from __future__ import print_function, division
from collections import defaultdict
from sympy.core import (Basic, S, Add, Mul, Pow,
Symbol, sympify, expand_mul, expand_func,
Function, Dummy, Expr, factor_terms,
symbols, expand_power_exp)
from sympy.core.compatibility import (iterable,
ordered, range, as_int)
from sympy.core.numbers import Float, I, pi, Rational, Integer
from sympy.core.function import expand_log, count_ops, _mexpand, _coeff_isneg
from sympy.core.rules import Transform
from sympy.core.evaluate import global_evaluate
from sympy.functions import (
gamma, exp, sqrt, log, exp_polar, piecewise_fold)
from sympy.functions.elementary.exponential import ExpBase
from sympy.functions.elementary.hyperbolic import HyperbolicFunction
from sympy.functions.elementary.integers import ceiling
from sympy.functions.elementary.complexes import unpolarify
from sympy.functions.elementary.trigonometric import TrigonometricFunction
from sympy.functions.combinatorial.factorials import CombinatorialFunction
from sympy.functions.special.bessel import besselj, besseli, besselk, jn, bessely
from sympy.utilities.iterables import has_variety
from sympy.simplify.radsimp import radsimp, fraction
from sympy.simplify.trigsimp import trigsimp, exptrigsimp
from sympy.simplify.powsimp import powsimp
from sympy.simplify.cse_opts import sub_pre, sub_post
from sympy.simplify.sqrtdenest import sqrtdenest
from sympy.simplify.combsimp import combsimp
from sympy.polys import (together, cancel, factor)
import mpmath
[docs]def separatevars(expr, symbols=[], dict=False, force=False):
"""
Separates variables in an expression, if possible. By
default, it separates with respect to all symbols in an
expression and collects constant coefficients that are
independent of symbols.
If dict=True then the separated terms will be returned
in a dictionary keyed to their corresponding symbols.
By default, all symbols in the expression will appear as
keys; if symbols are provided, then all those symbols will
be used as keys, and any terms in the expression containing
other symbols or non-symbols will be returned keyed to the
string 'coeff'. (Passing None for symbols will return the
expression in a dictionary keyed to 'coeff'.)
If force=True, then bases of powers will be separated regardless
of assumptions on the symbols involved.
Notes
=====
The order of the factors is determined by Mul, so that the
separated expressions may not necessarily be grouped together.
Although factoring is necessary to separate variables in some
expressions, it is not necessary in all cases, so one should not
count on the returned factors being factored.
Examples
========
>>> from sympy.abc import x, y, z, alpha
>>> from sympy import separatevars, sin
>>> separatevars((x*y)**y)
(x*y)**y
>>> separatevars((x*y)**y, force=True)
x**y*y**y
>>> e = 2*x**2*z*sin(y)+2*z*x**2
>>> separatevars(e)
2*x**2*z*(sin(y) + 1)
>>> separatevars(e, symbols=(x, y), dict=True)
{'coeff': 2*z, x: x**2, y: sin(y) + 1}
>>> separatevars(e, [x, y, alpha], dict=True)
{'coeff': 2*z, alpha: 1, x: x**2, y: sin(y) + 1}
If the expression is not really separable, or is only partially
separable, separatevars will do the best it can to separate it
by using factoring.
>>> separatevars(x + x*y - 3*x**2)
-x*(3*x - y - 1)
If the expression is not separable then expr is returned unchanged
or (if dict=True) then None is returned.
>>> eq = 2*x + y*sin(x)
>>> separatevars(eq) == eq
True
>>> separatevars(2*x + y*sin(x), symbols=(x, y), dict=True) == None
True
"""
expr = sympify(expr)
if dict:
return _separatevars_dict(_separatevars(expr, force), symbols)
else:
return _separatevars(expr, force)
def _separatevars(expr, force):
if len(expr.free_symbols) == 1:
return expr
# don't destroy a Mul since much of the work may already be done
if expr.is_Mul:
args = list(expr.args)
changed = False
for i, a in enumerate(args):
args[i] = separatevars(a, force)
changed = changed or args[i] != a
if changed:
expr = expr.func(*args)
return expr
# get a Pow ready for expansion
if expr.is_Pow:
expr = Pow(separatevars(expr.base, force=force), expr.exp)
# First try other expansion methods
expr = expr.expand(mul=False, multinomial=False, force=force)
_expr, reps = posify(expr) if force else (expr, {})
expr = factor(_expr).subs(reps)
if not expr.is_Add:
return expr
# Find any common coefficients to pull out
args = list(expr.args)
commonc = args[0].args_cnc(cset=True, warn=False)[0]
for i in args[1:]:
commonc &= i.args_cnc(cset=True, warn=False)[0]
commonc = Mul(*commonc)
commonc = commonc.as_coeff_Mul()[1] # ignore constants
commonc_set = commonc.args_cnc(cset=True, warn=False)[0]
# remove them
for i, a in enumerate(args):
c, nc = a.args_cnc(cset=True, warn=False)
c = c - commonc_set
args[i] = Mul(*c)*Mul(*nc)
nonsepar = Add(*args)
if len(nonsepar.free_symbols) > 1:
_expr = nonsepar
_expr, reps = posify(_expr) if force else (_expr, {})
_expr = (factor(_expr)).subs(reps)
if not _expr.is_Add:
nonsepar = _expr
return commonc*nonsepar
def _separatevars_dict(expr, symbols):
if symbols:
if not all((t.is_Atom for t in symbols)):
raise ValueError("symbols must be Atoms.")
