prjycq8up
/
image-handler
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity
You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
Branches Tags
${ item.name }
Create tag ${ searchTerm }
Create branch ${ searchTerm }
from '1cf58eadb3'
${ noResults }
image-handler/image-handler10.0/venv/lib/site-packages/sympy/solvers/solvers.py

3637 lines
133 KiB
Raw Normal View History Unescape

first commit
5 months ago
"""
This module contain solvers for all kinds of equations:
- algebraic or transcendental, use solve()
- recurrence, use rsolve()
- differential, use dsolve()
- nonlinear (numerically), use nsolve()
(you will need a good starting point)
"""
from __future__ import annotations
from sympy.core import (S, Add, Symbol, Dummy, Expr, Mul)
from sympy.core.assumptions import check_assumptions
from sympy.core.exprtools import factor_terms
from sympy.core.function import (expand_mul, expand_log, Derivative,
AppliedUndef, UndefinedFunction, nfloat,
Function, expand_power_exp, _mexpand, expand,
expand_func)
from sympy.core.logic import fuzzy_not
from sympy.core.numbers import ilcm, Float, Rational, _illegal
from sympy.core.power import integer_log, Pow
from sympy.core.relational import Eq, Ne
from sympy.core.sorting import ordered, default_sort_key
from sympy.core.sympify import sympify, _sympify
from sympy.core.traversal import preorder_traversal
from sympy.logic.boolalg import And, BooleanAtom
from sympy.functions import (log, exp, LambertW, cos, sin, tan, acos, asin, atan,
Abs, re, im, arg, sqrt, atan2)
from sympy.functions.combinatorial.factorials import binomial
from sympy.functions.elementary.hyperbolic import HyperbolicFunction
from sympy.functions.elementary.piecewise import piecewise_fold, Piecewise
from sympy.functions.elementary.trigonometric import TrigonometricFunction
from sympy.integrals.integrals import Integral
from sympy.ntheory.factor_ import divisors
from sympy.simplify import (simplify, collect, powsimp, posify, # type: ignore
powdenest, nsimplify, denom, logcombine, sqrtdenest, fraction,
separatevars)
from sympy.simplify.sqrtdenest import sqrt_depth
from sympy.simplify.fu import TR1, TR2i
from sympy.matrices.common import NonInvertibleMatrixError
from sympy.matrices import Matrix, zeros
from sympy.polys import roots, cancel, factor, Poly
from sympy.polys.polyerrors import GeneratorsNeeded, PolynomialError
from sympy.polys.solvers import sympy_eqs_to_ring, solve_lin_sys
from sympy.utilities.lambdify import lambdify
from sympy.utilities.misc import filldedent, debugf
from sympy.utilities.iterables import (connected_components,
generate_bell, uniq, iterable, is_sequence, subsets, flatten)
from sympy.utilities.decorator import conserve_mpmath_dps
from mpmath import findroot
from sympy.solvers.polysys import solve_poly_system
from types import GeneratorType
from collections import defaultdict
from itertools import combinations, product
import warnings
def recast_to_symbols(eqs, symbols):
"""
Return (e, s, d) where e and s are versions of *eqs* and
*symbols* in which any non-Symbol objects in *symbols* have
been replaced with generic Dummy symbols and d is a dictionary
that can be used to restore the original expressions.
Examples
========
>>> from sympy.solvers.solvers import recast_to_symbols
>>> from sympy import symbols, Function
>>> x, y = symbols('x y')
>>> fx = Function('f')(x)
>>> eqs, syms = [fx + 1, x, y], [fx, y]
>>> e, s, d = recast_to_symbols(eqs, syms); (e, s, d)
([_X0 + 1, x, y], [_X0, y], {_X0: f(x)})
The original equations and symbols can be restored using d:
>>> assert [i.xreplace(d) for i in eqs] == eqs
>>> assert [d.get(i, i) for i in s] == syms
"""
if not iterable(eqs) and iterable(symbols):
raise ValueError('Both eqs and symbols must be iterable')
orig = list(symbols)
symbols = list(ordered(symbols))
swap_sym = {}
i = 0
for j, s in enumerate(symbols):
if not isinstance(s, Symbol) and s not in swap_sym:
swap_sym[s] = Dummy('X%d' % i)
i += 1
new_f = []
for i in eqs:
isubs = getattr(i, 'subs', None)
if isubs is not None:
new_f.append(isubs(swap_sym))
else:
new_f.append(i)
restore = {v: k for k, v in swap_sym.items()}
return new_f, [swap_sym.get(i, i) for i in orig], restore
def _ispow(e):
"""Return True if e is a Pow or is exp."""
return isinstance(e, Expr) and (e.is_Pow or isinstance(e, exp))
def _simple_dens(f, symbols):
# when checking if a denominator is zero, we can just check the
# base of powers with nonzero exponents since if the base is zero
# the power will be zero, too. To keep it simple and fast, we
# limit simplification to exponents that are Numbers
dens = set()
for d in denoms(f, symbols):
if d.is_Pow and d.exp.is_Number:
if d.exp.is_zero:
continue # foo**0 is never 0
d = d.base
dens.add(d)
return dens
def denoms(eq, *symbols):
"""
Return (recursively) set of all denominators that appear in *eq*
that contain any symbol in *symbols*; if *symbols* are not
provided then all denominators will be returned.
Examples
========
>>> from sympy.solvers.solvers import denoms
>>> from sympy.abc import x, y, z
>>> denoms(x/y)
{y}
>>> denoms(x/(y*z))
{y, z}
>>> denoms(3/x + y/z)
{x, z}
>>> denoms(x/2 + y/z)
{2, z}
If *symbols* are provided then only denominators containing
those symbols will be returned:
>>> denoms(1/x + 1/y + 1/z, y, z)
{y, z}
"""
pot = preorder_traversal(eq)
dens = set()
for p in pot:
# Here p might be Tuple or Relational
# Expr subtrees (e.g. lhs and rhs) will be traversed after by pot
if not isinstance(p, Expr):
continue
den = denom(p)
if den is S.One:
continue
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]
return {d for d in dens if any(s in d.free_symbols for s in symbols)}
def checksol(f, symbol, sol=None, **flags):
"""
Checks whether sol is a solution of equation f == 0.
Explanation
===========
Input can be either a single symbol and corresponding value
or a dictionary of symbols and values. When given as a dictionary
and flag ``simplify=True``, the values in the dictionary will be
simplified. *f* can be a single equation or an iterable of equations.
A solution must satisfy all equations in *f* to be considered valid;
if a solution does not satisfy any equation, False is returned; if one or
more checks are inconclusive (and none are False) then None is returned.
Examples
========
>>> from sympy import checksol, symbols
>>> x, y = symbols('x,y')
>>> checksol(x**4 - 1, x, 1)
True
>>> checksol(x**4 - 1, x, 0)
False
>>> checksol(x**2 + y**2 - 5**2, {x: 3, y: 4})
True
To check if an expression is zero using ``checksol()``, pass it
as *f* and send an empty dictionary for *symbol*:
>>> checksol(x**2 + x - x*(x + 1), {})
True
None is returned if ``checksol()`` could not conclude.
flags:
'numerical=True (default)'
do a fast numerical check if ``f`` has only one symbol.
'minimal=True (default is False)'
a very fast, minimal testing.
'warn=True (default is False)'
show a warning if checksol() could not conclude.
'simplify=True (default)'
simplify solution before substituting into function and
simplify the function before trying specific simplifications
'force=True (default is False)'
make positive all symbols without assumptions regarding sign.
"""
from sympy.physics.units import Unit
minimal = flags.get('minimal', False)
if sol is not None:
sol = {symbol: sol}
elif isinstance(symbol, dict):
sol = symbol
else:
msg = 'Expecting (sym, val) or ({sym: val}, None) but got (%s, %s)'
raise ValueError(msg % (symbol, sol))
if iterable(f):
if not f:
raise ValueError('no functions to check')
rv = True
for fi in f:
check = checksol(fi, sol, **flags)
if check:
continue
if check is False:
return False
rv = None # don't return, wait to see if there's a False
return rv
f = _sympify(f)
if f.is_number:
return f.is_zero
if isinstance(f, Poly):
f = f.as_expr()
elif isinstance(f, (Eq, Ne)):
if f.rhs in (S.true, S.false):
f = f.reversed
B, E = f.args
if isinstance(B, BooleanAtom):
f = f.subs(sol)
if not f.is_Boolean:
return
else:
f = f.rewrite(Add, evaluate=False, deep=False)
if isinstance(f, BooleanAtom):
return bool(f)
elif not f.is_Relational and not f:
return True
illegal = set(_illegal)
if any(sympify(v).atoms() & illegal for k, v in sol.items()):
return False
attempt = -1
numerical = flags.get('numerical', True)
while 1:
attempt += 1
if attempt == 0:
val = f.subs(sol)
if isinstance(val, Mul):
val = val.as_independent(Unit)[0]
if val.atoms() & illegal:
return False
elif attempt == 1:
if not val.is_number:
if not val.is_constant(*list(sol.keys()), simplify=not minimal):
return False
# there are free symbols -- simple expansion might work
_, val = val.as_content_primitive()
val = _mexpand(val.as_numer_denom()[0], recursive=True)
elif attempt == 2:
if minimal:
return
if flags.get('simplify', True):
for k in sol:
sol[k] = simplify(sol[k])
# start over without the failed expanded form, possibly
# with a simplified solution
val = simplify(f.subs(sol))
if flags.get('force', True):
val, reps = posify(val)
# expansion may work now, so try again and check
exval = _mexpand(val, recursive=True)
if exval.is_number:
# we can decide now
val = exval
else:
# if there are no radicals and no functions then this can't be
# zero anymore -- can it?
pot = preorder_traversal(expand_mul(val))
seen = set()
saw_pow_func = False
for p in pot:
if p in seen:
continue
seen.add(p)
if p.is_Pow and not p.exp.is_Integer:
saw_pow_func = True
elif p.is_Function:
saw_pow_func = True
elif isinstance(p, UndefinedFunction):
saw_pow_func = True
if saw_pow_func:
break
if saw_pow_func is False:
return False
if flags.get('force', True):
# don't do a zero check with the positive assumptions in place
val = val.subs(reps)
nz = fuzzy_not(val.is_zero)
if nz is not None:
# issue 5673: nz may be True even when False
# so these are just hacks to keep a false positive
# from being returned
# HACK 1: LambertW (issue 5673)
if val.is_number and val.has(LambertW):
# don't eval this to verify solution since if we got here,
# numerical must be False
return None
# add other HACKs here if necessary, otherwise we assume
# the nz value is correct
return not nz
break
if val.is_Rational:
return val == 0
if numerical and val.is_number:
return (abs(val.n(18).n(12, chop=True)) < 1e-9) is S.true
if flags.get('warn', False):
warnings.warn("\n\tWarning: could not verify solution %s." % sol)
# returns None if it can't conclude
# TODO: improve solution testing
def solve(f, *symbols, **flags):
r"""
Algebraically solves equations and systems of equations.
Explanation
===========
Currently supported:
- polynomial
- transcendental
- piecewise combinations of the above
- systems of linear and polynomial equations
- systems containing relational expressions
- systems implied by undetermined coefficients
Examples
========
The default output varies according to the input and might
be a list (possibly empty), a dictionary, a list of
dictionaries or tuples, or an expression involving relationals.
For specifics regarding different forms of output that may appear, see :ref:`solve_output`.
Let it suffice here to say that to obtain a uniform output from
`solve` use ``dict=True`` or ``set=True`` (see below).
>>> from sympy import solve, Poly, Eq, Matrix, Symbol
>>> from sympy.abc import x, y, z, a, b
The expressions that are passed can be Expr, Equality, or Poly
classes (or lists of the same); a Matrix is considered to be a
list of all the elements of the matrix:
>>> solve(x - 3, x)
[3]
>>> solve(Eq(x, 3), x)
[3]
>>> solve(Poly(x - 3), x)
[3]
>>> solve(Matrix([[x, x + y]]), x, y) == solve([x, x + y], x, y)
True
If no symbols are indicated to be of interest and the equation is
univariate, a list of values is returned; otherwise, the keys in
a dictionary will indicate which (of all the variables used in
the expression(s)) variables and solutions were found:
>>> solve(x**2 - 4)
[-2, 2]
>>> solve((x - a)*(y - b))
[{a: x}, {b: y}]
>>> solve([x - 3, y - 1])
{x: 3, y: 1}
>>> solve([x - 3, y**2 - 1])
[{x: 3, y: -1}, {x: 3, y: 1}]
If you pass symbols for which solutions are sought, the output will vary
depending on the number of symbols you passed, whether you are passing
a list of expressions or not, and whether a linear system was solved.
Uniform output is attained by using ``dict=True`` or ``set=True``.
>>> #### *** feel free to skip to the stars below *** ####
>>> from sympy import TableForm
>>> h = [None, ';|;'.join(['e', 's', 'solve(e, s)', 'solve(e, s, dict=True)',
... 'solve(e, s, set=True)']).split(';')]
>>> t = []
>>> for e, s in [
... (x - y, y),
... (x - y, [x, y]),
... (x**2 - y, [x, y]),
... ([x - 3, y -1], [x, y]),
... ]:
... how = [{}, dict(dict=True), dict(set=True)]
... res = [solve(e, s, **f) for f in how]
... t.append([e, '|', s, '|'] + [res[0], '|', res[1], '|', res[2]])
...
