def _sqrtdenest_rec(expr):
"""Helper that denests the square root of three or more surds.
It returns the denested expression; if it cannot be denested it
throws SqrtdenestStopIteration
Algorithm: expr.base is in the extension Q_m = Q(sqrt(r_1),..,sqrt(r_k));
split expr.base = a + b*sqrt(r_k), where `a` and `b` are on
Q_(m-1) = Q(sqrt(r_1),..,sqrt(r_(k-1))); then a**2 - b**2*r_k is
on Q_(m-1); denest sqrt(a**2 - b**2*r_k) and so on.
See [1], section 6.
Examples
========
>>> from diofant import sqrt
>>> from diofant.simplify.sqrtdenest import _sqrtdenest_rec
>>> _sqrtdenest_rec(sqrt(-72*sqrt(2) + 158*sqrt(5) + 498))
-sqrt(10) + sqrt(2) + 9 + 9*sqrt(5)
>>> w=-6*sqrt(55)-6*sqrt(35)-2*sqrt(22)-2*sqrt(14)+2*sqrt(77)+6*sqrt(10)+65
>>> _sqrtdenest_rec(sqrt(w))
-sqrt(11) - sqrt(7) + sqrt(2) + 3*sqrt(5)
"""
from diofant.simplify.radsimp import radsimp, rad_rationalize, split_surds
if not expr.is_Pow:
return sqrtdenest(expr)
if expr.base < 0:
return sqrt(-1) * _sqrtdenest_rec(sqrt(-expr.base))
g, a, b = split_surds(expr.base)
a = a * sqrt(g)
if a < b:
a, b = b, a
c2 = _mexpand(a**2 - b**2)
if len(c2.args) > 2:
g, a1, b1 = split_surds(c2)
a1 = a1 * sqrt(g)
if a1 < b1:
a1, b1 = b1, a1
c2_1 = _mexpand(a1**2 - b1**2)
c_1 = _sqrtdenest_rec(sqrt(c2_1))
d_1 = _sqrtdenest_rec(sqrt(a1 + c_1))
num, den = rad_rationalize(b1, d_1)
c = _mexpand(d_1 / sqrt(2) + num / (den * sqrt(2)))
else:
c = _sqrtdenest1(sqrt(c2))
if sqrt_depth(c) > 1:
raise SqrtdenestStopIteration
ac = a + c
if len(ac.args) >= len(expr.args):
if count_ops(ac) >= count_ops(expr.base):
raise SqrtdenestStopIteration
d = sqrtdenest(sqrt(ac))
if sqrt_depth(d) > 1:
raise SqrtdenestStopIteration
num, den = rad_rationalize(b, d)
r = d / sqrt(2) + num / (den * sqrt(2))
r = radsimp(r)
return _mexpand(r)
def _sqrtdenest1(expr, denester=True):
"""Return denested expr after denesting with simpler methods or, that
failing, using the denester."""
from diofant.simplify.simplify import radsimp
if not is_sqrt(expr):
return expr
a = expr.base
if a.is_Atom:
return expr
val = _sqrt_match(a)
if not val:
return expr
a, b, r = val
# try a quick numeric denesting
d2 = _mexpand(a**2 - b**2 * r)
if d2.is_Rational:
if d2.is_positive:
z = _sqrt_numeric_denest(a, b, r, d2)
if z is not None:
return z
else:
# fourth root case
# sqrtdenest(sqrt(3 + 2*sqrt(3))) =
# sqrt(2)*3**(1/4)/2 + sqrt(2)*3**(3/4)/2
dr2 = _mexpand(-d2 * r)
dr = sqrt(dr2)
if dr.is_Rational:
z = _sqrt_numeric_denest(_mexpand(b * r), a, r, dr2)
if z is not None:
return z / root(r, 4)
else:
z = _sqrt_symbolic_denest(a, b, r)
if z is not None:
return z
if not denester or not is_algebraic(expr):
return expr
res = sqrt_biquadratic_denest(expr, a, b, r, d2)
if res:
return res
# now call to the denester
av0 = [a, b, r, d2]
z = _denester([radsimp(expr**2)], av0, 0, sqrt_depth(expr))[0]
if av0[1] is None:
return expr
if z is not None:
if sqrt_depth(z) == sqrt_depth(
expr) and count_ops(z) > count_ops(expr):
return expr
return z
return expr
def _sqrt_numeric_denest(a, b, r, d2):
"""Helper that denest expr = a + b*sqrt(r), with d2 = a**2 - b**2*r > 0
or returns None if not denested.
