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, nfloat
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.core.sympify import _sympify
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, evaluate) 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, rational=False, inverse=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.
If rational=True, Floats will be recast as Rationals before simplification.
If rational=None, Floats will be recast as Rationals but the result will
be recast as Floats. If rational=False(default) then nothing will be done
to the Floats.
If inverse=True, it will be assumed that a composition of inverse
functions, such as sin and asin, can be cancelled in any order.
For example, ``asin(sin(x))`` will yield ``x`` without checking whether
x belongs to the set where this relation is true. The default is
False.
"""
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 inverse and expr.has(Function):
expr = inversecombine(expr)
if not expr.args: # simplified to atomic
return expr
if not isinstance(expr, (Add, Mul, Pow, ExpBase)):
return expr.func(*[simplify(x, ratio=ratio, measure=measure, rational=rational)
for x in expr.args])
if not expr.is_commutative:
expr = nc_simplify(expr)
# 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)
# rationalize Floats
floats = False
if rational is not False and expr.has(Float):
floats = True
expr = nsimplify(expr, rational=True)
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, HyperbolicFunction):
expr = trigsimp(expr, deep=True)
if expr.has(log):
expr = shorter(expand_log(expr, deep=True), logcombine(expr))
if expr.has(CombinatorialFunction, gamma):
# expression with gamma functions or non-integer arguments is
# automatically passed to gammasimp
expr = combsimp(expr)
if expr.has(Sum):
expr = sum_simplify(expr)
if expr.has(Product):
expr = product_simplify(expr)
from sympy.physics.units import Quantity
from sympy.physics.units.util import quantity_simplify
if expr.has(Quantity):
expr = quantity_simplify(expr)
short = shorter(powsimp(expr, combine='exp', deep=True), powsimp(expr), expr)
short = shorter(short, cancel(short))
short = shorter(short, factor_terms(short), expand_power_exp(expand_mul(short)))
if short.has(TrigonometricFunction, HyperbolicFunction, ExpBase):
short = exptrigsimp(short)
# 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
# restore floats
if floats and rational is None:
expr = nfloat(expr, exponent=False)
return expr
def sum_simplify(s):
"""Main function for Sum simplification"""
from sympy.concrete.summations import Sum
from sympy.core.function import expand
terms = Add.make_args(expand(s))
s_t = [] # Sum Terms
o_t = [] # Other Terms
for term in terms:
if isinstance(term, Mul):
other = 1
sum_terms = []
if not term.has(Sum):
o_t.append(term)
continue
mul_terms = Mul.make_args(term)
for mul_term in mul_terms:
if isinstance(mul_term, Sum):
r = mul_term._eval_simplify()
sum_terms.extend(Add.make_args(r))
else:
other = other * mul_term
if len(sum_terms):
#some simplification may have happened
#use if so
s_t.append(Mul(*sum_terms) * other)
else:
o_t.append(other)
elif isinstance(term, Sum):
#as above, we need to turn this into an add list
r = term._eval_simplify()
s_t.extend(Add.make_args(r))
else:
o_t.append(term)
result = Add(sum_combine(s_t), *o_t)
return result
def sum_combine(s_t):
"""Helper function for Sum simplification
Attempts to simplify a list of sums, by combining limits / sum function's
returns the simplified sum
"""
from sympy.concrete.summations import Sum
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) or isinstance(temp, Mul):
s_t[i] = temp
s_term1 = s_t[i]
used[j] = True
result = S.Zero
for i, s_term in enumerate(s_t):
if not used[i]:
result = Add(result, s_term)
return result
def factor_sum(self, limits=None, radical=False, clear=False, fraction=False, sign=True):
"""Helper function for Sum simplification
if limits is specified, "self" is the inner part of a sum
Returns the sum with constant factors brought outside
"""
from sympy.core.exprtools import factor_terms
from sympy.concrete.summations import Sum
result = self.function if limits is None else self