symbols = list(symbols)
elif symbols is None:
return {'coeff': expr}
else:
symbols = list(expr.free_symbols)
if not symbols:
return None
ret = dict(((i, []) for i in symbols + ['coeff']))
for i in Mul.make_args(expr):
expsym = i.free_symbols
intersection = set(symbols).intersection(expsym)
if len(intersection) > 1:
return None
if len(intersection) == 0:
# There are no symbols, so it is part of the coefficient
ret['coeff'].append(i)
else:
ret[intersection.pop()].append(i)
# rebuild
for k, v in ret.items():
ret[k] = Mul(*v)
return ret
def _is_sum_surds(p):
args = p.args if p.is_Add else [p]
for y in args:
if not ((y**2).is_Rational and y.is_real):
return False
return True
[docs]def posify(eq):
"""Return eq (with generic symbols made positive) and a
dictionary containing the mapping between the old and new
symbols.
Any symbol that has positive=None will be replaced with a positive dummy
symbol having the same name. This replacement will allow more symbolic
processing of expressions, especially those involving powers and
logarithms.
A dictionary that can be sent to subs to restore eq to its original
symbols is also returned.
>>> from sympy import posify, Symbol, log, solve
>>> from sympy.abc import x
>>> posify(x + Symbol('p', positive=True) + Symbol('n', negative=True))
(_x + n + p, {_x: x})
>>> eq = 1/x
>>> log(eq).expand()
log(1/x)
>>> log(posify(eq)[0]).expand()
-log(_x)
>>> p, rep = posify(eq)
>>> log(p).expand().subs(rep)
-log(x)
It is possible to apply the same transformations to an iterable
of expressions:
>>> eq = x**2 - 4
>>> solve(eq, x)
[-2, 2]
>>> eq_x, reps = posify([eq, x]); eq_x
[_x**2 - 4, _x]
>>> solve(*eq_x)
[2]
"""
eq = sympify(eq)
if iterable(eq):
f = type(eq)
eq = list(eq)
syms = set()
for e in eq:
syms = syms.union(e.atoms(Symbol))
reps = {}
for s in syms:
reps.update(dict((v, k) for k, v in posify(s)[1].items()))
for i, e in enumerate(eq):
eq[i] = e.subs(reps)
return f(eq), {r: s for s, r in reps.items()}
reps = dict([(s, Dummy(s.name, positive=True))
for s in eq.free_symbols if s.is_positive is None])
eq = eq.subs(reps)
return eq, {r: s for s, r in reps.items()}
[docs]def hypersimp(f, k):
"""Given combinatorial term f(k) simplify its consecutive term ratio
i.e. f(k+1)/f(k). The input term can be composed of functions and
integer sequences which have equivalent representation in terms
of gamma special function.
The algorithm performs three basic steps:
1. Rewrite all functions in terms of gamma, if possible.
2. Rewrite all occurrences of gamma in terms of products
of gamma and rising factorial with integer, absolute
constant exponent.
3. Perform simplification of nested fractions, powers
and if the resulting expression is a quotient of
polynomials, reduce their total degree.
If f(k) is hypergeometric then as result we arrive with a
quotient of polynomials of minimal degree. Otherwise None
is returned.
For more information on the implemented algorithm refer to:
1. W. Koepf, Algorithms for m-fold Hypergeometric Summation,
Journal of Symbolic Computation (1995) 20, 399-417
"""
f = sympify(f)
g = f.subs(k, k + 1) / f
g = g.rewrite(gamma)
g = expand_func(g)
g = powsimp(g, deep=True, combine='exp')
if g.is_rational_function(k):
return simplify(g, ratio=S.Infinity)
else:
return None
[docs]def hypersimilar(f, g, k):
"""Returns True if 'f' and 'g' are hyper-similar.
Similarity in hypergeometric sense means that a quotient of
f(k) and g(k) is a rational function in k. This procedure
is useful in solving recurrence relations.
For more information see hypersimp().
"""
f, g = list(map(sympify, (f, g)))
h = (f/g).rewrite(gamma)
h = h.expand(func=True, basic=False)
return h.is_rational_function(k)
def signsimp(expr, evaluate=None):
"""Make all Add sub-expressions canonical wrt sign.
