 + 3*cos(x) - 4), x) == [Abs(x), x**2 + 3*x - 4, cos(x)] + assert decompogen(sin(x)**2 + sin(x) - sqrt(3)/2, x) == [x**2 + x - sqrt(3)/2, sin(x)] + assert decompogen(Abs(cos(y)**2 + 3*cos(x) - 4), x) == [Abs(x), 3*x + cos(y)**2 - 4, cos(x)] + assert decompogen(x, y) == [x] + assert decompogen(1, x) == [1] + assert decompogen(Max(3, x), x) == [Max(3, x)] + raises(TypeError, lambda: decompogen(x < 5, x)) + u = 2*x + 3 + assert decompogen(Max(sqrt(u),(u)**2), x) == [Max(sqrt(x), x**2), u] + assert decompogen(Max(u, u**2, y), x) == [Max(x, x**2, y), u] + assert decompogen(Max(sin(x), u), x) == [Max(2*x + 3, sin(x))] + + +def test_decompogen_poly(): + assert decompogen(x**4 + 2*x**2 + 1, x) == [x**2 + 2*x + 1, x**2] + assert decompogen(x**4 + 2*x**3 - x - 1, x) == [x**2 - x - 1, x**2 + x] + + +@XFAIL +def test_decompogen_fails(): + A = lambda x: x**2 + 2*x + 3 + B = lambda x: 4*x**2 + 5*x + 6 + assert decompogen(A(x*exp(x)), x) == [x**2 + 2*x + 3, x*exp(x)] + assert decompogen(A(B(x)), x) == [x**2 + 2*x + 3, 4*x**2 + 5*x + 6] + assert decompogen(A(1/x + 1/x**2), x) == [x**2 + 2*x + 3, 1/x + 1/x**2] + assert decompogen(A(1/x + 2/(x + 1)), x) == [x**2 + 2*x + 3, 1/x + 2/(x + 1)] + + +def test_compogen(): + assert compogen([sin(x), cos(x)], x) == sin(cos(x)) + assert compogen([x**2 + x + 1, sin(x)], x) == sin(x)**2 + sin(x) + 1 + assert compogen([sqrt(x), 6*x**2 - 5], x) == sqrt(6*x**2 - 5) + assert compogen([sin(x), sqrt(x), cos(x), x**2 + 1], x) == sin(sqrt( + cos(x**2 + 1))) + assert compogen([Abs(x), x**2 + 3*x - 4, cos(x)], x) == Abs(cos(x)**2 + + 3*cos(x) - 4) + assert compogen([x**2 + x - sqrt(3)/2, sin(x)], x) == (sin(x)**2 + sin(x) - + sqrt(3)/2) + assert compogen([Abs(x), 3*x + cos(y)**2 - 4, cos(x)], x) == \ + Abs(3*cos(x) + cos(y)**2 - 4) + assert compogen([x**2 + 2*x + 1, x**2], x) == x**4 + 2*x**2 + 1 + # the result is in unsimplified form + assert compogen([x**2 - x - 1, x**2 + x], x) == -x**2 - x + (x**2 + x)**2 - 1 diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_inequalities.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_inequalities.py new file mode 100644 index 0000000000000000000000000000000000000000..6ce6f4520b52d8714102c95457c90d44543c685c --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_inequalities.py @@ -0,0 +1,500 @@ +"""Tests for tools for solving inequalities and systems of inequalities. """ + +from sympy.concrete.summations import Sum +from sympy.core.function import Function +from sympy.core.numbers import I, Rational, oo, pi +from sympy.core.relational import Eq, Ge, Gt, Le, Lt, Ne +from sympy.core.singleton import S +from sympy.core.symbol import (Dummy, Symbol) +from sympy.functions.elementary.complexes import Abs +from sympy.functions.elementary.exponential import exp, log +from sympy.functions.elementary.miscellaneous import root, sqrt +from sympy.functions.elementary.piecewise import Piecewise +from sympy.functions.elementary.trigonometric import cos, sin, tan +from sympy.integrals.integrals import Integral +from sympy.logic.boolalg import And, Or +from sympy.polys.polytools import Poly, PurePoly +from sympy.sets.sets import FiniteSet, Interval, Union +from sympy.solvers.inequalities import (reduce_inequalities, + solve_poly_inequality as psolve, + reduce_rational_inequalities, + solve_univariate_inequality as isolve, + reduce_abs_inequality, + _solve_inequality) +from sympy.polys.rootoftools import rootof +from sympy.solvers.solvers import solve +from sympy.solvers.solveset import solveset +from sympy.core.mod import Mod +from sympy.abc import x, y + +from sympy.testing.pytest import raises, XFAIL + + +inf = oo.evalf() + + +def test_solve_poly_inequality(): + assert psolve(Poly(0, x), '==') == [S.Reals] + assert psolve(Poly(1, x), '==') == [S.EmptySet] + assert psolve(PurePoly(x + 1, x), ">") == [Interval(-1, oo, True, False)] + + +def test_reduce_poly_inequalities_real_interval(): + assert reduce_rational_inequalities( + [[Eq(x**2, 0)]], x, relational=False) == FiniteSet(0) + assert reduce_rational_inequalities( + [[Le(x**2, 0)]], x, relational=False) == FiniteSet(0) + assert reduce_rational_inequalities( + [[Lt(x**2, 0)]], x, relational=False) == S.EmptySet + assert reduce_rational_inequalities( + [[Ge(x**2, 0)]], x, relational=False) == \ + S.Reals if x.is_real else Interval(-oo, oo) + assert reduce_rational_inequalities( + [[Gt(x**2, 0)]], x, relational=False) == \ + FiniteSet(0).complement(S.Reals) + assert reduce_rational_inequalities( + [[Ne(x**2, 0)]], x, relational=False) == \ + FiniteSet(0).complement(S.Reals) + + assert reduce_rational_inequalities( + [[Eq(x**2, 1)]], x, relational=False) == FiniteSet(-1, 1) + assert reduce_rational_inequalities( + [[Le(x**2, 1)]], x, relational=False) == Interval(-1, 1) + assert reduce_rational_inequalities( + [[Lt(x**2, 1)]], x, relational=False) == Interval(-1, 1, True, True) + assert reduce_rational_inequalities( + [[Ge(x**2, 1)]], x, relational=False) == \ + Union(Interval(-oo, -1), Interval(1, oo)) + assert reduce_rational_inequalities( + [[Gt(x**2, 1)]], x, relational=False) == \ + Interval(-1, 1).complement(S.Reals) + assert reduce_rational_inequalities( + [[Ne(x**2, 1)]], x, relational=False) == \ + FiniteSet(-1, 1).complement(S.Reals) + assert reduce_rational_inequalities([[Eq( + x**2, 1.0)]], x, relational=False) == FiniteSet(-1.0, 1.0).evalf() + assert reduce_rational_inequalities( + [[Le(x**2, 1.0)]], x, relational=False) == Interval(-1.0, 1.0) + assert reduce_rational_inequalities([[Lt( + x**2, 1.0)]], x, relational=False) == Interval(-1.0, 1.0, True, True) + assert reduce_rational_inequalities( + [[Ge(x**2, 1.0)]], x, relational=False) == \ + Union(Interval(-inf, -1.0), Interval(1.0, inf)) + assert reduce_rational_inequalities( + [[Gt(x**2, 1.0)]], x, relational=False) == \ + Union(Interval(-inf, -1.0, right_open=True), + Interval(1.0, inf, left_open=True)) + assert reduce_rational_inequalities([[Ne( + x**2, 1.0)]], x, relational=False) == \ + FiniteSet(-1.0, 1.0).complement(S.Reals) + + s = sqrt(2) + + assert reduce_rational_inequalities([[Lt( + x**2 - 1, 0), Gt(x**2 - 1, 0)]], x, relational=False) == S.EmptySet + assert reduce_rational_inequalities([[Le(x**2 - 1, 0), Ge( + x**2 - 1, 0)]], x, relational=False) == FiniteSet(-1, 1) + assert reduce_rational_inequalities( + [[Le(x**2 - 2, 0), Ge(x**2 - 1, 0)]], x, relational=False + ) == Union(Interval(-s, -1, False, False), Interval(1, s, False, False)) + assert reduce_rational_inequalities( + [[Le(x**2 - 2, 0), Gt(x**2 - 1, 0)]], x, relational=False + ) == Union(Interval(-s, -1, False, True), Interval(1, s, True, False)) + assert reduce_rational_inequalities( + [[Lt(x**2 - 2, 0), Ge(x**2 - 1, 0)]], x, relational=False + ) == Union(Interval(-s, -1, True, False), Interval(1, s, False, True)) + assert reduce_rational_inequalities( + [[Lt(x**2 - 2, 0), Gt(x**2 - 1, 0)]], x, relational=False + ) == Union(Interval(-s, -1, True, True), Interval(1, s, True, True)) + assert reduce_rational_inequalities( + [[Lt(x**2 - 2, 0), Ne(x**2 - 1, 0)]], x, relational=False + ) == Union(Interval(-s, -1, True, True), Interval(-1, 1, True, True), + Interval(1, s, True, True)) + + assert reduce_rational_inequalities([[Lt(x**2, -1.)]], x) is S.false + + +def test_reduce_poly_inequalities_complex_relational(): + assert reduce_rational_inequalities( + [[Eq(x**2, 0)]], x, relational=True) == Eq(x, 0) + assert reduce_rational_inequalities( + [[Le(x**2, 0)]], x, relational=True) == Eq(x, 0) + assert reduce_rational_inequalities( + [[Lt(x**2, 0)]], x, relational=True) == False + assert reduce_rational_inequalities( + [[Ge(x**2, 0)]], x, relational=True) == And(Lt(-oo, x), Lt(x, oo)) + assert reduce_rational_inequalities( + [[Gt(x**2, 0)]], x, relational=True) == \ + And(Gt(x, -oo), Lt(x, oo), Ne(x, 0)) + assert reduce_rational_inequalities( + [[Ne(x**2, 0)]], x, relational=True) == \ + And(Gt(x, -oo), Lt(x, oo), Ne(x, 0)) + + for one in (S.One, S(1.0)): + inf = one*oo + assert reduce_rational_inequalities( + [[Eq(x**2, one)]], x, relational=True) == \ + Or(Eq(x, -one), Eq(x, one)) + assert reduce_rational_inequalities( + [[Le(x**2, one)]], x, relational=True) == \ + And(And(Le(-one, x), Le(x, one))) + assert reduce_rational_inequalities( + [[Lt(x**2, one)]], x, relational=True) == \ + And(And(Lt(-one, x), Lt(x, one))) + assert reduce_rational_inequalities( + [[Ge(x**2, one)]], x, relational=True) == \ + And(Or(And(Le(one, x), Lt(x, inf)), And(Le(x, -one), Lt(-inf, x)))) + assert reduce_rational_inequalities( + [[Gt(x**2, one)]], x, relational=True) == \ + And(Or(And(Lt(-inf, x), Lt(x, -one)), And(Lt(one, x), Lt(x, inf)))) + assert reduce_rational_inequalities( + [[Ne(x**2, one)]], x, relational=True) == \ + Or(And(Lt(-inf, x), Lt(x, -one)), + And(Lt(-one, x), Lt(x, one)), + And(Lt(one, x), Lt(x, inf))) + + +def test_reduce_rational_inequalities_real_relational(): + assert reduce_rational_inequalities([], x) == False + assert reduce_rational_inequalities( + [[(x**2 + 3*x + 2)/(x**2 - 16) >= 0]], x, relational=False) == \ + Union(Interval.open(-oo, -4), Interval(-2, -1), Interval.open(4, oo)) + + assert reduce_rational_inequalities( + [[((-2*x - 10)*(3 - x))/((x**2 + 5)*(x - 2)**2) < 0]], x, + relational=False) == \ + Union(Interval.open(-5, 2), Interval.open(2, 3)) + + assert reduce_rational_inequalities([[(x + 1)/(x - 5) <= 0]], x, + relational=False) == \ + Interval.Ropen(-1, 5) + + assert reduce_rational_inequalities([[(x**2 + 4*x + 3)/(x - 1) > 0]], x, + relational=False) == \ + Union(Interval.open(-3, -1), Interval.open(1, oo)) + + assert reduce_rational_inequalities([[(x**2 - 16)/(x - 1)**2 < 0]], x, + relational=False) == \ + Union(Interval.open(-4, 1), Interval.open(1, 4)) + + assert reduce_rational_inequalities([[(3*x + 1)/(x + 4) >= 1]], x, + relational=False) == \ + Union(Interval.open(-oo, -4), Interval.Ropen(Rational(3, 2), oo)) + + assert reduce_rational_inequalities([[(x - 8)/x <= 3 - x]], x, + relational=False) == \ + Union(Interval.Lopen(-oo, -2), Interval.Lopen(0, 4)) + + # issue sympy/sympy#10237 + assert reduce_rational_inequalities( + [[x < oo, x >= 0, -oo < x]], x, relational=False) == Interval(0, oo) + + +def test_reduce_abs_inequalities(): + e = abs(x - 5) < 3 + ans = And(Lt(2, x), Lt(x, 8)) + assert reduce_inequalities(e) == ans + assert reduce_inequalities(e, x) == ans + assert reduce_inequalities(abs(x - 5)) == Eq(x, 5) + assert reduce_inequalities( + abs(2*x + 3) >= 8) == Or(And(Le(Rational(5, 2), x), Lt(x, oo)), + And(Le(x, Rational(-11, 2)), Lt(-oo, x))) + assert reduce_inequalities(abs(x - 4) + abs( + 3*x - 5) < 7) == And(Lt(S.Half, x), Lt(x, 4)) + assert reduce_inequalities(abs(x - 4) + abs(3*abs(x) - 5) < 7) == \ + Or(And(S(-2) < x, x < -1), And(S.Half < x, x < 4)) + + nr = Symbol('nr', extended_real=False) + raises(TypeError, lambda: reduce_inequalities(abs(nr - 5) < 3)) + assert reduce_inequalities(x < 3, symbols=[x, nr]) == And(-oo < x, x < 3) + + +def test_reduce_inequalities_general(): + assert reduce_inequalities(Ge(sqrt(2)*x, 1)) == And(sqrt(2)/2 <= x, x < oo) + assert reduce_inequalities(x + 1 > 0) == And(S.NegativeOne < x, x < oo) + + +def test_reduce_inequalities_boolean(): + assert reduce_inequalities( + [Eq(x**2, 0), True]) == Eq(x, 0) + assert reduce_inequalities([Eq(x**2, 0), False]) == False + assert reduce_inequalities(x**2 >= 0) is S.true # issue 10196 + + +def test_reduce_inequalities_multivariate(): + assert reduce_inequalities([Ge(x**2, 1), Ge(y**2, 1)]) == And( + Or(And(Le(S.One, x), Lt(x, oo)), And(Le(x, -1), Lt(-oo, x))), + Or(And(Le(S.One, y), Lt(y, oo)), And(Le(y, -1), Lt(-oo, y)))) + + +def test_reduce_inequalities_errors(): + raises(NotImplementedError, lambda: reduce_inequalities(Ge(sin(x) + x, 1))) + raises(NotImplementedError, lambda: reduce_inequalities(Ge(x**2*y + y, 1))) + + +def test__solve_inequalities(): + assert reduce_inequalities(x + y < 1, symbols=[x]) == (x < 1 - y) + assert reduce_inequalities(x + y >= 1, symbols=[x]) == (x < oo) & (x >= -y + 1) + assert reduce_inequalities(Eq(0, x - y), symbols=[x]) == Eq(x, y) + assert reduce_inequalities(Ne(0, x - y), symbols=[x]) == Ne(x, y) + + +def test_issue_6343(): + eq = -3*x**2/2 - x*Rational(45, 4) + Rational(33, 2) > 0 + assert reduce_inequalities(eq) == \ + And(x < Rational(-15, 4) + sqrt(401)/4, -sqrt(401)/4 - Rational(15, 4) < x) + + +def test_issue_8235(): + assert reduce_inequalities(x**2 - 1 < 0) == \ + And(S.NegativeOne < x, x < 1) + assert reduce_inequalities(x**2 - 1 <= 0) == \ + And(S.NegativeOne <= x, x <= 1) + assert reduce_inequalities(x**2 - 1 > 0) == \ + Or(And(-oo < x, x < -1), And(x < oo, S.One < x)) + assert reduce_inequalities(x**2 - 1 >= 0) == \ + Or(And(-oo < x, x <= -1), And(S.One <= x, x < oo)) + + eq = x**8 + x - 9 # we want CRootOf solns here + sol = solve(eq >= 0) + tru = Or(And(rootof(eq, 1) <= x, x < oo), And(-oo < x, x <= rootof(eq, 0))) + assert sol == tru + + # recast vanilla as real + assert solve(sqrt((-x + 1)**2) < 1) == And(S.Zero < x, x < 2) + + +def test_issue_5526(): + assert reduce_inequalities(0 <= + x + Integral(y**2, (y, 1, 3)) - 1, [x]) == \ + (x >= -Integral(y**2, (y, 1, 3)) + 1) + f = Function('f') + e = Sum(f(x), (x, 1, 3)) + assert reduce_inequalities(0 <= x + e + y**2, [x]) == \ + (x >= -y**2 - Sum(f(x), (x, 1, 3))) + + +def test_solve_univariate_inequality(): + assert isolve(x**2 >= 4, x, relational=False) == Union(Interval(-oo, -2), + Interval(2, oo)) + assert isolve(x**2 >= 4, x) == Or(And(Le(2, x), Lt(x, oo)), And(Le(x, -2), + Lt(-oo, x))) + assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x, relational=False) == \ + Union(Interval(1, 2), Interval(3, oo)) + assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x) == \ + Or(And(Le(1, x), Le(x, 2)), And(Le(3, x), Lt(x, oo))) + assert isolve((x - 1)*(x - 2)*(x - 4) < 0, x, domain = FiniteSet(0, 3)) == \ + Or(Eq(x, 0), Eq(x, 3)) + # issue 2785: + assert isolve(x**3 - 2*x - 1 > 0, x, relational=False) == \ + Union(Interval(-1, -sqrt(5)/2 + S.Half, True, True), + Interval(S.Half + sqrt(5)/2, oo, True, True)) + # issue 2794: + assert isolve(x**3 - x**2 + x - 1 > 0, x, relational=False) == \ + Interval(1, oo, True) + #issue 13105 + assert isolve((x + I)*(x + 2*I) < 0, x) == Eq(x, 0) + assert isolve(((x - 1)*(x - 2) + I)*((x - 1)*(x - 2) + 2*I) < 0, x) == Or(Eq(x, 1), Eq(x, 2)) + assert isolve((((x - 1)*(x - 2) + I)*((x - 1)*(x - 2) + 2*I))/(x - 2) > 0, x) == Eq(x, 1) + raises (ValueError, lambda: isolve((x**2 - 3*x*I + 2)/x < 0, x)) + + # numerical testing in valid() is needed + assert isolve(x**7 - x - 2 > 0, x) == \ + And(rootof(x**7 - x - 2, 0) < x, x < oo) + + # handle numerator and denominator; although these would be handled as + # rational inequalities, these test confirm that the right thing is done + # when the domain is EX (e.g. when 2 is replaced with sqrt(2)) + assert isolve(1/(x - 2) > 0, x) == And(S(2) < x, x < oo) + den = ((x - 1)*(x - 2)).expand() + assert isolve((x - 1)/den <= 0, x) == \ + (x > -oo) & (x < 2) & Ne(x, 1) + + n = Dummy('n') + raises(NotImplementedError, lambda: isolve(Abs(x) <= n, x, relational=False)) + c1 = Dummy("c1", positive=True) + raises(NotImplementedError, lambda: isolve(n/c1 < 0, c1)) + n = Dummy('n', negative=True) + assert isolve(n/c1 > -2, c1) == (-n/2 < c1) + assert isolve(n/c1 < 0, c1) == True + assert isolve(n/c1 > 0, c1) == False + + zero = cos(1)**2 + sin(1)**2 - 1 + raises(NotImplementedError, lambda: isolve(x**2 < zero, x)) + raises(NotImplementedError, lambda: isolve( + x**2 < zero*I, x)) + raises(NotImplementedError, lambda: isolve(1/(x - y) < 2, x)) + raises(NotImplementedError, lambda: isolve(1/(x - y) < 0, x)) + raises(TypeError, lambda: isolve(x - I < 0, x)) + + zero = x**2 + x - x*(x + 1) + assert isolve(zero < 0, x, relational=False) is S.EmptySet + assert isolve(zero <= 0, x, relational=False) is S.Reals + + # make sure iter_solutions gets a default value + raises(NotImplementedError, lambda: isolve( + Eq(cos(x)**2 + sin(x)**2, 1), x)) + + +def test_trig_inequalities(): + # all the inequalities are solved in a periodic interval. + assert isolve(sin(x) < S.Half, x, relational=False) == \ + Union(Interval(0, pi/6, False, True), Interval.open(pi*Rational(5, 6), 2*pi)) + assert isolve(sin(x) > S.Half, x, relational=False) == \ + Interval(pi/6, pi*Rational(5, 6), True, True) + assert isolve(cos(x) < S.Zero, x, relational=False) == \ + Interval(pi/2, pi*Rational(3, 2), True, True) + assert isolve(cos(x) >= S.Zero, x, relational=False) == \ + Union(Interval(0, pi/2), Interval.Ropen(pi*Rational(3, 2), 2*pi)) + + assert isolve(tan(x) < S.One, x, relational=False) == \ + Union(Interval.Ropen(0, pi/4), Interval.open(pi/2, pi)) + + assert isolve(sin(x) <= S.Zero, x, relational=False) == \ + Union(FiniteSet(S.Zero), Interval.Ropen(pi, 2*pi)) + + assert isolve(sin(x) <= S.One, x, relational=False) == S.Reals + assert isolve(cos(x) < S(-2), x, relational=False) == S.EmptySet + assert isolve(sin(x) >= S.NegativeOne, x, relational=False) == S.Reals + assert isolve(cos(x) > S.One, x, relational=False) == S.EmptySet + + +def test_issue_9954(): + assert isolve(x**2 >= 0, x, relational=False) == S.Reals + assert isolve(x**2 >= 0, x, relational=True) == S.Reals.as_relational(x) + assert isolve(x**2 < 0, x, relational=False) == S.EmptySet + assert isolve(x**2 < 0, x, relational=True) == S.EmptySet.as_relational(x) + + +@XFAIL +def test_slow_general_univariate(): + r = rootof(x**5 - x**2 + 1, 0) + assert solve(sqrt(x) + 1/root(x, 3) > 1) == \ + Or(And(0 < x, x < r**6), And(r**6 < x, x < oo)) + + +def test_issue_8545(): + eq = 1 - x - abs(1 - x) + ans = And(Lt(1, x), Lt(x, oo)) + assert reduce_abs_inequality(eq, '<', x) == ans + eq = 1 - x - sqrt((1 - x)**2) + assert reduce_inequalities(eq < 0) == ans + + +def test_issue_8974(): + assert isolve(-oo < x, x) == And(-oo < x, x < oo) + assert isolve(oo > x, x) == And(-oo < x, x < oo) + + +def test_issue_10198(): + assert reduce_inequalities( + -1 + 1/abs(1/x - 1) < 0) == (x > -oo) & (x < S(1)/2) & Ne(x, 0) + + assert reduce_inequalities(abs(1/sqrt(x)) - 1, x) == Eq(x, 1) + assert reduce_abs_inequality(-3 + 1/abs(1 - 1/x), '<', x) == \ + Or(And(-oo < x, x < 0), + And(S.Zero < x, x < Rational(3, 4)), And(Rational(3, 2) < x, x < oo)) + raises(ValueError,lambda: reduce_abs_inequality(-3 + 1/abs( + 1 - 1/sqrt(x)), '<', x)) + + +def test_issue_10047(): + # issue 10047: this must remain an inequality, not True, since if x + # is not real the inequality is invalid + # assert solve(sin(x) < 2) == (x <= oo) + + # with PR 16956, (x <= oo) autoevaluates when x is extended_real + # which is assumed in the current implementation of inequality solvers + assert solve(sin(x) < 2) == True + assert solveset(sin(x) < 2, domain=S.Reals) == S.Reals + + +def test_issue_10268(): + assert solve(log(x) < 1000) == And(S.Zero < x, x < exp(1000)) + + +@XFAIL +def test_isolve_Sets(): + n = Dummy('n') + assert isolve(Abs(x) <= n, x, relational=False) == \ + Piecewise((S.EmptySet, n < 0), (Interval(-n, n), True)) + + +def test_integer_domain_relational_isolve(): + + dom = FiniteSet(0, 3) + x = Symbol('x',zero=False) + assert isolve((x - 1)*(x - 2)*(x - 4) < 0, x, domain=dom) == Eq(x, 3) + + x = Symbol('x') + assert isolve(x + 2 < 0, x, domain=S.Integers) == \ + (x <= -3) & (x > -oo) & Eq(Mod(x, 1), 0) + assert isolve(2 * x + 3 > 0, x, domain=S.Integers) == \ + (x >= -1) & (x < oo) & Eq(Mod(x, 1), 0) + assert isolve((x ** 2 + 3 * x - 2) < 0, x, domain=S.Integers) == \ + (x >= -3) & (x <= 0) & Eq(Mod(x, 1), 0) + assert isolve((x ** 2 + 3 * x - 2) > 0, x, domain=S.Integers) == \ + ((x >= 1) & (x < oo) & Eq(Mod(x, 1), 0)) | ( + (x <= -4) & (x > -oo) & Eq(Mod(x, 1), 0)) + + +def test_issue_10671_12466(): + assert solveset(sin(y), y, Interval(0, pi)) == FiniteSet(0, pi) + i = Interval(1, 10) + assert solveset((1/x).diff(x) < 0, x, i) == i + assert solveset((log(x - 6)/x) <= 0, x, S.Reals) == \ + Interval.Lopen(6, 7) + + +def test__solve_inequality(): + for op in (Gt, Lt, Le, Ge, Eq, Ne): + assert _solve_inequality(op(x, 1), x).lhs == x + assert _solve_inequality(op(S.One, x), x).lhs == x + # don't get tricked by symbol on right: solve it + assert _solve_inequality(Eq(2*x - 1, x), x) == Eq(x, 1) + ie = Eq(S.One, y) + assert _solve_inequality(ie, x) == ie + for fx in (x**2, exp(x), sin(x) + cos(x), x*(1 + x)): + for c in (0, 1): + e = 2*fx - c > 0 + assert _solve_inequality(e, x, linear=True) == ( + fx > c/S(2)) + assert _solve_inequality(2*x**2 + 2*x - 1 < 0, x, linear=True) == ( + x*(x + 1) < S.Half) + assert _solve_inequality(Eq(x*y, 1), x) == Eq(x*y, 1) + nz = Symbol('nz', nonzero=True) + assert _solve_inequality(Eq(x*nz, 1), x) == Eq(x, 1/nz) + assert _solve_inequality(x*nz < 1, x) == (x*nz < 1) + a = Symbol('a', positive=True) + assert _solve_inequality(a/x > 1, x) == (S.Zero < x) & (x < a) + assert _solve_inequality(a/x > 1, x, linear=True) == (1/x > 1/a) + # make sure to include conditions under which solution is valid + e = Eq(1 - x, x*(1/x - 1)) + assert _solve_inequality(e, x) == Ne(x, 0) + assert _solve_inequality(x < x*(1/x - 1), x) == (x < S.Half) & Ne(x, 0) + + +def test__pt(): + from sympy.solvers.inequalities import _pt + assert _pt(-oo, oo) == 0 + assert _pt(S.One, S(3)) == 2 + assert _pt(S.One, oo) == _pt(oo, S.One) == 2 + assert _pt(S.One, -oo) == _pt(-oo, S.One) == S.Half + assert _pt(S.NegativeOne, oo) == _pt(oo, S.NegativeOne) == Rational(-1, 2) + assert _pt(S.NegativeOne, -oo) == _pt(-oo, S.NegativeOne) == -2 + assert _pt(x, oo) == _pt(oo, x) == x + 1 + assert _pt(x, -oo) == _pt(-oo, x) == x - 1 + raises(ValueError, lambda: _pt(Dummy('i', infinite=True), S.One)) + + +def test_issue_25697(): + assert _solve_inequality(log(x, 3) <= 2, x) == (x <= 9) & (S.Zero < x) + + +def test_issue_25738(): + assert reduce_inequalities(3 < abs(x) + ) == reduce_inequalities(pi < abs(x)).subs(pi, 3) + + +def test_issue_25983(): + assert(reduce_inequalities(pi/Abs(x) <= 1) == ((pi <= x) & (x < oo)) | ((-oo < x) & (x <= -pi))) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_numeric.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_numeric.py new file mode 100644 index 0000000000000000000000000000000000000000..f40bab6965233b82984148960a62ed57a7ddb178 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_numeric.py @@ -0,0 +1,139 @@ +from sympy.core.function import nfloat +from sympy.core.numbers import (Float, I, Rational, pi) +from sympy.core.relational import Eq +from sympy.core.symbol import (Symbol, symbols) +from sympy.functions.elementary.miscellaneous import sqrt +from sympy.functions.elementary.piecewise import Piecewise +from sympy.functions.elementary.trigonometric import sin +from sympy.integrals.integrals import Integral +from sympy.matrices.dense import Matrix +from mpmath import mnorm, mpf +from sympy.solvers import nsolve +from sympy.utilities.lambdify import lambdify +from sympy.testing.pytest import raises, XFAIL +from sympy.utilities.decorator import conserve_mpmath_dps + +@XFAIL +def test_nsolve_fail(): + x = symbols('x') + # Sometimes it is better to use the numerator (issue 4829) + # but sometimes it is not (issue 11768) so leave this to + # the discretion of the user + ans = nsolve(x**2/(1 - x)/(1 - 2*x)**2 - 100, x, 0) + assert ans > 0.46 and ans < 0.47 + + +def test_nsolve_denominator(): + x = symbols('x') + # Test that nsolve uses the full expression (numerator and denominator). + ans = nsolve((x**2 + 3*x + 2)/(x + 2), -2.1) + # The root -2 was divided out, so make sure we don't find it. + assert ans == -1.0 + +def test_nsolve(): + # onedimensional + x = Symbol('x') + assert nsolve(sin(x), 2) - pi.evalf() < 1e-15 + assert nsolve(Eq(2*x, 2), x, -10) == nsolve(2*x - 2, -10) + # Testing checks on number of inputs + raises(TypeError, lambda: nsolve(Eq(2*x, 2))) + raises(TypeError, lambda: nsolve(Eq(2*x, 2), x, 1, 2)) + # multidimensional + x1 = Symbol('x1') + x2 = Symbol('x2') + f1 = 3 * x1**2 - 2 * x2**2 - 1 + f2 = x1**2 - 2 * x1 + x2**2 + 2 * x2 - 8 + f = Matrix((f1, f2)).T + F = lambdify((x1, x2), f.T, modules='mpmath') + for x0 in [(-1, 1), (1, -2), (4, 4), (-4, -4)]: + x = nsolve(f, (x1, x2), x0, tol=1.e-8) + assert mnorm(F(*x), 1) <= 1.e-10 + # The Chinese mathematician Zhu Shijie was the very first to solve this + # nonlinear system 700 years ago (z was added to make it 3-dimensional) + x = Symbol('x') + y = Symbol('y') + z = Symbol('z') + f1 = -x + 2*y + f2 = (x**2 + x*(y**2 - 2) - 4*y) / (x + 4) + f3 = sqrt(x**2 + y**2)*z + f = Matrix((f1, f2, f3)).T + F = lambdify((x, y, z), f.T, modules='mpmath') + + def getroot(x0): + root = nsolve(f, (x, y, z), x0) + assert mnorm(F(*root), 1) <= 1.e-8 + return root + assert list(map(round, getroot((1, 1, 1)))) == [2, 1, 0] + assert nsolve([Eq( + f1, 0), Eq(f2, 0), Eq(f3, 0)], [x, y, z], (1, 1, 1)) # just see that it works + a = Symbol('a') + assert abs(nsolve(1/(0.001 + a)**3 - 6/(0.9 - a)**3, a, 0.3) - + mpf('0.31883011387318591')) < 1e-15 + + +def test_issue_6408(): + x = Symbol('x') + assert nsolve(Piecewise((x, x < 1), (x**2, True)), x, 2) == 0.0 + + +def test_issue_6408_integral(): + x, y = symbols('x y') + assert nsolve(Integral(x*y, (x, 0, 5)), y, 2) == 0.0 + + +@conserve_mpmath_dps +def test_increased_dps(): + # Issue 8564 + import mpmath + mpmath.mp.dps = 128 + x = Symbol('x') + e1 = x**2 - pi + q = nsolve(e1, x, 3.0) + + assert abs(sqrt(pi).evalf(128) - q) < 1e-128 + +def test_nsolve_precision(): + x, y = symbols('x y') + sol = nsolve(x**2 - pi, x, 3, prec=128) + assert abs(sqrt(pi).evalf(128) - sol) < 1e-128 + assert isinstance(sol, Float) + + sols = nsolve((y**2 - x, x**2 - pi), (x, y), (3, 3), prec=128) + assert isinstance(sols, Matrix) + assert sols.shape == (2, 1) + assert abs(sqrt(pi).evalf(128) - sols[0]) < 1e-128 + assert abs(sqrt(sqrt(pi)).evalf(128) - sols[1]) < 1e-128 + assert all(isinstance(i, Float) for i in sols) + +def test_nsolve_complex(): + x, y = symbols('x y') + + assert nsolve(x**2 + 2, 1j) == sqrt(2.)*I + assert nsolve(x**2 + 2, I) == sqrt(2.)*I + + assert nsolve([x**2 + 2, y**2 + 2], [x, y], [I, I]) == Matrix([sqrt(2.)*I, sqrt(2.)*I]) + assert nsolve([x**2 + 2, y**2 + 2], [x, y], [I, I]) == Matrix([sqrt(2.)*I, sqrt(2.)*I]) + +def test_nsolve_dict_kwarg(): + x, y = symbols('x y') + # one variable + assert nsolve(x**2 - 2, 1, dict = True) == \ + [{x: sqrt(2.)}] + # one variable with complex solution + assert nsolve(x**2 + 2, I, dict = True) == \ + [{x: sqrt(2.)*I}] + # two variables + assert nsolve([x**2 + y**2 - 5, x**2 - y**2 + 1], [x, y], [1, 1], dict = True) == \ + [{x: sqrt(2.), y: sqrt(3.)}] + +def test_nsolve_rational(): + x = symbols('x') + assert nsolve(x - Rational(1, 3), 0, prec=100) == Rational(1, 3).evalf(100) + + +def test_issue_14950(): + x = Matrix(symbols('t s')) + x0 = Matrix([17, 23]) + eqn = x + x0 + assert nsolve(eqn, x, x0) == nfloat(-x0) + assert nsolve(eqn.T, x.T, x0.T) == nfloat(-x0) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_pde.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_pde.py new file mode 100644 index 0000000000000000000000000000000000000000..948d90c7be21a9e0e03753e723ef04f1fb08a5d6 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_pde.py @@ -0,0 +1,239 @@ +from sympy.core.function import (Derivative as D, Function) +from sympy.core.relational import Eq +from sympy.core.symbol import (Symbol, symbols) +from sympy.functions.elementary.exponential import (exp, log) +from sympy.functions.elementary.trigonometric import (cos, sin) +from sympy.core import S +from sympy.solvers.pde import (pde_separate, pde_separate_add, pde_separate_mul, + pdsolve, classify_pde, checkpdesol) +from sympy.testing.pytest import raises + + +a, b, c, x, y = symbols('a b c x y') + +def test_pde_separate_add(): + x, y, z, t = symbols("x,y,z,t") + F, T, X, Y, Z, u = map(Function, 'FTXYZu') + + eq = Eq(D(u(x, t), x), D(u(x, t), t)*exp(u(x, t))) + res = pde_separate_add(eq, u(x, t), [X(x), T(t)]) + assert res == [D(X(x), x)*exp(-X(x)), D(T(t), t)*exp(T(t))] + + +def test_pde_separate(): + x, y, z, t = symbols("x,y,z,t") + F, T, X, Y, Z, u = map(Function, 'FTXYZu') + + eq = Eq(D(u(x, t), x), D(u(x, t), t)*exp(u(x, t))) + raises(ValueError, lambda: pde_separate(eq, u(x, t), [X(x), T(t)], 'div')) + + +def test_pde_separate_mul(): + x, y, z, t = symbols("x,y,z,t") + c = Symbol("C", real=True) + Phi = Function('Phi') + F, R, T, X, Y, Z, u = map(Function, 'FRTXYZu') + r, theta, z = symbols('r,theta,z') + + # Something simple :) + eq = Eq(D(F(x, y, z), x) + D(F(x, y, z), y) + D(F(x, y, z), z), 0) + + # Duplicate arguments in functions + raises( + ValueError, lambda: pde_separate_mul(eq, F(x, y, z), [X(x), u(z, z)])) + # Wrong number of arguments + raises(ValueError, lambda: pde_separate_mul(eq, F(x, y, z), [X(x), Y(y)])) + # Wrong variables: [x, y] -> [x, z] + raises( + ValueError, lambda: pde_separate_mul(eq, F(x, y, z), [X(t), Y(x, y)])) + + assert pde_separate_mul(eq, F(x, y, z), [Y(y), u(x, z)]) == \ + [D(Y(y), y)/Y(y), -D(u(x, z), x)/u(x, z) - D(u(x, z), z)/u(x, z)] + assert pde_separate_mul(eq, F(x, y, z), [X(x), Y(y), Z(z)]) == \ + [D(X(x), x)/X(x), -D(Z(z), z)/Z(z) - D(Y(y), y)/Y(y)] + + # wave equation + wave = Eq(D(u(x, t), t, t), c**2*D(u(x, t), x, x)) + res = pde_separate_mul(wave, u(x, t), [X(x), T(t)]) + assert res == [D(X(x), x, x)/X(x), D(T(t), t, t)/(c**2*T(t))] + + # Laplace equation in cylindrical coords + eq = Eq(1/r * D(Phi(r, theta, z), r) + D(Phi(r, theta, z), r, 2) + + 1/r**2 * D(Phi(r, theta, z), theta, 2) + D(Phi(r, theta, z), z, 2), 0) + # Separate z + res = pde_separate_mul(eq, Phi(r, theta, z), [Z(z), u(theta, r)]) + assert res == [D(Z(z), z, z)/Z(z), + -D(u(theta, r), r, r)/u(theta, r) - + D(u(theta, r), r)/(r*u(theta, r)) - + D(u(theta, r), theta, theta)/(r**2*u(theta, r))] + # Lets use the result to create a new equation... + eq = Eq(res[1], c) + # ...and separate theta... + res = pde_separate_mul(eq, u(theta, r), [T(theta), R(r)]) + assert res == [D(T(theta), theta, theta)/T(theta), + -r*D(R(r), r)/R(r) - r**2*D(R(r), r, r)/R(r) - c*r**2] + # ...or r... + res = pde_separate_mul(eq, u(theta, r), [R(r), T(theta)]) + assert res == [r*D(R(r), r)/R(r) + r**2*D(R(r), r, r)/R(r) + c*r**2, + -D(T(theta), theta, theta)/T(theta)] + + +def test_issue_11726(): + x, t = symbols("x t") + f = symbols("f", cls=Function) + X, T = symbols("X T", cls=Function) + + u = f(x, t) + eq = u.diff(x, 2) - u.diff(t, 2) + res = pde_separate(eq, u, [T(x), X(t)]) + assert res == [D(T(x), x, x)/T(x),D(X(t), t, t)/X(t)] + + +def test_pde_classify(): + # When more number of hints are added, add tests for classifying here. + f = Function('f') + eq1 = a*f(x,y) + b*f(x,y).diff(x) + c*f(x,y).diff(y) + eq2 = 3*f(x,y) + 2*f(x,y).diff(x) + f(x,y).diff(y) + eq3 = a*f(x,y) + b*f(x,y).diff(x) + 2*f(x,y).diff(y) + eq4 = x*f(x,y) + f(x,y).diff(x) + 3*f(x,y).diff(y) + eq5 = x**2*f(x,y) + x*f(x,y).diff(x) + x*y*f(x,y).diff(y) + eq6 = y*x**2*f(x,y) + y*f(x,y).diff(x) + f(x,y).diff(y) + for eq in [eq1, eq2, eq3]: + assert classify_pde(eq) == ('1st_linear_constant_coeff_homogeneous',) + for eq in [eq4, eq5, eq6]: + assert classify_pde(eq) == ('1st_linear_variable_coeff',) + + +def test_checkpdesol(): + f, F = map(Function, ['f', 'F']) + eq1 = a*f(x,y) + b*f(x,y).diff(x) + c*f(x,y).diff(y) + eq2 = 3*f(x,y) + 2*f(x,y).diff(x) + f(x,y).diff(y) + eq3 = a*f(x,y) + b*f(x,y).diff(x) + 2*f(x,y).diff(y) + for eq in [eq1, eq2, eq3]: + assert checkpdesol(eq, pdsolve(eq))[0] + eq4 = x*f(x,y) + f(x,y).diff(x) + 3*f(x,y).diff(y) + eq5 = 2*f(x,y) + 1*f(x,y).diff(x) + 3*f(x,y).diff(y) + eq6 = f(x,y) + 1*f(x,y).diff(x) + 3*f(x,y).diff(y) + assert checkpdesol(eq4, [pdsolve(eq5), pdsolve(eq6)]) == [ + (False, (x - 2)*F(3*x - y)*exp(-x/S(5) - 3*y/S(5))), + (False, (x - 1)*F(3*x - y)*exp(-x/S(10) - 3*y/S(10)))] + for eq in [eq4, eq5, eq6]: + assert checkpdesol(eq, pdsolve(eq))[0] + sol = pdsolve(eq4) + sol4 = Eq(sol.lhs - sol.rhs, 0) + raises(NotImplementedError, lambda: + checkpdesol(eq4, sol4, solve_for_func=False)) + + +def test_solvefun(): + f, F, G, H = map(Function, ['f', 'F', 'G', 'H']) + eq1 = f(x,y) + f(x,y).diff(x) + f(x,y).diff(y) + assert pdsolve(eq1) == Eq(f(x, y), F(x - y)*exp(-x/2 - y/2)) + assert pdsolve(eq1, solvefun=G) == Eq(f(x, y), G(x - y)*exp(-x/2 - y/2)) + assert pdsolve(eq1, solvefun=H) == Eq(f(x, y), H(x - y)*exp(-x/2 - y/2)) + + +def test_pde_1st_linear_constant_coeff_homogeneous(): + f, F = map(Function, ['f', 'F']) + u = f(x, y) + eq = 2*u + u.diff(x) + u.diff(y) + assert classify_pde(eq) == ('1st_linear_constant_coeff_homogeneous',) + sol = pdsolve(eq) + assert sol == Eq(u, F(x - y)*exp(-x - y)) + assert checkpdesol(eq, sol)[0] + + eq = 4 + (3*u.diff(x)/u) + (2*u.diff(y)/u) + assert classify_pde(eq) == ('1st_linear_constant_coeff_homogeneous',) + sol = pdsolve(eq) + assert sol == Eq(u, F(2*x - 3*y)*exp(-S(12)*x/13 - S(8)*y/13)) + assert checkpdesol(eq, sol)[0] + + eq = u + (6*u.diff(x)) + (7*u.diff(y)) + assert classify_pde(eq) == ('1st_linear_constant_coeff_homogeneous',) + sol = pdsolve(eq) + assert sol == Eq(u, F(7*x - 6*y)*exp(-6*x/S(85) - 7*y/S(85))) + assert checkpdesol(eq, sol)[0] + + eq = a*u + b*u.diff(x) + c*u.diff(y) + sol = pdsolve(eq) + assert checkpdesol(eq, sol)[0] + + +def test_pde_1st_linear_constant_coeff(): + f, F = map(Function, ['f', 'F']) + u = f(x,y) + eq = -2*u.diff(x) + 4*u.diff(y) + 5*u - exp(x + 3*y) + sol = pdsolve(eq) + assert sol == Eq(f(x,y), + (F(4*x + 2*y)*exp(x/2) + exp(x + 4*y)/15)*exp(-y)) + assert classify_pde(eq) == ('1st_linear_constant_coeff', + '1st_linear_constant_coeff_Integral') + assert checkpdesol(eq, sol)[0] + + eq = (u.diff(x)/u) + (u.diff(y)/u) + 1 - (exp(x + y)/u) + sol = pdsolve(eq) + assert sol == Eq(f(x, y), F(x - y)*exp(-x/2 - y/2) + exp(x + y)/3) + assert classify_pde(eq) == ('1st_linear_constant_coeff', + '1st_linear_constant_coeff_Integral') + assert checkpdesol(eq, sol)[0] + + eq = 2*u + -u.diff(x) + 3*u.diff(y) + sin(x) + sol = pdsolve(eq) + assert sol == Eq(f(x, y), + F(3*x + y)*exp(x/5 - 3*y/5) - 2*sin(x)/5 - cos(x)/5) + assert classify_pde(eq) == ('1st_linear_constant_coeff', + '1st_linear_constant_coeff_Integral') + assert checkpdesol(eq, sol)[0] + + eq = u + u.diff(x) + u.diff(y) + x*y + sol = pdsolve(eq) + assert sol.expand() == Eq(f(x, y), + x + y + (x - y)**2/4 - (x + y)**2/4 + F(x - y)*exp(-x/2 - y/2) - 2).expand() + assert classify_pde(eq) == ('1st_linear_constant_coeff', + '1st_linear_constant_coeff_Integral') + assert checkpdesol(eq, sol)[0] + eq = u + u.diff(x) + u.diff(y) + log(x) + assert classify_pde(eq) == ('1st_linear_constant_coeff', + '1st_linear_constant_coeff_Integral') + + +def test_pdsolve_all(): + f, F = map(Function, ['f', 'F']) + u = f(x,y) + eq = u + u.diff(x) + u.diff(y) + x**2*y + sol = pdsolve(eq, hint = 'all') + keys = ['1st_linear_constant_coeff', + '1st_linear_constant_coeff_Integral', 'default', 'order'] + assert sorted(sol.keys()) == keys + assert sol['order'] == 1 + assert sol['default'] == '1st_linear_constant_coeff' + assert sol['1st_linear_constant_coeff'].expand() == Eq(f(x, y), + -x**2*y + x**2 + 2*x*y - 4*x - 2*y + F(x - y)*exp(-x/2 - y/2) + 6).expand() + + +def test_pdsolve_variable_coeff(): + f, F = map(Function, ['f', 'F']) + u = f(x, y) + eq = x*(u.diff(x)) - y*(u.diff(y)) + y**2*u - y**2 + sol = pdsolve(eq, hint="1st_linear_variable_coeff") + assert sol == Eq(u, F(x*y)*exp(y**2/2) + 1) + assert checkpdesol(eq, sol)[0] + + eq = x**2*u + x*u.diff(x) + x*y*u.diff(y) + sol = pdsolve(eq, hint='1st_linear_variable_coeff') + assert sol == Eq(u, F(y*exp(-x))*exp(-x**2/2)) + assert checkpdesol(eq, sol)[0] + + eq = y*x**2*u + y*u.diff(x) + u.diff(y) + sol = pdsolve(eq, hint='1st_linear_variable_coeff') + assert sol == Eq(u, F(-2*x + y**2)*exp(-x**3/3)) + assert checkpdesol(eq, sol)[0] + + eq = exp(x)**2*(u.diff(x)) + y + sol = pdsolve(eq, hint='1st_linear_variable_coeff') + assert sol == Eq(u, y*exp(-2*x)/2 + F(y)) + assert checkpdesol(eq, sol)[0] + + eq = exp(2*x)*(u.diff(y)) + y*u - u + sol = pdsolve(eq, hint='1st_linear_variable_coeff') + assert sol == Eq(u, F(x)*exp(-y*(y - 2)*exp(-2*x)/2)) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_polysys.