""" 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 sympy import divisors, binomial, expand_func from sympy.core.assumptions import check_assumptions from sympy.core.compatibility import (iterable, is_sequence, ordered, default_sort_key) from sympy.core.sympify import sympify from sympy.core import (S, Add, Symbol, Equality, Dummy, Expr, Mul, Pow, Unequality) 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) from sympy.integrals.integrals import Integral from sympy.core.numbers import ilcm, Float, Rational from sympy.core.relational import Relational from sympy.core.logic import fuzzy_not from sympy.core.power import integer_log from sympy.logic.boolalg import And, Or, BooleanAtom from sympy.core.basic import preorder_traversal from sympy.functions import (log, exp, LambertW, cos, sin, tan, acos, asin, atan, Abs, re, im, arg, sqrt, atan2) from sympy.functions.elementary.trigonometric import (TrigonometricFunction, HyperbolicFunction) 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 from sympy.matrices.common import NonInvertibleMatrixError from sympy.matrices import Matrix, zeros from sympy.polys import roots, cancel, factor, Poly from sympy.polys.polyerrors import GeneratorsNeeded, PolynomialError from sympy.polys.solvers import sympy_eqs_to_ring, solve_lin_sys from sympy.functions.elementary.piecewise import piecewise_fold, Piecewise from sympy.utilities.lambdify import lambdify from sympy.utilities.misc import filldedent from sympy.utilities.iterables import (cartes, connected_components, generate_bell, uniq) from sympy.utilities.decorator import conserve_mpmath_dps from mpmath import findroot from sympy.solvers.polysys import solve_poly_system from sympy.solvers.inequalities import reduce_inequalities from types import GeneratorType from collections import defaultdict 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') new_symbols = list(symbols) swap_sym = {} for i, s in enumerate(symbols): if not isinstance(s, Symbol) and s not in swap_sym: swap_sym[s] = Dummy('X%d' % i) new_symbols[i] = swap_sym[s] 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) swap_sym = {v: k for k, v in swap_sym.items()} return new_f, new_symbols, swap_sym 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 for d in Mul.make_args(den): dens.add(d) if not symbols: return dens elif len(symbols) == 1: if iterable(symbols[0]): symbols = symbols[0] rv = [] for d in dens: free = d.free_symbols if any(s in free for s in symbols): rv.append(d) return set(rv) 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 symbols >>> from sympy.solvers import checksol >>> x, y = symbols('x,y') >>> checksol(x**4 - 1, x, 1) True >>> checksol(x**4 - 1, x, 0) False >>> checksol(x**2 + y**2 - 5**2, {x: 3, y: 4}) True 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 if isinstance(f, Poly): f = f.as_expr() elif isinstance(f, (Equality, Unequality)): 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 else: f = f.rewrite(Add, evaluate=False) if isinstance(f, BooleanAtom): return bool(f) elif not f.is_Relational and not f: return True if sol and not f.free_symbols & set(sol.keys()): # if f(y) == 0, x=3 does not set f(y) to zero...nor does it not return None illegal = {S.NaN, S.ComplexInfinity, S.Infinity, S.NegativeInfinity} if any(sympify(v).atoms() & illegal for k, v in sol.items()): return False was = f 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 == was: continue elif 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 was = val 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 Examples ======== The output varies according to the input and can be seen by example: >>> from sympy import solve, Poly, Eq, Function, exp >>> from sympy.abc import x, y, z, a, b >>> f = Function('f') Boolean or univariate Relational: >>> solve(x < 3) (-oo < x) & (x < 3) To always get a list of solution mappings, use flag dict=True: >>> solve(x - 3, dict=True) [{x: 3}] >>> sol = solve([x - 3, y - 1], dict=True) >>> sol [{x: 3, y: 1}] >>> sol[0][x] 3 >>> sol[0][y] 1 To get a list of *symbols* and set of solution(s) use flag set=True: >>> solve([x**2 - 3, y - 1], set=True) ([x, y], {(-sqrt(3), 1), (sqrt(3), 1)}) Single expression and single symbol that is in the expression: >>> solve(x - y, x) [y] >>> solve(x - 3, x) [3] >>> solve(Eq(x, 3), x) [3] >>> solve(Poly(x - 3), x) [3] >>> solve(x**2 - y**2, x, set=True) ([x], {(-y,), (y,)}) >>> solve(x**4 - 1, x, set=True) ([x], {(-1,), (1,), (-I,), (I,)}) Single expression with no symbol that is in the expression: >>> solve(3, x) [] >>> solve(x - 3, y) [] Single expression with no symbol given. In this case, all free *symbols* will be selected as potential *symbols* to