If an Add subexpression, ``a``, can have a sign extracted,
as determined by could_extract_minus_sign, it is replaced
with Mul(-1, a, evaluate=False). This allows signs to be
extracted from powers and products.
Examples
========
>>> from sympy import signsimp, exp, symbols
>>> from sympy.abc import x, y
>>> i = symbols('i', odd=True)
>>> n = -1 + 1/x
>>> n/x/(-n)**2 - 1/n/x
(-1 + 1/x)/(x*(1 - 1/x)**2) - 1/(x*(-1 + 1/x))
>>> signsimp(_)
0
>>> x*n + x*-n
x*(-1 + 1/x) + x*(1 - 1/x)
>>> signsimp(_)
0
Since powers automatically handle leading signs
>>> (-2)**i
-2**i
signsimp can be used to put the base of a power with an integer
exponent into canonical form:
>>> n**i
(-1 + 1/x)**i
By default, signsimp doesn't leave behind any hollow simplification:
if making an Add canonical wrt sign didn't change the expression, the
original Add is restored. If this is not desired then the keyword
``evaluate`` can be set to False:
>>> e = exp(y - x)
>>> signsimp(e) == e
True
>>> signsimp(e, evaluate=False)
exp(-(x - y))
"""
if evaluate is None:
evaluate = global_evaluate[0]
expr = sympify(expr)
if not isinstance(expr, Expr) or expr.is_Atom:
return expr
e = sub_post(sub_pre(expr))
if not isinstance(e, Expr) or e.is_Atom:
return e
if e.is_Add:
return e.func(*[signsimp(a) for a in e.args])
if evaluate:
e = e.xreplace({m: -(-m) for m in e.atoms(Mul) if -(-m) != m})
return e
[docs]def simplify(expr, ratio=1.7, measure=count_ops, fu=False):
"""
Simplifies the given expression.
Simplification is not a well defined term and the exact strategies
this function tries can change in the future versions of SymPy. If
your algorithm relies on "simplification" (whatever it is), try to
determine what you need exactly - is it powsimp()?, radsimp()?,
together()?, logcombine()?, or something else? And use this particular
function directly, because those are well defined and thus your algorithm
will be robust.
Nonetheless, especially for interactive use, or when you don't know
anything about the structure of the expression, simplify() tries to apply
intelligent heuristics to make the input expression "simpler". For
example:
>>> from sympy import simplify, cos, sin
>>> from sympy.abc import x, y
>>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
>>> a
(x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
>>> simplify(a)
x + 1
Note that we could have obtained the same result by using specific
simplification functions:
>>> from sympy import trigsimp, cancel
>>> trigsimp(a)
(x**2 + x)/x
>>> cancel(_)
x + 1
In some cases, applying :func:`simplify` may actually result in some more
complicated expression. The default ``ratio=1.7`` prevents more extreme
cases: if (result length)/(input length) > ratio, then input is returned
unmodified. The ``measure`` parameter lets you specify the function used
to determine how complex an expression is. The function should take a
single argument as an expression and return a number such that if
expression ``a`` is more complex than expression ``b``, then
``measure(a) > measure(b)``. The default measure function is
:func:`count_ops`, which returns the total number of operations in the
expression.
For example, if ``ratio=1``, ``simplify`` output can't be longer
than input.
::
>>> from sympy import sqrt, simplify, count_ops, oo
>>> root = 1/(sqrt(2)+3)
Since ``simplify(root)`` would result in a slightly longer expression,
root is returned unchanged instead::
>>> simplify(root, ratio=1) == root
True
If ``ratio=oo``, simplify will be applied anyway::
>>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
True
Note that the shortest expression is not necessary the simplest, so
setting ``ratio`` to 1 may not be a good idea.
Heuristically, the default value ``ratio=1.7`` seems like a reasonable
choice.
You can easily define your own measure function based on what you feel
should represent the "size" or "complexity" of the input expression. Note
that some choices, such as ``lambda expr: len(str(expr))`` may appear to be
good metrics, but have other problems (in this case, the measure function
may slow down simplify too much for very large expressions). If you don't
know what a good metric would be, the default, ``count_ops``, is a good
one.
For example:
>>> from sympy import symbols, log
>>> a, b = symbols('a b', positive=True)
>>> g = log(a) + log(b) + log(a)*log(1/b)
>>> h = simplify(g)
>>> h
log(a*b**(-log(a) + 1))
>>> count_ops(g)
8
>>> count_ops(h)
5
So you can see that ``h`` is simpler than ``g`` using the count_ops metric.