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_polysys.py new file mode 100644 index 0000000000000000000000000000000000000000..9f0a70c89cd94e9f03cb7cc5a009bc8209a21178 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_polysys.py @@ -0,0 +1,178 @@ +"""Tests for solvers of systems of polynomial equations. """ +from sympy.core.numbers import (I, Integer, Rational) +from sympy.core.singleton import S +from sympy.core.symbol import symbols +from sympy.functions.elementary.miscellaneous import sqrt +from sympy.polys.domains.rationalfield import QQ +from sympy.polys.polyerrors import UnsolvableFactorError +from sympy.polys.polyoptions import Options +from sympy.polys.polytools import Poly +from sympy.solvers.solvers import solve +from sympy.utilities.iterables import flatten +from sympy.abc import x, y, z +from sympy.polys import PolynomialError +from sympy.solvers.polysys import (solve_poly_system, + solve_triangulated, + solve_biquadratic, SolveFailed, + solve_generic) +from sympy.polys.polytools import parallel_poly_from_expr +from sympy.testing.pytest import raises + + +def test_solve_poly_system(): + assert solve_poly_system([x - 1], x) == [(S.One,)] + + assert solve_poly_system([y - x, y - x - 1], x, y) is None + + assert solve_poly_system([y - x**2, y + x**2], x, y) == [(S.Zero, S.Zero)] + + assert solve_poly_system([2*x - 3, y*Rational(3, 2) - 2*x, z - 5*y], x, y, z) == \ + [(Rational(3, 2), Integer(2), Integer(10))] + + assert solve_poly_system([x*y - 2*y, 2*y**2 - x**2], x, y) == \ + [(0, 0), (2, -sqrt(2)), (2, sqrt(2))] + + assert solve_poly_system([y - x**2, y + x**2 + 1], x, y) == \ + [(-I*sqrt(S.Half), Rational(-1, 2)), (I*sqrt(S.Half), Rational(-1, 2))] + + f_1 = x**2 + y + z - 1 + f_2 = x + y**2 + z - 1 + f_3 = x + y + z**2 - 1 + + a, b = sqrt(2) - 1, -sqrt(2) - 1 + + assert solve_poly_system([f_1, f_2, f_3], x, y, z) == \ + [(0, 0, 1), (0, 1, 0), (1, 0, 0), (a, a, a), (b, b, b)] + + solution = [(1, -1), (1, 1)] + + assert solve_poly_system([Poly(x**2 - y**2), Poly(x - 1)]) == solution + assert solve_poly_system([x**2 - y**2, x - 1], x, y) == solution + assert solve_poly_system([x**2 - y**2, x - 1]) == solution + + assert solve_poly_system( + [x + x*y - 3, y + x*y - 4], x, y) == [(-3, -2), (1, 2)] + + raises(NotImplementedError, lambda: solve_poly_system([x**3 - y**3], x, y)) + raises(NotImplementedError, lambda: solve_poly_system( + [z, -2*x*y**2 + x + y**2*z, y**2*(-z - 4) + 2])) + raises(PolynomialError, lambda: solve_poly_system([1/x], x)) + + raises(NotImplementedError, lambda: solve_poly_system( + [x-1,], (x, y))) + raises(NotImplementedError, lambda: solve_poly_system( + [y-1,], (x, y))) + + # solve_poly_system should ideally construct solutions using + # CRootOf for the following four tests + assert solve_poly_system([x**5 - x + 1], [x], strict=False) == [] + raises(UnsolvableFactorError, lambda: solve_poly_system( + [x**5 - x + 1], [x], strict=True)) + + assert solve_poly_system([(x - 1)*(x**5 - x + 1), y**2 - 1], [x, y], + strict=False) == [(1, -1), (1, 1)] + raises(UnsolvableFactorError, + lambda: solve_poly_system([(x - 1)*(x**5 - x + 1), y**2-1], + [x, y], strict=True)) + + +def test_solve_generic(): + NewOption = Options((x, y), {'domain': 'ZZ'}) + assert solve_generic([x**2 - 2*y**2, y**2 - y + 1], NewOption) == \ + [(-sqrt(-1 - sqrt(3)*I), Rational(1, 2) - sqrt(3)*I/2), + (sqrt(-1 - sqrt(3)*I), Rational(1, 2) - sqrt(3)*I/2), + (-sqrt(-1 + sqrt(3)*I), Rational(1, 2) + sqrt(3)*I/2), + (sqrt(-1 + sqrt(3)*I), Rational(1, 2) + sqrt(3)*I/2)] + + # solve_generic should ideally construct solutions using + # CRootOf for the following two tests + assert solve_generic( + [2*x - y, (y - 1)*(y**5 - y + 1)], NewOption, strict=False) == \ + [(Rational(1, 2), 1)] + raises(UnsolvableFactorError, lambda: solve_generic( + [2*x - y, (y - 1)*(y**5 - y + 1)], NewOption, strict=True)) + + +def test_solve_biquadratic(): + x0, y0, x1, y1, r = symbols('x0 y0 x1 y1 r') + + f_1 = (x - 1)**2 + (y - 1)**2 - r**2 + f_2 = (x - 2)**2 + (y - 2)**2 - r**2 + s = sqrt(2*r**2 - 1) + a = (3 - s)/2 + b = (3 + s)/2 + assert solve_poly_system([f_1, f_2], x, y) == [(a, b), (b, a)] + + f_1 = (x - 1)**2 + (y - 2)**2 - r**2 + f_2 = (x - 1)**2 + (y - 1)**2 - r**2 + + assert solve_poly_system([f_1, f_2], x, y) == \ + [(1 - sqrt((2*r - 1)*(2*r + 1))/2, Rational(3, 2)), + (1 + sqrt((2*r - 1)*(2*r + 1))/2, Rational(3, 2))] + + query = lambda expr: expr.is_Pow and expr.exp is S.Half + + f_1 = (x - 1 )**2 + (y - 2)**2 - r**2 + f_2 = (x - x1)**2 + (y - 1)**2 - r**2 + + result = solve_poly_system([f_1, f_2], x, y) + + assert len(result) == 2 and all(len(r) == 2 for r in result) + assert all(r.count(query) == 1 for r in flatten(result)) + + f_1 = (x - x0)**2 + (y - y0)**2 - r**2 + f_2 = (x - x1)**2 + (y - y1)**2 - r**2 + + result = solve_poly_system([f_1, f_2], x, y) + + assert len(result) == 2 and all(len(r) == 2 for r in result) + assert all(len(r.find(query)) == 1 for r in flatten(result)) + + s1 = (x*y - y, x**2 - x) + assert solve(s1) == [{x: 1}, {x: 0, y: 0}] + s2 = (x*y - x, y**2 - y) + assert solve(s2) == [{y: 1}, {x: 0, y: 0}] + gens = (x, y) + for seq in (s1, s2): + (f, g), opt = parallel_poly_from_expr(seq, *gens) + raises(SolveFailed, lambda: solve_biquadratic(f, g, opt)) + seq = (x**2 + y**2 - 2, y**2 - 1) + (f, g), opt = parallel_poly_from_expr(seq, *gens) + assert solve_biquadratic(f, g, opt) == [ + (-1, -1), (-1, 1), (1, -1), (1, 1)] + ans = [(0, -1), (0, 1)] + seq = (x**2 + y**2 - 1, y**2 - 1) + (f, g), opt = parallel_poly_from_expr(seq, *gens) + assert solve_biquadratic(f, g, opt) == ans + seq = (x**2 + y**2 - 1, x**2 - x + y**2 - 1) + (f, g), opt = parallel_poly_from_expr(seq, *gens) + assert solve_biquadratic(f, g, opt) == ans + + +def test_solve_triangulated(): + f_1 = x**2 + y + z - 1 + f_2 = x + y**2 + z - 1 + f_3 = x + y + z**2 - 1 + + a, b = sqrt(2) - 1, -sqrt(2) - 1 + + assert solve_triangulated([f_1, f_2, f_3], x, y, z) == \ + [(0, 0, 1), (0, 1, 0), (1, 0, 0)] + + dom = QQ.algebraic_field(sqrt(2)) + + assert solve_triangulated([f_1, f_2, f_3], x, y, z, domain=dom) == \ + [(0, 0, 1), (0, 1, 0), (1, 0, 0), (a, a, a), (b, b, b)] + + +def test_solve_issue_3686(): + roots = solve_poly_system([((x - 5)**2/250000 + (y - Rational(5, 10))**2/250000) - 1, x], x, y) + assert roots == [(0, S.Half - 15*sqrt(1111)), (0, S.Half + 15*sqrt(1111))] + + roots = solve_poly_system([((x - 5)**2/250000 + (y - 5.0/10)**2/250000) - 1, x], x, y) + # TODO: does this really have to be so complicated?! + assert len(roots) == 2 + assert roots[0][0] == 0 + assert roots[0][1].epsilon_eq(-499.474999374969, 1e12) + assert roots[1][0] == 0 + assert roots[1][1].epsilon_eq(500.474999374969, 1e12) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_recurr.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_recurr.py new file mode 100644 index 0000000000000000000000000000000000000000..5a6306b51a5cf33ccd9fae131430a24690d540a7 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_recurr.py @@ -0,0 +1,295 @@ +from sympy.core.function import (Function, Lambda, expand) +from sympy.core.numbers import (I, Rational) +from sympy.core.relational import Eq +from sympy.core.singleton import S +from sympy.core.symbol import (Symbol, symbols) +from sympy.functions.combinatorial.factorials import (rf, binomial, factorial) +from sympy.functions.elementary.complexes import Abs +from sympy.functions.elementary.miscellaneous import sqrt +from sympy.functions.elementary.trigonometric import (cos, sin) +from sympy.polys.polytools import factor +from sympy.solvers.recurr import rsolve, rsolve_hyper, rsolve_poly, rsolve_ratio +from sympy.testing.pytest import raises, slow, XFAIL +from sympy.abc import a, b + +y = Function('y') +n, k = symbols('n,k', integer=True) +C0, C1, C2 = symbols('C0,C1,C2') + + +def test_rsolve_poly(): + assert rsolve_poly([-1, -1, 1], 0, n) == 0 + assert rsolve_poly([-1, -1, 1], 1, n) == -1 + + assert rsolve_poly([-1, n + 1], n, n) == 1 + assert rsolve_poly([-1, 1], n, n) == C0 + (n**2 - n)/2 + assert rsolve_poly([-n - 1, n], 1, n) == C0*n - 1 + assert rsolve_poly([-4*n - 2, 1], 4*n + 1, n) == -1 + + assert rsolve_poly([-1, 1], n**5 + n**3, n) == \ + C0 - n**3 / 2 - n**5 / 2 + n**2 / 6 + n**6 / 6 + 2*n**4 / 3 + + +def test_rsolve_ratio(): + solution = rsolve_ratio([-2*n**3 + n**2 + 2*n - 1, 2*n**3 + n**2 - 6*n, + -2*n**3 - 11*n**2 - 18*n - 9, 2*n**3 + 13*n**2 + 22*n + 8], 0, n) + assert solution == C0*(2*n - 3)/(n**2 - 1)/2 + + +def test_rsolve_hyper(): + assert rsolve_hyper([-1, -1, 1], 0, n) in [ + C0*(S.Half - S.Half*sqrt(5))**n + C1*(S.Half + S.Half*sqrt(5))**n, + C1*(S.Half - S.Half*sqrt(5))**n + C0*(S.Half + S.Half*sqrt(5))**n, + ] + + assert rsolve_hyper([n**2 - 2, -2*n - 1, 1], 0, n) in [ + C0*rf(sqrt(2), n) + C1*rf(-sqrt(2), n), + C1*rf(sqrt(2), n) + C0*rf(-sqrt(2), n), + ] + + assert rsolve_hyper([n**2 - k, -2*n - 1, 1], 0, n) in [ + C0*rf(sqrt(k), n) + C1*rf(-sqrt(k), n), + C1*rf(sqrt(k), n) + C0*rf(-sqrt(k), n), + ] + + assert rsolve_hyper( + [2*n*(n + 1), -n**2 - 3*n + 2, n - 1], 0, n) == C1*factorial(n) + C0*2**n + + assert rsolve_hyper( + [n + 2, -(2*n + 3)*(17*n**2 + 51*n + 39), n + 1], 0, n) == 0 + + assert rsolve_hyper([-n - 1, -1, 1], 0, n) == 0 + + assert rsolve_hyper([-1, 1], n, n).expand() == C0 + n**2/2 - n/2 + + assert rsolve_hyper([-1, 1], 1 + n, n).expand() == C0 + n**2/2 + n/2 + + assert rsolve_hyper([-1, 1], 3*(n + n**2), n).expand() == C0 + n**3 - n + + assert rsolve_hyper([-a, 1],0,n).expand() == C0*a**n + + assert rsolve_hyper([-a, 0, 1], 0, n).expand() == (-1)**n*C1*a**(n/2) + C0*a**(n/2) + + assert rsolve_hyper([1, 1, 1], 0, n).expand() == \ + C0*(Rational(-1, 2) - sqrt(3)*I/2)**n + C1*(Rational(-1, 2) + sqrt(3)*I/2)**n + + assert rsolve_hyper([1, -2*n/a - 2/a, 1], 0, n) == 0 + + +@XFAIL +def test_rsolve_ratio_missed(): + # this arises during computation + # assert rsolve_hyper([-1, 1], 3*(n + n**2), n).expand() == C0 + n**3 - n + assert rsolve_ratio([-n, n + 2], n, n) is not None + + +def recurrence_term(c, f): + """Compute RHS of recurrence in f(n) with coefficients in c.""" + return sum(c[i]*f.subs(n, n + i) for i in range(len(c))) + + +def test_rsolve_bulk(): + """Some bulk-generated tests.""" + funcs = [ n, n + 1, n**2, n**3, n**4, n + n**2, 27*n + 52*n**2 - 3* + n**3 + 12*n**4 - 52*n**5 ] + coeffs = [ [-2, 1], [-2, -1, 1], [-1, 1, 1, -1, 1], [-n, 1], [n**2 - + n + 12, 1] ] + for p in funcs: + # compute difference + for c in coeffs: + q = recurrence_term(c, p) + if p.is_polynomial(n): + assert rsolve_poly(c, q, n) == p + # See issue 3956: + if p.is_hypergeometric(n) and len(c) <= 3: + assert rsolve_hyper(c, q, n).subs(zip(symbols('C:3'), [0, 0, 0])).expand() == p + + +def test_rsolve_0_sol_homogeneous(): + # fixed by cherry-pick from + # https://github.com/diofant/diofant/commit/e1d2e52125199eb3df59f12e8944f8a5f24b00a5 + assert rsolve_hyper([n**2 - n + 12, 1], n*(n**2 - n + 12) + n + 1, n) == n + + +def test_rsolve(): + f = y(n + 2) - y(n + 1) - y(n) + h = sqrt(5)*(S.Half + S.Half*sqrt(5))**n \ + - sqrt(5)*(S.Half - S.Half*sqrt(5))**n + + assert rsolve(f, y(n)) in [ + C0*(S.Half - S.Half*sqrt(5))**n + C1*(S.Half + S.Half*sqrt(5))**n, + C1*(S.Half - S.Half*sqrt(5))**n + C0*(S.Half + S.Half*sqrt(5))**n, + ] + + assert rsolve(f, y(n), [0, 5]) == h + assert rsolve(f, y(n), {0: 0, 1: 5}) == h + assert rsolve(f, y(n), {y(0): 0, y(1): 5}) == h + assert rsolve(y(n) - y(n - 1) - y(n - 2), y(n), [0, 5]) == h + assert rsolve(Eq(y(n), y(n - 1) + y(n - 2)), y(n), [0, 5]) == h + + assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0 + + f = (n - 1)*y(n + 2) - (n**2 + 3*n - 2)*y(n + 1) + 2*n*(n + 1)*y(n) + g = C1*factorial(n) + C0*2**n + h = -3*factorial(n) + 3*2**n + + assert rsolve(f, y(n)) == g + assert rsolve(f, y(n), []) == g + assert rsolve(f, y(n), {}) == g + + assert rsolve(f, y(n), [0, 3]) == h + assert rsolve(f, y(n), {0: 0, 1: 3}) == h + assert rsolve(f, y(n), {y(0): 0, y(1): 3}) == h + + assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0 + + f = y(n) - y(n - 1) - 2 + + assert rsolve(f, y(n), {y(0): 0}) == 2*n + assert rsolve(f, y(n), {y(0): 1}) == 2*n + 1 + assert rsolve(f, y(n), {y(0): 0, y(1): 1}) is None + + assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0 + + f = 3*y(n - 1) - y(n) - 1 + + assert rsolve(f, y(n), {y(0): 0}) == -3**n/2 + S.Half + assert rsolve(f, y(n), {y(0): 1}) == 3**n/2 + S.Half + assert rsolve(f, y(n), {y(0): 2}) == 3*3**n/2 + S.Half + + assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0 + + f = y(n) - 1/n*y(n - 1) + assert rsolve(f, y(n)) == C0/factorial(n) + assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0 + + f = y(n) - 1/n*y(n - 1) - 1 + assert rsolve(f, y(n)) is None + + f = 2*y(n - 1) + (1 - n)*y(n)/n + + assert rsolve(f, y(n), {y(1): 1}) == 2**(n - 1)*n + assert rsolve(f, y(n), {y(1): 2}) == 2**(n - 1)*n*2 + assert rsolve(f, y(n), {y(1): 3}) == 2**(n - 1)*n*3 + + assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0 + + f = (n - 1)*(n - 2)*y(n + 2) - (n + 1)*(n + 2)*y(n) + + assert rsolve(f, y(n), {y(3): 6, y(4): 24}) == n*(n - 1)*(n - 2) + assert rsolve( + f, y(n), {y(3): 6, y(4): -24}) == -n*(n - 1)*(n - 2)*(-1)**(n) + + assert f.subs(y, Lambda(k, rsolve(f, y(n)).subs(n, k))).simplify() == 0 + + assert rsolve(Eq(y(n + 1), a*y(n)), y(n), {y(1): a}).simplify() == a**n + + assert rsolve(y(n) - a*y(n-2),y(n), \ + {y(1): sqrt(a)*(a + b), y(2): a*(a - b)}).simplify() == \ + a**(n/2 + 1) - b*(-sqrt(a))**n + + f = (-16*n**2 + 32*n - 12)*y(n - 1) + (4*n**2 - 12*n + 9)*y(n) + + yn = rsolve(f, y(n), {y(1): binomial(2*n + 1, 3)}) + sol = 2**(2*n)*n*(2*n - 1)**2*(2*n + 1)/12 + assert factor(expand(yn, func=True)) == sol + + sol = rsolve(y(n) + a*(y(n + 1) + y(n - 1))/2, y(n)) + assert str(sol) == 'C0*((-sqrt(1 - a**2) - 1)/a)**n + C1*((sqrt(1 - a**2) - 1)/a)**n' + + assert rsolve((k + 1)*y(k), y(k)) is None + assert (rsolve((k + 1)*y(k) + (k + 3)*y(k + 1) + (k + 5)*y(k + 2), y(k)) + is None) + + assert rsolve(y(n) + y(n + 1) + 2**n + 3**n, y(n)) == (-1)**n*C0 - 2**n/3 - 3**n/4 + + +def test_rsolve_raises(): + x = Function('x') + raises(ValueError, lambda: rsolve(y(n) - y(k + 1), y(n))) + raises(ValueError, lambda: rsolve(y(n) - y(n + 1), x(n))) + raises(ValueError, lambda: rsolve(y(n) - x(n + 1), y(n))) + raises(ValueError, lambda: rsolve(y(n) - sqrt(n)*y(n + 1), y(n))) + raises(ValueError, lambda: rsolve(y(n) - y(n + 1), y(n), {x(0): 0})) + raises(ValueError, lambda: rsolve(y(n) + y(n + 1) + 2**n + cos(n), y(n))) + + +def test_issue_6844(): + f = y(n + 2) - y(n + 1) + y(n)/4 + assert rsolve(f, y(n)) == 2**(-n + 1)*C1*n + 2**(-n)*C0 + assert rsolve(f, y(n), {y(0): 0, y(1): 1}) == 2**(1 - n)*n + + +def test_issue_18751(): + r = Symbol('r', positive=True) + theta = Symbol('theta', real=True) + f = y(n) - 2 * r * cos(theta) * y(n - 1) + r**2 * y(n - 2) + assert rsolve(f, y(n)) == \ + C0*(r*(cos(theta) - I*Abs(sin(theta))))**n + C1*(r*(cos(theta) + I*Abs(sin(theta))))**n + + +def test_constant_naming(): + #issue 8697 + assert rsolve(y(n+3) - y(n+2) - y(n+1) + y(n), y(n)) == (-1)**n*C1 + C0 + C2*n + assert rsolve(y(n+3)+3*y(n+2)+3*y(n+1)+y(n), y(n)).expand() == (-1)**n*C0 - (-1)**n*C1*n - (-1)**n*C2*n**2 + assert rsolve(y(n) - 2*y(n - 3) + 5*y(n - 2) - 4*y(n - 1),y(n),[1,3,8]) == 3*2**n - n - 2 + + #issue 19630 + assert rsolve(y(n+3) - 3*y(n+1) + 2*y(n), y(n), {y(1):0, y(2):8, y(3):-2}) == (-2)**n + 2*n + + +@slow +def test_issue_15751(): + f = y(n) + 21*y(n + 1) - 273*y(n + 2) - 1092*y(n + 3) + 1820*y(n + 4) + 1092*y(n + 5) - 273*y(n + 6) - 21*y(n + 7) + y(n + 8) + assert rsolve(f, y(n)) is not None + + +def test_issue_17990(): + f = -10*y(n) + 4*y(n + 1) + 6*y(n + 2) + 46*y(n + 3) + sol = rsolve(f, y(n)) + expected = C0*((86*18**(S(1)/3)/69 + (-12 + (-1 + sqrt(3)*I)*(290412 + + 3036*sqrt(9165))**(S(1)/3))*(1 - sqrt(3)*I)*(24201 + 253*sqrt(9165))** + (S(1)/3)/276)/((1 - sqrt(3)*I)*(24201 + 253*sqrt(9165))**(S(1)/3)) + )**n + C1*((86*18**(S(1)/3)/69 + (-12 + (-1 - sqrt(3)*I)*(290412 + 3036 + *sqrt(9165))**(S(1)/3))*(1 + sqrt(3)*I)*(24201 + 253*sqrt(9165))** + (S(1)/3)/276)/((1 + sqrt(3)*I)*(24201 + 253*sqrt(9165))**(S(1)/3)) + )**n + C2*(-43*18**(S(1)/3)/(69*(24201 + 253*sqrt(9165))**(S(1)/3)) - + S(1)/23 + (290412 + 3036*sqrt(9165))**(S(1)/3)/138)**n + assert sol == expected + e = sol.subs({C0: 1, C1: 1, C2: 1, n: 1}).evalf() + assert abs(e + 0.130434782608696) < 1e-13 + + +def test_issue_8697(): + a = Function('a') + eq = a(n + 3) - a(n + 2) - a(n + 1) + a(n) + assert rsolve(eq, a(n)) == (-1)**n*C1 + C0 + C2*n + eq2 = a(n + 3) + 3*a(n + 2) + 3*a(n + 1) + a(n) + assert (rsolve(eq2, a(n)) == + (-1)**n*C0 + (-1)**(n + 1)*C1*n + (-1)**(n + 1)*C2*n**2) + + assert rsolve(a(n) - 2*a(n - 3) + 5*a(n - 2) - 4*a(n - 1), + a(n), {a(0): 1, a(1): 3, a(2): 8}) == 3*2**n - n - 2 + + # From issue thread (but fixed by https://github.com/diofant/diofant/commit/da9789c6cd7d0c2ceeea19fbf59645987125b289): + assert rsolve(a(n) - 2*a(n - 1) - n, a(n), {a(0): 1}) == 3*2**n - n - 2 + + +def test_diofantissue_294(): + f = y(n) - y(n - 1) - 2*y(n - 2) - 2*n + assert rsolve(f, y(n)) == (-1)**n*C0 + 2**n*C1 - n - Rational(5, 2) + # issue sympy/sympy#11261 + assert rsolve(f, y(n), {y(0): -1, y(1): 1}) == (-(-1)**n/2 + 2*2**n - + n - Rational(5, 2)) + # issue sympy/sympy#7055 + assert rsolve(-2*y(n) + y(n + 1) + n - 1, y(n)) == 2**n*C0 + n + + +def test_issue_15553(): + f = Function("f") + assert rsolve(Eq(f(n), 2*f(n - 1) + n), f(n)) == 2**n*C0 - n - 2 + assert rsolve(Eq(f(n + 1), 2*f(n) + n**2 + 1), f(n)) == 2**n*C0 - n**2 - 2*n - 4 + assert rsolve(Eq(f(n + 1), 2*f(n) + n**2 + 1), f(n), {f(1): 0}) == 7*2**n/2 - n**2 - 2*n - 4 + assert rsolve(Eq(f(n), 2*f(n - 1) + 3*n**2), f(n)) == 2**n*C0 - 3*n**2 - 12*n - 18 + assert rsolve(Eq(f(n), 2*f(n - 1) + n**2), f(n)) == 2**n*C0 - n**2 - 4*n - 6 + assert rsolve(Eq(f(n), 2*f(n - 1) + n), f(n), {f(0): 1}) == 3*2**n - n - 2 diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_simplex.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_simplex.py new file mode 100644 index 0000000000000000000000000000000000000000..611205f5df009a6d0de6e687501695b63bb932c9 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_simplex.py @@ -0,0 +1,254 @@ +from sympy.core.numbers import Rational +from sympy.core.relational import Eq, Ne +from sympy.core.symbol import symbols +from sympy.core.sympify import sympify +from sympy.core.singleton import S +from sympy.core.random import random, choice +from sympy.functions.elementary.miscellaneous import sqrt +from sympy.ntheory.generate import randprime +from sympy.matrices.dense import Matrix +from sympy.solvers.solveset import linear_eq_to_matrix +from sympy.solvers.simplex import (_lp as lp, _primal_dual, + UnboundedLPError, InfeasibleLPError, lpmin, lpmax, + _m, _abcd, _simplex, linprog) + +from sympy.external.importtools import import_module + +from sympy.testing.pytest import raises + +from sympy.abc import x, y, z + + +np = import_module("numpy") +scipy = import_module("scipy") + + +def test_lp(): + r1 = y + 2*z <= 3 + r2 = -x - 3*z <= -2 + r3 = 2*x + y + 7*z <= 5 + constraints = [r1, r2, r3, x >= 0, y >= 0, z >= 0] + objective = -x - y - 5 * z + ans = optimum, argmax = lp(max, objective, constraints) + assert ans == lpmax(objective, constraints) + assert objective.subs(argmax) == optimum + for constr in constraints: + assert constr.subs(argmax) == True + + r1 = x - y + 2*z <= 3 + r2 = -x + 2*y - 3*z <= -2 + r3 = 2*x + y - 7*z <= -5 + constraints = [r1, r2, r3, x >= 0, y >= 0, z >= 0] + objective = -x - y - 5*z + ans = optimum, argmax = lp(max, objective, constraints) + assert ans == lpmax(objective, constraints) + assert objective.subs(argmax) == optimum + for constr in constraints: + assert constr.subs(argmax) == True + + r1 = x - y + 2*z <= -4 + r2 = -x + 2*y - 3*z <= 8 + r3 = 2*x + y - 7*z <= 10 + constraints = [r1, r2, r3, x >= 0, y >= 0, z >= 0] + const = 2 + objective = -x-y-5*z+const # has constant term + ans = optimum, argmax = lp(max, objective, constraints) + assert ans == lpmax(objective, constraints) + assert objective.subs(argmax) == optimum + for constr in constraints: + assert constr.subs(argmax) == True + + # Section 4 Problem 1 from + # http://web.tecnico.ulisboa.pt/mcasquilho/acad/or/ftp/FergusonUCLA_LP.pdf + # answer on page 55 + v = x1, x2, x3, x4 = symbols('x1 x2 x3 x4') + r1 = x1 - x2 - 2*x3 - x4 <= 4 + r2 = 2*x1 + x3 -4*x4 <= 2 + r3 = -2*x1 + x2 + x4 <= 1 + objective, constraints = x1 - 2*x2 - 3*x3 - x4, [r1, r2, r3] + [ + i >= 0 for i in v] + ans = optimum, argmax = lp(max, objective, constraints) + assert ans == lpmax(objective, constraints) + assert ans == (4, {x1: 7, x2: 0, x3: 0, x4: 3}) + + # input contains Floats + r1 = x - y + 2.0*z <= -4 + r2 = -x + 2*y - 3.0*z <= 8 + r3 = 2*x + y - 7*z <= 10 + constraints = [r1, r2, r3] + [i >= 0 for i in (x, y, z)] + objective = -x-y-5*z + optimum, argmax = lp(max, objective, constraints) + assert objective.subs(argmax) == optimum + for constr in constraints: + assert constr.subs(argmax) == True + + # input contains non-float or non-Rational + r1 = x - y + sqrt(2) * z <= -4 + r2 = -x + 2*y - 3*z <= 8 + r3 = 2*x + y - 7*z <= 10 + raises(TypeError, lambda: lp(max, -x-y-5*z, [r1, r2, r3])) + + r1 = x >= 0 + raises(UnboundedLPError, lambda: lp(max, x, [r1])) + r2 = x <= -1 + raises(InfeasibleLPError, lambda: lp(max, x, [r1, r2])) + + # strict inequalities are not allowed + r1 = x > 0 + raises(TypeError, lambda: lp(max, x, [r1])) + + # not equals not allowed + r1 = Ne(x, 0) + raises(TypeError, lambda: lp(max, x, [r1])) + + def make_random_problem(nvar=2, num_constraints=2, sparsity=.1): + def rand(): + if random() < sparsity: + return sympify(0) + int1, int2 = [randprime(0, 200) for _ in range(2)] + return Rational(int1, int2)*choice([-1, 1]) + variables = symbols('x1:%s' % (nvar + 1)) + constraints = [(sum(rand()*x for x in variables) <= rand()) + for _ in range(num_constraints)] + objective = sum(rand() * x for x in variables) + return objective, constraints, variables + + # equality + r1 = Eq(x, y) + r2 = Eq(y, z) + r3 = z <= 3 + constraints = [r1, r2, r3] + objective = x + ans = optimum, argmax = lp(max, objective, constraints) + assert ans == lpmax(objective, constraints) + assert objective.subs(argmax) == optimum + for constr in constraints: + assert constr.subs(argmax) == True + + +def test_simplex(): + L = [ + [[1, 1], [-1, 1], [0, 1], [-1, 0]], + [5, 1, 2, -1], + [[1, 1]], + [-1]] + A, B, C, D = _abcd(_m(*L), list=False) + assert _simplex(A, B, -C, -D) == (-6, [3, 2], [1, 0, 0, 0]) + assert _simplex(A, B, -C, -D, dual=True) == (-6, + [1, 0, 0, 0], [5, 0]) + + assert _simplex([[]],[],[[1]],[0]) == (0, [0], []) + + # handling of Eq (or Eq-like x<=y, x>=y conditions) + assert lpmax(x - y, [x <= y + 2, x >= y + 2, x >= 0, y >= 0] + ) == (2, {x: 2, y: 0}) + assert lpmax(x - y, [x <= y + 2, Eq(x, y + 2), x >= 0, y >= 0] + ) == (2, {x: 2, y: 0}) + assert lpmax(x - y, [x <= y + 2, Eq(x, 2)]) == (2, {x: 2, y: 0}) + assert lpmax(y, [Eq(y, 2)]) == (2, {y: 2}) + + # the conditions are equivalent to Eq(x, y + 2) + assert lpmin(y, [x <= y + 2, x >= y + 2, y >= 0] + ) == (0, {x: 2, y: 0}) + # equivalent to Eq(y, -2) + assert lpmax(y, [0 <= y + 2, 0 >= y + 2]) == (-2, {y: -2}) + assert lpmax(y, [0 <= y + 2, 0 >= y + 2, y <= 0] + ) == (-2, {y: -2}) + + # extra symbols symbols + assert lpmin(x, [y >= 1, x >= y]) == (1, {x: 1, y: 1}) + assert lpmin(x, [y >= 1, x >= y + z, x >= 0, z >= 0] + ) == (1, {x: 1, y: 1, z: 0}) + + # detect oscillation + # o1 + v = x1, x2, x3, x4 = symbols('x1 x2 x3 x4') + raises(InfeasibleLPError, lambda: lpmin( + 9*x2 - 8*x3 + 3*x4 + 6, + [5*x2 - 2*x3 <= 0, + -x1 - 8*x2 + 9*x3 <= -3, + 10*x1 - x2+ 9*x4 <= -4] + [i >= 0 for i in v])) + # o2 - equations fed to lpmin are changed into a matrix + # system that doesn't oscillate and has the same solution + # as below + M = linear_eq_to_matrix + f = 5*x2 + x3 + 4*x4 - x1 + L = 5*x2 + 2*x3 + 5*x4 - (x1 + 5) + cond = [L <= 0] + [Eq(3*x2 + x4, 2), Eq(-x1 + x3 + 2*x4, 1)] + c, d = M(f, v) + a, b = M(L, v) + aeq, beq = M(cond[1:], v) + ans = (S(9)/2, [0, S(1)/2, 0, S(1)/2]) + assert linprog(c, a, b, aeq, beq, bounds=(0, 1)) == ans + lpans = lpmin(f, cond + [x1 >= 0, x1 <= 1, + x2 >= 0, x2 <= 1, x3 >= 0, x3 <= 1, x4 >= 0, x4 <= 1]) + assert (lpans[0], list(lpans[1].values())) == ans + + +def test_lpmin_lpmax(): + v = x1, x2, y1, y2 = symbols('x1 x2 y1 y2') + L = [[1, -1]], [1], [[1, 1]], [2] + a, b, c, d = [Matrix(i) for i in L] + m = Matrix([[a, b], [c, d]]) + f, constr = _primal_dual(m)[0] + ans = lpmin(f, constr + [i >= 0 for i in v[:2]]) + assert ans == (-1, {x1: 1, x2: 0}),ans + + L = [[1, -1], [1, 1]], [1, 1], [[1, 1]], [2] + a, b, c, d = [Matrix(i) for i in L] + m = Matrix([[a, b], [c, d]]) + f, constr = _primal_dual(m)[1] + ans = lpmax(f, constr + [i >= 0 for i in v[-2:]]) + assert ans == (-1, {y1: 1, y2: 0}) + + +def test_linprog(): + for do in range(2): + if not do: + M = lambda a, b: linear_eq_to_matrix(a, b) + else: + # check matrices as list + M = lambda a, b: tuple([ + i.tolist() for i in linear_eq_to_matrix(a, b)]) + + v = x, y, z = symbols('x1:4') + f = x + y - 2*z + c = M(f, v)[0] + ineq = [7*x + 4*y - 7*z <= 3, + 3*x - y + 10*z <= 6, + x >= 0, y >= 0, z >= 0] + ab = M([i.lts - i.gts for i in ineq], v) + ans = (-S(6)/5, [0, 0, S(3)/5]) + assert lpmin(f, ineq) == (ans[0], dict(zip(v, ans[1]))) + assert linprog(c, *ab) == ans + + f += 1 + c = M(f, v)[0] + eq = [Eq(y - 9*x, 1)] + abeq = M([i.lhs - i.rhs for i in eq], v) + ans = (1 - S(2)/5, [0, 1, S(7)/10]) + assert lpmin(f, ineq + eq) == (ans[0], dict(zip(v, ans[1]))) + assert linprog(c, *ab, *abeq) == (ans[0] - 1, ans[1]) + + eq = [z - y <= S.Half] + abeq = M([i.lhs - i.rhs for i in eq], v) + ans = (1 - S(10)/9, [0, S(1)/9, S(11)/18]) + assert lpmin(f, ineq + eq) == (ans[0], dict(zip(v, ans[1]))) + assert linprog(c, *ab, *abeq) == (ans[0] - 1, ans[1]) + + bounds = [(0, None), (0, None), (None, S.Half)] + ans = (0, [0, 0, S.Half]) + assert lpmin(f, ineq + [z <= S.Half]) == ( + ans[0], dict(zip(v, ans[1]))) + assert linprog(c, *ab, bounds=bounds) == (ans[0] - 1, ans[1]) + assert linprog(c, *ab, bounds={v.index(z): bounds[-1]} + ) == (ans[0] - 1, ans[1]) + eq = [z - y <= S.Half] + + assert linprog([[1]], [], [], bounds=(2, 3)) == (2, [2]) + assert linprog([1], [], [], bounds=(2, 3)) == (2, [2]) + assert linprog([1], bounds=(2, 3)) == (2, [2]) + assert linprog([1, -1], [[1, 1]], [2], bounds={1:(None, None)} + ) == (-2, [0, 2]) + assert linprog([1, -1], [[1, 1]], [5], bounds={1:(3, None)} + ) == (-5, [0, 5]) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_solvers.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_solvers.py new file mode 100644 index 0000000000000000000000000000000000000000..c3ef819bbecd171adebdad76d9e52dead4f9fe31 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_solvers.py @@ -0,0 +1,2703 @@ +from sympy.assumptions.ask import (Q, ask) +from sympy.core.add import Add +from sympy.core.containers import Tuple +from sympy.core.function import (Derivative, Function, diff) +from sympy.core.mod import Mod +from sympy.core.mul import Mul +from sympy.core import (GoldenRatio, TribonacciConstant) +from sympy.core.numbers import (E, Float, I, Rational, oo, pi) +from sympy.core.relational import (Eq, Gt, Lt, Ne) +from sympy.core.singleton import S +from sympy.core.symbol import (Dummy, Symbol, Wild, symbols) +from sympy.core.sympify import sympify +from sympy.functions.combinatorial.factorials import binomial +from sympy.functions.elementary.complexes import (Abs, arg, conjugate, im, re) +from sympy.functions.elementary.exponential import (LambertW, exp, log) +from sympy.functions.elementary.hyperbolic import (atanh, cosh, sinh, tanh) +from sympy.functions.elementary.integers import floor +from sympy.functions.elementary.miscellaneous import (cbrt, root, sqrt) +from sympy.functions.elementary.piecewise import Piecewise +from sympy.functions.elementary.trigonometric import (acos, asin, atan, atan2, cos, sec, sin, tan) +from sympy.functions.special.error_functions import (erf, erfc, erfcinv, erfinv) +from sympy.integrals.integrals import Integral +from sympy.logic.boolalg import (And, Or) +from sympy.matrices.dense import Matrix +from sympy.matrices import SparseMatrix +from sympy.polys.polytools import Poly +from sympy.printing.str import sstr +from sympy.simplify.radsimp import denom +from sympy.solvers.solvers import (nsolve, solve, solve_linear) + +from sympy.core.function import nfloat +from sympy.solvers import solve_linear_system, solve_linear_system_LU, \ + solve_undetermined_coeffs +from sympy.solvers.bivariate import _filtered_gens, _solve_lambert, _lambert +from sympy.solvers.solvers import _invert, unrad, checksol, posify, _ispow, \ + det_quick, det_perm, det_minor, _simple_dens, denoms + +from sympy.physics.units import cm +from sympy.polys.rootoftools import CRootOf + +from sympy.testing.pytest import slow, XFAIL, SKIP, raises +from sympy.core.random import verify_numerically as tn + +from sympy.abc import a, b, c, d, e, k, h, p, x, y, z, t, q, m, R + + +def NS(e, n=15, **options): + return sstr(sympify(e).evalf(n, **options), full_prec=True) + + +def test_swap_back(): + f, g = map(Function, 'fg') + fx, gx = f(x), g(x) + assert solve([fx + y - 2, fx - gx - 5], fx, y, gx) == \ + {fx: gx + 5, y: -gx - 3} + assert solve(fx + gx*x - 2, [fx, gx], dict=True) == [{fx: 2, gx: 0}] + assert solve(fx + gx**2*x - y, [fx, gx], dict=True) == [{fx: y, gx: 0}] + assert solve([f(1) - 2, x + 2], dict=True) == [{x: -2, f(1): 2}] + + +def guess_solve_strategy(eq, symbol): + try: + solve(eq, symbol) + return True + except (TypeError, NotImplementedError): + return False + + +def test_guess_poly(): + # polynomial equations + assert guess_solve_strategy( S(4), x ) # == GS_POLY + assert guess_solve_strategy( x, x ) # == GS_POLY + assert guess_solve_strategy( x + a, x ) # == GS_POLY + assert guess_solve_strategy( 2*x, x ) # == GS_POLY + assert guess_solve_strategy( x + sqrt(2), x) # == GS_POLY + assert guess_solve_strategy( x + 2**Rational(1, 4), x) # == GS_POLY + assert guess_solve_strategy( x**2 + 1, x ) # == GS_POLY + assert guess_solve_strategy( x**2 - 1, x ) # == GS_POLY + assert guess_solve_strategy( x*y + y, x ) # == GS_POLY + assert guess_solve_strategy( x*exp(y) + y, x) # == GS_POLY + assert guess_solve_strategy( + (x - y**3)/(y**2*sqrt(1 - y**2)), x) # == GS_POLY + + +def test_guess_poly_cv(): + # polynomial equations via a change of variable + assert guess_solve_strategy( sqrt(x) + 1, x ) # == GS_POLY_CV_1 + assert guess_solve_strategy( + x**Rational(1, 3) + sqrt(x) + 1, x ) # == GS_POLY_CV_1 + assert guess_solve_strategy( 4*x*(1 - sqrt(x)), x ) # == GS_POLY_CV_1 + + # polynomial equation multiplying both sides by x**n + assert guess_solve_strategy( x + 1/x + y, x ) # == GS_POLY_CV_2 + + +def test_guess_rational_cv(): + # rational functions + assert guess_solve_strategy( (x + 1)/(x**2 + 2), x) # == GS_RATIONAL + assert guess_solve_strategy( + (x - y**3)/(y**2*sqrt(1 - y**2)), y) # == GS_RATIONAL_CV_1 + + # rational functions via the change of variable y -> x**n + assert guess_solve_strategy( (sqrt(x) + 1)/(x**Rational(1, 3) + sqrt(x) + 1), x ) \ + #== GS_RATIONAL_CV_1 + + +def test_guess_transcendental(): + #transcendental functions + assert guess_solve_strategy( exp(x) + 1, x ) # == GS_TRANSCENDENTAL + assert guess_solve_strategy( 2*cos(x) - y, x ) # == GS_TRANSCENDENTAL + assert guess_solve_strategy( + exp(x) + exp(-x) - y, x ) # == GS_TRANSCENDENTAL + assert guess_solve_strategy(3**x - 10, x) # == GS_TRANSCENDENTAL + assert guess_solve_strategy(-3**x + 10, x) # == GS_TRANSCENDENTAL + + assert guess_solve_strategy(a*x**b - y, x) # == GS_TRANSCENDENTAL + + +def test_solve_args(): + # equation container, issue 5113 + ans = {x: -3, y: 1} + eqs = (x + 5*y - 2, -3*x + 6*y - 15) + assert all(solve(container(eqs), x, y) == ans for container in + (tuple, list, set, frozenset)) + assert solve(Tuple(*eqs), x, y) == ans + # implicit symbol to solve for + assert set(solve(x**2 - 4)) == {S(2), -S(2)} + assert solve([x + y - 3, x - y - 5]) == {x: 4, y: -1} + assert solve(x - exp(x), x, implicit=True) == [exp(x)] + # no symbol to solve for + assert solve(42) == solve(42, x) == [] + assert solve([1, 2]) == [] + assert solve([sqrt(2)],[x]) == [] + # duplicate symbols raises + raises(ValueError, lambda: solve((x - 3, y + 2), x, y, x)) + raises(ValueError, lambda: solve(x, x, x)) + # no error in exclude + assert solve(x, x, exclude=[y, y]) == [0] + # duplicate symbols raises + raises(ValueError, lambda: solve((x - 3, y + 2), x, y, x)) + raises(ValueError, lambda: solve(x, x, x)) + # no error in exclude + assert solve(x, x, exclude=[y, y]) == [0] + # unordered symbols + # only 1 + assert solve(y - 3, {y}) == [3] + # more than 1 + assert solve(y - 3, {x, y}) == [{y: 3}] + # multiple symbols: take the first linear solution+ + # - return as tuple with values for all requested symbols + assert solve(x + y - 3, [x, y]) == [(3 - y, y)] + # - unless dict is True + assert solve(x + y - 3, [x, y], dict=True) == [{x: 3 - y}] + # - or no symbols are given + assert solve(x + y - 3) == [{x: 3 - y}] + # multiple symbols might represent an undetermined coefficients system + assert solve(a + b*x - 2, [a, b]) == {a: 2, b: 0} + assert solve((a + b)*x + b - c, [a, b]) == {a: -c, b: c} + eq = a*x**2 + b*x + c - ((x - h)**2 + 4*p*k)/4/p + # - check that flags are obeyed + sol = solve(eq, [h, p, k], exclude=[a, b, c]) + assert sol == {h: -b/(2*a), k: (4*a*c - b**2)/(4*a), p: 1/(4*a)} + assert solve(eq, [h, p, k], dict=True) == [sol] + assert solve(eq, [h, p, k], set=True) == \ + ([h, p, k], {(-b/(2*a), 1/(4*a), (4*a*c - b**2)/(4*a))}) + # issue 23889 - polysys not simplified + assert solve(eq, [h, p, k], exclude=[a, b, c], simplify=False) == \ + {h: -b/(2*a), k: (4*a*c - b**2)/(4*a), p: 1/(4*a)} + # but this only happens when system has a single solution + args = (a + b)*x - b**2 + 2, a, b + assert solve(*args) == [((b**2 - b*x - 2)/x, b)] + # and if the system has a solution; the following doesn't so + # an algebraic solution is returned + assert solve(a*x + b**2/(x + 4) - 3*x - 4/x, a, b, dict=True) == \ + [{a: (-b**2*x + 3*x**3 + 12*x**2 + 4*x + 16)/(x**2*(x + 4))}] + # failed single equation + assert solve(1/(1/x - y + exp(y))) == [] + raises( + NotImplementedError, lambda: solve(exp(x) + sin(x) + exp(y) + sin(y))) + # failed system + # -- when no symbols given, 1 fails + assert solve([y, exp(x) + x]) == [{x: -LambertW(1), y: 0}] + # both fail + assert solve( + (exp(x) - x, exp(y) - y)) == [{x: -LambertW(-1), y: -LambertW(-1)}] + # -- when symbols given + assert solve([y, exp(x) + x], x, y) == [(-LambertW(1), 0)] + # symbol is a number + assert solve(x**2 - pi, pi) == [x**2] + # no equations + assert solve([], [x]) == [] + # nonlinear system + assert solve((x**2 - 4, y - 2), x, y) == [(-2, 2), (2, 2)] + assert solve((x**2 - 4, y - 2), y, x) == [(2, -2), (2, 2)] + assert solve((x**2 - 4 + z, y - 2 - z), a, z, y, x, set=True + ) == ([a, z, y, x], { + (a, z, z + 2, -sqrt(4 - z)), + (a, z, z + 2, sqrt(4 - z))}) + # overdetermined system + # - nonlinear + assert solve([(x + y)**2 - 4, x + y - 2]) == [{x: -y + 2}] + # - linear + assert solve((x + y - 2, 2*x + 2*y - 4)) == {x: -y + 2} + # When one or more args are Boolean + assert solve(Eq(x**2, 0.0)) == [0.0] # issue 19048 + assert solve([True, Eq(x, 0)], [x], dict=True) == [{x: 0}] + assert solve([Eq(x, x), Eq(x, 0), Eq(x, x+1)], [x], dict=True) == [] + assert not solve([Eq(x, x+1), x < 2], x) + assert solve([Eq(x, 0), x+1<2]) == Eq(x, 0) + assert solve([Eq(x, x), Eq(x, x+1)], x) == [] + assert solve(True, x) == [] + assert solve([x - 1, False], [x], set=True) == ([], set()) + assert solve([-y*(x + y - 1)/2, (y - 1)/x/y + 1/y], + set=True, check=False) == ([x, y], {(1 - y, y), (x, 0)}) + # ordering should be canonical, fastest to order by keys instead + # of by size + assert list(solve((y - 1, x - sqrt(3)*z)).keys()) == [x, y] + # as set always returns as symbols, set even if no solution + assert solve([x - 1, x], (y, x), set=True) == ([y, x], set()) + assert solve([x - 1, x], {y, x}, set=True) == ([x, y], set()) + + +def test_solve_polynomial1(): + assert solve(3*x - 2, x) == [Rational(2, 3)] + assert solve(Eq(3*x, 2), x) == [Rational(2, 3)] + + assert set(solve(x**2 - 1, x)) == {-S.One, S.One} + assert set(solve(Eq(x**2, 1), x)) == {-S.One, S.One} + + assert solve(x - y**3, x) == [y**3] + rx = root(x, 3) + assert solve(x - y**3, y) == [ + rx, -rx/2 - sqrt(3)*I*rx/2, -rx/2 + sqrt(3)*I*rx/2] + a11, a12, a21, a22, b1, b2 = symbols('a11,a12,a21,a22,b1,b2') + + assert solve([a11*x + a12*y - b1, a21*x + a22*y - b2], x, y) == \ + { + x: (a22*b1 - a12*b2)/(a11*a22 - a12*a21), + y: (a11*b2 - a21*b1)/(a11*a22 - a12*a21), + } + + solution = {x: S.Zero, y: S.Zero} + + assert solve((x - y, x + y), x, y ) == solution + assert solve((x - y, x + y), (x, y)) == solution + assert solve((x - y, x + y), [x, y]) == solution + + assert set(solve(x**3 - 15*x - 4, x)) == { + -2 + 3**S.Half, + S(4), + -2 - 3**S.Half + } + + assert set(solve((x**2 - 1)**2 - a, x)) == \ + {sqrt(1 + sqrt(a)), -sqrt(1 + sqrt(a)), + sqrt(1 - sqrt(a)), -sqrt(1 - sqrt(a))} + + +def test_solve_polynomial2(): + assert solve(4, x) == [] + + +def test_solve_polynomial_cv_1a(): + """ + Test for solving on equations that can be converted to a polynomial equation + using the change of variable y -> x**Rational(p, q) + """ + assert solve( sqrt(x) - 1, x) == [1] + assert solve( sqrt(x) - 2, x) == [4] + assert solve( x**Rational(1, 4) - 2, x) == [16] + assert solve( x**Rational(1, 3) - 3, x) == [27] + assert solve(sqrt(x) + x**Rational(1, 3) + x**Rational(1, 4), x) == [0] + + +def test_solve_polynomial_cv_1b(): + assert set(solve(4*x*(1 - a*sqrt(x)), x)) == {S.Zero, 1/a**2} + assert set(solve(x*(root(x, 3) - 3), x)) == {S.Zero, S(27)} + + +def test_solve_polynomial_cv_2(): + """ + Test for solving on equations that can be converted to a polynomial equation + multiplying both sides of the equation by x**m + """ + assert solve(x + 1/x - 1, x) in \ + [[ S.Half + I*sqrt(3)/2, S.Half - I*sqrt(3)/2], + [ S.Half - I*sqrt(3)/2, S.Half + I*sqrt(3)/2]] + + +def test_quintics_1(): + f = x**5 - 110*x**3 - 55*x**2 + 2310*x + 979 + s = solve(f, check=False) + for r in s: + res = f.subs(x, r.n()).n() + assert tn(res, 0) + + f = x**5 - 15*x**3 - 5*x**2 + 10*x + 20 + s = solve(f) + for r in s: + assert r.func == CRootOf + + # if one uses solve to get the roots of a polynomial that has a CRootOf + # solution, make sure that the use of nfloat during the solve process + # doesn't fail. Note: if you want numerical solutions to a polynomial + # it is *much* faster to use nroots to get them than to solve the + # equation only to get RootOf solutions which are then numerically + # evaluated. So for eq = x**5 + 3*x + 7 do Poly(eq).nroots() rather + # than [i.n() for i in solve(eq)] to get the numerical roots of eq. + assert nfloat(solve(x**5 + 3*x**3 + 7)[0], exponent=False) == \ + CRootOf(x**5 + 3*x**3 + 7, 0).n() + + +def test_quintics_2(): + f = x**5 + 15*x + 12 + s = solve(f, check=False) + for r in s: + res = f.subs(x, r.n()).n() + assert tn(res, 0) + + f = x**5 - 15*x**3 - 5*x**2 + 10*x + 20 + s = solve(f) + for r in s: + assert r.func == CRootOf + + assert solve(x**5 - 6*x**3 - 6*x**2 + x - 6) == [ + CRootOf(x**5 - 6*x**3 - 6*x**2 + x - 6, 0), + CRootOf(x**5 - 6*x**3 - 6*x**2 + x - 6, 1), + CRootOf(x**5 - 6*x**3 - 6*x**2 + x - 6, 2), + CRootOf(x**5 - 6*x**3 - 6*x**2 + x - 6, 3), + CRootOf(x**5 - 6*x**3 - 6*x**2 + x - 6, 4)] + + +def test_quintics_3(): + y = x**5 + x**3 - 2**Rational(1, 3) + assert solve(y) == solve(-y) == [] + + +def test_highorder_poly(): + # just testing that the uniq generator is unpacked + sol = solve(x**6 - 2*x + 2) + assert all(isinstance(i, CRootOf) for i in sol) and len(sol) == 6 + + +def test_solve_rational(): + """Test solve for rational functions""" + assert solve( ( x - y**3 )/( (y**2)*sqrt(1 - y**2) ), x) == [y**3] + + +def test_solve_conjugate(): + """Test solve for simple conjugate functions""" + assert solve(conjugate(x) -3 + I) == [3 + I] + + +def test_solve_nonlinear(): + assert solve(x**2 - y**2, x, y, dict=True) == [{x: -y}, {x: y}] + assert solve(x**2 - y**2/exp(x), y, x, dict=True) == [{y: -x*sqrt(exp(x))}, + {y: x*sqrt(exp(x))}] + + +def test_issue_8666(): + x = symbols('x') + assert solve(Eq(x**2 - 1/(x**2 - 4), 4 - 1/(x**2 - 4)), x) == [] + assert solve(Eq(x + 1/x, 1/x), x) == [] + + +def test_issue_7228(): + assert solve(4**(2*(x**2) + 2*x) - 8, x) == [Rational(-3, 2), S.Half] + + +def test_issue_7190(): + assert solve(log(x-3) + log(x+3), x) == [sqrt(10)] + + +def test_issue_21004(): + x = symbols('x') + f = x/sqrt(x**2+1) + f_diff = f.diff(x) + assert solve(f_diff, x) == [] + + +def test_issue_24650(): + x = symbols('x') + r = solve(Eq(Piecewise((x, Eq(x, 0) | (x > 1))), 0)) + assert r == [0] + r = checksol(Eq(Piecewise((x, Eq(x, 0) | (x > 1))), 0), x, sol=0) + assert r is True + + +def test_linear_system(): + x, y, z, t, n = symbols('x, y, z, t, n') + + assert solve([x - 1, x - y, x - 2*y, y - 1], [x, y]) == [] + + assert solve([x - 1, x - y, x - 2*y, x - 1], [x, y]) == [] + assert solve([x - 1, x - 1, x - y, x - 2*y], [x, y]) == [] + + assert solve([x + 5*y - 2, -3*x + 6*y - 15], x, y) == {x: -3, y: 1} + + M = Matrix([[0, 0, n*(n + 1), (n + 1)**2, 0], + [n + 1, n + 1, -2*n - 1, -(n + 1), 0], + [-1, 0, 1, 0, 0]]) + + assert solve_linear_system(M, x, y, z, t) == \ + {x: t*(-n-1)/n, y: 0, z: t*(-n-1)/n} + + assert solve([x + y + z + t, -z - t], x, y, z, t) == {x: -y, z: -t} + + +@XFAIL +def test_linear_system_xfail(): + # https://github.com/sympy/sympy/issues/6420 + M = Matrix([[0, 15.0, 10.0, 700.0], + [1, 1, 1, 100.0], + [0, 10.0, 5.0, 200.0], + [-5.0, 0, 0, 0 ]]) + + assert solve_linear_system(M, x, y, z) == {x: 0, y: -60.0, z: 160.0} + + +def test_linear_system_function(): + a = Function('a') + assert solve([a(0, 0) + a(0, 1) + a(1, 0) + a(1, 1), -a(1, 0) - a(1, 1)], + a(0, 0), a(0, 1), a(1, 0), a(1, 1)) == {a(1, 0): -a(1, 1), a(0, 0): -a(0, 1)} + + +def test_linear_system_symbols_doesnt_hang_1(): + + def _mk_eqs(wy): + # Equations for fitting a wy*2 - 1 degree polynomial between two points, + # at end points derivatives are known up to order: wy - 1 + order = 2*wy - 1 + x, x0, x1 = symbols('x, x0, x1', real=True) + y0s = symbols('y0_:{}'.format(wy), real=True) + y1s = symbols('y1_:{}'.format(wy), real=True) + c = symbols('c_:{}'.format(order+1), real=True) + + expr = sum(coeff*x**o for o, coeff in enumerate(c)) + eqs = [] + for i in range(wy): + eqs.append(expr.diff(x, i).subs({x: x0}) - y0s[i]) + eqs.append(expr.diff(x, i).subs({x: x1}) - y1s[i]) + return eqs, c + + # + # The purpose of this test is just to see that these calls don't hang. The + # expressions returned are complicated so are not included here. Testing + # their correctness takes longer than solving the system. + # + + for n in range(1, 7+1): + eqs, c = _mk_eqs(n) + solve(eqs, c) + + +def test_linear_system_symbols_doesnt_hang_2(): + + M = Matrix([ + [66, 24, 39, 50, 88, 40, 37, 96, 16, 65, 31, 11, 37, 72, 16, 19, 55, 37, 28, 76], + [10, 93, 34, 98, 59, 44, 67, 74, 74, 94, 71, 61, 60, 23, 6, 2, 57, 8, 29, 78], + [19, 91, 57, 13, 64, 65, 24, 53, 77, 34, 85, 58, 87, 39, 39, 7, 36, 67, 91, 3], + [74, 70, 15, 53, 68, 43, 86, 83, 81, 72, 25, 46, 67, 17, 59, 25, 78, 39, 63, 6], + [69, 40, 67, 21, 67, 40, 17, 13, 93, 44, 46, 89, 62, 31, 30, 38, 18, 20, 12, 81], + [50, 22, 74, 76, 34, 45, 19, 76, 28, 28, 11, 99, 97, 82, 8, 46, 99, 57, 68, 35], + [58, 18, 45, 88, 10, 64, 9, 34, 90, 82, 17, 41, 43, 81, 45, 83, 22, 88, 24, 39], + [42, 21, 70, 68, 6, 33, 64, 81, 83, 15, 86, 75, 86, 17, 77, 34, 62, 72, 20, 24], + [ 7, 8, 2, 72, 71, 52, 96, 5, 32, 51, 31, 36, 79, 88, 25, 77, 29, 26, 33, 13], + [19, 31, 30, 85, 81, 39, 63, 28, 19, 12, 16, 49, 37, 66, 38, 13, 3, 71, 61, 51], + [29, 82, 80, 49, 26, 85, 1, 37, 2, 74, 54, 82, 26, 47, 54, 9, 35, 0, 99, 40], + [15, 49, 82, 91, 93, 57, 45, 25, 45, 97, 15, 98, 48, 52, 66, 24, 62, 54, 97, 37], + [62, 23, 73, 53, 52, 86, 28, 38, 0, 74, 92, 38, 97, 70, 71, 29, 26, 90, 67, 45], + [ 2, 32, 23, 24, 71, 37, 25, 71, 5, 41, 97, 65, 93, 13, 65, 45, 25, 88, 69, 50], + [40, 56, 1, 29, 79, 98, 79, 62, 37, 28, 45, 47, 3, 1, 32, 74, 98, 35, 84, 32], + [33, 15, 87, 79, 65, 9, 14, 63, 24, 19, 46, 28, 74, 20, 29, 96, 84, 91, 93, 1], + [97, 18, 12, 52, 1, 2, 50, 14, 52, 76, 19, 82, 41, 73, 51, 79, 13, 3, 82, 96], + [40, 28, 52, 10, 10, 71, 56, 78, 82, 5, 29, 48, 1, 26, 16, 18, 50, 76, 86, 52], + [38, 89, 83, 43, 29, 52, 90, 77, 57, 0, 67, 20, 81, 88, 48, 96, 88, 58, 14, 3]]) + + syms = x0,x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12,x13,x14,x15,x16,x17,x18 = symbols('x:19') + + sol = { + x0: -S(1967374186044955317099186851240896179)/3166636564687820453598895768302256588, + x1: -S(84268280268757263347292368432053826)/791659141171955113399723942075564147, + x2: -S(229962957341664730974463872411844965)/1583318282343910226799447884151128294, + x3: S(990156781744251750886760432229180537)/6333273129375640907197791536604513176, + x4: -S(2169830351210066092046760299593096265)/18999819388126922721593374609813539528, + x5: S(4680868883477577389628494526618745355)/9499909694063461360796687304906769764, + x6: -S(1590820774344371990683178396480879213)/3166636564687820453598895768302256588, + x7: -S(54104723404825537735226491634383072)/339282489073695048599881689460956063, + x8: S(3182076494196560075964847771774733847)/6333273129375640907197791536604513176, + x9: -S(10870817431029210431989147852497539675)/18999819388126922721593374609813539528, + x10: -S(13118019242576506476316318268573312603)/18999819388126922721593374609813539528, + x11: -S(5173852969886775824855781403820641259)/4749954847031730680398343652453384882, + x12: S(4261112042731942783763341580651820563)/4749954847031730680398343652453384882, + x13: -S(821833082694661608993818117038209051)/6333273129375640907197791536604513176, + x14: S(906881575107250690508618713632090559)/904753304196520129599684505229216168, + x15: -S(732162528717458388995329317371283987)/6333273129375640907197791536604513176, + x16: S(4524215476705983545537087360959896817)/9499909694063461360796687304906769764, + x17: -S(3898571347562055611881270844646055217)/6333273129375640907197791536604513176, + x18: S(7513502486176995632751685137907442269)/18999819388126922721593374609813539528 + } + + eqs = list(M * Matrix(syms + (1,))) + assert solve(eqs, syms) == sol + + y = Symbol('y') + eqs = list(y * M * Matrix(syms + (1,))) + assert solve(eqs, syms) == sol + + +def test_linear_systemLU(): + n = Symbol('n') + + M = Matrix([[1, 2, 0, 1], [1, 3, 2*n, 1], [4, -1, n**2, 1]]) + + assert solve_linear_system_LU(M, [x, y, z]) == {z: -3/(n**2 + 18*n), + x: 1 - 12*n/(n**2 + 18*n), + y: 6*n/(n**2 + 18*n)} + +# Note: multiple solutions exist for some of these equations, so the tests +# should be expected to break if the implementation of the solver changes +# in such a way that a different branch is chosen + +@slow +def test_solve_transcendental(): + from sympy.abc import a, b + + assert solve(exp(x) - 3, x) == [log(3)] + assert set(solve((a*x + b)*(exp(x) - 3), x)) == {-b/a, log(3)} + assert solve(cos(x) - y, x) == [-acos(y) + 2*pi, acos(y)] + assert solve(2*cos(x) - y, x) == [-acos(y/2) + 2*pi, acos(y/2)] + assert solve(Eq(cos(x), sin(x)), x) == [pi/4] + + assert set(solve(exp(x) + exp(-x) - y, x)) in [{ + log(y/2 - sqrt(y**2 - 4)/2), + log(y/2 + sqrt(y**2 - 4)/2), + }, { + log(y - sqrt(y**2 - 4)) - log(2), + log(y + sqrt(y**2 - 4)) - log(2)}, + { + log(y/2 - sqrt((y - 2)*(y + 2))/2), + log(y/2 + sqrt((y - 2)*(y + 2))/2)}] + assert solve(exp(x) - 3, x) == [log(3)] + assert solve(Eq(exp(x), 3), x) == [log(3)] + assert solve(log(x) - 3, x) == [exp(3)] + assert solve(sqrt(3*x) - 4, x) == [Rational(16, 3)] + assert solve(3**(x + 2), x) == [] + assert solve(3**(2 - x), x) == [] + assert solve(x + 2**x, x) == [-LambertW(log(2))/log(2)] + assert solve(2*x + 5 + log(3*x - 2), x) == \ + [Rational(2, 3) + LambertW(2*exp(Rational(-19, 3))/3)/2] + assert solve(3*x + log(4*x), x) == [LambertW(Rational(3, 4))/3] + assert set(solve((2*x + 8)*(8 + exp(x)), x)) == {S(-4), log(8) + pi*I} + eq = 2*exp(3*x + 4) - 3 + ans = solve(eq, x) # this generated a failure in flatten + assert len(ans) == 3 and all(eq.subs(x, a).n(chop=True) == 0 for a in ans) + assert solve(2*log(3*x + 4) - 3, x) == [(exp(Rational(3, 2)) - 4)/3] + assert solve(exp(x) + 1, x) == [pi*I] + + eq = 2*(3*x + 4)**5 - 6*7**(3*x + 9) + result = solve(eq, x) + x0 = -log(2401) + x1 = 3**Rational(1, 5) + x2 = log(7**(7*x1/20)) + x3 = sqrt(2) + x4 = sqrt(5) + x5 = x3*sqrt(x4 - 5) + x6 = x4 + 1 + x7 = 1/(3*log(7)) + x8 = -x4 + x9 = x3*sqrt(x8 - 5) + x10 = x8 + 1 + ans = [x7*(x0 - 5*LambertW(x2*(-x5 + x6))), + x7*(x0 - 5*LambertW(x2*(x5 + x6))), + x7*(x0 - 5*LambertW(x2*(x10 - x9))), + x7*(x0 - 5*LambertW(x2*(x10 + x9))), + x7*(x0 - 5*LambertW(-log(7**(7*x1/5))))] + assert result == ans, result + # it works if expanded, too + assert solve(eq.expand(), x) == result + + assert solve(z*cos(x) - y, x) == [-acos(y/z) + 2*pi, acos(y/z)] + assert solve(z*cos(2*x) - y, x) == [-acos(y/z)/2 + pi, acos(y/z)/2] + assert solve(z*cos(sin(x)) - y, x) == [ + pi - asin(acos(y/z)), asin(acos(y/z) - 2*pi) + pi, + -asin(acos(y/z) - 2*pi), asin(acos(y/z))] + + assert solve(z*cos(x), x) == [pi/2, pi*Rational(3, 2)] + + # issue 4508 + assert solve(y - b*x/(a + x), x) in [[-a*y/(y - b)], [a*y/(b - y)]] + assert solve(y - b*exp(a/x), x) == [a/log(y/b)] + # issue 4507 + assert solve(y - b/(1 + a*x), x) in [[(b - y)/(a*y)], [-((y - b)/(a*y))]] + # issue 4506 + assert solve(y - a*x**b, x) == [(y/a)**(1/b)] + # issue 4505 + assert solve(z**x - y, x) == [log(y)/log(z)] + # issue 4504 + assert solve(2**x - 10, x) == [1 + log(5)/log(2)] + # issue 6744 + assert solve(x*y) == [{x: 0}, {y: 0}] + assert solve([x*y]) == [{x: 0}, {y: 0}] + assert solve(x**y - 1) == [{x: 1}, {y: 0}] + assert solve([x**y - 1]) == [{x: 1}, {y: 0}] + assert solve(x*y*(x**2 - y**2)) == [{x: 0}, {x: -y}, {x: y}, {y: 0}] + assert solve([x*y*(x**2 - y**2)]) == [{x: 0}, {x: -y}, {x: y}, {y: 0}] + # issue 4739 + assert solve(exp(log(5)*x) - 2**x, x) == [0] + # issue 14791 + assert solve(exp(log(5)*x) - exp(log(2)*x), x) == [0] + f = Function('f') + assert solve(y*f(log(5)*x) - y*f(log(2)*x), x) == [0] + assert solve(f(x) - f(0), x) == [0] + assert solve(f(x) - f(2 - x), x) == [1] + raises(NotImplementedError, lambda: solve(f(x, y) - f(1, 2), x)) + raises(NotImplementedError, lambda: solve(f(x, y) - f(2 - x, 2), x)) + raises(ValueError, lambda: solve(f(x, y) - f(1 - x), x)) + raises(ValueError, lambda: solve(f(x, y) - f(1), x)) + + # misc + # make sure that the right variables is picked up in tsolve + # shouldn't generate a GeneratorsNeeded error in _tsolve when the NaN is generated + # for eq_down. Actual answers, as determined numerically are approx. +/- 0.83 + raises(NotImplementedError, lambda: + solve(sinh(x)*sinh(sinh(x)) + cosh(x)*cosh(sinh(x)) - 3)) + + # watch out for recursive loop in tsolve + raises(NotImplementedError, lambda: solve((x + 2)**y*x - 3, x)) + + # issue 7245 + assert solve(sin(sqrt(x))) == [0, pi**2] + + # issue 7602 + a, b = symbols('a, b', real=True, negative=False) + assert str(solve(Eq(a, 0.5 - cos(pi*b)/2), b)) == \ + '[2.0 - 0.318309886183791*acos(1.0 - 2.0*a), 0.318309886183791*acos(1.0 - 2.0*a)]' + + # issue 15325 + assert solve(y**(1/x) - z, x) == [log(y)/log(z)] + + # issue 25685 (basic trig identies should give simple solutions) + for yi in [cos(2*x),sin(2*x),cos(x - pi/3)]: + sol = solve([cos(x) - S(3)/5, yi - y]) + assert (sol[0][y] + sol[1][y]).is_Rational, (yi,sol) + # don't allow massive expansion + assert solve(cos(1000*x) - S.Half) == [pi/3000, pi/600] + assert solve(cos(x - 1000*y) - 1, x) == [1000*y, 1000*y + 2*pi] + assert solve(cos(x + y + z) - 1, x) == [-y - z, -y - z + 2*pi] + + # issue 26008 + assert solve(sin(x + pi/6)) == [-pi/6, 5*pi/6] + + +def test_solve_for_functions_derivatives(): + t = Symbol('t') + x = Function('x')(t) + y = Function('y')(t) + a11, a12, a21, a22, b1, b2 = symbols('a11,a12,a21,a22,b1,b2') + + soln = solve([a11*x + a12*y - b1, a21*x + a22*y - b2], x, y) + assert soln == { + x: (a22*b1 - a12*b2)/(a11*a22 - a12*a21), + y: (a11*b2 - a21*b1)/(a11*a22 - a12*a21), + } + + assert solve(x - 1, x) == [1] + assert solve(3*x - 2, x) == [Rational(2, 3)] + + soln = solve([a11*x.diff(t) + a12*y.diff(t) - b1, a21*x.diff(t) + + a22*y.diff(t) - b2], x.diff(t), y.diff(t)) + assert soln == { y.diff(t): (a11*b2 - a21*b1)/(a11*a22 - a12*a21), + x.diff(t): (a22*b1 - a12*b2)/(a11*a22 - a12*a21) } + + assert solve(x.diff(t) - 1, x.diff(t)) == [1] + assert solve(3*x.diff(t) - 2, x.diff(t)) == [Rational(2, 3)] + + eqns = {3*x - 1, 2*y - 4} + assert solve(eqns, {x, y}) == { x: Rational(1, 3), y: 2 } + x = Symbol('x') + f = Function('f') + F = x**2 + f(x)**2 - 4*x - 1 + assert solve(F.diff(x), diff(f(x), x)) == [(-x + 2)/f(x)] + + # Mixed cased with a Symbol and a Function + x = Symbol('x') + y = Function('y')(t) + + soln = solve([a11*x + a12*y.diff(t) - b1, a21*x + + a22*y.diff(t) - b2], x, y.diff(t)) + assert soln == { y.diff(t): (a11*b2 - a21*b1)/(a11*a22 - a12*a21), + x: (a22*b1 - a12*b2)/(a11*a22 - a12*a21) } + + # issue 13263 + x = Symbol('x') + f = Function('f') + soln = solve([f(x).diff(x) + f(x).diff(x, 2) - 1, f(x).diff(x) - f(x).diff(x, 2)], + f(x).diff(x), f(x).diff(x, 2)) + assert soln == { f(x).diff(x, 2): S(1)/2, f(x).diff(x): S(1)/2 } + + soln = solve([f(x).diff(x, 2) + f(x).diff(x, 3) - 1, 1 - f(x).diff(x, 2) - + f(x).diff(x, 3), 1 - f(x).diff(x,3)], f(x).diff(x, 2), f(x).diff(x, 3)) + assert soln == { f(x).diff(x, 2): 0, f(x).diff(x, 3): 1 } + + +def test_issue_3725(): + f = Function('f') + F = x**2 + f(x)**2 - 4*x - 1 + e = F.diff(x) + assert solve(e, f(x).diff(x)) in [[(2 - x)/f(x)], [-((x - 2)/f(x))]] + + +def test_issue_3870(): + a, b, c, d = symbols('a b c d') + A = Matrix(2, 2, [a, b, c, d]) + B = Matrix(2, 2, [0, 2, -3, 0]) + C = Matrix(2, 2, [1, 2, 3, 4]) + + assert solve(A*B - C, [a, b, c, d]) == {a: 1, b: Rational(-1, 3), c: 2, d: -1} + assert solve([A*B - C], [a, b, c, d]) == {a: 1, b: Rational(-1, 3), c: 2, d: -1} + assert solve(Eq(A*B, C), [a, b, c, d]) == {a: 1, b: Rational(-1, 3), c: 2, d: -1} + + assert solve([A*B - B*A], [a, b, c, d]) == {a: d, b: Rational(-2, 3)*c} + assert solve([A*C - C*A], [a, b, c, d]) == {a: d - c, b: Rational(2, 3)*c} + assert solve([A*B - B*A, A*C - C*A], [a, b, c, d]) == {a: d, b: 0, c: 0} + + assert solve([Eq(A*B, B*A)], [a, b, c, d]) == {a: d, b: Rational(-2, 3)*c} + assert solve([Eq(A*C, C*A)], [a, b, c, d]) == {a: d - c, b: Rational(2, 3)*c} + assert solve([Eq(A*B, B*A), Eq(A*C, C*A)], [a, b, c, d]) == {a: d, b: 0, c: 0} + + +def test_solve_linear(): + w = Wild('w') + assert solve_linear(x, x) == (0, 1) + assert solve_linear(x, exclude=[x]) == (0, 1) + assert solve_linear(x, symbols=[w]) == (0, 1) + assert solve_linear(x, y - 2*x) in [(x, y/3), (y, 3*x)] + assert solve_linear(x, y - 2*x, exclude=[x]) == (y, 3*x) + assert solve_linear(3*x - y, 0) in [(x, y/3), (y, 3*x)] + assert solve_linear(3*x - y, 0, [x]) == (x, y/3) + assert solve_linear(3*x - y, 0, [y]) == (y, 3*x) + assert solve_linear(x**2/y, 1) == (y, x**2) + assert solve_linear(w, x) in [(w, x), (x, w)] + assert solve_linear(cos(x)**2 + sin(x)**2 + 2 + y) == \ + (y, -2 - cos(x)**2 - sin(x)**2) + assert solve_linear(cos(x)**2 + sin(x)**2 + 2 + y, symbols=[x]) == (0, 1) + assert solve_linear(Eq(x, 3)) == (x, 3) + assert solve_linear(1/(1/x - 2)) == (0, 0) + assert solve_linear((x + 1)*exp(-x), symbols=[x]) == (x, -1) + assert solve_linear((x + 1)*exp(x), symbols=[x]) == ((x + 1)*exp(x), 1) + assert solve_linear(x*exp(-x**2), symbols=[x]) == (x, 0) + assert solve_linear(0**x - 1) == (0**x - 1, 1) + assert solve_linear(1 + 1/(x - 1)) == (x, 0) + eq = y*cos(x)**2 + y*sin(x)**2 - y # = y*(1 - 1) = 0 + assert solve_linear(eq) == (0, 1) + eq = cos(x)**2 + sin(x)**2 # = 1 + assert solve_linear(eq) == (0, 1) + raises(ValueError, lambda: solve_linear(Eq(x, 3), 3)) + + +def test_solve_undetermined_coeffs(): + assert solve_undetermined_coeffs( + a*x**2 + b*x**2 + b*x + 2*c*x + c + 1, [a, b, c], x + ) == {a: -2, b: 2, c: -1} + # Test that rational functions work + assert solve_undetermined_coeffs(a/x + b/(x + 1) + - (2*x + 1)/(x**2 + x), [a, b], x) == {a: 1, b: 1} + # Test cancellation in rational functions + assert solve_undetermined_coeffs( + ((c + 1)*a*x**2 + (c + 1)*b*x**2 + + (c + 1)*b*x + (c + 1)*2*c*x + (c + 1)**2)/(c + 1), + [a, b, c], x) == \ + {a: -2, b: 2, c: -1} + # multivariate + X, Y, Z = y, x**y, y*x**y + eq = a*X + b*Y + c*Z - X - 2*Y - 3*Z + coeffs = a, b, c + syms = x, y + assert solve_undetermined_coeffs(eq, coeffs) == { + a: 1, b: 2, c: 3} + assert solve_undetermined_coeffs(eq, coeffs, syms) == { + a: 1, b: 2, c: 3} + assert solve_undetermined_coeffs(eq, coeffs, *syms) == { + a: 1, b: 2, c: 3} + # check output format + assert solve_undetermined_coeffs(a*x + a - 2, [a]) == [] + assert solve_undetermined_coeffs(a**2*x - 4*x, [a]) == [ + {a: -2}, {a: 2}] + assert solve_undetermined_coeffs(0, [a]) == [] + assert solve_undetermined_coeffs(0, [a], dict=True) == [] + assert solve_undetermined_coeffs(0, [a], set=True) == ([], {}) + assert solve_undetermined_coeffs(1, [a]) == [] + abeq = a*x - 2*x + b - 3 + s = {b, a} + assert solve_undetermined_coeffs(abeq, s, x) == {a: 2, b: 3} + assert solve_undetermined_coeffs(abeq, s, x, set=True) == ([a, b], {(2, 3)}) + assert solve_undetermined_coeffs(sin(a*x) - sin(2*x), (a,)) is None + assert solve_undetermined_coeffs(a*x + b*x - 2*x, (a, b)) == {a: 2 - b} + + +def test_solve_inequalities(): + x = Symbol('x') + sol = And(S.Zero < x, x < oo) + assert solve(x + 1 > 1) == sol + assert solve([x + 1 > 1]) == sol + assert solve([x + 1 > 1], x) == sol + assert solve([x + 1 > 1], [x]) == sol + + system = [Lt(x**2 - 2, 0), Gt(x**2 - 1, 0)] + assert solve(system) == \ + And(Or(And(Lt(-sqrt(2), x), Lt(x, -1)), + And(Lt(1, x), Lt(x, sqrt(2)))), Eq(0, 0)) + + x = Symbol('x', real=True) + system = [Lt(x**2 - 2, 0), Gt(x**2 - 1, 0)] + assert solve(system) == \ + Or(And(Lt(-sqrt(2), x), Lt(x, -1)), And(Lt(1, x), Lt(x, sqrt(2)))) + + # issues 6627, 3448 + assert solve((x - 3)/(x - 2) < 0, x) == And(Lt(2, x), Lt(x, 3)) + assert solve(x/(x + 1) > 1, x) == And(Lt(-oo, x), Lt(x, -1)) + + assert solve(sin(x) > S.Half) == And(pi/6 < x, x < pi*Rational(5, 6)) + + assert solve(Eq(False, x < 1)) == (S.One <= x) & (x < oo) + assert solve(Eq(True, x < 1)) == (-oo < x) & (x < 1) + assert solve(Eq(x < 1, False)) == (S.One <= x) & (x < oo) + assert solve(Eq(x < 1, True)) == (-oo < x) & (x < 1) + + assert solve(Eq(False, x)) == False + assert solve(Eq(0, x)) == [0] + assert solve(Eq(True, x)) == True + assert solve(Eq(1, x)) == [1] + assert solve(Eq(False, ~x)) == True + assert solve(Eq(True, ~x)) == False + assert solve(Ne(True, x)) == False + assert solve(Ne(1, x)) == (x > -oo) & (x < oo) & Ne(x, 1) + + +def test_issue_4793(): + assert solve(1/x) == [] + assert solve(x*(1 - 5/x)) == [5] + assert solve(x + sqrt(x) - 2) == [1] + assert solve(-(1 + x)/(2 + x)**2 + 1/(2 + x)) == [] + assert solve(-x**2 - 2*x + (x + 1)**2 - 1) == [] + assert solve((x/(x + 1) + 3)**(-2)) == [] + assert solve(x/sqrt(x**2 + 1), x) == [0] + assert solve(exp(x) - y, x) == [log(y)] + assert solve(exp(x)) == [] + assert solve(x**2 + x + sin(y)**2 + cos(y)**2 - 1, x) in [[0, -1], [-1, 0]] + eq = 4*3**(5*x + 2) - 7 + ans = solve(eq, x) + assert len(ans) == 5 and all(eq.subs(x, a).n(chop=True) == 0 for a in ans) + assert solve(log(x**2) - y**2/exp(x), x, y, set=True) == ( + [x, y], + {(x, sqrt(exp(x) * log(x ** 2))), (x, -sqrt(exp(x) * log(x ** 2)))}) + assert solve(x**2*z**2 - z**2*y**2) == [{x: -y}, {x: y}, {z: 0}] + assert solve((x - 1)/(1 + 1/(x - 1))) == [] + assert solve(x**(y*z) - x, x) == [1] + raises(NotImplementedError, lambda: solve(log(x) - exp(x), x)) + raises(NotImplementedError, lambda: solve(2**x - exp(x) - 3)) + + +def test_PR1964(): + # issue 5171 + assert solve(sqrt(x)) == solve(sqrt(x**3)) == [0] + assert solve(sqrt(x - 1)) == [1] + # issue 4462 + a = Symbol('a') + assert solve(-3*a/sqrt(x), x) == [] + # issue 4486 + assert solve(2*x/(x + 2) - 1, x) == [2] + # issue 4496 + assert set(solve((x**2/(7 - x)).diff(x))) == {S.Zero, S(14)} + # issue 4695 + f = Function('f') + assert solve((3 - 5*x/f(x))*f(x), f(x)) == [x*Rational(5, 3)] + # issue 4497 + assert solve(1/root(5 + x, 5) - 9, x) == [Rational(-295244, 59049)] + + assert solve(sqrt(x) + sqrt(sqrt(x)) - 4) == [(Rational(-1, 2) + sqrt(17)/2)**4] + assert set(solve(Poly(sqrt(exp(x)) + sqrt(exp(-x)) - 4))) in \ + [ + {log((-sqrt(3) + 2)**2), log((sqrt(3) + 2)**2)}, + {2*log(-sqrt(3) + 2), 2*log(sqrt(3) + 2)}, + {log(-4*sqrt(3) + 7), log(4*sqrt(3) + 7)}, + ] + assert set(solve(Poly(exp(x) + exp(-x) - 4))) == \ + {log(-sqrt(3) + 2), log(sqrt(3) + 2)} + assert set(solve(x**y + x**(2*y) - 1, x)) == \ + {(Rational(-1, 2) + sqrt(5)/2)**(1/y), (Rational(-1, 2) - sqrt(5)/2)**(1/y)} + + assert solve(exp(x/y)*exp(-z/y) - 2, y) == [(x - z)/log(2)] + assert solve( + x**z*y**z - 2, z) in [[log(2)/(log(x) + log(y))], [log(2)/(log(x*y))]] + # if you do inversion too soon then multiple roots (as for the following) + # will be missed, e.g. if exp(3*x) = exp(3) -> 3*x = 3 + E = S.Exp1 + assert solve(exp(3*x) - exp(3), x) in [ + [1, log(E*(Rational(-1, 2) - sqrt(3)*I/2)), log(E*(Rational(-1, 2) + sqrt(3)*I/2))], + [1, log(-E/2 - sqrt(3)*E*I/2), log(-E/2 + sqrt(3)*E*I/2)], + ] + + # coverage test + p = Symbol('p', positive=True) + assert solve((1/p + 1)**(p + 1)) == [] + + +def test_issue_5197(): + x = Symbol('x', real=True) + assert solve(x**2 + 1, x) == [] + n = Symbol('n', integer=True, positive=True) + assert solve((n - 1)*(n + 2)*(2*n - 1), n) == [1] + x = Symbol('x', positive=True) + y = Symbol('y') + assert solve([x + 5*y - 2, -3*x + 6*y - 15], x, y) == [] + # not {x: -3, y: 1} b/c x is positive + # The solution following should not contain (-sqrt(2), sqrt(2)) + assert solve([(x + y), 2 - y**2], x, y) == [(sqrt(2), -sqrt(2))] + y = Symbol('y', positive=True) + # The solution following should not contain {y: -x*exp(x/2)} + assert solve(x**2 - y**2/exp(x), y, x, dict=True) == [{y: x*exp(x/2)}] + x, y, z = symbols('x y z', positive=True) + assert solve(z**2*x**2 - z**2*y**2/exp(x), y, x, z, dict=True) == [{y: x*exp(x/2)}] + + +def test_checking(): + assert set( + solve(x*(x - y/x), x, check=False)) == {sqrt(y), S.Zero, -sqrt(y)} + assert set(solve(x*(x - y/x), x, check=True)) == {sqrt(y), -sqrt(y)} + # {x: 0, y: 4} sets denominator to 0 in the following so system should return None + assert solve((1/(1/x + 2), 1/(y - 3) - 1)) == [] + # 0 sets denominator of 1/x to zero so None is returned + assert solve(1/(1/x + 2)) == [] + + +def test_issue_4671_4463_4467(): + assert solve(sqrt(x**2 - 1) - 2) in ([sqrt(5), -sqrt(5)], + [-sqrt(5), sqrt(5)]) + assert solve((2**exp(y**2/x) + 2)/(x**2 + 15), y) == [ + -sqrt(x*log(1 + I*pi/log(2))), sqrt(x*log(1 + I*pi/log(2)))] + + C1, C2 = symbols('C1 C2') + f = Function('f') + assert solve(C1 + C2/x**2 - exp(-f(x)), f(x)) == [log(x**2/(C1*x**2 + C2))] + a = Symbol('a') + E = S.Exp1 + assert solve(1 - log(a + 4*x**2), x) in ( + [-sqrt(-a + E)/2, sqrt(-a + E)/2], + [sqrt(-a + E)/2, -sqrt(-a + E)/2] + ) + assert solve(log(a**(-3) - x**2)/a, x) in ( + [-sqrt(-1 + a**(-3)), sqrt(-1 + a**(-3))], + [sqrt(-1 + a**(-3)), -sqrt(-1 + a**(-3))],) + assert solve(1 - log(a + 4*x**2), x) in ( + [-sqrt(-a + E)/2, sqrt(-a + E)/2], + [sqrt(-a + E)/2, -sqrt(-a + E)/2],) + assert solve((a**2 + 1)*(sin(a*x) + cos(a*x)), x) == [-pi/(4*a)] + assert solve(3 - (sinh(a*x) + cosh(a*x)), x) == [log(3)/a] + assert set(solve(3 - (sinh(a*x) + cosh(a*x)**2), x)) == \ + {log(-2 + sqrt(5))/a, log(-sqrt(2) + 1)/a, + log(-sqrt(5) - 2)/a, log(1 + sqrt(2))/a} + assert solve(atan(x) - 1) == [tan(1)] + + +def test_issue_5132(): + r, t = symbols('r,t') + assert set(solve([r - x**2 - y**2, tan(t) - y/x], [x, y])) == \ + {( + -sqrt(r*cos(t)**2), -1*sqrt(r*cos(t)**2)*tan(t)), + (sqrt(r*cos(t)**2), sqrt(r*cos(t)**2)*tan(t))} + assert solve([exp(x) - sin(y), 1/y - 3], [x, y]) == \ + [(log(sin(Rational(1, 3))), Rational(1, 3))] + assert solve([exp(x) - sin(y), 1/exp(y) - 3], [x, y]) == \ + [(log(-sin(log(3))), -log(3))] + assert set(solve([exp(x) - sin(y), y**2 - 4], [x, y])) == \ + {(log(-sin(2)), -S(2)), (log(sin(2)), S(2))} + eqs = [exp(x)**2 - sin(y) + z**2, 1/exp(y) - 3] + assert solve(eqs, set=True) == \ + ([y, z], { + (-log(3), sqrt(-exp(2*x) - sin(log(3)))), + (-log(3), -sqrt(-exp(2*x) - sin(log(3))))}) + assert solve(eqs, x, z, set=True) == ( + [x, z], + {(x, sqrt(-exp(2*x) + sin(y))), (x, -sqrt(-exp(2*x) + sin(y)))}) + assert set(solve(eqs, x, y)) == \ + { + (log(-sqrt(-z**2 - sin(log(3)))), -log(3)), + (log(-z**2 - sin(log(3)))/2, -log(3))} + assert set(solve(eqs, y, z)) == \ + { + (-log(3), -sqrt(-exp(2*x) - sin(log(3)))), + (-log(3), sqrt(-exp(2*x) - sin(log(3))))} + eqs = [exp(x)**2 - sin(y) + z, 1/exp(y) - 3] + assert solve(eqs, set=True) == ([y, z], { + (-log(3), -exp(2*x) - sin(log(3)))}) + assert solve(eqs, x, z, set=True) == ( + [x, z], {(x, -exp(2*x) + sin(y))}) + assert set(solve(eqs, x, y)) == { + (log(-sqrt(-z - sin(log(3)))), -log(3)), + (log(-z - sin(log(3)))/2, -log(3))} + assert solve(eqs, z, y) == \ + [(-exp(2*x) - sin(log(3)), -log(3))] + assert solve((sqrt(x**2 + y**2) - sqrt(10), x + y - 4), set=True) == ( + [x, y], {(S.One, S(3)), (S(3), S.One)}) + assert set(solve((sqrt(x**2 + y**2) - sqrt(10), x + y - 4), x, y)) == \ + {(S.One, S(3)), (S(3), S.One)} + + +def test_issue_5335(): + lam, a0, conc = symbols('lam a0 conc') + a = 0.005 + b = 0.743436700916726 + eqs = [lam + 2*y - a0*(1 - x/2)*x - a*x/2*x, + a0*(1 - x/2)*x - 1*y - b*y, + x + y - conc] + sym = [x, y, a0] + # there are 4 solutions obtained manually but only two are valid + assert len(solve(eqs, sym, manual=True, minimal=True)) == 2 + assert len(solve(eqs, sym)) == 2 # cf below with rational=False + + +@SKIP("Hangs") +def _test_issue_5335_float(): + # gives ZeroDivisionError: polynomial division + lam, a0, conc = symbols('lam a0 conc') + a = 0.005 + b = 0.743436700916726 + eqs = [lam + 2*y - a0*(1 - x/2)*x - a*x/2*x, + a0*(1 - x/2)*x - 1*y - b*y, + x + y - conc] + sym = [x, y, a0] + assert len(solve(eqs, sym, rational=False)) == 2 + + +def test_issue_5767(): + assert set(solve([x**2 + y + 4], [x])) == \ + {(-sqrt(-y - 4),), (sqrt(-y - 4),)} + + +def _make_example_24609(): + D, R, H, B_g, V, D_c = symbols("D, R, H, B_g, V, D_c", real=True, positive=True) + Sigma_f, Sigma_a, nu = symbols("Sigma_f, Sigma_a, nu", real=True, positive=True) + x = symbols("x", real=True, positive=True) + eq = ( + 2**(S(2)/3)*pi**(S(2)/3)*D_c*(S(231361)/10000 + pi**2/x**2) + /(6*V**(S(2)/3)*x**(S(1)/3)) + - 2**(S(2)/3)*pi**(S(8)/3)*D_c/(2*V**(S(2)/3)*x**(S(7)/3)) + ) + expected = 100*sqrt(2)*pi/481 + return eq, expected, x + + +def test_issue_24609(): + # https://github.com/sympy/sympy/issues/24609 + eq, expected, x = _make_example_24609() + assert solve(eq, x, simplify=True) == [expected] + [solapprox] = solve(eq.n(), x) + assert abs(solapprox - expected.n()) < 1e-14 + + +@XFAIL +def test_issue_24609_xfail(): + # + # This returns 5 solutions when it should be 1 (with x positive). + # Simplification reveals all solutions to be equivalent. It is expected + # that solve without simplify=True returns duplicate solutions in some + # cases but the core of this equation is a simple quadratic that can easily + # be solved without introducing any redundant solutions: + # + # >>> print(factor_terms(eq.as_numer_denom()[0])) + # 2**(2/3)*pi**(2/3)*D_c*V**(2/3)*x**(7/3)*(231361*x**2 - 20000*pi**2) + # + eq, expected, x = _make_example_24609() + assert len(solve(eq, x)) == [expected] + # + # We do not want to pass this test just by using simplify so if the above + # passes then uncomment the additional test below: + # + # assert len(solve(eq, x, simplify=False)) == 1 + + +def test_polysys(): + assert set(solve([x**2 + 2/y - 2, x + y - 3], [x, y])) == \ + {(S.One, S(2)), (1 + sqrt(5), 2 - sqrt(5)), + (1 - sqrt(5), 2 + sqrt(5))} + assert solve([x**2 + y - 2, x**2 + y]) == [] + # the ordering should be whatever the user requested + assert solve([x**2 + y - 3, x - y - 4], (x, y)) != solve([x**2 + + y - 3, x - y - 4], (y, x)) + + +@slow +def test_unrad1(): + raises(NotImplementedError, lambda: + unrad(sqrt(x) + sqrt(x + 1) + sqrt(1 - sqrt(x)) + 3)) + raises(NotImplementedError, lambda: + unrad(sqrt(x) + (x + 1)**Rational(1, 3) + 2*sqrt(y))) + + s = symbols('s', cls=Dummy) + + # checkers to deal with possibility of answer coming + # back with a sign change (cf issue 5203) + def check(rv, ans): + assert bool(rv[1]) == bool(ans[1]) + if ans[1]: + return s_check(rv, ans) + e = rv[0].expand() + a = ans[0].expand() + return e in [a, -a] and rv[1] == ans[1] + + def s_check(rv, ans): + # get the dummy + rv = list(rv) + d = rv[0].atoms(Dummy) + reps = list(zip(d, [s]*len(d))) + # replace s with this dummy + rv = (rv[0].subs(reps).expand(), [rv[1][0].subs(reps), rv[1][1].subs(reps)]) + ans = (ans[0].subs(reps).expand(), [ans[1][0].subs(reps), ans[1][1].subs(reps)]) + return str(rv[0]) in [str(ans[0]), str(-ans[0])] and \ + str(rv[1]) == str(ans[1]) + + assert unrad(1) is None + assert check(unrad(sqrt(x)), + (x, [])) + assert check(unrad(sqrt(x) + 1), + (x - 1, [])) + assert check(unrad(sqrt(x) + root(x, 3) + 2), + (s**3 + s**2 + 2, [s, s**6 - x])) + assert check(unrad(sqrt(x)*root(x, 3) + 2), + (x**5 - 64, [])) + assert check(unrad(sqrt(x) + (x + 1)**Rational(1, 3)), + (x**3 - (x + 1)**2, [])) + assert check(unrad(sqrt(x) + sqrt(x + 1) + sqrt(2*x)), + (-2*sqrt(2)*x - 2*x + 1, [])) + assert check(unrad(sqrt(x) + sqrt(x + 1) + 2), + (16*x - 9, [])) + assert check(unrad(sqrt(x) + sqrt(x + 1) + sqrt(1 - x)), + (5*x**2 - 4*x, [])) + assert check(unrad(a*sqrt(x) + b*sqrt(x) + c*sqrt(y) + d*sqrt(y)), + ((a*sqrt(x) + b*sqrt(x))**2 - (c*sqrt(y) + d*sqrt(y))**2, [])) + assert check(unrad(sqrt(x) + sqrt(1 - x)), + (2*x - 1, [])) + assert check(unrad(sqrt(x) + sqrt(1 - x) - 3), + (x**2 - x + 16, [])) + assert check(unrad(sqrt(x) + sqrt(1 - x) + sqrt(2 + x)), + (5*x**2 - 2*x + 1, [])) + assert unrad(sqrt(x) + sqrt(1 - x) + sqrt(2 + x) - 3) in [ + (25*x**4 + 376*x**3 + 1256*x**2 - 2272*x + 784, []), + (25*x**8 - 476*x**6 + 2534*x**4 - 1468*x**2 + 169, [])] + assert unrad(sqrt(x) + sqrt(1 - x) + sqrt(2 + x) - sqrt(1 - 2*x)) == \ + (41*x**4 + 40*x**3 + 232*x**2 - 160*x + 16, []) # orig root at 0.487 + assert check(unrad(sqrt(x) + sqrt(x + 1)), (S.One, [])) + + eq = sqrt(x) + sqrt(x + 1) + sqrt(1 - sqrt(x)) + assert check(unrad(eq), + (16*x**2 - 9*x, [])) + assert set(solve(eq, check=False)) == {S.Zero, Rational(9, 16)} + assert solve(eq) == [] + # but this one really does have those solutions + assert set(solve(sqrt(x) - sqrt(x + 1) + sqrt(1 - sqrt(x)))) == \ + {S.Zero, Rational(9, 16)} + + assert check(unrad(sqrt(x) + root(x + 1, 3) + 2*sqrt(y), y), + (S('2*sqrt(x)*(x + 1)**(1/3) + x - 4*y + (x + 1)**(2/3)'), [])) + assert check(unrad(sqrt(x/(1 - x)) + (x + 1)**Rational(1, 3)), + (x**5 - x**4 - x**3 + 2*x**2 + x - 1, [])) + assert check(unrad(sqrt(x/(1 - x)) + 2*sqrt(y), y), + (4*x*y + x - 4*y, [])) + assert check(unrad(sqrt(x)*sqrt(1 - x) + 2, x), + (x**2 - x + 4, [])) + + # http://tutorial.math.lamar.edu/ + # Classes/Alg/SolveRadicalEqns.aspx#Solve_Rad_Ex2_a + assert solve(Eq(x, sqrt(x + 6))) == [3] + assert solve(Eq(x + sqrt(x - 4), 4)) == [4] + assert solve(Eq(1, x + sqrt(2*x - 3))) == [] + assert set(solve(Eq(sqrt(5*x + 6) - 2, x))) == {-S.One, S(2)} + assert set(solve(Eq(sqrt(2*x - 1) - sqrt(x - 4), 2))) == {S(5), S(13)} + assert solve(Eq(sqrt(x + 7) + 2, sqrt(3 - x))) == [-6] + # http://www.purplemath.com/modules/solverad.htm + assert solve((2*x - 5)**Rational(1, 3) - 3) == [16] + assert set(solve(x + 1 - root(x**4 + 4*x**3 - x, 4))) == \ + {Rational(-1, 2), Rational(-1, 3)} + assert set(solve(sqrt(2*x**2 - 7) - (3 - x))) == {-S(8), S(2)} + assert solve(sqrt(2*x + 9) - sqrt(x + 1) - sqrt(x + 4)) == [0] + assert solve(sqrt(x + 4) + sqrt(2*x - 1) - 3*sqrt(x - 1)) == [5] + assert solve(sqrt(x)*sqrt(x - 7) - 12) == [16] + assert solve(sqrt(x - 3) + sqrt(x) - 3) == [4] + assert solve(sqrt(9*x**2 + 4) - (3*x + 2)) == [0] + assert solve(sqrt(x) - 2 - 5) == [49] + assert solve(sqrt(x - 3) - sqrt(x) - 3) == [] + assert solve(sqrt(x - 1) - x + 7) == [10] + assert solve(sqrt(x - 2) - 5) == [27] + assert solve(sqrt(17*x - sqrt(x**2 - 5)) - 7) == [3] + assert solve(sqrt(x) - sqrt(x - 1) + sqrt(sqrt(x))) == [] + + # don't posify the expression in unrad and do use _mexpand + z = sqrt(2*x + 1)/sqrt(x) - sqrt(2 + 1/x) + p = posify(z)[0] + assert solve(p) == [] + assert solve(z) == [] + assert solve(z + 6*I) == [Rational(-1, 11)] + assert solve(p + 6*I) == [] + # issue 8622 + assert unrad(root(x + 1, 5) - root(x, 3)) == ( + -(x**5 - x**3 - 3*x**2 - 3*x - 1), []) + # issue #8679 + assert check(unrad(x + root(x, 3) + root(x, 3)**2 + sqrt(y), x), + (s**3 + s**2 + s + sqrt(y), [s, s**3 - x])) + + # for coverage + assert check(unrad(sqrt(x) + root(x, 3) + y), + (s**3 + s**2 + y, [s, s**6 - x])) + assert solve(sqrt(x) + root(x, 3) - 2) == [1] + raises(NotImplementedError, lambda: + solve(sqrt(x) + root(x, 3) + root(x + 1, 5) - 2)) + # fails through a different code path + raises(NotImplementedError, lambda: solve(-sqrt(2) + cosh(x)/x)) + # unrad some + assert solve(sqrt(x + root(x, 3))+root(x - y, 5), y) == [ + x + (x**Rational(1, 3) + x)**Rational(5, 2)] + assert check(unrad(sqrt(x) - root(x + 1, 3)*sqrt(x + 2) + 2), + (s**10 + 8*s**8 + 24*s**6 - 12*s**5 - 22*s**4 - 160*s**3 - 212*s**2 - + 192*s - 56, [s, s**2 - x])) + e = root(x + 1, 3) + root(x, 3) + assert unrad(e) == (2*x + 1, []) + eq = (sqrt(x) + sqrt(x + 1) + sqrt(1 - x) - 6*sqrt(5)/5) + assert check(unrad(eq), + (15625*x**4 + 173000*x**3 + 355600*x**2 - 817920*x + 331776, [])) + assert check(unrad(root(x, 4) + root(x, 4)**3 - 1), + (s**3 + s - 1, [s, s**4 - x])) + assert check(unrad(root(x, 2) + root(x, 2)**3 - 1), + (x**3 + 2*x**2 + x - 1, [])) + assert unrad(x**0.5) is None + assert check(unrad(t + root(x + y, 5) + root(x + y, 5)**3), + (s**3 + s + t, [s, s**5 - x - y])) + assert check(unrad(x + root(x + y, 5) + root(x + y, 5)**3, y), + (s**3 + s + x, [s, s**5 - x - y])) + assert check(unrad(x + root(x + y, 5) + root(x + y, 5)**3, x), + (s**5 + s**3 + s - y, [s, s**5 - x - y])) + assert check(unrad(root(x - 1, 3) + root(x + 1, 5) + root(2, 5)), + (s**5 + 5*2**Rational(1, 5)*s**4 + s**3 + 10*2**Rational(2, 5)*s**3 + + 10*2**Rational(3, 5)*s**2 + 5*2**Rational(4, 5)*s + 4, [s, s**3 - x + 1])) + raises(NotImplementedError, lambda: + unrad((root(x, 2) + root(x, 3) + root(x, 4)).subs(x, x**5 - x + 1))) + + # the simplify flag should be reset to False for unrad results; + # if it's not then this next test will take a long time + assert solve(root(x, 3) + root(x, 5) - 2) == [1] + eq = (sqrt(x) + sqrt(x + 1) + sqrt(1 - x) - 6*sqrt(5)/5) + assert check(unrad(eq), + ((5*x - 4)*(3125*x**3 + 37100*x**2 + 100800*x - 82944), [])) + ans = S(''' + [4/5, -1484/375 + 172564/(140625*(114*sqrt(12657)/78125 + + 12459439/52734375)**(1/3)) + + 4*(114*sqrt(12657)/78125 + 12459439/52734375)**(1/3)]''') + assert solve(eq) == ans + # duplicate radical handling + assert check(unrad(sqrt(x + root(x + 1, 3)) - root(x + 1, 3) - 2), + (s**3 - s**2 - 3*s - 5, [s, s**3 - x - 1])) + # cov post-processing + e = root(x**2 + 1, 3) - root(x**2 - 1, 5) - 2 + assert check(unrad(e), + (s**5 - 10*s**4 + 39*s**3 - 80*s**2 + 80*s - 30, + [s, s**3 - x**2 - 1])) + + e = sqrt(x + root(x + 1, 2)) - root(x + 1, 3) - 2 + assert check(unrad(e), + (s**6 - 2*s**5 - 7*s**4 - 3*s**3 + 26*s**2 + 40*s + 25, + [s, s**3 - x - 1])) + assert check(unrad(e, _reverse=True), + (s**6 - 14*s**5 + 73*s**4 - 187*s**3 + 276*s**2 - 228*s + 89, + [s, s**2 - x - sqrt(x + 1)])) + # this one needs r0, r1 reversal to work + assert check(unrad(sqrt(x + sqrt(root(x, 3) - 1)) - root(x, 6) - 2), + (s**12 - 2*s**8 - 8*s**7 - 8*s**6 + s**4 + 8*s**3 + 23*s**2 + + 32*s + 17, [s, s**6 - x])) + + # why does this pass + assert unrad(root(cosh(x), 3)/x*root(x + 1, 5) - 1) == ( + -(x**15 - x**3*cosh(x)**5 - 3*x**2*cosh(x)**5 - 3*x*cosh(x)**5 + - cosh(x)**5), []) + # and this fail? + #assert unrad(sqrt(cosh(x)/x) + root(x + 1, 3)*sqrt(x) - 1) == ( + # -s**6 + 6*s**5 - 15*s**4 + 20*s**3 - 15*s**2 + 6*s + x**5 + + # 2*x**4 + x**3 - 1, [s, s**2 - cosh(x)/x]) + + # watch for symbols in exponents + assert unrad(S('(x+y)**(2*y/3) + (x+y)**(1/3) + 1')) is None + assert check(unrad(S('(x+y)**(2*y/3) + (x+y)**(1/3) + 1'), x), + (s**(2*y) + s + 1, [s, s**3 - x - y])) + # should _Q be so lenient? + assert unrad(x**(S.Half/y) + y, x) == (x**(1/y) - y**2, []) + + # This tests two things: that if full unrad is attempted and fails + # the solution should still be found; also it tests that the use of + # composite + assert len(solve(sqrt(y)*x + x**3 - 1, x)) == 3 + assert len(solve(-512*y**3 + 1344*(x + 2)**Rational(1, 3)*y**2 - + 1176*(x + 2)**Rational(2, 3)*y - 169*x + 686, y, _unrad=False)) == 3 + + # watch out for when the cov doesn't involve the symbol of interest + eq = S('-x + (7*y/8 - (27*x/2 + 27*sqrt(x**2)/2)**(1/3)/3)**3 - 1') + assert solve(eq, y) == [ + 2**(S(2)/3)*(27*x + 27*sqrt(x**2))**(S(1)/3)*S(4)/21 + (512*x/343 + + S(512)/343)**(S(1)/3)*(-S(1)/2 - sqrt(3)*I/2), 2**(S(2)/3)*(27*x + + 27*sqrt(x**2))**(S(1)/3)*S(4)/21 + (512*x/343 + + S(512)/343)**(S(1)/3)*(-S(1)/2 + sqrt(3)*I/2), 2**(S(2)/3)*(27*x + + 27*sqrt(x**2))**(S(1)/3)*S(4)/21 + (512*x/343 + S(512)/343)**(S(1)/3)] + + eq = root(x + 1, 3) - (root(x, 3) + root(x, 5)) + assert check(unrad(eq), + (3*s**13 + 3*s**11 + s**9 - 1, [s, s**15 - x])) + assert check(unrad(eq - 2), + (3*s**13 + 3*s**11 + 6*s**10 + s**9 + 12*s**8 + 6*s**6 + 12*s**5 + + 12*s**3 + 7, [s, s**15 - x])) + assert check(unrad(root(x, 3) - root(x + 1, 4)/2 + root(x + 2, 3)), + (s*(4096*s**9 + 960*s**8 + 48*s**7 - s**6 - 1728), + [s, s**4 - x - 1])) # orig expr has two real roots: -1, -.389 + assert check(unrad(root(x, 3) + root(x + 1, 4) - root(x + 2, 3)/2), + (343*s**13 + 2904*s**12 + 1344*s**11 + 512*s**10 - 1323*s**9 - + 3024*s**8 - 1728*s**7 + 1701*s**5 + 216*s**4 - 729*s, [s, s**4 - x - + 1])) # orig expr has one real root: -0.048 + assert check(unrad(root(x, 3)/2 - root(x + 1, 4) + root(x + 2, 3)), + (729*s**13 - 216*s**12 + 1728*s**11 - 512*s**10 + 1701*s**9 - + 3024*s**8 + 1344*s**7 + 1323*s**5 - 2904*s**4 + 343*s, [s, s**4 - x - + 1])) # orig expr has 2 real roots: -0.91, -0.15 + assert check(unrad(root(x, 3)/2 - root(x + 1, 4) + root(x + 2, 3) - 2), + (729*s**13 + 1242*s**12 + 18496*s**10 + 129701*s**9 + 388602*s**8 + + 453312*s**7 - 612864*s**6 - 3337173*s**5 - 6332418*s**4 - 7134912*s**3 + - 5064768*s**2 - 2111913*s - 398034, [s, s**4 - x - 1])) + # orig expr has 1 real root: 19.53 + + ans = solve(sqrt(x) + sqrt(x + 1) - + sqrt(1 - x) - sqrt(2 + x)) + assert len(ans) == 1 and NS(ans[0])[:4] == '0.73' + # the fence optimization problem + # https://github.com/sympy/sympy/issues/4793#issuecomment-36994519 + F = Symbol('F') + eq = F - (2*x + 2*y + sqrt(x**2 + y**2)) + ans = F*Rational(2, 7) - sqrt(2)*F/14 + X = solve(eq, x, check=False) + for xi in reversed(X): # reverse since currently, ans is the 2nd one + Y = solve((x*y).subs(x, xi).diff(y), y, simplify=False, check=False) + if