# Source code for sympy.solvers.solvers

```
"""
This module contain solvers for all kinds of equations:
- algebraic or transcendental, use solve()
- recurrence, use rsolve()
- differential, use dsolve()
- nonlinear (numerically), use nsolve()
(you will need a good starting point)
"""
from __future__ import print_function, division
from sympy.core.compatibility import (iterable, is_sequence, ordered,
default_sort_key, range)
from sympy.core.sympify import sympify
from sympy.core import S, Add, Symbol, Equality, Dummy, Expr, Mul, Pow
from sympy.core.exprtools import factor_terms
from sympy.core.function import (expand_mul, expand_multinomial, expand_log,
Derivative, AppliedUndef, UndefinedFunction, nfloat,
Function, expand_power_exp, Lambda, _mexpand)
from sympy.integrals.integrals import Integral
from sympy.core.numbers import ilcm, Float
from sympy.core.relational import Relational, Ge
from sympy.core.logic import fuzzy_not
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, powdenest,
nsimplify, denom, logcombine)
from sympy.simplify.sqrtdenest import sqrt_depth
from sympy.simplify.fu import TR1
from sympy.matrices import Matrix, zeros
from sympy.polys import roots, cancel, factor, Poly, together, degree
from sympy.polys.polyerrors import GeneratorsNeeded, PolynomialError
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 uniq, generate_bell, flatten
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 _ispow(e):
"""Return True if e is a Pow or is exp."""
return isinstance(e, Expr) and (e.is_Pow or e.func is 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
>>> from sympy import sqrt
>>> 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:
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)
[docs]def checksol(f, symbol, sol=None, **flags):
"""Checks whether sol is a solution of equation f == 0.
Input can be either a single symbol and corresponding value
or a dictionary of symbols and values. 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):
f = f.lhs - f.rhs
if not f:
return True
if sol and not f.has(*list(sol.keys())):
# if f(y) == 0, x=3 does not set f(y) to zero...nor does it not
return None
illegal = set([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 val.free_symbols:
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 = expand_mul(expand_multinomial(val))
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 = expand_mul(expand_multinomial(val))
if exval.is_number or not exval.free_symbols:
# 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 not val.free_symbols:
return bool(abs(val.n(18).n(12, chop=True)) < 1e-9)
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
[docs]def check_assumptions(expr, against=None, **assumptions):
"""Checks whether expression `expr` satisfies all assumptions.
`assumptions` is a dict of assumptions: {'assumption': True|False, ...}.
Examples
========
>>> from sympy import Symbol, pi, I, exp, check_assumptions
>>> check_assumptions(-5, integer=True)
True
>>> check_assumptions(pi, real=True, integer=False)
True
>>> check_assumptions(pi, real=True, negative=True)
False
>>> check_assumptions(exp(I*pi/7), real=False)
True
>>> x = Symbol('x', real=True, positive=True)
>>> check_assumptions(2*x + 1, real=True, positive=True)
True
>>> check_assumptions(-2*x - 5, real=True, positive=True)
False
To check assumptions of ``expr`` against another variable or expression,
pass the expression or variable as ``against``.
>>> check_assumptions(2*x + 1, x)
True
`None` is returned if check_assumptions() could not conclude.
>>> check_assumptions(2*x - 1, real=True, positive=True)
>>> z = Symbol('z')
>>> check_assumptions(z, real=True)
"""
if against is not None:
assumptions = against.assumptions0
expr = sympify(expr)
result = True
for key, expected in assumptions.items():
if expected is None:
continue
test = getattr(expr, 'is_' + key, None)
if test is expected:
continue
elif test is not None:
return False
result = None # Can't conclude, unless an other test fails.
return result
[docs]def solve(f, *symbols, **flags):
r"""
Algebraically solves equations and systems of equations.
Currently supported are:
- polynomial,
- transcendental
- piecewise combinations of the above
- systems of linear and polynomial equations
- systems containing relational expressions.
Input is formed as:
* f
- a single Expr or Poly that must be zero,
- an Equality
- a Relational expression or 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)'
don't 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, don't do any testing of solutions. This can be
useful if one wants 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, ....