However, we may not like how ``simplify`` (in this case, using
``logcombine``) has created the ``b**(log(1/a) + 1)`` term. A simple way
to reduce this would be to give more weight to powers as operations in
``count_ops``. We can do this by using the ``visual=True`` option:
>>> print(count_ops(g, visual=True))
2*ADD + DIV + 4*LOG + MUL
>>> print(count_ops(h, visual=True))
2*LOG + MUL + POW + SUB
>>> from sympy import Symbol, S
>>> def my_measure(expr):
... POW = Symbol('POW')
... # Discourage powers by giving POW a weight of 10
... count = count_ops(expr, visual=True).subs(POW, 10)
... # Every other operation gets a weight of 1 (the default)
... count = count.replace(Symbol, type(S.One))
... return count
>>> my_measure(g)
8
>>> my_measure(h)
14
>>> 15./8 > 1.7 # 1.7 is the default ratio
True
>>> simplify(g, measure=my_measure)
-log(a)*log(b) + log(a) + log(b)
Note that because ``simplify()`` internally tries many different
simplification strategies and then compares them using the measure
function, we get a completely different result that is still different
from the input expression by doing this.
"""
expr = sympify(expr)
try:
return expr._eval_simplify(ratio=ratio, measure=measure)
except AttributeError:
pass
original_expr = expr = signsimp(expr)
from sympy.simplify.hyperexpand import hyperexpand
from sympy.functions.special.bessel import BesselBase
from sympy import Sum, Product
if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack
return expr
if not isinstance(expr, (Add, Mul, Pow, ExpBase)):
if isinstance(expr, Function) and hasattr(expr, "inverse"):
if len(expr.args) == 1 and len(expr.args[0].args) == 1 and \
isinstance(expr.args[0], expr.inverse(argindex=1)):
return simplify(expr.args[0].args[0], ratio=ratio,
measure=measure, fu=fu)
return expr.func(*[simplify(x, ratio=ratio, measure=measure, fu=fu)
for x in expr.args])
# TODO: Apply different strategies, considering expression pattern:
# is it a purely rational function? Is there any trigonometric function?...
# See also https://github.com/sympy/sympy/pull/185.
def shorter(*choices):
'''Return the choice that has the fewest ops. In case of a tie,
the expression listed first is selected.'''
if not has_variety(choices):
return choices[0]
return min(choices, key=measure)
expr = bottom_up(expr, lambda w: w.normal())
expr = Mul(*powsimp(expr).as_content_primitive())
_e = cancel(expr)
expr1 = shorter(_e, _mexpand(_e).cancel()) # issue 6829
expr2 = shorter(together(expr, deep=True), together(expr1, deep=True))
if ratio is S.Infinity:
expr = expr2
else:
expr = shorter(expr2, expr1, expr)
if not isinstance(expr, Basic): # XXX: temporary hack
return expr
expr = factor_terms(expr, sign=False)
# hyperexpand automatically only works on hypergeometric terms
expr = hyperexpand(expr)
expr = piecewise_fold(expr)
if expr.has(BesselBase):
expr = besselsimp(expr)
if expr.has(TrigonometricFunction) and not fu or expr.has(
HyperbolicFunction):
expr = trigsimp(expr, deep=True)
if expr.has(log):
expr = shorter(expand_log(expr, deep=True), logcombine(expr))
if expr.has(CombinatorialFunction, gamma):
expr = combsimp(expr)
if expr.has(Sum):
expr = sum_simplify(expr)
if expr.has(Product):
expr = product_simplify(expr)
short = shorter(powsimp(expr, combine='exp', deep=True), powsimp(expr), expr)
short = shorter(short, factor_terms(short), expand_power_exp(expand_mul(short)))
if short.has(TrigonometricFunction, HyperbolicFunction, ExpBase):
short = exptrigsimp(short, simplify=False)
# get rid of hollow 2-arg Mul factorization
hollow_mul = Transform(
lambda x: Mul(*x.args),
lambda x:
x.is_Mul and
len(x.args) == 2 and
x.args[0].is_Number and
x.args[1].is_Add and
x.is_commutative)
expr = short.xreplace(hollow_mul)
numer, denom = expr.as_numer_denom()
if denom.is_Add:
n, d = fraction(radsimp(1/denom, symbolic=False, max_terms=1))
if n is not S.One:
expr = (numer*n).expand()/d
if expr.could_extract_minus_sign():
n, d = fraction(expr)
if d != 0:
expr = signsimp(-n/(-d))
if measure(expr) > ratio*measure(original_expr):
expr = original_expr
return expr
def sum_simplify(s):
"""Main function for Sum simplification"""
from sympy.concrete.summations import Sum