>>> # ******************************************************* #
>>> TableForm(t, headings=h, alignments="<")
e | s | solve(e, s) | solve(e, s, dict=True) | solve(e, s, set=True)
---------------------------------------------------------------------------------------
x - y | y | [x] | [{y: x}] | ([y], {(x,)})
x - y | [x, y] | [(y, y)] | [{x: y}] | ([x, y], {(y, y)})
x**2 - y | [x, y] | [(x, x**2)] | [{y: x**2}] | ([x, y], {(x, x**2)})
[x - 3, y - 1] | [x, y] | {x: 3, y: 1} | [{x: 3, y: 1}] | ([x, y], {(3, 1)})
* If any equation does not depend on the symbol(s) given, it will be
eliminated from the equation set and an answer may be given
implicitly in terms of variables that were not of interest:
>>> solve([x - y, y - 3], x)
{x: y}
When you pass all but one of the free symbols, an attempt
is made to find a single solution based on the method of
undetermined coefficients. If it succeeds, a dictionary of values
is returned. If you want an algebraic solutions for one
or more of the symbols, pass the expression to be solved in a list:
>>> e = a*x + b - 2*x - 3
>>> solve(e, [a, b])
{a: 2, b: 3}
>>> solve([e], [a, b])
{a: -b/x + (2*x + 3)/x}
When there is no solution for any given symbol which will make all
expressions zero, the empty list is returned (or an empty set in
the tuple when ``set=True``):
>>> from sympy import sqrt
>>> solve(3, x)
[]
>>> solve(x - 3, y)
[]
>>> solve(sqrt(x) + 1, x, set=True)
([x], set())
When an object other than a Symbol is given as a symbol, it is
isolated algebraically and an implicit solution may be obtained.
This is mostly provided as a convenience to save you from replacing
the object with a Symbol and solving for that Symbol. It will only
work if the specified object can be replaced with a Symbol using the
subs method:
>>> from sympy import exp, Function
>>> f = Function('f')
>>> solve(f(x) - x, f(x))
[x]
>>> solve(f(x).diff(x) - f(x) - x, f(x).diff(x))
[x + f(x)]
>>> solve(f(x).diff(x) - f(x) - x, f(x))
[-x + Derivative(f(x), x)]
>>> solve(x + exp(x)**2, exp(x), set=True)
([exp(x)], {(-sqrt(-x),), (sqrt(-x),)})
>>> from sympy import Indexed, IndexedBase, Tuple
>>> A = IndexedBase('A')
>>> eqs = Tuple(A[1] + A[2] - 3, A[1] - A[2] + 1)
>>> solve(eqs, eqs.atoms(Indexed))
{A[1]: 1, A[2]: 2}
* To solve for a function within a derivative, use :func:`~.dsolve`.
To solve for a symbol implicitly, use implicit=True:
>>> solve(x + exp(x), x)
[-LambertW(1)]
>>> solve(x + exp(x), x, implicit=True)
[-exp(x)]
It is possible to solve for anything in an expression that can be
replaced with a symbol using :obj:`~sympy.core.basic.Basic.subs`:
>>> solve(x + 2 + sqrt(3), x + 2)
[-sqrt(3)]
>>> solve((x + 2 + sqrt(3), x + 4 + y), y, x + 2)
{y: -2 + sqrt(3), x + 2: -sqrt(3)}
* Nothing heroic is done in this implicit solving so you may end up
with a symbol still in the solution:
>>> eqs = (x*y + 3*y + sqrt(3), x + 4 + y)
>>> solve(eqs, y, x + 2)
{y: -sqrt(3)/(x + 3), x + 2: -2*x/(x + 3) - 6/(x + 3) + sqrt(3)/(x + 3)}
>>> solve(eqs, y*x, x)
{x: -y - 4, x*y: -3*y - sqrt(3)}
* If you attempt to solve for a number, remember that the number
you have obtained does not necessarily mean that the value is
equivalent to the expression obtained:
>>> solve(sqrt(2) - 1, 1)
[sqrt(2)]
>>> solve(x - y + 1, 1) # /!\ -1 is targeted, too
[x/(y - 1)]
>>> [_.subs(z, -1) for _ in solve((x - y + 1).subs(-1, z), 1)]
[-x + y]
**Additional Examples**
``solve()`` with check=True (default) will run through the symbol tags to
eliminate unwanted solutions. If no assumptions are included, all possible
solutions will be returned:
>>> x = Symbol("x")
>>> solve(x**2 - 1)
[-1, 1]
By setting the ``positive`` flag, only one solution will be returned:
>>> pos = Symbol("pos", positive=True)
>>> solve(pos**2 - 1)
[1]
When the solutions are checked, those that make any denominator zero
are automatically excluded. If you do not want to exclude such solutions,
then use the check=False option:
>>> from sympy import sin, limit
>>> solve(sin(x)/x) # 0 is excluded
[pi]
If ``check=False``, then a solution to the numerator being zero is found
but the value of $x = 0$ is a spurious solution since $\sin(x)/x$ has the well
known limit (without discontinuity) of 1 at $x = 0$:
>>> solve(sin(x)/x, check=False)
[0, pi]
In the following case, however, the limit exists and is equal to the
value of $x = 0$ that is excluded when check=True:
>>> eq = x**2*(1/x - z**2/x)
>>> solve(eq, x)
[]
>>> solve(eq, x, check=False)
[0]
>>> limit(eq, x, 0, '-')
0
>>> limit(eq, x, 0, '+')
0
**Solving Relationships**
When one or more expressions passed to ``solve`` is a relational,
a relational result is returned (and the ``dict`` and ``set`` flags
are ignored):
>>> solve(x < 3)
(-oo < x) & (x < 3)
>>> solve([x < 3, x**2 > 4], x)
((-oo < x) & (x < -2)) | ((2 < x) & (x < 3))
>>> solve([x + y - 3, x > 3], x)
(3 < x) & (x < oo) & Eq(x, 3 - y)
Although checking of assumptions on symbols in relationals
is not done, setting assumptions will affect how certain
relationals might automatically simplify:
>>> solve(x**2 > 4)
((-oo < x) & (x < -2)) | ((2 < x) & (x < oo))
>>> r = Symbol('r', real=True)
>>> solve(r**2 > 4)
(2 < r) | (r < -2)
There is currently no algorithm in SymPy that allows you to use
relationships to resolve more than one variable. So the following
does not determine that ``q < 0`` (and trying to solve for ``r``
and ``q`` will raise an error):
>>> from sympy import symbols
>>> r, q = symbols('r, q', real=True)
>>> solve([r + q - 3, r > 3], r)
(3 < r) & Eq(r, 3 - q)
You can directly call the routine that ``solve`` calls
when it encounters a relational: :func:`~.reduce_inequalities`.
It treats Expr like Equality.
>>> from sympy import reduce_inequalities
>>> reduce_inequalities([x**2 - 4])
Eq(x, -2) | Eq(x, 2)
If each relationship contains only one symbol of interest,
the expressions can be processed for multiple symbols:
>>> reduce_inequalities([0 <= x - 1, y < 3], [x, y])
(-oo < y) & (1 <= x) & (x < oo) & (y < 3)
But an error is raised if any relationship has more than one
symbol of interest:
>>> reduce_inequalities([0 <= x*y - 1, y < 3], [x, y])
Traceback (most recent call last):
...
NotImplementedError:
inequality has more than one symbol of interest.
**Disabling High-Order Explicit Solutions**
When solving polynomial expressions, you might not want explicit solutions
(which can be quite long). If the expression is univariate, ``CRootOf``
instances will be returned instead:
>>> solve(x**3 - x + 1)
[-1/((-1/2 - sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)) -
(-1/2 - sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)/3,
-(-1/2 + sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)/3 -
1/((-1/2 + sqrt(3)*I/2)*(3*sqrt(69)/2 + 27/2)**(1/3)),
-(3*sqrt(69)/2 + 27/2)**(1/3)/3 -
1/(3*sqrt(69)/2 + 27/2)**(1/3)]
>>> solve(x**3 - x + 1, cubics=False)
[CRootOf(x**3 - x + 1, 0),
CRootOf(x**3 - x + 1, 1),
CRootOf(x**3 - x + 1, 2)]
If the expression is multivariate, no solution might be returned:
>>> solve(x**3 - x + a, x, cubics=False)
[]
Sometimes solutions will be obtained even when a flag is False because the
expression could be factored. In the following example, the equation can
be factored as the product of a linear and a quadratic factor so explicit
solutions (which did not require solving a cubic expression) are obtained:
>>> eq = x**3 + 3*x**2 + x - 1
>>> solve(eq, cubics=False)
[-1, -1 + sqrt(2), -sqrt(2) - 1]
**Solving Equations Involving Radicals**
Because of SymPy's use of the principle root, some solutions
to radical equations will be missed unless check=False:
>>> from sympy import root
>>> eq = root(x**3 - 3*x**2, 3) + 1 - x
>>> solve(eq)
[]
>>> solve(eq, check=False)
[1/3]
In the above example, there is only a single solution to the
equation. Other expressions will yield spurious roots which
must be checked manually; roots which give a negative argument
to odd-powered radicals will also need special checking:
>>> from sympy import real_root, S
>>> eq = root(x, 3) - root(x, 5) + S(1)/7
>>> solve(eq) # this gives 2 solutions but misses a 3rd
[CRootOf(7*x**5 - 7*x**3 + 1, 1)**15,
CRootOf(7*x**5 - 7*x**3 + 1, 2)**15]
>>> sol = solve(eq, check=False)
>>> [abs(eq.subs(x,i).n(2)) for i in sol]
[0.48, 0.e-110, 0.e-110, 0.052, 0.052]
The first solution is negative so ``real_root`` must be used to see that it
satisfies the expression:
>>> abs(real_root(eq.subs(x, sol[0])).n(2))
0.e-110
If the roots of the equation are not real then more care will be
necessary to find the roots, especially for higher order equations.
Consider the following expression:
>>> expr = root(x, 3) - root(x, 5)
We will construct a known value for this expression at x = 3 by selecting
the 1-th root for each radical:
>>> expr1 = root(x, 3, 1) - root(x, 5, 1)
>>> v = expr1.subs(x, -3)
The ``solve`` function is unable to find any exact roots to this equation:
>>> eq = Eq(expr, v); eq1 = Eq(expr1, v)
>>> solve(eq, check=False), solve(eq1, check=False)
([], [])
The function ``unrad``, however, can be used to get a form of the equation
for which numerical roots can be found:
>>> from sympy.solvers.solvers import unrad
>>> from sympy import nroots
>>> e, (p, cov) = unrad(eq)
>>> pvals = nroots(e)
>>> inversion = solve(cov, x)[0]
>>> xvals = [inversion.subs(p, i) for i in pvals]
Although ``eq`` or ``eq1`` could have been used to find ``xvals``, the
solution can only be verified with ``expr1``:
>>> z = expr - v
>>> [xi.n(chop=1e-9) for xi in xvals if abs(z.subs(x, xi).n()) < 1e-9]
[]
>>> z1 = expr1 - v
>>> [xi.n(chop=1e-9) for xi in xvals if abs(z1.subs(x, xi).n()) < 1e-9]
[-3.0]
Parameters
==========
f :
- a single Expr or Poly that must be zero
- an Equality
- a Relational expression
- a Boolean
- iterable of one or more of the above
symbols : (object(s) to solve for) specified as
- none given (other non-numeric objects will be used)
- single symbol
- denested list of symbols
(e.g., ``solve(f, x, y)``)
- ordered iterable of symbols
(e.g., ``solve(f, [x, y])``)
flags :
dict=True (default is False)
Return list (perhaps empty) of solution mappings.
set=True (default is False)
Return list of symbols and set of tuple(s) of solution(s).
exclude=[] (default)
Do not try to solve for any of the free symbols in exclude;
if expressions are given, the free symbols in them will
be extracted automatically.
check=True (default)
If False, do not do any testing of solutions. This can be
useful if you want to include solutions that make any
denominator zero.
numerical=True (default)
Do a fast numerical check if *f* has only one symbol.
minimal=True (default is False)
A very fast, minimal testing.
warn=True (default is False)
Show a warning if ``checksol()`` could not conclude.
simplify=True (default)
Simplify all but polynomials of order 3 or greater before
returning them and (if check is not False) use the
general simplify function on the solutions and the
expression obtained when they are substituted into the
function which should be zero.
force=True (default is False)
Make positive all symbols without assumptions regarding sign.
rational=True (default)
Recast Floats as Rational; if this option is not used, the
system containing Floats may fail to solve because of issues
with polys. If rational=None, Floats will be recast as
rationals but the answer will be recast as Floats. If the
flag is False then nothing will be done to the Floats.
manual=True (default is False)
Do not use the polys/matrix method to solve a system of
equations, solve them one at a time as you might "manually."
implicit=True (default is False)
Allows ``solve`` to return a solution for a pattern in terms of
other functions that contain that pattern; this is only
needed if the pattern is inside of some invertible function
like cos, exp, ect.
particular=True (default is False)
Instructs ``solve`` to try to find a particular solution to
a linear system with as many zeros as possible; this is very
expensive.
quick=True (default is False; ``particular`` must be True)
Selects a fast heuristic to find a solution with many zeros
whereas a value of False uses the very slow method guaranteed
to find the largest number of zeros possible.
cubics=True (default)
Return explicit solutions when cubic expressions are encountered.
When False, quartics and quintics are disabled, too.
quartics=True (default)
Return explicit solutions when quartic expressions are encountered.
When False, quintics are disabled, too.
quintics=True (default)
Return explicit solutions (if possible) when quintic expressions
are encountered.
See Also
========
rsolve: For solving recurrence relationships
dsolve: For solving differential equations
"""
from .inequalities import reduce_inequalities
# checking/recording flags
###########################################################################
# set solver types explicitly; as soon as one is False
# all the rest will be False
hints = ('cubics', 'quartics', 'quintics')
default = True
for k in hints:
default = flags.setdefault(k, bool(flags.get(k, default)))
# allow solution to contain symbol if True:
implicit = flags.get('implicit', False)
# record desire to see warnings
warn = flags.get('warn', False)
# this flag will be needed for quick exits below, so record
# now -- but don't record `dict` yet since it might change
as_set = flags.get('set', False)
# keeping track of how f was passed
bare_f = not iterable(f)
# check flag usage for particular/quick which should only be used
# with systems of equations
if flags.get('quick', None) is not None:
if not flags.get('particular', None):
raise ValueError('when using `quick`, `particular` should be True')
if flags.get('particular', False) and bare_f:
raise ValueError(filldedent("""
The 'particular/quick' flag is usually used with systems of
equations. Either pass your equation in a list or
consider using a solver like `diophantine` if you are
looking for a solution in integers."""))