solve for. If the equation is univariate then a list of solutions is returned; otherwise - as is the case when *symbols* are given as an iterable of length greater than 1 - a list of mappings will be returned: >>> solve(x - 3) [3] >>> solve(x**2 - y**2) [{x: -y}, {x: y}] >>> solve(z**2*x**2 - z**2*y**2) [{x: -y}, {x: y}, {z: 0}] >>> solve(z**2*x - z**2*y**2) [{x: y**2}, {z: 0}] 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: >>> 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, sqrt >>> 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 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 that can be targeted with 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] * To solve for a function within a derivative, use ``dsolve``. Single expression and more than one symbol: * When there is a linear solution: >>> solve(x - y**2, x, y) [(y**2, y)] >>> solve(x**2 - y, x, y) [(x, x**2)] >>> solve(x**2 - y, x, y, dict=True) [{y: x**2}] * When undetermined coefficients are identified: * That are linear: >>> solve((a + b)*x - b + 2, a, b) {a: -2, b: 2} * That are nonlinear: >>> solve((a + b)*x - b**2 + 2, a, b, set=True) ([a, b], {(-sqrt(2), sqrt(2)), (sqrt(2), -sqrt(2))}) * If there is no linear solution, then the first successful attempt for a nonlinear solution will be returned: >>> solve(x**2 - y**2, x, y, dict=True) [{x: -y}, {x: y}] >>> solve(x**2 - y**2/exp(x), x, y, dict=True) [{x: 2*LambertW(-y/2)}, {x: 2*LambertW(y/2)}] >>> solve(x**2 - y**2/exp(x), y, x) [(-x*sqrt(exp(x)), x), (x*sqrt(exp(x)), x)] Iterable of one or more of the above: * Involving relationals or bools: >>> solve([x < 3, x - 2]) Eq(x, 2) >>> solve([x > 3, x - 2]) False * When the system is linear: * With a solution: >>> solve([x - 3], x) {x: 3} >>> solve((x + 5*y - 2, -3*x + 6*y - 15), x, y) {x: -3, y: 1} >>> solve((x + 5*y - 2, -3*x + 6*y - 15), x, y, z) {x: -3, y: 1} >>> solve((x + 5*y - 2, -3*x + 6*y - z), z, x, y) {x: 2 - 5*y, z: 21*y - 6} * Without a solution: >>> solve([x + 3, x - 3]) [] * When the system is not linear: >>> solve([x**2 + y -2, y**2 - 4], x, y, set=True) ([x, y], {(-2, -2), (0, 2), (2, -2)}) * If no *symbols* are given, all free *symbols* will be selected and a list of mappings returned: >>> solve([x - 2, x**2 + y]) [{x: 2, y: -4}] >>> solve([x - 2, x**2 + f(x)], {f(x), x}) [{x: 2, f(x): -4}] * 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} **Additional Examples** ``solve()`` with check=True (default) will run through the symbol tags to elimate unwanted solutions. If no assumptions are included, all possible solutions will be returned: >>> from sympy import Symbol, solve >>> x = Symbol("x") >>> solve(x**2 - 1) [-1, 1] By using the positive tag, only one solution will be returned: >>> pos = Symbol("pos", positive=True) >>> solve(pos**2 - 1) [1] Assumptions are not checked when ``solve()`` input involves relationals or bools. 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: x = 0. In this case, this is a spurious solution since $\sin(x)/x$ has the well known limit (without dicontinuity) 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 **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) When using particular=True, use a fast heuristic to find a solution with many zeros (instead of using the very slow method guaranteed to find the largest number of zeros possible). cubics=True (default) Return explicit solutions when cubic expressions are encountered. quartics=True (default) Return explicit solutions when quartic expressions are encountered. 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 """ # keeping track of how f was passed since if it is a list # a dictionary of results will be returned. ########################################################################### def _sympified_list(w): return list(map(sympify, w if iterable(w) else [w])) bare_f = not iterable(f) ordered_symbols = (symbols and symbols[0] and (isinstance(symbols[0], Symbol) or is_sequence(symbols[0], include=GeneratorType) ) ) f, symbols = (_sympified_list(w) for w in [f, symbols]) if isinstance(f, list): f = [s for s in f if s is not S.true and s is not True] implicit = flags.get('implicit', False) # preprocess symbol(s) ########################################################################### 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): flags['dict'] = True # better show symbols symbols.add(p) pot.skip() # don't go any deeper symbols = list(symbols) ordered_symbols = False elif len(symbols) == 1 and iterable(symbols[0]): symbols = symbols[0] # remove symbols the user is not interested in 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) ########################################################################### for