limits = self.limits if limits is None else limits
#avoid any confusion w/ as_independent
if result == 0:
return S.Zero
#get the summation variables
sum_vars = set([limit.args[0] for limit in limits])
#finally we try to factor out any common terms
#and remove the from the sum if independent
retv = factor_terms(result, radical=radical, clear=clear, fraction=fraction, sign=sign)
#avoid doing anything bad
if not result.is_commutative:
return Sum(result, *limits)
i, d = retv.as_independent(*sum_vars)
if isinstance(retv, Add):
return i * Sum(1, *limits) + Sum(d, *limits)
else:
return i * Sum(d, *limits)
def sum_add(self, other, method=0):
"""Helper function for Sum simplification"""
from sympy.concrete.summations import Sum
from sympy import Mul
#we know this is something in terms of a constant * a sum
#so we temporarily put the constants inside for simplification
#then simplify the result
def __refactor(val):
args = Mul.make_args(val)
sumv = next(x for x in args if isinstance(x, Sum))
constant = Mul(*[x for x in args if x != sumv])
return Sum(constant * sumv.function, *sumv.limits)
if isinstance(self, Mul):
rself = __refactor(self)
else:
rself = self
if isinstance(other, Mul):
rother = __refactor(other)
else:
rother = other
if type(rself) == type(rother):
if method == 0:
if rself.limits == rother.limits:
return factor_sum(Sum(rself.function + rother.function, *rself.limits))
elif method == 1:
if simplify(rself.function - rother.function) == 0:
if len(rself.limits) == len(rother.limits) == 1:
i = rself.limits[0][0]
x1 = rself.limits[0][1]
y1 = rself.limits[0][2]
j = rother.limits[0][0]
x2 = rother.limits[0][1]
y2 = rother.limits[0][2]
if i == j:
if x2 == y1 + 1:
return factor_sum(Sum(rself.function, (i, x1, y2)))
elif x1 == y2 + 1:
return factor_sum(Sum(rself.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
sympy.core.function.expand_log: expand the logarithms of products
and powers; the opposite of logcombine
"""
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 isinstance(a, 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 isinstance(ai, 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 isinstance(l, 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), evaluate=False)
# 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],
evaluate=False))
else:
other.append(k*log1.pop(k))
return Add(*other)
return bottom_up(expr, f)
def inversecombine(expr):
"""Simplify the composition of a function and its inverse.
No attention is paid to whether the inverse is a left inverse or a
right inverse; thus, the result will in general not be equivalent
to the original expression.
Examples
========
>>> from sympy.simplify.simplify import inversecombine
>>> from sympy import asin, sin, log, exp
>>> from sympy.abc import x
>>> inversecombine(asin(sin(x)))
x
>>> inversecombine(2*log(exp(3*x)))
6*x
"""
def f(rv):
if rv.is_Function and hasattr(rv, "inverse"):
if (len(rv.args) == 1 and len(rv.args[0].args) == 1 and
isinstance(rv.args[0], rv.inverse(argindex=1))):
rv = rv.args[0].args[0]
return rv
return bottom_up(expr, f)
def walk(e, *target):
"""iterate through the args that are the given types (target) and
return a list of the args that were traversed; arguments
that are not of the specified types are not traversed.
Examples
========
>>> from sympy.simplify.simplify import walk
>>> from sympy import Min, Max
>>> from sympy.abc import x, y, z
>>> list(walk(Min(x, Max(y, Min(1, z))), Min))
[Min(x, Max(y, Min(1, z)))]
>>> list(walk(Min(x, Max(y, Min(1, z))), Min, Max))
[Min(x, Max(y, Min(1, z))), Max(y, Min(1, z)), Min(1, z)]
See Also
========
bottom_up
"""
if isinstance(e, target):
yield e
for i in e.args:
for w in walk(i, *target):
yield w
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 simplify(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,
rational_conversion='base10'):
"""
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).
When converting to rational, if rational_conversion='base10' (the default), then
convert floats to rationals using their base-10 (string) representation.
When rational_conversion='exact' it uses the exact, base-2 representation.