any((a - ans).expand().is_zero for a in Y): + break + else: + assert None # no answer was found + assert solve(sqrt(x + 1) + root(x, 3) - 2) == S(''' + [(-11/(9*(47/54 + sqrt(93)/6)**(1/3)) + 1/3 + (47/54 + + sqrt(93)/6)**(1/3))**3]''') + assert solve(sqrt(sqrt(x + 1)) + x**Rational(1, 3) - 2) == S(''' + [(-sqrt(-2*(-1/16 + sqrt(6913)/16)**(1/3) + 6/(-1/16 + + sqrt(6913)/16)**(1/3) + 17/2 + 121/(4*sqrt(-6/(-1/16 + + sqrt(6913)/16)**(1/3) + 2*(-1/16 + sqrt(6913)/16)**(1/3) + 17/4)))/2 + + sqrt(-6/(-1/16 + sqrt(6913)/16)**(1/3) + 2*(-1/16 + + sqrt(6913)/16)**(1/3) + 17/4)/2 + 9/4)**3]''') + assert solve(sqrt(x) + root(sqrt(x) + 1, 3) - 2) == S(''' + [(-(81/2 + 3*sqrt(741)/2)**(1/3)/3 + (81/2 + 3*sqrt(741)/2)**(-1/3) + + 2)**2]''') + eq = S(''' + -x + (1/2 - sqrt(3)*I/2)*(3*x**3/2 - x*(3*x**2 - 34)/2 + sqrt((-3*x**3 + + x*(3*x**2 - 34) + 90)**2/4 - 39304/27) - 45)**(1/3) + 34/(3*(1/2 - + sqrt(3)*I/2)*(3*x**3/2 - x*(3*x**2 - 34)/2 + sqrt((-3*x**3 + x*(3*x**2 + - 34) + 90)**2/4 - 39304/27) - 45)**(1/3))''') + assert check(unrad(eq), + (s*-(-s**6 + sqrt(3)*s**6*I - 153*2**Rational(2, 3)*3**Rational(1, 3)*s**4 + + 51*12**Rational(1, 3)*s**4 - 102*2**Rational(2, 3)*3**Rational(5, 6)*s**4*I - 1620*s**3 + + 1620*sqrt(3)*s**3*I + 13872*18**Rational(1, 3)*s**2 - 471648 + + 471648*sqrt(3)*I), [s, s**3 - 306*x - sqrt(3)*sqrt(31212*x**2 - + 165240*x + 61484) + 810])) + + assert solve(eq) == [] # not other code errors + eq = root(x, 3) - root(y, 3) + root(x, 5) + assert check(unrad(eq), + (s**15 + 3*s**13 + 3*s**11 + s**9 - y, [s, s**15 - x])) + eq = root(x, 3) + root(y, 3) + root(x*y, 4) + assert check(unrad(eq), + (s*y*(-s**12 - 3*s**11*y - 3*s**10*y**2 - s**9*y**3 - + 3*s**8*y**2 + 21*s**7*y**3 - 3*s**6*y**4 - 3*s**4*y**4 - + 3*s**3*y**5 - y**6), [s, s**4 - x*y])) + raises(NotImplementedError, + lambda: unrad(root(x, 3) + root(y, 3) + root(x*y, 5))) + + # Test unrad with an Equality + eq = Eq(-x**(S(1)/5) + x**(S(1)/3), -3**(S(1)/3) - (-1)**(S(3)/5)*3**(S(1)/5)) + assert check(unrad(eq), + (-s**5 + s**3 - 3**(S(1)/3) - (-1)**(S(3)/5)*3**(S(1)/5), [s, s**15 - x])) + + # make sure buried radicals are exposed + s = sqrt(x) - 1 + assert unrad(s**2 - s**3) == (x**3 - 6*x**2 + 9*x - 4, []) + # make sure numerators which are already polynomial are rejected + assert unrad((x/(x + 1) + 3)**(-2), x) is None + + # https://github.com/sympy/sympy/issues/23707 + eq = sqrt(x - y)*exp(t*sqrt(x - y)) - exp(t*sqrt(x - y)) + assert solve(eq, y) == [x - 1] + assert unrad(eq) is None + + +@slow +def test_unrad_slow(): + # this has roots with multiplicity > 1; there should be no + # repeats in roots obtained, however + eq = (sqrt(1 + sqrt(1 - 4*x**2)) - x*(1 + sqrt(1 + 2*sqrt(1 - 4*x**2)))) + assert solve(eq) == [S.Half] + + +@XFAIL +def test_unrad_fail(): + # this only works if we check real_root(eq.subs(x, Rational(1, 3))) + # but checksol doesn't work like that + assert solve(root(x**3 - 3*x**2, 3) + 1 - x) == [Rational(1, 3)] + assert solve(root(x + 1, 3) + root(x**2 - 2, 5) + 1) == [ + -1, -1 + CRootOf(x**5 + x**4 + 5*x**3 + 8*x**2 + 10*x + 5, 0)**3] + + +def test_checksol(): + x, y, r, t = symbols('x, y, r, t') + eq = r - x**2 - y**2 + dict_var_soln = {y: - sqrt(r) / sqrt(tan(t)**2 + 1), + x: -sqrt(r)*tan(t)/sqrt(tan(t)**2 + 1)} + assert checksol(eq, dict_var_soln) == True + assert checksol(Eq(x, False), {x: False}) is True + assert checksol(Ne(x, False), {x: False}) is False + assert checksol(Eq(x < 1, True), {x: 0}) is True + assert checksol(Eq(x < 1, True), {x: 1}) is False + assert checksol(Eq(x < 1, False), {x: 1}) is True + assert checksol(Eq(x < 1, False), {x: 0}) is False + assert checksol(Eq(x + 1, x**2 + 1), {x: 1}) is True + assert checksol([x - 1, x**2 - 1], x, 1) is True + assert checksol([x - 1, x**2 - 2], x, 1) is False + assert checksol(Poly(x**2 - 1), x, 1) is True + assert checksol(0, {}) is True + assert checksol([1e-10, x - 2], x, 2) is False + assert checksol([0.5, 0, x], x, 0) is False + assert checksol(y, x, 2) is False + assert checksol(x+1e-10, x, 0, numerical=True) is True + assert checksol(x+1e-10, x, 0, numerical=False) is False + assert checksol(exp(92*x), {x: log(sqrt(2)/2)}) is False + assert checksol(exp(92*x), {x: log(sqrt(2)/2) + I*pi}) is False + assert checksol(1/x**5, x, 1000) is False + raises(ValueError, lambda: checksol(x, 1)) + raises(ValueError, lambda: checksol([], x, 1)) + + +def test__invert(): + assert _invert(x - 2) == (2, x) + assert _invert(2) == (2, 0) + assert _invert(exp(1/x) - 3, x) == (1/log(3), x) + assert _invert(exp(1/x + a/x) - 3, x) == ((a + 1)/log(3), x) + assert _invert(a, x) == (a, 0) + + +def test_issue_4463(): + assert solve(-a*x + 2*x*log(x), x) == [exp(a/2)] + assert solve(x**x) == [] + assert solve(x**x - 2) == [exp(LambertW(log(2)))] + assert solve(((x - 3)*(x - 2))**((x - 3)*(x - 4))) == [2] + +@slow +def test_issue_5114_solvers(): + a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r = symbols('a:r') + + # there is no 'a' in the equation set but this is how the + # problem was originally posed + syms = a, b, c, f, h, k, n + eqs = [b + r/d - c/d, + c*(1/d + 1/e + 1/g) - f/g - r/d, + f*(1/g + 1/i + 1/j) - c/g - h/i, + h*(1/i + 1/l + 1/m) - f/i - k/m, + k*(1/m + 1/o + 1/p) - h/m - n/p, + n*(1/p + 1/q) - k/p] + assert len(solve(eqs, syms, manual=True, check=False, simplify=False)) == 1 + + +def test_issue_5849(): + # + # XXX: This system does not have a solution for most values of the + # parameters. Generally solve returns the empty set for systems that are + # generically inconsistent. + # + I1, I2, I3, I4, I5, I6 = symbols('I1:7') + dI1, dI4, dQ2, dQ4, Q2, Q4 = symbols('dI1,dI4,dQ2,dQ4,Q2,Q4') + + e = ( + I1 - I2 - I3, + I3 - I4 - I5, + I4 + I5 - I6, + -I1 + I2 + I6, + -2*I1 - 2*I3 - 2*I5 - 3*I6 - dI1/2 + 12, + -I4 + dQ4, + -I2 + dQ2, + 2*I3 + 2*I5 + 3*I6 - Q2, + I4 - 2*I5 + 2*Q4 + dI4 + ) + + ans = [{ + I1: I2 + I3, + dI1: -4*I2 - 8*I3 - 4*I5 - 6*I6 + 24, + I4: I3 - I5, + dQ4: I3 - I5, + Q4: -I3/2 + 3*I5/2 - dI4/2, + dQ2: I2, + Q2: 2*I3 + 2*I5 + 3*I6}] + + v = I1, I4, Q2, Q4, dI1, dI4, dQ2, dQ4 + assert solve(e, *v, manual=True, check=False, dict=True) == ans + assert solve(e, *v, manual=True, check=False) == [ + tuple([a.get(i, i) for i in v]) for a in ans] + assert solve(e, *v, manual=True) == [] + assert solve(e, *v) == [] + + # the matrix solver (tested below) doesn't like this because it produces + # a zero row in the matrix. Is this related to issue 4551? + assert [ei.subs( + ans[0]) for ei in e] == [0, 0, I3 - I6, -I3 + I6, 0, 0, 0, 0, 0] + + +def test_issue_5849_matrix(): + '''Same as test_issue_5849 but solved with the matrix solver. + + A solution only exists if I3 == I6 which is not generically true, + but `solve` does not return conditions under which the solution is + valid, only a solution that is canonical and consistent with the input. + ''' + # a simple example with the same issue + # assert solve([x+y+z, x+y], [x, y]) == {x: y} + # the longer example + I1, I2, I3, I4, I5, I6 = symbols('I1:7') + dI1, dI4, dQ2, dQ4, Q2, Q4 = symbols('dI1,dI4,dQ2,dQ4,Q2,Q4') + + e = ( + I1 - I2 - I3, + I3 - I4 - I5, + I4 + I5 - I6, + -I1 + I2 + I6, + -2*I1 - 2*I3 - 2*I5 - 3*I6 - dI1/2 + 12, + -I4 + dQ4, + -I2 + dQ2, + 2*I3 + 2*I5 + 3*I6 - Q2, + I4 - 2*I5 + 2*Q4 + dI4 + ) + assert solve(e, I1, I4, Q2, Q4, dI1, dI4, dQ2, dQ4) == [] + + +def test_issue_21882(): + + a, b, c, d, f, g, k = unknowns = symbols('a, b, c, d, f, g, k') + + equations = [ + -k*a + b + 5*f/6 + 2*c/9 + 5*d/6 + 4*a/3, + -k*f + 4*f/3 + d/2, + -k*d + f/6 + d, + 13*b/18 + 13*c/18 + 13*a/18, + -k*c + b/2 + 20*c/9 + a, + -k*b + b + c/18 + a/6, + 5*b/3 + c/3 + a, + 2*b/3 + 2*c + 4*a/3, + -g, + ] + + answer = [ + {a: 0, f: 0, b: 0, d: 0, c: 0, g: 0}, + {a: 0, f: -d, b: 0, k: S(5)/6, c: 0, g: 0}, + {a: -2*c, f: 0, b: c, d: 0, k: S(13)/18, g: 0}] + # but not {a: 0, f: 0, b: 0, k: S(3)/2, c: 0, d: 0, g: 0} + # since this is already covered by the first solution + got = solve(equations, unknowns, dict=True) + assert got == answer, (got,answer) + + +def test_issue_5901(): + f, g, h = map(Function, 'fgh') + a = Symbol('a') + D = Derivative(f(x), x) + G = Derivative(g(a), a) + assert solve(f(x) + f(x).diff(x), f(x)) == \ + [-D] + assert solve(f(x) - 3, f(x)) == \ + [3] + assert solve(f(x) - 3*f(x).diff(x), f(x)) == \ + [3*D] + assert solve([f(x) - 3*f(x).diff(x)], f(x)) == \ + {f(x): 3*D} + assert solve([f(x) - 3*f(x).diff(x), f(x)**2 - y + 4], f(x), y) == \ + [(3*D, 9*D**2 + 4)] + assert solve(-f(a)**2*g(a)**2 + f(a)**2*h(a)**2 + g(a).diff(a), + h(a), g(a), set=True) == \ + ([h(a), g(a)], { + (-sqrt(f(a)**2*g(a)**2 - G)/f(a), g(a)), + (sqrt(f(a)**2*g(a)**2 - G)/f(a), g(a))}), solve(-f(a)**2*g(a)**2 + f(a)**2*h(a)**2 + g(a).diff(a), + h(a), g(a), set=True) + args = [[f(x).diff(x, 2)*(f(x) + g(x)), 2 - g(x)**2], f(x), g(x)] + assert solve(*args, set=True)[1] == \ + {(-sqrt(2), sqrt(2)), (sqrt(2), -sqrt(2))} + eqs = [f(x)**2 + g(x) - 2*f(x).diff(x), g(x)**2 - 4] + assert solve(eqs, f(x), g(x), set=True) == \ + ([f(x), g(x)], { + (-sqrt(2*D - 2), S(2)), + (sqrt(2*D - 2), S(2)), + (-sqrt(2*D + 2), -S(2)), + (sqrt(2*D + 2), -S(2))}) + + # the underlying problem was in solve_linear that was not masking off + # anything but a Mul or Add; it now raises an error if it gets anything + # but a symbol and solve handles the substitutions necessary so solve_linear + # won't make this error + raises( + ValueError, lambda: solve_linear(f(x) + f(x).diff(x), symbols=[f(x)])) + assert solve_linear(f(x) + f(x).diff(x), symbols=[x]) == \ + (f(x) + Derivative(f(x), x), 1) + assert solve_linear(f(x) + Integral(x, (x, y)), symbols=[x]) == \ + (f(x) + Integral(x, (x, y)), 1) + assert solve_linear(f(x) + Integral(x, (x, y)) + x, symbols=[x]) == \ + (x + f(x) + Integral(x, (x, y)), 1) + assert solve_linear(f(y) + Integral(x, (x, y)) + x, symbols=[x]) == \ + (x, -f(y) - Integral(x, (x, y))) + assert solve_linear(x - f(x)/a + (f(x) - 1)/a, symbols=[x]) == \ + (x, 1/a) + assert solve_linear(x + Derivative(2*x, x)) == \ + (x, -2) + assert solve_linear(x + Integral(x, y), symbols=[x]) == \ + (x, 0) + assert solve_linear(x + Integral(x, y) - 2, symbols=[x]) == \ + (x, 2/(y + 1)) + + assert set(solve(x + exp(x)**2, exp(x))) == \ + {-sqrt(-x), sqrt(-x)} + assert solve(x + exp(x), x, implicit=True) == \ + [-exp(x)] + assert solve(cos(x) - sin(x), x, implicit=True) == [] + assert solve(x - sin(x), x, implicit=True) == \ + [sin(x)] + assert solve(x**2 + x - 3, x, implicit=True) == \ + [-x**2 + 3] + assert solve(x**2 + x - 3, x**2, implicit=True) == \ + [-x + 3] + + +def test_issue_5912(): + assert set(solve(x**2 - x - 0.1, rational=True)) == \ + {S.Half + sqrt(35)/10, -sqrt(35)/10 + S.Half} + ans = solve(x**2 - x - 0.1, rational=False) + assert len(ans) == 2 and all(a.is_Number for a in ans) + ans = solve(x**2 - x - 0.1) + assert len(ans) == 2 and all(a.is_Number for a in ans) + + +def test_float_handling(): + def test(e1, e2): + return len(e1.atoms(Float)) == len(e2.atoms(Float)) + assert solve(x - 0.5, rational=True)[0].is_Rational + assert solve(x - 0.5, rational=False)[0].is_Float + assert solve(x - S.Half, rational=False)[0].is_Rational + assert solve(x - 0.5, rational=None)[0].is_Float + assert solve(x - S.Half, rational=None)[0].is_Rational + assert test(nfloat(1 + 2*x), 1.0 + 2.0*x) + for contain in [list, tuple, set]: + ans = nfloat(contain([1 + 2*x])) + assert type(ans) is contain and test(list(ans)[0], 1.0 + 2.0*x) + k, v = list(nfloat({2*x: [1 + 2*x]}).items())[0] + assert test(k, 2*x) and test(v[0], 1.0 + 2.0*x) + assert test(nfloat(cos(2*x)), cos(2.0*x)) + assert test(nfloat(3*x**2), 3.0*x**2) + assert test(nfloat(3*x**2, exponent=True), 3.0*x**2.0) + assert test(nfloat(exp(2*x)), exp(2.0*x)) + assert test(nfloat(x/3), x/3.0) + assert test(nfloat(x**4 + 2*x + cos(Rational(1, 3)) + 1), + x**4 + 2.0*x + 1.94495694631474) + # don't call nfloat if there is no solution + tot = 100 + c + z + t + assert solve(((.7 + c)/tot - .6, (.2 + z)/tot - .3, t/tot - .1)) == [] + + +def test_check_assumptions(): + x = symbols('x', positive=True) + assert solve(x**2 - 1) == [1] + + +def test_issue_6056(): + assert solve(tanh(x + 3)*tanh(x - 3) - 1) == [] + assert solve(tanh(x - 1)*tanh(x + 1) + 1) == \ + [I*pi*Rational(-3, 4), -I*pi/4, I*pi/4, I*pi*Rational(3, 4)] + assert solve((tanh(x + 3)*tanh(x - 3) + 1)**2) == \ + [I*pi*Rational(-3, 4), -I*pi/4, I*pi/4, I*pi*Rational(3, 4)] + + +def test_issue_5673(): + eq = -x + exp(exp(LambertW(log(x)))*LambertW(log(x))) + assert checksol(eq, x, 2) is True + assert checksol(eq, x, 2, numerical=False) is None + + +def test_exclude(): + R, C, Ri, Vout, V1, Vminus, Vplus, s = \ + symbols('R, C, Ri, Vout, V1, Vminus, Vplus, s') + Rf = symbols('Rf', positive=True) # to eliminate Rf = 0 soln + eqs = [C*V1*s + Vplus*(-2*C*s - 1/R), + Vminus*(-1/Ri - 1/Rf) + Vout/Rf, + C*Vplus*s + V1*(-C*s - 1/R) + Vout/R, + -Vminus + Vplus] + assert solve(eqs, exclude=s*C*R) == [ + { + Rf: Ri*(C*R*s + 1)**2/(C*R*s), + Vminus: Vplus, + V1: 2*Vplus + Vplus/(C*R*s), + Vout: C*R*Vplus*s + 3*Vplus + Vplus/(C*R*s)}, + { + Vplus: 0, + Vminus: 0, + V1: 0, + Vout: 0}, + ] + + # TODO: Investigate why currently solution [0] is preferred over [1]. + assert solve(eqs, exclude=[Vplus, s, C]) in [[{ + Vminus: Vplus, + V1: Vout/2 + Vplus/2 + sqrt((Vout - 5*Vplus)*(Vout - Vplus))/2, + R: (Vout - 3*Vplus - sqrt(Vout**2 - 6*Vout*Vplus + 5*Vplus**2))/(2*C*Vplus*s), + Rf: Ri*(Vout - Vplus)/Vplus, + }, { + Vminus: Vplus, + V1: Vout/2 + Vplus/2 - sqrt((Vout - 5*Vplus)*(Vout - Vplus))/2, + R: (Vout - 3*Vplus + sqrt(Vout**2 - 6*Vout*Vplus + 5*Vplus**2))/(2*C*Vplus*s), + Rf: Ri*(Vout - Vplus)/Vplus, + }], [{ + Vminus: Vplus, + Vout: (V1**2 - V1*Vplus - Vplus**2)/(V1 - 2*Vplus), + Rf: Ri*(V1 - Vplus)**2/(Vplus*(V1 - 2*Vplus)), + R: Vplus/(C*s*(V1 - 2*Vplus)), + }]] + + +def test_high_order_roots(): + s = x**5 + 4*x**3 + 3*x**2 + Rational(7, 4) + assert set(solve(s)) == set(Poly(s*4, domain='ZZ').all_roots()) + + +def test_minsolve_linear_system(): + pqt = {"quick": True, "particular": True} + pqf = {"quick": False, "particular": True} + assert solve([x + y - 5, 2*x - y - 1], **pqt) == {x: 2, y: 3} + assert solve([x + y - 5, 2*x - y - 1], **pqf) == {x: 2, y: 3} + def count(dic): + return len([x for x in dic.values() if x == 0]) + assert count(solve([x + y + z, y + z + a + t], **pqt)) == 3 + assert count(solve([x + y + z, y + z + a + t], **pqf)) == 3 + assert count(solve([x + y + z, y + z + a], **pqt)) == 1 + assert count(solve([x + y + z, y + z + a], **pqf)) == 2 + # issue 22718 + A = Matrix([ + [ 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0], + [ 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, -1, -1, 0, 0], + [-1, -1, 0, 0, -1, 0, 0, 0, 0, 0, 1, 1, 0, 1], + [ 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, -1, 0, -1, 0], + [-1, 0, -1, 0, 0, -1, 0, 0, 0, 0, 1, 0, 1, 1], + [-1, 0, 0, -1, 0, 0, -1, 0, 0, 0, -1, 0, 0, -1], + [ 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, -1, -1, 0], + [ 0, -1, -1, 0, 0, 0, 0, -1, 0, 0, 0, 1, 1, 1], + [ 0, -1, 0, -1, 0, 0, 0, 0, -1, 0, 0, -1, 0, -1], + [ 0, 0, -1, -1, 0, 0, 0, 0, 0, -1, 0, 0, -1, -1], + [ 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0], + [ 0, 0, 0, 0, -1, -1, 0, -1, 0, 0, 0, 0, 0, 0]]) + v = Matrix(symbols("v:14", integer=True)) + B = Matrix([[2], [-2], [0], [0], [0], [0], [0], [0], [0], + [0], [0], [0]]) + eqs = A@v-B + assert solve(eqs) == [] + assert solve(eqs, particular=True) == [] # assumption violated + assert all(v for v in solve([x + y + z, y + z + a]).values()) + for _q in (True, False): + assert not all(v for v in solve( + [x + y + z, y + z + a], quick=_q, + particular=True).values()) + # raise error if quick used w/o particular=True + raises(ValueError, lambda: solve([x + 1], quick=_q)) + raises(ValueError, lambda: solve([x + 1], quick=_q, particular=False)) + # and give a good error message if someone tries to use + # particular with a single equation + raises(ValueError, lambda: solve(x + 1, particular=True)) + + +def test_real_roots(): + # cf. issue 6650 + x = Symbol('x', real=True) + assert len(solve(x**5 + x**3 + 1)) == 1 + + +def test_issue_6528(): + eqs = [ + 327600995*x**2 - 37869137*x + 1809975124*y**2 - 9998905626, + 895613949*x**2 - 273830224*x*y + 530506983*y**2 - 10000000000] + # two expressions encountered are > 1400 ops long so if this hangs + # it is likely because simplification is being done + assert len(solve(eqs, y, x, check=False)) == 4 + + +def test_overdetermined(): + x = symbols('x', real=True) + eqs = [Abs(4*x - 7) - 5, Abs(3 - 8*x) - 1] + assert solve(eqs, x) == [(S.Half,)] + assert solve(eqs, x, manual=True) == [(S.Half,)] + assert solve(eqs, x, manual=True, check=False) == [(S.Half,), (S(3),)] + + +def test_issue_6605(): + x = symbols('x') + assert solve(4**(x/2) - 2**(x/3)) == [0, 3*I*pi/log(2)] + # while the first one passed, this one failed + x = symbols('x', real=True) + assert solve(5**(x/2) - 2**(x/3)) == [0] + b = sqrt(6)*sqrt(log(2))/sqrt(log(5)) + assert solve(5**(x/2) - 2**(3/x)) == [-b, b] + + +def test__ispow(): + assert _ispow(x**2) + assert not _ispow(x) + assert not _ispow(True) + + +def test_issue_6644(): + eq = -sqrt((m - q)**2 + (-m/(2*q) + S.Half)**2) + sqrt((-m**2/2 - sqrt( + 4*m**4 - 4*m**2 + 8*m + 1)/4 - Rational(1, 4))**2 + (m**2/2 - m - sqrt( + 4*m**4 - 4*m**2 + 8*m + 1)/4 - Rational(1, 4))**2) + sol = solve(eq, q, simplify=False, check=False) + assert len(sol) == 5 + + +def test_issue_6752(): + assert solve([a**2 + a, a - b], [a, b]) == [(-1, -1), (0, 0)] + assert solve([a**2 + a*c, a - b], [a, b]) == [(0, 0), (-c, -c)] + + +def test_issue_6792(): + assert solve(x*(x - 1)**2*(x + 1)*(x**6 - x + 1)) == [ + -1, 0, 1, CRootOf(x**6 - x + 1, 0), CRootOf(x**6 - x + 1, 1), + CRootOf(x**6 - x + 1, 2), CRootOf(x**6 - x + 1, 3), + CRootOf(x**6 - x + 1, 4), CRootOf(x**6 - x + 1, 5)] + + +def test_issues_6819_6820_6821_6248_8692_25777_25779(): + # issue 6821 + x, y = symbols('x y', real=True) + assert solve(abs(x + 3) - 2*abs(x - 3)) == [1, 9] + assert solve([abs(x) - 2, arg(x) - pi], x) == [(-2,)] + assert set(solve(abs(x - 7) - 8)) == {-S.One, S(15)} + + # issue 8692 + assert solve(Eq(Abs(x + 1) + Abs(x**2 - 7), 9), x) == [ + Rational(-1, 2) + sqrt(61)/2, -sqrt(69)/2 + S.Half] + + # issue 7145 + assert solve(2*abs(x) - abs(x - 1)) == [-1, Rational(1, 3)] + + # 25777 + assert solve(abs(x**3 + x + 2)/(x + 1)) == [] + + # 25779 + assert solve(abs(x)) == [0] + assert solve(Eq(abs(x**2 - 2*x), 4), x) == [ + 1 - sqrt(5), 1 + sqrt(5)] + nn = symbols('nn', nonnegative=True) + assert solve(abs(sqrt(nn))) == [0] + nz = symbols('nz', nonzero=True) + assert solve(Eq(Abs(4 + 1 / (4*nz)), 0)) == [-Rational(1, 16)] + + x = symbols('x') + assert solve([re(x) - 1, im(x) - 2], x) == [ + {x: 1 + 2*I, re(x): 1, im(x): 2}] + + # check for 'dict' handling of solution + eq = sqrt(re(x)**2 + im(x)**2) - 3 + assert solve(eq) == solve(eq, x) + + i = symbols('i', imaginary=True) + assert solve(abs(i) - 3) == [-3*I, 3*I] + raises(NotImplementedError, lambda: solve(abs(x) - 3)) + + w = symbols('w', integer=True) + assert solve(2*x**w - 4*y**w, w) == solve((x/y)**w - 2, w) + + x, y = symbols('x y', real=True) + assert solve(x + y*I + 3) == {y: 0, x: -3} + # issue 2642 + assert solve(x*(1 + I)) == [0] + + x, y = symbols('x y', imaginary=True) + assert solve(x + y*I + 3 + 2*I) == {x: -2*I, y: 3*I} + + x = symbols('x', real=True) + assert solve(x + y + 3 + 2*I) == {x: -3, y: -2*I} + + # issue 6248 + f = Function('f') + assert solve(f(x + 1) - f(2*x - 1)) == [2] + assert solve(log(x + 1) - log(2*x - 1)) == [2] + + x = symbols('x') + assert solve(2**x + 4**x) == [I*pi/log(2)] + +def test_issue_17638(): + + assert solve(((2-exp(2*x))*exp(x))/(exp(2*x)+2)**2 > 0, x) == (-oo < x) & (x < log(2)/2) + assert solve(((2-exp(2*x)+2)*exp(x+2))/(exp(x)+2)**2 > 0, x) == (-oo < x) & (x < log(4)/2) + assert solve((exp(x)+2+x**2)*exp(2*x+2)/(exp(x)+2)**2 > 0, x) == (-oo < x) & (x < oo) + + + +def test_issue_14607(): + # issue 14607 + s, tau_c, tau_1, tau_2, phi, K = symbols( + 's, tau_c, tau_1, tau_2, phi, K') + + target = (s**2*tau_1*tau_2 + s*tau_1 + s*tau_2 + 1)/(K*s*(-phi + tau_c)) + + K_C, tau_I, tau_D = symbols('K_C, tau_I, tau_D', + positive=True, nonzero=True) + PID = K_C*(1 + 1/(tau_I*s) + tau_D*s) + + eq = (target - PID).together() + eq *= denom(eq).simplify() + eq = Poly(eq, s) + c = eq.coeffs() + + vars = [K_C, tau_I, tau_D] + s = solve(c, vars, dict=True) + + assert len(s) == 1 + + knownsolution = {K_C: -(tau_1 + tau_2)/(K*(phi - tau_c)), + tau_I: tau_1 + tau_2, + tau_D: tau_1*tau_2/(tau_1 + tau_2)} + + for var in vars: + assert s[0][var].simplify() == knownsolution[var].simplify() + + +def test_lambert_multivariate(): + from sympy.abc import x, y + assert _filtered_gens(Poly(x + 1/x + exp(x) + y), x) == {x, exp(x)} + assert _lambert(x, x) == [] + assert solve((x**2 - 2*x + 1).subs(x, log(x) + 3*x)) == [LambertW(3*S.Exp1)/3] + assert solve((x**2 - 2*x + 1).subs(x, (log(x) + 3*x)**2 - 1)) == \ + [LambertW(3*exp(-sqrt(2)))/3, LambertW(3*exp(sqrt(2)))/3] + assert solve((x**2 - 2*x - 2).subs(x, log(x) + 3*x)) == \ + [LambertW(3*exp(1 - sqrt(3)))/3, LambertW(3*exp(1 + sqrt(3)))/3] + eq = (x*exp(x) - 3).subs(x, x*exp(x)) + assert solve(eq) == [LambertW(3*exp(-LambertW(3)))] + # coverage test + raises(NotImplementedError, lambda: solve(x - sin(x)*log(y - x), x)) + ans = [3, -3*LambertW(-log(3)/3)/log(3)] # 3 and 2.478... + assert solve(x**3 - 3**x, x) == ans + assert set(solve(3*log(x) - x*log(3))) == set(ans) + assert solve(LambertW(2*x) - y, x) == [y*exp(y)/2] + + +@XFAIL +def test_other_lambert(): + assert solve(3*sin(x) - x*sin(3), x) == [3] + assert set(solve(x**a - a**x), x) == { + a, -a*LambertW(-log(a)/a)/log(a)} + + +@slow +def test_lambert_bivariate(): + # tests passing current implementation + assert solve((x**2 + x)*exp(x**2 + x) - 1) == [ + Rational(-1, 2) + sqrt(1 + 4*LambertW(1))/2, + Rational(-1, 2) - sqrt(1 + 4*LambertW(1))/2] + assert solve((x**2 + x)*exp((x**2 + x)*2) - 1) == [ + Rational(-1, 2) + sqrt(1 + 2*LambertW(2))/2, + Rational(-1, 2) - sqrt(1 + 2*LambertW(2))/2] + assert solve(a/x + exp(x/2), x) == [2*LambertW(-a/2)] + assert solve((a/x + exp(x/2)).diff(x), x) == \ + [4*LambertW(-sqrt(2)*sqrt(a)/4), 4*LambertW(sqrt(2)*sqrt(a)/4)] + assert solve((1/x + exp(x/2)).diff(x), x) == \ + [4*LambertW(-sqrt(2)/4), + 4*LambertW(sqrt(2)/4), # nsimplifies as 2*2**(141/299)*3**(206/299)*5**(205/299)*7**(37/299)/21 + 4*LambertW(-sqrt(2)/4, -1)] + assert solve(x*log(x) + 3*x + 1, x) == \ + [exp(-3 + LambertW(-exp(3)))] + assert solve(-x**2 + 2**x, x) == [2, 4, -2*LambertW(log(2)/2)/log(2)] + assert solve(x**2 - 2**x, x) == [2, 4, -2*LambertW(log(2)/2)/log(2)] + ans = solve(3*x + 5 + 2**(-5*x + 3), x) + assert len(ans) == 1 and ans[0].expand() == \ + Rational(-5, 3) + LambertW(-10240*root(2, 3)*log(2)/3)/(5*log(2)) + assert solve(5*x - 1 + 3*exp(2 - 7*x), x) == \ + [Rational(1, 5) + LambertW(-21*exp(Rational(3, 5))/5)/7] + assert solve((log(x) + x).subs(x, x**2 + 1)) == [ + -I*sqrt(-LambertW(1) + 1), sqrt(-1 + LambertW(1))] + # check collection + ax = a**(3*x + 5) + ans = solve(3*log(ax) + b*log(ax) + ax, x) + x0 = 1/log(a) + x1 = sqrt(3)*I + x2 = b + 3 + x3 = x2*LambertW(1/x2)/a**5 + x4 = x3**Rational(1, 3)/2 + assert ans == [ + x0*log(x4*(-x1 - 1)), + x0*log(x4*(x1 - 1)), + x0*log(x3)/3] + x1 = LambertW(Rational(1, 3)) + x2 = a**(-5) + x3 = -3**Rational(1, 3) + x4 = 3**Rational(5, 6)*I + x5 = x1**Rational(1, 3)*x2**Rational(1, 3)/2 + ans = solve(3*log(ax) + ax, x) + assert ans == [ + x0*log(3*x1*x2)/3, + x0*log(x5*(x3 - x4)), + x0*log(x5*(x3 + x4))] + # coverage + p = symbols('p', positive=True) + eq = 4*2**(2*p + 3) - 2*p - 3 + assert _solve_lambert(eq, p, _filtered_gens(Poly(eq), p)) == [ + Rational(-3, 2) - LambertW(-4*log(2))/(2*log(2))] + assert set(solve(3**cos(x) - cos(x)**3)) == { + acos(3), acos(-3*LambertW(-log(3)/3)/log(3))} + # should give only one solution after using `uniq` + assert solve(2*log(x) - 2*log(z) + log(z + log(x) + log(z)), x) == [ + exp(-z + LambertW(2*z**4*exp(2*z))/2)/z] + # cases when p != S.One + # issue 4271 + ans = solve((a/x + exp(x/2)).diff(x, 2), x) + x0 = (-a)**Rational(1, 3) + x1 = sqrt(3)*I + x2 = x0/6 + assert ans == [ + 6*LambertW(x0/3), + 6*LambertW(x2*(-x1 - 1)), + 6*LambertW(x2*(x1 - 1))] + assert solve((1/x + exp(x/2)).diff(x, 2), x) == \ + [6*LambertW(Rational(-1, 3)), 6*LambertW(Rational(1, 6) - sqrt(3)*I/6), \ + 6*LambertW(Rational(1, 6) + sqrt(3)*I/6), 6*LambertW(Rational(-1, 3), -1)] + assert solve(x**2 - y**2/exp(x), x, y, dict=True) == \ + [{x: 2*LambertW(-y/2)}, {x: 2*LambertW(y/2)}] + # this is slow but not exceedingly slow + assert solve((x**3)**(x/2) + pi/2, x) == [ + exp(LambertW(-2*log(2)/3 + 2*log(pi)/3 + I*pi*Rational(2, 3)))] + + # issue 23253 + assert solve((1/log(sqrt(x) + 2)**2 - 1/x)) == [ + (LambertW(-exp(-2), -1) + 2)**2] + assert solve((1/log(1/sqrt(x) + 2)**2 - x)) == [ + (LambertW(-exp(-2), -1) + 2)**-2] + assert solve((1/log(x**2 + 2)**2 - x**-4)) == [ + -I*sqrt(2 - LambertW(exp(2))), + -I*sqrt(LambertW(-exp(-2)) + 2), + sqrt(-2 - LambertW(-exp(-2))), + sqrt(-2 + LambertW(exp(2))), + -sqrt(-2 - LambertW(-exp(-2), -1)), + sqrt(-2 - LambertW(-exp(-2), -1))] + + +def test_rewrite_trig(): + assert solve(sin(x) + tan(x)) == [0, -pi, pi, 2*pi] + assert solve(sin(x) + sec(x)) == [ + -2*atan(Rational(-1, 2) + sqrt(2)*sqrt(1 - sqrt(3)*I)/2 + sqrt(3)*I/2), + 2*atan(S.Half - sqrt(2)*sqrt(1 + sqrt(3)*I)/2 + sqrt(3)*I/2), 2*atan(S.Half + + sqrt(2)*sqrt(1 + sqrt(3)*I)/2 + sqrt(3)*I/2), 2*atan(S.Half - + sqrt(3)*I/2 + sqrt(2)*sqrt(1 - sqrt(3)*I)/2)] + assert solve(sinh(x) + tanh(x)) == [0, I*pi] + + # issue 6157 + assert solve(2*sin(x) - cos(x), x) == [atan(S.Half)] + + +@XFAIL +def test_rewrite_trigh(): + # if this import passes then the test below should also pass + from sympy.functions.elementary.hyperbolic import sech + assert solve(sinh(x) + sech(x)) == [ + 2*atanh(Rational(-1, 2) + sqrt(5)/2 - sqrt(-2*sqrt(5) + 2)/2), + 2*atanh(Rational(-1, 2) + sqrt(5)/2 + sqrt(-2*sqrt(5) + 2)/2), + 2*atanh(-sqrt(5)/2 - S.Half + sqrt(2 + 2*sqrt(5))/2), + 2*atanh(-sqrt(2 + 2*sqrt(5))/2 - sqrt(5)/2 - S.Half)] + + +def test_uselogcombine(): + eq = z - log(x) + log(y/(x*(-1 + y**2/x**2))) + assert solve(eq, x, force=True) == [-sqrt(y*(y - exp(z))), sqrt(y*(y - exp(z)))] + assert solve(log(x + 3) + log(1 + 3/x) - 3) in [ + [-3 + sqrt(-12 + exp(3))*exp(Rational(3, 2))/2 + exp(3)/2, + -sqrt(-12 + exp(3))*exp(Rational(3, 2))/2 - 3 + exp(3)/2], + [-3 + sqrt(-36 + (-exp(3) + 6)**2)/2 + exp(3)/2, + -3 - sqrt(-36 + (-exp(3) + 6)**2)/2 + exp(3)/2], + ] + assert solve(log(exp(2*x) + 1) + log(-tanh(x) + 1) - log(2)) == [] + + +def test_atan2(): + assert solve(atan2(x, 2) - pi/3, x) == [2*sqrt(3)] + + +def test_errorinverses(): + assert solve(erf(x) - y, x) == [erfinv(y)] + assert solve(erfinv(x) - y, x) == [erf(y)] + assert solve(erfc(x) - y, x) == [erfcinv(y)] + assert solve(erfcinv(x) - y, x) == [erfc(y)] + + +def test_issue_2725(): + R = Symbol('R') + eq = sqrt(2)*R*sqrt(1/(R + 1)) + (R + 1)*(sqrt(2)*sqrt(1/(R + 1)) - 1) + sol = solve(eq, R, set=True)[1] + assert sol == {(Rational(5, 3) + (Rational(-1, 2) - sqrt(3)*I/2)*(Rational(251, 27) + + sqrt(111)*I/9)**Rational(1, 3) + 40/(9*((Rational(-1, 2) - sqrt(3)*I/2)*(Rational(251, 27) + + sqrt(111)*I/9)**Rational(1, 3))),), (Rational(5, 3) + 40/(9*(Rational(251, 27) + + sqrt(111)*I/9)**Rational(1, 3)) + (Rational(251, 27) + sqrt(111)*I/9)**Rational(1, 3),)} + + +def test_issue_5114_6611(): + # See that it doesn't hang; this solves in about 2 seconds. + # Also check that the solution is relatively small. + # Note: the system in issue 6611 solves in about 5 seconds and has + # an op-count of 138336 (with simplify=False). + b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r = symbols('b:r') + eqs = Matrix([ + [b - c/d + r/d], [c*(1/g + 1/e + 1/d) - f/g - r/d], + [-c/g + f*(1/j + 1/i + 1/g) - h/i], [-f/i + h*(1/m + 1/l + 1/i) - k/m], + [-h/m + k*(1/p + 1/o + 1/m) - n/p], [-k/p + n*(1/q + 1/p)]]) + v = Matrix([f, h, k, n, b, c]) + ans = solve(list(eqs), list(v), simplify=False) + # If time is taken to simplify then then 2617 below becomes + # 1168 and the time is about 50 seconds instead of 2. + assert sum(s.count_ops() for s in ans.values()) <= 3270 + + +def test_det_quick(): + m = Matrix(3, 3, symbols('a:9')) + assert m.det() == det_quick(m) # calls det_perm + m[0, 0] = 1 + assert m.det() == det_quick(m) # calls det_minor + m = Matrix(3, 3, list(range(9))) + assert m.det() == det_quick(m) # defaults to .det() + # make sure they work with Sparse + s = SparseMatrix(2, 2, (1, 2, 1, 4)) + assert det_perm(s) == det_minor(s) == s.det() + + +def test_real_imag_splitting(): + a, b = symbols('a b', real=True) + assert solve(sqrt(a**2 + b**2) - 3, a) == \ + [-sqrt(-b**2 + 9), sqrt(-b**2 + 9)] + a, b = symbols('a b', imaginary=True) + assert solve(sqrt(a**2 + b**2) - 3, a) == [] + + +def test_issue_7110(): + y = -2*x**3 + 4*x**2 - 2*x + 5 + assert any(ask(Q.real(i)) for i in solve(y)) + + +def test_units(): + assert solve(1/x - 1/(2*cm)) == [2*cm] + + +def test_issue_7547(): + A, B, V = symbols('A,B,V') + eq1 = Eq(630.26*(V - 39.0)*V*(V + 39) - A + B, 0) + eq2 = Eq(B, 1.36*10**8*(V - 39)) + eq3 = Eq(A, 5.75*10**5*V*(V + 39.0)) + sol = Matrix(nsolve(Tuple(eq1, eq2, eq3), [A, B, V], (0, 0, 0))) + assert str(sol) == str(Matrix( + [['4442890172.68209'], + ['4289299466.1432'], + ['70.5389666628177']])) + + +def test_issue_7895(): + r = symbols('r', real=True) + assert solve(sqrt(r) - 2) == [4] + + +def test_issue_2777(): + # the equations represent two circles + x, y = symbols('x y', real=True) + e1, e2 = sqrt(x**2 + y**2) - 10, sqrt(y**2 + (-x + 10)**2) - 3 + a, b = Rational(191, 20), 3*sqrt(391)/20 + ans = [(a, -b), (a, b)] + assert solve((e1, e2), (x, y)) == ans + assert solve((e1, e2/(x - a)), (x, y)) == [] + # make the 2nd circle's radius be -3 + e2 += 6 + assert solve((e1, e2), (x, y)) == [] + assert solve((e1, e2), (x, y), check=False) == ans + + +def test_issue_7322(): + number = 5.62527e-35 + assert solve(x - number, x)[0] == number + + +def test_nsolve(): + raises(ValueError, lambda: nsolve(x, (-1, 1), method='bisect')) + raises(TypeError, lambda: nsolve((x - y + 3,x + y,z - y),(x,y,z),(-50,50))) + raises(TypeError, lambda: nsolve((x + y, x - y), (0, 1))) + raises(TypeError, lambda: nsolve(x < 0.5, x, 1)) + + +@slow +def test_high_order_multivariate(): + assert len(solve(a*x**3 - x + 1, x)) == 3 + assert len(solve(a*x**4 - x + 1, x)) == 4 + assert solve(a*x**5 - x + 1, x) == [] # incomplete solution allowed + raises(NotImplementedError, lambda: + solve(a*x**5 - x + 1, x, incomplete=False)) + + # result checking must always consider the denominator and CRootOf + # must be checked, too + d = x**5 - x + 1 + assert solve(d*(1 + 1/d)) == [CRootOf(d + 1, i) for i in range(5)] + d = x - 1 + assert solve(d*(2 + 1/d)) == [S.Half] + + +def test_base_0_exp_0(): + assert solve(0**x - 1) == [0] + assert solve(0**(x - 2) - 1) == [2] + assert solve(S('x*(1/x**0 - x)', evaluate=False)) == \ + [0, 1] + + +def test__simple_dens(): + assert _simple_dens(1/x**0, [x]) == set() + assert _simple_dens(1/x**y, [x]) == {x**y} + assert _simple_dens(1/root(x, 3), [x]) == {x} + + +def test_issue_8755(): + # This tests two things: that if full unrad is attempted and fails + # the solution should still be found; also it tests the use of + # keyword `composite`. + assert len(solve(sqrt(y)*x + x**3 - 1, x)) == 3 + assert len(solve(-512*y**3 + 1344*(x + 2)**Rational(1, 3)*y**2 - + 1176*(x + 2)**Rational(2, 3)*y - 169*x + 686, y, _unrad=False)) == 3 + + +@slow +def test_issue_8828(): + x1 = 0 + y1 = -620 + r1 = 920 + x2 = 126 + y2 = 276 + x3 = 51 + y3 = 205 + r3 = 104 + v = x, y, z + + f1 = (x - x1)**2 + (y - y1)**2 - (r1 - z)**2 + f2 = (x - x2)**2 + (y - y2)**2 - z**2 + f3 = (x - x3)**2 + (y - y3)**2 - (r3 - z)**2 + F = f1,f2,f3 + + g1 = sqrt((x - x1)**2 + (y - y1)**2) + z - r1 + g2 = f2 + g3 = sqrt((x - x3)**2 + (y - y3)**2) + z - r3 + G = g1,g2,g3 + + A = solve(F, v) + B = solve(G, v) + C = solve(G, v, manual=True) + + p, q, r = [{tuple(i.evalf(2) for i in j) for j in R} for R in [A, B, C]] + assert p == q == r + + +def test_issue_2840_8155(): + # with parameter-free solutions (i.e. no `n`), we want to avoid + # excessive periodic solutions + assert solve(sin(3*x) + sin(6*x)) == [0, -2*pi/9, 2*pi/9] + assert solve(sin(300*x) + sin(600*x)) == [0, -pi/450, pi/450] + assert solve(2*sin(x) - 2*sin(2*x)) == [0, -pi/3, pi/3] + + +def test_issue_9567(): + assert solve(1 + 1/(x - 1)) == [0] + + +def test_issue_11538(): + assert solve(x + E) == [-E] + assert solve(x**2 + E) == [-I*sqrt(E), I*sqrt(E)] + assert solve(x**3 + 2*E) == [ + -cbrt(2 * E), + cbrt(2)*cbrt(E)/2 - cbrt(2)*sqrt(3)*I*cbrt(E)/2, + cbrt(2)*cbrt(E)/2 + cbrt(2)*sqrt(3)*I*cbrt(E)/2] + assert solve([x + 4, y + E], x, y) == {x: -4, y: -E} + assert solve([x**2 + 4, y + E], x, y) == [ + (-2*I, -E), (2*I, -E)] + + e1 = x - y**3 + 4 + e2 = x + y + 4 + 4 * E + assert len(solve([e1, e2], x, y)) == 3 + + +@slow +def test_issue_12114(): + a, b, c, d, e, f, g = symbols('a,b,c,d,e,f,g') + terms = [1 + a*b + d*e, 1 + a*c + d*f, 1 + b*c + e*f, + g - a**2 - d**2, g - b**2 - e**2, g - c**2 - f**2] + sol = solve(terms, [a, b, c, d, e, f, g], dict=True) + s = sqrt(-f**2 - 1) + s2 = sqrt(2 - f**2) + s3 = sqrt(6 - 3*f**2) + s4 = sqrt(3)*f + s5 = sqrt(3)*s2 + assert sol == [ + {a: -s, b: -s, c: -s, d: f, e: f, g: -1}, + {a: s, b: s, c: s, d: f, e: f, g: -1}, + {a: -s4/2 - s2/2, b: s4/2 - s2/2, c: s2, + d: -f/2 + s3/2, e: -f/2 - s5/2, g: 2}, + {a: -s4/2 + s2/2, b: s4/2 + s2/2, c: -s2, + d: -f/2 - s3/2, e: -f/2 + s5/2, g: 2}, + {a: s4/2 - s2/2, b: -s4/2 - s2/2, c: s2, + d: -f/2 - s3/2, e: -f/2 + s5/2, g: 2}, + {a: s4/2 + s2/2, b: -s4/2 + s2/2, c: -s2, + d: -f/2 + s3/2, e: -f/2 - s5/2, g: 2}] + + +def test_inf(): + assert solve(1 - oo*x) == [] + assert solve(oo*x, x) == [] + assert solve(oo*x - oo, x) == [] + + +def test_issue_12448(): + f = Function('f') + fun = [f(i) for i in range(15)] + sym = symbols('x:15') + reps = dict(zip(fun, sym)) + + (x, y, z), c = sym[:3], sym[3:] + ssym = solve([c[4*i]*x + c[4*i + 1]*y + c[4*i + 2]*z + c[4*i + 3] + for i in range(3)], (x, y, z)) + + (x, y, z), c = fun[:3], fun[3:] + sfun = solve([c[4*i]*x + c[4*i + 1]*y + c[4*i + 2]*z + c[4*i + 3] + for i in range(3)], (x, y, z)) + + assert sfun[fun[0]].xreplace(reps).count_ops() == \ + ssym[sym[0]].count_ops() + + +def test_denoms(): + assert denoms(x/2 + 1/y) == {2, y} + assert denoms(x/2 + 1/y, y) == {y} + assert denoms(x/2 + 1/y, [y]) == {y} + assert denoms(1/x + 1/y + 1/z, [x, y]) == {x, y} + assert denoms(1/x + 1/y + 1/z, x, y) == {x, y} + assert denoms(1/x + 1/y + 1/z, {x, y}) == {x, y} + + +def test_issue_12476(): + x0, x1, x2, x3, x4, x5 = symbols('x0 x1 x2 x3 x4 x5') + eqns = [x0**2 - x0, x0*x1 - x1, x0*x2 - x2, x0*x3 - x3, x0*x4 - x4, x0*x5 - x5, + x0*x1 - x1, -x0/3 + x1**2 - 2*x2/3, x1*x2 - x1/3 - x2/3 - x3/3, + x1*x3 - x2/3 - x3/3 - x4/3, x1*x4 - 2*x3/3 - x5/3, x1*x5 - x4, x0*x2 - x2, + x1*x2 - x1/3 - x2/3 - x3/3, -x0/6 - x1/6 + x2**2 - x2/6 - x3/3 - x4/6, + -x1/6 + x2*x3 - x2/3 - x3/6 - x4/6 - x5/6, x2*x4 - x2/3 - x3/3 - x4/3, + x2*x5 - x3, x0*x3 - x3, x1*x3 - x2/3 - x3/3 - x4/3, + -x1/6 + x2*x3 - x2/3 - x3/6 - x4/6 - x5/6, + -x0/6 - x1/6 - x2/6 + x3**2 - x3/3 - x4/6, -x1/3 - x2/3 + x3*x4 - x3/3, + -x2 + x3*x5, x0*x4 - x4, x1*x4 - 2*x3/3 - x5/3, x2*x4 - x2/3 - x3/3 - x4/3, + -x1/3 - x2/3 + x3*x4 - x3/3, -x0/3 - 2*x2/3 + x4**2, -x1 + x4*x5, x0*x5 - x5, + x1*x5 - x4, x2*x5 - x3, -x2 + x3*x5, -x1 + x4*x5, -x0 + x5**2, x0 - 1] + sols = [{x0: 1, x3: Rational(1, 6), x2: Rational(1, 6), x4: Rational(-2, 3), x1: Rational(-2, 3), x5: 1}, + {x0: 1, x3: S.Half, x2: Rational(-1, 2), x4: 0, x1: 0, x5: -1}, + {x0: 1, x3: Rational(-1, 3), x2: Rational(-1, 3), x4: Rational(1, 3), x1: Rational(1, 3), x5: 1}, + {x0: 1, x3: 1, x2: 1, x4: 1, x1: 1, x5: 1}, + {x0: 1, x3: Rational(-1, 3), x2: Rational(1, 3), x4: sqrt(5)/3, x1: -sqrt(5)/3, x5: -1}, + {x0: 1, x3: Rational(-1, 3), x2: Rational(1, 3), x4: -sqrt(5)/3, x1: sqrt(5)/3, x5: -1}] + + assert solve(eqns) == sols + + +def test_issue_13849(): + t = symbols('t') + assert solve((t*(sqrt(5) + sqrt(2)) - sqrt(2), t), t) == [] + + +def test_issue_14860(): + from sympy.physics.units import newton, kilo + assert solve(8*kilo*newton + x + y, x) == [-8000*newton - y] + + +def test_issue_14721(): + k, h, a, b = symbols(':4') + assert solve([ + -1 + (-k + 1)**2/b**2 + (-h - 1)**2/a**2, + -1 + (-k + 1)**2/b**2 + (-h + 1)**2/a**2, + h, k + 2], h, k, a, b) == [ + (0, -2, -b*sqrt(1/(b**2 - 9)), b), + (0, -2, b*sqrt(1/(b**2 - 9)), b)] + assert solve([ + h, h/a + 1/b**2 - 2, -h/2 + 1/b**2 - 2], a, h, b) == [ + (a, 0, -sqrt(2)/2), (a, 0, sqrt(2)/2)] + assert solve((a + b**2 - 1, a + b**2 - 2)) == [] + + +def test_issue_14779(): + x = symbols('x', real=True) + assert solve(sqrt(x**4 - 130*x**2 + 1089) + sqrt(x**4 - 130*x**2 + + 3969) - 96*Abs(x)/x,x) == [sqrt(130)] + + +def test_issue_15307(): + assert solve((y - 2, Mul(x + 3,x - 2, evaluate=False))) == \ + [{x: -3, y: 2}, {x: 2, y: 2}] + assert solve((y - 2, Mul(3, x - 2, evaluate=False))) == \ + {x: 2, y: 2} + assert solve((y - 2, Add(x + 4, x - 2, evaluate=False))) == \ + {x: -1, y: 2} + eq1 = Eq(12513*x + 2*y - 219093, -5726*x - y) + eq2 = Eq(-2*x + 8, 2*x - 40) + assert solve([eq1, eq2]) == {x:12, y:75} + + +def test_issue_15415(): + assert solve(x - 3, x) == [3] + assert solve([x - 3], x) == {x:3} + assert solve(Eq(y + 3*x**2/2, y + 3*x), y) == [] + assert solve([Eq(y + 3*x**2/2, y + 3*x)], y) == [] + assert solve([Eq(y + 3*x**2/2, y + 3*x), Eq(x, 1)], y) == [] + + +@slow +def test_issue_15731(): + # f(x)**g(x)=c + assert solve(Eq((x**2 - 7*x + 11)**(x**2 - 13*x + 42), 1)) == [2, 3, 4, 5, 6, 7] + assert solve((x)**(x + 4) - 4) == [-2] + assert solve((-x)**(-x + 4) - 4) == [2] + assert solve((x**2 - 6)**(x**2 - 2) - 4) == [-2, 2] + assert solve((x**2 - 2*x - 1)**(x**2 - 3) - 1/(1 - 2*sqrt(2))) == [sqrt(2)] + assert solve(x**(x + S.Half) - 4*sqrt(2)) == [S(2)] + assert solve((x**2 + 1)**x - 25) == [2] + assert solve(x**(2/x) - 2) == [2, 4] + assert solve((x/2)**(2/x) - sqrt(2)) == [4, 8] + assert solve(x**(x + S.Half) - Rational(9, 4)) == [Rational(3, 2)] + # a**g(x)=c + assert solve((-sqrt(sqrt(2)))**x - 2) == [4, log(2)/(log(2**Rational(1, 4)) + I*pi)] + assert solve((sqrt(2))**x - sqrt(sqrt(2))) == [S.Half] + assert solve((-sqrt(2))**x + 2*(sqrt(2))) == [3, + (3*log(2)**2 + 4*pi**2 - 4*I*pi*log(2))/(log(2)**2 + 4*pi**2)] + assert solve((sqrt(2))**x - 2*(sqrt(2))) == [3] + assert solve(I**x + 1) == [2] + assert solve((1 + I)**x - 2*I) == [2] + assert solve((sqrt(2) + sqrt(3))**x - (2*sqrt(6) + 5)**Rational(1, 3)) == [Rational(2, 3)] + # bases of both sides are equal + b = Symbol('b') + assert solve(b**x - b**2, x) == [2] + assert solve(b**x - 1/b, x) == [-1] + assert solve(b**x - b, x) == [1] + b = Symbol('b', positive=True) + assert solve(b**x - b**2, x) == [2] + assert solve(b**x - 1/b, x) == [-1] + + +def test_issue_10933(): + assert solve(x**4 + y*(x + 0.1), x) # doesn't fail + assert solve(I*x**4 + x**3 + x**2 + 1.) # doesn't fail + + +def test_Abs_handling(): + x = symbols('x', real=True) + assert solve(abs(x/y), x) == [0] + + +def test_issue_7982(): + x = Symbol('x') + # Test that no exception happens + assert solve([2*x**2 + 5*x + 20 <= 0, x >= 1.5], x) is S.false + # From #8040 + assert solve([x**3 - 8.08*x**2 - 56.48*x/5 - 106 >= 0, x - 1 <= 0], [x]) is S.false + + +def test_issue_14645(): + x, y = symbols('x y') + assert solve([x*y - x - y, x*y - x - y], [x, y]) == [(y/(y - 1), y)] + + +def test_issue_12024(): + x, y = symbols('x y') + assert solve(Piecewise((0.0, x < 0.1), (x, x >= 0.1)) - y) == \ + [{y: Piecewise((0.0, x < 0.1), (x, True))}] + + +def test_issue_17452(): + assert solve((7**x)**x + pi, x) == [-sqrt(log(pi) + I*pi)/sqrt(log(7)), + sqrt(log(pi) + I*pi)/sqrt(log(7))] + assert solve(x**(x/11) + pi/11, x) == [exp(LambertW(-11*log(11) + 11*log(pi) + 11*I*pi))] + + +def test_issue_17799(): + assert solve(-erf(x**(S(1)/3))**pi + I, x) == [] + + +def test_issue_17650(): + x = Symbol('x', real=True) + assert solve(abs(abs(x**2 - 1) - x) - x) == [1, -1 + sqrt(2), 1 + sqrt(2)] + + +def test_issue_17882(): + eq = -8*x**2/(9*(x**2 - 1)**(S(4)/3)) + 4/(3*(x**2 - 1)**(S(1)/3)) + assert unrad(eq) is None + + +def test_issue_17949(): + assert solve(exp(+x+x**2), x) == [] + assert solve(exp(-x+x**2), x) == [] + assert solve(exp(+x-x**2), x) == [] + assert solve(exp(-x-x**2), x) == [] + + +def test_issue_10993(): + assert solve(Eq(binomial(x, 2), 3)) == [-2, 3] + assert solve(Eq(pow(x, 2) + binomial(x, 3), x)) == [-4, 0, 1] + assert solve(Eq(binomial(x, 2), 0)) == [0, 1] + assert solve(a+binomial(x, 3), a) == [-binomial(x, 3)] + assert solve(x-binomial(a, 3) + binomial(y, 2) + sin(a), x) == [-sin(a) + binomial(a, 3) - binomial(y, 2)] + assert solve((x+1)-binomial(x+1, 3), x) == [-2, -1, 3] + + +def test_issue_11553(): + eq1 = x + y + 1 + eq2 = x + GoldenRatio + assert solve([eq1, eq2], x, y) == {x: -GoldenRatio, y: -1 + GoldenRatio} + eq3 = x + 2 + TribonacciConstant + assert solve([eq1, eq3], x, y) == {x: -2 - TribonacciConstant, y: 1 + TribonacciConstant} + + +def test_issue_19113_19102(): + t = S(1)/3 + solve(cos(x)**5-sin(x)**5) + assert solve(4*cos(x)**3 - 2*sin(x)**3) == [ + atan(2**(t)), -atan(2**(t)*(1 - sqrt(3)*I)/2), + -atan(2**(t)*(1 + sqrt(3)*I)/2)] + h = S.Half + assert solve(cos(x)**2 + sin(x)) == [ + 2*atan(-h + sqrt(5)/2 + sqrt(2)*sqrt(1 - sqrt(5))/2), + -2*atan(h + sqrt(5)/2 + sqrt(2)*sqrt(1 + sqrt(5))/2), + -2*atan(-sqrt(5)/2 + h + sqrt(2)*sqrt(1 - sqrt(5))/2), + -2*atan(-sqrt(2)*sqrt(1 + sqrt(5))/2 + h + sqrt(5)/2)] + assert solve(3*cos(x) - sin(x)) == [atan(3)] + + +def test_issue_19509(): + a = S(3)/4 + b = S(5)/8 + c = sqrt(5)/8 + d = sqrt(5)/4 + assert solve(1/(x -1)**5 - 1) == [2, + -d + a - sqrt(-b + c), + -d + a + sqrt(-b + c), + d + a - sqrt(-b - c), + d + a + sqrt(-b - c)] + +def test_issue_20747(): + THT, HT, DBH, dib, c0, c1, c2, c3, c4 = symbols('THT HT DBH dib c0 c1 c2 c3 c4') + f = DBH*c3 + THT*c4 + c2 + rhs = 1 - ((HT - 1)/(THT - 1))**c1*(1 - exp(c0/f)) + eq = dib - DBH*(c0 - f*log(rhs)) + term = ((1 - exp((DBH*c0 - dib)/(DBH*(DBH*c3 + THT*c4 + c2)))) + / (1 - exp(c0/(DBH*c3 + THT*c4 + c2)))) + sol = [THT*term**(1/c1) - term**(1/c1) + 1] + assert solve(eq, HT) == sol + + +def test_issue_20902(): + f = (t / ((1 + t) ** 2)) + assert solve(f.subs({t: 3 * x + 2}).diff(x) > 0, x) == (S(-1) < x) & (x < S(-1)/3) + assert solve(f.subs({t: 3 * x + 3}).diff(x) > 0, x) == (S(-4)/3 < x) & (x < S(-2)/3) + assert solve(f.subs({t: 3 * x + 4}).diff(x) > 0, x) == (S(-5)/3 < x) & (x < S(-1)) + assert solve(f.subs({t: 3 * x + 2}).diff(x) > 0, x) == (S(-1) < x) & (x < S(-1)/3) + + +def test_issue_21034(): + a = symbols('a', real=True) + system = [x - cosh(cos(4)), y - sinh(cos(a)), z - tanh(x)] + # constants inside hyperbolic functions should not be rewritten in terms of exp + assert solve(system, x, y, z) == [(cosh(cos(4)), sinh(cos(a)), tanh(cosh(cos(4))))] + # but if the variable of interest is present in a hyperbolic function, + # then it should be rewritten in terms of exp and solved further + newsystem = [(exp(x) - exp(-x)) - tanh(x)*(exp(x) + exp(-x)) + x - 5] + assert solve(newsystem, x) == {x: 5} + + +def test_issue_4886(): + z = a*sqrt(R**2*a**2 + R**2*b**2 - c**2)/(a**2 + b**2) + t = b*c/(a**2 + b**2) + sol = [((b*(t - z) - c)/(-a), t - z), ((b*(t + z) - c)/(-a), t + z)] + assert solve([x**2 + y**2 - R**2, a*x + b*y - c], x, y) == sol + + +def test_issue_6819(): + a, b, c, d = symbols('a b c d', positive=True) + assert solve(a*b**x - c*d**x, x) == [log(c/a)/log(b/d)] + + +def test_issue_17454(): + x = Symbol('x') + assert solve((1 - x - I)**4, x) == [1 - I] + + +def test_issue_21852(): + solution = [21 - 21*sqrt(2)/2] + assert solve(2*x + sqrt(2*x**2) - 21) == solution + + +def test_issue_21942(): + eq = -d + (a*c**(1 - e) + b**(1 - e)*(1 - a))**(1/(1 - e)) + sol = solve(eq, c, simplify=False, check=False) + assert sol == [((a*b**(1 - e) - b**(1 - e) + + d**(1 - e))/a)**(1/(1 - e))] + + +def test_solver_flags(): + root = solve(x**5 + x**2 - x - 1, cubics=False) + rad = solve(x**5 + x**2 - x - 1, cubics=True) + assert root != rad + + +def test_issue_22768(): + eq = 2*x**3 - 16*(y - 1)**6*z**3 + assert solve(eq.expand(), x, simplify=False + ) == [2*z*(y - 1)**2, z*(-1 + sqrt(3)*I)*(y - 1)**2, + -z*(1 + sqrt(3)*I)*(y - 1)**2] + + +def test_issue_22717(): + assert solve((-y**2 + log(y**2/x) + 2, -2*x*y + 2*x/y)) == [ + {y: -1, x: E}, {y: 1, x: E}] + + +def test_issue_25176(): + eq = (x - 5)**-8 - 3 + sol = solve(eq) + assert not any(eq.subs(x, i) for i in sol) + + +def test_issue_10169(): + eq = S(-8*a - x**5*(a + b + c + e) - x**4*(4*a - 2**Rational(3,4)*c + 4*c + + d + 2**Rational(3,4)*e + 4*e + k) - x**3*(-4*2**Rational(3,4)*c + sqrt(2)*c - + 2**Rational(3,4)*d + 4*d + sqrt(2)*e + 4*2**Rational(3,4)*e + 2**Rational(3,4)*k + 4*k) - + x**2*(4*sqrt(2)*c - 4*2**Rational(3,4)*d + sqrt(2)*d + 4*sqrt(2)*e + + sqrt(2)*k + 4*2**Rational(3,4)*k) - x*(2*a + 2*b + 4*sqrt(2)*d + + 4*sqrt(2)*k) + 5) + assert solve_undetermined_coeffs(eq, [a, b, c, d, e, k], x) == { + a: Rational(5,8), + b: Rational(-5,1032), + c: Rational(-40,129) - 5*2**Rational(3,4)/129 + 5*2**Rational(1,4)/1032, + d: -20*2**Rational(3,4)/129 - 10*sqrt(2)/129 - 5*2**Rational(1,4)/258, + e: Rational(-40,129) - 5*2**Rational(1,4)/1032 + 5*2**Rational(3,4)/129, + k: -10*sqrt(2)/129 + 5*2**Rational(1,4)/258 + 20*2**Rational(3,4)/129 + } + + +def test_solve_undetermined_coeffs_issue_23927(): + A, B, r, phi = symbols('A, B, r, phi') + e = Eq(A*sin(t) + B*cos(t), r*sin(t - phi)) + eq = (e.lhs - e.rhs).expand(trig=True) + soln = solve_undetermined_coeffs(eq, (r, phi), t) + assert soln == [{ + phi: 2*atan((A - sqrt(A**2 + B**2))/B), + r: (-A**2 + A*sqrt(A**2 + B**2) - B**2)/(A - sqrt(A**2 + B**2)) + }, { + phi: 2*atan((A + sqrt(A**2 + B**2))/B), + r: (A**2 + A*sqrt(A**2 + B**2) + B**2)/(A + sqrt(A**2 + B**2))/-1 + }] + +def test_issue_24368(): + # Ideally these would produce a solution, but for now just check that they + # don't fail with a RuntimeError + raises(NotImplementedError, lambda: solve(Mod(x**2, 49), x)) + s2 = Symbol('s2', integer=True, positive=True) + f = floor(s2/2 - S(1)/2) + raises(NotImplementedError, lambda: solve((Mod(f**2/(f + 1) + 2*f/(f + 1) + 1/(f + 1), 1))*f + Mod(f**2/(f + 1) + 2*f/(f + 1) + 1/(f + 1), 1), s2)) + + +def test_solve_Piecewise(): + assert [S(10)/3] == solve(3*Piecewise( + (S.NaN, x <= 0), + (20*x - 3*(x - 6)**2/2 - 176, (x >= 0) & (x >= 2) & (x>= 4) & (x >= 6) & (x < 10)), + (100 - 26*x, (x >= 0) & (x >= 2) & (x >= 4) & (x < 10)), + (16*x - 3*(x - 6)**2/2 - 176, (x >= 2) & (x >= 4) & (x >= 6) & (x < 10)), + (100 - 30*x, (x >= 2) & (x >= 4) & (x < 10)), + (30*x - 3*(x - 6)**2/2 - 196, (x>= 0) & (x >= 4) & (x >= 6) & (x < 10)), + (80 - 16*x, (x >= 0) & (x >= 4) & (x < 10)), + (26*x - 3*(x - 6)**2/2 - 196, (x >= 4) & (x >= 6) & (x < 10)), + (80 - 20*x, (x >= 4) & (x < 10)), + (40*x - 3*(x - 6)**2/2 - 256, (x >= 0) & (x >= 2) & (x >= 6) & (x < 10)), + (20 - 6*x, (x >= 0) & (x >= 2) & (x < 10)), + (36*x - 3*(x - 6)**2/2 - 256, (x >= 2) & (x >= 6) & (x < 10)), + (20 - 10*x, (x >= 2) & (x < 10)), + (50*x - 3*(x - 6)**2/2 - 276, (x >= 0) & (x >= 6) & (x < 10)), + (4*x, (x >= 0) & (x < 10)), + (46*x - 3*(x - 6)**2/2 - 276, (x >= 6) & (x < 10)), + (0, x < 10), # this will simplify away + (S.NaN,True))) diff --git a/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_solveset.py b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_solveset.py new file mode 100644 index 0000000000000000000000000000000000000000..a1ba7a11e68ed518c4d83c050947b78756ade181 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/solvers/tests/test_solveset.py @@ -0,0 +1,3548 @@ +from math import isclose + +from sympy.calculus.util import stationary_points +from sympy.core.containers import Tuple +from sympy.core.function import (Function, Lambda, nfloat, diff) +from sympy.core.mod import Mod +from sympy.core.numbers import (E, I, Rational, oo, pi, Integer, all_close) +from sympy.core.relational import (Eq, Gt, Ne, Ge) +from sympy.core.singleton import S +from sympy.core.sorting import ordered +from sympy.core.symbol import (Dummy, Symbol, symbols) +from sympy.core.sympify import sympify +from sympy.functions.elementary.complexes import (Abs, arg, im, re, sign, conjugate) +from sympy.functions.elementary.exponential import (LambertW, exp, log) +from sympy.functions.elementary.hyperbolic import (HyperbolicFunction, + sinh, cosh, tanh, coth, sech, csch, asinh, acosh, atanh, acoth, asech, acsch) +from sympy.functions.elementary.miscellaneous import sqrt, Min, Max +from sympy.functions.elementary.piecewise import Piecewise +from sympy.functions.elementary.trigonometric import ( + TrigonometricFunction, acos, acot, acsc, asec, asin, atan, atan2, + cos, cot, csc, sec, sin, tan) +from sympy.functions.special.error_functions import (erf, erfc, + erfcinv, erfinv) +from sympy.logic.boolalg import And +from sympy.matrices.dense import MutableDenseMatrix as Matrix +from sympy.matrices.immutable import ImmutableDenseMatrix +from sympy.polys.polytools import Poly +from sympy.polys.rootoftools import CRootOf +from sympy.sets.contains import Contains +from sympy.sets.conditionset import ConditionSet +from sympy.sets.fancysets import ImageSet, Range +from sympy.sets.sets import (Complement, FiniteSet, + Intersection, Interval, Union, imageset, ProductSet) +from sympy.simplify import simplify +from sympy.tensor.indexed import Indexed +from sympy.utilities.iterables import numbered_symbols + +from sympy.testing.pytest import (XFAIL, raises, skip, slow, SKIP, _both_exp_pow) +from sympy.core.random import verify_numerically as tn +from sympy.physics.units import cm + +from sympy.solvers import solve +from sympy.solvers.solveset import ( + solveset_real, domain_check, solveset_complex, linear_eq_to_matrix, + linsolve, _is_function_class_equation, invert_real, invert_complex, + _invert_trig_hyp_real, solveset, solve_decomposition, substitution, + nonlinsolve, solvify, + _is_finite_with_finite_vars, _transolve, _is_exponential, + _solve_exponential, _is_logarithmic, _is_lambert, + _solve_logarithm, _term_factors, _is_modular, NonlinearError) + +from sympy.abc import (a, b, c, d, e, f, g, h, i, j, k, l, m, n, q, r, + t, w, x, y, z) + + +def dumeq(i, j): + if type(i) in (list, tuple): + return all(dumeq(i, j) for i, j in zip(i, j)) + return i == j or i.dummy_eq(j) + + +def assert_close_ss(sol1, sol2): + """Test solutions with floats from solveset are close""" + sol1 = sympify(sol1) + sol2 = sympify(sol2) + assert isinstance(sol1, FiniteSet) + assert isinstance(sol2, FiniteSet) + assert len(sol1) == len(sol2) + assert all(isclose(v1, v2) for v1, v2 in zip(sol1, sol2)) + + +def assert_close_nl(sol1, sol2): + """Test solutions with floats from nonlinsolve are close""" + sol1 = sympify(sol1) + sol2 = sympify(sol2) + assert isinstance(sol1, FiniteSet) + assert isinstance(sol2, FiniteSet) + assert len(sol1) == len(sol2) + for s1, s2 in zip(sol1, sol2): + assert len(s1) == len(s2) + assert all(isclose(v1, v2) for v1, v2 in zip(s1, s2)) + + +@_both_exp_pow +def test_invert_real(): + x = Symbol('x', real=True) + + def ireal(x, s=S.Reals): + return Intersection(s, x) + + assert invert_real(exp(x), z, x) == (x, ireal(FiniteSet(log(z)))) + + y = Symbol('y', positive=True) + n = Symbol('n', real=True) + assert invert_real(x + 3, y, x) == (x, FiniteSet(y - 3)) + assert invert_real(x*3, y, x) == (x, FiniteSet(y / 3)) + + assert invert_real(exp(x), y, x) == (x, FiniteSet(log(y))) + assert invert_real(exp(3*x), y, x) == (x, FiniteSet(log(y) / 3)) + assert invert_real(exp(x + 3), y, x) == (x, FiniteSet(log(y) - 3)) + + assert invert_real(exp(x) + 3, y, x) == (x, ireal(FiniteSet(log(y - 3)))) + assert invert_real(exp(x)*3, y, x) == (x, FiniteSet(log(y / 3))) + + assert invert_real(log(x), y, x) == (x, FiniteSet(exp(y))) + assert invert_real(log(3*x), y, x) == (x, FiniteSet(exp(y) / 3)) + assert invert_real(log(x + 3), y, x) == (x, FiniteSet(exp(y) - 3)) + + assert invert_real(Abs(x), y, x) == (x, FiniteSet(y, -y)) + + assert invert_real(2**x, y, x) == (x, FiniteSet(log(y)/log(2))) + assert invert_real(2**exp(x), y, x) == (x, ireal(FiniteSet(log(log(y)/log(2))))) + + assert invert_real(x**2, y, x) == (x, FiniteSet(sqrt(y), -sqrt(y))) + assert invert_real(x**S.Half, y, x) == (x, FiniteSet(y**2)) + + raises(ValueError, lambda: invert_real(x, x, x)) + + # issue 21236 + assert invert_real(x**pi, y, x) == (x, FiniteSet(y**(1/pi))) + assert invert_real(x**pi, -E, x) == (x, S.EmptySet) + assert invert_real(x**Rational(3/2), 1000, x) == (x, FiniteSet(100)) + assert invert_real(x**1.0, 1, x) == (x**1.0, FiniteSet(1)) + + raises(ValueError, lambda: invert_real(S.One, y, x)) + + assert invert_real(x**31 + x, y, x) == (x**31 + x, FiniteSet(y)) + + lhs = x**31 + x + base_values = FiniteSet(y - 1, -y - 1) + assert invert_real(Abs(x**31 + x + 1), y, x) == (lhs, base_values) + + assert dumeq(invert_real(sin(x), y, x), (x, + ConditionSet(x, (S(-1) <= y) & (y <= S(1)), Union( + ImageSet(Lambda(n, 2*n*pi + asin(y)), S.Integers), + ImageSet(Lambda(n, pi*2*n + pi - asin(y)), S.Integers))))) + + assert dumeq(invert_real(sin(exp(x)), y, x), (x, + ConditionSet(x, (S(-1) <= y) & (y <= S(1)), Union( + ImageSet(Lambda(n, log(2*n*pi + asin(y))), S.Integers), + ImageSet(Lambda(n, log(pi*2*n + pi - asin(y))), S.Integers))))) + + assert dumeq(invert_real(csc(x), y, x), (x, + ConditionSet(x, ((S(1) <= y) & (y < oo)) | ((-oo < y) & (y <= S(-1))), + Union(ImageSet(Lambda(n, 2*n*pi + acsc(y)), S.Integers), + ImageSet(Lambda(n, 2*n*pi - acsc(y) + pi), S.Integers))))) + + assert dumeq(invert_real(csc(exp(x)), y, x), (x, + ConditionSet(x, ((S(1) <= y) & (y < oo)) | ((-oo < y) & (y <= S(-1))), + Union(ImageSet(Lambda(n, log(2*n*pi + acsc(y))), S.Integers), + ImageSet(Lambda(n, log(2*n*pi - acsc(y) + pi)), S.Integers))))) + + assert dumeq(invert_real(cos(x), y, x), (x, + ConditionSet(x, (S(-1) <= y) & (y <= S(1)), Union( + ImageSet(Lambda(n, 2*n*pi + acos(y)), S.Integers), + ImageSet(Lambda(n, 2*n*pi - acos(y)), S.Integers))))) + + assert dumeq(invert_real(cos(exp(x)), y, x), (x, + ConditionSet(x, (S(-1) <= y) & (y <= S(1)), Union( + ImageSet(Lambda(n, log(2*n*pi + acos(y))), S.Integers), + ImageSet(Lambda(n, log(2*n*pi - acos(y))), S.Integers))))) + + assert dumeq(invert_real(sec(x), y, x), (x, + ConditionSet(x, ((S(1) <= y) & (y < oo)) | ((-oo < y) & (y <= S(-1))), + Union(ImageSet(Lambda(n, 2*n*pi + asec(y)), S.Integers), \ + ImageSet(Lambda(n, 2*n*pi - asec(y)), S.Integers))))) + + assert dumeq(invert_real(sec(exp(x)), y, x), (x, + ConditionSet(x, ((S(1) <= y) & (y < oo)) | ((-oo < y) & (y <= S(-1))), + Union(ImageSet(Lambda(n, log(2*n*pi - asec(y))), S.Integers), + ImageSet(Lambda(n, log(2*n*pi + asec(y))), S.Integers))))) + + assert dumeq(invert_real(tan(x), y, x), (x, + ConditionSet(x, (-oo < y) & (y < oo), + ImageSet(Lambda(n, n*pi + atan(y)), S.Integers)))) + + assert dumeq(invert_real(tan(exp(x)), y, x), (x, + ConditionSet(x, (-oo < y) & (y < oo), + ImageSet(Lambda(n, log(n*pi + atan(y))), S.Integers)))) + + assert dumeq(invert_real(cot(x), y, x), (x, + ConditionSet(x, (-oo < y) & (y < oo), + ImageSet(Lambda(n, n*pi + acot(y)), S.Integers)))) + + assert dumeq(invert_real(cot(exp(x)), y, x), (x, + ConditionSet(x, (-oo < y) & (y < oo), + ImageSet(Lambda(n, log(n*pi + acot(y))), S.Integers)))) + + assert dumeq(invert_real(tan(tan(x)), y, x), + (x, ConditionSet(x, Eq(tan(tan(x)), y), S.Reals))) + # slight regression compared to previous result: + # (tan(x), imageset(Lambda(n, n*pi + atan(y)), S.Integers))) + + x = Symbol('x', positive=True) + assert invert_real(x**pi, y, x) == (x, FiniteSet(y**(1/pi))) + + r = Symbol('r', real=True) + p = Symbol('p', positive=True) + assert invert_real(sinh(x), r, x) == (x, FiniteSet(asinh(r))) + assert invert_real(sinh(log(x)), p, x) == (x, FiniteSet(exp(asinh(p)))) + + assert invert_real(cosh(x), r, x) == (x, Intersection( + FiniteSet(-acosh(r), acosh(r)), S.Reals)) + assert invert_real(cosh(x), p + 1, x) == (x, + FiniteSet(-acosh(p + 1), acosh(p + 1))) + + assert invert_real(tanh(x), r, x) == (x, Intersection(FiniteSet(atanh(r)), S.Reals)) + assert invert_real(coth(x), p+1, x) == (x, FiniteSet(acoth(p+1))) + assert invert_real(sech(x), r, x) == (x, Intersection( + FiniteSet(-asech(r), asech(r)), S.Reals)) + assert invert_real(csch(x), p, x) == (x, FiniteSet(acsch(p))) + + assert dumeq(invert_real(tanh(sin(x)), r, x), (x, + ConditionSet(x, (S(-1) <= atanh(r)) & (atanh(r) <= S(1)), Union( + ImageSet(Lambda(n, 2*n*pi + asin(atanh(r))), S.Integers), + ImageSet(Lambda(n, 2*n*pi - asin(atanh(r)) + pi), S.Integers))))) + + +def test_invert_trig_hyp_real(): + # check some codepaths that are not as easily reached otherwise + n = Dummy('n') + assert _invert_trig_hyp_real(cosh(x), Range(-5, 10, 1), x)[1].dummy_eq(Union( + ImageSet(Lambda(n, -acosh(n)), Range(1, 10, 1)), + ImageSet(Lambda(n, acosh(n)), Range(1, 10, 1)))) + assert _invert_trig_hyp_real(coth(x), Interval(-3, 2), x) == (x, Union( + Interval(-oo, -acoth(3)), Interval(acoth(2), oo))) + assert _invert_trig_hyp_real(tanh(x), Interval(-S.Half, 1), x) == (x, + Interval(-atanh(S.Half), oo)) + assert _invert_trig_hyp_real(sech(x), imageset(n, S.Half + n/3, S.Naturals0), x) == \ + (x, FiniteSet(-asech(S(1)/2), asech(S(1)/2), -asech(S(5)/6), asech(S(5)/6))) + assert _invert_trig_hyp_real(csch(x), S.Reals, x) == (x, + Union(Interval.open(-oo, 0), Interval.open(0, oo))) + + +def test_invert_complex(): + assert invert_complex(x + 3, y, x) == (x, FiniteSet(y - 3)) + assert invert_complex(x*3, y, x) == (x, FiniteSet(y / 3)) + assert invert_complex((x - 1)**3, 0, x) == (x, FiniteSet(1)) + + assert dumeq(invert_complex(exp(x), y, x), + (x, imageset(Lambda(n, I*(2*pi*n + arg(y)) + log(Abs(y))), S.Integers))) + + assert invert_complex(log(x), y, x) == (x, FiniteSet(exp(y))) + + raises(ValueError, lambda: invert_real(1, y, x)) + raises(ValueError, lambda: invert_complex(x, x, x)) + raises(ValueError, lambda: invert_complex(x, x, 1)) + + assert dumeq(invert_complex(sin(x), I, x), (x, Union( + ImageSet(Lambda(n, 2*n*pi + I*log(1 + sqrt(2))), S.Integers), + ImageSet(Lambda(n, 2*n*pi + pi - I*log(1 + sqrt(2))), S.Integers)))) + assert dumeq(invert_complex(cos(x), 1+I, x), (x, Union( + ImageSet(Lambda(n, 2*n*pi - acos(1 + I)), S.Integers), + ImageSet(Lambda(n, 2*n*pi + acos(1 + I)), S.Integers)))) + assert dumeq(invert_complex(tan(2*x), 1, x), (x, + ImageSet(Lambda(n, n*pi/2 + pi/8), S.Integers))) + assert dumeq(invert_complex(cot(x), 2*I, x), (x, + ImageSet(Lambda(n, n*pi - I*acoth(2)), S.Integers))) + + assert dumeq(invert_complex(sinh(x), 0, x), (x, Union( + ImageSet(Lambda(n, 2*n*I*pi), S.Integers), + ImageSet(Lambda(n, 2*n*I*pi + I*pi), S.Integers)))) + assert dumeq(invert_complex(cosh(x), 0, x), (x, Union( + ImageSet(Lambda(n, 2*n*I*pi + I*pi/2), S.Integers), + ImageSet(Lambda(n, 2*n*I*pi + 3*I*pi/2), S.Integers)))) + assert invert_complex(tanh(x), 1, x) == (x, S.EmptySet) + assert dumeq(invert_complex(tanh(x), a, x), (x, + ConditionSet(x, Ne(a, -1) & Ne(a, 1), + ImageSet(Lambda(n, n*I*pi + atanh(a)), S.Integers)))) + assert invert_complex(coth(x), 1, x) == (x, S.EmptySet) + assert dumeq(invert_complex(coth(x), a, x), (x, + ConditionSet(x, Ne(a, -1) & Ne(a, 1), + ImageSet(Lambda(n, n*I*pi + acoth(a)), S.Integers)))) + assert dumeq(invert_complex(sech(x), 2, x), (x, Union( + ImageSet(Lambda(n, 2*n*I*pi + I*pi/3), S.Integers), + ImageSet(Lambda(n, 2*n*I*pi + 5*I*pi/3), S.Integers)))) + + +def test_domain_check(): + assert domain_check(1/(1 + (1/(x+1))**2), x, -1) is False + assert domain_check(x**2, x, 0) is True + assert domain_check(x, x, oo) is False + assert domain_check(0, x, oo) is False + + +def test_issue_11536(): + assert solveset(0**x - 100, x, S.Reals) == S.EmptySet + assert solveset(0**x - 1, x, S.Reals) == FiniteSet(0) + + +def test_issue_17479(): + f = (x**2 + y**2)**2 + (x**2 + z**2)**2 - 2*(2*x**2 + y**2 + z**2) + fx = f.diff(x) + fy = f.diff(y) + fz = f.diff(z) + sol = nonlinsolve([fx, fy, fz], [x, y, z]) + assert len(sol) >= 4 and len(sol) <= 20 + # nonlinsolve has been giving a varying number of solutions + # (originally 18, then 20, now 19) due to various internal changes. + # Unfortunately not all the solutions are actually valid and some are + # redundant. Since the original issue was that an exception was raised, + # this first test only checks that nonlinsolve returns a "plausible" + # solution set. The next test checks the result for correctness. + + +@XFAIL +def test_issue_18449(): + x, y, z = symbols("x, y, z") + f = (x**2 + y**2)**2 + (x**2 + z**2)**2 - 2*(2*x**2 + y**2 + z**2) + fx = diff(f, x) + fy = diff(f, y) + fz = diff(f, z) + sol = nonlinsolve([fx, fy, fz], [x, y, z]) + for (xs, ys, zs) in sol: + d = {x: xs, y: ys, z: zs} + assert tuple(_.subs(d).simplify() for _ in (fx, fy, fz)) == (0, 0, 0) + # After simplification and removal of duplicate elements, there should + # only be 4 parametric solutions left: + # simplifiedsolutions = FiniteSet((sqrt(1 - z**2), z, z), + # (-sqrt(1 - z**2), z, z), + # (sqrt(1 - z**2), -z, z), + # (-sqrt(1 - z**2), -z, z)) + # TODO: Is the above solution set definitely complete? + + +def test_issue_21047(): + f = (2 - x)**2 + (sqrt(x - 1) - 1)**6 + assert solveset(f, x, S.Reals) == FiniteSet(2) + + f = (sqrt(x)-1)**2 + (sqrt(x)+1)**2 -2*x**2 + sqrt(2) + assert solveset(f, x, S.Reals) == FiniteSet( + S.Half - sqrt(2*sqrt(2) + 5)/2, S.Half + sqrt(2*sqrt(2) + 5)/2) + + +def test_is_function_class_equation(): + assert _is_function_class_equation(TrigonometricFunction, + tan(x), x) is True + assert _is_function_class_equation(TrigonometricFunction, + tan(x) - 1, x) is True + assert _is_function_class_equation(TrigonometricFunction, + tan(x) + sin(x), x) is True + assert _is_function_class_equation(TrigonometricFunction, + tan(x) + sin(x) - a, x) is True + assert _is_function_class_equation(TrigonometricFunction, + sin(x)*tan(x) + sin(x), x) is True + assert _is_function_class_equation(TrigonometricFunction, + sin(x)*tan(x + a) + sin(x), x) is True + assert _is_function_class_equation(TrigonometricFunction, + sin(x)*tan(x*a) + sin(x), x) is True + assert _is_function_class_equation(TrigonometricFunction, + a*tan(x) - 1, x) is True + assert _is_function_class_equation(TrigonometricFunction, + tan(x)**2 + sin(x) - 1, x) is True + assert _is_function_class_equation(TrigonometricFunction, + tan(x) + x, x) is False + assert _is_function_class_equation(TrigonometricFunction, + tan(x**2), x) is False + assert _is_function_class_equation(TrigonometricFunction, + tan(x**2) + sin(x), x) is False + assert _is_function_class_equation(TrigonometricFunction, + tan(x)**sin(x), x) is False + assert _is_function_class_equation(TrigonometricFunction, + tan(sin(x)) + sin(x), x) is False + assert _is_function_class_equation(HyperbolicFunction, + tanh(x), x) is True + assert _is_function_class_equation(HyperbolicFunction, + tanh(x) - 1, x) is True + assert _is_function_class_equation(HyperbolicFunction, + tanh(x) + sinh(x), x) is True + assert _is_function_class_equation(HyperbolicFunction, + tanh(x) + sinh(x) - a, x) is True + assert _is_function_class_equation(HyperbolicFunction, + sinh(x)*tanh(x) + sinh(x), x) is True + assert _is_function_class_equation(HyperbolicFunction, + sinh(x)*tanh(x + a) + sinh(x), x) is True + assert _is_function_class_equation(HyperbolicFunction, + sinh(x)*tanh(x*a) + sinh(x), x) is True + assert _is_function_class_equation(HyperbolicFunction, + a*tanh(x) - 1, x) is True + assert _is_function_class_equation(HyperbolicFunction, + tanh(x)**2 + sinh(x) - 1, x) is True + assert _is_function_class_equation(HyperbolicFunction, + tanh(x) + x, x) is False + assert _is_function_class_equation(HyperbolicFunction, + tanh(x**2), x) is False + assert _is_function_class_equation(HyperbolicFunction, + tanh(x**2) + sinh(x), x) is False + assert _is_function_class_equation(HyperbolicFunction, + tanh(x)**sinh(x), x) is False + assert _is_function_class_equation(HyperbolicFunction, + tanh(sinh(x)) + sinh(x), x) is False + + +def test_garbage_input(): + raises(ValueError, lambda: solveset_real([y], y)) + x = Symbol('x', real=True) + assert solveset_real(x, 1) == S.EmptySet + assert solveset_real(x - 1, 1) == FiniteSet(x) + assert solveset_real(x, pi) == S.EmptySet + assert solveset_real(x, x**2) == S.EmptySet + + raises(ValueError, lambda: solveset_complex([x], x)) + assert solveset_complex(x, pi) == S.EmptySet + + raises(ValueError, lambda: solveset((x, y), x)) + raises(ValueError, lambda: solveset(x + 1, S.Reals)) + raises(ValueError, lambda: solveset(x + 1, x, 2)) + + +def test_solve_mul(): + assert solveset_real((a*x + b)*(exp(x) - 3), x) == \ + Union({log(3)}, Intersection({-b/a}, S.Reals)) + anz = Symbol('anz', nonzero=True) + bb = Symbol('bb', real=True) + assert solveset_real((anz*x + bb)*(exp(x) - 3), x) == \ + FiniteSet(-bb/anz, log(3)) + assert solveset_real((2*x + 8)*(8 + exp(x)), x) == FiniteSet(S(-4)) + assert solveset_real(x/log(x), x) is S.EmptySet + + +def test_solve_invert(): + assert solveset_real(exp(x) - 3, x) == FiniteSet(log(3)) + assert solveset_real(log(x) - 3, x) == FiniteSet(exp(3)) + + assert solveset_real(3**(x + 2), x) == FiniteSet() + assert solveset_real(3**(2 - x), x) == FiniteSet() + + assert solveset_real(y - b*exp(a/x), x) == Intersection( + S.Reals, FiniteSet(a/log(y/b))) + + # issue 4504 + assert solveset_real(2**x - 10, x) == FiniteSet(1 + log(5)/log(2)) + + +def test_issue_25768(): + assert dumeq(solveset_real(sin(x) - S.Half, x), Union( + ImageSet(Lambda(n, pi*2*n + pi/6), S.Integers), + ImageSet(Lambda(n, pi*2*n + pi*5/6), S.Integers))) + n1 = solveset_real(sin(x) - 0.5, x).n(5) + n2 = solveset_real(sin(x) - S.Half, x).n(5) + # help pass despite fp differences + eq = [i.replace( + lambda x:x.is_Float, + lambda x:Rational(x).limit_denominator(1000)) for i in (n1, n2)] + assert dumeq(*eq),(n1,n2) + + +def test_errorinverses(): + assert solveset_real(erf(x) - S.Half, x) == \ + FiniteSet(erfinv(S.Half)) + assert solveset_real(erfinv(x) - 2, x) == \ + FiniteSet(erf(2)) + assert solveset_real(erfc(x) - S.One, x) == \ + FiniteSet(erfcinv(S.One)) + assert solveset_real(erfcinv(x) - 2, x) == FiniteSet(erfc(2)) + + +def test_solve_polynomial(): + x = Symbol('x', real=True) + y = Symbol('y', real=True) + assert solveset_real(3*x - 2, x) == FiniteSet(Rational(2, 3)) + + assert solveset_real(x**2 - 1, x) == FiniteSet(-S.One, S.One) + assert solveset_real(x - y**3, x) == FiniteSet(y ** 3) + + assert solveset_real(x**3 - 15*x - 4, x) == FiniteSet( + -2 + 3 ** S.Half, + S(4), + -2 - 3 ** S.Half) + + assert solveset_real(sqrt(x) - 1, x) == FiniteSet(1) + assert solveset_real(sqrt(x) - 2, x) == FiniteSet(4) + assert solveset_real(x**Rational(1, 4) - 2, x) == FiniteSet(16) + assert solveset_real(x**Rational(1, 3) - 3, x) == FiniteSet(27) + assert len(solveset_real(x**5 + x**3 + 1, x)) == 1 + assert len(solveset_real(-2*x**3 + 4*x**2 - 2*x + 6, x)) > 0 + assert solveset_real(x**6 + x**4 + I, x) is S.EmptySet + + +def test_return_root_of(): + f = x**5 - 15*x**3 - 5*x**2 + 10*x + 20 + s = list(solveset_complex(f, x)) + for root in s: + assert root.func == CRootOf + + # if one uses solve to get the roots of a polynomial that has a CRootOf + # solution, make sure that the use of nfloat during the solve process + # doesn't fail. Note: if you want numerical solutions to a polynomial + # it is *much* faster to use nroots to get them than to solve the + # equation only to get CRootOf solutions which are then numerically + # evaluated. So for eq = x**5 + 3*x + 7 do Poly(eq).nroots() rather + # than [i.n() for i in solve(eq)] to get the numerical roots of eq. + assert nfloat(list(solveset_complex(x**5 + 3*x**3 + 7, x))[0], + exponent=False) == CRootOf(x**5 + 3*x**3 + 7, 0).n() + + sol = list(solveset_complex(x**6 - 2*x + 2, x)) + assert all(isinstance(i, CRootOf) for i in sol) and len(sol) == 6 + + f = x**5 - 15*x**3 - 5*x**2 + 10*x + 20 + s = list(solveset_complex(f, x)) + for root in s: + assert root.func == CRootOf + + s = x**5 + 4*x**3 + 3*x**2 + Rational(7, 4) + assert solveset_complex(s, x) == \ + FiniteSet(*Poly(s*4, domain='ZZ').all_roots()) + + # Refer issue #7876 + eq = x*(x - 1)**2*(x + 1)*(x**6 - x + 1) + assert solveset_complex(eq, x) == \ + FiniteSet(-1, 0, 1, CRootOf(x**6 - x + 1, 0), + CRootOf(x**6 - x + 1, 1), + CRootOf(x**6 - x + 1, 2), + CRootOf(x**6 - x + 1, 3), + CRootOf(x**6 - x + 1, 4), + CRootOf(x**6 - x + 1, 5)) + + +def test_solveset_sqrt_1(): + assert solveset_real(sqrt(5*x + 6) - 2 - x, x) == \ + FiniteSet(-S.One, S(2)) + assert solveset_real(sqrt(x - 1) - x + 7, x) == FiniteSet(10) + assert solveset_real(sqrt(x - 2) - 5, x) == FiniteSet(27) + assert solveset_real(sqrt(x) - 2 - 5, x) == FiniteSet(49) + assert solveset_real(sqrt(x**3), x) == FiniteSet(0) + assert solveset_real(sqrt(x - 1), x) == FiniteSet(1) + assert solveset_real(sqrt((x-3)/x), x) == FiniteSet(3) + assert solveset_real(sqrt((x-3)/x)-Rational(1, 2), x) == \ + FiniteSet(4) + +def test_solveset_sqrt_2(): + x = Symbol('x', real=True) + y = Symbol('y', real=True) + # http://tutorial.math.lamar.edu/Classes/Alg/SolveRadicalEqns.aspx#Solve_Rad_Ex2_a + assert solveset_real(sqrt(2*x - 1) - sqrt(x - 4) - 2, x) == \ + FiniteSet(S(5), S(13)) + assert solveset_real(sqrt(x + 7) + 2 - sqrt(3 - x), x) == \ + FiniteSet(-6) + + # http://www.purplemath.com/modules/solverad.htm + assert solveset_real(sqrt(17*x - sqrt(x**2 - 5)) - 7, x) == \ + FiniteSet(3) + + eq = x + 1 - (x**4 + 4*x**3 - x)**Rational(1, 4) + assert solveset_real(eq, x) == FiniteSet(Rational(-1, 2), Rational(-1, 3)) + + eq = sqrt(2*x + 9) - sqrt(x + 1) - sqrt(x + 4) + assert solveset_real(eq, x) == FiniteSet(0) + + eq = sqrt(x + 4) + sqrt(2*x - 1) - 3*sqrt(x - 1) + assert solveset_real(eq, x) == FiniteSet(5) + + eq = sqrt(x)*sqrt(x - 7) - 12 + assert solveset_real(eq, x) == FiniteSet(16) + + eq = sqrt(x - 3) + sqrt(x) - 3 + assert solveset_real(eq, x) == FiniteSet(4) + + eq = sqrt(2*x**2 - 7) - (3 - x) + assert solveset_real(eq, x) == FiniteSet(-S(8), S(2)) + + # others + eq = sqrt(9*x**2 + 4) - (3*x + 2) + assert solveset_real(eq, x) == FiniteSet(0) + + assert solveset_real(sqrt(x - 3) - sqrt(x) - 3, x) == FiniteSet() + + eq = (2*x - 5)**Rational(1, 3) - 3 + assert solveset_real(eq, x) == FiniteSet(16) + + assert solveset_real(sqrt(x) + sqrt(sqrt(x)) - 4, x) == \ + FiniteSet((Rational(-1, 2) + sqrt(17)/2)**4) + + eq = sqrt(x) - sqrt(x - 1) + sqrt(sqrt(x)) + assert solveset_real(eq, x) == FiniteSet() + + eq = (x - 4)**2 + (sqrt(x) - 2)**4 + assert solveset_real(eq, x) == FiniteSet(-4, 4) + + eq = (sqrt(x) + sqrt(x + 1) + sqrt(1 - x) - 6*sqrt(5)/5) + ans = solveset_real(eq, x) + ra = S('''-1484/375 - 4*(-S(1)/2 + sqrt(3)*I/2)*(-12459439/52734375 + + 114*sqrt(12657)/78125)**(S(1)/3) - 172564/(140625*(-S(1)/2 + + sqrt(3)*I/2)*(-12459439/52734375 + 114*sqrt(12657)/78125)**(S(1)/3))''') + rb = Rational(4, 5) + assert all(abs(eq.subs(x, i).n()) < 1e-10 for i in (ra, rb)) and \ + len(ans) == 2 and \ + {i.n(chop=True) for i in ans} == \ + {i.n(chop=True) for i in (ra, rb)} + + assert solveset_real(sqrt(x) + x**Rational(1, 3) + + x**Rational(1, 4), x) == FiniteSet(0) + + assert solveset_real(x/sqrt(x**2 + 1), x) == FiniteSet(0) + + eq = (x - y**3)/((y**2)*sqrt(1 - y**2)) + assert solveset_real(eq, x) == FiniteSet(y**3) + + # issue 4497 + assert solveset_real(1/(5 + x)**Rational(1, 5) - 9, x) == \ + FiniteSet(Rational(-295244, 59049)) + + +@XFAIL +def test_solve_sqrt_fail(): + # this only works if we check real_root(eq.subs(x, Rational(1, 3))) + # but checksol doesn't work like that + eq = (x**3 - 3*x**2)**Rational(1, 3) + 1 - x + assert solveset_real(eq, x) == FiniteSet(Rational(1, 3)) + + +@slow +def test_solve_sqrt_3(): + R = Symbol('R') + eq = sqrt(2)*R*sqrt(1/(R + 1)) + (R + 1)*(sqrt(2)*sqrt(1/(R + 1)) - 1) + sol = solveset_complex(eq, R) + fset = [Rational(5, 3) + 4*sqrt(10)*cos(atan(3*sqrt(111)/251)/3)/3, + -sqrt(10)*cos(atan(3*sqrt(111)/251)/3)/3 + + 40*re(1/((Rational(-1, 2) - sqrt(3)*I/2)*(Rational(251, 27) + sqrt(111)*I/9)**Rational(1, 3)))/9 + + sqrt(30)*sin(atan(3*sqrt(111)/251)/3)/3 + Rational(5, 3) + + I*(-sqrt(30)*cos(atan(3*sqrt(111)/251)/3)/3 - + sqrt(10)*sin(atan(3*sqrt(111)/251)/3)/3 + + 40*im(1/((Rational(-1, 2) - sqrt(3)*I/2)*(Rational(251, 27) + sqrt(111)*I/9)**Rational(1, 3)))/9)] + cset = [40*re(1/((Rational(-1, 2) + sqrt(3)*I/2)*(Rational(251, 27) + sqrt(111)*I/9)**Rational(1, 3)))/9 - + sqrt(10)*cos(atan(3*sqrt(111)/251)/3)/3 - sqrt(30)*sin(atan(3*sqrt(111)/251)/3)/3 + + Rational(5, 3) + + I*(40*im(1/((Rational(-1, 2) + sqrt(3)*I/2)*(Rational(251, 27) + sqrt(111)*I/9)**Rational(1, 3)))/9 - + sqrt(10)*sin(atan(3*sqrt(111)/251)/3)/3 + + sqrt(30)*cos(atan(3*sqrt(111)/251)/3)/3)] + + fs = FiniteSet(*fset) + cs = ConditionSet(R, Eq(eq, 0), FiniteSet(*cset)) + assert sol == (fs - {-1}) | (cs - {-1}) + + # the number of real roots will depend on the value of m: for m=1 there are 4 + # and for m=-1 there are none. + eq = -sqrt((m - q)**2 + (-m/(2*q) + S.Half)**2) + sqrt((-m**2/2 - sqrt( + 4*m**4 - 4*m**2 + 8*m + 1)/4 - Rational(1, 4))**2 + (m**2/2 - m - sqrt( + 4*m**4 - 4*m**2 + 8*m + 1)/4 - Rational(1, 4))**2) + unsolved_object = ConditionSet(q, Eq(sqrt((m - q)**2 + (-m/(2*q) + S.Half)**2) - + sqrt((-m**2/2 - sqrt(4*m**4 - 4*m**2 + 8*m + 1)/4 - Rational(1, 4))**2 + (m**2/2 - m - + sqrt(4*m**4 - 4*m**2 + 8*m + 1)/4 - Rational(1, 4))**2), 0), S.Reals) + assert solveset_real(eq, q) == unsolved_object + + +def test_solve_polynomial_symbolic_param(): + assert solveset_complex((x**2 - 1)**2 - a, x) == \ + FiniteSet(sqrt(1 + sqrt(a)), -sqrt(1 + sqrt(a)), + sqrt(1 - sqrt(a)), -sqrt(1 - sqrt(a))) + + # issue 4507 + assert solveset_complex(y - b/(1 + a*x), x) == \ + FiniteSet((b/y - 1)/a) - FiniteSet(-1/a) + + # issue 4508 + assert solveset_complex(y - b*x/(a + x), x) == \ + FiniteSet(-a*y/(y - b)) - FiniteSet(-a) + + +def test_solve_rational(): + assert solveset_real(1/x + 1, x) == FiniteSet(-S.One) + assert solveset_real(1/exp(x) - 1, x) == FiniteSet(0) + assert solveset_real(x*(1 - 5/x), x) == FiniteSet(5) + assert solveset_real(2*x/(x + 2) - 1, x) == FiniteSet(2) + assert solveset_real((x**2/(7 - x)).diff(x), x) == \ + FiniteSet(S.Zero, S(14)) + + +def test_solveset_real_gen_is_pow(): + assert solveset_real(sqrt(1) + 1, x) is S.EmptySet + + +def test_no_sol(): + assert solveset(1 - oo*x) is S.EmptySet + assert solveset(oo*x, x) is S.EmptySet + assert solveset(oo*x - oo, x) is S.EmptySet + assert solveset_real(4, x) is S.EmptySet + assert solveset_real(exp(x), x) is S.EmptySet + assert solveset_real(x**2 + 1, x) is S.EmptySet + assert solveset_real(-3*a/sqrt(x), x) is S.EmptySet + assert solveset_real(1/x, x) is S.EmptySet + assert solveset_real(-(1 + x)/(2 + x)**2 + 1/(2 + x), x + ) is S.EmptySet + + +def test_sol_zero_real(): + assert solveset_real(0, x) == S.Reals + assert solveset(0, x, Interval(1, 2)) == Interval(1, 2) + assert solveset_real(-x**2 - 2*x + (x + 1)**2 - 1, x) == S.Reals + + +def test_no_sol_rational_extragenous(): + assert solveset_real((x/(x + 1) + 3)**(-2), x) is S.EmptySet + assert solveset_real((x - 1)/(1 + 1/(x - 1)), x) is S.EmptySet + + +def test_solve_polynomial_cv_1a(): + """ + Test for solving on equations that can be converted to + a polynomial equation using the change of variable y -> x**Rational(p, q) + """ + assert solveset_real(sqrt(x) - 1, x) == FiniteSet(1) + assert solveset_real(sqrt(x) - 2, x) == FiniteSet(4) + assert solveset_real(x**Rational(1, 4) - 2, x) == FiniteSet(16) + assert solveset_real(x**Rational(1, 3) - 3, x) == FiniteSet(27) + assert solveset_real(x*(x**(S.One / 3) - 3), x) == \ + FiniteSet(S.Zero, S(27)) + + +def test_solveset_real_rational(): + """Test solveset_real for rational functions""" + x = Symbol('x', real=True) + y = Symbol('y', real=True) + assert solveset_real((x - y**3) / ((y**2)*sqrt(1 - y**2)), x) \ + == FiniteSet(y**3) + # issue 4486 + assert solveset_real(2*x/(x + 2) - 1, x) == FiniteSet(2) + + +def test_solveset_real_log(): + assert solveset_real(log((x-1)*(x+1)), x) == \ + FiniteSet(sqrt(2), -sqrt(2)) + + +def test_poly_gens(): + assert solveset_real(4**(2*(x**2) + 2*x) - 8, x) == \ + FiniteSet(Rational(-3, 2), S.Half) + + +def test_solve_abs(): + n = Dummy('n') + raises(ValueError, lambda: solveset(Abs(x) - 1, x)) + assert solveset(Abs(x) - n, x, S.Reals).dummy_eq( + ConditionSet(x, Contains(n, Interval(0, oo)), {-n, n})) + assert solveset_real(Abs(x) - 2, x) == FiniteSet(-2, 2) + assert solveset_real(Abs(x) + 2, x) is S.EmptySet + assert solveset_real(Abs(x + 3) - 2*Abs(x - 3), x) == \ + FiniteSet(1, 9) + assert solveset_real(2*Abs(x) - Abs(x - 1), x) == \ + FiniteSet(-1, Rational(1, 3)) + + sol = ConditionSet( + x, + And( + Contains(b, Interval(0, oo)), + Contains(a + b, Interval(0, oo)), + Contains(a - b, Interval(0, oo))), + FiniteSet(-a - b - 3, -a + b - 3, a - b - 3, a + b - 3)) + eq = Abs(Abs(x + 3) - a) - b + assert invert_real(eq, 0, x)[1] == sol + reps = {a: 3, b: 1} + eqab = eq.subs(reps) + for si in sol.subs(reps): + assert not eqab.subs(x, si) + assert dumeq(solveset(Eq(sin(Abs(x)), 1), x, domain=S.Reals), Union( + Intersection(Interval(0, oo), Union( + Intersection(ImageSet(Lambda(n, 2*n*pi + 3*pi/2), S.Integers), + Interval(-oo, 0)), + Intersection(ImageSet(Lambda(n, 2*n*pi + pi/2), S.Integers), + Interval(0, oo)))))) + + +def test_issue_9824(): + assert dumeq(solveset(sin(x)**2 - 2*sin(x) + 1, x), ImageSet(Lambda(n, 2*n*pi + pi/2), S.Integers)) + assert dumeq(solveset(cos(x)**2 - 2*cos(x) + 1, x), ImageSet(Lambda(n, 2*n*pi), S.Integers)) + + +def test_issue_9565(): + assert solveset_real(Abs((x - 1)/(x - 5)) <= Rational(1, 3), x) == Interval(-1, 2) + + +def test_issue_10069(): + eq = abs(1/(x - 1)) - 1 > 0 + assert solveset_real(eq, x) == Union( + Interval.open(0, 1), Interval.open(1, 2)) + + +def test_real_imag_splitting(): + a, b = symbols('a b', real=True) + assert solveset_real(sqrt(a**2 - b**2) - 3, a) == \ + FiniteSet(-sqrt(b**2 + 9), sqrt(b**2 + 9)) + assert solveset_real(sqrt(a**2 + b**2) - 3, a) != \ + S.EmptySet + + +def test_units(): + assert solveset_real(1/x - 1/(2*cm), x) == FiniteSet(2*cm) + + +def test_solve_only_exp_1(): + y = Symbol('y', positive=True) + assert solveset_real(exp(x) - y, x) == FiniteSet(log(y)) + assert solveset_real(exp(x) + exp(-x) - 4, x) == \ + FiniteSet(log(-sqrt(3) + 2), log(sqrt(3) + 2)) + assert solveset_real(exp(x) + exp(-x) - y, x) != S.EmptySet + + +def test_atan2(): + # The .inverse() method on atan2 works only if x.is_real is True and the + # second argument is a real constant + assert solveset_real(atan2(x, 2) - pi/3, x) == FiniteSet(2*sqrt(3)) + + +def test_piecewise_solveset(): + eq = Piecewise((x - 2, Gt(x, 2)), (2 - x, True)) - 3 + assert set(solveset_real(eq, x)) == set(FiniteSet(-1, 5)) + + absxm3 = Piecewise( + (x - 3, 0 <= x - 3), + (3 - x, 0 > x - 3)) + y = Symbol('y', positive=True) + assert solveset_real(absxm3 - y, x) == FiniteSet(-y + 3, y + 3) + + f = Piecewise(((x - 2)**2, x >= 0), (0, True)) + assert solveset(f, x, domain=S.Reals) == Union(FiniteSet(2), Interval(-oo, 0, True, True)) + + assert solveset( + Piecewise((x + 1, x > 0), (I, True)) - I, x, S.Reals + ) == Interval(-oo, 0) + + assert solveset(Piecewise((x - 1, Ne(x, I)), (x, True)), x) == FiniteSet(1) + + # issue 19718 + g = Piecewise((1, x > 10), (0, True)) + assert solveset(g > 0, x, S.Reals) == Interval.open(10, oo) + + from sympy.logic.boolalg import BooleanTrue + f = BooleanTrue() + assert solveset(f, x, domain=Interval(-3, 10)) == Interval(-3, 10) + + # issue 20552 + f = Piecewise((0, Eq(x, 0)), (x**2/Abs(x), True)) + g = Piecewise((0, Eq(x, pi)), ((x - pi)/sin(x), True)) + assert solveset(f, x, domain=S.Reals) == FiniteSet(0) + assert solveset(g) == FiniteSet(pi) + + +def test_solveset_complex_polynomial(): + assert solveset_complex(a*x**2 + b*x + c, x) == \ + FiniteSet(-b/(2*a) - sqrt(-4*a*c + b**2)/(2*a), + -b/(2*a) + sqrt(-4*a*c + b**2)/(2*a)) + + assert solveset_complex(x - y**3, y) == FiniteSet( + (-x**Rational(1, 3))/2 + I*sqrt(3)*x**Rational(1, 3)/2, + x**Rational(1, 3), + (-x**Rational(1, 3))/2 - I*sqrt(3)*x**Rational(1, 3)/2) + + assert solveset_complex(x + 1/x - 1, x) == \ + FiniteSet(S.Half + I*sqrt(3)/2, S.Half - I*sqrt(3)/2) + + +def test_sol_zero_complex(): + assert solveset_complex(0, x) is S.Complexes + + +def test_solveset_complex_rational(): + assert solveset_complex((x - 1)*(x - I)/(x - 3), x) == \ + FiniteSet(1, I) + + assert solveset_complex((x - y**3)/((y**2)*sqrt(1 - y**2)), x) == \ + FiniteSet(y**3) + assert solveset_complex(-x**2 - I, x) == \ + FiniteSet(-sqrt(2)/2 + sqrt(2)*I/2, sqrt(2)/2 - sqrt(2)*I/2) + + +def test_solve_quintics(): + skip("This test is too slow") + f = x**5 - 110*x**3 - 55*x**2 + 2310*x + 979 + s = solveset_complex(f, x) + for root in s: + res = f.subs(x, root.n()).n() + assert tn(res, 0) + + f = x**5 + 15*x + 12 + s = solveset_complex(f, x) + for root in s: + res = f.subs(x, root.n()).n() + assert tn(res, 0) + + +def test_solveset_complex_exp(): + assert dumeq(solveset_complex(exp(x) - 1, x), + imageset(Lambda(n, I*2*n*pi), S.Integers)) + assert dumeq(solveset_complex(exp(x) - I, x), + imageset(Lambda(n, I*(2*n*pi + pi/2)), S.Integers)) + assert solveset_complex(1/exp(x), x) == S.EmptySet + assert dumeq(solveset_complex(sinh(x).rewrite(exp), x), + imageset(Lambda(n, n*pi*I), S.Integers)) + + +def test_solveset_real_exp(): + assert solveset(Eq((-2)**x, 4), x, S.Reals) == FiniteSet(2) + assert solveset(Eq(-2**x, 4), x, S.Reals) == S.EmptySet + assert solveset(Eq((-3)**x, 27), x, S.Reals) == S.EmptySet + assert solveset(Eq((-5)**(x+1), 625), x, S.Reals) == FiniteSet(3) + assert solveset(Eq(2**(x-3), -16), x, S.Reals) == S.EmptySet + assert solveset(Eq((-3)**(x - 3), -3**39), x, S.Reals) == FiniteSet(42) + assert solveset(Eq(2**x, y), x, S.Reals) == Intersection(S.Reals, FiniteSet(log(y)/log(2))) + + assert invert_real((-2)**(2*x) - 16, 0, x) == (x, FiniteSet(2)) + + +def test_solve_complex_log(): + assert solveset_complex(log(x), x) == FiniteSet(1) + assert solveset_complex(1 - log(a + 4*x**2), x) == \ + FiniteSet(-sqrt(-a + E)/2, sqrt(-a + E)/2) + + +def test_solve_complex_sqrt(): + assert solveset_complex(sqrt(5*x + 6) - 2 - x, x) == \ + FiniteSet(-S.One, S(2)) + assert solveset_complex(sqrt(5*x + 6) - (2 + 2*I) - x, x) == \ + FiniteSet(-S(2), 3 - 4*I) + assert solveset_complex(4*x*(1 - a * sqrt(x)), x) == \ + FiniteSet(S.Zero, 1 / a ** 2) + + +def test_solveset_complex_tan(): + s = solveset_complex(tan(x).rewrite(exp), x) + assert dumeq(s, imageset(Lambda(n, pi*n), S.Integers) - \ + imageset(Lambda(n, pi*n + pi/2), S.Integers)) + + +@_both_exp_pow +def test_solve_trig(): + assert dumeq(solveset_real(sin(x), x), + Union(imageset(Lambda(n, 2*pi*n), S.Integers), + imageset(Lambda(n, 2*pi*n + pi), S.Integers))) + + assert dumeq(solveset_real(sin(x) - 1, x), + imageset(Lambda(n, 2*pi*n + pi/2), S.Integers)) + + assert dumeq(solveset_real(cos(x), x), + Union(imageset(Lambda(n, 2*pi*n + pi/2), S.Integers), + imageset(Lambda(n, 2*pi*n + pi*Rational(3, 2)), S.Integers))) + + assert dumeq(solveset_real(sin(x) + cos(x), x), + Union(imageset(Lambda(n, 2*n*pi + pi*Rational(3, 4)), S.Integers), + imageset(Lambda(n, 2*n*pi + pi*Rational(7, 4)), S.Integers))) + + assert solveset_real(sin(x)**2 + cos(x)**2, x) == S.EmptySet + + assert dumeq(solveset_complex(cos(x) - S.Half, x), + Union(imageset(Lambda(n, 2*n*pi + pi*Rational(5, 3)), S.Integers), + imageset(Lambda(n, 2*n*pi + pi/3), S.Integers))) + + assert dumeq(solveset(sin(y + a) - sin(y), a, domain=S.Reals), + ConditionSet(a, (S(-1) <= sin(y)) & (sin(y) <= S(1)), Union( + ImageSet(Lambda(n, 2*n*pi - y + asin(sin(y))), S.Integers), + ImageSet(Lambda(n, 2*n*pi - y - asin(sin(y)) + pi), S.Integers)))) + + assert dumeq(solveset_real(sin(2*x)*cos(x) + cos(2*x)*sin(x)-1, x), + ImageSet(Lambda(n, n*pi*Rational(2, 3) + pi/6), S.Integers)) + + assert dumeq(solveset_real(2*tan(x)*sin(x) + 1, x), Union( + ImageSet(Lambda(n, 2*n*pi + atan(sqrt(2)*sqrt(-1 + sqrt(17))/ + (1 - sqrt(17))) + pi), S.Integers), + ImageSet(Lambda(n, 2*n*pi - atan(sqrt(2)*sqrt(-1 + sqrt(17))/ + (1 - sqrt(17))) + pi), S.Integers))) + + assert dumeq(solveset_real(cos(2*x)*cos(4*x) - 1, x), + ImageSet(Lambda(n, n*pi), S.Integers)) + + assert dumeq(solveset(sin(x/10) + Rational(3, 4)), Union( + ImageSet(Lambda(n, 20*n*pi - 10*asin(S(3)/4) + 20*pi), S.Integers), + ImageSet(Lambda(n, 20*n*pi + 10*asin(S(3)/4) + 10*pi), S.Integers))) + + assert dumeq(solveset(cos(x/15) + cos(x/5)), Union( + ImageSet(Lambda(n, 30*n*pi + 15*pi/2), S.Integers), + ImageSet(Lambda(n, 30*n*pi + 45*pi/2), S.Integers), + ImageSet(Lambda(n, 30*n*pi + 75*pi/4), S.Integers), + ImageSet(Lambda(n, 30*n*pi + 45*pi/4), S.Integers), + ImageSet(Lambda(n, 30*n*pi + 105*pi/4), S.Integers), + ImageSet(Lambda(n, 30*n*pi + 15*pi/4), S.Integers))) + + assert dumeq(solveset(sec(sqrt(2)*x/3) + 5), Union( + ImageSet(Lambda(n, 3*sqrt(2)*(2*n*pi - asec(-5))/2), S.Integers), + ImageSet(Lambda(n, 3*sqrt(2)*(2*n*pi + asec(-5))/2), S.Integers))) + + assert dumeq(simplify(solveset(tan(pi*x) - cot(pi/2*x))), Union( + ImageSet(Lambda(n, 4*n + 1), S.Integers), + ImageSet(Lambda(n, 4*n + 3), S.Integers), + ImageSet(Lambda(n, 4*n + Rational(7, 3)), S.Integers), + ImageSet(Lambda(n, 4*n + Rational(5, 3)), S.Integers), + ImageSet(Lambda(n, 4*n + Rational(11, 3)), S.Integers), + ImageSet(Lambda(n, 4*n + Rational(1, 3)), S.Integers))) + + assert dumeq(solveset(cos(9*x)), Union( + ImageSet(Lambda(n, 2*n*pi/9 + pi/18), S.Integers), + ImageSet(Lambda(n, 2*n*pi/9 + pi/6), S.Integers))) + + assert dumeq(solveset(sin(8*x) + cot(12*x), x, S.Reals), Union( + ImageSet(Lambda(n, n*pi/2 + pi/8), S.Integers), + ImageSet(Lambda(n, n*pi/2 + 3*pi/8), S.Integers), + ImageSet(Lambda(n, n*pi/2 + 5*pi/16), S.Integers), + ImageSet(Lambda(n, n*pi/2 + 3*pi/16), S.Integers), + ImageSet(Lambda(n, n*pi/2 + 7*pi/16), S.Integers), + ImageSet(Lambda(n, n*pi/2 + pi/16), S.Integers))) + + # This is the only remaining solveset test that actually ends up being solved + # by _solve_trig2(). All others are handled by the improved _solve_trig1. + assert dumeq(solveset_real(2*cos(x)*cos(2*x) - 1, x), + Union(ImageSet(Lambda(n, 2*n*pi + 2*atan(sqrt(-2*2**Rational(1, 3)*(67 + + 9*sqrt(57))**Rational(2, 3) + 8*2**Rational(2, 3) + 11*(67 + + 9*sqrt(57))**Rational(1, 3))/(3*(67 + 9*sqrt(57))**Rational(1, 6)))), S.Integers), + ImageSet(Lambda(n, 2*n*pi - 2*atan(sqrt(-2*2**Rational(1, 3)*(67 + + 9*sqrt(57))**Rational(2, 3) + 8*2**Rational(2, 3) + 11*(67 + + 9*sqrt(57))**Rational(1, 3))/(3*(67 + 9*sqrt(57))**Rational(1, 6))) + + 2*pi), S.Integers))) + + # issue #16870 + assert dumeq(simplify(solveset(sin(x/180*pi) - S.Half, x, S.Reals)), Union( + ImageSet(Lambda(n, 360*n + 150), S.Integers), + ImageSet(Lambda(n, 360*n + 30), S.Integers))) + + +def test_solve_trig_hyp_by_inversion(): + n = Dummy('n') + assert solveset_real(sin(2*x + 3) - S(1)/2, x).dummy_eq(Union( + ImageSet(Lambda(n, n*pi - S(3)/2 + 13*pi/12), S.Integers), + ImageSet(Lambda(n, n*pi - S(3)/2 + 17*pi/12), S.Integers))) + assert solveset_complex(sin(2*x + 3) - S(1)/2, x).dummy_eq(Union( + ImageSet(Lambda(n, n*pi - S(3)/2 + 13*pi/12), S.Integers), + ImageSet(Lambda(n, n*pi - S(3)/2 + 17*pi/12), S.Integers))) + assert solveset_real(tan(x) - tan(pi/10), x).dummy_eq( + ImageSet(Lambda(n, n*pi + pi/10), S.Integers)) + assert solveset_complex(tan(x) - tan(pi/10), x).dummy_eq( + ImageSet(Lambda(n, n*pi + pi/10), S.Integers)) + + assert solveset_real(3*cosh(2*x) - 5, x) == FiniteSet( + -acosh(S(5)/3)/2, acosh(S(5)/3)/2) + assert solveset_complex(3*cosh(2*x) - 5, x).dummy_eq(Union( + ImageSet(Lambda(n, n*I*pi - acosh(S(5)/3)/2), S.Integers), + ImageSet(Lambda(n, n*I*pi + acosh(S(5)/3)/2), S.Integers))) + assert solveset_real(sinh(x - 3) - 2, x) == FiniteSet( + asinh(2) + 3) + assert solveset_complex(sinh(x - 3) - 2, x).dummy_eq(Union( + ImageSet(Lambda(n, 2*n*I*pi + asinh(2) + 3), S.Integers), + ImageSet(Lambda(n, 2*n*I*pi - asinh(2) + 3 + I*pi), S.Integers))) + + assert solveset_real(cos(sinh(x))-cos(pi/12), x).dummy_eq(Union( + ImageSet(Lambda(n, asinh(2*n*pi + pi/12)), S.Integers), + ImageSet(Lambda(n, asinh(2*n*pi + 23*pi/12)), S.Integers))) + assert solveset(cos(sinh(x))-cos(pi/12), x, Interval(2,3)) == \ + FiniteSet(asinh(23*pi/12), asinh(25*pi/12)) + assert solveset_real(cosh(x**2-1)-2, x) == FiniteSet( + -sqrt(1 + acosh(2)), sqrt(1 + acosh(2))) + + assert solveset_real(sin(x) - 2, x) == S.EmptySet # issue #17334 + assert solveset_real(cos(x) + 2, x) == S.EmptySet + assert solveset_real(sec(x), x) == S.EmptySet + assert solveset_real(csc(x), x) == S.EmptySet + assert solveset_real(cosh(x) + 1, x) == S.EmptySet + assert solveset_real(coth(x), x) == S.EmptySet + assert solveset_real(sech(x) - 2, x) == S.EmptySet + assert solveset_real(sech(x), x) == S.EmptySet + assert solveset_real(tanh(x) + 1, x) == S.EmptySet + assert solveset_complex(tanh(x), 1) == S.EmptySet + assert solveset_complex(coth(x), -1) == S.EmptySet + assert solveset_complex(sech(x), 0) == S.EmptySet + assert solveset_complex(csch(x), 0) == S.EmptySet + + assert solveset_real(abs(csch(x)) - 3, x) == FiniteSet(-acsch(3), acsch(3)) + + assert solveset_real(tanh(x**2 - 1) - exp(-9), x) == FiniteSet( + -sqrt(atanh(exp(-9)) + 1), sqrt(atanh(exp(-9)) + 1)) + + assert solveset_real(coth(log(x)) + 2, x) == FiniteSet(exp(-acoth(2))) + assert solveset_real(coth(exp(x)) + 2, x) == S.EmptySet + + assert solveset_complex(sinh(x) - I/2, x).dummy_eq(Union( + ImageSet(Lambda(n, 2*I*pi*n + 5*I*pi/6), S.Integers), + ImageSet(Lambda(n, 2*I*pi*n + I*pi/6), S.Integers))) + assert solveset_complex(sinh(x/10) + Rational(3, 4), x).dummy_eq(Union( + ImageSet(Lambda(n, 20*n*I*pi - 10*asinh(S(3)/4)), S.Integers), + ImageSet(Lambda(n, 20*n*I*pi + 10*asinh(S(3)/4) + 10*I*pi), S.Integers))) + assert solveset_complex(sech(sqrt(2)*x/3) + 5, x).dummy_eq(Union( + ImageSet(Lambda(n, 3*sqrt(2)*(2*n*I*pi - asech(-5))/2), S.Integers), + ImageSet(Lambda(n, 3*sqrt(2)*(2*n*I*pi + asech(-5))/2), S.Integers))) + assert solveset_complex(cosh(9*x), x).dummy_eq(Union( + ImageSet(Lambda(n, 2*n*I*pi/9 + I*pi/18), S.Integers), + ImageSet(Lambda(n, 2*n*I*pi/9 + I*pi/6), S.Integers))) + + eq = (x**5 -4*x + 1).subs(x, coth(z)) + assert solveset(eq, z, S.Complexes).dummy_eq(Union( + ImageSet(Lambda(n, n*I*pi + acoth(CRootOf(x**5 -4*x + 1, 0))), S.Integers), + ImageSet(Lambda(n, n*I*pi + acoth(CRootOf(x**5 -4*x + 1, 1))), S.Integers), + ImageSet(Lambda(n, n*I*pi + acoth(CRootOf(x**5 -4*x + 1, 2))), S.Integers), + ImageSet(Lambda(n, n*I*pi + acoth(CRootOf(x**5 -4*x + 1, 3))), S.Integers), + ImageSet(Lambda(n, n*I*pi + acoth(CRootOf(x**5 -4*x + 1, 4))), S.Integers))) + assert solveset(eq, z, S.Reals) == FiniteSet( + acoth(CRootOf(x**5 - 4*x + 1, 0)), acoth(CRootOf(x**5 - 4*x + 1, 2))) + + eq = ((x-sqrt(3)/2)*(x+2)).expand().subs(x, cos(x)) + assert solveset(eq, x, S.Complexes).dummy_eq(Union( + ImageSet(Lambda(n, 2*n*pi - acos(-2)), S.Integers), + ImageSet(Lambda(n, 2*n*pi + acos(-2)), S.Integers), + ImageSet(Lambda(n, 2*n*pi + pi/6), S.Integers), + ImageSet(Lambda(n, 2*n*pi + 11*pi/6), S.Integers))) + assert solveset(eq, x, S.Reals).dummy_eq(Union( + ImageSet(Lambda(n, 2*n*pi + pi/6), S.Integers), + ImageSet(Lambda(n, 2*n*pi + 11*pi/6), S.Integers))) + + assert solveset((1+sec(sqrt(3)*x+4)**2)/(1-sec(sqrt(3)*x+4))).dummy_eq(Union( + ImageSet(Lambda(n, sqrt(3)*(2*n*pi - 4 - asec(I))/3), S.Integers), + ImageSet(Lambda(n, sqrt(3)*(2*n*pi - 4 + asec(I))/3), S.Integers), + ImageSet(Lambda(n, sqrt(3)*(2*n*pi - 4 - asec(-I))/3), S.Integers), + ImageSet(Lambda(n, sqrt(3)*(2*n*pi - 4 + asec(-I))/3), S.Integers))) + + assert all_close(solveset(tan(3.14*x)**(S(3)/2)-5.678, x, Interval(0, 3)), + FiniteSet(0.403301114561067, 0.403301114561067 + 0.318471337579618*pi, + 0.403301114561067 + 0.636942675159236*pi)) + + +def test_old_trig_issues(): + # issues #9606 / #9531: + assert solveset(sinh(x), x, S.Reals) == FiniteSet(0) + assert solveset(sinh(x), x, S.Complexes).dummy_eq(Union( + ImageSet(Lambda(n, 2*n*I*pi), S.Integers), + ImageSet(Lambda(n, 2*n*I*pi + I*pi), S.Integers))) + + # issues #11218 / #18427 + assert solveset(sin(pi*x), x, S.Reals).dummy_eq(Union( + ImageSet(Lambda(n, (2*n*pi + pi)/pi), S.Integers), + ImageSet(Lambda(n, 2*n), S.Integers))) + assert solveset(sin(pi*x), x).dummy_eq(Union( + ImageSet(Lambda(n, (2*n*pi + pi)/pi), S.Integers), + ImageSet(Lambda(n, 2*n), S.Integers))) + + # issue #17543 + assert solveset(I*cot(8*x - 8*E), x).dummy_eq( + ImageSet(Lambda(n, pi*n/8 - 13*pi/16 + E), S.Integers)) + + # issue #20798 + assert all_close(solveset(cos(2*x) - 0.5, x, Interval(0, 2*pi)), FiniteSet( + 0.523598775598299, -0.523598775598299 + pi, + -0.523598775598299 + 2*pi, 0.523598775598299 + pi)) + sol = Union(ImageSet(Lambda(n, n*pi - 0.523598775598299), S.Integers), + ImageSet(Lambda(n, n*pi + 0.523598775598299), S.Integers)) + ret = solveset(cos(2*x) - 0.5, x, S.Reals) + # replace Dummy n by the regular Symbol n to allow all_close comparison. + ret = ret.subs(ret.atoms(Dummy).pop(), n) + assert all_close(ret, sol) + ret = solveset(cos(2*x) - 0.5, x, S.Complexes) + ret = ret.subs(ret.atoms(Dummy).pop(), n) + assert all_close(ret, sol) + + # issue #21296 / #17667 + assert solveset(tan(x)-sqrt(2), x, Interval(0, pi/2)) == FiniteSet(atan(sqrt(2))) + assert solveset(tan(x)-pi, x, Interval(0, pi/2)) == FiniteSet(atan(pi)) + + # issue #17667 + # not yet working properly: + # solveset(cos(x)-y, x, Interval(0, pi)) + assert solveset(cos(x)-y, x, S.Reals).dummy_eq( + ConditionSet(x,(S(-1) <= y) & (y <= S(1)), Union( + ImageSet(Lambda(n, 2*n*pi - acos(y)), S.Integers), + ImageSet(Lambda(n, 2*n*pi + acos(y)), S.Integers)))) + + # issue #17579 + # Valid result, but the intersection could potentially be simplified. + assert solveset(sin(log(x)), x, Interval(0,1, True, False)).dummy_eq( + Union(Intersection(ImageSet(Lambda(n, exp(2*n*pi)), S.Integers), Interval.Lopen(0, 1)), + Intersection(ImageSet(Lambda(n, exp(2*n*pi + pi)), S.Integers), Interval.Lopen(0, 1)))) + + # issue #17334 + assert solveset(sin(x) - sin(1), x, S.Reals).dummy_eq(Union( + ImageSet(Lambda(n, 2*n*pi + 1), S.Integers), + ImageSet(Lambda(n, 2*n*pi - 1 + pi), S.Integers))) + assert solveset(sin(x) - sqrt(5)/3, x, S.Reals).dummy_eq(Union( + ImageSet(Lambda(n, 2*n*pi + asin(sqrt(5)/3)), S.Integers), + ImageSet(Lambda(n, 2*n*pi - asin(sqrt(5)/3) + pi), S.Integers))) + assert solveset(sinh(x)-cosh(2), x, S.Reals) == FiniteSet(asinh(cosh(2))) + + # issue 9825 + assert solveset(Eq(tan(x), y), x, domain=S.Reals).dummy_eq( + ConditionSet(x, (-oo < y) & (y < oo), + ImageSet(Lambda(n, n*pi + atan(y)), S.Integers))) + r = Symbol('r', real=True) + assert solveset(Eq(tan(x), r), x, domain=S.Reals).dummy_eq( + ImageSet(Lambda(n, n*pi + atan(r)), S.Integers)) + + +def test_solve_hyperbolic(): + # actual solver: _solve_trig1 + n = Dummy('n') + assert solveset(sinh(x) + cosh(x), x) == S.EmptySet + assert solveset(sinh(x) + cos(x), x) == ConditionSet(x, + Eq(cos(x) + sinh(x), 0), S.Complexes) + assert solveset_real(sinh(x) + sech(x), x) == FiniteSet( + log(sqrt(sqrt(5) - 2))) + assert solveset_real(cosh(2*x) + 2*sinh(x) - 5, x) == FiniteSet( + log(-2 + sqrt(5)), log(1 + sqrt(2))) + assert solveset_real((coth(x) + sinh(2*x))/cosh(x) - 3, x) == FiniteSet( + log(S.Half + sqrt(5)/2), log(1 + sqrt(2))) + assert solveset_real(cosh(x)*sinh(x) - 2, x) == FiniteSet( + log(4 + sqrt(17))/2) + assert solveset_real(sinh(x) + tanh(x) - 1, x) == FiniteSet( + log(sqrt(2)/2 + sqrt(-S(1)/2 + sqrt(2)))) + + assert dumeq(solveset_complex(sinh(x) + sech(x), x), Union( + ImageSet(Lambda(n, 2*n*I*pi + log(sqrt(-2 + sqrt(5)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi + pi/2) + log(sqrt(2 + sqrt(5)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi + pi) + log(sqrt(-2 + sqrt(5)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi - pi/2) + log(sqrt(2 + sqrt(5)))), S.Integers))) + + assert dumeq(solveset(cosh(x/15) + cosh(x/5)), Union( + ImageSet(Lambda(n, 15*I*(2*n*pi + pi/2)), S.Integers), + ImageSet(Lambda(n, 15*I*(2*n*pi - pi/2)), S.Integers), + ImageSet(Lambda(n, 15*I*(2*n*pi - 3*pi/4)), S.Integers), + ImageSet(Lambda(n, 15*I*(2*n*pi + 3*pi/4)), S.Integers), + ImageSet(Lambda(n, 15*I*(2*n*pi - pi/4)), S.Integers), + ImageSet(Lambda(n, 15*I*(2*n*pi + pi/4)), S.Integers))) + + assert dumeq(solveset(tanh(pi*x) - coth(pi/2*x)), Union( + ImageSet(Lambda(n, 2*I*(2*n*pi + pi/2)/pi), S.Integers), + ImageSet(Lambda(n, 2*I*(2*n*pi - pi/2)/pi), S.Integers))) + + # issues #18490 / #19489 + assert solveset(cosh(x) + cosh(3*x) - cosh(5*x), x, S.Reals + ).dummy_eq(ConditionSet(x, + Eq(cosh(x) + cosh(3*x) - cosh(5*x), 0), S.Reals)) + assert solveset(sinh(8*x) + coth(12*x)).dummy_eq( + ConditionSet(x, Eq(sinh(8*x) + coth(12*x), 0), S.Complexes)) + + +def test_solve_trig_hyp_symbolic(): + # actual solver: invert_trig_hyp + assert dumeq(solveset(sin(a*x), x), ConditionSet(x, Ne(a, 0), Union( + ImageSet(Lambda(n, (2*n*pi + pi)/a), S.Integers), + ImageSet(Lambda(n, 2*n*pi/a), S.Integers)))) + + assert dumeq(solveset(cosh(x/a), x), ConditionSet(x, Ne(a, 0), Union( + ImageSet(Lambda(n, a*(2*n*I*pi + I*pi/2)), S.Integers), + ImageSet(Lambda(n, a*(2*n*I*pi + 3*I*pi/2)), S.Integers)))) + + assert dumeq(solveset(sin(2*sqrt(3)/3*a**2/(b*pi)*x) + + cos(4*sqrt(3)/3*a**2/(b*pi)*x), x), + ConditionSet(x, Ne(b, 0) & Ne(a**2, 0), Union( + ImageSet(Lambda(n, sqrt(3)*pi*b*(2*n*pi + pi/2)/(2*a**2)), S.Integers), + ImageSet(Lambda(n, sqrt(3)*pi*b*(2*n*pi - 5*pi/6)/(2*a**2)), S.Integers), + ImageSet(Lambda(n, sqrt(3)*pi*b*(2*n*pi - pi/6)/(2*a**2)), S.Integers)))) + + assert dumeq(solveset(cosh((a**2 + 1)*x) - 3, x), ConditionSet( + x, Ne(a**2 + 1, 0), Union( + ImageSet(Lambda(n, (2*n*I*pi - acosh(3))/(a**2 + 1)), S.Integers), + ImageSet(Lambda(n, (2*n*I*pi + acosh(3))/(a**2 + 1)), S.Integers)))) + + ar = Symbol('ar', real=True) + assert solveset(cosh((ar**2 + 1)*x) - 2, x, S.Reals) == FiniteSet( + -acosh(2)/(ar**2 + 1), acosh(2)/(ar**2 + 1)) + + # actual solver: _solve_trig1 + assert dumeq(simplify(solveset(cot((1 + I)*x) - cot((3 + 3*I)*x), x)), Union( + ImageSet(Lambda(n, pi*(1 - I)*(4*n + 1)/4), S.Integers), + ImageSet(Lambda(n, pi*(1 - I)*(4*n - 1)/4), S.Integers))) + + +def test_issue_9616(): + assert dumeq(solveset(sinh(x) + tanh(x) - 1, x), Union( + ImageSet(Lambda(n, 2*n*I*pi + log(sqrt(2)/2 + sqrt(-S.Half + sqrt(2)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi - atan(sqrt(2)*sqrt(S.Half + sqrt(2))) + pi) + + log(sqrt(1 + sqrt(2)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi + pi) + log(-sqrt(2)/2 + sqrt(-S.Half + sqrt(2)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi - pi + atan(sqrt(2)*sqrt(S.Half + sqrt(2)))) + + log(sqrt(1 + sqrt(2)))), S.Integers))) + f1 = (sinh(x)).rewrite(exp) + f2 = (tanh(x)).rewrite(exp) + assert dumeq(solveset(f1 + f2 - 1, x), Union( + Complement(ImageSet( + Lambda(n, I*(2*n*pi + pi) + log(-sqrt(2)/2 + sqrt(-S.Half + sqrt(2)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi + pi)/2), S.Integers)), + Complement(ImageSet(Lambda(n, I*(2*n*pi - pi + atan(sqrt(2)*sqrt(S.Half + sqrt(2)))) + + log(sqrt(1 + sqrt(2)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi + pi)/2), S.Integers)), + Complement(ImageSet(Lambda(n, I*(2*n*pi - atan(sqrt(2)*sqrt(S.Half + sqrt(2))) + pi) + + log(sqrt(1 + sqrt(2)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi + pi)/2), S.Integers)), + Complement( + ImageSet(Lambda(n, 2*n*I*pi + log(sqrt(2)/2 + sqrt(-S.Half + sqrt(2)))), S.Integers), + ImageSet(Lambda(n, I*(2*n*pi + pi)/2), S.Integers)))) + + +def test_solve_invalid_sol(): + assert 0 not in solveset_real(sin(x)/x, x) + assert 0 not in solveset_complex((exp(x) - 1)/x, x) + + +@XFAIL +def test_solve_trig_simplified(): + n = Dummy('n') + assert dumeq(solveset_real(sin(x), x), + imageset(Lambda(n, n*pi), S.Integers)) + + assert dumeq(solveset_real(cos(x), x), + imageset(Lambda(n, n*pi + pi/2), S.Integers)) + + assert dumeq(solveset_real(cos(x) + sin(x), x), + imageset(Lambda(n, n*pi - pi/4), S.Integers)) + + +@XFAIL +def test_solve_lambert(): + assert solveset_real(x*exp(x) - 1, x) == FiniteSet(LambertW(1)) + assert solveset_real(exp(x) + x, x) == FiniteSet(-LambertW(1)) + assert solveset_real(x + 2**x, x) == \ + FiniteSet(-LambertW(log(2))/log(2)) + + # issue 4739 + ans = solveset_real(3*x + 5 + 2**(-5*x + 3), x) + assert ans == FiniteSet(Rational(-5, 3) + + LambertW(-10240*2**Rational(1, 3)*log(2)/3)/(5*log(2))) + + eq = 2*(3*x + 4)**5 - 6*7**(3*x + 9) + result = solveset_real(eq, x) + ans = FiniteSet((log(2401) + + 5*LambertW(-log(7**(7*3**Rational(1, 5)/5))))/(3*log(7))/-1) + assert result == ans + assert solveset_real(eq.expand(), x) == result + + assert solveset_real(5*x - 1 + 3*exp(2 - 7*x), x) == \ + FiniteSet(Rational(1, 5) + LambertW(-21*exp(Rational(3, 5))/5)/7) + + assert solveset_real(2*x + 5 + log(3*x - 2), x) == \ + FiniteSet(Rational(2, 3) + LambertW(2*exp(Rational(-19, 3))/3)/2) + + assert solveset_real(3*x + log(4*x), x) == \ + FiniteSet(LambertW(Rational(3, 4))/3) + + assert solveset_real(x**x - 2) == FiniteSet(exp(LambertW(log(2)))) + + a = Symbol('a') + assert solveset_real(-a*x + 2*x*log(x), x) == FiniteSet(exp(a/2)) + a = Symbol('a', real=True) + assert solveset_real(a/x + exp(x/2), x) == \ + FiniteSet(2*LambertW(-a/2)) + assert solveset_real((a/x + exp(x/2)).diff(x), x) == \ + FiniteSet(4*LambertW(sqrt(2)*sqrt(a)/4)) + + # coverage test + assert solveset_real(tanh(x + 3)*tanh(x - 3) - 1, x) is S.EmptySet + + assert solveset_real((x**2 - 2*x + 1).subs(x, log(x) + 3*x), x) == \ + FiniteSet(LambertW(3*S.Exp1)/3) + assert solveset_real((x**2 - 2*x + 1).subs(x, (log(x) + 3*x)**2 - 1), x) == \ + FiniteSet(LambertW(3*exp(-sqrt(2)))/3, LambertW(3*exp(sqrt(2)))/3) + assert solveset_real((x**2 - 2*x - 2).subs(x, log(x) + 3*x), x) == \ + FiniteSet(LambertW(3*exp(1 + sqrt(3)))/3, LambertW(3*exp(-sqrt(3) + 1))/3) + assert solveset_real(x*log(x) + 3*x + 1, x) == \ + FiniteSet(exp(-3 + LambertW(-exp(3)))) + eq = (x*exp(x) - 3).subs(x, x*exp(x)) + assert solveset_real(eq, x) == \ + FiniteSet(LambertW(3*exp(-LambertW(3)))) + + assert solveset_real(3*log(a**(3*x + 5)) + a**(3*x + 5), x) == \ + FiniteSet(-((log(a**5) + LambertW(Rational(1, 3)))/(3*log(a)))) + p = symbols('p', positive=True) + assert solveset_real(3*log(p**(3*x + 5)) + p**(3*x + 5), x) == \ + FiniteSet( + log((-3**Rational(1, 3) - 3**Rational(5, 6)*I)*LambertW(Rational(1, 3))**Rational(1, 3)/(2*p**Rational(5, 3)))/log(p), + log((-3**Rational(1, 3) + 3**Rational(5, 6)*I)*LambertW(Rational(1, 3))**Rational(1, 3)/(2*p**Rational(5, 3)))/log(p), + log((3*LambertW(Rational(1, 3))/p**5)**(1/(3*log(p)))),) # checked numerically + # check collection + b = Symbol('b') + eq = 3*log(a**(3*x + 5)) + b*log(a**(3*x + 5)) + a**(3*x + 5) + assert solveset_real(eq, x) == FiniteSet( + -((log(a**5) + LambertW(1/(b + 3)))/(3*log(a)))) + + # issue 4271 + assert solveset_real((a/x + exp(x/2)).diff(x, 2), x) == FiniteSet( + 6*LambertW((-1)**Rational(1, 3)*a**Rational(1, 3)/3)) + + assert solveset_real(x**3 - 3**x, x) == \ + FiniteSet(-3/log(3)*LambertW(-log(3)/3)) + assert solveset_real(3**cos(x) - cos(x)**3) == FiniteSet( + acos(-3*LambertW(-log(3)/3)/log(3))) + + assert solveset_real(x**2 - 2**x, x) == \ + solveset_real(-x**2 + 2**x, x) + + assert solveset_real(3*log(x) - x*log(3)) == FiniteSet( + -3*LambertW(-log(3)/3)/log(3), + -3*LambertW(-log(3)/3, -1)/log(3)) + + assert solveset_real(LambertW(2*x) - y) == FiniteSet( + y*exp(y)/2) + + +@XFAIL +def test_other_lambert(): + a = Rational(6, 5) + assert solveset_real(x**a - a**x, x) == FiniteSet( + a, -a*LambertW(-log(a)/a)/log(a)) + + +@_both_exp_pow +def test_solveset(): + f = Function('f') + raises(ValueError, lambda: solveset(x + y)) + assert solveset(x, 1) == S.EmptySet + assert solveset(f(1)**2 + y + 1, f(1) + ) == FiniteSet(-sqrt(-y - 1), sqrt(-y - 1)) + assert solveset(f(1)**2 - 1, f(1), S.Reals) == FiniteSet(-1, 1) + assert solveset(f(1)**2 + 1, f(1)) == FiniteSet(-I, I) + assert solveset(x - 1, 1) == FiniteSet(x) + assert solveset(sin(x) - cos(x), sin(x)) == FiniteSet(cos(x)) + + assert solveset(0, domain=S.Reals) == S.Reals + assert solveset(1) == S.EmptySet + assert solveset(True, domain=S.Reals) == S.Reals # issue 10197 + assert solveset(False, domain=S.Reals) == S.EmptySet + + assert solveset(exp(x) - 1, domain=S.Reals) == FiniteSet(0) + assert solveset(exp(x) - 1, x, S.Reals) == FiniteSet(0) + assert solveset(Eq(exp(x), 1), x, S.Reals) == FiniteSet(0) + assert solveset(exp(x) - 1, exp(x), S.Reals) == FiniteSet(1) + A = Indexed('A', x) + assert solveset(A - 1, A, S.Reals) == FiniteSet(1) + + assert solveset(x - 1 >= 0, x, S.Reals) == Interval(1, oo) + assert solveset(exp(x) - 1 >= 0, x, S.Reals) == Interval(0, oo) + + assert dumeq(solveset(exp(x) - 1, x), imageset(Lambda(n, 2*I*pi*n), S.Integers)) + assert dumeq(solveset(Eq(exp(x), 1), x), imageset(Lambda(n, 2*I*pi*n), + S.Integers)) + # issue 13825 + assert solveset(x**2 + f(0) + 1, x) == {-sqrt(-f(0) - 1), sqrt(-f(0) - 1)} + + # issue 19977 + assert solveset(atan(log(x)) > 0, x, domain=Interval.open(0, oo)) == Interval.open(1, oo) + + +@_both_exp_pow +def test_multi_exp(): + k1, k2, k3 = symbols('k1, k2, k3') + assert dumeq(solveset(exp(exp(x)) - 5, x),\ + imageset(Lambda(((k1, n),), I*(2*k1*pi + arg(2*n*I*pi + log(5))) + log(Abs(2*n*I*pi + log(5)))),\ + ProductSet(S.Integers, S.Integers))) + assert dumeq(solveset((d*exp(exp(a*x + b)) + c), x),\ + imageset(Lambda(x, (-b + x)/a), ImageSet(Lambda(((k1, n),), \ + I*(2*k1*pi + arg(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))) + log(Abs(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d))))), \ + ProductSet(S.Integers, S.Integers)))) + + assert dumeq(solveset((d*exp(exp(exp(a*x + b))) + c), x),\ + imageset(Lambda(x, (-b + x)/a), ImageSet(Lambda(((k2, k1, n),), \ + I*(2*k2*pi + arg(I*(2*k1*pi + arg(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))) + \ + log(Abs(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))))) + log(Abs(I*(2*k1*pi + arg(I*(2*n*pi + arg(-c/d)) + \ + log(Abs(c/d)))) + log(Abs(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d))))))), \ + ProductSet(S.Integers, S.Integers, S.Integers)))) + + assert dumeq(solveset((d*exp(exp(exp(exp(a*x + b)))) + c), x),\ + ImageSet(Lambda(x, (-b + x)/a), ImageSet(Lambda(((k3, k2, k1, n),), \ + I*(2*k3*pi + arg(I*(2*k2*pi + arg(I*(2*k1*pi + arg(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))) + \ + log(Abs(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))))) + log(Abs(I*(2*k1*pi + arg(I*(2*n*pi + arg(-c/d)) + \ + log(Abs(c/d)))) + log(Abs(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))))))) + log(Abs(I*(2*k2*pi + \ + arg(I*(2*k1*pi + arg(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))) + log(Abs(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))))) + \ + log(Abs(I*(2*k1*pi + arg(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d)))) + log(Abs(I*(2*n*pi + arg(-c/d)) + log(Abs(c/d))))))))), \ + ProductSet(S.Integers, S.Integers, S.Integers, S.Integers)))) + + +def test__solveset_multi(): + from sympy.solvers.solveset import _solveset_multi + from sympy.sets import Reals + + # Basic univariate case: + assert _solveset_multi([x**2-1], [x], [S.Reals]) == FiniteSet((1,), (-1,)) + + # Linear systems of two equations + assert _solveset_multi([x+y, x+1], [x, y], [Reals, Reals]) == FiniteSet((-1, 1)) + assert _solveset_multi([x+y, x+1], [y, x], [Reals, Reals]) == FiniteSet((1, -1)) + assert _solveset_multi([x+y, x-y-1], [x, y], [Reals, Reals]) == FiniteSet((S(1)/2, -S(1)/2)) + assert _solveset_multi([x-1, y-2], [x, y], [Reals, Reals]) == FiniteSet((1, 2)) + # assert dumeq(_solveset_multi([x+y], [x, y], [Reals, Reals]), ImageSet(Lambda(x, (x, -x)), Reals)) + assert dumeq(_solveset_multi([x+y], [x, y], [Reals, Reals]), Union( + ImageSet(Lambda(((x,),), (x, -x)), ProductSet(Reals)), + ImageSet(Lambda(((y,),), (-y, y)), ProductSet(Reals)))) + assert _solveset_multi([x+y, x+y+1], [x, y], [Reals, Reals]) == S.EmptySet + assert _solveset_multi([x+y, x-y, x-1], [x, y], [Reals, Reals]) == S.EmptySet + assert _solveset_multi([x+y, x-y, x-1], [y, x], [Reals, Reals]) == S.EmptySet + + # Systems of three equations: + assert _solveset_multi([x+y+z-1, x+y-z-2, x-y-z-3], [x, y, z], [Reals, + Reals, Reals]) == FiniteSet((2, -S.Half, -S.Half)) + + # Nonlinear systems: + from sympy.abc import theta + assert _solveset_multi([x**2+y**2-2, x+y], [x, y], [Reals, Reals]) == FiniteSet((-1, 1), (1, -1)) + assert _solveset_multi([x**2-1, y], [x, y], [Reals, Reals]) == FiniteSet((1, 0), (-1, 0)) + #assert _solveset_multi([x**2-y**2], [x, y], [Reals, Reals]) == Union( + # ImageSet(Lambda(x, (x, -x)), Reals), ImageSet(Lambda(x, (x, x)), Reals)) + assert dumeq(_solveset_multi([x**2-y**2], [x, y], [Reals, Reals]), Union( + ImageSet(Lambda(((x,),), (x, -Abs(x))), ProductSet(Reals)), + ImageSet(Lambda(((x,),), (x, Abs(x))), ProductSet(Reals)), + ImageSet(Lambda(((y,),), (-Abs(y), y)), ProductSet(Reals)), + ImageSet(Lambda(((y,),), (Abs(y), y)), ProductSet(Reals)))) + assert _solveset_multi([r*cos(theta)-1, r*sin(theta)], [theta, r], + [Interval(0, pi), Interval(-1, 1)]) == FiniteSet((0, 1), (pi, -1)) + assert _solveset_multi([r*cos(theta)-1, r*sin(theta)], [r, theta], + [Interval(0, 1), Interval(0, pi)]) == FiniteSet((1, 0)) + assert _solveset_multi([r*cos(theta)-r, r*sin(theta)], [r, theta], + [Interval(0, 1), Interval(0, pi)]) == Union( + ImageSet(Lambda(((r,),), (r, 0)), + ImageSet(Lambda(r, (r,)), Interval(0, 1))), + ImageSet(Lambda(((theta,),), (0, theta)), + ImageSet(Lambda(theta, (theta,)), Interval(0, pi)))) + + +def test_conditionset(): + assert solveset(Eq(sin(x)**2 + cos(x)**2, 1), x, domain=S.Reals + ) is S.Reals + + assert solveset(Eq(x**2 + x*sin(x), 1), x, domain=S.Reals + ).dummy_eq(ConditionSet(x, Eq(x**2 + x*sin(x) - 1, 0), S.Reals)) + + assert dumeq(solveset(Eq(-I*(exp(I*x) - exp(-I*x))/2, 1), x + ), imageset(Lambda(n, 2*n*pi + pi/2), S.Integers)) + + assert solveset(x + sin(x) > 1, x, domain=S.Reals + ).dummy_eq(ConditionSet(x, x + sin(x) > 1, S.Reals)) + + assert solveset(Eq(sin(Abs(x)), x), x, domain=S.Reals + ).dummy_eq(ConditionSet(x, Eq(-x + sin(Abs(x)), 0), S.Reals)) + + assert solveset(y**x-z, x, S.Reals + ).dummy_eq(ConditionSet(x, Eq(y**x - z, 0), S.Reals)) + + +@XFAIL +def test_conditionset_equality(): + ''' Checking equality of different representations of ConditionSet''' + assert solveset(Eq(tan(x), y), x) == ConditionSet(x, Eq(tan(x), y), S.Complexes) + + +def test_solveset_domain(): + assert solveset(x**2 - x - 6, x, Interval(0, oo)) == FiniteSet(3) + assert solveset(x**2 - 1, x, Interval(0, oo)) == FiniteSet(1) + assert solveset(x**4 - 16, x, Interval(0, 10)) == FiniteSet(2) + + +def test_improve_coverage(): + solution = solveset(exp(x) + sin(x), x, S.Reals) + unsolved_object = ConditionSet(x, Eq(exp(x) + sin(x), 0), S.Reals) + assert solution.dummy_eq(unsolved_object) + + +def test_issue_9522(): + expr1 = Eq(1/(x**2 - 4) + x, 1/(x**2 - 4) + 2) + expr2 = Eq(1/x + x, 1/x) + + assert solveset(expr1, x, S.Reals) is S.EmptySet + assert solveset(expr2, x, S.Reals) is S.EmptySet + + +def test_solvify(): + assert solvify(x**2 + 10, x, S.Reals) == [] + assert solvify(x**3 + 1, x, S.Complexes) == [-1, S.Half - sqrt(3)*I/2, + S.Half + sqrt(3)*I/2] + assert solvify(log(x), x, S.Reals) == [1] + assert solvify(cos(x), x, S.Reals) == [pi/2, pi*Rational(3, 2)] + assert solvify(sin(x) + 1, x, S.Reals) == [pi*Rational(3, 2)] + raises(NotImplementedError, lambda: solvify(sin(exp(x)), x, S.Complexes)) + + +def test_solvify_piecewise(): + p1 = Piecewise((0, x < -1), (x**2, x <= 1), (log(x), True)) + p2 = Piecewise((0, x < -10), (x**2 + 5*x - 6, x >= -9)) + p3 = Piecewise((0, Eq(x, 0)), (x**2/Abs(x), True)) + p4 = Piecewise((0, Eq(x, pi)), ((x - pi)/sin(x), True)) + + # issue 21079 + assert solvify(p1, x, S.Reals) == [0] + assert solvify(p2, x, S.Reals) == [-6, 1] + assert solvify(p3, x, S.Reals) == [0] + assert solvify(p4, x, S.Reals) == [pi] + + +def test_abs_invert_solvify(): + + x = Symbol('x',positive=True) + assert solvify(sin(Abs(x)), x, S.Reals) == [0, pi] + x = Symbol('x') + assert solvify(sin(Abs(x)), x, S.Reals) is None + + +def test_linear_eq_to_matrix(): + assert linear_eq_to_matrix(0, x) == (Matrix([[0]]), Matrix([[0]])) + assert linear_eq_to_matrix(1, x) == (Matrix([[0]]), Matrix([[-1]])) + + # integer coefficients + eqns1 = [2*x + y - 2*z - 3, x - y - z, x + y + 3*z - 12] + eqns2 = [Eq(3*x + 2*y - z, 1), Eq(2*x - 2*y + 4*z, -2), -2*x + y - 2*z] + + A, B = linear_eq_to_matrix(eqns1, x, y, z) + assert A == Matrix([[2, 1, -2], [1, -1, -1], [1, 1, 3]]) + assert B == Matrix([[3], [0], [12]]) + + A, B = linear_eq_to_matrix(eqns2, x, y, z) + assert A == Matrix([[3, 2, -1], [2, -2, 4], [-2, 1, -2]]) + assert B == Matrix([[1], [-2], [0]]) + + # Pure symbolic coefficients + eqns3 = [a*b*x + b*y + c*z - d, e*x + d*x + f*y + g*z - h, i*x + j*y + k*z - l] + A, B = linear_eq_to_matrix(eqns3, x, y, z) + assert A == Matrix([[a*b, b, c], [d + e, f, g], [i, j, k]]) + assert B == Matrix([[d], [h], [l]]) + + # raise Errors if + # 1) no symbols are given + raises(ValueError, lambda: linear_eq_to_matrix(eqns3)) + # 2) there are duplicates + raises(ValueError, lambda: linear_eq_to_matrix(eqns3, [x, x, y])) + # 3) a nonlinear term is detected in the original expression + raises(NonlinearError, lambda: linear_eq_to_matrix(Eq(1/x + x, 1/x), [x])) + raises(NonlinearError, lambda: linear_eq_to_matrix([x**2], [x])) + raises(NonlinearError, lambda: linear_eq_to_matrix([x*y], [x, y])) + # 4) Eq being used to represent equations autoevaluates + # (use unevaluated Eq instead) + raises(ValueError, lambda: linear_eq_to_matrix(Eq(x, x), x)) + raises(ValueError, lambda: linear_eq_to_matrix(Eq(x, x + 1), x)) + + + # if non-symbols are passed, the user is responsible for interpreting + assert linear_eq_to_matrix([x], [1/x]) == (Matrix([[0]]), Matrix([[-x]])) + + # issue 15195 + assert linear_eq_to_matrix(x + y*(z*(3*x + 2) + 3), x) == ( + Matrix([[3*y*z + 1]]), Matrix([[-y*(2*z + 3)]])) + assert linear_eq_to_matrix(Matrix( + [[a*x + b*y - 7], [5*x + 6*y - c]]), x, y) == ( + Matrix([[a, b], [5, 6]]), Matrix([[7], [c]])) + + # issue 15312 + assert linear_eq_to_matrix(Eq(x + 2, 1), x) == ( + Matrix([[1]]), Matrix([[-1]])) + + # issue 25423 + raises(TypeError, lambda: linear_eq_to_matrix([], {x, y})) + raises(TypeError, lambda: linear_eq_to_matrix([x + y], {x, y})) + raises(ValueError, lambda: linear_eq_to_matrix({x + y}, (x, y))) + + +def test_issue_16577(): + assert linear_eq_to_matrix(Eq(a*(2*x + 3*y) + 4*y, 5), x, y) == ( + Matrix([[2*a, 3*a + 4]]), Matrix([[5]])) + + +def test_issue_10085(): + assert invert_real(exp(x),0,x) == (x, S.EmptySet) + + +def test_linsolve(): + x1, x2, x3, x4 = symbols('x1, x2, x3, x4') + + # Test for different input forms + + M = Matrix([[1, 2, 1, 1, 7], [1, 2, 2, -1, 12], [2, 4, 0, 6, 4]]) + system1 = A, B = M[:, :-1], M[:, -1] + Eqns = [x1 + 2*x2 + x3 + x4 - 7, x1 + 2*x2 + 2*x3 - x4 - 12, + 2*x1 + 4*x2 + 6*x4 - 4] + + sol = FiniteSet((-2*x2 - 3*x4 + 2, x2, 2*x4 + 5, x4)) + assert linsolve(Eqns, (x1, x2, x3, x4)) == sol + assert linsolve(Eqns, *(x1, x2, x3, x4)) == sol + assert