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image-handler/image-handler10.0/venv/lib/site-packages/mpmath/identification.py

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"""
Implements the PSLQ algorithm for integer relation detection,
and derivative algorithms for constant recognition.
"""
from .libmp.backend import xrange
from .libmp import int_types, sqrt_fixed
# round to nearest integer (can be done more elegantly...)
def round_fixed(x, prec):
return ((x + (1<<(prec-1))) >> prec) << prec
class IdentificationMethods(object):
pass
def pslq(ctx, x, tol=None, maxcoeff=1000, maxsteps=100, verbose=False):
r"""
Given a vector of real numbers `x = [x_0, x_1, ..., x_n]`, ``pslq(x)``
uses the PSLQ algorithm to find a list of integers
`[c_0, c_1, ..., c_n]` such that
.. math ::
|c_1 x_1 + c_2 x_2 + ... + c_n x_n| < \mathrm{tol}
and such that `\max |c_k| < \mathrm{maxcoeff}`. If no such vector
exists, :func:`~mpmath.pslq` returns ``None``. The tolerance defaults to
3/4 of the working precision.
**Examples**
Find rational approximations for `\pi`::
>>> from mpmath import *
>>> mp.dps = 15; mp.pretty = True
>>> pslq([-1, pi], tol=0.01)
[22, 7]
>>> pslq([-1, pi], tol=0.001)
[355, 113]
>>> mpf(22)/7; mpf(355)/113; +pi
3.14285714285714
3.14159292035398
3.14159265358979
Pi is not a rational number with denominator less than 1000::
>>> pslq([-1, pi])
>>>
To within the standard precision, it can however be approximated
by at least one rational number with denominator less than `10^{12}`::
>>> p, q = pslq([-1, pi], maxcoeff=10**12)
>>> print(p); print(q)
238410049439
75888275702
>>> mpf(p)/q
3.14159265358979
The PSLQ algorithm can be applied to long vectors. For example,
we can investigate the rational (in)dependence of integer square
roots::
>>> mp.dps = 30
>>> pslq([sqrt(n) for n in range(2, 5+1)])
>>>
>>> pslq([sqrt(n) for n in range(2, 6+1)])
>>>
>>> pslq([sqrt(n) for n in range(2, 8+1)])
[2, 0, 0, 0, 0, 0, -1]
**Machin formulas**
A famous formula for `\pi` is Machin's,
.. math ::
\frac{\pi}{4} = 4 \operatorname{acot} 5 - \operatorname{acot} 239
There are actually infinitely many formulas of this type. Two
others are
.. math ::
\frac{\pi}{4} = \operatorname{acot} 1
\frac{\pi}{4} = 12 \operatorname{acot} 49 + 32 \operatorname{acot} 57
+ 5 \operatorname{acot} 239 + 12 \operatorname{acot} 110443
We can easily verify the formulas using the PSLQ algorithm::
>>> mp.dps = 30
>>> pslq([pi/4, acot(1)])
[1, -1]
>>> pslq([pi/4, acot(5), acot(239)])
[1, -4, 1]
>>> pslq([pi/4, acot(49), acot(57), acot(239), acot(110443)])
[1, -12, -32, 5, -12]
We could try to generate a custom Machin-like formula by running
the PSLQ algorithm with a few inverse cotangent values, for example
acot(2), acot(3) ... acot(10). Unfortunately, there is a linear
dependence among these values, resulting in only that dependence
being detected, with a zero coefficient for `\pi`::
>>> pslq([pi] + [acot(n) for n in range(2,11)])
[0, 1, -1, 0, 0, 0, -1, 0, 0, 0]
We get better luck by removing linearly dependent terms::
>>> pslq([pi] + [acot(n) for n in range(2,11) if n not in (3, 5)])
[1, -8, 0, 0, 4, 0, 0, 0]
In other words, we found the following formula::
>>> 8*acot(2) - 4*acot(7)
3.14159265358979323846264338328
>>> +pi
3.14159265358979323846264338328
**Algorithm**
This is a fairly direct translation to Python of the pseudocode given by
David Bailey, "The PSLQ Integer Relation Algorithm":
http://www.cecm.sfu.ca/organics/papers/bailey/paper/html/node3.html
The present implementation uses fixed-point instead of floating-point
arithmetic, since this is significantly (about 7x) faster.