# sympify everything, creating list of expressions and list of symbols
###########################################################################
def _sympified_list(w):
return list(map(sympify, w if iterable(w) else [w]))
f, symbols = (_sympified_list(w) for w in [f, symbols])
# preprocess symbol(s)
###########################################################################
ordered_symbols = None # were the symbols in a well defined order?
if not symbols:
# get symbols from equations
symbols = set().union(*[fi.free_symbols for fi in f])
if len(symbols) < len(f):
for fi in f:
pot = preorder_traversal(fi)
for p in pot:
if isinstance(p, AppliedUndef):
if not as_set:
flags['dict'] = True # better show symbols
symbols.add(p)
pot.skip() # don't go any deeper
ordered_symbols = False
symbols = list(ordered(symbols)) # to make it canonical
else:
if len(symbols) == 1 and iterable(symbols[0]):
symbols = symbols[0]
ordered_symbols = symbols and is_sequence(symbols,
include=GeneratorType)
_symbols = list(uniq(symbols))
if len(_symbols) != len(symbols):
ordered_symbols = False
symbols = list(ordered(symbols))
else:
symbols = _symbols
# check for duplicates
if len(symbols) != len(set(symbols)):
raise ValueError('duplicate symbols given')
# remove those not of interest
exclude = flags.pop('exclude', set())
if exclude:
if isinstance(exclude, Expr):
exclude = [exclude]
exclude = set().union(*[e.free_symbols for e in sympify(exclude)])
symbols = [s for s in symbols if s not in exclude]
# preprocess equation(s)
###########################################################################
# automatically ignore True values
if isinstance(f, list):
f = [s for s in f if s is not S.true]
# handle canonicalization of equation types
for i, fi in enumerate(f):
if isinstance(fi, (Eq, Ne)):
if 'ImmutableDenseMatrix' in [type(a).__name__ for a in fi.args]:
fi = fi.lhs - fi.rhs
else:
L, R = fi.args
if isinstance(R, BooleanAtom):
L, R = R, L
if isinstance(L, BooleanAtom):
if isinstance(fi, Ne):
L = ~L
if R.is_Relational:
fi = ~R if L is S.false else R
elif R.is_Symbol:
return L
elif R.is_Boolean and (~R).is_Symbol:
return ~L
else:
raise NotImplementedError(filldedent('''
Unanticipated argument of Eq when other arg
is True or False.
'''))
else:
fi = fi.rewrite(Add, evaluate=False, deep=False)
f[i] = fi
# *** dispatch and handle as a system of relationals
# **************************************************
if fi.is_Relational:
if len(symbols) != 1:
raise ValueError("can only solve for one symbol at a time")
if warn and symbols[0].assumptions0:
warnings.warn(filldedent("""
\tWarning: assumptions about variable '%s' are
not handled currently.""" % symbols[0]))
return reduce_inequalities(f, symbols=symbols)
# convert Poly to expression
if isinstance(fi, Poly):
f[i] = fi.as_expr()
# rewrite hyperbolics in terms of exp if they have symbols of
# interest
f[i] = f[i].replace(lambda w: isinstance(w, HyperbolicFunction) and \
w.has_free(*symbols), lambda w: w.rewrite(exp))
# if we have a Matrix, we need to iterate over its elements again
if f[i].is_Matrix:
bare_f = False
f.extend(list(f[i]))
f[i] = S.Zero
# if we can split it into real and imaginary parts then do so
freei = f[i].free_symbols
if freei and all(s.is_extended_real or s.is_imaginary for s in freei):
fr, fi = f[i].as_real_imag()
# accept as long as new re, im, arg or atan2 are not introduced
had = f[i].atoms(re, im, arg, atan2)
if fr and fi and fr != fi and not any(
i.atoms(re, im, arg, atan2) - had for i in (fr, fi)):
if bare_f:
bare_f = False
f[i: i + 1] = [fr, fi]
# real/imag handling -----------------------------
if any(isinstance(fi, (bool, BooleanAtom)) for fi in f):
if as_set:
return [], set()
return []
for i, fi in enumerate(f):
# Abs
while True:
was = fi
fi = fi.replace(Abs, lambda arg:
separatevars(Abs(arg)).rewrite(Piecewise) if arg.has(*symbols)
else Abs(arg))
if was == fi:
break
for e in fi.find(Abs):
if e.has(*symbols):
raise NotImplementedError('solving %s when the argument '
'is not real or imaginary.' % e)
# arg
fi = fi.replace(arg, lambda a: arg(a).rewrite(atan2).rewrite(atan))
# save changes
f[i] = fi
# see if re(s) or im(s) appear
freim = [fi for fi in f if fi.has(re, im)]
if freim:
irf = []
for s in symbols:
if s.is_real or s.is_imaginary:
continue # neither re(x) nor im(x) will appear
# if re(s) or im(s) appear, the auxiliary equation must be present
if any(fi.has(re(s), im(s)) for fi in freim):
irf.append((s, re(s) + S.ImaginaryUnit*im(s)))
if irf:
for s, rhs in irf:
f = [fi.xreplace({s: rhs}) for fi in f] + [s - rhs]
symbols.extend([re(s), im(s)])
if bare_f:
bare_f = False
flags['dict'] = True
# end of real/imag handling -----------------------------
# we can solve for non-symbol entities by replacing them with Dummy symbols
f, symbols, swap_sym = recast_to_symbols(f, symbols)
# this set of symbols (perhaps recast) is needed below
symset = set(symbols)
# get rid of equations that have no symbols of interest; we don't
# try to solve them because the user didn't ask and they might be
# hard to solve; this means that solutions may be given in terms
# of the eliminated equations e.g. solve((x-y, y-3), x) -> {x: y}
newf = []
for fi in f:
# let the solver handle equations that..
# - have no symbols but are expressions
# - have symbols of interest
# - have no symbols of interest but are constant
# but when an expression is not constant and has no symbols of
# interest, it can't change what we obtain for a solution from
# the remaining equations so we don't include it; and if it's
# zero it can be removed and if it's not zero, there is no
# solution for the equation set as a whole
#
# The reason for doing this filtering is to allow an answer
# to be obtained to queries like solve((x - y, y), x); without
# this mod the return value is []
ok = False
if fi.free_symbols & symset:
ok = True
else:
if fi.is_number:
if fi.is_Number:
if fi.is_zero:
continue
return []
ok = True
else:
if fi.is_constant():
ok = True
if ok:
newf.append(fi)
if not newf:
if as_set:
return symbols, set()
return []
f = newf
del newf
# mask off any Object that we aren't going to invert: Derivative,
# Integral, etc... so that solving for anything that they contain will
# give an implicit solution
seen = set()
non_inverts = set()
for fi in f:
pot = preorder_traversal(fi)
for p in pot:
if not isinstance(p, Expr) or isinstance(p, Piecewise):
pass
elif (isinstance(p, bool) or
not p.args or
p in symset or
p.is_Add or p.is_Mul or
p.is_Pow and not implicit or
p.is_Function and not implicit) and p.func not in (re, im):
continue
elif p not in seen:
seen.add(p)
if p.free_symbols & symset:
non_inverts.add(p)
else:
continue
pot.skip()
del seen
non_inverts = dict(list(zip(non_inverts, [Dummy() for _ in non_inverts])))
f = [fi.subs(non_inverts) for fi in f]
# Both xreplace and subs are needed below: xreplace to force substitution
# inside Derivative, subs to handle non-straightforward substitutions
non_inverts = [(v, k.xreplace(swap_sym).subs(swap_sym)) for k, v in non_inverts.items()]
# rationalize Floats
floats = False
if flags.get('rational', True) is not False:
for i, fi in enumerate(f):
if fi.has(Float):
floats = True
f[i] = nsimplify(fi, rational=True)
# capture any denominators before rewriting since
# they may disappear after the rewrite, e.g. issue 14779
flags['_denominators'] = _simple_dens(f[0], symbols)
# Any embedded piecewise functions need to be brought out to the
# top level so that the appropriate strategy gets selected.
# However, this is necessary only if one of the piecewise
# functions depends on one of the symbols we are solving for.
def _has_piecewise(e):
if e.is_Piecewise:
return e.has(*symbols)
return any(_has_piecewise(a) for a in e.args)
for i, fi in enumerate(f):
if _has_piecewise(fi):
f[i] = piecewise_fold(fi)
#
# try to get a solution
###########################################################################
if bare_f:
solution = None
if len(symbols) != 1:
solution = _solve_undetermined(f[0], symbols, flags)
if not solution:
solution = _solve(f[0], *symbols, **flags)
else:
linear, solution = _solve_system(f, symbols, **flags)
assert type(solution) is list
assert not solution or type(solution[0]) is dict, solution
#
# postprocessing
###########################################################################
# capture as_dict flag now (as_set already captured)
as_dict = flags.get('dict', False)
# define how solution will get unpacked
tuple_format = lambda s: [tuple([i.get(x, x) for x in symbols]) for i in s]
if as_dict or as_set:
unpack = None
elif bare_f:
if len(symbols) == 1:
unpack = lambda s: [i[symbols[0]] for i in s]
elif len(solution) == 1 and len(solution[0]) == len(symbols):
# undetermined linear coeffs solution
unpack = lambda s: s[0]
elif ordered_symbols:
unpack = tuple_format
else:
unpack = lambda s: s
else:
if solution:
if linear and len(solution) == 1:
# if you want the tuple solution for the linear
# case, use `set=True`
unpack = lambda s: s[0]
elif ordered_symbols:
unpack = tuple_format
else:
unpack = lambda s: s
else:
unpack = None
# Restore masked-off objects
if non_inverts and type(solution) is list:
solution = [{k: v.subs(non_inverts) for k, v in s.items()}
for s in solution]
# Restore original "symbols" if a dictionary is returned.
# This is not necessary for
# - the single univariate equation case
# since the symbol will have been removed from the solution;
# - the nonlinear poly_system since that only supports zero-dimensional
# systems and those results come back as a list
#
# ** unless there were Derivatives with the symbols, but those were handled
# above.
if swap_sym:
symbols = [swap_sym.get(k, k) for k in symbols]
for i, sol in enumerate(solution):
solution[i] = {swap_sym.get(k, k): v.subs(swap_sym)
for k, v in sol.items()}
# Get assumptions about symbols, to filter solutions.
# Note that if assumptions about a solution can't be verified, it is still
# returned.
check = flags.get('check', True)
# restore floats
if floats and solution and flags.get('rational', None) is None:
solution = nfloat(solution, exponent=False)
# nfloat might reveal more duplicates
solution = _remove_duplicate_solutions(solution)
if check and solution: # assumption checking
warn = flags.get('warn', False)
got_None = [] # solutions for which one or more symbols gave None
no_False = [] # solutions for which no symbols gave False
for sol in solution:
a_None = False
for symb, val in sol.items():
test = check_assumptions(val, **symb.assumptions0)
if test:
continue
if test is False:
break
a_None = True
else:
no_False.append(sol)
if a_None:
got_None.append(sol)
solution = no_False
if warn and got_None:
warnings.warn(filldedent("""
\tWarning: assumptions concerning following solution(s)
cannot be checked:""" + '\n\t' +
', '.join(str(s) for s in got_None)))
#
# done
###########################################################################
if not solution:
if as_set:
return symbols, set()
return []
# make orderings canonical for list of dictionaries
if not as_set: # for set, no point in ordering
solution = [{k: s[k] for k in ordered(s)} for s in solution]
solution.sort(key=default_sort_key)
if not (as_set or as_dict):
return unpack(solution)
if as_dict:
return solution
# set output: (symbols, {t1, t2, ...}) from list of dictionaries;
# include all symbols for those that like a verbose solution
# and to resolve any differences in dictionary keys.
#
# The set results can easily be used to make a verbose dict as
# k, v = solve(eqs, syms, set=True)
# sol = [dict(zip(k,i)) for i in v]
#
if ordered_symbols:
k = symbols # keep preferred order
else:
# just unify the symbols for which solutions were found
k = list(ordered(set(flatten(tuple(i.keys()) for i in solution))))
return k, {tuple([s.get(ki, ki) for ki in k]) for s in solution}
def _solve_undetermined(g, symbols, flags):
"""solve helper to return a list with one dict (solution) else None
A direct call to solve_undetermined_coeffs is more flexible and
can return both multiple solutions and handle more than one independent
variable. Here, we have to be more cautious to keep from solving
something that does not look like an undetermined coeffs system --
to minimize the surprise factor since singularities that cancel are not
prohibited in solve_undetermined_coeffs.
"""
if g.free_symbols - set(symbols):
sol = solve_undetermined_coeffs(g, symbols, **dict(flags, dict=True, set=None))
if len(sol) == 1:
return sol
def _solve(f, *symbols, **flags):
"""Return a checked solution for *f* in terms of one or more of the
symbols in the form of a list of dictionaries.
If no method is implemented to solve the equation, a NotImplementedError
will be raised. In the case that conversion of an expression to a Poly
gives None a ValueError will be raised.