linsolve(system1, (x1, x2, x3, x4)) == sol + assert linsolve(system1, *(x1, x2, x3, x4)) == sol + # issue 9667 - symbols can be Dummy symbols + x1, x2, x3, x4 = symbols('x:4', cls=Dummy) + assert linsolve(system1, x1, x2, x3, x4) == FiniteSet( + (-2*x2 - 3*x4 + 2, x2, 2*x4 + 5, x4)) + + # raise ValueError for garbage value + raises(ValueError, lambda: linsolve(Eqns)) + raises(ValueError, lambda: linsolve(x1)) + raises(ValueError, lambda: linsolve(x1, x2)) + raises(ValueError, lambda: linsolve((A,), x1, x2)) + raises(ValueError, lambda: linsolve(A, B, x1, x2)) + raises(ValueError, lambda: linsolve([x1], x1, x1)) + raises(ValueError, lambda: linsolve([x1], (i for i in (x1, x1)))) + + #raise ValueError if equations are non-linear in given variables + raises(NonlinearError, lambda: linsolve([x + y - 1, x ** 2 + y - 3], [x, y])) + raises(NonlinearError, lambda: linsolve([cos(x) + y, x + y], [x, y])) + assert linsolve([x + z - 1, x ** 2 + y - 3], [z, y]) == {(-x + 1, -x**2 + 3)} + + # Fully symbolic test + A = Matrix([[a, b], [c, d]]) + B = Matrix([[e], [g]]) + system2 = (A, B) + sol = FiniteSet(((-b*g + d*e)/(a*d - b*c), (a*g - c*e)/(a*d - b*c))) + assert linsolve(system2, [x, y]) == sol + + # No solution + A = Matrix([[1, 2, 3], [2, 4, 6], [3, 6, 9]]) + B = Matrix([0, 0, 1]) + assert linsolve((A, B), (x, y, z)) is S.EmptySet + + # Issue #10056 + A, B, J1, J2 = symbols('A B J1 J2') + Augmatrix = Matrix([ + [2*I*J1, 2*I*J2, -2/J1], + [-2*I*J2, -2*I*J1, 2/J2], + [0, 2, 2*I/(J1*J2)], + [2, 0, 0], + ]) + + assert linsolve(Augmatrix, A, B) == FiniteSet((0, I/(J1*J2))) + + # Issue #10121 - Assignment of free variables + Augmatrix = Matrix([[0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0]]) + assert linsolve(Augmatrix, a, b, c, d, e) == FiniteSet((a, 0, c, 0, e)) + #raises(IndexError, lambda: linsolve(Augmatrix, a, b, c)) + + x0, x1, x2, _x0 = symbols('tau0 tau1 tau2 _tau0') + assert linsolve(Matrix([[0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, _x0]]) + ) == FiniteSet((x0, 0, x1, _x0, x2)) + x0, x1, x2, _x0 = symbols('tau00 tau01 tau02 tau0') + assert linsolve(Matrix([[0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, _x0]]) + ) == FiniteSet((x0, 0, x1, _x0, x2)) + x0, x1, x2, _x0 = symbols('tau00 tau01 tau02 tau1') + assert linsolve(Matrix([[0, 1, 0, 0, 0, 0], [0, 0, 0, 1, 0, _x0]]) + ) == FiniteSet((x0, 0, x1, _x0, x2)) + # symbols can be given as generators + x0, x2, x4 = symbols('x0, x2, x4') + assert linsolve(Augmatrix, numbered_symbols('x') + ) == FiniteSet((x0, 0, x2, 0, x4)) + Augmatrix[-1, -1] = x0 + # use Dummy to avoid clash; the names may clash but the symbols + # will not + Augmatrix[-1, -1] = symbols('_x0') + assert len(linsolve( + Augmatrix, numbered_symbols('x', cls=Dummy)).free_symbols) == 4 + + # Issue #12604 + f = Function('f') + assert linsolve([f(x) - 5], f(x)) == FiniteSet((5,)) + + # Issue #14860 + from sympy.physics.units import meter, newton, kilo + kN = kilo*newton + Eqns = [8*kN + x + y, 28*kN*meter + 3*x*meter] + assert linsolve(Eqns, x, y) == { + (kilo*newton*Rational(-28, 3), kN*Rational(4, 3))} + + # linsolve does not allow expansion (real or implemented) + # to remove singularities, but it will cancel linear terms + assert linsolve([Eq(x, x + y)], [x, y]) == {(x, 0)} + assert linsolve([Eq(x + x*y, 1 + y)], [x]) == {(1,)} + assert linsolve([Eq(1 + y, x + x*y)], [x]) == {(1,)} + raises(NonlinearError, lambda: + linsolve([Eq(x**2, x**2 + y)], [x, y])) + + # corner cases + # + # XXX: The case below should give the same as for [0] + # assert linsolve([], [x]) == {(x,)} + assert linsolve([], [x]) is S.EmptySet + assert linsolve([0], [x]) == {(x,)} + assert linsolve([x], [x, y]) == {(0, y)} + assert linsolve([x, 0], [x, y]) == {(0, y)} + + +def test_linsolve_large_sparse(): + # + # This is mainly a performance test + # + + def _mk_eqs_sol(n): + xs = symbols('x:{}'.format(n)) + ys = symbols('y:{}'.format(n)) + syms = xs + ys + eqs = [] + sol = (-S.Half,) * n + (S.Half,) * n + for xi, yi in zip(xs, ys): + eqs.extend([xi + yi, xi - yi + 1]) + return eqs, syms, FiniteSet(sol) + + n = 500 + eqs, syms, sol = _mk_eqs_sol(n) + assert linsolve(eqs, syms) == sol + + +def test_linsolve_immutable(): + A = ImmutableDenseMatrix([[1, 1, 2], [0, 1, 2], [0, 0, 1]]) + B = ImmutableDenseMatrix([2, 1, -1]) + assert linsolve([A, B], (x, y, z)) == FiniteSet((1, 3, -1)) + + A = ImmutableDenseMatrix([[1, 1, 7], [1, -1, 3]]) + assert linsolve(A) == FiniteSet((5, 2)) + + +def test_solve_decomposition(): + n = Dummy('n') + + f1 = exp(3*x) - 6*exp(2*x) + 11*exp(x) - 6 + f2 = sin(x)**2 - 2*sin(x) + 1 + f3 = sin(x)**2 - sin(x) + f4 = sin(x + 1) + f5 = exp(x + 2) - 1 + f6 = 1/log(x) + f7 = 1/x + + s1 = ImageSet(Lambda(n, 2*n*pi), S.Integers) + s2 = ImageSet(Lambda(n, 2*n*pi + pi), S.Integers) + s3 = ImageSet(Lambda(n, 2*n*pi + pi/2), S.Integers) + s4 = ImageSet(Lambda(n, 2*n*pi - 1), S.Integers) + s5 = ImageSet(Lambda(n, 2*n*pi - 1 + pi), S.Integers) + + assert solve_decomposition(f1, x, S.Reals) == FiniteSet(0, log(2), log(3)) + assert dumeq(solve_decomposition(f2, x, S.Reals), s3) + assert dumeq(solve_decomposition(f3, x, S.Reals), Union(s1, s2, s3)) + assert dumeq(solve_decomposition(f4, x, S.Reals), Union(s4, s5)) + assert solve_decomposition(f5, x, S.Reals) == FiniteSet(-2) + assert solve_decomposition(f6, x, S.Reals) == S.EmptySet + assert solve_decomposition(f7, x, S.Reals) == S.EmptySet + assert solve_decomposition(x, x, Interval(1, 2)) == S.EmptySet + + +# nonlinsolve testcases +def test_nonlinsolve_basic(): + assert nonlinsolve([],[]) == S.EmptySet + assert nonlinsolve([],[x, y]) == S.EmptySet + + system = [x, y - x - 5] + assert nonlinsolve([x],[x, y]) == FiniteSet((0, y)) + assert nonlinsolve(system, [y]) == S.EmptySet + soln = (ImageSet(Lambda(n, 2*n*pi + pi/2), S.Integers),) + assert dumeq(nonlinsolve([sin(x) - 1], [x]), FiniteSet(tuple(soln))) + soln = ((ImageSet(Lambda(n, 2*n*pi + pi), S.Integers), 1), + (ImageSet(Lambda(n, 2*n*pi), S.Integers), 1)) + assert dumeq(nonlinsolve([sin(x), y - 1], [x, y]), FiniteSet(*soln)) + assert nonlinsolve([x**2 - 1], [x]) == FiniteSet((-1,), (1,)) + + soln = FiniteSet((y, y)) + assert nonlinsolve([x - y, 0], x, y) == soln + assert nonlinsolve([0, x - y], x, y) == soln + assert nonlinsolve([x - y, x - y], x, y) == soln + assert nonlinsolve([x, 0], x, y) == FiniteSet((0, y)) + f = Function('f') + assert nonlinsolve([f(x), 0], f(x), y) == FiniteSet((0, y)) + assert nonlinsolve([f(x), 0], f(x), f(y)) == FiniteSet((0, f(y))) + A = Indexed('A', x) + assert nonlinsolve([A, 0], A, y) == FiniteSet((0, y)) + assert nonlinsolve([x**2 -1], [sin(x)]) == FiniteSet((S.EmptySet,)) + assert nonlinsolve([x**2 -1], sin(x)) == FiniteSet((S.EmptySet,)) + assert nonlinsolve([x**2 -1], 1) == FiniteSet((x**2,)) + assert nonlinsolve([x**2 -1], x + y) == FiniteSet((S.EmptySet,)) + assert nonlinsolve([Eq(1, x + y), Eq(1, -x + y - 1), Eq(1, -x + y - 1)], x, y) == FiniteSet( + (-S.Half, 3*S.Half)) + + +def test_nonlinsolve_abs(): + soln = FiniteSet((y, y), (-y, y)) + assert nonlinsolve([Abs(x) - y], x, y) == soln + + +def test_raise_exception_nonlinsolve(): + raises(IndexError, lambda: nonlinsolve([x**2 -1], [])) + raises(ValueError, lambda: nonlinsolve([x**2 -1])) + + +def test_trig_system(): + # TODO: add more simple testcases when solveset returns + # simplified soln for Trig eq + assert nonlinsolve([sin(x) - 1, cos(x) -1 ], x) == S.EmptySet + soln1 = (ImageSet(Lambda(n, 2*n*pi + pi/2), S.Integers),) + soln = FiniteSet(soln1) + assert dumeq(nonlinsolve([sin(x) - 1, cos(x)], x), soln) + + +@XFAIL +def test_trig_system_fail(): + # fails because solveset trig solver is not much smart. + sys = [x + y - pi/2, sin(x) + sin(y) - 1] + # solveset returns conditionset for sin(x) + sin(y) - 1 + soln_1 = (ImageSet(Lambda(n, n*pi + pi/2), S.Integers), + ImageSet(Lambda(n, n*pi), S.Integers)) + soln_1 = FiniteSet(soln_1) + soln_2 = (ImageSet(Lambda(n, n*pi), S.Integers), + ImageSet(Lambda(n, n*pi+ pi/2), S.Integers)) + soln_2 = FiniteSet(soln_2) + soln = soln_1 + soln_2 + assert dumeq(nonlinsolve(sys, [x, y]), soln) + + # Add more cases from here + # http://www.vitutor.com/geometry/trigonometry/equations_systems.html#uno + sys = [sin(x) + sin(y) - (sqrt(3)+1)/2, sin(x) - sin(y) - (sqrt(3) - 1)/2] + soln_x = Union(ImageSet(Lambda(n, 2*n*pi + pi/3), S.Integers), + ImageSet(Lambda(n, 2*n*pi + pi*Rational(2, 3)), S.Integers)) + soln_y = Union(ImageSet(Lambda(n, 2*n*pi + pi/6), S.Integers), + ImageSet(Lambda(n, 2*n*pi + pi*Rational(5, 6)), S.Integers)) + assert dumeq(nonlinsolve(sys, [x, y]), FiniteSet((soln_x, soln_y))) + + +def test_nonlinsolve_positive_dimensional(): + x, y, a, b, c, d = symbols('x, y, a, b, c, d', extended_real=True) + assert nonlinsolve([x*y, x*y - x], [x, y]) == FiniteSet((0, y)) + + system = [a**2 + a*c, a - b] + assert nonlinsolve(system, [a, b]) == FiniteSet((0, 0), (-c, -c)) + # here (a= 0, b = 0) is independent soln so both is printed. + # if symbols = [a, b, c] then only {a : -c ,b : -c} + + eq1 = a + b + c + d + eq2 = a*b + b*c + c*d + d*a + eq3 = a*b*c + b*c*d + c*d*a + d*a*b + eq4 = a*b*c*d - 1 + system = [eq1, eq2, eq3, eq4] + sol1 = (-1/d, -d, 1/d, FiniteSet(d) - FiniteSet(0)) + sol2 = (1/d, -d, -1/d, FiniteSet(d) - FiniteSet(0)) + soln = FiniteSet(sol1, sol2) + assert nonlinsolve(system, [a, b, c, d]) == soln + + assert nonlinsolve([x**4 - 3*x**2 + y*x, x*z**2, y*z - 1], [x, y, z]) == \ + {(0, 1/z, z)} + + +def test_nonlinsolve_polysys(): + x, y, z = symbols('x, y, z', real=True) + assert nonlinsolve([x**2 + y - 2, x**2 + y], [x, y]) == S.EmptySet + + s = (-y + 2, y) + assert nonlinsolve([(x + y)**2 - 4, x + y - 2], [x, y]) == FiniteSet(s) + + system = [x**2 - y**2] + soln_real = FiniteSet((-y, y), (y, y)) + soln_complex = FiniteSet((-Abs(y), y), (Abs(y), y)) + soln =soln_real + soln_complex + assert nonlinsolve(system, [x, y]) == soln + + system = [x**2 - y**2] + soln_real= FiniteSet((y, -y), (y, y)) + soln_complex = FiniteSet((y, -Abs(y)), (y, Abs(y))) + soln = soln_real + soln_complex + assert nonlinsolve(system, [y, x]) == soln + + system = [x**2 + y - 3, x - y - 4] + assert nonlinsolve(system, (x, y)) != nonlinsolve(system, (y, x)) + + assert nonlinsolve([-x**2 - y**2 + z, -2*x, -2*y, S.One], [x, y, z]) == S.EmptySet + assert nonlinsolve([x + y + z, S.One, S.One, S.One], [x, y, z]) == S.EmptySet + + system = [-x**2*z**2 + x*y*z + y**4, -2*x*z**2 + y*z, x*z + 4*y**3, -2*x**2*z + x*y] + assert nonlinsolve(system, [x, y, z]) == FiniteSet((0, 0, z), (x, 0, 0)) + + +def test_nonlinsolve_using_substitution(): + x, y, z, n = symbols('x, y, z, n', real = True) + system = [(x + y)*n - y**2 + 2] + s_x = (n*y - y**2 + 2)/n + soln = (-s_x, y) + assert nonlinsolve(system, [x, y]) == FiniteSet(soln) + + system = [z**2*x**2 - z**2*y**2/exp(x)] + soln_real_1 = (y, x, 0) + soln_real_2 = (-exp(x/2)*Abs(x), x, z) + soln_real_3 = (exp(x/2)*Abs(x), x, z) + soln_complex_1 = (-x*exp(x/2), x, z) + soln_complex_2 = (x*exp(x/2), x, z) + syms = [y, x, z] + soln = FiniteSet(soln_real_1, soln_complex_1, soln_complex_2,\ + soln_real_2, soln_real_3) + assert nonlinsolve(system,syms) == soln + + +def test_nonlinsolve_complex(): + n = Dummy('n') + assert dumeq(nonlinsolve([exp(x) - sin(y), 1/y - 3], [x, y]), { + (ImageSet(Lambda(n, 2*n*I*pi + log(sin(Rational(1, 3)))), S.Integers), Rational(1, 3))}) + + system = [exp(x) - sin(y), 1/exp(y) - 3] + assert dumeq(nonlinsolve(system, [x, y]), { + (ImageSet(Lambda(n, I*(2*n*pi + pi) + + log(sin(log(3)))), S.Integers), -log(3)), + (ImageSet(Lambda(n, I*(2*n*pi + arg(sin(2*n*I*pi - log(3)))) + + log(Abs(sin(2*n*I*pi - log(3))))), S.Integers), + ImageSet(Lambda(n, 2*n*I*pi - log(3)), S.Integers))}) + + system = [exp(x) - sin(y), y**2 - 4] + assert dumeq(nonlinsolve(system, [x, y]), { + (ImageSet(Lambda(n, I*(2*n*pi + pi) + log(sin(2))), S.Integers), -2), + (ImageSet(Lambda(n, 2*n*I*pi + log(sin(2))), S.Integers), 2)}) + + system = [exp(x) - 2, y ** 2 - 2] + assert dumeq(nonlinsolve(system, [x, y]), { + (log(2), -sqrt(2)), (log(2), sqrt(2)), + (ImageSet(Lambda(n, 2*n*I*pi + log(2)), S.Integers), -sqrt(2)), + (ImageSet(Lambda(n, 2 * n * I * pi + log(2)), S.Integers), sqrt(2))}) + + +def test_nonlinsolve_radical(): + assert nonlinsolve([sqrt(y) - x - z, y - 1], [x, y, z]) == {(1 - z, 1, z)} + + +def test_nonlinsolve_inexact(): + sol = [(-1.625, -1.375), (1.625, 1.375)] + res = nonlinsolve([(x + y)**2 - 9, x**2 - y**2 - 0.75], [x, y]) + assert all(abs(res.args[i][j]-sol[i][j]) < 1e-9 + for i in range(2) for j in range(2)) + + assert nonlinsolve([(x + y)**2 - 9, (x + y)**2 - 0.75], [x, y]) == S.EmptySet + + assert nonlinsolve([y**2 + (x - 0.5)**2 - 0.0625, 2*x - 1.0, 2*y], [x, y]) == \ + S.EmptySet + + res = nonlinsolve([x**2 + y - 0.5, (x + y)**2, log(z)], [x, y, z]) + sol = [(-0.366025403784439, 0.366025403784439, 1), + (-0.366025403784439, 0.366025403784439, 1), + (1.36602540378444, -1.36602540378444, 1)] + assert all(abs(res.args[i][j]-sol[i][j]) < 1e-9 + for i in range(3) for j in range(3)) + + res = nonlinsolve([y - x**2, x**5 - x + 1.0], [x, y]) + sol = [(-1.16730397826142, 1.36259857766493), + (-0.181232444469876 - 1.08395410131771*I, + -1.14211129483496 + 0.392895302949911*I), + (-0.181232444469876 + 1.08395410131771*I, + -1.14211129483496 - 0.392895302949911*I), + (0.764884433600585 - 0.352471546031726*I, + 0.460812006002492 - 0.539199997693599*I), + (0.764884433600585 + 0.352471546031726*I, + 0.460812006002492 + 0.539199997693599*I)] + assert all(abs(res.args[i][j] - sol[i][j]) < 1e-9 + for i in range(5) for j in range(2)) + +@XFAIL +def test_solve_nonlinear_trans(): + # After the transcendental equation solver these will work + x, y = symbols('x, y', real=True) + soln1 = FiniteSet((2*LambertW(y/2), y)) + soln2 = FiniteSet((-x*sqrt(exp(x)), y), (x*sqrt(exp(x)), y)) + soln3 = FiniteSet((x*exp(x/2), x)) + soln4 = FiniteSet(2*LambertW(y/2), y) + assert nonlinsolve([x**2 - y**2/exp(x)], [x, y]) == soln1 + assert nonlinsolve([x**2 - y**2/exp(x)], [y, x]) == soln2 + assert nonlinsolve([x**2 - y**2/exp(x)], [y, x]) == soln3 + assert nonlinsolve([x**2 - y**2/exp(x)], [x, y]) == soln4 + + +def test_nonlinsolve_issue_25182(): + a1, b1, c1, ca, cb, cg = symbols('a1, b1, c1, ca, cb, cg') + eq1 = a1*a1 + b1*b1 - 2.*a1*b1*cg - c1*c1 + eq2 = a1*a1 + c1*c1 - 2.*a1*c1*cb - b1*b1 + eq3 = b1*b1 + c1*c1 - 2.*b1*c1*ca - a1*a1 + assert nonlinsolve([eq1, eq2, eq3], [c1, cb, cg]) == FiniteSet( + (1.0*b1*ca - 1.0*sqrt(a1**2 + b1**2*ca**2 - b1**2), + -1.0*sqrt(a1**2 + b1**2*ca**2 - b1**2)/a1, + -1.0*b1*(ca - 1)*(ca + 1)/a1 + 1.0*ca*sqrt(a1**2 + b1**2*ca**2 - b1**2)/a1), + (1.0*b1*ca + 1.0*sqrt(a1**2 + b1**2*ca**2 - b1**2), + 1.0*sqrt(a1**2 + b1**2*ca**2 - b1**2)/a1, + -1.0*b1*(ca - 1)*(ca + 1)/a1 - 1.0*ca*sqrt(a1**2 + b1**2*ca**2 - b1**2)/a1)) + + +def test_issue_14642(): + x = Symbol('x') + n1 = 0.5*x**3+x**2+0.5+I #add I in the Polynomials + solution = solveset(n1, x) + assert abs(solution.args[0] - (-2.28267560928153 - 0.312325580497716*I)) <= 1e-9 + assert abs(solution.args[1] - (-0.297354141679308 + 1.01904778618762*I)) <= 1e-9 + assert abs(solution.args[2] - (0.580029750960839 - 0.706722205689907*I)) <= 1e-9 + + # Symbolic + n1 = S.Half*x**3+x**2+S.Half+I + res = FiniteSet(-((3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49) + /2)/2)**2)**(S(1)/6)*cos(atan((27 + 3*sqrt(3)*31985**(S(1)/4)* + cos(atan(S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan( + S(172)/49)/2)/2 + S(43)/2))/3)/3 - S(2)/3 - 4*cos(atan((27 + + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/2)/2)/(3*sqrt(3)* + 31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + S(43)/2))/3)/(3*((3* + sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + S(43)/2)**2 + + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/2)/2)**2)**(S(1)/ + 6)) + I*(-((3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/ + 2)/2)**2)**(S(1)/6)*sin(atan((27 + 3*sqrt(3)*31985**(S(1)/4)*cos( + atan(S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49) + /2)/2 + S(43)/2))/3)/3 + 4*sin(atan((27 + 3*sqrt(3)*31985**(S(1)/4)* + cos(atan(S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172) + /49)/2)/2 + S(43)/2))/3)/(3*((3*sqrt(3)*31985**(S(1)/4)*sin(atan( + S(172)/49)/2)/2 + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)* + cos(atan(S(172)/49)/2)/2)**2)**(S(1)/6))), -S(2)/3 - sqrt(3)*((3* + sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + S(43)/2)**2 + + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/2)/2)**2)**(S(1) + /6)*sin(atan((27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/2) + /2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + S(43)/2)) + /3)/6 - 4*re(1/((-S(1)/2 - sqrt(3)*I/2)*(S(43)/2 + 27*I + sqrt(-256 + + (43 + 54*I)**2)/2)**(S(1)/3)))/3 + ((3*sqrt(3)*31985**(S(1)/4)*sin( + atan(S(172)/49)/2)/2 + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)* + cos(atan(S(172)/49)/2)/2)**2)**(S(1)/6)*cos(atan((27 + 3*sqrt(3)* + 31985**(S(1)/4)*cos(atan(S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)* + sin(atan(S(172)/49)/2)/2 + S(43)/2))/3)/6 + I*(-4*im(1/((-S(1)/2 - + sqrt(3)*I/2)*(S(43)/2 + 27*I + sqrt(-256 + (43 + 54*I)**2)/2)**(S(1)/ + 3)))/3 + ((3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/2) + /2)**2)**(S(1)/6)*sin(atan((27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan( + S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2))/3)/6 + sqrt(3)*((3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/ + 49)/2)/2 + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan( + S(172)/49)/2)/2)**2)**(S(1)/6)*cos(atan((27 + 3*sqrt(3)*31985**(S(1)/ + 4)*cos(atan(S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan( + S(172)/49)/2)/2 + S(43)/2))/3)/6), -S(2)/3 - 4*re(1/((-S(1)/2 + + sqrt(3)*I/2)*(S(43)/2 + 27*I + sqrt(-256 + (43 + 54*I)**2)/2)**(S(1) + /3)))/3 + sqrt(3)*((3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/2) + /2)**2)**(S(1)/6)*sin(atan((27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan( + S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2))/3)/6 + ((3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan(S(172)/49)/2) + /2)**2)**(S(1)/6)*cos(atan((27 + 3*sqrt(3)*31985**(S(1)/4)*cos(atan( + S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin(atan(S(172)/49)/2)/2 + + S(43)/2))/3)/6 + I*(-sqrt(3)*((3*sqrt(3)*31985**(S(1)/4)*sin(atan( + S(172)/49)/2)/2 + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)*cos( + atan(S(172)/49)/2)/2)**2)**(S(1)/6)*cos(atan((27 + 3*sqrt(3)*31985**( + S(1)/4)*cos(atan(S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin( + atan(S(172)/49)/2)/2 + S(43)/2))/3)/6 + ((3*sqrt(3)*31985**(S(1)/4)* + sin(atan(S(172)/49)/2)/2 + S(43)/2)**2 + (27 + 3*sqrt(3)*31985**(S(1)/4)* + cos(atan(S(172)/49)/2)/2)**2)**(S(1)/6)*sin(atan((27 + 3*sqrt(3)*31985**( + S(1)/4)*cos(atan(S(172)/49)/2)/2)/(3*sqrt(3)*31985**(S(1)/4)*sin( + atan(S(172)/49)/2)/2 + S(43)/2))/3)/6 - 4*im(1/((-S(1)/2 + sqrt(3)*I/2)* + (S(43)/2 + 27*I + sqrt(-256 + (43 + 54*I)**2)/2)**(S(1)/3)))/3)) + + assert solveset(n1, x) == res + + +def test_issue_13961(): + V = (ax, bx, cx, gx, jx, lx, mx, nx, q) = symbols('ax bx cx gx jx lx mx nx q') + S = (ax*q - lx*q - mx, ax - gx*q - lx, bx*q**2 + cx*q - jx*q - nx, q*(-ax*q + lx*q + mx), q*(-ax + gx*q + lx)) + + sol = FiniteSet((lx + mx/q, (-cx*q + jx*q + nx)/q**2, cx, mx/q**2, jx, lx, mx, nx, Complement({q}, {0})), + (lx + mx/q, (cx*q - jx*q - nx)/q**2*-1, cx, mx/q**2, jx, lx, mx, nx, Complement({q}, {0}))) + assert nonlinsolve(S, *V) == sol + # The two solutions are in fact identical, so even better if only one is returned + + +def test_issue_14541(): + solutions = solveset(sqrt(-x**2 - 2.0), x) + assert abs(solutions.args[0]+1.4142135623731*I) <= 1e-9 + assert abs(solutions.args[1]-1.4142135623731*I) <= 1e-9 + + +def test_issue_13396(): + expr = -2*y*exp(-x**2 - y**2)*Abs(x) + sol = FiniteSet(0) + + assert solveset(expr, y, domain=S.Reals) == sol + + # Related type of equation also solved here + assert solveset(atan(x**2 - y**2)-pi/2, y, S.Reals) is S.EmptySet + + +def test_issue_12032(): + sol = FiniteSet(-sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)))/2 + + sqrt(Abs(-2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)) + + 2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2/sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)))))/2, + -sqrt(Abs(-2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)) + + 2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2/sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)))))/2 - + sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)))/2, + sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)))/2 - + I*sqrt(Abs(-2/sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) - + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)) + + 2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)))))/2, + sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)))/2 + + I*sqrt(Abs(-2/sqrt(-2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) + + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3))) - + 2*(Rational(1, 16) + sqrt(849)/144)**(Rational(1, 3)) + + 2/(3*(Rational(1, 16) + sqrt(849)/144)**(Rational(1,3)))))/2) + assert solveset(x**4 + x - 1, x) == sol + + +def test_issue_10876(): + assert solveset(1/sqrt(x), x) == S.EmptySet + + +def test_issue_19050(): + # test_issue_19050 --> TypeError removed + assert dumeq(nonlinsolve([x + y, sin(y)], [x, y]), + FiniteSet((ImageSet(Lambda(n, -2*n*pi), S.Integers), ImageSet(Lambda(n, 2*n*pi), S.Integers)),\ + (ImageSet(Lambda(n, -2*n*pi - pi), S.Integers), ImageSet(Lambda(n, 2*n*pi + pi), S.Integers)))) + assert dumeq(nonlinsolve([x + y, sin(y) + cos(y)], [x, y]), + FiniteSet((ImageSet(Lambda(n, -2*n*pi - 3*pi/4), S.Integers), ImageSet(Lambda(n, 2*n*pi + 3*pi/4), S.Integers)), \ + (ImageSet(Lambda(n, -2*n*pi - 7*pi/4), S.Integers), ImageSet(Lambda(n, 2*n*pi + 7*pi/4), S.Integers)))) + + +def test_issue_16618(): + eqn = [sin(x)*sin(y), cos(x)*cos(y) - 1] + # nonlinsolve's answer is still suspicious since it contains only three + # distinct Dummys instead of 4. (Both 'x' ImageSets share the same Dummy.) + ans = FiniteSet((ImageSet(Lambda(n, 2*n*pi), S.Integers), ImageSet(Lambda(n, 2*n*pi), S.Integers)), + (ImageSet(Lambda(n, 2*n*pi + pi), S.Integers), ImageSet(Lambda(n, 2*n*pi + pi), S.Integers))) + sol = nonlinsolve(eqn, [x, y]) + + for i0, j0 in zip(ordered(sol), ordered(ans)): + assert len(i0) == len(j0) == 2 + assert all(a.dummy_eq(b) for a, b in zip(i0, j0)) + assert len(sol) == len(ans) + + +def test_issue_17566(): + assert nonlinsolve([32*(2**x)/2**(-y) - 4**y, 27*(3**x) - S(1)/3**y], x, y) ==\ + FiniteSet((-log(81)/log(3), 1)) + + +def test_issue_16643(): + n = Dummy('n') + assert solveset(x**2*sin(x), x).dummy_eq(Union(ImageSet(Lambda(n, 2*n*pi + pi), S.Integers), + ImageSet(Lambda(n, 2*n*pi), S.Integers))) + + +def test_issue_19587(): + n,m = symbols('n m') + assert nonlinsolve([32*2**m*2**n - 4**n, 27*3**m - 3**(-n)], m, n) ==\ + FiniteSet((-log(81)/log(3), 1)) + + +def test_issue_5132_1(): + system = [sqrt(x**2 + y**2) - sqrt(10), x + y - 4] + assert nonlinsolve(system, [x, y]) == FiniteSet((1, 3), (3, 1)) + + n = Dummy('n') + eqs = [exp(x)**2 - sin(y) + z**2, 1/exp(y) - 3] + s_real_y = -log(3) + s_real_z = sqrt(-exp(2*x) - sin(log(3))) + soln_real = FiniteSet((s_real_y, s_real_z), (s_real_y, -s_real_z)) + lam = Lambda(n, 2*n*I*pi + -log(3)) + s_complex_y = ImageSet(lam, S.Integers) + lam = Lambda(n, sqrt(-exp(2*x) + sin(2*n*I*pi + -log(3)))) + s_complex_z_1 = ImageSet(lam, S.Integers) + lam = Lambda(n, -sqrt(-exp(2*x) + sin(2*n*I*pi + -log(3)))) + s_complex_z_2 = ImageSet(lam, S.Integers) + soln_complex = FiniteSet( + (s_complex_y, s_complex_z_1), + (s_complex_y, s_complex_z_2) + ) + soln = soln_real + soln_complex + assert dumeq(nonlinsolve(eqs, [y, z]), soln) + + +def test_issue_5132_2(): + x, y = symbols('x, y', real=True) + eqs = [exp(x)**2 - sin(y) + z**2] + n = Dummy('n') + soln_real = (log(-z**2 + sin(y))/2, z) + lam = Lambda( n, I*(2*n*pi + arg(-z**2 + sin(y)))/2 + log(Abs(z**2 - sin(y)))/2) + img = ImageSet(lam, S.Integers) + # not sure about the complex soln. But it looks correct. + soln_complex = (img, z) + soln = FiniteSet(soln_real, soln_complex) + assert dumeq(nonlinsolve(eqs, [x, z]), soln) + + system = [r - x**2 - y**2, tan(t) - y/x] + s_x = sqrt(r/(tan(t)**2 + 1)) + s_y = sqrt(r/(tan(t)**2 + 1))*tan(t) + soln = FiniteSet((s_x, s_y), (-s_x, -s_y)) + assert nonlinsolve(system, [x, y]) == soln + + +def test_issue_6752(): + a, b = symbols('a, b', real=True) + assert nonlinsolve([a**2 + a, a - b], [a, b]) == {(-1, -1), (0, 0)} + + +@SKIP("slow") +def test_issue_5114_solveset(): + # slow testcase + from sympy.abc import o, p + + # there is no 'a' in the equation set but this is how the + # problem was originally posed + syms = [a, b, c, f, h, k, n] + eqs = [b + r/d - c/d, + c*(1/d + 1/e + 1/g) - f/g - r/d, + f*(1/g + 1/i + 1/j) - c/g - h/i, + h*(1/i + 1/l + 1/m) - f/i - k/m, + k*(1/m + 1/o + 1/p) - h/m - n/p, + n*(1/p + 1/q) - k/p] + assert len(nonlinsolve(eqs, syms)) == 1 + + +@SKIP("Hangs") +def _test_issue_5335(): + # Not able to check zero dimensional system. + # is_zero_dimensional Hangs + lam, a0, conc = symbols('lam a0 conc') + eqs = [lam + 2*y - a0*(1 - x/2)*x - 0.005*x/2*x, + a0*(1 - x/2)*x - 1*y - 0.743436700916726*y, + x + y - conc] + sym = [x, y, a0] + # there are 4 solutions but only two are valid + assert len(nonlinsolve(eqs, sym)) == 2 + # float + eqs = [lam + 2*y - a0*(1 - x/2)*x - 0.005*x/2*x, + a0*(1 - x/2)*x - 1*y - 0.743436700916726*y, + x + y - conc] + sym = [x, y, a0] + assert len(nonlinsolve(eqs, sym)) == 2 + + +def test_issue_2777(): + # the equations represent two circles + x, y = symbols('x y', real=True) + e1, e2 = sqrt(x**2 + y**2) - 10, sqrt(y**2 + (-x + 10)**2) - 3 + a, b = Rational(191, 20), 3*sqrt(391)/20 + ans = {(a, -b), (a, b)} + assert nonlinsolve((e1, e2), (x, y)) == ans + assert nonlinsolve((e1, e2/(x - a)), (x, y)) == S.EmptySet + # make the 2nd circle's radius be -3 + e2 += 6 + assert nonlinsolve((e1, e2), (x, y)) == S.EmptySet + + +def test_issue_8828(): + x1 = 0 + y1 = -620 + r1 = 920 + x2 = 126 + y2 = 276 + x3 = 51 + y3 = 205 + r3 = 104 + v = [x, y, z] + + f1 = (x - x1)**2 + (y - y1)**2 - (r1 - z)**2 + f2 = (x2 - x)**2 + (y2 - y)**2 - z**2 + f3 = (x - x3)**2 + (y - y3)**2 - (r3 - z)**2 + F = [f1, f2, f3] + + g1 = sqrt((x - x1)**2 + (y - y1)**2) + z - r1 + g2 = f2 + g3 = sqrt((x - x3)**2 + (y - y3)**2) + z - r3 + G = [g1, g2, g3] + + # both soln same + A = nonlinsolve(F, v) + B = nonlinsolve(G, v) + assert A == B + + +def test_nonlinsolve_conditionset(): + # when solveset failed to solve all the eq + # return conditionset + f = Function('f') + f1 = f(x) - pi/2 + f2 = f(y) - pi*Rational(3, 2) + intermediate_system = Eq(2*f(x) - pi, 0) & Eq(2*f(y) - 3*pi, 0) + syms = Tuple(x, y) + soln = ConditionSet( + syms, + intermediate_system, + S.Complexes**2) + assert nonlinsolve([f1, f2], [x, y]) == soln + + +def test_substitution_basic(): + assert substitution([], [x, y]) == S.EmptySet + assert substitution([], []) == S.EmptySet + system = [2*x**2 + 3*y**2 - 30, 3*x**2 - 2*y**2 - 19] + soln = FiniteSet((-3, -2), (-3, 2), (3, -2), (3, 2)) + assert substitution(system, [x, y]) == soln + + soln = FiniteSet((-1, 1)) + assert substitution([x + y], [x], [{y: 1}], [y], set(), [x, y]) == soln + assert substitution( + [x + y], [x], [{y: 1}], [y], + {x + 1}, [y, x]) == S.EmptySet + + +def test_substitution_incorrect(): + # the solutions in the following two tests are incorrect. The + # correct result is EmptySet in both cases. + assert substitution([h - 1, k - 1, f - 2, f - 4, -2 * k], + [h, k, f]) == {(1, 1, f)} + assert substitution([x + y + z, S.One, S.One, S.One], [x, y, z]) == \ + {(-y - z, y, z)} + + # the correct result in the test below is {(-I, I, I, -I), + # (I, -I, -I, I)} + assert substitution([a - d, b + d, c + d, d**2 + 1], [a, b, c, d]) == \ + {(d, -d, -d, d)} + + # the result in the test below is incomplete. The complete result + # is {(0, b), (log(2), 2)} + assert substitution([a*(a - log(b)), a*(b - 2)], [a, b]) == \ + {(0, b)} + + # The system in the test below is zero-dimensional, so the result + # should have no free symbols + assert substitution([-k*y + 6*x - 4*y, -81*k + 49*y**2 - 270, + -3*k*z + k + z**3, k**2 - 2*k + 4], + [x, y, z, k]).free_symbols == {z} + + +def test_substitution_redundant(): + # the third and fourth solutions are redundant in the test below + assert substitution([x**2 - y**2, z - 1], [x, z]) == \ + {(-y, 1), (y, 1), (-sqrt(y**2), 1), (sqrt(y**2), 1)} + + # the system below has three solutions. Two of the solutions + # returned by substitution are redundant. + res = substitution([x - y, y**3 - 3*y**2 + 1], [x, y]) + assert len(res) == 5 + + +def test_issue_5132_substitution(): + x, y, z, r, t = symbols('x, y, z, r, t', real=True) + system = [r - x**2 - y**2, tan(t) - y/x] + s_x_1 = Complement(FiniteSet(-sqrt(r/(tan(t)**2 + 1))), FiniteSet(0)) + s_x_2 = Complement(FiniteSet(sqrt(r/(tan(t)**2 + 1))), FiniteSet(0)) + s_y = sqrt(r/(tan(t)**2 + 1))*tan(t) + soln = FiniteSet((s_x_2, s_y)) + FiniteSet((s_x_1, -s_y)) + assert substitution(system, [x, y]) == soln + + n = Dummy('n') + eqs = [exp(x)**2 - sin(y) + z**2, 1/exp(y) - 3] + s_real_y = -log(3) + s_real_z = sqrt(-exp(2*x) - sin(log(3))) + soln_real = FiniteSet((s_real_y, s_real_z), (s_real_y, -s_real_z)) + lam = Lambda(n, 2*n*I*pi + -log(3)) + s_complex_y = ImageSet(lam, S.Integers) + lam = Lambda(n, sqrt(-exp(2*x) + sin(2*n*I*pi + -log(3)))) + s_complex_z_1 = ImageSet(lam, S.Integers) + lam = Lambda(n, -sqrt(-exp(2*x) + sin(2*n*I*pi + -log(3)))) + s_complex_z_2 = ImageSet(lam, S.Integers) + soln_complex = FiniteSet( + (s_complex_y, s_complex_z_1), + (s_complex_y, s_complex_z_2)) + soln = soln_real + soln_complex + assert dumeq(substitution(eqs, [y, z]), soln) + + +def test_raises_substitution(): + raises(ValueError, lambda: substitution([x**2 -1], [])) + raises(TypeError, lambda: substitution([x**2 -1])) + raises(ValueError, lambda: substitution([x**2 -1], [sin(x)])) + raises(TypeError, lambda: substitution([x**2 -1], x)) + raises(TypeError, lambda: substitution([x**2 -1], 1)) + + +def test_issue_21022(): + from sympy.core.sympify import sympify + + eqs = [ + 'k-16', + 'p-8', + 'y*y+z*z-x*x', + 'd - x + p', + 'd*d+k*k-y*y', + 'z*z-p*p-k*k', + 'abc-efg', + ] + efg = Symbol('efg') + eqs = [sympify(x) for x in eqs] + + syb = list(ordered(set.union(*[x.free_symbols for x in eqs]))) + res = nonlinsolve(eqs, syb) + + ans = FiniteSet( + (efg, 32, efg, 16, 8, 40, -16*sqrt(5), -8*sqrt(5)), + (efg, 32, efg, 16, 8, 40, -16*sqrt(5), 8*sqrt(5)), + (efg, 32, efg, 16, 8, 40, 16*sqrt(5), -8*sqrt(5)), + (efg, 32, efg, 16, 8, 40, 16*sqrt(5), 8*sqrt(5)), + ) + assert len(res) == len(ans) == 4 + assert res == ans + for result in res.args: + assert len(result) == 8 + + +def test_issue_17940(): + n = Dummy('n') + k1 = Dummy('k1') + sol = ImageSet(Lambda(((k1, n),), I*(2*k1*pi + arg(2*n*I*pi + log(5))) + + log(Abs(2*n*I*pi + log(5)))), + ProductSet(S.Integers, S.Integers)) + assert solveset(exp(exp(x)) - 5, x).dummy_eq(sol) + + +def test_issue_17906(): + assert solveset(7**(x**2 - 80) - 49**x, x) == FiniteSet(-8, 10) + + +@XFAIL +def test_issue_17933(): + eq1 = x*sin(45) - y*cos(q) + eq2 = x*cos(45) - y*sin(q) + eq3 = 9*x*sin(45)/10 + y*cos(q) + eq4 = 9*x*cos(45)/10 + y*sin(z) - z + assert nonlinsolve([eq1, eq2, eq3, eq4], x, y, z, q) ==\ + FiniteSet((0, 0, 0, q)) + +def test_issue_17933_bis(): + # nonlinsolve's result depends on the 'default_sort_key' ordering of + # the unknowns. + eq1 = x*sin(45) - y*cos(q) + eq2 = x*cos(45) - y*sin(q) + eq3 = 9*x*sin(45)/10 + y*cos(q) + eq4 = 9*x*cos(45)/10 + y*sin(z) - z + zz = Symbol('zz') + eqs = [e.subs(q, zz) for e in (eq1, eq2, eq3, eq4)] + assert nonlinsolve(eqs, x, y, z, zz) == FiniteSet((0, 0, 0, zz)) + + +def test_issue_14565(): + # removed redundancy + assert dumeq(nonlinsolve([k + m, k + m*exp(-2*pi*k)], [k, m]) , + FiniteSet((-n*I, ImageSet(Lambda(n, n*I), S.Integers)))) + + +# end of tests for nonlinsolve + + +def test_issue_9556(): + b = Symbol('b', positive=True) + + assert solveset(Abs(x) + 1, x, S.Reals) is S.EmptySet + assert solveset(Abs(x) + b, x, S.Reals) is S.EmptySet + assert solveset(Eq(b, -1), b, S.Reals) is S.EmptySet + + +def test_issue_9611(): + assert solveset(Eq(x - x + a, a), x, S.Reals) == S.Reals + assert solveset(Eq(y - y + a, a), y) == S.Complexes + + +def test_issue_9557(): + assert solveset(x**2 + a, x, S.Reals) == Intersection(S.Reals, + FiniteSet(-sqrt(-a), sqrt(-a))) + + +def test_issue_9778(): + x = Symbol('x', real=True) + y = Symbol('y', real=True) + assert solveset(x**3 + 1, x, S.Reals) == FiniteSet(-1) + assert solveset(x**Rational(3, 5) + 1, x, S.Reals) == S.EmptySet + assert solveset(x**3 + y, x, S.Reals) == \ + FiniteSet(-Abs(y)**Rational(1, 3)*sign(y)) + + +def test_issue_10214(): + assert solveset(x**Rational(3, 2) + 4, x, S.Reals) == S.EmptySet + assert solveset(x**(Rational(-3, 2)) + 4, x, S.Reals) == S.EmptySet + + ans = FiniteSet(-2**Rational(2, 3)) + assert solveset(x**(S(3)) + 4, x, S.Reals) == ans + assert (x**(S(3)) + 4).subs(x,list(ans)[0]) == 0 # substituting ans and verifying the result. + assert (x**(S(3)) + 4).subs(x,-(-2)**Rational(2, 3)) == 0 + + +def test_issue_9849(): + assert solveset(Abs(sin(x)) + 1, x, S.Reals) == S.EmptySet + + +def test_issue_9953(): + assert linsolve([ ], x) == S.EmptySet + + +def test_issue_9913(): + assert solveset(2*x + 1/(x - 10)**2, x, S.Reals) == \ + FiniteSet(-(3*sqrt(24081)/4 + Rational(4027, 4))**Rational(1, 3)/3 - 100/ + (3*(3*sqrt(24081)/4 + Rational(4027, 4))**Rational(1, 3)) + Rational(20, 3)) + + +def test_issue_10397(): + assert solveset(sqrt(x), x, S.Complexes) == FiniteSet(0) + + +def test_issue_14987(): + raises(ValueError, lambda: linear_eq_to_matrix( + [x**2], x)) + raises(ValueError, lambda: linear_eq_to_matrix( + [x*(-3/x + 1) + 2*y - a], [x, y])) + raises(ValueError, lambda: linear_eq_to_matrix( + [(x**2 - 3*x)/(x - 3) - 3], x)) + raises(ValueError, lambda: linear_eq_to_matrix( + [(x + 1)**3 - x**3 - 3*x**2 + 7], x)) + raises(ValueError, lambda: linear_eq_to_matrix( + [x*(1/x + 1) + y], [x, y])) + raises(ValueError, lambda: linear_eq_to_matrix( + [(x + 1)*y], [x, y])) + raises(ValueError, lambda: linear_eq_to_matrix( + [Eq(1/x, 1/x + y)], [x, y])) + raises(ValueError, lambda: linear_eq_to_matrix( + [Eq(y/x, y/x + y)], [x, y])) + raises(ValueError, lambda: linear_eq_to_matrix( + [Eq(x*(x + 1), x**2 + y)], [x, y])) + + +def test_simplification(): + eq = x + (a - b)/(-2*a + 2*b) + assert solveset(eq, x) == FiniteSet(S.Half) + assert solveset(eq, x, S.Reals) == Intersection({-((a - b)/(-2*a + 2*b))}, S.Reals) + # So that ap - bn is not zero: + ap = Symbol('ap', positive=True) + bn = Symbol('bn', negative=True) + eq = x + (ap - bn)/(-2*ap + 2*bn) + assert solveset(eq, x) == FiniteSet(S.Half) + assert solveset(eq, x, S.Reals) == FiniteSet(S.Half) + + +def test_integer_domain_relational(): + eq1 = 2*x + 3 > 0 + eq2 = x**2 + 3*x - 2 >= 0 + eq3 = x + 1/x > -2 + 1/x + eq4 = x + sqrt(x**2 - 5) > 0 + eq = x + 1/x > -2 + 1/x + eq5 = eq.subs(x,log(x)) + eq6 = log(x)/x <= 0 + eq7 = log(x)/x < 0 + eq8 = x/(x-3) < 3 + eq9 = x/(x**2-3) < 3 + + assert solveset(eq1, x, S.Integers) == Range(-1, oo, 1) + assert solveset(eq2, x, S.Integers) == Union(Range(-oo, -3, 1), Range(1, oo, 1)) + assert solveset(eq3, x, S.Integers) == Union(Range(-1, 0, 1), Range(1, oo, 1)) + assert solveset(eq4, x, S.Integers) == Range(3, oo, 1) + assert solveset(eq5, x, S.Integers) == Range(2, oo, 1) + assert solveset(eq6, x, S.Integers) == Range(1, 2, 1) + assert solveset(eq7, x, S.Integers) == S.EmptySet + assert solveset(eq8, x, domain=Range(0,5)) == Range(0, 3, 1) + assert solveset(eq9, x, domain=Range(0,5)) == Union(Range(0, 2, 1), Range(2, 5, 1)) + + # test_issue_19794 + assert solveset(x + 2 < 0, x, S.Integers) == Range(-oo, -2, 1) + + +def test_issue_10555(): + f = Function('f') + g = Function('g') + assert solveset(f(x) - pi/2, x, S.Reals).dummy_eq( + ConditionSet(x, Eq(f(x) - pi/2, 0), S.Reals)) + assert solveset(f(g(x)) - pi/2, g(x), S.Reals).dummy_eq( + ConditionSet(g(x), Eq(f(g(x)) - pi/2, 0), S.Reals)) + + +def test_issue_8715(): + eq = x + 1/x > -2 + 1/x + assert solveset(eq, x, S.Reals) == \ + (Interval.open(-2, oo) - FiniteSet(0)) + assert solveset(eq.subs(x,log(x)), x, S.Reals) == \ + Interval.open(exp(-2), oo) - FiniteSet(1) + + +def test_issue_11174(): + eq = z**2 + exp(2*x) - sin(y) + soln = Intersection(S.Reals, FiniteSet(log(-z**2 + sin(y))/2)) + assert solveset(eq, x, S.Reals) == soln + + eq = sqrt(r)*Abs(tan(t))/sqrt(tan(t)**2 + 1) + x*tan(t) + s = -sqrt(r)*Abs(tan(t))/(sqrt(tan(t)**2 + 1)*tan(t)) + soln = Intersection(S.Reals, FiniteSet(s)) + assert solveset(eq, x, S.Reals) == soln + + +def test_issue_11534(): + # eq1 and eq2 should not have the same solutions because squaring both + # sides of the radical equation introduces a spurious solution branch. + # The equations have a symbolic parameter y and it is easy to see that for + # y != 0 the solution s1 will not be valid for eq1. + x = Symbol('x', real=True) + y = Symbol('y', real=True) + eq1 = -y + x/sqrt(-x**2 + 1) + eq2 = -y**2 + x**2/(-x**2 + 1) + + # We get a ConditionSet here because s1 works in eq1 if y is equal to zero + # although not for any other value of y. That case is redundant though + # because if y=0 then s1=s2 so the solution for eq1 could just be returned + # as s2 - {-1, 1}. In fact we have + # |y/sqrt(y**2 + 1)| < 1 + # So the complements are not needed either. The ideal output here would be + # sol1 = s2 + # sol2 = s1 | s2. + s1, s2 = FiniteSet(-y/sqrt(y**2 + 1)), FiniteSet(y/sqrt(y**2 + 1)) + cset = ConditionSet(x, Eq(eq1, 0), s1) + sol1 = (s2 - {-1, 1}) | (cset - {-1, 1}) + sol2 = (s1 | s2) - {-1, 1} + + assert solveset(eq1, x, S.Reals) == sol1 + assert solveset(eq2, x, S.Reals) == sol2 + + +def test_issue_10477(): + assert solveset((x**2 + 4*x - 3)/x < 2, x, S.Reals) == \ + Union(Interval.open(-oo, -3), Interval.open(0, 1)) + + +def test_issue_10671(): + assert solveset(sin(y), y, Interval(0, pi)) == FiniteSet(0, pi) + i = Interval(1, 10) + assert solveset((1/x).diff(x) < 0, x, i) == i + + +def test_issue_11064(): + eq = x + sqrt(x**2 - 5) + assert solveset(eq > 0, x, S.Reals) == \ + Interval(sqrt(5), oo) + assert solveset(eq < 0, x, S.Reals) == \ + Interval(-oo, -sqrt(5)) + assert solveset(eq > sqrt(5), x, S.Reals) == \ + Interval.Lopen(sqrt(5), oo) + + +def