terms = Add.make_args(s)
s_t = [] # Sum Terms
o_t = [] # Other Terms
for term in terms:
if isinstance(term, Mul):
constant = 1
other = 1
s = 0
n_sum_terms = 0
for j in range(len(term.args)):
if isinstance(term.args[j], Sum):
s = term.args[j]
n_sum_terms = n_sum_terms + 1
elif term.args[j].is_number == True:
constant = constant * term.args[j]
else:
other = other * term.args[j]
if other == 1 and n_sum_terms == 1:
# Insert the constant inside the Sum
s_t.append(Sum(constant * s.function, *s.limits))
elif other != 1 and n_sum_terms == 1:
o_t.append(other * Sum(constant * s.function, *s.limits))
else:
o_t.append(term)
elif isinstance(term, Sum):
s_t.append(term)
else:
o_t.append(term)
used = [False] * len(s_t)
for method in range(2):
for i, s_term1 in enumerate(s_t):
if not used[i]:
for j, s_term2 in enumerate(s_t):
if not used[j] and i != j:
temp = sum_add(s_term1, s_term2, method)
if isinstance(temp, Sum):
s_t[i] = temp
s_term1 = s_t[i]
used[j] = True
result = Add(*o_t)
for i, s_term in enumerate(s_t):
if not used[i]:
result = Add(result, s_term)
return result
def sum_add(self, other, method=0):
"""Helper function for Sum simplification"""
from sympy.concrete.summations import Sum
if type(self) == type(other):
if method == 0:
if self.limits == other.limits:
return Sum(self.function + other.function, *self.limits)
elif method == 1:
if simplify(self.function - other.function) == 0:
if len(self.limits) == len(other.limits) == 1:
i = self.limits[0][0]
x1 = self.limits[0][1]
y1 = self.limits[0][2]
j = other.limits[0][0]
x2 = other.limits[0][1]
y2 = other.limits[0][2]
if i == j:
if x2 == y1 + 1:
return Sum(self.function, (i, x1, y2))
elif x1 == y2 + 1:
return Sum(self.function, (i, x2, y1))
return Add(self, other)
def product_simplify(s):
"""Main function for Product simplification"""
from sympy.concrete.products import Product
terms = Mul.make_args(s)
p_t = [] # Product Terms
o_t = [] # Other Terms
for term in terms:
if isinstance(term, Product):
p_t.append(term)
else:
o_t.append(term)
used = [False] * len(p_t)
for method in range(2):
for i, p_term1 in enumerate(p_t):
if not used[i]:
for j, p_term2 in enumerate(p_t):
if not used[j] and i != j:
if isinstance(product_mul(p_term1, p_term2, method), Product):
p_t[i] = product_mul(p_term1, p_term2, method)
used[j] = True
result = Mul(*o_t)
for i, p_term in enumerate(p_t):
if not used[i]:
result = Mul(result, p_term)
return result
def product_mul(self, other, method=0):
"""Helper function for Product simplification"""
from sympy.concrete.products import Product
if type(self) == type(other):
if method == 0:
if self.limits == other.limits:
return Product(self.function * other.function, *self.limits)
elif method == 1:
if simplify(self.function - other.function) == 0:
if len(self.limits) == len(other.limits) == 1:
i = self.limits[0][0]
x1 = self.limits[0][1]
y1 = self.limits[0][2]
j = other.limits[0][0]
x2 = other.limits[0][1]
y2 = other.limits[0][2]
if i == j:
if x2 == y1 + 1:
return Product(self.function, (i, x1, y2))
elif x1 == y2 + 1:
return Product(self.function, (i, x2, y1))
return Mul(self, other)
def _nthroot_solve(p, n, prec):
"""
helper function for ``nthroot``
It denests ``p**Rational(1, n)`` using its minimal polynomial
"""
from sympy.polys.numberfields import _minimal_polynomial_sq
from sympy.solvers import solve
while n % 2 == 0:
p = sqrtdenest(sqrt(p))
n = n // 2
if n == 1:
return p
pn = p**Rational(1, n)
x = Symbol('x')
f = _minimal_polynomial_sq(p, n, x)
if f is None:
return None
sols = solve(f, x)
for sol in sols:
if abs(sol - pn).n() < 1./10**prec:
sol = sqrtdenest(sol)
if _mexpand(sol**n) == p:
return sol
[docs]def logcombine(expr, force=False):
"""
Takes logarithms and combines them using the following rules:
- log(x) + log(y) == log(x*y) if both are not negative
- a*log(x) == log(x**a) if x is positive and a is real
If ``force`` is True then the assumptions above will be assumed to hold if
there is no assumption already in place on a quantity. For example, if
``a`` is imaginary or the argument negative, force will not perform a
combination but if ``a`` is a symbol with no assumptions the change will
take place.