"""
not_impl_msg = "No algorithms are implemented to solve equation %s"
if len(symbols) != 1:
# look for solutions for desired symbols that are independent
# of symbols already solved for, e.g. if we solve for x = y
# then no symbol having x in its solution will be returned.
# First solve for linear symbols (since that is easier and limits
# solution size) and then proceed with symbols appearing
# in a non-linear fashion. Ideally, if one is solving a single
# expression for several symbols, they would have to be
# appear in factors of an expression, but we do not here
# attempt factorization. XXX perhaps handling a Mul
# should come first in this routine whether there is
# one or several symbols.
nonlin_s = []
got_s = set()
rhs_s = set()
result = []
for s in symbols:
xi, v = solve_linear(f, symbols=[s])
if xi == s:
# no need to check but we should simplify if desired
if flags.get('simplify', True):
v = simplify(v)
vfree = v.free_symbols
if vfree & got_s:
# was linear, but has redundant relationship
# e.g. x - y = 0 has y == x is redundant for x == y
# so ignore
continue
rhs_s |= vfree
got_s.add(xi)
result.append({xi: v})
elif xi: # there might be a non-linear solution if xi is not 0
nonlin_s.append(s)
if not nonlin_s:
return result
for s in nonlin_s:
try:
soln = _solve(f, s, **flags)
for sol in soln:
if sol[s].free_symbols & got_s:
# depends on previously solved symbols: ignore
continue
got_s.add(s)
result.append(sol)
except NotImplementedError:
continue
if got_s:
return result
else:
raise NotImplementedError(not_impl_msg % f)
# solve f for a single variable
symbol = symbols[0]
# expand binomials only if it has the unknown symbol
f = f.replace(lambda e: isinstance(e, binomial) and e.has(symbol),
lambda e: expand_func(e))
# checking will be done unless it is turned off before making a
# recursive call; the variables `checkdens` and `check` are
# captured here (for reference below) in case flag value changes
flags['check'] = checkdens = check = flags.pop('check', True)
# build up solutions if f is a Mul
if f.is_Mul:
result = set()
for m in f.args:
if m in {S.NegativeInfinity, S.ComplexInfinity, S.Infinity}:
result = set()
break
soln = _vsolve(m, symbol, **flags)
result.update(set(soln))
result = [{symbol: v} for v in result]
if check:
# all solutions have been checked but now we must
# check that the solutions do not set denominators
# in any factor to zero
dens = flags.get('_denominators', _simple_dens(f, symbols))
result = [s for s in result if
not any(checksol(den, s, **flags) for den in
dens)]
# set flags for quick exit at end; solutions for each
# factor were already checked and simplified
check = False
flags['simplify'] = False
elif f.is_Piecewise:
result = set()
for i, (expr, cond) in enumerate(f.args):
if expr.is_zero:
raise NotImplementedError(
'solve cannot represent interval solutions')
candidates = _vsolve(expr, symbol, **flags)
# the explicit condition for this expr is the current cond
# and none of the previous conditions
args = [~c for _, c in f.args[:i]] + [cond]
cond = And(*args)
for candidate in candidates:
if candidate in result:
# an unconditional value was already there
continue
try:
v = cond.subs(symbol, candidate)
_eval_simplify = getattr(v, '_eval_simplify', None)
if _eval_simplify is not None:
# unconditionally take the simplification of v
v = _eval_simplify(ratio=2, measure=lambda x: 1)
except TypeError:
# incompatible type with condition(s)
continue
if v == False:
continue
if v == True:
result.add(candidate)
else:
result.add(Piecewise(
(candidate, v),
(S.NaN, True)))
# solutions already checked and simplified
# ****************************************
return [{symbol: r} for r in result]
else:
# first see if it really depends on symbol and whether there
# is only a linear solution
f_num, sol = solve_linear(f, symbols=symbols)
if f_num.is_zero or sol is S.NaN:
return []
elif f_num.is_Symbol:
# no need to check but simplify if desired
if flags.get('simplify', True):
sol = simplify(sol)
return [{f_num: sol}]
poly = None
# check for a single Add generator
if not f_num.is_Add:
add_args = [i for i in f_num.atoms(Add)
if symbol in i.free_symbols]
if len(add_args) == 1:
gen = add_args[0]
spart = gen.as_independent(symbol)[1].as_base_exp()[0]
if spart == symbol:
try:
poly = Poly(f_num, spart)
except PolynomialError:
pass
result = False # no solution was obtained
msg = '' # there is no failure message
# Poly is generally robust enough to convert anything to
# a polynomial and tell us the different generators that it
# contains, so we will inspect the generators identified by
# polys to figure out what to do.
# try to identify a single generator that will allow us to solve this
# as a polynomial, followed (perhaps) by a change of variables if the
# generator is not a symbol
try:
if poly is None:
poly = Poly(f_num)
if poly is None:
raise ValueError('could not convert %s to Poly' % f_num)
except GeneratorsNeeded:
simplified_f = simplify(f_num)
if simplified_f != f_num:
return _solve(simplified_f, symbol, **flags)
raise ValueError('expression appears to be a constant')
gens = [g for g in poly.gens if g.has(symbol)]
def _as_base_q(x):
"""Return (b**e, q) for x = b**(p*e/q) where p/q is the leading
Rational of the exponent of x, e.g. exp(-2*x/3) -> (exp(x), 3)
"""
b, e = x.as_base_exp()
if e.is_Rational:
return b, e.q
if not e.is_Mul:
return x, 1
c, ee = e.as_coeff_Mul()
if c.is_Rational and c is not S.One: # c could be a Float
return b**ee, c.q
return x, 1
if len(gens) > 1:
# If there is more than one generator, it could be that the
# generators have the same base but different powers, e.g.
# >>> Poly(exp(x) + 1/exp(x))
# Poly(exp(-x) + exp(x), exp(-x), exp(x), domain='ZZ')
#
# If unrad was not disabled then there should be no rational
# exponents appearing as in
# >>> Poly(sqrt(x) + sqrt(sqrt(x)))
# Poly(sqrt(x) + x**(1/4), sqrt(x), x**(1/4), domain='ZZ')
bases, qs = list(zip(*[_as_base_q(g) for g in gens]))
bases = set(bases)
if len(bases) > 1 or not all(q == 1 for q in qs):
funcs = {b for b in bases if b.is_Function}
trig = {_ for _ in funcs if
isinstance(_, TrigonometricFunction)}
other = funcs - trig
if not other and len(funcs.intersection(trig)) > 1:
newf = None
if f_num.is_Add and len(f_num.args) == 2:
# check for sin(x)**p = cos(x)**p
_args = f_num.args
t = a, b = [i.atoms(Function).intersection(
trig) for i in _args]
if all(len(i) == 1 for i in t):
a, b = [i.pop() for i in t]
if isinstance(a, cos):
a, b = b, a
_args = _args[::-1]
if isinstance(a, sin) and isinstance(b, cos
) and a.args[0] == b.args[0]:
# sin(x) + cos(x) = 0 -> tan(x) + 1 = 0
newf, _d = (TR2i(_args[0]/_args[1]) + 1
).as_numer_denom()
if not _d.is_Number:
newf = None
if newf is None:
newf = TR1(f_num).rewrite(tan)
if newf != f_num:
# don't check the rewritten form --check
# solutions in the un-rewritten form below
flags['check'] = False
result = _solve(newf, symbol, **flags)
flags['check'] = check
# just a simple case - see if replacement of single function
# clears all symbol-dependent functions, e.g.
# log(x) - log(log(x) - 1) - 3 can be solved even though it has
# two generators.
if result is False and funcs:
funcs = list(ordered(funcs)) # put shallowest function first
f1 = funcs[0]
t = Dummy('t')
# perform the substitution
ftry = f_num.subs(f1, t)
# if no Functions left, we can proceed with usual solve
if not ftry.has(symbol):
cv_sols = _solve(ftry, t, **flags)
cv_inv = list(ordered(_vsolve(t - f1, symbol, **flags)))[0]
result = [{symbol: cv_inv.subs(sol)} for sol in cv_sols]
if result is False:
msg = 'multiple generators %s' % gens
else:
# e.g. case where gens are exp(x), exp(-x)
u = bases.pop()
t = Dummy('t')
inv = _vsolve(u - t, symbol, **flags)
if isinstance(u, (Pow, exp)):
# this will be resolved by factor in _tsolve but we might
# as well try a simple expansion here to get things in
# order so something like the following will work now without
# having to factor:
#
# >>> eq = (exp(I*(-x-2))+exp(I*(x+2)))
# >>> eq.subs(exp(x),y) # fails
# exp(I*(-x - 2)) + exp(I*(x + 2))
# >>> eq.expand().subs(exp(x),y) # works
# y**I*exp(2*I) + y**(-I)*exp(-2*I)
def _expand(p):
b, e = p.as_base_exp()
e = expand_mul(e)
return expand_power_exp(b**e)
ftry = f_num.replace(
lambda w: w.is_Pow or isinstance(w, exp),
_expand).subs(u, t)
if not ftry.has(symbol):
soln = _solve(ftry, t, **flags)
result = [{symbol: i.subs(s)} for i in inv for s in soln]
elif len(gens) == 1:
# There is only one generator that we are interested in, but
# there may have been more than one generator identified by
# polys (e.g. for symbols other than the one we are interested
# in) so recast the poly in terms of our generator of interest.
# Also use composite=True with f_num since Poly won't update
# poly as documented in issue 8810.
poly = Poly(f_num, gens[0], composite=True)
# if we aren't on the tsolve-pass, use roots
if not flags.pop('tsolve', False):
soln = None
deg = poly.degree()
flags['tsolve'] = True
hints = ('cubics', 'quartics', 'quintics')
solvers = {h: flags.get(h) for h in hints}
soln = roots(poly, **solvers)
if sum(soln.values()) < deg:
# e.g. roots(32*x**5 + 400*x**4 + 2032*x**3 +
# 5000*x**2 + 6250*x + 3189) -> {}
# so all_roots is used and RootOf instances are
# returned *unless* the system is multivariate
# or high-order EX domain.
try:
soln = poly.all_roots()
except NotImplementedError:
if not flags.get('incomplete', True):
raise NotImplementedError(
filldedent('''
Neither high-order multivariate polynomials
nor sorting of EX-domain polynomials is supported.
If you want to see any results, pass keyword incomplete=True to
solve; to see numerical values of roots
for univariate expressions, use nroots.
'''))
else:
pass
else:
soln = list(soln.keys())
if soln is not None:
u = poly.gen
if u != symbol:
try:
t = Dummy('t')
inv = _vsolve(u - t, symbol, **flags)
soln = {i.subs(t, s) for i in inv for s in soln}
except NotImplementedError:
# perhaps _tsolve can handle f_num
soln = None
else:
check = False # only dens need to be checked
if soln is not None:
if len(soln) > 2:
# if the flag wasn't set then unset it since high-order
# results are quite long. Perhaps one could base this
# decision on a certain critical length of the
# roots. In addition, wester test M2 has an expression
# whose roots can be shown to be real with the
# unsimplified form of the solution whereas only one of
# the simplified forms appears to be real.
flags['simplify'] = flags.get('simplify', False)
if soln is not None:
result = [{symbol: v} for v in soln]
# fallback if above fails
# -----------------------
if result is False:
# try unrad
if flags.pop('_unrad', True):
try:
u = unrad(f_num, symbol)
except (ValueError, NotImplementedError):
u = False
if u:
eq, cov = u
if cov:
isym, ieq = cov
inv = _vsolve(ieq, symbol, **flags)[0]
rv = {inv.subs(xi) for xi in _solve(eq, isym, **flags)}
else:
try:
rv = set(_vsolve(eq, symbol, **flags))
except NotImplementedError:
rv = None
if rv is not None:
result = [{symbol: v} for v in rv]
# if the flag wasn't set then unset it since unrad results
# can be quite long or of very high order
flags['simplify'] = flags.get('simplify', False)
else:
pass # for coverage
# try _tsolve
if result is False:
flags.pop('tsolve', None) # allow tsolve to be used on next pass
try:
soln = _tsolve(f_num, symbol, **flags)
if soln is not None:
result = [{symbol: v} for v in soln]
except PolynomialError:
pass
# ----------- end of fallback ----------------------------
if result is False:
raise NotImplementedError('\n'.join([msg, not_impl_msg % f]))
result = _remove_duplicate_solutions(result)
if flags.get('simplify', True):
result = [{k: d[k].simplify() for k in d} for d in result]
# Simplification might reveal more duplicates
result = _remove_duplicate_solutions(result)
# we just simplified the solution so we now set the flag to
# False so the simplification doesn't happen again in checksol()
flags['simplify'] = False
if checkdens:
# reject any result that makes any denom. affirmatively 0;
# if in doubt, keep it
dens = _simple_dens(f, symbols)
result = [r for r in result if
not any(checksol(d, r, **flags)
for d in dens)]
if check:
# keep only results if the check is not False
result = [r for r in result if
checksol(f_num, r, **flags) is not False]
return result
def _remove_duplicate_solutions(solutions: list[dict[Expr, Expr]]
) -> list[dict[Expr, Expr]]:
"""Remove duplicates from a list of dicts"""
solutions_set = set()
solutions_new = []
for sol in solutions:
solset = frozenset(sol.items())
if solset not in solutions_set:
solutions_new.append(sol)
solutions_set.add(solset)
return solutions_new
def _solve_system(exprs, symbols, **flags):
"""return ``(linear, solution)`` where ``linear`` is True
if the system was linear, else False; ``solution``
is a list of dictionaries giving solutions for the symbols
"""
if not exprs:
return False, []
if flags.pop('_split', True):
# Split the system into connected components
V = exprs
symsset = set(symbols)
exprsyms = {e: e.free_symbols & symsset for e in exprs}
E = []
sym_indices = {sym: i for i, sym in enumerate(symbols)}
for n, e1 in enumerate(exprs):
for e2 in exprs[:n]:
# Equations are connected if they share a symbol
if exprsyms[e1] & exprsyms[e2]:
E.append((e1, e2))
G = V, E
subexprs = connected_components(G)
if len(subexprs) > 1:
subsols = []
linear = True
for subexpr in subexprs:
subsyms = set()
for e in subexpr:
subsyms |= exprsyms[e]
subsyms = sorted(subsyms, key = lambda x: sym_indices[x])
flags['_split'] = False # skip split step
_linear, subsol = _solve_system(subexpr, subsyms, **flags)
if linear:
linear = linear and _linear
if not isinstance(subsol, list):
subsol = [subsol]
subsols.append(subsol)
# Full solution is cartesion product of subsystems
sols = []
for soldicts in product(*subsols):
sols.append(dict(item for sd in soldicts
for item in sd.items()))
return linear, sols
polys = []
dens = set()
failed = []
result = []
solved_syms = []
linear = True
manual = flags.get('manual', False)
checkdens = check = flags.get('check', True)
for j, g in enumerate(exprs):
dens.update(_simple_dens(g, symbols))
i, d = _invert(g, *symbols)
if d in symbols:
if linear:
linear = solve_linear(g, 0, [d])[0] == d
g = d - i
g = g.as_numer_denom()[0]
if manual:
failed.append(g)
continue
poly = g.as_poly(*symbols, extension=True)
if poly is not None:
polys.append(poly)
else:
failed.append(g)
if polys:
if all(p.is_linear for p in polys):
n, m = len(polys), len(symbols)
matrix = zeros(n, m + 1)
for i, poly in enumerate(polys):
for monom, coeff in poly.terms():
try:
j = monom.index(1)
matrix[i, j] = coeff
except ValueError:
matrix[i, m] = -coeff
# returns a dictionary ({symbols: values}) or None
if flags.pop('particular', False):
result = minsolve_linear_system(matrix, *symbols, **flags)
else:
result = solve_linear_system(matrix, *symbols, **flags)
result = [result] if result else []
if failed:
if result:
solved_syms = list(result[0].keys()) # there is only one result dict
else:
solved_syms = []
# linear doesn't change
else:
linear = False
if len(symbols) > len(polys):
free = set().union(*[p.free_symbols for p in polys])
free = list(ordered(free.intersection(symbols)))
got_s = set()
result = []
for syms in subsets(free, len(polys)):
try:
# returns [], None or list of tuples
res = solve_poly_system(polys, *syms)
if res:
for r in set(res):
skip = False
for r1 in r:
if got_s and any(ss in r1.free_symbols
for ss in got_s):
# sol depends on previously
# solved symbols: discard it
skip = True
if not skip:
got_s.update(syms)
result.append(dict(list(zip(syms, r))))
except NotImplementedError:
pass
if got_s:
solved_syms = list(got_s)
else:
raise NotImplementedError('no valid subset found')
else:
try:
result = solve_poly_system(polys, *symbols)
if result:
solved_syms = symbols
result = [dict(list(zip(solved_syms, r))) for r in set(result)]
except NotImplementedError:
failed.extend([g.as_expr() for g in polys])
solved_syms = []
# convert None or [] to [{}]
result = result or [{}]
if failed:
linear = False
# For each failed equation, see if we can solve for one of the
# remaining symbols from that equation. If so, we update the
# solution set and continue with the next failed equation,
# repeating until we are done or we get an equation that can't
# be solved.
def _ok_syms(e, sort=False):