i, fi in enumerate(f): if isinstance(fi, (Equality, Unequality)): 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, Unequality): 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. ''')) else: fi = fi.rewrite(Add, evaluate=False) f[i] = fi if fi.is_Relational: return reduce_inequalities(f, symbols=symbols) if isinstance(fi, Poly): f[i] = fi.as_expr() # rewrite hyperbolics in terms of exp f[i] = f[i].replace(lambda w: isinstance(w, HyperbolicFunction) and \ (len(w.free_symbols & set(symbols)) > 0), 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 flags.get('set', False): 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: for i, fi in enumerate(f): f[i] = fi.xreplace({s: rhs}) f.append(s - rhs) symbols.extend([re(s), im(s)]) if bare_f: bare_f = False flags['dict'] = True # end of real/imag handling ----------------------------- symbols = list(uniq(symbols)) if not ordered_symbols: # we do this to make the results returned canonical in case f # contains a system of nonlinear equations; all other cases should # be unambiguous symbols = sorted(symbols, key=default_sort_key) # we can solve for non-symbol entities by replacing them with Dummy symbols f, symbols, swap_sym = recast_to_symbols(f, symbols) # this is needed in the next two events 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: 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 not p 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) # # try to get a solution ########################################################################### if bare_f: solution = _solve(f[0], *symbols, **flags) else: solution = _solve_system(f, symbols, **flags) # # postprocessing ########################################################################### # Restore masked-off objects if non_inverts: def _do_dict(solution): return {k: v.subs(non_inverts) for k, v in solution.items()} for i in range(1): if isinstance(solution, dict): solution = _do_dict(solution) break elif solution and isinstance(solution, list): if isinstance(solution[0], dict): solution = [_do_dict(s) for s in solution] break elif isinstance(solution[0], tuple): solution = [tuple([v.subs(non_inverts) for v in s]) for s in solution] break else: solution = [v.subs(non_inverts) for v in solution] break elif not solution: break else: raise NotImplementedError(filldedent(''' no handling of %s was implemented''' % 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] if isinstance(solution, dict): solution = {swap_sym.get(k, k): v.subs(swap_sym) for k, v in solution.items()} elif solution and isinstance(solution, list) and isinstance(solution[0], dict): for i, sol in enumerate(solution): solution[i] = {swap_sym.get(k, k): v.subs(swap_sym) for k, v in sol.items()} # undo the dictionary solutions returned when the system was only partially # solved with poly-system if all symbols are present if ( not flags.get('dict', False) and solution and ordered_symbols and not isinstance(solution, dict) and all(isinstance(sol, dict) for sol in solution) ): solution = [tuple([r.get(s, s) for s in symbols]) for r in solution] # 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) 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 if isinstance(solution, tuple): # this has already been checked and is in as_set form return solution elif isinstance(solution, list): if isinstance(solution[0], tuple): for sol in solution: for symb, val in zip(symbols, sol): test = check_assumptions(val, **symb.assumptions0) if test is False: break if test is None: got_None.append(sol) else: no_False.append(sol) elif isinstance(solution[0], dict): 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) else: # list of expressions for sol in solution: test = check_assumptions(sol, **symbols[0].assumptions0) if test is False: continue no_False.append(sol) if test is None: got_None.append(sol) elif isinstance(solution, dict): a_None = False for symb, val in solution.items(): test = check_assumptions(val, **symb.assumptions0) if test: continue if test is False: no_False = None break a_None = True else: no_False = solution if a_None: got_None.append(solution) elif isinstance(solution, (Relational, And, Or)): if len(symbols) != 1: raise ValueError("Length should be 1") if warn and symbols[0].assumptions0: warnings.warn(filldedent(""" \tWarning: assumptions about variable '%s' are not handled currently.""" % symbols[0])) # TODO: check also variable assumptions for inequalities else: raise TypeError('Unrecognized solution') # improve the checker solution = no_False if warn and got_None: warnings.warn(filldedent(""" \tWarning: assumptions concerning