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 simplify(expr, ratio=1.7, measure=count_ops, rational=False):
# type: (object, object, object, object) -> object
"""
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.
"""
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, rational=rational)
return expr.func(*[simplify(x, ratio=ratio, measure=measure, rational=rational)
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)
# 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)
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