rv = e.free_symbols & legal
# Solve first for symbols that have lower degree in the equation.
# Ideally we want to solve firstly for symbols that appear linearly
# with rational coefficients e.g. if e = x*y + z then we should
# solve for z first.
def key(sym):
ep = e.as_poly(sym)
if ep is None:
complexity = (S.Infinity, S.Infinity, S.Infinity)
else:
coeff_syms = ep.LC().free_symbols
complexity = (ep.degree(), len(coeff_syms & rv), len(coeff_syms))
return complexity + (default_sort_key(sym),)
if sort:
rv = sorted(rv, key=key)
return rv
legal = set(symbols) # what we are interested in
# sort so equation with the fewest potential symbols is first
u = Dummy() # used in solution checking
for eq in ordered(failed, lambda _: len(_ok_syms(_))):
newresult = []
bad_results = []
hit = False
for r in result:
got_s = set()
# update eq with everything that is known so far
eq2 = eq.subs(r)
# if check is True then we see if it satisfies this
# equation, otherwise we just accept it
if check and r:
b = checksol(u, u, eq2, minimal=True)
if b is not None:
# this solution is sufficient to know whether
# it is valid or not so we either accept or
# reject it, then continue
if b:
newresult.append(r)
else:
bad_results.append(r)
continue
# search for a symbol amongst those available that
# can be solved for
ok_syms = _ok_syms(eq2, sort=True)
if not ok_syms:
if r:
newresult.append(r)
break # skip as it's independent of desired symbols
for s in ok_syms:
try:
soln = _vsolve(eq2, s, **flags)
except NotImplementedError:
continue
# put each solution in r and append the now-expanded
# result in the new result list; use copy since the
# solution for s is being added in-place
for sol in soln:
if got_s and any(ss in sol.free_symbols for ss in got_s):
# sol depends on previously solved symbols: discard it
continue
rnew = r.copy()
for k, v in r.items():
rnew[k] = v.subs(s, sol)
# and add this new solution
rnew[s] = sol
# check that it is independent of previous solutions
iset = set(rnew.items())
for i in newresult:
if len(i) < len(iset) and not set(i.items()) - iset:
# this is a superset of a known solution that
# is smaller
break
else:
# keep it
newresult.append(rnew)
hit = True
got_s.add(s)
if not hit:
raise NotImplementedError('could not solve %s' % eq2)
else:
result = newresult
for b in bad_results:
if b in result:
result.remove(b)
if not result:
return False, []
# rely on linear/polynomial system solvers to simplify
# XXX the following tests show that the expressions
# returned are not the same as they would be if simplify
# were applied to this:
# sympy/solvers/ode/tests/test_systems/test__classify_linear_system
# sympy/solvers/tests/test_solvers/test_issue_4886
# so the docs should be updated to reflect that or else
# the following should be `bool(failed) or not linear`
default_simplify = bool(failed)
if flags.get('simplify', default_simplify):
for r in result:
for k in r:
r[k] = simplify(r[k])
flags['simplify'] = False # don't need to do so in checksol now
if checkdens:
result = [r for r in result
if not any(checksol(d, r, **flags) for d in dens)]
if check and not linear:
result = [r for r in result
if not any(checksol(e, r, **flags) is False for e in exprs)]
result = [r for r in result if r]
return linear, result
def solve_linear(lhs, rhs=0, symbols=[], exclude=[]):
r"""
Return a tuple derived from ``f = lhs - rhs`` that is one of
the following: ``(0, 1)``, ``(0, 0)``, ``(symbol, solution)``, ``(n, d)``.
Explanation
===========
``(0, 1)`` meaning that ``f`` is independent of the symbols in *symbols*
that are not in *exclude*.
``(0, 0)`` meaning that there is no solution to the equation amongst the
symbols given. If the first element of the tuple is not zero, then the
function is guaranteed to be dependent on a symbol in *symbols*.
``(symbol, solution)`` where symbol appears linearly in the numerator of
``f``, is in *symbols* (if given), and is not in *exclude* (if given). No
simplification is done to ``f`` other than a ``mul=True`` expansion, so the
solution will correspond strictly to a unique solution.
``(n, d)`` where ``n`` and ``d`` are the numerator and denominator of ``f``
when the numerator was not linear in any symbol of interest; ``n`` will
never be a symbol unless a solution for that symbol was found (in which case
the second element is the solution, not the denominator).
Examples
========
>>> from sympy import cancel, Pow
``f`` is independent of the symbols in *symbols* that are not in
*exclude*:
>>> from sympy import cos, sin, solve_linear
>>> from sympy.abc import x, y, z
>>> eq = y*cos(x)**2 + y*sin(x)**2 - y # = y*(1 - 1) = 0
>>> solve_linear(eq)
(0, 1)
>>> eq = cos(x)**2 + sin(x)**2 # = 1
>>> solve_linear(eq)
(0, 1)
>>> solve_linear(x, exclude=[x])
(0, 1)
The variable ``x`` appears as a linear variable in each of the
following:
>>> solve_linear(x + y**2)
(x, -y**2)
>>> solve_linear(1/x - y**2)
(x, y**(-2))
When not linear in ``x`` or ``y`` then the numerator and denominator are
returned:
>>> solve_linear(x**2/y**2 - 3)
(x**2 - 3*y**2, y**2)
If the numerator of the expression is a symbol, then ``(0, 0)`` is
returned if the solution for that symbol would have set any
denominator to 0:
>>> eq = 1/(1/x - 2)
>>> eq.as_numer_denom()
(x, 1 - 2*x)
>>> solve_linear(eq)
(0, 0)
But automatic rewriting may cause a symbol in the denominator to
appear in the numerator so a solution will be returned:
>>> (1/x)**-1
x
>>> solve_linear((1/x)**-1)
(x, 0)
Use an unevaluated expression to avoid this:
>>> solve_linear(Pow(1/x, -1, evaluate=False))
(0, 0)
If ``x`` is allowed to cancel in the following expression, then it
appears to be linear in ``x``, but this sort of cancellation is not
done by ``solve_linear`` so the solution will always satisfy the
original expression without causing a division by zero error.
>>> eq = x**2*(1/x - z**2/x)
>>> solve_linear(cancel(eq))
(x, 0)
>>> solve_linear(eq)
(x**2*(1 - z**2), x)
A list of symbols for which a solution is desired may be given:
>>> solve_linear(x + y + z, symbols=[y])
(y, -x - z)
A list of symbols to ignore may also be given:
>>> solve_linear(x + y + z, exclude=[x])
(y, -x - z)
(A solution for ``y`` is obtained because it is the first variable
from the canonically sorted list of symbols that had a linear
solution.)
"""
if isinstance(lhs, Eq):
if rhs:
raise ValueError(filldedent('''
If lhs is an Equality, rhs must be 0 but was %s''' % rhs))
rhs = lhs.rhs
lhs = lhs.lhs
dens = None
eq = lhs - rhs
n, d = eq.as_numer_denom()
if not n:
return S.Zero, S.One
free = n.free_symbols
if not symbols:
symbols = free
else:
bad = [s for s in symbols if not s.is_Symbol]
if bad:
if len(bad) == 1:
bad = bad[0]
if len(symbols) == 1:
eg = 'solve(%s, %s)' % (eq, symbols[0])
else:
eg = 'solve(%s, *%s)' % (eq, list(symbols))
raise ValueError(filldedent('''
solve_linear only handles symbols, not %s. To isolate
non-symbols use solve, e.g. >>> %s <<<.
''' % (bad, eg)))
symbols = free.intersection(symbols)
symbols = symbols.difference(exclude)
if not symbols:
return S.Zero, S.One
# derivatives are easy to do but tricky to analyze to see if they
# are going to disallow a linear solution, so for simplicity we
# just evaluate the ones that have the symbols of interest
derivs = defaultdict(list)
for der in n.atoms(Derivative):
csym = der.free_symbols & symbols
for c in csym:
derivs[c].append(der)
all_zero = True
for xi in sorted(symbols, key=default_sort_key): # canonical order
# if there are derivatives in this var, calculate them now
if isinstance(derivs[xi], list):
derivs[xi] = {der: der.doit() for der in derivs[xi]}
newn = n.subs(derivs[xi])
dnewn_dxi = newn.diff(xi)
# dnewn_dxi can be nonzero if it survives differentation by any
# of its free symbols
free = dnewn_dxi.free_symbols
if dnewn_dxi and (not free or any(dnewn_dxi.diff(s) for s in free) or free == symbols):
all_zero = False
if dnewn_dxi is S.NaN:
break
if xi not in dnewn_dxi.free_symbols:
vi = -1/dnewn_dxi*(newn.subs(xi, 0))
if dens is None:
dens = _simple_dens(eq, symbols)
if not any(checksol(di, {xi: vi}, minimal=True) is True
for di in dens):
# simplify any trivial integral
irep = [(i, i.doit()) for i in vi.atoms(Integral) if
i.function.is_number]
# do a slight bit of simplification
vi = expand_mul(vi.subs(irep))
return xi, vi
if all_zero:
return S.Zero, S.One
if n.is_Symbol: # no solution for this symbol was found
return S.Zero, S.Zero
return n, d
def minsolve_linear_system(system, *symbols, **flags):
r"""
Find a particular solution to a linear system.
Explanation
===========
In particular, try to find a solution with the minimal possible number
of non-zero variables using a naive algorithm with exponential complexity.
If ``quick=True``, a heuristic is used.
"""
quick = flags.get('quick', False)
# Check if there are any non-zero solutions at all
s0 = solve_linear_system(system, *symbols, **flags)
if not s0 or all(v == 0 for v in s0.values()):
return s0
if quick:
# We just solve the system and try to heuristically find a nice
# solution.
s = solve_linear_system(system, *symbols)
def update(determined, solution):
delete = []
for k, v in solution.items():
solution[k] = v.subs(determined)
if not solution[k].free_symbols:
delete.append(k)
determined[k] = solution[k]
for k in delete:
del solution[k]
determined = {}
update(determined, s)
while s:
# NOTE sort by default_sort_key to get deterministic result
k = max((k for k in s.values()),
key=lambda x: (len(x.free_symbols), default_sort_key(x)))
kfree = k.free_symbols
x = next(reversed(list(ordered(kfree))))
if len(kfree) != 1:
determined[x] = S.Zero
else:
val = _vsolve(k, x, check=False)[0]
if not val and not any(v.subs(x, val) for v in s.values()):
determined[x] = S.One
else:
determined[x] = val
update(determined, s)
return determined
else:
# We try to select n variables which we want to be non-zero.
# All others will be assumed zero. We try to solve the modified system.
# If there is a non-trivial solution, just set the free variables to
# one. If we do this for increasing n, trying all combinations of
# variables, we will find an optimal solution.
# We speed up slightly by starting at one less than the number of
# variables the quick method manages.
N = len(symbols)
bestsol = minsolve_linear_system(system, *symbols, quick=True)
n0 = len([x for x in bestsol.values() if x != 0])
for n in range(n0 - 1, 1, -1):
debugf('minsolve: %s', n)
thissol = None
for nonzeros in combinations(range(N), n):
subm = Matrix([system.col(i).T for i in nonzeros] + [system.col(-1).T]).T
s = solve_linear_system(subm, *[symbols[i] for i in nonzeros])
if s and not all(v == 0 for v in s.values()):
subs = [(symbols[v], S.One) for v in nonzeros]
for k, v in s.items():
s[k] = v.subs(subs)
for sym in symbols:
if sym not in s:
if symbols.index(sym) in nonzeros:
s[sym] = S.One
else:
s[sym] = S.Zero
thissol = s
break
if thissol is None:
break
bestsol = thissol
return bestsol
def solve_linear_system(system, *symbols, **flags):
r"""
Solve system of $N$ linear equations with $M$ variables, which means
both under- and overdetermined systems are supported.
Explanation
===========
The possible number of solutions is zero, one, or infinite. Respectively,
this procedure will return None or a dictionary with solutions. In the
case of underdetermined systems, all arbitrary parameters are skipped.
This may cause a situation in which an empty dictionary is returned.
In that case, all symbols can be assigned arbitrary values.
Input to this function is a $N\times M + 1$ matrix, which means it has
to be in augmented form. If you prefer to enter $N$ equations and $M$
unknowns then use ``solve(Neqs, *Msymbols)`` instead. Note: a local
copy of the matrix is made by this routine so the matrix that is
passed will not be modified.
The algorithm used here is fraction-free Gaussian elimination,
which results, after elimination, in an upper-triangular matrix.
Then solutions are found using back-substitution. This approach
is more efficient and compact than the Gauss-Jordan method.