"""
n = len(x)
if n < 2:
raise ValueError("n cannot be less than 2")
# At too low precision, the algorithm becomes meaningless
prec = ctx.prec
if prec < 53:
raise ValueError("prec cannot be less than 53")
if verbose and prec // max(2,n) < 5:
print("Warning: precision for PSLQ may be too low")
target = int(prec * 0.75)
if tol is None:
tol = ctx.mpf(2)**(-target)
else:
tol = ctx.convert(tol)
extra = 60
prec += extra
if verbose:
print("PSLQ using prec %i and tol %s" % (prec, ctx.nstr(tol)))
tol = ctx.to_fixed(tol, prec)
assert tol
# Convert to fixed-point numbers. The dummy None is added so we can
# use 1-based indexing. (This just allows us to be consistent with
# Bailey's indexing. The algorithm is 100 lines long, so debugging
# a single wrong index can be painful.)
x = [None] + [ctx.to_fixed(ctx.mpf(xk), prec) for xk in x]
# Sanity check on magnitudes
minx = min(abs(xx) for xx in x[1:])
if not minx:
raise ValueError("PSLQ requires a vector of nonzero numbers")
if minx < tol//100:
if verbose:
print("STOPPING: (one number is too small)")
return None
g = sqrt_fixed((4<<prec)//3, prec)
A = {}
B = {}
H = {}
# Initialization
# step 1
for i in xrange(1, n+1):
for j in xrange(1, n+1):
A[i,j] = B[i,j] = (i==j) << prec
H[i,j] = 0
# step 2
s = [None] + [0] * n
for k in xrange(1, n+1):
t = 0
for j in xrange(k, n+1):
t += (x[j]**2 >> prec)
s[k] = sqrt_fixed(t, prec)
t = s[1]
y = x[:]
for k in xrange(1, n+1):
y[k] = (x[k] << prec) // t
s[k] = (s[k] << prec) // t
# step 3
for i in xrange(1, n+1):
for j in xrange(i+1, n):
H[i,j] = 0
if i <= n-1:
if s[i]:
H[i,i] = (s[i+1] << prec) // s[i]
else:
H[i,i] = 0
for j in range(1, i):
sjj1 = s[j]*s[j+1]
if sjj1:
H[i,j] = ((-y[i]*y[j])<<prec)//sjj1
else:
H[i,j] = 0
# step 4
for i in xrange(2, n+1):
for j in xrange(i-1, 0, -1):
#t = floor(H[i,j]/H[j,j] + 0.5)
if H[j,j]:
t = round_fixed((H[i,j] << prec)//H[j,j], prec)
else:
#t = 0
continue
y[j] = y[j] + (t*y[i] >> prec)
for k in xrange(1, j+1):
H[i,k] = H[i,k] - (t*H[j,k] >> prec)
for k in xrange(1, n+1):
A[i,k] = A[i,k] - (t*A[j,k] >> prec)
B[k,j] = B[k,j] + (t*B[k,i] >> prec)
# Main algorithm
for REP in range(maxsteps):
# Step 1
m = -1
szmax = -1
for i in range(1, n):
h = H[i,i]
sz = (g**i * abs(h)) >> (prec*(i-1))
if sz > szmax:
m = i
szmax = sz
# Step 2
y[m], y[m+1] = y[m+1], y[m]
for i in xrange(1,n+1): H[m,i], H[m+1,i] = H[m+1,i], H[m,i]
for i in xrange(1,n+1): A[m,i], A[m+1,i] = A[m+1,i], A[m,i]
for i in xrange(1,n+1): B[i,m], B[i,m+1] = B[i,m+1], B[i,m]
# Step 3
if m <= n - 2:
t0 = sqrt_fixed((H[m,m]**2 + H[m,m+1]**2)>>prec, prec)
# A zero element probably indicates that the precision has
# been exhausted. XXX: this could be spurious, due to
# using fixed-point arithmetic
if not t0:
break
t1 = (H[m,m] << prec) // t0
t2 = (H[m,m+1] << prec) // t0
for i in xrange(m, n+1):
t3 = H[i,m]
t4 = H[i,m+1]
H[i,m] = (t1*t3+t2*t4) >> prec
H[i,m+1] = (-t2*t3+t1*t4) >> prec
# Step 4
for i in xrange(m+1, n+1):
for j in xrange(min(i-1, m+1), 0, -1):
try:
t = round_fixed((H[i,j] << prec)//H[j,j], prec)
# Precision probably exhausted
except ZeroDivisionError:
break
y[j] = y[j] + ((t*y[i]) >> prec)
for k in xrange(1, j+1):
H[i,k] = H[i,k] - (t*H[j,k] >> prec)
for k in xrange(1, n+1):
A[i,k] = A[i,k] - (t*A[j,k] >> prec)
B[k,j] = B[k,j] + (t*B[k,i] >> prec)