Examples
========
>>> from sympy import Symbol, symbols, log, logcombine, I
>>> from sympy.abc import a, x, y, z
>>> logcombine(a*log(x) + log(y) - log(z))
a*log(x) + log(y) - log(z)
>>> logcombine(a*log(x) + log(y) - log(z), force=True)
log(x**a*y/z)
>>> x,y,z = symbols('x,y,z', positive=True)
>>> a = Symbol('a', real=True)
>>> logcombine(a*log(x) + log(y) - log(z))
log(x**a*y/z)
The transformation is limited to factors and/or terms that
contain logs, so the result depends on the initial state of
expansion:
>>> eq = (2 + 3*I)*log(x)
>>> logcombine(eq, force=True) == eq
True
>>> logcombine(eq.expand(), force=True)
log(x**2) + I*log(x**3)
See Also
========
posify: replace all symbols with symbols having positive assumptions
"""
def f(rv):
if not (rv.is_Add or rv.is_Mul):
return rv
def gooda(a):
# bool to tell whether the leading ``a`` in ``a*log(x)``
# could appear as log(x**a)
return (a is not S.NegativeOne and # -1 *could* go, but we disallow
(a.is_real or force and a.is_real is not False))
def goodlog(l):
# bool to tell whether log ``l``'s argument can combine with others
a = l.args[0]
return a.is_positive or force and a.is_nonpositive is not False
other = []
logs = []
log1 = defaultdict(list)
for a in Add.make_args(rv):
if a.func is log and goodlog(a):
log1[()].append(([], a))
elif not a.is_Mul:
other.append(a)
else:
ot = []
co = []
lo = []
for ai in a.args:
if ai.is_Rational and ai < 0:
ot.append(S.NegativeOne)
co.append(-ai)
elif ai.func is log and goodlog(ai):
lo.append(ai)
elif gooda(ai):
co.append(ai)
else:
ot.append(ai)
if len(lo) > 1:
logs.append((ot, co, lo))
elif lo:
log1[tuple(ot)].append((co, lo[0]))
else:
other.append(a)
# if there is only one log at each coefficient and none have
# an exponent to place inside the log then there is nothing to do
if not logs and all(len(log1[k]) == 1 and log1[k][0] == [] for k in log1):
return rv
# collapse multi-logs as far as possible in a canonical way
# TODO: see if x*log(a)+x*log(a)*log(b) -> x*log(a)*(1+log(b))?
# -- in this case, it's unambiguous, but if it were were a log(c) in
# each term then it's arbitrary whether they are grouped by log(a) or
# by log(c). So for now, just leave this alone; it's probably better to
# let the user decide
for o, e, l in logs:
l = list(ordered(l))
e = log(l.pop(0).args[0]**Mul(*e))
while l:
li = l.pop(0)
e = log(li.args[0]**e)
c, l = Mul(*o), e
if l.func is log: # it should be, but check to be sure
log1[(c,)].append(([], l))
else:
other.append(c*l)
# logs that have the same coefficient can multiply
for k in list(log1.keys()):
log1[Mul(*k)] = log(logcombine(Mul(*[
l.args[0]**Mul(*c) for c, l in log1.pop(k)]),
force=force))
# logs that have oppositely signed coefficients can divide
for k in ordered(list(log1.keys())):
if not k in log1: # already popped as -k
continue
if -k in log1:
# figure out which has the minus sign; the one with
# more op counts should be the one
num, den = k, -k
if num.count_ops() > den.count_ops():
num, den = den, num
other.append(num*log(log1.pop(num).args[0]/log1.pop(den).args[0]))
else:
other.append(k*log1.pop(k))
return Add(*other)
return bottom_up(expr, f)
def bottom_up(rv, F, atoms=False, nonbasic=False):
"""Apply ``F`` to all expressions in an expression tree from the
bottom up. If ``atoms`` is True, apply ``F`` even if there are no args;
if ``nonbasic`` is True, try to apply ``F`` to non-Basic objects.
"""
try:
if rv.args:
args = tuple([bottom_up(a, F, atoms, nonbasic)
for a in rv.args])
if args != rv.args:
rv = rv.func(*args)
rv = F(rv)
elif atoms:
rv = F(rv)
except AttributeError:
if nonbasic:
try:
rv = F(rv)
except TypeError:
pass
return rv
[docs]def besselsimp(expr):
"""
Simplify bessel-type functions.
This routine tries to simplify bessel-type functions. Currently it only
works on the Bessel J and I functions, however. It works by looking at all
such functions in turn, and eliminating factors of "I" and "-1" (actually
their polar equivalents) in front of the argument. Then, functions of
half-integer order are rewritten using strigonometric functions and
functions of integer order (> 1) are rewritten using functions
of low order. Finally, if the expression was changed, compute
factorization of the result with factor().
>>> from sympy import besselj, besseli, besselsimp, polar_lift, I, S
>>> from sympy.abc import z, nu
>>> besselsimp(besselj(nu, z*polar_lift(-1)))
exp(I*pi*nu)*besselj(nu, z)
>>> besselsimp(besseli(nu, z*polar_lift(-I)))
exp(-I*pi*nu/2)*besselj(nu, z)
>>> besselsimp(besseli(S(-1)/2, z))
sqrt(2)*cosh(z)/(sqrt(pi)*sqrt(z))
>>> besselsimp(z*besseli(0, z) + z*(besseli(2, z))/2 + besseli(1, z))
3*z*besseli(0, z)/2
"""