'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 instead 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
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 > 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 one 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 - 6 + 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 1 symbol
* when there is a linear solution
>>> solve(x - y**2, x, y)
[{x: y**2}]
>>> solve(x**2 - y, x, y)
[{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)
[{x: -y}, {x: y}]
>>> solve(x**2 - y**2/exp(x), x, y)
[{x: 2*LambertW(y/2)}]
>>> solve(x**2 - y**2/exp(x), y, x)
[{y: -x*sqrt(exp(x))}, {y: x*sqrt(exp(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: -5*y + 2, 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 doesn't 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}
Notes
=====
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 aren't 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 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, one 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 (issue #8789), 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*_p**5 - 7*_p**3 + 1, 1)**15,
CRootOf(7*_p**5 - 7*_p**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]
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])
implicit = flags.get('implicit', False)
# preprocess equation(s)
###########################################################################
for i, fi in enumerate(f):
if isinstance(fi, Equality):
if 'ImmutableDenseMatrix' in [type(a).__name__ for a in fi.args]:
f[i] = fi.lhs - fi.rhs
else:
f[i] = Add(fi.lhs, -fi.rhs, evaluate=False)
elif isinstance(fi, Poly):
f[i] = fi.as_expr()
elif isinstance(fi, (bool, BooleanAtom)) or fi.is_Relational:
return reduce_inequalities(f, symbols=symbols)
# rewrite hyperbolics in terms of exp
f[i] = f[i].replace(lambda w: isinstance(w, HyperbolicFunction),
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_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]
# 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]
# real/imag handling -----------------------------
w = Dummy('w')
piece = Lambda(w, Piecewise((w, Ge(w, 0)), (-w, True)))
for i, fi in enumerate(f):
# Abs
reps = []
for a in fi.atoms(Abs):
if not a.has(*symbols):
continue
if a.args[0].is_real is None:
raise NotImplementedError('solving %s when the argument '
'is not real or imaginary.' % a)
reps.append((a, piece(a.args[0]) if a.args[0].is_real else \
piece(a.args[0]*S.ImaginaryUnit)))
fi = fi.subs(reps)
# arg
_arg = [a for a in fi.atoms(arg) if a.has(*symbols)]
fi = fi.xreplace(dict(list(zip(_arg,
[atan(im(a.args[0])/re(a.args[0])) for a in _arg]))))
# save changes
f[i] = fi
# see if re(s) or im(s) appear
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 f):
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
symbols_new = []
symbol_swapped = False
for i, s in enumerate(symbols):
if s.is_Symbol:
s_new = s
else:
symbol_swapped = True
s_new = Dummy('X%d' % i)
symbols_new.append(s_new)
if symbol_swapped:
swap_sym = list(zip(symbols, symbols_new))
f = [fi.subs(swap_sym) for fi in f]
symbols = symbols_new
swap_sym = {v: k for k, v in swap_sym}
else:
swap_sym = {}
# 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.has(*symset):
ok = True
else:
free = fi.free_symbols
if not free:
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 d in non_inverts])))
f = [fi.subs(non_inverts) for fi in f]
non_inverts = [(v, k.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)
# 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 dict([(k, v.subs(non_inverts)) for k, v in
solution.items()])
for i in range(1):
if type(solution) is dict:
solution = _do_dict(solution)
break
elif solution and type(solution) is list:
if type(solution[0]) is dict:
solution = [_do_dict(s) for s in solution]
break
elif type(solution[0]) is 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 symbol_swapped:
symbols = [swap_sym[k] for k in symbols]
if type(solution) is dict:
solution = dict([(swap_sym[k], v.subs(swap_sym))
for k, v in solution.items()])
elif solution and type(solution) is list and type(solution[0]) is dict:
for i, sol in enumerate(solution):
solution[i] = dict([(swap_sym[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
type(solution) is not dict and
type(solution[0]) is dict and
all(s in solution[0] for s in symbols)
):
solution = [tuple([r[s].subs(r) 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 type(solution) is tuple:
# this has already been checked and is in as_set form
return solution
elif type(solution) is list:
if type(solution[0]) is 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 type(solution[0]) is 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 type(solution) is 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 type(soln) is dict:
for k in soln:
soln[k] = simplify(soln[k])
elif type(soln) is list:
if type(soln[0]) is dict:
for d in soln:
for k in d:
d[k] = simplify(d[k])
elif type(soln[0]) is tuple:
soln = [tuple(simplify(i) for i in j) for j in soln]
else:
raise TypeError('unrecognized args in list')
elif type(soln) is 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]
# /!\ capture this flag then set it to False so that no checking in
# recursive calls will be done; only the final answer is checked
checkdens = check = flags.pop('check', True)
flags['check'] = False
# build up solutions if f is a Mul
if f.is_Mul:
result = set()
for m in f.args:
if m in set([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 = _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
check = False
flags['simplify'] = False
elif f.is_Piecewise:
result = set()
for n, (expr, cond) in enumerate(f.args):
candidates = _solve(piecewise_fold(expr), symbol, **flags)
for candidate in candidates:
if candidate in result:
continue
try:
v = (cond == True) or cond.subs(symbol, candidate)
except:
v = False
if v != False:
# Only include solutions that do not match the condition
# of any previous pieces.
matches_other_piece = False
for other_n, (other_expr, other_cond) in enumerate(f.args):
if other_n == n:
break
if other_cond == False:
continue
try:
if other_cond.subs(symbol, candidate) == True:
matches_other_piece = True
break
except:
pass
if not matches_other_piece:
v = v == True or v.doit()
if isinstance(v, Relational):
v = v.canonical
result.add(Piecewise(
(candidate, v),
(S.NaN, True)
))
check = 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 S.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]
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:
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 = set(b for b in bases if b.is_Function)
trig = set([_ for _ in funcs if
isinstance(_, TrigonometricFunction)])
other = funcs - trig
if not other and len(funcs.intersection(trig)) > 1:
newf = TR1(f_num).rewrite(tan)
if newf != f_num:
result = _solve(newf, symbol, **flags)
# 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 = dict([(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 []
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)
solved_syms = symbols
except NotImplementedError:
failed.extend([g.as_expr() for g in polys])
solved_syms = []
if result:
# 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]
if result:
if type(result) is 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
if sort:
rv = list(rv)
rv.sort(key=default_sort_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
for eq in ordered(failed, lambda _: len(_ok_syms(_))):
u = Dummy() # used in solution checking
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
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
[docs]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) meaning that ``f`` is independent of the symbols in
``symbols`` that aren't in ``exclude``, e.g::
>>> 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)
(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
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, -2*x + 1)
>>> 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*(-z**2 + 1), 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
dfree = d.free_symbols
# 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 type(derivs[xi]) is 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)):
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.
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(0)
else:
val = solve(k)[0]
if val == 0 and all(v.subs(x, val) == 0 for v in s.values()):
determined[x] = S(1)
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(1)) 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(1)
else:
s[sym] = S(0)
thissol = s
break
if thissol is None:
break
bestsol = thissol
return bestsol
[docs]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. 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 functions is a Nx(M+1) matrix, which means it has
to be in augmented form. If you prefer to enter N equations and M
unknowns then use `solve(Neqs, *Msymbols)` instead. Note: a local
copy of the matrix is made by this routine so the matrix that is
passed will not be modified.
The algorithm used here is fraction-free Gaussian elimination,
which results, after elimination, in an upper-triangular matrix.
Then solutions are found using back-substitution. This approach
is more efficient and compact than the Gauss-Jordan method.
>>> from sympy import Matrix, solve_linear_system
>>> from sympy.abc import x, y
Solve the following system::
x + 4 y == 2
-2 x + y == 14
>>> system = Matrix(( (1, 4, 2), (-2, 1, 14)))
>>> solve_linear_system(system, x, y)
{x: -6, y: 2}
A degenerate system returns an empty dictionary.
>>> system = Matrix(( (0,0,0), (0,0,0) ))
>>> solve_linear_system(system, x, y)
{}
"""
do_simplify = flags.get('simplify', True)
if system.rows == system.cols - 1 == len(symbols):
try:
# well behaved n-equations and n-unknowns
inv = inv_quick(system[:, :-1])
rv = dict(zip(symbols, inv*system[:, -1]))
if do_simplify:
for k, v in rv.items():
rv[k] = simplify(v)
if not all(i.is_zero for i in rv.values()):
# non-trivial solution
return rv
except ValueError:
pass
matrix = system[:, :]
syms = list(symbols)
i, m = 0, matrix.cols - 1 # don't count augmentation
while i < matrix.rows:
if i == m:
# an overdetermined system
if any(matrix[i:, m]):
return None # no solutions
else:
# remove trailing rows
matrix = matrix[:i, :]
break
if not matrix[i, i]:
# there is no pivot in current column
# so try to find one in other columns
for k in range(i + 1, m):
if matrix[i, k]:
break
else:
if matrix[i, m]:
# We need to know if this is always zero or not. We
# assume that if there are free symbols that it is not
# identically zero (or that there is more than one way
# to make this zero). Otherwise, if there are none, this
# is a constant and we assume that it does not simplify
# to zero XXX are there better (fast) ways to test this?
# The .equals(0) method could be used but that can be
# slow; numerical testing is prone to errors of scaling.
if not matrix[i, m].free_symbols:
return None # no solution
# A row of zeros with a non-zero rhs can only be accepted
# if there is another equivalent row. Any such rows will
# be deleted.
nrows = matrix.rows
rowi = matrix.row(i)
ip = None
j = i + 1
while j < matrix.rows:
# do we need to see if the rhs of j
# is a constant multiple of i's rhs?
rowj = matrix.row(j)
if rowj == rowi:
matrix.row_del(j)
elif rowj[:-1] == rowi[:-1]:
if ip is None:
_, ip = rowi[-1].as_content_primitive()
_, jp = rowj[-1].as_content_primitive()
if not (simplify(jp - ip) or simplify(jp + ip)):
matrix.row_del(j)
j += 1
if nrows == matrix.rows:
# no solution
return None
# zero row or was a linear combination of
# other rows or was a row with a symbolic
# expression that matched other rows, e.g. [0, 0, x - y]
# so now we can safely skip it
matrix.row_del(i)
if not matrix:
# every choice of variable values is a solution
# so we return an empty dict instead of None
return dict()
continue
# we want to change the order of colums so
# the order of variables must also change
syms[i], syms[k] = syms[k], syms[i]
matrix.col_swap(i, k)
pivot_inv = S.One/matrix[i, i]
# divide all elements in the current row by the pivot
matrix.row_op(i, lambda x, _: x * pivot_inv)
for k in range(i + 1, matrix.rows):
if matrix[k, i]:
coeff = matrix[k, i]
# subtract from the current row the row containing
# pivot and multiplied by extracted coefficient
matrix.row_op(k, lambda x, j: simplify(x - matrix[i, j]*coeff))
i += 1
# if there weren't any problems, augmented matrix is now
# in row-echelon form so we can check how many solutions
# there are and extract them using back substitution
if len(syms) == matrix.rows:
# this system is Cramer equivalent so there is
# exactly one solution to this system of equations
k, solutions = i - 1, {}
while k >= 0:
content = matrix[k, m]
# run back-substitution for variables
for j in range(k + 1, m):
content -= matrix[k, j]*solutions[syms[j]]
if do_simplify:
solutions[syms[k]] = simplify(content)
else:
solutions[syms[k]] = content
k -= 1
return solutions
elif len(syms) > matrix.rows:
# this system will have infinite number of solutions
# dependent on exactly len(syms) - i parameters
k, solutions = i - 1, {}
while k >= 0:
content = matrix[k, m]
# run back-substitution for variables
for j in range(k + 1, i):
content -= matrix[k, j]*solutions[syms[j]]
# run back-substitution for parameters
for j in range(i, m):
content -= matrix[k, j]*syms[j]
if do_simplify:
solutions[syms[k]] = simplify(content)
else:
solutions[syms[k]] = content
k -= 1
return solutions
else:
return [] # no solutions
[docs]def solve_undetermined_coeffs(equ, coeffs, sym, **flags):
"""Solve equation of a type p(x; a_1, ..., a_k) == q(x) where both
p, q are univariate polynomials and f depends on k parameters.
The result of this functions is a dictionary with symbolic
values of those parameters with respect to coefficients in q.
This functions accepts both Equations class instances and ordinary
SymPy expressions. Specification of parameters and variable is
obligatory for efficiency and simplicity reason.
>>> from sympy import Eq
>>> from sympy.abc import a, b, c, x
>>> from sympy.solvers import solve_undetermined_coeffs
>>> solve_undetermined_coeffs(Eq(2*a*x + a+b, x), [a, b], x)
{a: 1/2, b: -1/2}
>>> solve_undetermined_coeffs(Eq(a*c*x + a+b, x), [a, b], x)
{a: 1/c, b: -1/c}
"""
if isinstance(equ, Equality):
# got equation, so move all the
# terms to the left hand side
equ = equ.lhs - equ.rhs
equ = cancel(equ).as_numer_denom()[0]
system = 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
[docs]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*.
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
========
sympy.matrices.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 det(``M``) by using permutations to select factors.