# Until a relation is found, the error typically decreases
# slowly (e.g. a factor 1-10) with each step TODO: we could
# compare err from two successive iterations. If there is a
# large drop (several orders of magnitude), that indicates a
# "high quality" relation was detected. Reporting this to
# the user somehow might be useful.
best_err = maxcoeff<<prec
for i in xrange(1, n+1):
err = abs(y[i])
# Maybe we are done?
if err < tol:
# We are done if the coefficients are acceptable
vec = [int(round_fixed(B[j,i], prec) >> prec) for j in \
range(1,n+1)]
if max(abs(v) for v in vec) < maxcoeff:
if verbose:
print("FOUND relation at iter %i/%i, error: %s" % \
(REP, maxsteps, ctx.nstr(err / ctx.mpf(2)**prec, 1)))
return vec
best_err = min(err, best_err)
# Calculate a lower bound for the norm. We could do this
# more exactly (using the Euclidean norm) but there is probably
# no practical benefit.
recnorm = max(abs(h) for h in H.values())
if recnorm:
norm = ((1 << (2*prec)) // recnorm) >> prec
norm //= 100
else:
norm = ctx.inf
if verbose:
print("%i/%i: Error: %8s Norm: %s" % \
(REP, maxsteps, ctx.nstr(best_err / ctx.mpf(2)**prec, 1), norm))
if norm >= maxcoeff:
break
if verbose:
print("CANCELLING after step %i/%i." % (REP, maxsteps))
print("Could not find an integer relation. Norm bound: %s" % norm)
return None
def findpoly(ctx, x, n=1, **kwargs):
r"""
``findpoly(x, n)`` returns the coefficients of an integer
polynomial `P` of degree at most `n` such that `P(x) \approx 0`.
If no polynomial having `x` as a root can be found,
:func:`~mpmath.findpoly` returns ``None``.
:func:`~mpmath.findpoly` works by successively calling :func:`~mpmath.pslq` with
the vectors `[1, x]`, `[1, x, x^2]`, `[1, x, x^2, x^3]`, ...,
`[1, x, x^2, .., x^n]` as input. Keyword arguments given to
:func:`~mpmath.findpoly` are forwarded verbatim to :func:`~mpmath.pslq`. In
particular, you can specify a tolerance for `P(x)` with ``tol``
and a maximum permitted coefficient size with ``maxcoeff``.
For large values of `n`, it is recommended to run :func:`~mpmath.findpoly`
at high precision; preferably 50 digits or more.
**Examples**
By default (degree `n = 1`), :func:`~mpmath.findpoly` simply finds a linear
polynomial with a rational root::
>>> from mpmath import *
>>> mp.dps = 15; mp.pretty = True
>>> findpoly(0.7)
[-10, 7]
The generated coefficient list is valid input to ``polyval`` and
``polyroots``::
>>> nprint(polyval(findpoly(phi, 2), phi), 1)
-2.0e-16
>>> for r in polyroots(findpoly(phi, 2)):
... print(r)
...