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
>>> nsimplify(0.333333333333333, rational=True, rational_conversion='exact')
6004799503160655/18014398509481984
>>> nsimplify(0.333333333333333, rational=True)
1/3
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, rational_conversion)
# 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, rational_conversion=rational_conversion)
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, rational_conversion=rational_conversion)
def _real_to_rational(expr, tolerance=None, rational_conversion='base10'):
"""
Replace all reals in expr with rationals.
>>> from sympy import Rational
>>> from sympy.simplify.simplify import _real_to_rational
>>> from sympy.abc import x
>>> _real_to_rational(.76 + .1*x**.5)
sqrt(x)/10 + 19/25
If rational_conversion='base10', this uses the base-10 string. If
rational_conversion='exact', the exact, base-2 representation is used.
>>> _real_to_rational(0.333333333333333, rational_conversion='exact')
6004799503160655/18014398509481984
>>> _real_to_rational(0.333333333333333)
1/3
"""
expr = _sympify(expr)
inf = Float('inf')
p = expr
reps = {}
reduce_num = None
if tolerance is not None and tolerance < 1:
reduce_num = ceiling(1/tolerance)
for fl in p.atoms(Float):
key = fl
if reduce_num is not None:
r = Rational(fl).limit_denominator(reduce_num)
elif (tolerance is not None and tolerance >= 1 and
fl.is_Integer is False):
r = Rational(tolerance*round(fl/tolerance)
).limit_denominator(int(tolerance))
else:
if rational_conversion == 'exact':
r = Rational(fl)
reps[key] = r
continue
elif rational_conversion != 'base10':
raise ValueError("rational_conversion must be 'base10' or 'exact'")
r = nsimplify(fl, rational=False)
# e.g. log(3).n() -> log(3) instead of a Rational
if fl and not r:
r = Rational(fl)
elif not r.is_Rational:
if fl == inf or fl == -inf:
r = S.ComplexInfinity
elif fl < 0:
fl = -fl
d = Pow(10, int((mpmath.log(fl)/mpmath.log(10))))
r = -Rational(str(fl/d))*d
elif fl > 0:
d = Pow(10, int((mpmath.log(fl)/mpmath.log(10))))
r = Rational(str(fl/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
def nc_simplify(expr, deep=True):
'''
Simplify a non-commutative expression composed of multiplication
and raising to a power by grouping repeated subterms into one power.
Priority is given to simplifications that give the fewest number
of arguments in the end (for example, in a*b*a*b*c*a*b*c simplifying
to (a*b)**2*c*a*b*c gives 5 arguments while a*b*(a*b*c)**2 has 3).
If `expr` is a sum of such terms, the sum of the simplified terms
is returned.
Keyword argument `deep` controls whether or not subexpressions
nested deeper inside the main expression are simplified. See examples
below. Setting `deep` to `False` can save time on nested expressions
that don't need simplifying on all levels.
Examples
========
>>> from sympy import symbols
>>> from sympy.simplify.simplify import nc_simplify
>>> a, b, c = symbols("a b c", commutative=False)
>>> nc_simplify(a*b*a*b*c*a*b*c)
a*b*(a*b*c)**2
>>> expr = a**2*b*a**4*b*a**4
>>> nc_simplify(expr)
a**2*(b*a**4)**2
>>> nc_simplify(a*b*a*b*c**2*(a*b)**2*c**2)
((a*b)**2*c**2)**2
>>> nc_simplify(a*b*a*b + 2*a*c*a**2*c*a**2*c*a)
(a*b)**2 + 2*(a*c*a)**3
>>> nc_simplify(b**-1*a**-1*(a*b)**2)
a*b
>>> nc_simplify(a**-1*b**-1*c*a)
(b*a)**(-1)*c*a
>>> expr = (a*b*a*b)**2*a*c*a*c
>>> nc_simplify(expr)
(a*b)**4*(a*c)**2
>>> nc_simplify(expr, deep=False)
(a*b*a*b)**2*(a*c)**2
'''
# =========== Auxiliary functions ========================
def _overlaps(args):
