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--- /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)