-0.618033988749895
1.61803398874989
Numbers of the form `m + n \sqrt p` for integers `(m, n, p)` are
solutions to quadratic equations. As we find here, `1+\sqrt 2`
is a root of the polynomial `x^2 - 2x - 1`::
>>> findpoly(1+sqrt(2), 2)
[1, -2, -1]
>>> findroot(lambda x: x**2 - 2*x - 1, 1)
2.4142135623731
Despite only containing square roots, the following number results
in a polynomial of degree 4::
>>> findpoly(sqrt(2)+sqrt(3), 4)
[1, 0, -10, 0, 1]
In fact, `x^4 - 10x^2 + 1` is the *minimal polynomial* of
`r = \sqrt 2 + \sqrt 3`, meaning that a rational polynomial of
lower degree having `r` as a root does not exist. Given sufficient
precision, :func:`~mpmath.findpoly` will usually find the correct
minimal polynomial of a given algebraic number.
**Non-algebraic numbers**
If :func:`~mpmath.findpoly` fails to find a polynomial with given
coefficient size and tolerance constraints, that means no such
polynomial exists.
We can verify that `\pi` is not an algebraic number of degree 3 with
coefficients less than 1000::
>>> mp.dps = 15
>>> findpoly(pi, 3)
>>>
It is always possible to find an algebraic approximation of a number
using one (or several) of the following methods:
1. Increasing the permitted degree
2. Allowing larger coefficients
3. Reducing the tolerance
One example of each method is shown below::
>>> mp.dps = 15
>>> findpoly(pi, 4)
[95, -545, 863, -183, -298]
>>> findpoly(pi, 3, maxcoeff=10000)
[836, -1734, -2658, -457]
>>> findpoly(pi, 3, tol=1e-7)
[-4, 22, -29, -2]
It is unknown whether Euler's constant is transcendental (or even
irrational). We can use :func:`~mpmath.findpoly` to check that if is
an algebraic number, its minimal polynomial must have degree
at least 7 and a coefficient of magnitude at least 1000000::
>>> mp.dps = 200
>>> findpoly(euler, 6, maxcoeff=10**6, tol=1e-100, maxsteps=1000)
>>>
Note that the high precision and strict tolerance is necessary
for such high-degree runs, since otherwise unwanted low-accuracy
approximations will be detected. It may also be necessary to set
maxsteps high to prevent a premature exit (before the coefficient
bound has been reached). Running with ``verbose=True`` to get an
idea what is happening can be useful.
"""
x = ctx.mpf(x)
if n < 1:
raise ValueError("n cannot be less than 1")
if x == 0:
return [1, 0]
xs = [ctx.mpf(1)]
for i in range(1,n+1):
xs.append(x**i)
a = ctx.pslq(xs, **kwargs)
if a is not None:
return a[::-1]
def fracgcd(p, q):
x, y = p, q
while y:
x, y = y, x % y
if x != 1:
p //= x
q //= x
if q == 1:
return p
return p, q
def pslqstring(r, constants):
q = r[0]
r = r[1:]
s = []
for i in range(len(r)):
p = r[i]
if p:
z = fracgcd(-p,q)
cs = constants[i][1]
if cs == '1':
cs = ''
else:
cs = '*' + cs
if isinstance(z, int_types):
if z > 0: term = str(z) + cs
else: term = ("(%s)" % z) + cs
else:
term = ("(%s/%s)" % z) + cs
s.append(term)
s = ' + '.join(s)
if '+' in s or '*' in s:
s = '(' + s + ')'
return s or '0'
def prodstring(r, constants):
q = r[0]
r = r[1:]
num = []
den = []
for i in range(len(r)):
p = r[i]
if p:
z = fracgcd(-p,q)
cs = constants[i][1]
if isinstance(z, int_types):
if abs(z) == 1: t = cs
else: t = '%s**%s' % (cs, abs(z))
([num,den][z<0]).append(t)
else:
t = '%s**(%s/%s)' % (cs, abs(z[0]), z[1])
([num,den][z[0]<0]).append(t)
num = '*'.join(num)