following solution(s) can't be checked:""" + '\n\t' + ', '.join(str(s) for s in got_None))) # # done ########################################################################### as_dict = flags.get('dict', False) as_set = flags.get('set', False) if not as_set and isinstance(solution, list): # Make sure that a list of solutions is ordered in a canonical way. solution.sort(key=default_sort_key) if not as_dict and not as_set: return solution or [] # return a list of mappings or [] if not solution: solution = [] else: if isinstance(solution, dict): solution = [solution] elif iterable(solution[0]): solution = [dict(list(zip(symbols, s))) for s in solution] elif isinstance(solution[0], dict): pass else: if len(symbols) != 1: raise ValueError("Length should be 1") solution = [{symbols[0]: s} for s in solution] if as_dict: return solution assert as_set if not solution: return [], set() k = list(ordered(solution[0].keys())) return k, {tuple([s[ki] for ki in k]) for s in solution} def _solve(f, *symbols, **flags): """ Return a checked solution for *f* in terms of one or more of the symbols. A list should be returned except for the case when a linear undetermined-coefficients equation is encountered (in which case a dictionary is returned). 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: soln = None free = f.free_symbols ex = free - set(symbols) if len(ex) != 1: ind, dep = f.as_independent(*symbols) ex = ind.free_symbols & dep.free_symbols if len(ex) == 1: ex = ex.pop() try: # soln may come back as dict, list of dicts or tuples, or # tuple of symbol list and set of solution tuples soln = solve_undetermined_coeffs(f, symbols, ex, **flags) except NotImplementedError: pass if soln: if flags.get('simplify', True): if isinstance(soln, dict): for k in soln: soln[k] = simplify(soln[k]) elif isinstance(soln, list): if isinstance(soln[0], dict): for d in soln: for k in d: d[k] = simplify(d[k]) elif isinstance(soln[0], tuple): soln = [tuple(simplify(i) for i in j) for j in soln] else: raise TypeError('unrecognized args in list') elif isinstance(soln, tuple): sym, sols = soln soln = sym, {tuple(simplify(i) for i in j) for j in sols} else: raise TypeError('unrecognized solution type') return soln # find first successful solution failed = [] got_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 got_s and any([ss in vfree for ss in got_s]): # sol depends on previously solved symbols: discard it continue got_s.add(xi) result.append({xi: v}) elif xi: # there might be a non-linear solution if xi is not 0 failed.append(s) if not failed: return result for s in failed: try: soln = _solve(f, s, **flags) 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 got_s.add(s) result.append({s: sol}) except NotImplementedError: continue if got_s: return result else: raise NotImplementedError(not_impl_msg % f) 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)) # /!\ capture this flag then set it to False so that no checking in # recursive calls will be done; only the final answer is checked 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 = _solve(m, symbol, **flags) result.update(set(soln)) result = list(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 all(not checksol(den, {symbol: 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() for i, (expr, cond) in enumerate(f.args): if expr.is_zero: raise NotImplementedError( 'solve cannot represent interval solutions') candidates = _solve(expr, symbol, **flags) # the explicit condition for this expr is the current cond # and none of the previous conditions args = [~c for _, c in f.args[:i]] + [cond] cond = And(*args) 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 simpification 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))) # set flags for quick exit at end; solutions for each # piece were already checked and simplified check = False flags['simplify'] = False 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 [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 = _solve(t - f1, symbol, **flags)[0] sols = list() for sol in cv_sols: sols.append(cv_inv.subs(t, sol)) result = list(ordered(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 = _solve(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) sols = list() for sol in soln: for i in inv: sols.append(i.subs(t, sol)) result = list(ordered(sols)) 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 solvers = {k: flags.get(k, True) for k in ('cubics', 'quartics', 'quintics')} 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') iv = _solve(u - t, symbol, **flags) soln = list(ordered({i.subs(t, s) for i in iv 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) result = 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 = _solve(ieq, symbol, **flags)[0] rv = {inv.subs(isym, xi) for xi in _solve(eq, isym, **flags)} else: try: rv = set(_solve(eq, symbol, **flags)) except NotImplementedError: rv = None if rv is not None: result = list(ordered(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 = soln except PolynomialError: pass # ----------- end of fallback ---------------------------- if result is False: raise NotImplementedError('\n'.join([msg, not_impl_msg % f])) if flags.get('simplify', True): result = list(map(simplify, 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 = [s for s in result if all(not checksol(d, {symbol: s}, **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, {symbol: r}, **flags) is not False] return result def _solve_system(exprs, symbols, **flags): if not exprs: return [] 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 = [] for subexpr in subexprs: subsyms = set() for e in subexpr: subsyms |= exprsyms[e] subsyms = list(sorted(subsyms, key = lambda x: sym_indices[x])) flags['_split'] = False # skip split step subsol = _solve_system(subexpr, subsyms, **flags) if not isinstance(subsol, list): subsol = [subsol] subsols.append(subsol) # Full solution is cartesion product of subsystems sols = [] for soldicts in cartes(*subsols): sols.append(dict(item for sd in soldicts for item in sd.items())) # Return a list with one dict as just the dict if len(sols) == 1: return sols[0] return sols polys = [] dens = set() failed = [] result = False linear = False 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) 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 not polys: solved_syms = [] else: 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) if failed: if result: solved_syms = list(result.keys()) else: solved_syms = [] else: linear = True else: if len(symbols) > len(polys): from sympy.utilities.iterables import subsets 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 [] or list of tuples of solutions for syms res = solve_poly_system(polys, *syms) if res: for r in 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.extend([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 # we don't know here if the symbols provided # were given or not, so let solve resolve that. # A list of dictionaries is going to always be # returned from here. result = [dict(list(zip(solved_syms, r))) for r in result] except NotImplementedError: failed.extend([g.as_expr() for g in polys]) solved_syms = [] result = None if result: if isinstance(result, dict): result = [result] else: result = [{}] if failed: # 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 - solved_syms) & 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 solved_syms = set(solved_syms) # set of symbols we have solved for 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 = [] got_s = set() hit = False for r in result: # 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 = _solve(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 in 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) default_simplify = bool(failed) # rely on system-solvers to simplify 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] if linear and result: result = result[0] return 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.core.power import Pow >>> from sympy.polys.polytools import cancel ``f`` is independent of the symbols in *symbols* that are not in *exclude*: >>> from sympy.solvers.solvers import solve_linear >>> from sympy.abc import x, y, z >>> from sympy import cos, sin >>> 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, Equality): 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))) x = max(k.free_symbols, key=default_sort_key) if len(k.free_symbols) != 1: determined[x] = S.Zero else: val = solve(k)[0] if val == 0 and all(v.subs(x, val) == 0 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. from itertools import combinations from sympy.utilities.misc import debug 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): debug('minsolve: %s' % n) thissol = None for nonzeros in combinations(list(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, sym, **flags): r""" Solve equation of a type $p(x; a_1, \ldots, a_k) = q(x)$ where both $p$ and $q$ are univariate polynomials that depend on $k$ parameters. Explanation =========== The result of this function is a dictionary with symbolic values of those parameters with respect to coefficients in $q$. This function accepts both equations class instances and ordinary SymPy expressions. Specification of parameters and variables is obligatory for efficiency and simplicity reasons. Examples ======== >>> from sympy import Eq >>> from sympy.abc import a, b, c, x >>> from sympy.solvers import solve_undetermined_coeffs >>> solve_undetermined_coeffs(Eq(2*a*x + a+b, x), [a, b], x) {a: 1/2, b: -1/2} >>> solve_undetermined_coeffs(Eq(a*c*x + a+b, x), [a, b], x) {a: 1/c, b: -1/c} """ if isinstance(equ, Equality): # got equation, so move all the # terms to the left hand side equ = equ.lhs - equ.rhs equ = cancel(equ).as_numer_denom()[0] system = list(collect(equ.expand(), sym, evaluate=False).values()) if not any(equ.has(sym) for equ in system): # consecutive powers in the input expressions have # been successfully collected, so solve remaining # system using Gaussian elimination algorithm return solve(system, *coeffs, **flags) else: return None # no solutions def solve_linear_system_LU(matrix, syms): """ 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 >>> from sympy.abc import x, y, z >>> from sympy.solvers.solvers import solve_linear_system_LU >>> 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. """ from sympy.matrices import zeros 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 _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 >>> from sympy.solvers.solvers import _tsolve as tsolve >>> from sympy.abc import x >>> 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 _solve(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 _solve(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 _solve(lhs - rhs, sym, **flags) if sym not in lhs.exp.free_symbols: return _solve(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 = _solve(lhs.base, sym, **flags) return [s for s in sol_base if lhs.exp.subs(sym, s) != 0] # a ** g(x) == b if not lhs.base.has(sym): if lhs.base == 0: return _solve(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 = _solve(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 sol = _solve(exp(log(lhs.base)*lhs.exp)-exp(log(rhs)), 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) 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(_solve(lhs.exp - e_rat, sym, **flags)) return list(ordered(set(sol))) # f(x) ** g(x) == c else: sol = [] logform = lhs.exp*log(lhs.base) - log(rhs) if logform != lhs - rhs: try: sol.extend(_solve(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(_solve(lhs.exp, sym, **flags)) check.extend(_solve(lhs.base - 1, sym, **flags)) check.extend(_solve(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(_solve(lhs.base - r, sym, **flags)) check.extend(_solve(lhs.base + r, sym, **flags)) check.extend(_solve(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 (_solve(lhs.base - b, sym, **flags))] check.extend([sqrtdenest(i) for i in (_solve(lhs.exp - e, sym, **flags))]) if e_l*d != 1: check.extend(_solve(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(ordered(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[lhs.func](rhs): soln.extend(_solve(lhs.args[0] - i, sym, **flags)) return list(ordered(soln)) elif lhs.func == LambertW: return _solve(lhs.args[0] - rhs*exp(rhs), sym, **flags) rewrite = lhs.rewrite(exp) if rewrite != lhs: return _solve(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 g = _filtered_gens(eq.as_poly(), sym) up_or_log = set() for gi in g: if isinstance(gi, exp) or (gi.is_Pow and gi.base == S.Exp1) or isinstance(gi, log): 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 = _solve(p, u, **flags) return list(ordered({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 = _solve(pos, u, **flags) return list(ordered([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: prec = kwargs.pop('prec') import mpmath mpmath.mp.dps = prec else: prec = None # 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, Equality): 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, Equality): f = f.lhs - f.rhs 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) (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: # a = -b lhs = powsimp(powdenest(ad/bd)) 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]) """ from sympy import Equality as Eq uflags = dict(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, _solve(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(dict(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) # - 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(list(zip(rads, list(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 = _solve(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 = _solve(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, _solve(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(dict(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 from sympy.solvers.bivariate import ( bivariate_type, _solve_lambert, _filtered_gens)