Examples
========
>>> from sympy import Matrix, solve_linear_system
>>> from sympy.abc import x, y
Solve the following system::
x + 4 y == 2
-2 x + y == 14
>>> system = Matrix(( (1, 4, 2), (-2, 1, 14)))
>>> solve_linear_system(system, x, y)
{x: -6, y: 2}
A degenerate system returns an empty dictionary:
>>> system = Matrix(( (0,0,0), (0,0,0) ))
>>> solve_linear_system(system, x, y)
{}
"""
assert system.shape[1] == len(symbols) + 1
# This is just a wrapper for solve_lin_sys
eqs = list(system * Matrix(symbols + (-1,)))
eqs, ring = sympy_eqs_to_ring(eqs, symbols)
sol = solve_lin_sys(eqs, ring, _raw=False)
if sol is not None:
sol = {sym:val for sym, val in sol.items() if sym != val}
return sol
def solve_undetermined_coeffs(equ, coeffs, *syms, **flags):
r"""
Solve a system of equations in $k$ parameters that is formed by
matching coefficients in variables ``coeffs`` that are on
factors dependent on the remaining variables (or those given
explicitly by ``syms``.
Explanation
===========
The result of this function is a dictionary with symbolic values of those
parameters with respect to coefficients in $q$ -- empty if there
is no solution or coefficients do not appear in the equation -- else
None (if the system was not recognized). If there is more than one
solution, the solutions are passed as a list. The output can be modified using
the same semantics as for `solve` since the flags that are passed are sent
directly to `solve` so, for example the flag ``dict=True`` will always return a list
of solutions as dictionaries.
This function accepts both Equality and Expr class instances.
The solving process is most efficient when symbols are specified
in addition to parameters to be determined, but an attempt to
determine them (if absent) will be made. If an expected solution is not
obtained (and symbols were not specified) try specifying them.
Examples
========
>>> from sympy import Eq, solve_undetermined_coeffs
>>> from sympy.abc import a, b, c, h, p, k, x, y
>>> solve_undetermined_coeffs(Eq(a*x + a + b, x/2), [a, b], x)
{a: 1/2, b: -1/2}
>>> solve_undetermined_coeffs(a - 2, [a])
{a: 2}
The equation can be nonlinear in the symbols:
>>> X, Y, Z = y, x**y, y*x**y
>>> eq = a*X + b*Y + c*Z - X - 2*Y - 3*Z
>>> coeffs = a, b, c
>>> syms = x, y
>>> solve_undetermined_coeffs(eq, coeffs, syms)
{a: 1, b: 2, c: 3}
And the system can be nonlinear in coefficients, too, but if
there is only a single solution, it will be returned as a
dictionary:
>>> eq = a*x**2 + b*x + c - ((x - h)**2 + 4*p*k)/4/p
>>> solve_undetermined_coeffs(eq, (h, p, k), x)
{h: -b/(2*a), k: (4*a*c - b**2)/(4*a), p: 1/(4*a)}
Multiple solutions are always returned in a list:
>>> solve_undetermined_coeffs(a**2*x + b - x, [a, b], x)
[{a: -1, b: 0}, {a: 1, b: 0}]
Using flag ``dict=True`` (in keeping with semantics in :func:`~.solve`)
will force the result to always be a list with any solutions
as elements in that list.
>>> solve_undetermined_coeffs(a*x - 2*x, [a], dict=True)
[{a: 2}]
"""
if not (coeffs and all(i.is_Symbol for i in coeffs)):
raise ValueError('must provide symbols for coeffs')
if isinstance(equ, Eq):
eq = equ.lhs - equ.rhs
else:
eq = equ
ceq = cancel(eq)
xeq = _mexpand(ceq.as_numer_denom()[0], recursive=True)
free = xeq.free_symbols
coeffs = free & set(coeffs)
if not coeffs:
return ([], {}) if flags.get('set', None) else [] # solve(0, x) -> []
if not syms:
# e.g. A*exp(x) + B - (exp(x) + y) separated into parts that
# don't/do depend on coeffs gives
# -(exp(x) + y), A*exp(x) + B
# then see what symbols are common to both
# {x} = {x, A, B} - {x, y}
ind, dep = xeq.as_independent(*coeffs, as_Add=True)
dfree = dep.free_symbols
syms = dfree & ind.free_symbols
if not syms:
# but if the system looks like (a + b)*x + b - c
# then {} = {a, b, x} - c
# so calculate {x} = {a, b, x} - {a, b}
syms = dfree - set(coeffs)
if not syms:
syms = [Dummy()]
else:
if len(syms) == 1 and iterable(syms[0]):
syms = syms[0]
e, s, _ = recast_to_symbols([xeq], syms)
xeq = e[0]
syms = s
# find the functional forms in which symbols appear
gens = set(xeq.as_coefficients_dict(*syms).keys()) - {1}
cset = set(coeffs)
if any(g.has_xfree(cset) for g in gens):
return # a generator contained a coefficient symbol
# make sure we are working with symbols for generators
e, gens, _ = recast_to_symbols([xeq], list(gens))
xeq = e[0]
# collect coefficients in front of generators
system = list(collect(xeq, gens, evaluate=False).values())
# get a solution
soln = solve(system, coeffs, **flags)
# unpack unless told otherwise if length is 1
settings = flags.get('dict', None) or flags.get('set', None)
if type(soln) is dict or settings or len(soln) != 1:
return soln
return soln[0]
def solve_linear_system_LU(matrix, syms):
"""
Solves the augmented matrix system using ``LUsolve`` and returns a
dictionary in which solutions are keyed to the symbols of *syms* as ordered.
Explanation
===========
The matrix must be invertible.
Examples
========
>>> from sympy import Matrix, solve_linear_system_LU
>>> from sympy.abc import x, y, z
>>> solve_linear_system_LU(Matrix([
... [1, 2, 0, 1],
... [3, 2, 2, 1],
... [2, 0, 0, 1]]), [x, y, z])
{x: 1/2, y: 1/4, z: -1/2}
See Also
========
LUsolve
"""
if matrix.rows != matrix.cols - 1:
raise ValueError("Rows should be equal to columns - 1")
A = matrix[:matrix.rows, :matrix.rows]
b = matrix[:, matrix.cols - 1:]
soln = A.LUsolve(b)
solutions = {}
for i in range(soln.rows):
solutions[syms[i]] = soln[i, 0]
return solutions
def det_perm(M):
"""
Return the determinant of *M* by using permutations to select factors.
Explanation
===========
For sizes larger than 8 the number of permutations becomes prohibitively
large, or if there are no symbols in the matrix, it is better to use the
standard determinant routines (e.g., ``M.det()``.)
See Also
========
det_minor
det_quick
"""
args = []
s = True
n = M.rows
list_ = M.flat()
for perm in generate_bell(n):
fac = []
idx = 0
for j in perm:
fac.append(list_[idx + j])
idx += n
term = Mul(*fac) # disaster with unevaluated Mul -- takes forever for n=7
args.append(term if s else -term)
s = not s
return Add(*args)
def det_minor(M):
"""
Return the ``det(M)`` computed from minors without
introducing new nesting in products.
See Also
========
det_perm
det_quick
"""
n = M.rows
if n == 2:
return M[0, 0]*M[1, 1] - M[1, 0]*M[0, 1]
else:
return sum([(1, -1)[i % 2]*Add(*[M[0, i]*d for d in
Add.make_args(det_minor(M.minor_submatrix(0, i)))])
if M[0, i] else S.Zero for i in range(n)])
def det_quick(M, method=None):
"""
Return ``det(M)`` assuming that either
there are lots of zeros or the size of the matrix
is small. If this assumption is not met, then the normal
Matrix.det function will be used with method = ``method``.
See Also
========
det_minor
det_perm
"""
if any(i.has(Symbol) for i in M):
if M.rows < 8 and all(i.has(Symbol) for i in M):
return det_perm(M)
return det_minor(M)
else:
return M.det(method=method) if method else M.det()
def inv_quick(M):
"""Return the inverse of ``M``, assuming that either
there are lots of zeros or the size of the matrix
is small.
"""
if not all(i.is_Number for i in M):
if not any(i.is_Number for i in M):
det = lambda _: det_perm(_)
else:
det = lambda _: det_minor(_)
else:
return M.inv()
n = M.rows
d = det(M)
if d == S.Zero:
raise NonInvertibleMatrixError("Matrix det == 0; not invertible")
ret = zeros(n)
s1 = -1
for i in range(n):
s = s1 = -s1
for j in range(n):
di = det(M.minor_submatrix(i, j))
ret[j, i] = s*di/d
s = -s
return ret
# these are functions that have multiple inverse values per period
multi_inverses = {
sin: lambda x: (asin(x), S.Pi - asin(x)),
cos: lambda x: (acos(x), 2*S.Pi - acos(x)),
}
def _vsolve(e, s, **flags):
"""return list of scalar values for the solution of e for symbol s"""
return [i[s] for i in _solve(e, s, **flags)]
def _tsolve(eq, sym, **flags):
"""
Helper for ``_solve`` that solves a transcendental equation with respect
to the given symbol. Various equations containing powers and logarithms,
can be solved.
There is currently no guarantee that all solutions will be returned or
that a real solution will be favored over a complex one.
Either a list of potential solutions will be returned or None will be
returned (in the case that no method was known to get a solution
for the equation). All other errors (like the inability to cast an
expression as a Poly) are unhandled.
Examples
========
>>> from sympy import log, ordered
>>> from sympy.solvers.solvers import _tsolve as tsolve
>>> from sympy.abc import x
>>> list(ordered(tsolve(3**(2*x + 5) - 4, x)))
[-5/2 + log(2)/log(3), (-5*log(3)/2 + log(2) + I*pi)/log(3)]
>>> tsolve(log(x) + 2*x, x)
[LambertW(2)/2]
"""
if 'tsolve_saw' not in flags:
flags['tsolve_saw'] = []
if eq in flags['tsolve_saw']:
return None
else:
flags['tsolve_saw'].append(eq)
rhs, lhs = _invert(eq, sym)
if lhs == sym:
return [rhs]
try:
if lhs.is_Add:
# it's time to try factoring; powdenest is used
# to try get powers in standard form for better factoring
f = factor(powdenest(lhs - rhs))
if f.is_Mul:
return _vsolve(f, sym, **flags)
if rhs:
f = logcombine(lhs, force=flags.get('force', True))
if f.count(log) != lhs.count(log):
if isinstance(f, log):
return _vsolve(f.args[0] - exp(rhs), sym, **flags)
return _tsolve(f - rhs, sym, **flags)
elif lhs.is_Pow:
if lhs.exp.is_Integer:
if lhs - rhs != eq:
return _vsolve(lhs - rhs, sym, **flags)
if sym not in lhs.exp.free_symbols:
return _vsolve(lhs.base - rhs**(1/lhs.exp), sym, **flags)
# _tsolve calls this with Dummy before passing the actual number in.
if any(t.is_Dummy for t in rhs.free_symbols):
raise NotImplementedError # _tsolve will call here again...
# a ** g(x) == 0
if not rhs:
# f(x)**g(x) only has solutions where f(x) == 0 and g(x) != 0 at
# the same place
sol_base = _vsolve(lhs.base, sym, **flags)
return [s for s in sol_base if lhs.exp.subs(sym, s) != 0] # XXX use checksol here?
# a ** g(x) == b
if not lhs.base.has(sym):
if lhs.base == 0:
return _vsolve(lhs.exp, sym, **flags) if rhs != 0 else []
# Gets most solutions...
if lhs.base == rhs.as_base_exp()[0]:
# handles case when bases are equal
sol = _vsolve(lhs.exp - rhs.as_base_exp()[1], sym, **flags)
else:
# handles cases when bases are not equal and exp
# may or may not be equal
f = exp(log(lhs.base)*lhs.exp) - exp(log(rhs))
sol = _vsolve(f, sym, **flags)
# Check for duplicate solutions
def equal(expr1, expr2):
_ = Dummy()
eq = checksol(expr1 - _, _, expr2)
if eq is None:
if nsimplify(expr1) != nsimplify(expr2):
return False
# they might be coincidentally the same
# so check more rigorously
eq = expr1.equals(expr2) # XXX expensive but necessary?
return eq
# Guess a rational exponent
e_rat = nsimplify(log(abs(rhs))/log(abs(lhs.base)))
e_rat = simplify(posify(e_rat)[0])
n, d = fraction(e_rat)
if expand(lhs.base**n - rhs**d) == 0:
sol = [s for s in sol if not equal(lhs.exp.subs(sym, s), e_rat)]
sol.extend(_vsolve(lhs.exp - e_rat, sym, **flags))
return list(set(sol))
# f(x) ** g(x) == c
else:
sol = []
logform = lhs.exp*log(lhs.base) - log(rhs)
if logform != lhs - rhs:
try:
sol.extend(_vsolve(logform, sym, **flags))
except NotImplementedError:
pass
# Collect possible solutions and check with substitution later.
check = []
if rhs == 1:
# f(x) ** g(x) = 1 -- g(x)=0 or f(x)=+-1
check.extend(_vsolve(lhs.exp, sym, **flags))
check.extend(_vsolve(lhs.base - 1, sym, **flags))
check.extend(_vsolve(lhs.base + 1, sym, **flags))
elif rhs.is_Rational:
for d in (i for i in divisors(abs(rhs.p)) if i != 1):
e, t = integer_log(rhs.p, d)
if not t:
continue # rhs.p != d**b
for s in divisors(abs(rhs.q)):
if s**e== rhs.q:
r = Rational(d, s)
check.extend(_vsolve(lhs.base - r, sym, **flags))
check.extend(_vsolve(lhs.base + r, sym, **flags))
check.extend(_vsolve(lhs.exp - e, sym, **flags))
elif rhs.is_irrational:
b_l, e_l = lhs.base.as_base_exp()
n, d = (e_l*lhs.exp).as_numer_denom()
b, e = sqrtdenest(rhs).as_base_exp()
check = [sqrtdenest(i) for i in (_vsolve(lhs.base - b, sym, **flags))]
check.extend([sqrtdenest(i) for i in (_vsolve(lhs.exp - e, sym, **flags))])
if e_l*d != 1:
check.extend(_vsolve(b_l**n - rhs**(e_l*d), sym, **flags))
for s in check:
ok = checksol(eq, sym, s)
if ok is None:
ok = eq.subs(sym, s).equals(0)
if ok:
sol.append(s)
return list(set(sol))
elif lhs.is_Function and len(lhs.args) == 1:
if lhs.func in multi_inverses:
# sin(x) = 1/3 -> x - asin(1/3) & x - (pi - asin(1/3))
soln = []
for i in multi_inverses[type(lhs)](rhs):
soln.extend(_vsolve(lhs.args[0] - i, sym, **flags))
return list(set(soln))
elif lhs.func == LambertW:
return _vsolve(lhs.args[0] - rhs*exp(rhs), sym, **flags)
rewrite = lhs.rewrite(exp)
if rewrite != lhs:
return _vsolve(rewrite - rhs, sym, **flags)
except NotImplementedError:
pass
# maybe it is a lambert pattern
if flags.pop('bivariate', True):
# lambert forms may need some help being recognized, e.g. changing
# 2**(3*x) + x**3*log(2)**3 + 3*x**2*log(2)**2 + 3*x*log(2) + 1
# to 2**(3*x) + (x*log(2) + 1)**3
# make generator in log have exponent of 1
logs = eq.atoms(log)
spow = min(
{i.exp for j in logs for i in j.atoms(Pow)
if i.base == sym} or {1})
if spow != 1:
p = sym**spow
u = Dummy('bivariate-cov')
ueq = eq.subs(p, u)
if not ueq.has_free(sym):
sol = _vsolve(ueq, u, **flags)
inv = _vsolve(p - u, sym)
return [i.subs(u, s) for i in inv for s in sol]
g = _filtered_gens(eq.as_poly(), sym)
up_or_log = set()
for gi in g:
if isinstance(gi, (exp, log)) or (gi.is_Pow and gi.base == S.Exp1):
up_or_log.add(gi)
elif gi.is_Pow:
gisimp = powdenest(expand_power_exp(gi))
if gisimp.is_Pow and sym in gisimp.exp.free_symbols:
up_or_log.add(gi)
eq_down = expand_log(expand_power_exp(eq)).subs(
dict(list(zip(up_or_log, [0]*len(up_or_log)))))
eq = expand_power_exp(factor(eq_down, deep=True) + (eq - eq_down))
rhs, lhs = _invert(eq, sym)
if lhs.has(sym):
try:
poly = lhs.as_poly()
g = _filtered_gens(poly, sym)
_eq = lhs - rhs
sols = _solve_lambert(_eq, sym, g)
# use a simplified form if it satisfies eq
# and has fewer operations
for n, s in enumerate(sols):
ns = nsimplify(s)
if ns != s and ns.count_ops() <= s.count_ops():
ok = checksol(_eq, sym, ns)
if ok is None:
ok = _eq.subs(sym, ns).equals(0)
if ok:
sols[n] = ns
return sols
except NotImplementedError:
# maybe it's a convoluted function
if len(g) == 2:
try:
gpu = bivariate_type(lhs - rhs, *g)
if gpu is None:
raise NotImplementedError
g, p, u = gpu
flags['bivariate'] = False
inversion = _tsolve(g - u, sym, **flags)
if inversion:
sol = _vsolve(p, u, **flags)
return list({i.subs(u, s)
for i in inversion for s in sol})
except NotImplementedError:
pass
else:
pass
if flags.pop('force', True):
flags['force'] = False
pos, reps = posify(lhs - rhs)
if rhs == S.ComplexInfinity:
return []
for u, s in reps.items():
if s == sym:
break
else:
u = sym
if pos.has(u):
try:
soln = _vsolve(pos, u, **flags)
return [s.subs(reps) for s in soln]
except NotImplementedError:
pass
else:
pass # here for coverage
return # here for coverage
# TODO: option for calculating J numerically
@conserve_mpmath_dps
def nsolve(*args, dict=False, **kwargs):
r"""
Solve a nonlinear equation system numerically: ``nsolve(f, [args,] x0,
modules=['mpmath'], **kwargs)``.
Explanation
===========
``f`` is a vector function of symbolic expressions representing the system.
*args* are the variables. If there is only one variable, this argument can
be omitted. ``x0`` is a starting vector close to a solution.
Use the modules keyword to specify which modules should be used to
evaluate the function and the Jacobian matrix. Make sure to use a module
that supports matrices. For more information on the syntax, please see the
docstring of ``lambdify``.
If the keyword arguments contain ``dict=True`` (default is False) ``nsolve``
will return a list (perhaps empty) of solution mappings. This might be
especially useful if you want to use ``nsolve`` as a fallback to solve since
using the dict argument for both methods produces return values of
consistent type structure. Please note: to keep this consistent with
``solve``, the solution will be returned in a list even though ``nsolve``
(currently at least) only finds one solution at a time.
Overdetermined systems are supported.
Examples
========
>>> from sympy import Symbol, nsolve
>>> import mpmath
>>> mpmath.mp.dps = 15
>>> x1 = Symbol('x1')
>>> x2 = Symbol('x2')
>>> f1 = 3 * x1**2 - 2 * x2**2 - 1
>>> f2 = x1**2 - 2 * x1 + x2**2 + 2 * x2 - 8
>>> print(nsolve((f1, f2), (x1, x2), (-1, 1)))
Matrix([[-1.19287309935246], [1.27844411169911]])
For one-dimensional functions the syntax is simplified:
>>> from sympy import sin, nsolve
>>> from sympy.abc import x
>>> nsolve(sin(x), x, 2)
3.14159265358979
>>> nsolve(sin(x), 2)
3.14159265358979
To solve with higher precision than the default, use the prec argument:
>>> from sympy import cos
>>> nsolve(cos(x) - x, 1)
0.739085133215161
>>> nsolve(cos(x) - x, 1, prec=50)
0.73908513321516064165531208767387340401341175890076
>>> cos(_)
0.73908513321516064165531208767387340401341175890076
To solve for complex roots of real functions, a nonreal initial point
must be specified:
>>> from sympy import I
>>> nsolve(x**2 + 2, I)
1.4142135623731*I
``mpmath.findroot`` is used and you can find their more extensive
documentation, especially concerning keyword parameters and
available solvers. Note, however, that functions which are very
steep near the root, the verification of the solution may fail. In
this case you should use the flag ``verify=False`` and
independently verify the solution.
>>> from sympy import cos, cosh
>>> f = cos(x)*cosh(x) - 1
>>> nsolve(f, 3.14*100)
Traceback (most recent call last):
...
ValueError: Could not find root within given tolerance. (1.39267e+230 > 2.1684e-19)
>>> ans = nsolve(f, 3.14*100, verify=False); ans
312.588469032184
>>> f.subs(x, ans).n(2)
2.1e+121
>>> (f/f.diff(x)).subs(x, ans).n(2)
7.4e-15
One might safely skip the verification if bounds of the root are known
and a bisection method is used:
>>> bounds = lambda i: (3.14*i, 3.14*(i + 1))
>>> nsolve(f, bounds(100), solver='bisect', verify=False)
315.730061685774
Alternatively, a function may be better behaved when the
denominator is ignored. Since this is not always the case, however,
the decision of what function to use is left to the discretion of
the user.
>>> eq = x**2/(1 - x)/(1 - 2*x)**2 - 100
>>> nsolve(eq, 0.46)
Traceback (most recent call last):
...
ValueError: Could not find root within given tolerance. (10000 > 2.1684e-19)
Try another starting point or tweak arguments.
>>> nsolve(eq.as_numer_denom()[0], 0.46)
0.46792545969349058
"""
# there are several other SymPy functions that use method= so
# guard against that here
if 'method' in kwargs:
raise ValueError(filldedent('''
Keyword "method" should not be used in this context. When using
some mpmath solvers directly, the keyword "method" is
used, but when using nsolve (and findroot) the keyword to use is
"solver".'''))
if 'prec' in kwargs:
import mpmath
mpmath.mp.dps = kwargs.pop('prec')
# keyword argument to return result as a dictionary
as_dict = dict
from builtins import dict # to unhide the builtin
# interpret arguments
if len(args) == 3:
f = args[0]
fargs = args[1]
x0 = args[2]
if iterable(fargs) and iterable(x0):
if len(x0) != len(fargs):
raise TypeError('nsolve expected exactly %i guess vectors, got %i'
% (len(fargs), len(x0)))
elif len(args) == 2:
f = args[0]
fargs = None
x0 = args[1]
if iterable(f):
raise TypeError('nsolve expected 3 arguments, got 2')
elif len(args) < 2:
raise TypeError('nsolve expected at least 2 arguments, got %i'
% len(args))
else:
raise TypeError('nsolve expected at most 3 arguments, got %i'
% len(args))
modules = kwargs.get('modules', ['mpmath'])
if iterable(f):
f = list(f)
for i, fi in enumerate(f):
if isinstance(fi, Eq):
f[i] = fi.lhs - fi.rhs
f = Matrix(f).T
if iterable(x0):
x0 = list(x0)
if not isinstance(f, Matrix):
# assume it's a SymPy expression
if isinstance(f, Eq):
f = f.lhs - f.rhs
syms = f.free_symbols
if fargs is None:
fargs = syms.copy().pop()
if not (len(syms) == 1 and (fargs in syms or fargs[0] in syms)):
raise ValueError(filldedent('''
expected a one-dimensional and numerical function'''))
# the function is much better behaved if there is no denominator
# but sending the numerator is left to the user since sometimes
# the function is better behaved when the denominator is present
# e.g., issue 11768
f = lambdify(fargs, f, modules)
x = sympify(findroot(f, x0, **kwargs))
if as_dict:
return [{fargs: x}]
return x
if len(fargs) > f.cols:
raise NotImplementedError(filldedent('''
need at least as many equations as variables'''))
verbose = kwargs.get('verbose', False)
if verbose:
print('f(x):')
print(f)
# derive Jacobian
J = f.jacobian(fargs)
if verbose:
print('J(x):')
print(J)
# create functions
f = lambdify(fargs, f.T, modules)
J = lambdify(fargs, J, modules)
# solve the system numerically
x = findroot(f, x0, J=J, **kwargs)
if as_dict:
return [dict(zip(fargs, [sympify(xi) for xi in x]))]
return Matrix(x)
def _invert(eq, *symbols, **kwargs):
"""
Return tuple (i, d) where ``i`` is independent of *symbols* and ``d``
contains symbols.
Explanation
===========
``i`` and ``d`` are obtained after recursively using algebraic inversion
until an uninvertible ``d`` remains. If there are no free symbols then
``d`` will be zero. Some (but not necessarily all) solutions to the
expression ``i - d`` will be related to the solutions of the original
expression.
Examples
========
>>> from sympy.solvers.solvers import _invert as invert
>>> from sympy import sqrt, cos
>>> from sympy.abc import x, y
>>> invert(x - 3)
(3, x)
>>> invert(3)
(3, 0)
>>> invert(2*cos(x) - 1)
(1/2, cos(x))
>>> invert(sqrt(x) - 3)
(3, sqrt(x))
>>> invert(sqrt(x) + y, x)
(-y, sqrt(x))
>>> invert(sqrt(x) + y, y)
(-sqrt(x), y)
>>> invert(sqrt(x) + y, x, y)
(0, sqrt(x) + y)
If there is more than one symbol in a power's base and the exponent
is not an Integer, then the principal root will be used for the
inversion:
>>> invert(sqrt(x + y) - 2)
(4, x + y)
>>> invert(sqrt(x + y) - 2)
(4, x + y)
If the exponent is an Integer, setting ``integer_power`` to True
will force the principal root to be selected:
>>> invert(x**2 - 4, integer_power=True)
(2, x)
"""
eq = sympify(eq)
if eq.args:
# make sure we are working with flat eq
eq = eq.func(*eq.args)
free = eq.free_symbols
if not symbols:
symbols = free
if not free & set(symbols):
return eq, S.Zero
dointpow = bool(kwargs.get('integer_power', False))
lhs = eq
rhs = S.Zero
while True:
was = lhs
while True:
indep, dep = lhs.as_independent(*symbols)
# dep + indep == rhs
if lhs.is_Add:
# this indicates we have done it all
if indep.is_zero:
break
lhs = dep
rhs -= indep
# dep * indep == rhs
else:
# this indicates we have done it all
if indep is S.One:
break
lhs = dep
rhs /= indep
# collect like-terms in symbols
if lhs.is_Add:
terms = {}
for a in lhs.args:
i, d = a.as_independent(*symbols)
terms.setdefault(d, []).append(i)
if any(len(v) > 1 for v in terms.values()):
args = []
for d, i in terms.items():
if len(i) > 1:
args.append(Add(*i)*d)
else:
args.append(i[0]*d)
lhs = Add(*args)
# if it's a two-term Add with rhs = 0 and two powers we can get the
# dependent terms together, e.g. 3*f(x) + 2*g(x) -> f(x)/g(x) = -2/3
if lhs.is_Add and not rhs and len(lhs.args) == 2 and \
not lhs.is_polynomial(*symbols):
a, b = ordered(lhs.args)
ai, ad = a.as_independent(*symbols)
bi, bd = b.as_independent(*symbols)
if any(_ispow(i) for i in (ad, bd)):
a_base, a_exp = ad.as_base_exp()
b_base, b_exp = bd.as_base_exp()
if a_base == b_base:
# a = -b
lhs = powsimp(powdenest(ad/bd))
rhs = -bi/ai
else:
rat = ad/bd
_lhs = powsimp(ad/bd)
if _lhs != rat:
lhs = _lhs
rhs = -bi/ai
elif ai == -bi:
if isinstance(ad, Function) and ad.func == bd.func:
if len(ad.args) == len(bd.args) == 1:
lhs = ad.args[0] - bd.args[0]
elif len(ad.args) == len(bd.args):
# should be able to solve
# f(x, y) - f(2 - x, 0) == 0 -> x == 1
raise NotImplementedError(
'equal function with more than 1 argument')
else:
raise ValueError(
'function with different numbers of args')
elif lhs.is_Mul and any(_ispow(a) for a in lhs.args):
lhs = powsimp(powdenest(lhs))
if lhs.is_Function:
if hasattr(lhs, 'inverse') and lhs.inverse() is not None and len(lhs.args) == 1:
# -1
# f(x) = g -> x = f (g)
#
# /!\ inverse should not be defined if there are multiple values
# for the function -- these are handled in _tsolve
#
rhs = lhs.inverse()(rhs)
lhs = lhs.args[0]
elif isinstance(lhs, atan2):
y, x = lhs.args
lhs = 2*atan(y/(sqrt(x**2 + y**2) + x))
elif lhs.func == rhs.func:
if len(lhs.args) == len(rhs.args) == 1:
lhs = lhs.args[0]
rhs = rhs.args[0]
elif len(lhs.args) == len(rhs.args):
# should be able to solve
# f(x, y) == f(2, 3) -> x == 2
# f(x, x + y) == f(2, 3) -> x == 2
raise NotImplementedError(
'equal function with more than 1 argument')
else:
raise ValueError(
'function with different numbers of args')
if rhs and lhs.is_Pow and lhs.exp.is_Integer and lhs.exp < 0:
lhs = 1/lhs
rhs = 1/rhs
# base**a = b -> base = b**(1/a) if
# a is an Integer and dointpow=True (this gives real branch of root)
# a is not an Integer and the equation is multivariate and the
# base has more than 1 symbol in it
# The rationale for this is that right now the multi-system solvers
# doesn't try to resolve generators to see, for example, if the whole
# system is written in terms of sqrt(x + y) so it will just fail, so we
# do that step here.
if lhs.is_Pow and (
lhs.exp.is_Integer and dointpow or not lhs.exp.is_Integer and
len(symbols) > 1 and len(lhs.base.free_symbols & set(symbols)) > 1):
rhs = rhs**(1/lhs.exp)
lhs = lhs.base
if lhs == was:
break
return rhs, lhs
def unrad(eq, *syms, **flags):
"""
Remove radicals with symbolic arguments and return (eq, cov),
None, or raise an error.