For size 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
try:
list = M._mat
except AttributeError:
list = flatten(M.tolist())
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 is S.Zero:
raise ValueError("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 f.func is log:
return _solve(f.args[0] - exp(rhs), sym, **flags)
return _tsolve(f - rhs, sym)
elif lhs.is_Pow:
if lhs.exp.is_Integer:
if lhs - rhs != eq:
return _solve(lhs - rhs, sym, **flags)
elif sym not in lhs.exp.free_symbols:
return _solve(lhs.base - rhs**(1/lhs.exp), sym, **flags)
elif not rhs and sym in lhs.exp.free_symbols:
# 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)
if not sol_base:
return sol_base # no solutions to remove so return now
return list(ordered(set(sol_base) - set(
_solve(lhs.exp, sym, **flags))))
elif (rhs is not S.Zero and
lhs.base.is_positive and
lhs.exp.is_real):
return _solve(lhs.exp*log(lhs.base) - log(rhs), sym, **flags)
elif lhs.base == 0 and rhs == 1:
return _solve(lhs.exp, sym, **flags)
elif lhs.is_Mul and rhs.is_positive:
llhs = expand_log(log(lhs))
if llhs.is_Add:
return _solve(llhs - log(rhs), sym, **flags)
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 gi.func is exp or gi.func is 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)
down = g.difference(up_or_log)
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)
return _solve_lambert(lhs - rhs, sym, g)
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(set([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)
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
[docs]def nsolve(*args, **kwargs):
r"""
Solve a nonlinear equation system numerically::
nsolve(f, [args,] x0, modules=['mpmath'], **kwargs)
f is a vector function of symbolic expressions representing the system.
args are the variables. If there is only one variable, this argument can
be omitted.
x0 is a starting vector close to a solution.
Use the modules keyword to specify which modules should be used to
evaluate the function and the Jacobian matrix. Make sure to use a module
that supports matrices. For more information on the syntax, please see the
docstring of lambdify.
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 consistency 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.
>>> from sympy import Symbol, nsolve
>>> import sympy
>>> 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 there 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
>>> from sympy.abc import i
>>> 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 = kwargs.pop('dict', False)
# 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 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 [dict([(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. ``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)
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 S.Zero:
break
lhs = dep
rhs -= indep
# dep * indep == rhs
else:
# this indicates we have done it all
if indep is S.One:
break
lhs = dep
rhs /= indep
# 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
if ai*bi is S.NegativeOne:
if all(
isinstance(i, Function) for i in (ad, bd)) and \
ad.func == bd.func and len(ad.args) == len(bd.args):
if len(ad.args) == 1:
lhs = ad.args[0] - bd.args[0]
else:
# should be able to solve
# f(x, y) == f(2, 3) -> x == 2
# f(x, x + y) == f(2, 3) -> x == 2 or x == 3 - y
raise NotImplementedError('equal function with more than 1 argument')
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 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 lhs.func is atan2:
y, x = lhs.args
lhs = 2*atan(y/(sqrt(x**2 + y**2) + x))
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:
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 re-written 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 4 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, real_roots, solve
>>> 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])
"""
_inv_error = 'cannot get an analytical solution for the inversion'
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 and _take(f, True):
args.append(f.base)
else:
args.append(f)
eq = Mul(*args) # leave as Mul for more efficient solving
# make the sign canonical
free = eq.free_symbols
if len(free) == 1:
if eq.coeff(free.pop()**degree(eq)).could_extract_minus_sign():
eq = -eq
elif eq.could_extract_minus_sign():
eq = -eq
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, take_int_pow):
# 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_Symbol or pow.is_Pow):
continue
b, e = pow.as_base_exp()
if not b.has(*syms):
continue
if not take_int_pow and _Q(pow) == 1:
continue
free = pow.free_symbols
if free.intersection(syms):
return True
return False
_take = flags.setdefault('_take', _take)
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, d = eq.as_numer_denom()
eq = _mexpand(eq, recursive=True)
if eq.is_number:
return
syms = set(syms) or eq.free_symbols
poly = eq.as_poly()
gens = [g for g in poly.gens if _take(g, True)]
if not gens:
return
# check for trivial case
# - already a polynomial in integer powers
if all(_Q(g) == 1 for g in gens):
return
# - an exponent has a symbol of interest (don't handle)
if any(g.as_base_exp()[1].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:
if not _take(g, False):
continue
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)
if not rads:
return
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
gens = [g for g in gens if g.free_symbols & syms]
# 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, False):
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:
free = b.free_symbols
x = {g for g in gens if g.is_Symbol} & free
if not x:
x = free
x = ordered(x)
else:
x = syms
x = list(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
else:
# no longer consider integer powers as generators
gens = [g for g in gens if _Q(g) != 1]
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)
```