test_issue_12478(): + eq = sqrt(x - 2) + 2 + soln = solveset_real(eq, x) + assert soln is S.EmptySet + assert solveset(eq < 0, x, S.Reals) is S.EmptySet + assert solveset(eq > 0, x, S.Reals) == Interval(2, oo) + + +def test_issue_12429(): + eq = solveset(log(x)/x <= 0, x, S.Reals) + sol = Interval.Lopen(0, 1) + assert eq == sol + + +def test_issue_19506(): + eq = arg(x + I) + C = Dummy('C') + assert solveset(eq).dummy_eq(Intersection(ConditionSet(C, Eq(im(C) + 1, 0), S.Complexes), + ConditionSet(C, re(C) > 0, S.Complexes))) + + +def test_solveset_arg(): + assert solveset(arg(x), x, S.Reals) == Interval.open(0, oo) + assert solveset(arg(4*x -3), x, S.Reals) == Interval.open(Rational(3, 4), oo) + + +def test__is_finite_with_finite_vars(): + f = _is_finite_with_finite_vars + # issue 12482 + assert all(f(1/x) is None for x in ( + Dummy(), Dummy(real=True), Dummy(complex=True))) + assert f(1/Dummy(real=False)) is True # b/c it's finite but not 0 + + +def test_issue_13550(): + assert solveset(x**2 - 2*x - 15, symbol = x, domain = Interval(-oo, 0)) == FiniteSet(-3) + + +def test_issue_13849(): + assert nonlinsolve((t*(sqrt(5) + sqrt(2)) - sqrt(2), t), t) is S.EmptySet + + +def test_issue_14223(): + assert solveset((Abs(x + Min(x, 2)) - 2).rewrite(Piecewise), x, + S.Reals) == FiniteSet(-1, 1) + assert solveset((Abs(x + Min(x, 2)) - 2).rewrite(Piecewise), x, + Interval(0, 2)) == FiniteSet(1) + assert solveset(x, x, FiniteSet(1, 2)) is S.EmptySet + + +def test_issue_10158(): + dom = S.Reals + assert solveset(x*Max(x, 15) - 10, x, dom) == FiniteSet(Rational(2, 3)) + assert solveset(x*Min(x, 15) - 10, x, dom) == FiniteSet(-sqrt(10), sqrt(10)) + assert solveset(Max(Abs(x - 3) - 1, x + 2) - 3, x, dom) == FiniteSet(-1, 1) + assert solveset(Abs(x - 1) - Abs(y), x, dom) == FiniteSet(-Abs(y) + 1, Abs(y) + 1) + assert solveset(Abs(x + 4*Abs(x + 1)), x, dom) == FiniteSet(Rational(-4, 3), Rational(-4, 5)) + assert solveset(2*Abs(x + Abs(x + Max(3, x))) - 2, x, S.Reals) == FiniteSet(-1, -2) + dom = S.Complexes + raises(ValueError, lambda: solveset(x*Max(x, 15) - 10, x, dom)) + raises(ValueError, lambda: solveset(x*Min(x, 15) - 10, x, dom)) + raises(ValueError, lambda: solveset(Max(Abs(x - 3) - 1, x + 2) - 3, x, dom)) + raises(ValueError, lambda: solveset(Abs(x - 1) - Abs(y), x, dom)) + raises(ValueError, lambda: solveset(Abs(x + 4*Abs(x + 1)), x, dom)) + + +def test_issue_14300(): + f = 1 - exp(-18000000*x) - y + a1 = FiniteSet(-log(-y + 1)/18000000) + + assert solveset(f, x, S.Reals) == \ + Intersection(S.Reals, a1) + assert dumeq(solveset(f, x), + ImageSet(Lambda(n, -I*(2*n*pi + arg(-y + 1))/18000000 - + log(Abs(y - 1))/18000000), S.Integers)) + + +def test_issue_14454(): + number = CRootOf(x**4 + x - 1, 2) + raises(ValueError, lambda: invert_real(number, 0, x)) + assert invert_real(x**2, number, x) # no error + + +def test_issue_17882(): + assert solveset(-8*x**2/(9*(x**2 - 1)**(S(4)/3)) + 4/(3*(x**2 - 1)**(S(1)/3)), x, S.Complexes) == \ + FiniteSet(sqrt(3), -sqrt(3)) + + +def test_term_factors(): + assert list(_term_factors(3**x - 2)) == [-2, 3**x] + expr = 4**(x + 1) + 4**(x + 2) + 4**(x - 1) - 3**(x + 2) - 3**(x + 3) + assert set(_term_factors(expr)) == { + 3**(x + 2), 4**(x + 2), 3**(x + 3), 4**(x - 1), -1, 4**(x + 1)} + + +#################### tests for transolve and its helpers ############### + +def test_transolve(): + + assert _transolve(3**x, x, S.Reals) == S.EmptySet + assert _transolve(3**x - 9**(x + 5), x, S.Reals) == FiniteSet(-10) + + +def test_issue_21276(): + eq = (2*x*(y - z) - y*erf(y - z) - y + z*erf(y - z) + z)**2 + assert solveset(eq.expand(), y) == FiniteSet(z, z + erfinv(2*x - 1)) + + +# exponential tests +def test_exponential_real(): + from sympy.abc import y + + e1 = 3**(2*x) - 2**(x + 3) + e2 = 4**(5 - 9*x) - 8**(2 - x) + e3 = 2**x + 4**x + e4 = exp(log(5)*x) - 2**x + e5 = exp(x/y)*exp(-z/y) - 2 + e6 = 5**(x/2) - 2**(x/3) + e7 = 4**(x + 1) + 4**(x + 2) + 4**(x - 1) - 3**(x + 2) - 3**(x + 3) + e8 = -9*exp(-2*x + 5) + 4*exp(3*x + 1) + e9 = 2**x + 4**x + 8**x - 84 + e10 = 29*2**(x + 1)*615**(x) - 123*2726**(x) + + assert solveset(e1, x, S.Reals) == FiniteSet( + -3*log(2)/(-2*log(3) + log(2))) + assert solveset(e2, x, S.Reals) == FiniteSet(Rational(4, 15)) + assert solveset(e3, x, S.Reals) == S.EmptySet + assert solveset(e4, x, S.Reals) == FiniteSet(0) + assert solveset(e5, x, S.Reals) == Intersection( + S.Reals, FiniteSet(y*log(2*exp(z/y)))) + assert solveset(e6, x, S.Reals) == FiniteSet(0) + assert solveset(e7, x, S.Reals) == FiniteSet(2) + assert solveset(e8, x, S.Reals) == FiniteSet(-2*log(2)/5 + 2*log(3)/5 + Rational(4, 5)) + assert solveset(e9, x, S.Reals) == FiniteSet(2) + assert solveset(e10,x, S.Reals) == FiniteSet((-log(29) - log(2) + log(123))/(-log(2726) + log(2) + log(615))) + + assert solveset_real(-9*exp(-2*x + 5) + 2**(x + 1), x) == FiniteSet( + -((-5 - 2*log(3) + log(2))/(log(2) + 2))) + assert solveset_real(4**(x/2) - 2**(x/3), x) == FiniteSet(0) + b = sqrt(6)*sqrt(log(2))/sqrt(log(5)) + assert solveset_real(5**(x/2) - 2**(3/x), x) == FiniteSet(-b, b) + + # coverage test + C1, C2 = symbols('C1 C2') + f = Function('f') + assert solveset_real(C1 + C2/x**2 - exp(-f(x)), f(x)) == Intersection( + S.Reals, FiniteSet(-log(C1 + C2/x**2))) + y = symbols('y', positive=True) + assert solveset_real(x**2 - y**2/exp(x), y) == Intersection( + S.Reals, FiniteSet(-sqrt(x**2*exp(x)), sqrt(x**2*exp(x)))) + p = Symbol('p', positive=True) + assert solveset_real((1/p + 1)**(p + 1), p).dummy_eq( + ConditionSet(x, Eq((1 + 1/x)**(x + 1), 0), S.Reals)) + assert solveset(2**x - 4**x + 12, x, S.Reals) == {2} + assert solveset(2**x - 2**(2*x) + 12, x, S.Reals) == {2} + + +@XFAIL +def test_exponential_complex(): + n = Dummy('n') + + assert dumeq(solveset_complex(2**x + 4**x, x),imageset( + Lambda(n, I*(2*n*pi + pi)/log(2)), S.Integers)) + assert solveset_complex(x**z*y**z - 2, z) == FiniteSet( + log(2)/(log(x) + log(y))) + assert dumeq(solveset_complex(4**(x/2) - 2**(x/3), x), imageset( + Lambda(n, 3*n*I*pi/log(2)), S.Integers)) + assert dumeq(solveset(2**x + 32, x), imageset( + Lambda(n, (I*(2*n*pi + pi) + 5*log(2))/log(2)), S.Integers)) + + eq = (2**exp(y**2/x) + 2)/(x**2 + 15) + a = sqrt(x)*sqrt(-log(log(2)) + log(log(2) + 2*n*I*pi)) + assert solveset_complex(eq, y) == FiniteSet(-a, a) + + union1 = imageset(Lambda(n, I*(2*n*pi - pi*Rational(2, 3))/log(2)), S.Integers) + union2 = imageset(Lambda(n, I*(2*n*pi + pi*Rational(2, 3))/log(2)), S.Integers) + assert dumeq(solveset(2**x + 4**x + 8**x, x), Union(union1, union2)) + + eq = 4**(x + 1) + 4**(x + 2) + 4**(x - 1) - 3**(x + 2) - 3**(x + 3) + res = solveset(eq, x) + num = 2*n*I*pi - 4*log(2) + 2*log(3) + den = -2*log(2) + log(3) + ans = imageset(Lambda(n, num/den), S.Integers) + assert dumeq(res, ans) + + +def test_expo_conditionset(): + + f1 = (exp(x) + 1)**x - 2 + f2 = (x + 2)**y*x - 3 + f3 = 2**x - exp(x) - 3 + f4 = log(x) - exp(x) + f5 = 2**x + 3**x - 5**x + + assert solveset(f1, x, S.Reals).dummy_eq(ConditionSet( + x, Eq((exp(x) + 1)**x - 2, 0), S.Reals)) + assert solveset(f2, x, S.Reals).dummy_eq(ConditionSet( + x, Eq(x*(x + 2)**y - 3, 0), S.Reals)) + assert solveset(f3, x, S.Reals).dummy_eq(ConditionSet( + x, Eq(2**x - exp(x) - 3, 0), S.Reals)) + assert solveset(f4, x, S.Reals).dummy_eq(ConditionSet( + x, Eq(-exp(x) + log(x), 0), S.Reals)) + assert solveset(f5, x, S.Reals).dummy_eq(ConditionSet( + x, Eq(2**x + 3**x - 5**x, 0), S.Reals)) + + +def test_exponential_symbols(): + x, y, z = symbols('x y z', positive=True) + xr, zr = symbols('xr, zr', real=True) + + assert solveset(z**x - y, x, S.Reals) == Intersection( + S.Reals, FiniteSet(log(y)/log(z))) + + f1 = 2*x**w - 4*y**w + f2 = (x/y)**w - 2 + sol1 = Intersection({log(2)/(log(x) - log(y))}, S.Reals) + sol2 = Intersection({log(2)/log(x/y)}, S.Reals) + assert solveset(f1, w, S.Reals) == sol1, solveset(f1, w, S.Reals) + assert solveset(f2, w, S.Reals) == sol2, solveset(f2, w, S.Reals) + + assert solveset(x**x, x, Interval.Lopen(0,oo)).dummy_eq( + ConditionSet(w, Eq(w**w, 0), Interval.open(0, oo))) + assert solveset(x**y - 1, y, S.Reals) == FiniteSet(0) + assert solveset(exp(x/y)*exp(-z/y) - 2, y, S.Reals) == \ + Complement(ConditionSet(y, Eq(im(x)/y, 0) & Eq(im(z)/y, 0), \ + Complement(Intersection(FiniteSet((x - z)/log(2)), S.Reals), FiniteSet(0))), FiniteSet(0)) + assert solveset(exp(xr/y)*exp(-zr/y) - 2, y, S.Reals) == \ + Complement(FiniteSet((xr - zr)/log(2)), FiniteSet(0)) + + assert solveset(a**x - b**x, x).dummy_eq(ConditionSet( + w, Ne(a, 0) & Ne(b, 0), FiniteSet(0))) + + +def test_ignore_assumptions(): + # make sure assumptions are ignored + xpos = symbols('x', positive=True) + x = symbols('x') + assert solveset_complex(xpos**2 - 4, xpos + ) == solveset_complex(x**2 - 4, x) + + +@XFAIL +def test_issue_10864(): + assert solveset(x**(y*z) - x, x, S.Reals) == FiniteSet(1) + + +@XFAIL +def test_solve_only_exp_2(): + assert solveset_real(sqrt(exp(x)) + sqrt(exp(-x)) - 4, x) == \ + FiniteSet(2*log(-sqrt(3) + 2), 2*log(sqrt(3) + 2)) + + +def test_is_exponential(): + assert _is_exponential(y, x) is False + assert _is_exponential(3**x - 2, x) is True + assert _is_exponential(5**x - 7**(2 - x), x) is True + assert _is_exponential(sin(2**x) - 4*x, x) is False + assert _is_exponential(x**y - z, y) is True + assert _is_exponential(x**y - z, x) is False + assert _is_exponential(2**x + 4**x - 1, x) is True + assert _is_exponential(x**(y*z) - x, x) is False + assert _is_exponential(x**(2*x) - 3**x, x) is False + assert _is_exponential(x**y - y*z, y) is False + assert _is_exponential(x**y - x*z, y) is True + + +def test_solve_exponential(): + assert _solve_exponential(3**(2*x) - 2**(x + 3), 0, x, S.Reals) == \ + FiniteSet(-3*log(2)/(-2*log(3) + log(2))) + assert _solve_exponential(2**y + 4**y, 1, y, S.Reals) == \ + FiniteSet(log(Rational(-1, 2) + sqrt(5)/2)/log(2)) + assert _solve_exponential(2**y + 4**y, 0, y, S.Reals) == \ + S.EmptySet + assert _solve_exponential(2**x + 3**x - 5**x, 0, x, S.Reals) == \ + ConditionSet(x, Eq(2**x + 3**x - 5**x, 0), S.Reals) + +# end of exponential tests + + +# logarithmic tests +def test_logarithmic(): + assert solveset_real(log(x - 3) + log(x + 3), x) == FiniteSet( + -sqrt(10), sqrt(10)) + assert solveset_real(log(x + 1) - log(2*x - 1), x) == FiniteSet(2) + assert solveset_real(log(x + 3) + log(1 + 3/x) - 3, x) == FiniteSet( + -3 + sqrt(-12 + exp(3))*exp(Rational(3, 2))/2 + exp(3)/2, + -sqrt(-12 + exp(3))*exp(Rational(3, 2))/2 - 3 + exp(3)/2) + + eq = z - log(x) + log(y/(x*(-1 + y**2/x**2))) + assert solveset_real(eq, x) == \ + Intersection(S.Reals, FiniteSet(-sqrt(y**2 - y*exp(z)), + sqrt(y**2 - y*exp(z)))) - \ + Intersection(S.Reals, FiniteSet(-sqrt(y**2), sqrt(y**2))) + assert solveset_real( + log(3*x) - log(-x + 1) - log(4*x + 1), x) == FiniteSet(Rational(-1, 2), S.Half) + assert solveset(log(x**y) - y*log(x), x, S.Reals) == S.Reals + +@XFAIL +def test_uselogcombine_2(): + eq = log(exp(2*x) + 1) + log(-tanh(x) + 1) - log(2) + assert solveset_real(eq, x) is S.EmptySet + eq = log(8*x) - log(sqrt(x) + 1) - 2 + assert solveset_real(eq, x) is S.EmptySet + + +def test_is_logarithmic(): + assert _is_logarithmic(y, x) is False + assert _is_logarithmic(log(x), x) is True + assert _is_logarithmic(log(x) - 3, x) is True + assert _is_logarithmic(log(x)*log(y), x) is True + assert _is_logarithmic(log(x)**2, x) is False + assert _is_logarithmic(log(x - 3) + log(x + 3), x) is True + assert _is_logarithmic(log(x**y) - y*log(x), x) is True + assert _is_logarithmic(sin(log(x)), x) is False + assert _is_logarithmic(x + y, x) is False + assert _is_logarithmic(log(3*x) - log(1 - x) + 4, x) is True + assert _is_logarithmic(log(x) + log(y) + x, x) is False + assert _is_logarithmic(log(log(x - 3)) + log(x - 3), x) is True + assert _is_logarithmic(log(log(3) + x) + log(x), x) is True + assert _is_logarithmic(log(x)*(y + 3) + log(x), y) is False + + +def test_solve_logarithm(): + y = Symbol('y') + assert _solve_logarithm(log(x**y) - y*log(x), 0, x, S.Reals) == S.Reals + y = Symbol('y', positive=True) + assert _solve_logarithm(log(x)*log(y), 0, x, S.Reals) == FiniteSet(1) + +# end of logarithmic tests + + +# lambert tests +def test_is_lambert(): + a, b, c = symbols('a,b,c') + assert _is_lambert(x**2, x) is False + assert _is_lambert(a**x**2+b*x+c, x) is True + assert _is_lambert(E**2, x) is False + assert _is_lambert(x*E**2, x) is False + assert _is_lambert(3*log(x) - x*log(3), x) is True + assert _is_lambert(log(log(x - 3)) + log(x-3), x) is True + assert _is_lambert(5*x - 1 + 3*exp(2 - 7*x), x) is True + assert _is_lambert((a/x + exp(x/2)).diff(x, 2), x) is True + assert _is_lambert((x**2 - 2*x + 1).subs(x, (log(x) + 3*x)**2 - 1), x) is True + assert _is_lambert(x*sinh(x) - 1, x) is True + assert _is_lambert(x*cos(x) - 5, x) is True + assert _is_lambert(tanh(x) - 5*x, x) is True + assert _is_lambert(cosh(x) - sinh(x), x) is False + +# end of lambert tests + + +def test_linear_coeffs(): + from sympy.solvers.solveset import linear_coeffs + assert linear_coeffs(0, x) == [0, 0] + assert all(i is S.Zero for i in linear_coeffs(0, x)) + assert linear_coeffs(x + 2*y + 3, x, y) == [1, 2, 3] + assert linear_coeffs(x + 2*y + 3, y, x) == [2, 1, 3] + assert linear_coeffs(x + 2*x**2 + 3, x, x**2) == [1, 2, 3] + raises(ValueError, lambda: + linear_coeffs(x + 2*x**2 + x**3, x, x**2)) + raises(ValueError, lambda: + linear_coeffs(1/x*(x - 1) + 1/x, x)) + raises(ValueError, lambda: + linear_coeffs(x, x, x)) + assert linear_coeffs(a*(x + y), x, y) == [a, a, 0] + assert linear_coeffs(1.0, x, y) == [0, 0, 1.0] + # don't include coefficients of 0 + assert linear_coeffs(Eq(x, x + y), x, y, dict=True) == {y: -1} + assert linear_coeffs(0, x, y, dict=True) == {} + + +def test_is_modular(): + assert _is_modular(y, x) is False + assert _is_modular(Mod(x, 3) - 1, x) is True + assert _is_modular(Mod(x**3 - 3*x**2 - x + 1, 3) - 1, x) is True + assert _is_modular(Mod(exp(x + y), 3) - 2, x) is True + assert _is_modular(Mod(exp(x + y), 3) - log(x), x) is True + assert _is_modular(Mod(x, 3) - 1, y) is False + assert _is_modular(Mod(x, 3)**2 - 5, x) is False + assert _is_modular(Mod(x, 3)**2 - y, x) is False + assert _is_modular(exp(Mod(x, 3)) - 1, x) is False + assert _is_modular(Mod(3, y) - 1, y) is False + + +def test_invert_modular(): + n = Dummy('n', integer=True) + from sympy.solvers.solveset import _invert_modular as invert_modular + + # no solutions + assert invert_modular(Mod(x, 12), S(1)/2, n, x) == (x, S.EmptySet) + # non invertible cases + assert invert_modular(Mod(sin(x), 7), S(5), n, x) == (Mod(sin(x), 7), 5) + assert invert_modular(Mod(exp(x), 7), S(5), n, x) == (Mod(exp(x), 7), 5) + assert invert_modular(Mod(log(x), 7), S(5), n, x) == (Mod(log(x), 7), 5) + # a is symbol + assert dumeq(invert_modular(Mod(x, 7), S(5), n, x), + (x, ImageSet(Lambda(n, 7*n + 5), S.Integers))) + # a.is_Add + assert dumeq(invert_modular(Mod(x + 8, 7), S(5), n, x), + (x, ImageSet(Lambda(n, 7*n + 4), S.Integers))) + assert invert_modular(Mod(x**2 + x, 7), S(5), n, x) == \ + (Mod(x**2 + x, 7), 5) + # a.is_Mul + assert dumeq(invert_modular(Mod(3*x, 7), S(5), n, x), + (x, ImageSet(Lambda(n, 7*n + 4), S.Integers))) + assert invert_modular(Mod((x + 1)*(x + 2), 7), S(5), n, x) == \ + (Mod((x + 1)*(x + 2), 7), 5) + # a.is_Pow + assert invert_modular(Mod(x**4, 7), S(5), n, x) == \ + (x, S.EmptySet) + assert dumeq(invert_modular(Mod(3**x, 4), S(3), n, x), + (x, ImageSet(Lambda(n, 2*n + 1), S.Naturals0))) + assert dumeq(invert_modular(Mod(2**(x**2 + x + 1), 7), S(2), n, x), + (x**2 + x + 1, ImageSet(Lambda(n, 3*n + 1), S.Naturals0))) + assert invert_modular(Mod(sin(x)**4, 7), S(5), n, x) == (x, S.EmptySet) + + +def test_solve_modular(): + n = Dummy('n', integer=True) + # if rhs has symbol (need to be implemented in future). + assert solveset(Mod(x, 4) - x, x, S.Integers + ).dummy_eq( + ConditionSet(x, Eq(-x + Mod(x, 4), 0), + S.Integers)) + # when _invert_modular fails to invert + assert solveset(3 - Mod(sin(x), 7), x, S.Integers + ).dummy_eq( + ConditionSet(x, Eq(Mod(sin(x), 7) - 3, 0), S.Integers)) + assert solveset(3 - Mod(log(x), 7), x, S.Integers + ).dummy_eq( + ConditionSet(x, Eq(Mod(log(x), 7) - 3, 0), S.Integers)) + assert solveset(3 - Mod(exp(x), 7), x, S.Integers + ).dummy_eq(ConditionSet(x, Eq(Mod(exp(x), 7) - 3, 0), + S.Integers)) + # EmptySet solution definitely + assert solveset(7 - Mod(x, 5), x, S.Integers) is S.EmptySet + assert solveset(5 - Mod(x, 5), x, S.Integers) is S.EmptySet + # Negative m + assert dumeq(solveset(2 + Mod(x, -3), x, S.Integers), + ImageSet(Lambda(n, -3*n - 2), S.Integers)) + assert solveset(4 + Mod(x, -3), x, S.Integers) is S.EmptySet + # linear expression in Mod + assert dumeq(solveset(3 - Mod(x, 5), x, S.Integers), + ImageSet(Lambda(n, 5*n + 3), S.Integers)) + assert dumeq(solveset(3 - Mod(5*x - 8, 7), x, S.Integers), + ImageSet(Lambda(n, 7*n + 5), S.Integers)) + assert dumeq(solveset(3 - Mod(5*x, 7), x, S.Integers), + ImageSet(Lambda(n, 7*n + 2), S.Integers)) + # higher degree expression in Mod + assert dumeq(solveset(Mod(x**2, 160) - 9, x, S.Integers), + Union(ImageSet(Lambda(n, 160*n + 3), S.Integers), + ImageSet(Lambda(n, 160*n + 13), S.Integers), + ImageSet(Lambda(n, 160*n + 67), S.Integers), + ImageSet(Lambda(n, 160*n + 77), S.Integers), + ImageSet(Lambda(n, 160*n + 83), S.Integers), + ImageSet(Lambda(n, 160*n + 93), S.Integers), + ImageSet(Lambda(n, 160*n + 147), S.Integers), + ImageSet(Lambda(n, 160*n + 157), S.Integers))) + assert solveset(3 - Mod(x**4, 7), x, S.Integers) is S.EmptySet + assert dumeq(solveset(Mod(x**4, 17) - 13, x, S.Integers), + Union(ImageSet(Lambda(n, 17*n + 3), S.Integers), + ImageSet(Lambda(n, 17*n + 5), S.Integers), + ImageSet(Lambda(n, 17*n + 12), S.Integers), + ImageSet(Lambda(n, 17*n + 14), S.Integers))) + # a.is_Pow tests + assert dumeq(solveset(Mod(7**x, 41) - 15, x, S.Integers), + ImageSet(Lambda(n, 40*n + 3), S.Naturals0)) + assert dumeq(solveset(Mod(12**x, 21) - 18, x, S.Integers), + ImageSet(Lambda(n, 6*n + 2), S.Naturals0)) + assert dumeq(solveset(Mod(3**x, 4) - 3, x, S.Integers), + ImageSet(Lambda(n, 2*n + 1), S.Naturals0)) + assert dumeq(solveset(Mod(2**x, 7) - 2 , x, S.Integers), + ImageSet(Lambda(n, 3*n + 1), S.Naturals0)) + assert dumeq(solveset(Mod(3**(3**x), 4) - 3, x, S.Integers), + Intersection(ImageSet(Lambda(n, Intersection({log(2*n + 1)/log(3)}, + S.Integers)), S.Naturals0), S.Integers)) + # Implemented for m without primitive root + assert solveset(Mod(x**3, 7) - 2, x, S.Integers) is S.EmptySet + assert dumeq(solveset(Mod(x**3, 8) - 1, x, S.Integers), + ImageSet(Lambda(n, 8*n + 1), S.Integers)) + assert dumeq(solveset(Mod(x**4, 9) - 4, x, S.Integers), + Union(ImageSet(Lambda(n, 9*n + 4), S.Integers), + ImageSet(Lambda(n, 9*n + 5), S.Integers))) + # domain intersection + assert dumeq(solveset(3 - Mod(5*x - 8, 7), x, S.Naturals0), + Intersection(ImageSet(Lambda(n, 7*n + 5), S.Integers), S.Naturals0)) + # Complex args + assert solveset(Mod(x, 3) - I, x, S.Integers) == \ + S.EmptySet + assert solveset(Mod(I*x, 3) - 2, x, S.Integers + ).dummy_eq( + ConditionSet(x, Eq(Mod(I*x, 3) - 2, 0), S.Integers)) + assert solveset(Mod(I + x, 3) - 2, x, S.Integers + ).dummy_eq( + ConditionSet(x, Eq(Mod(x + I, 3) - 2, 0), S.Integers)) + + # issue 17373 (https://github.com/sympy/sympy/issues/17373) + assert dumeq(solveset(Mod(x**4, 14) - 11, x, S.Integers), + Union(ImageSet(Lambda(n, 14*n + 3), S.Integers), + ImageSet(Lambda(n, 14*n + 11), S.Integers))) + assert dumeq(solveset(Mod(x**31, 74) - 43, x, S.Integers), + ImageSet(Lambda(n, 74*n + 31), S.Integers)) + + # issue 13178 + n = symbols('n', integer=True) + a = 742938285 + b = 1898888478 + m = 2**31 - 1 + c = 20170816 + assert dumeq(solveset(c - Mod(a**n*b, m), n, S.Integers), + ImageSet(Lambda(n, 2147483646*n + 100), S.Naturals0)) + assert dumeq(solveset(c - Mod(a**n*b, m), n, S.Naturals0), + Intersection(ImageSet(Lambda(n, 2147483646*n + 100), S.Naturals0), + S.Naturals0)) + assert dumeq(solveset(c - Mod(a**(2*n)*b, m), n, S.Integers), + Intersection(ImageSet(Lambda(n, 1073741823*n + 50), S.Naturals0), + S.Integers)) + assert solveset(c - Mod(a**(2*n + 7)*b, m), n, S.Integers) is S.EmptySet + assert dumeq(solveset(c - Mod(a**(n - 4)*b, m), n, S.Integers), + Intersection(ImageSet(Lambda(n, 2147483646*n + 104), S.Naturals0), + S.Integers)) + +# end of modular tests + +def test_issue_17276(): + assert nonlinsolve([Eq(x, 5**(S(1)/5)), Eq(x*y, 25*sqrt(5))], x, y) == \ + FiniteSet((5**(S(1)/5), 25*5**(S(3)/10))) + + +def test_issue_10426(): + x = Dummy('x') + a = Symbol('a') + n = Dummy('n') + assert (solveset(sin(x + a) - sin(x), a)).dummy_eq(Dummy('x')) == (Union( + ImageSet(Lambda(n, 2*n*pi), S.Integers), + Intersection(S.Complexes, ImageSet(Lambda(n, -I*(I*(2*n*pi + arg(-exp(-2*I*x))) + 2*im(x))), + S.Integers)))).dummy_eq(Dummy('x,n')) + + +def test_solveset_conjugate(): + """Test solveset for simple conjugate functions""" + assert solveset(conjugate(x) -3 + I) == FiniteSet(3 + I) + + +def test_issue_18208(): + variables = symbols('x0:16') + symbols('y0:12') + x0, x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15,\ + y0, y1, y2, y3, y4, y5, y6, y7, y8, y9, y10, y11 = variables + + eqs = [x0 + x1 + x2 + x3 - 51, + x0 + x1 + x4 + x5 - 46, + x2 + x3 + x6 + x7 - 39, + x0 + x3 + x4 + x7 - 50, + x1 + x2 + x5 + x6 - 35, + x4 + x5 + x6 + x7 - 34, + x4 + x5 + x8 + x9 - 46, + x10 + x11 + x6 + x7 - 23, + x11 + x4 + x7 + x8 - 25, + x10 + x5 + x6 + x9 - 44, + x10 + x11 + x8 + x9 - 35, + x12 + x13 + x8 + x9 - 35, + x10 + x11 + x14 + x15 - 29, + x11 + x12 + x15 + x8 - 35, + x10 + x13 + x14 + x9 - 29, + x12 + x13 + x14 + x15 - 29, + y0 + y1 + y2 + y3 - 55, + y0 + y1 + y4 + y5 - 53, + y2 + y3 + y6 + y7 - 56, + y0 + y3 + y4 + y7 - 57, + y1 + y2 + y5 + y6 - 52, + y4 + y5 + y6 + y7 - 54, + y4 + y5 + y8 + y9 - 48, + y10 + y11 + y6 + y7 - 60, + y11 + y4 + y7 + y8 - 51, + y10 + y5 + y6 + y9 - 57, + y10 + y11 + y8 + y9 - 54, + x10 - 2, + x11 - 5, + x12 - 1, + x13 - 6, + x14 - 1, + x15 - 21, + y0 - 12, + y1 - 20] + + expected = [38 - x3, x3 - 10, 23 - x3, x3, 12 - x7, x7 + 6, 16 - x7, x7, + 8, 20, 2, 5, 1, 6, 1, 21, 12, 20, -y11 + y9 + 2, y11 - y9 + 21, + -y11 - y7 + y9 + 24, y11 + y7 - y9 - 3, 33 - y7, y7, 27 - y9, y9, + 27 - y11, y11] + + A, b = linear_eq_to_matrix(eqs, variables) + + # solve + solve_expected = {v:eq for v, eq in zip(variables, expected) if v != eq} + + assert solve(eqs, variables) == solve_expected + + # linsolve + linsolve_expected = FiniteSet(Tuple(*expected)) + + assert linsolve(eqs, variables) == linsolve_expected + assert linsolve((A, b), variables) == linsolve_expected + + # gauss_jordan_solve + gj_solve, new_vars = A.gauss_jordan_solve(b) + gj_solve = list(gj_solve) + + gj_expected = linsolve_expected.subs(zip([x3, x7, y7, y9, y11], new_vars)) + + assert FiniteSet(Tuple(*gj_solve)) == gj_expected + + # nonlinsolve + # The solution set of nonlinsolve is currently equivalent to linsolve and is + # also correct. However, we would prefer to use the same symbols as parameters + # for the solution to the underdetermined system in all cases if possible. + # We want a solution that is not just equivalent but also given in the same form. + # This test may be changed should nonlinsolve be modified in this way. + + nonlinsolve_expected = FiniteSet((38 - x3, x3 - 10, 23 - x3, x3, 12 - x7, x7 + 6, + 16 - x7, x7, 8, 20, 2, 5, 1, 6, 1, 21, 12, 20, + -y5 + y7 - 1, y5 - y7 + 24, 21 - y5, y5, 33 - y7, + y7, 27 - y9, y9, -y5 + y7 - y9 + 24, y5 - y7 + y9 + 3)) + + assert nonlinsolve(eqs, variables) == nonlinsolve_expected + + +def test_substitution_with_infeasible_solution(): + a00, a01, a10, a11, l0, l1, l2, l3, m0, m1, m2, m3, m4, m5, m6, m7, c00, c01, c10, c11, p00, p01, p10, p11 = symbols( + 'a00, a01, a10, a11, l0, l1, l2, l3, m0, m1, m2, m3, m4, m5, m6, m7, c00, c01, c10, c11, p00, p01, p10, p11' + ) + solvefor = [p00, p01, p10, p11, c00, c01, c10, c11, m0, m1, m3, l0, l1, l2, l3] + system = [ + -l0 * c00 - l1 * c01 + m0 + c00 + c01, + -l0 * c10 - l1 * c11 + m1, + -l2 * c00 - l3 * c01 + c00 + c01, + -l2 * c10 - l3 * c11 + m3, + -l0 * p00 - l2 * p10 + p00 + p10, + -l1 * p00 - l3 * p10 + p00 + p10, + -l0 * p01 - l2 * p11, + -l1 * p01 - l3 * p11, + -a00 + c00 * p00 + c10 * p01, + -a01 + c01 * p00 + c11 * p01, + -a10 + c00 * p10 + c10 * p11, + -a11 + c01 * p10 + c11 * p11, + -m0 * p00, + -m1 * p01, + -m2 * p10, + -m3 * p11, + -m4 * c00, + -m5 * c01, + -m6 * c10, + -m7 * c11, + m2, + m4, + m5, + m6, + m7 + ] + sol = FiniteSet( + (0, Complement(FiniteSet(p01), FiniteSet(0)), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, l2, l3), + (p00, Complement(FiniteSet(p01), FiniteSet(0)), 0, p11, 0, 0, 0, 0, 0, 0, 0, 1, 1, -p01/p11, -p01/p11), + (0, Complement(FiniteSet(p01), FiniteSet(0)), 0, p11, 0, 0, 0, 0, 0, 0, 0, 1, -l3*p11/p01, -p01/p11, l3), + (0, Complement(FiniteSet(p01), FiniteSet(0)), 0, p11, 0, 0, 0, 0, 0, 0, 0, -l2*p11/p01, -l3*p11/p01, l2, l3), + ) + assert sol != nonlinsolve(system, solvefor) + + +def test_issue_20097(): + assert solveset(1/sqrt(x)) is S.EmptySet + + +def test_issue_15350(): + assert solveset(diff(sqrt(1/x+x))) == FiniteSet(-1, 1) + + +def test_issue_18359(): + c1 = Piecewise((0, x < 0), (Min(1, x)/2 - Min(2, x)/2 + Min(3, x)/2, True)) + c2 = Piecewise((Piecewise((0, x < 0), (Min(1, x)/2 - Min(2, x)/2 + Min(3, x)/2, True)), x >= 0), (0, True)) + correct_result = Interval(1, 2) + result1 = solveset(c1 - Rational(1, 2), x, Interval(0, 3)) + result2 = solveset(c2 - Rational(1, 2), x, Interval(0, 3)) + assert result1 == correct_result + assert result2 == correct_result + + +def test_issue_17604(): + lhs = -2**(3*x/11)*exp(x/11) + pi**(x/11) + assert _is_exponential(lhs, x) + assert _solve_exponential(lhs, 0, x, S.Complexes) == FiniteSet(0) + + +def test_issue_17580(): + assert solveset(1/(1 - x**3)**2, x, S.Reals) is S.EmptySet + + +def test_issue_17566_actual(): + sys = [2**x + 2**y - 3, 4**x + 9**y - 5] + # Not clear this is the correct result, but at least no recursion error + assert nonlinsolve(sys, x, y) == FiniteSet((log(3 - 2**y)/log(2), y)) + + +def test_issue_17565(): + eq = Ge(2*(x - 2)**2/(3*(x + 1)**(Integer(1)/3)) + 2*(x - 2)*(x + 1)**(Integer(2)/3), 0) + res = Union(Interval.Lopen(-1, -Rational(1, 4)), Interval(2, oo)) + assert solveset(eq, x, S.Reals) == res + + +def test_issue_15024(): + function = (x + 5)/sqrt(-x**2 - 10*x) + assert solveset(function, x, S.Reals) == FiniteSet(Integer(-5)) + + +def test_issue_16877(): + assert dumeq(nonlinsolve([x - 1, sin(y)], x, y), + FiniteSet((1, ImageSet(Lambda(n, 2*n*pi), S.Integers)), + (1, ImageSet(Lambda(n, 2*n*pi + pi), S.Integers)))) + # Even better if (1, ImageSet(Lambda(n, n*pi), S.Integers)) is obtained + + +def test_issue_16876(): + assert dumeq(nonlinsolve([sin(x), 2*x - 4*y], x, y), + FiniteSet((ImageSet(Lambda(n, 2*n*pi), S.Integers), + ImageSet(Lambda(n, n*pi), S.Integers)), + (ImageSet(Lambda(n, 2*n*pi + pi), S.Integers), + ImageSet(Lambda(n, n*pi + pi/2), S.Integers)))) + # Even better if (ImageSet(Lambda(n, n*pi), S.Integers), + # ImageSet(Lambda(n, n*pi/2), S.Integers)) is obtained + +def test_issue_21236(): + x, z = symbols("x z") + y = symbols('y', rational=True) + assert solveset(x**y - z, x, S.Reals) == ConditionSet(x, Eq(x**y - z, 0), S.Reals) + e1, e2 = symbols('e1 e2', even=True) + y = e1/e2 # don't know if num or den will be odd and the other even + assert solveset(x**y - z, x, S.Reals) == ConditionSet(x, Eq(x**y - z, 0), S.Reals) + + +def test_issue_21908(): + assert nonlinsolve([(x**2 + 2*x - y**2)*exp(x), -2*y*exp(x)], x, y + ) == {(-2, 0), (0, 0)} + + +def test_issue_19144(): + # test case 1 + expr1 = [x + y - 1, y**2 + 1] + eq1 = [Eq(i, 0) for i in expr1] + soln1 = {(1 - I, I), (1 + I, -I)} + soln_expr1 = nonlinsolve(expr1, [x, y]) + soln_eq1 = nonlinsolve(eq1, [x, y]) + assert soln_eq1 == soln_expr1 == soln1 + # test case 2 - with denoms + expr2 = [x/y - 1, y**2 + 1] + eq2 = [Eq(i, 0) for i in expr2] + soln2 = {(-I, -I), (I, I)} + soln_expr2 = nonlinsolve(expr2, [x, y]) + soln_eq2 = nonlinsolve(eq2, [x, y]) + assert soln_eq2 == soln_expr2 == soln2 + # denominators that cancel in expression + assert nonlinsolve([Eq(x + 1/x, 1/x)], [x]) == FiniteSet((S.EmptySet,)) + + +def test_issue_22413(): + res = nonlinsolve((4*y*(2*x + 2*exp(y) + 1)*exp(2*x), + 4*x*exp(2*x) + 4*y*exp(2*x + y) + 4*exp(2*x + y) + 1), + x, y) + # First solution is not correct, but the issue was an exception + sols = FiniteSet((x, S.Zero), (-exp(y) - S.Half, y)) + assert res == sols + + +def test_issue_23318(): + eqs_eq = [ + Eq(53.5780461486929, x * log(y / (5.0 - y) + 1) / y), + Eq(x, 0.0015 * z), + Eq(0.0015, 7845.32 * y / z), + ] + eqs_expr = [eq.lhs - eq.rhs for eq in eqs_eq] + + sol = {(266.97755814852, 0.0340301680681629, 177985.03876568)} + + assert_close_nl(nonlinsolve(eqs_eq, [x, y, z]), sol) + assert_close_nl(nonlinsolve(eqs_expr, [x, y, z]), sol) + + logterm = log(1.91196789933362e-7*z/(5.0 - 1.91196789933362e-7*z) + 1) + eq = -0.0015*z*logterm + 1.02439504345316e-5*z + assert_close_ss(solveset(eq, z), {0, 177985.038765679}) + + +def test_issue_19814(): + assert nonlinsolve([ 2**m - 2**(2*n), 4*2**m - 2**(4*n)], m, n + ) == FiniteSet((log(2**(2*n))/log(2), S.Complexes)) + + +def test_issue_22058(): + sol = solveset(-sqrt(t)*x**2 + 2*x + sqrt(t), x, S.Reals) + # doesn't fail (and following numerical check) + assert sol.xreplace({t: 1}) == {1 - sqrt(2), 1 + sqrt(2)}, sol.xreplace({t: 1}) + + +def test_issue_11184(): + assert solveset(20*sqrt(y**2 + (sqrt(-(y - 10)*(y + 10)) + 10)**2) - 60, y, S.Reals) is S.EmptySet + + +def test_issue_21890(): + e = S(2)/3 + assert nonlinsolve([4*x**3*y**4 - 2*y, 4*x**4*y**3 - 2*x], x, y) == { + (2**e/(2*y), y), ((-2**e/4 - 2**e*sqrt(3)*I/4)/y, y), + ((-2**e/4 + 2**e*sqrt(3)*I/4)/y, y)} + assert nonlinsolve([(1 - 4*x**2)*exp(-2*x**2 - 2*y**2), + -4*x*y*exp(-2*x**2)*exp(-2*y**2)], x, y) == {(-S(1)/2, 0), (S(1)/2, 0)} + rx, ry = symbols('x y', real=True) + sol = nonlinsolve([4*rx**3*ry**4 - 2*ry, 4*rx**4*ry**3 - 2*rx], rx, ry) + ans = {(2**(S(2)/3)/(2*ry), ry), + ((-2**(S(2)/3)/4 - 2**(S(2)/3)*sqrt(3)*I/4)/ry, ry), + ((-2**(S(2)/3)/4 + 2**(S(2)/3)*sqrt(3)*I/4)/ry, ry)} + assert sol == ans + + +def test_issue_22628(): + assert nonlinsolve([h - 1, k - 1, f - 2, f - 4, -2*k], h, k, f) == S.EmptySet + assert nonlinsolve([x**3 - 1, x + y, x**2 - 4], [x, y]) == S.EmptySet + + +def test_issue_25781(): + assert solve(sqrt(x/2) - x) == [0, S.Half] + + +def test_issue_26077(): + _n = Symbol('_n') + function = x*cot(5*x) + critical_points = stationary_points(function, x, S.Reals) + excluded_points = Union( + ImageSet(Lambda(_n, 2*_n*pi/5), S.Integers), + ImageSet(Lambda(_n, 2*_n*pi/5 + pi/5), S.Integers) + ) + solution = ConditionSet(x, + Eq(x*(-5*cot(5*x)**2 - 5) + cot(5*x), 0), + Complement(S.Reals, excluded_points) + ) + assert solution.as_dummy() == critical_points.as_dummy() diff --git 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b/pllava/lib/python3.10/site-packages/sympy/tensor/array/expressions/tests/test_convert_array_to_indexed.py @@ -0,0 +1,61 @@ +from sympy import Sum, Dummy, sin +from sympy.tensor.array.expressions import ArraySymbol, ArrayTensorProduct, ArrayContraction, PermuteDims, \ + ArrayDiagonal, ArrayAdd, OneArray, ZeroArray, convert_indexed_to_array, ArrayElementwiseApplyFunc, Reshape +from sympy.tensor.array.expressions.from_array_to_indexed import convert_array_to_indexed + +from sympy.abc import i, j, k, l, m, n, o + + +def test_convert_array_to_indexed_main(): + A = ArraySymbol("A", (3, 3, 3)) + B = ArraySymbol("B", (3, 3)) + C = ArraySymbol("C", (3, 3)) + + d_ = Dummy("d_") + + assert convert_array_to_indexed(A, [i, j, k]) == A[i, j, k] + + expr = ArrayTensorProduct(A, B, C) + conv = convert_array_to_indexed(expr, [i,j,k,l,m,n,o]) + assert conv == A[i,j,k]*B[l,m]*C[n,o] + assert convert_indexed_to_array(conv, [i,j,k,l,m,n,o]) == expr + + expr = ArrayContraction(A, (0, 2)) + assert convert_array_to_indexed(expr, [i]).dummy_eq(Sum(A[d_, i, d_], (d_, 0, 2))) + + expr = ArrayDiagonal(A, (0, 2)) + assert convert_array_to_indexed(expr, [i, j]) == A[j, i, j] + + expr = PermuteDims(A, [1, 2, 0]) + conv = convert_array_to_indexed(expr, [i, j, k]) + assert conv == A[k, i, j] + assert convert_indexed_to_array(conv, [i, j, k]) == expr + + expr = ArrayAdd(B, C, PermuteDims(C, [1, 0])) + conv = convert_array_to_indexed(expr, [i, j]) + assert conv == B[i, j] + C[i, j] + C[j, i] + assert convert_indexed_to_array(conv, [i, j]) == expr + + expr = ArrayElementwiseApplyFunc(sin, A) + conv = convert_array_to_indexed(expr, [i, j, k]) + assert conv == sin(A[i, j, k]) + assert convert_indexed_to_array(conv, [i, j, k]).dummy_eq(expr) + + assert convert_array_to_indexed(OneArray(3, 3), [i, j]) == 1 + assert convert_array_to_indexed(ZeroArray(3, 3), [i, j]) == 0 + + expr = Reshape(A, (27,)) + assert convert_array_to_indexed(expr, [i]) == A[i // 9, i // 3 % 3, i % 3] + + X = ArraySymbol("X", (2, 3, 4, 5, 6)) + expr = Reshape(X, (2*3*4*5*6,)) + assert convert_array_to_indexed(expr, [i]) == X[i // 360, i // 120 % 3, i // 30 % 4, i // 6 % 5, i % 6] + + expr = Reshape(X, (4, 9, 2, 2, 5)) + one_index = 180*i + 20*j + 10*k + 5*l + m + expected = X[one_index // (3*4*5*6), one_index // (4*5*6) % 3, one_index // (5*6) % 4, one_index // 6 % 5, one_index % 6] + assert convert_array_to_indexed(expr, [i, j, k, l, m]) == expected + + X = ArraySymbol("X", (2*3*5,)) + expr = Reshape(X, (2, 3, 5)) + assert convert_array_to_indexed(expr, [i, j, k]) == X[15*i + 5*j + k] diff --git a/pllava/lib/python3.10/site-packages/sympy/tensor/array/expressions/tests/test_convert_matrix_to_array.py b/pllava/lib/python3.10/site-packages/sympy/tensor/array/expressions/tests/test_convert_matrix_to_array.py new file mode 100644 index 0000000000000000000000000000000000000000..142585882588df6aa0e4648d9d8881ea755f42a0 --- /dev/null +++ b/pllava/lib/python3.10/site-packages/sympy/tensor/array/expressions/tests/test_convert_matrix_to_array.py @@ -0,0 +1,128 @@ +from sympy import Lambda, KroneckerProduct +from sympy.core.symbol import symbols, Dummy +from sympy.matrices.expressions.hadamard import (HadamardPower, HadamardProduct) +from sympy.matrices.expressions.inverse import Inverse +from sympy.matrices.expressions.matexpr import MatrixSymbol +from sympy.matrices.expressions.matpow import MatPow +from sympy.matrices.expressions.special import Identity +from sympy.matrices.expressions.trace import Trace +from sympy.matrices.expressions.transpose import Transpose +from sympy.tensor.array.expressions.array_expressions import ArrayTensorProduct, ArrayContraction, \ + PermuteDims, ArrayDiagonal, ArrayElementwiseApplyFunc, _array_contraction, _array_tensor_product, Reshape +from sympy.tensor.array.expressions.from_array_to_matrix import convert_array_to_matrix +from sympy.tensor.array.expressions.from_matrix_to_array import convert_matrix_to_array + +i, j, k, l, m, n = symbols("i j k l m n") + +I = Identity(k) + +M = MatrixSymbol("M", k, k) +N = MatrixSymbol("N", k, k) +P = MatrixSymbol("P", k, k) +Q = MatrixSymbol("Q", k, k) + +A = MatrixSymbol("A", k, k) +B = MatrixSymbol("B", k, k) +C = MatrixSymbol("C", k, k) +D = MatrixSymbol("D", k, k) + +X = MatrixSymbol("X", k, k) +Y = MatrixSymbol("Y", k, k) + +a = MatrixSymbol("a", k, 1) +b = MatrixSymbol("b", k, 1) +c = MatrixSymbol("c", k, 1) +d = MatrixSymbol("d", k, 1) + + +def test_arrayexpr_convert_matrix_to_array(): + + expr = M*N + result = ArrayContraction(ArrayTensorProduct(M, N), (1, 2)) + assert convert_matrix_to_array(expr) == result + + expr = M*N*M + result = _array_contraction(ArrayTensorProduct(M, N, M), (1, 2), (3, 4)) + assert convert_matrix_to_array(expr) == result + + expr = Transpose(M) + assert convert_matrix_to_array(expr) == PermuteDims(M, [1, 0]) + + expr = M*Transpose(N) + assert convert_matrix_to_array(expr) == _array_contraction(_array_tensor_product(M, PermuteDims(N, [1, 0])), (1, 2)) + + expr = 3*M*N + res = convert_matrix_to_array(expr) + rexpr = convert_array_to_matrix(res) + assert expr == rexpr + + expr = 3*M + N*M.T*M + 4*k*N + res = convert_matrix_to_array(expr) + rexpr = convert_array_to_matrix(res) + assert expr == rexpr + + expr = Inverse(M)*N + rexpr = convert_array_to_matrix(convert_matrix_to_array(expr)) + assert expr == rexpr + + expr = M**2 + rexpr = convert_array_to_matrix(convert_matrix_to_array(expr)) + assert expr == rexpr + + expr = M*(2*N + 3*M) + res = convert_matrix_to_array(expr) + rexpr = convert_array_to_matrix(res) + assert expr == rexpr + + expr = Trace(M) + result = ArrayContraction(M, (0, 1)) + assert convert_matrix_to_array(expr) == result + + expr = 3*Trace(M) + result = ArrayContraction(ArrayTensorProduct(3, M), (0, 1)) + assert convert_matrix_to_array(expr) == result + + expr = 3*Trace(Trace(M) * M) + result = ArrayContraction(ArrayTensorProduct(3, M, M), (0, 1), (2, 3)) + assert convert_matrix_to_array(expr) == result + + expr = 3*Trace(M)**2 + result = ArrayContraction(ArrayTensorProduct(3, M, M), (0, 1), (2, 3)) + assert convert_matrix_to_array(expr) == result + + expr = HadamardProduct(M, N) + result = ArrayDiagonal(ArrayTensorProduct(M, N), (0, 2), (1, 3)) + assert convert_matrix_to_array(expr) == result + + expr = HadamardProduct(M*N, N*M) + result = ArrayDiagonal(ArrayContraction(ArrayTensorProduct(M, N, N, M), (1, 2), (5, 6)), (0, 2), (1, 3)) + assert convert_matrix_to_array(expr) == result + + expr = HadamardPower(M, 2) + result = ArrayDiagonal(ArrayTensorProduct(M, M), (0, 2), (1, 3)) + assert convert_matrix_to_array(expr) == result + + expr = HadamardPower(M*N, 2) + result = ArrayDiagonal(ArrayContraction(ArrayTensorProduct(M, N, M, N), (1, 2), (5, 6)), (0, 2), (1, 3)) + assert convert_matrix_to_array(expr) == result + + expr = HadamardPower(M, n) + d0 = Dummy("d0") + result = ArrayElementwiseApplyFunc(Lambda(d0, d0**n), M) + assert convert_matrix_to_array(expr).dummy_eq(result) + + expr = M**2 + assert isinstance(expr, MatPow) + assert convert_matrix_to_array(expr) == ArrayContraction(ArrayTensorProduct(M, M), (1, 2)) + + expr = a.T*b + cg = convert_matrix_to_array(expr) + assert cg == ArrayContraction(ArrayTensorProduct(a, b), (0, 2)) + + expr = KroneckerProduct(A, B) + cg = convert_matrix_to_array(expr) + assert cg == Reshape(PermuteDims(ArrayTensorProduct(A, B), [0, 2, 1, 3]), (k**2, k**2)) + + expr = KroneckerProduct(A, B, C, D) + cg = convert_matrix_to_array(expr) + assert cg == Reshape(PermuteDims(ArrayTensorProduct(A, B, C, D), [0, 2, 4, 6, 1, 3, 5, 7]), (k**4, k**4))