"""
from diofant.simplify.simplify import radsimp
depthr = sqrt_depth(r)
d = sqrt(d2)
vad = a + d
# sqrt_depth(res) <= sqrt_depth(vad) + 1
# sqrt_depth(expr) = depthr + 2
# there is denesting if sqrt_depth(vad)+1 < depthr + 2
# if vad**2 is Number there is a fourth root
if sqrt_depth(vad) < depthr + 1 or (vad**2).is_Rational:
vad1 = radsimp(1 / vad)
return (sqrt(vad / 2) + sign(b) * sqrt(
(b**2 * r * vad1 / 2).expand())).expand()
def _denester(nested, av0, h, max_depth_level):
"""Denests a list of expressions that contain nested square roots.
Algorithm based on <http://www.almaden.ibm.com/cs/people/fagin/symb85.pdf>.
It is assumed that all of the elements of 'nested' share the same
bottom-level radicand. (This is stated in the paper, on page 177, in
the paragraph immediately preceding the algorithm.)
When evaluating all of the arguments in parallel, the bottom-level
radicand only needs to be denested once. This means that calling
_denester with x arguments results in a recursive invocation with x+1
arguments; hence _denester has polynomial complexity.
However, if the arguments were evaluated separately, each call would
result in two recursive invocations, and the algorithm would have
exponential complexity.
This is discussed in the paper in the middle paragraph of page 179.
"""
from diofant.simplify.simplify import radsimp
if h > max_depth_level:
return None, None
if av0[1] is None:
return None, None
if (av0[0] is None
and all(n.is_Number for n in nested)): # no arguments are nested
for f in _subsets(len(nested)): # test subset 'f' of nested
p = _mexpand(Mul(*[nested[i] for i in range(len(f)) if f[i]]))
if f.count(1) > 1 and f[-1]:
p = -p
sqp = sqrt(p)
if sqp.is_Rational:
return sqp, f # got a perfect square so return its square root.
# Otherwise, return the radicand from the previous invocation.
return sqrt(nested[-1]), [0] * len(nested)
else:
R = None
if av0[0] is not None:
values = [av0[:2]]
R = av0[2]
nested2 = [av0[3], R]
av0[0] = None
else:
values = list(filter(None, [_sqrt_match(expr) for expr in nested]))
for v in values:
if v[2]: # Since if b=0, r is not defined
if R is not None:
if R != v[2]:
av0[1] = None
return None, None
else:
R = v[2]
if R is None:
# return the radicand from the previous invocation
return sqrt(nested[-1]), [0] * len(nested)
nested2 = [
_mexpand(v[0]**2) - _mexpand(R * v[1]**2) for v in values
] + [R]
d, f = _denester(nested2, av0, h + 1, max_depth_level)
if not f:
return None, None
if not any(f[i] for i in range(len(nested))):
v = values[-1]
return sqrt(v[0] + _mexpand(v[1] * d)), f
else:
p = Mul(*[nested[i] for i in range(len(nested)) if f[i]])
v = _sqrt_match(p)
if 1 in f and f.index(1) < len(nested) - 1 and f[len(nested) - 1]:
v[0] = -v[0]
v[1] = -v[1]
if not f[len(nested)]: # Solution denests with square roots
vad = _mexpand(v[0] + d)
if vad <= 0:
# return the radicand from the previous invocation.
return sqrt(nested[-1]), [0] * len(nested)
if not (sqrt_depth(vad) <= sqrt_depth(R) + 1 or
(vad**2).is_Number):
av0[1] = None
return None, None
sqvad = _sqrtdenest1(sqrt(vad), denester=False)
if not (sqrt_depth(sqvad) <= sqrt_depth(R) + 1):
av0[1] = None
return None, None
sqvad1 = radsimp(1 / sqvad)
res = _mexpand(sqvad / sqrt(2) +
(v[1] * sqrt(R) * sqvad1 / sqrt(2)))
return res, f
else: # Solution requires a fourth root
s2 = _mexpand(v[1] * R) + d
if s2 <= 0:
return sqrt(nested[-1]), [0] * len(nested)
FR, s = root(_mexpand(R), 4), sqrt(s2)
return _mexpand(s / (sqrt(2) * FR) + v[0] * FR /
(sqrt(2) * s)), f
def sqrt_biquadratic_denest(expr, a, b, r, d2):
"""denest expr = sqrt(a + b*sqrt(r))
where a, b, r are linear combinations of square roots of
positive rationals on the rationals (SQRR) and r > 0, b != 0,
d2 = a**2 - b**2*r > 0
If it cannot denest it returns None.
ALGORITHM
Search for a solution A of type SQRR of the biquadratic equation
4*A**4 - 4*a*A**2 + b**2*r = 0 (1)
sqd = sqrt(a**2 - b**2*r)
Choosing the sqrt to be positive, the possible solutions are
A = sqrt(a/2 +/- sqd/2)
Since a, b, r are SQRR, then a**2 - b**2*r is a SQRR,
so if sqd can be denested, it is done by
_sqrtdenest_rec, and the result is a SQRR.
Similarly for A.
Examples of solutions (in both cases a and sqd are positive):
Example of expr with solution sqrt(a/2 + sqd/2) but not
solution sqrt(a/2 - sqd/2):
expr = sqrt(-sqrt(15) - sqrt(2)*sqrt(-sqrt(5) + 5) - sqrt(3) + 8)
a = -sqrt(15) - sqrt(3) + 8; sqd = -2*sqrt(5) - 2 + 4*sqrt(3)
Example of expr with solution sqrt(a/2 - sqd/2) but not
solution sqrt(a/2 + sqd/2):
w = 2 + r2 + r3 + (1 + r3)*sqrt(2 + r2 + 5*r3)
expr = sqrt((w**2).expand())
a = 4*sqrt(6) + 8*sqrt(2) + 47 + 28*sqrt(3)
sqd = 29 + 20*sqrt(3)
Define B = b/2*A; eq.(1) implies a = A**2 + B**2*r; then
expr**2 = a + b*sqrt(r) = (A + B*sqrt(r))**2
Examples
========
>>> from diofant import sqrt
>>> from diofant.simplify.sqrtdenest import _sqrt_match, sqrt_biquadratic_denest
>>> z = sqrt((2*sqrt(2) + 4)*sqrt(2 + sqrt(2)) + 5*sqrt(2) + 8)
>>> a, b, r = _sqrt_match(z**2)
>>> d2 = a**2 - b**2*r
>>> sqrt_biquadratic_denest(z, a, b, r, d2)
sqrt(2) + sqrt(sqrt(2) + 2) + 2
"""
from diofant.simplify.radsimp import radsimp, rad_rationalize
if r <= 0 or d2 < 0 or not b or sqrt_depth(expr.base) < 2:
return
for x in (a, b, r):
for y in x.args:
y2 = y**2
if not y2.is_Integer or not y2.is_positive:
return
sqd = _mexpand(sqrtdenest(sqrt(radsimp(d2))))
if sqrt_depth(sqd) > 1:
return
x1, x2 = [a / 2 + sqd / 2, a / 2 - sqd / 2]
# look for a solution A with depth 1
for x in (x1, x2):
A = sqrtdenest(sqrt(x))
if sqrt_depth(A) > 1:
continue
Bn, Bd = rad_rationalize(b, _mexpand(2 * A))
B = Bn / Bd
z = A + B * sqrt(r)
if z < 0:
z = -z
return _mexpand(z)
return
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 Diofant. 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 diofant import simplify, cos, sin
>>> from diofant.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 diofant 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:`~diofant.core.function.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 diofant 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 diofant 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 diofant 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 diofant.simplify.hyperexpand import hyperexpand
from diofant.functions.special.bessel import BesselBase
from diofant import Sum, Product
if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack
return expr
if not isinstance(expr, (Add, Mul, Pow, exp_polar)):
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, exp_polar)
or any(a.base is S.Exp1 for a in short.atoms(Pow))):
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