den = '*'.join(den)
if num and den: return "(%s)/(%s)" % (num, den)
if num: return num
if den: return "1/(%s)" % den
def quadraticstring(ctx,t,a,b,c):
if c < 0:
a,b,c = -a,-b,-c
u1 = (-b+ctx.sqrt(b**2-4*a*c))/(2*c)
u2 = (-b-ctx.sqrt(b**2-4*a*c))/(2*c)
if abs(u1-t) < abs(u2-t):
if b: s = '((%s+sqrt(%s))/%s)' % (-b,b**2-4*a*c,2*c)
else: s = '(sqrt(%s)/%s)' % (-4*a*c,2*c)
else:
if b: s = '((%s-sqrt(%s))/%s)' % (-b,b**2-4*a*c,2*c)
else: s = '(-sqrt(%s)/%s)' % (-4*a*c,2*c)
return s
# Transformation y = f(x,c), with inverse function x = f(y,c)
# The third entry indicates whether the transformation is
# redundant when c = 1
transforms = [
(lambda ctx,x,c: x*c, '$y/$c', 0),
(lambda ctx,x,c: x/c, '$c*$y', 1),
(lambda ctx,x,c: c/x, '$c/$y', 0),
(lambda ctx,x,c: (x*c)**2, 'sqrt($y)/$c', 0),
(lambda ctx,x,c: (x/c)**2, '$c*sqrt($y)', 1),
(lambda ctx,x,c: (c/x)**2, '$c/sqrt($y)', 0),
(lambda ctx,x,c: c*x**2, 'sqrt($y)/sqrt($c)', 1),
(lambda ctx,x,c: x**2/c, 'sqrt($c)*sqrt($y)', 1),
(lambda ctx,x,c: c/x**2, 'sqrt($c)/sqrt($y)', 1),
(lambda ctx,x,c: ctx.sqrt(x*c), '$y**2/$c', 0),
(lambda ctx,x,c: ctx.sqrt(x/c), '$c*$y**2', 1),
(lambda ctx,x,c: ctx.sqrt(c/x), '$c/$y**2', 0),
(lambda ctx,x,c: c*ctx.sqrt(x), '$y**2/$c**2', 1),
(lambda ctx,x,c: ctx.sqrt(x)/c, '$c**2*$y**2', 1),
(lambda ctx,x,c: c/ctx.sqrt(x), '$c**2/$y**2', 1),
(lambda ctx,x,c: ctx.exp(x*c), 'log($y)/$c', 0),
(lambda ctx,x,c: ctx.exp(x/c), '$c*log($y)', 1),
(lambda ctx,x,c: ctx.exp(c/x), '$c/log($y)', 0),
(lambda ctx,x,c: c*ctx.exp(x), 'log($y/$c)', 1),
(lambda ctx,x,c: ctx.exp(x)/c, 'log($c*$y)', 1),
(lambda ctx,x,c: c/ctx.exp(x), 'log($c/$y)', 0),
(lambda ctx,x,c: ctx.ln(x*c), 'exp($y)/$c', 0),
(lambda ctx,x,c: ctx.ln(x/c), '$c*exp($y)', 1),
(lambda ctx,x,c: ctx.ln(c/x), '$c/exp($y)', 0),
(lambda ctx,x,c: c*ctx.ln(x), 'exp($y/$c)', 1),
(lambda ctx,x,c: ctx.ln(x)/c, 'exp($c*$y)', 1),
(lambda ctx,x,c: c/ctx.ln(x), 'exp($c/$y)', 0),
]
def identify(ctx, x, constants=[], tol=None, maxcoeff=1000, full=False,
verbose=False):
r"""
Given a real number `x`, ``identify(x)`` attempts to find an exact
formula for `x`. This formula is returned as a string. If no match
is found, ``None`` is returned. With ``full=True``, a list of
matching formulas is returned.
As a simple example, :func:`~mpmath.identify` will find an algebraic
formula for the golden ratio::
>>> from mpmath import *
>>> mp.dps = 15; mp.pretty = True
>>> identify(phi)
'((1+sqrt(5))/2)'
:func:`~mpmath.identify` can identify simple algebraic numbers and simple
combinations of given base constants, as well as certain basic
transformations thereof. More specifically, :func:`~mpmath.identify`
looks for the following:
1. Fractions
2. Quadratic algebraic numbers
3. Rational linear combinations of the base constants
4. Any of the above after first transforming `x` into `f(x)` where
`f(x)` is `1/x`, `\sqrt x`, `x^2`, `\log x` or `\exp x`, either
directly or with `x` or `f(x)` multiplied or divided by one of
the base constants
5. Products of fractional powers of the base constants and
small integers
Base constants can be given as a list of strings representing mpmath
expressions (:func:`~mpmath.identify` will ``eval`` the strings to numerical
values and use the original strings for the output), or as a dict of
formula:value pairs.