Explanation
===========
None is returned if there are no radicals to remove.
NotImplementedError is raised if there are radicals and they cannot be
removed or if the relationship between the original symbols and the
change of variable needed to rewrite the system as a polynomial cannot
be solved.
Otherwise the tuple, ``(eq, cov)``, is returned where:
*eq*, ``cov``
*eq* is an equation without radicals (in the symbol(s) of
interest) whose solutions are a superset of the solutions to the
original expression. *eq* might be rewritten in terms of a new
variable; the relationship to the original variables is given by
``cov`` which is a list containing ``v`` and ``v**p - b`` where
``p`` is the power needed to clear the radical and ``b`` is the
radical now expressed as a polynomial in the symbols of interest.
For example, for sqrt(2 - x) the tuple would be
``(c, c**2 - 2 + x)``. The solutions of *eq* will contain
solutions to the original equation (if there are any).
*syms*
An iterable of symbols which, if provided, will limit the focus of
radical removal: only radicals with one or more of the symbols of
interest will be cleared. All free symbols are used if *syms* is not
set.
*flags* are used internally for communication during recursive calls.
Two options are also recognized:
``take``, when defined, is interpreted as a single-argument function
that returns True if a given Pow should be handled.
Radicals can be removed from an expression if:
* All bases of the radicals are the same; a change of variables is
done in this case.
* If all radicals appear in one term of the expression.
* There are only four terms with sqrt() factors or there are less than
four terms having sqrt() factors.
* There are only two terms with radicals.
Examples
========
>>> from sympy.solvers.solvers import unrad
>>> from sympy.abc import x
>>> from sympy import sqrt, Rational, root
>>> unrad(sqrt(x)*x**Rational(1, 3) + 2)
(x**5 - 64, [])
>>> unrad(sqrt(x) + root(x + 1, 3))
(-x**3 + x**2 + 2*x + 1, [])
>>> eq = sqrt(x) + root(x, 3) - 2
>>> unrad(eq)
(_p**3 + _p**2 - 2, [_p, _p**6 - x])
"""
uflags = {"check": False, "simplify": False}
def _cov(p, e):
if cov:
# XXX - uncovered
oldp, olde = cov
if Poly(e, p).degree(p) in (1, 2):
cov[:] = [p, olde.subs(oldp, _vsolve(e, p, **uflags)[0])]
else:
raise NotImplementedError
else:
cov[:] = [p, e]
def _canonical(eq, cov):
if cov:
# change symbol to vanilla so no solutions are eliminated
p, e = cov
rep = {p: Dummy(p.name)}
eq = eq.xreplace(rep)
cov = [p.xreplace(rep), e.xreplace(rep)]
# remove constants and powers of factors since these don't change
# the location of the root; XXX should factor or factor_terms be used?
eq = factor_terms(_mexpand(eq.as_numer_denom()[0], recursive=True), clear=True)
if eq.is_Mul:
args = []
for f in eq.args:
if f.is_number:
continue
if f.is_Pow:
args.append(f.base)
else:
args.append(f)
eq = Mul(*args) # leave as Mul for more efficient solving
# make the sign canonical
margs = list(Mul.make_args(eq))
changed = False
for i, m in enumerate(margs):
if m.could_extract_minus_sign():
margs[i] = -m
changed = True
if changed:
eq = Mul(*margs, evaluate=False)
return eq, cov
def _Q(pow):
# return leading Rational of denominator of Pow's exponent
c = pow.as_base_exp()[1].as_coeff_Mul()[0]
if not c.is_Rational:
return S.One
return c.q
# define the _take method that will determine whether a term is of interest
def _take(d):
# return True if coefficient of any factor's exponent's den is not 1
for pow in Mul.make_args(d):
if not pow.is_Pow:
continue
if _Q(pow) == 1:
continue
if pow.free_symbols & syms:
return True
return False
_take = flags.setdefault('_take', _take)
if isinstance(eq, Eq):
eq = eq.lhs - eq.rhs # XXX legacy Eq as Eqn support
elif not isinstance(eq, Expr):
return
cov, nwas, rpt = [flags.setdefault(k, v) for k, v in
sorted({"cov": [], "n": None, "rpt": 0}.items())]
# preconditioning
eq = powdenest(factor_terms(eq, radical=True, clear=True))
eq = eq.as_numer_denom()[0]
eq = _mexpand(eq, recursive=True)
if eq.is_number:
return
# see if there are radicals in symbols of interest
syms = set(syms) or eq.free_symbols # _take uses this
poly = eq.as_poly()
gens = [g for g in poly.gens if _take(g)]
if not gens:
return
# recast poly in terms of eigen-gens
poly = eq.as_poly(*gens)
# not a polynomial e.g. 1 + sqrt(x)*exp(sqrt(x)) with gen sqrt(x)
if poly is None:
return
# - an exponent has a symbol of interest (don't handle)
if any(g.exp.has(*syms) for g in gens):
return
def _rads_bases_lcm(poly):
# if all the bases are the same or all the radicals are in one
# term, `lcm` will be the lcm of the denominators of the
# exponents of the radicals
lcm = 1
rads = set()
bases = set()
for g in poly.gens:
q = _Q(g)
if q != 1:
rads.add(g)
lcm = ilcm(lcm, q)
bases.add(g.base)
return rads, bases, lcm
rads, bases, lcm = _rads_bases_lcm(poly)
covsym = Dummy('p', nonnegative=True)
# only keep in syms symbols that actually appear in radicals;
# and update gens
newsyms = set()
for r in rads:
newsyms.update(syms & r.free_symbols)
if newsyms != syms:
syms = newsyms
# get terms together that have common generators
drad = dict(zip(rads, range(len(rads))))
rterms = {(): []}
args = Add.make_args(poly.as_expr())
for t in args:
if _take(t):
common = set(t.as_poly().gens).intersection(rads)
key = tuple(sorted([drad[i] for i in common]))
else:
key = ()
rterms.setdefault(key, []).append(t)
others = Add(*rterms.pop(()))
rterms = [Add(*rterms[k]) for k in rterms.keys()]
# the output will depend on the order terms are processed, so
# make it canonical quickly
rterms = list(reversed(list(ordered(rterms))))
ok = False # we don't have a solution yet
depth = sqrt_depth(eq)
if len(rterms) == 1 and not (rterms[0].is_Add and lcm > 2):
eq = rterms[0]**lcm - ((-others)**lcm)
ok = True
else:
if len(rterms) == 1 and rterms[0].is_Add:
rterms = list(rterms[0].args)
if len(bases) == 1:
b = bases.pop()
if len(syms) > 1:
x = b.free_symbols
else:
x = syms
x = list(ordered(x))[0]
try:
inv = _vsolve(covsym**lcm - b, x, **uflags)
if not inv:
raise NotImplementedError
eq = poly.as_expr().subs(b, covsym**lcm).subs(x, inv[0])
_cov(covsym, covsym**lcm - b)
return _canonical(eq, cov)
except NotImplementedError:
pass
if len(rterms) == 2:
if not others:
eq = rterms[0]**lcm - (-rterms[1])**lcm
ok = True
elif not log(lcm, 2).is_Integer:
# the lcm-is-power-of-two case is handled below
r0, r1 = rterms
if flags.get('_reverse', False):
r1, r0 = r0, r1
i0 = _rads0, _bases0, lcm0 = _rads_bases_lcm(r0.as_poly())
i1 = _rads1, _bases1, lcm1 = _rads_bases_lcm(r1.as_poly())
for reverse in range(2):
if reverse:
i0, i1 = i1, i0
r0, r1 = r1, r0
_rads1, _, lcm1 = i1
_rads1 = Mul(*_rads1)
t1 = _rads1**lcm1
c = covsym**lcm1 - t1
for x in syms:
try:
sol = _vsolve(c, x, **uflags)
if not sol:
raise NotImplementedError
neweq = r0.subs(x, sol[0]) + covsym*r1/_rads1 + \
others
tmp = unrad(neweq, covsym)
if tmp:
eq, newcov = tmp
if newcov:
newp, newc = newcov
_cov(newp, c.subs(covsym,
_vsolve(newc, covsym, **uflags)[0]))
else:
_cov(covsym, c)
else:
eq = neweq
_cov(covsym, c)
ok = True
break
except NotImplementedError:
if reverse:
raise NotImplementedError(
'no successful change of variable found')
else:
pass
if ok:
break
elif len(rterms) == 3:
# two cube roots and another with order less than 5
# (so an analytical solution can be found) or a base
# that matches one of the cube root bases
info = [_rads_bases_lcm(i.as_poly()) for i in rterms]
RAD = 0
BASES = 1
LCM = 2
if info[0][LCM] != 3:
info.append(info.pop(0))
rterms.append(rterms.pop(0))
elif info[1][LCM] != 3:
info.append(info.pop(1))
rterms.append(rterms.pop(1))
if info[0][LCM] == info[1][LCM] == 3:
if info[1][BASES] != info[2][BASES]:
info[0], info[1] = info[1], info[0]
rterms[0], rterms[1] = rterms[1], rterms[0]
if info[1][BASES] == info[2][BASES]:
eq = rterms[0]**3 + (rterms[1] + rterms[2] + others)**3
ok = True
elif info[2][LCM] < 5:
# a*root(A, 3) + b*root(B, 3) + others = c
a, b, c, d, A, B = [Dummy(i) for i in 'abcdAB']
# zz represents the unraded expression into which the
# specifics for this case are substituted
zz = (c - d)*(A**3*a**9 + 3*A**2*B*a**6*b**3 -
3*A**2*a**6*c**3 + 9*A**2*a**6*c**2*d - 9*A**2*a**6*c*d**2 +
3*A**2*a**6*d**3 + 3*A*B**2*a**3*b**6 + 21*A*B*a**3*b**3*c**3 -
63*A*B*a**3*b**3*c**2*d + 63*A*B*a**3*b**3*c*d**2 -
21*A*B*a**3*b**3*d**3 + 3*A*a**3*c**6 - 18*A*a**3*c**5*d +
45*A*a**3*c**4*d**2 - 60*A*a**3*c**3*d**3 + 45*A*a**3*c**2*d**4 -
18*A*a**3*c*d**5 + 3*A*a**3*d**6 + B**3*b**9 - 3*B**2*b**6*c**3 +
9*B**2*b**6*c**2*d - 9*B**2*b**6*c*d**2 + 3*B**2*b**6*d**3 +
3*B*b**3*c**6 - 18*B*b**3*c**5*d + 45*B*b**3*c**4*d**2 -
60*B*b**3*c**3*d**3 + 45*B*b**3*c**2*d**4 - 18*B*b**3*c*d**5 +
3*B*b**3*d**6 - c**9 + 9*c**8*d - 36*c**7*d**2 + 84*c**6*d**3 -
126*c**5*d**4 + 126*c**4*d**5 - 84*c**3*d**6 + 36*c**2*d**7 -
9*c*d**8 + d**9)
def _t(i):
b = Mul(*info[i][RAD])
return cancel(rterms[i]/b), Mul(*info[i][BASES])
aa, AA = _t(0)
bb, BB = _t(1)
cc = -rterms[2]
dd = others
eq = zz.xreplace(dict(zip(
(a, A, b, B, c, d),
(aa, AA, bb, BB, cc, dd))))
ok = True
# handle power-of-2 cases
if not ok:
if log(lcm, 2).is_Integer and (not others and
len(rterms) == 4 or len(rterms) < 4):
def _norm2(a, b):
return a**2 + b**2 + 2*a*b
if len(rterms) == 4:
# (r0+r1)**2 - (r2+r3)**2
r0, r1, r2, r3 = rterms
eq = _norm2(r0, r1) - _norm2(r2, r3)
ok = True
elif len(rterms) == 3:
# (r1+r2)**2 - (r0+others)**2
r0, r1, r2 = rterms
eq = _norm2(r1, r2) - _norm2(r0, others)
ok = True
elif len(rterms) == 2:
# r0**2 - (r1+others)**2
r0, r1 = rterms
eq = r0**2 - _norm2(r1, others)
ok = True
new_depth = sqrt_depth(eq) if ok else depth
rpt += 1 # XXX how many repeats with others unchanging is enough?
if not ok or (
nwas is not None and len(rterms) == nwas and
new_depth is not None and new_depth == depth and
rpt > 3):
raise NotImplementedError('Cannot remove all radicals')
flags.update({"cov": cov, "n": len(rterms), "rpt": rpt})
neq = unrad(eq, *syms, **flags)
if neq:
eq, cov = neq
eq, cov = _canonical(eq, cov)
return eq, cov
# delayed imports
from sympy.solvers.bivariate import (
bivariate_type, _solve_lambert, _filtered_gens)