# TODO
# - better algorithm?
# - simplify (cos(pi*b)*besselj(b,z) - besselj(-b,z))/sin(pi*b) ...
# - use contiguity relations?
def replacer(fro, to, factors):
factors = set(factors)
def repl(nu, z):
if factors.intersection(Mul.make_args(z)):
return to(nu, z)
return fro(nu, z)
return repl
def torewrite(fro, to):
def tofunc(nu, z):
return fro(nu, z).rewrite(to)
return tofunc
def tominus(fro):
def tofunc(nu, z):
return exp(I*pi*nu)*fro(nu, exp_polar(-I*pi)*z)
return tofunc
orig_expr = expr
ifactors = [I, exp_polar(I*pi/2), exp_polar(-I*pi/2)]
expr = expr.replace(
besselj, replacer(besselj,
torewrite(besselj, besseli), ifactors))
expr = expr.replace(
besseli, replacer(besseli,
torewrite(besseli, besselj), ifactors))
minusfactors = [-1, exp_polar(I*pi)]
expr = expr.replace(
besselj, replacer(besselj, tominus(besselj), minusfactors))
expr = expr.replace(
besseli, replacer(besseli, tominus(besseli), minusfactors))
z0 = Dummy('z')
def expander(fro):
def repl(nu, z):
if (nu % 1) == S(1)/2:
return exptrigsimp(trigsimp(unpolarify(
fro(nu, z0).rewrite(besselj).rewrite(jn).expand(
func=True)).subs(z0, z)))
elif nu.is_Integer and nu > 1:
return fro(nu, z).expand(func=True)
return fro(nu, z)
return repl
expr = expr.replace(besselj, expander(besselj))
expr = expr.replace(bessely, expander(bessely))
expr = expr.replace(besseli, expander(besseli))
expr = expr.replace(besselk, expander(besselk))
if expr != orig_expr:
expr = expr.factor()
return expr
[docs]def nthroot(expr, n, max_len=4, prec=15):
"""
compute a real nth-root of a sum of surds
Parameters
==========
expr : sum of surds
n : integer
max_len : maximum number of surds passed as constants to ``nsimplify``
Algorithm
=========
First ``nsimplify`` is used to get a candidate root; if it is not a
root the minimal polynomial is computed; the answer is one of its
roots.
Examples
========
>>> from sympy.simplify.simplify import nthroot
>>> from sympy import Rational, sqrt
>>> nthroot(90 + 34*sqrt(7), 3)
sqrt(7) + 3
"""
expr = sympify(expr)
n = sympify(n)
p = expr**Rational(1, n)
if not n.is_integer:
return p
if not _is_sum_surds(expr):
return p
surds = []
coeff_muls = [x.as_coeff_Mul() for x in expr.args]
for x, y in coeff_muls:
if not x.is_rational:
return p
if y is S.One:
continue
if not (y.is_Pow and y.exp == S.Half and y.base.is_integer):
return p
surds.append(y)
surds.sort()
surds = surds[:max_len]
if expr < 0 and n % 2 == 1:
p = (-expr)**Rational(1, n)
a = nsimplify(p, constants=surds)
res = a if _mexpand(a**n) == _mexpand(-expr) else p
return -res
a = nsimplify(p, constants=surds)
if _mexpand(a) is not _mexpand(p) and _mexpand(a**n) == _mexpand(expr):
return _mexpand(a)
expr = _nthroot_solve(expr, n, prec)
if expr is None:
return p
return expr
[docs]def nsimplify(expr, constants=[], tolerance=None, full=False, rational=None):
"""
Find a simple representation for a number or, if there are free symbols or
if rational=True, then replace Floats with their Rational equivalents. If
no change is made and rational is not False then Floats will at least be
converted to Rationals.
For numerical expressions, a simple formula that numerically matches the
given numerical expression is sought (and the input should be possible
to evalf to a precision of at least 30 digits).
Optionally, a list of (rationally independent) constants to
include in the formula may be given.
A lower tolerance may be set to find less exact matches. If no tolerance
is given then the least precise value will set the tolerance (e.g. Floats
default to 15 digits of precision, so would be tolerance=10**-15).
With full=True, a more extensive search is performed
(this is useful to find simpler numbers when the tolerance
is set low).
Examples
========
>>> from sympy import nsimplify, sqrt, GoldenRatio, exp, I, exp, pi
>>> nsimplify(4/(1+sqrt(5)), [GoldenRatio])
-2 + 2*GoldenRatio
>>> nsimplify((1/(exp(3*pi*I/5)+1)))
1/2 - I*sqrt(sqrt(5)/10 + 1/4)
>>> nsimplify(I**I, [pi])
exp(-pi/2)
>>> nsimplify(pi, tolerance=0.01)
22/7
See Also
========
sympy.core.function.nfloat
"""
try:
return sympify(as_int(expr))
except (TypeError, ValueError):
pass
expr = sympify(expr).xreplace({
Float('inf'): S.Infinity,
Float('-inf'): S.NegativeInfinity,
})
if expr is S.Infinity or expr is S.NegativeInfinity:
return expr
if rational or expr.free_symbols:
return _real_to_rational(expr, tolerance)
# SymPy's default tolerance for Rationals is 15; other numbers may have
# lower tolerances set, so use them to pick the largest tolerance if None
# was given
if tolerance is None:
tolerance = 10**-min([15] +
[mpmath.libmp.libmpf.prec_to_dps(n._prec)
for n in expr.atoms(Float)])