# Calculate a list of lists m such that m[i][j] contains the lengths
# of all possible overlaps between args[:i+1] and args[i+1+j:].
# An overlap is a suffix of the prefix that matches a prefix
# of the suffix.
# For example, let expr=c*a*b*a*b*a*b*a*b. Then m[3][0] contains
# the lengths of overlaps of c*a*b*a*b with a*b*a*b. The overlaps
# are a*b*a*b, a*b and the empty word so that m[3][0]=[4,2,0].
# All overlaps rather than only the longest one are recorded
# because this information helps calculate other overlap lengths.
m = [[([1, 0] if a == args[0] else [0]) for a in args[1:]]]
for i in range(1, len(args)):
overlaps = []
j = 0
for j in range(len(args) - i - 1):
overlap = []
for v in m[i-1][j+1]:
if j + i + 1 + v < len(args) and args[i] == args[j+i+1+v]:
overlap.append(v + 1)
overlap += [0]
overlaps.append(overlap)
m.append(overlaps)
return m
def _reduce_inverses(_args):
# replace consecutive negative powers by an inverse
# of a product of positive powers, e.g. a**-1*b**-1*c
# will simplify to (a*b)**-1*c;
# return that new args list and the number of negative
# powers in it (inv_tot)
inv_tot = 0 # total number of inverses
inverses = []
args = []
for arg in _args:
if isinstance(arg, Pow) and arg.args[1] < 0:
inverses = [arg**-1] + inverses
inv_tot += 1
else:
if inverses:
args.append(Pow(Mul(*inverses), -1))
inv_tot -= len(inverses) - 1
inverses = []
args.append(arg)
if inverses:
args.append(Pow(Mul(*inverses), -1))
inv_tot -= len(inverses) - 1
return inv_tot, tuple(args)
# ========================================================
if not isinstance(expr, (Add, Mul, Pow)):
return expr
args = expr.args[:]
if isinstance(expr, Pow):
if deep:
return Pow(nc_simplify(args[0]), args[1])
else:
return expr
elif isinstance(expr, Add):
return Add(*[nc_simplify(a, deep=deep) for a in args])
inv_tot, args = _reduce_inverses(args)
# if most arguments are negative, work with the inverse
# of the expression, e.g. a**-1*b*a**-1*c**-1 will become
# (c*a*b**-1*a)**-1 at the end so can work with c*a*b**-1*a
invert = False
if inv_tot > len(args)/2:
invert = True
args = [a**-1 for a in args[::-1]]
if deep:
args = tuple(nc_simplify(a) for a in args)
m = _overlaps(args)
# simps will be {subterm: end} where `end` is the ending
# index of a sequence of repetitions of subterm;
# this is for not wasting time with subterms that are part
# of longer, already considered sequences
simps = {}
post = 1
pre = 1
# the simplification coefficient is the number of
# arguments by which contracting a given sequence
# would reduce the word; e.g. in a*b*a*b*c*a*b*c,
# contracting a*b*a*b to (a*b)**2 removes 3 arguments
# while a*b*c*a*b*c to (a*b*c)**2 removes 6. It's
# better to contract the latter so simplification
# with a maximum simplification coefficient will be chosen
max_simp_coeff = 0
simp = None # information about future simplification
for i in range(1, len(args)):
simp_coeff = 0
l = 0 # length of a subterm
p = 0 # the power of a subterm
if i < len(args) - 1:
rep = m[i][0]
start = i # starting index of the repeated sequence
end = i+1 # ending index of the repeated sequence
if i == len(args)-1 or rep == [0]:
# no subterm is repeated at this stage, at least as
# far as the arguments are concerned - there may be
# a repetition if powers are taken into account
if isinstance(args[i], Pow) and len(args[i].args[0].args) > 0:
subterm = args[i].args[0].args
l = len(subterm)
if args[i-l:i] == subterm:
# e.g. a*b in a*b*(a*b)**2 is not repeated
# in args (= [a, b, (a*b)**2]) but it
# can be matched here
p += 1
start -= l
if args[i+1:i+1+l] == subterm:
# e.g. a*b in (a*b)**2*a*b
p += 1
end += l
if p:
p += args[i].args[1]
else:
continue
else:
l = rep[0] # length of the longest repeated subterm at this point
start -= l - 1
subterm = args[start:end]
p = 2
end += l
if subterm in simps and simps[subterm] >= start:
# the subterm is part of a sequence that
# has already been considered
continue
# count how many times it's repeated
while end < len(args):
if l in m[end-1][0]:
p += 1
end += l
elif isinstance(args[end], Pow) and args[end].args[0].args == subterm:
# for cases like a*b*a*b*(a*b)**2*a*b
p += args[end].args[1]
end += 1
else:
break
# see if another match can be made, e.g.
# for b*a**2 in b*a**2*b*a**3 or a*b in
# a**2*b*a*b
pre_exp = 0
pre_arg = 1
if start - l >= 0 and args[start-l+1:start] == subterm[1:]:
if isinstance(subterm[0], Pow):
pre_arg = subterm[0].args[0]
exp = subterm[0].args[1]
else:
pre_arg = subterm[0]
exp = 1
if isinstance(args[start-l], Pow) and args[start-l].args[0] == pre_arg:
pre_exp = args[start-l].args[1] - exp
start -= l
p += 1
elif args[start-l] == pre_arg:
pre_exp = 1 - exp
start -= l
p += 1
post_exp = 0
post_arg = 1
if end + l - 1 < len(args) and args[end:end+l-1] == subterm[:-1]:
if isinstance(subterm[-1], Pow):
post_arg = subterm[-1].args[0]
exp = subterm[-1].args[1]
else:
post_arg = subterm[-1]
exp = 1
if isinstance(args[end+l-1], Pow) and args[end+l-1].args[0] == post_arg:
post_exp = args[end+l-1].args[1] - exp
end += l
p += 1
elif args[end+l-1] == post_arg:
post_exp = 1 - exp
end += l
p += 1
# Consider a*b*a**2*b*a**2*b*a:
# b*a**2 is explicitly repeated, but note
# that in this case a*b*a is also repeated
# so there are two possible simplifications:
# a*(b*a**2)**3*a**-1 or (a*b*a)**3
# The latter is obviously simpler.
# But in a*b*a**2*b**2*a**2 the simplifications are
# a*(b*a**2)**2 and (a*b*a)**3*a in which case
# it's better to stick with the shorter subterm
if post_exp and exp % 2 == 0 and start > 0:
exp = exp/2
_pre_exp = 1
_post_exp = 1
if isinstance(args[start-1], Pow) and args[start-1].args[0] == post_arg:
_post_exp = post_exp + exp
_pre_exp = args[start-1].args[1] - exp
elif args[start-1] == post_arg:
_post_exp = post_exp + exp
_pre_exp = 1 - exp
if _pre_exp == 0 or _post_exp == 0:
if not pre_exp:
start -= 1
post_exp = _post_exp
pre_exp = _pre_exp
pre_arg = post_arg
subterm = (post_arg**exp,) + subterm[:-1] + (post_arg**exp,)
simp_coeff += end-start
if post_exp:
simp_coeff -= 1
if pre_exp:
simp_coeff -= 1
simps[subterm] = end
if simp_coeff > max_simp_coeff:
max_simp_coeff = simp_coeff
simp = (start, Mul(*subterm), p, end, l)
pre = pre_arg**pre_exp
post = post_arg**post_exp
if simp:
subterm = Pow(nc_simplify(simp[1], deep=deep), simp[2])
pre = nc_simplify(Mul(*args[:simp[0]])*pre, deep=deep)
post = post*nc_simplify(Mul(*args[simp[3]:]), deep=deep)
simp = pre*subterm*post
if pre != 1 or post != 1:
# new simplifications may be possible but no need
# to recurse over arguments
simp = nc_simplify(simp, deep=False)
else:
simp = Mul(*args)
if invert:
simp = simp**-1
return simp