# XXX should prec be set independent of tolerance or should it be computed
# from tolerance?
prec = 30
bprec = int(prec*3.33)
constants_dict = {}
for constant in constants:
constant = sympify(constant)
v = constant.evalf(prec)
if not v.is_Float:
raise ValueError("constants must be real-valued")
constants_dict[str(constant)] = v._to_mpmath(bprec)
exprval = expr.evalf(prec, chop=True)
re, im = exprval.as_real_imag()
# safety check to make sure that this evaluated to a number
if not (re.is_Number and im.is_Number):
return expr
def nsimplify_real(x):
orig = mpmath.mp.dps
xv = x._to_mpmath(bprec)
try:
# We'll be happy with low precision if a simple fraction
if not (tolerance or full):
mpmath.mp.dps = 15
rat = mpmath.pslq([xv, 1])
if rat is not None:
return Rational(-int(rat[1]), int(rat[0]))
mpmath.mp.dps = prec
newexpr = mpmath.identify(xv, constants=constants_dict,
tol=tolerance, full=full)
if not newexpr:
raise ValueError
if full:
newexpr = newexpr[0]
expr = sympify(newexpr)
if x and not expr: # don't let x become 0
raise ValueError
if expr.is_finite is False and not xv in [mpmath.inf, mpmath.ninf]:
raise ValueError
return expr
finally:
# even though there are returns above, this is executed
# before leaving
mpmath.mp.dps = orig
try:
if re:
re = nsimplify_real(re)
if im:
im = nsimplify_real(im)
except ValueError:
if rational is None:
return _real_to_rational(expr)
return expr
rv = re + im*S.ImaginaryUnit
# if there was a change or rational is explicitly not wanted
# return the value, else return the Rational representation
if rv != expr or rational is False:
return rv
return _real_to_rational(expr)
def _real_to_rational(expr, tolerance=None):
"""
Replace all reals in expr with rationals.
>>> from sympy import nsimplify
>>> from sympy.abc import x
>>> nsimplify(.76 + .1*x**.5, rational=True)
sqrt(x)/10 + 19/25
"""
inf = Float('inf')
p = expr
reps = {}
reduce_num = None
if tolerance is not None and tolerance < 1:
reduce_num = ceiling(1/tolerance)
for float in p.atoms(Float):
key = float
if reduce_num is not None:
r = Rational(float).limit_denominator(reduce_num)
elif (tolerance is not None and tolerance >= 1 and
float.is_Integer is False):
r = Rational(tolerance*round(float/tolerance)
).limit_denominator(int(tolerance))
else:
r = nsimplify(float, rational=False)
# e.g. log(3).n() -> log(3) instead of a Rational
if float and not r:
r = Rational(float)
elif not r.is_Rational:
if float == inf or float == -inf:
r = S.ComplexInfinity
elif float < 0:
float = -float
d = Pow(10, int((mpmath.log(float)/mpmath.log(10))))
r = -Rational(str(float/d))*d
elif float > 0:
d = Pow(10, int((mpmath.log(float)/mpmath.log(10))))
r = Rational(str(float/d))*d
else:
r = Integer(0)
reps[key] = r
return p.subs(reps, simultaneous=True)
def clear_coefficients(expr, rhs=S.Zero):
"""Return `p, r` where `p` is the expression obtained when Rational
additive and multiplicative coefficients of `expr` have been stripped
away in a naive fashion (i.e. without simplification). The operations
needed to remove the coefficients will be applied to `rhs` and returned
as `r`.
Examples
========
>>> from sympy.simplify.simplify import clear_coefficients
>>> from sympy.abc import x, y
>>> from sympy import Dummy
>>> expr = 4*y*(6*x + 3)
>>> clear_coefficients(expr - 2)
(y*(2*x + 1), 1/6)
When solving 2 or more expressions like `expr = a`,
`expr = b`, etc..., it is advantageous to provide a Dummy symbol
for `rhs` and simply replace it with `a`, `b`, etc... in `r`.
>>> rhs = Dummy('rhs')
>>> clear_coefficients(expr, rhs)
(y*(2*x + 1), _rhs/12)
>>> _[1].subs(rhs, 2)
1/6
"""
was = None
free = expr.free_symbols
if expr.is_Rational:
return (S.Zero, rhs - expr)
while expr and was != expr:
was = expr
m, expr = (
expr.as_content_primitive()
if free else
factor_terms(expr).as_coeff_Mul(rational=True))
rhs /= m
c, expr = expr.as_coeff_Add(rational=True)
rhs -= c
expr = signsimp(expr, evaluate = False)
if _coeff_isneg(expr):
expr = -expr
rhs = -rhs
return expr, rhs