In order not to produce spurious results, :func:`~mpmath.identify` should
be used with high precision; preferably 50 digits or more.
**Examples**
Simple identifications can be performed safely at standard
precision. Here the default recognition of rational, algebraic,
and exp/log of algebraic numbers is demonstrated::
>>> mp.dps = 15
>>> identify(0.22222222222222222)
'(2/9)'
>>> identify(1.9662210973805663)
'sqrt(((24+sqrt(48))/8))'
>>> identify(4.1132503787829275)
'exp((sqrt(8)/2))'
>>> identify(0.881373587019543)
'log(((2+sqrt(8))/2))'
By default, :func:`~mpmath.identify` does not recognize `\pi`. At standard
precision it finds a not too useful approximation. At slightly
increased precision, this approximation is no longer accurate
enough and :func:`~mpmath.identify` more correctly returns ``None``::
>>> identify(pi)
'(2**(176/117)*3**(20/117)*5**(35/39))/(7**(92/117))'
>>> mp.dps = 30
>>> identify(pi)
>>>
Numbers such as `\pi`, and simple combinations of user-defined
constants, can be identified if they are provided explicitly::
>>> identify(3*pi-2*e, ['pi', 'e'])
'(3*pi + (-2)*e)'
Here is an example using a dict of constants. Note that the
constants need not be "atomic"; :func:`~mpmath.identify` can just
as well express the given number in terms of expressions
given by formulas::
>>> identify(pi+e, {'a':pi+2, 'b':2*e})
'((-2) + 1*a + (1/2)*b)'
Next, we attempt some identifications with a set of base constants.
It is necessary to increase the precision a bit.
>>> mp.dps = 50
>>> base = ['sqrt(2)','pi','log(2)']
>>> identify(0.25, base)
'(1/4)'
>>> identify(3*pi + 2*sqrt(2) + 5*log(2)/7, base)
'(2*sqrt(2) + 3*pi + (5/7)*log(2))'
>>> identify(exp(pi+2), base)
'exp((2 + 1*pi))'
>>> identify(1/(3+sqrt(2)), base)
'((3/7) + (-1/7)*sqrt(2))'
>>> identify(sqrt(2)/(3*pi+4), base)
'sqrt(2)/(4 + 3*pi)'
>>> identify(5**(mpf(1)/3)*pi*log(2)**2, base)
'5**(1/3)*pi*log(2)**2'
An example of an erroneous solution being found when too low
precision is used::
>>> mp.dps = 15
>>> identify(1/(3*pi-4*e+sqrt(8)), ['pi', 'e', 'sqrt(2)'])
'((11/25) + (-158/75)*pi + (76/75)*e + (44/15)*sqrt(2))'
>>> mp.dps = 50
>>> identify(1/(3*pi-4*e+sqrt(8)), ['pi', 'e', 'sqrt(2)'])
'1/(3*pi + (-4)*e + 2*sqrt(2))'
**Finding approximate solutions**
The tolerance ``tol`` defaults to 3/4 of the working precision.
Lowering the tolerance is useful for finding approximate matches.
We can for example try to generate approximations for pi::
>>> mp.dps = 15
>>> identify(pi, tol=1e-2)
'(22/7)'
>>> identify(pi, tol=1e-3)
'(355/113)'
>>> identify(pi, tol=1e-10)
'(5**(339/269))/(2**(64/269)*3**(13/269)*7**(92/269))'
With ``full=True``, and by supplying a few base constants,
``identify`` can generate almost endless lists of approximations
for any number (the output below has been truncated to show only
the first few)::
>>> for p in identify(pi, ['e', 'catalan'], tol=1e-5, full=True):
... print(p)
... # doctest: +ELLIPSIS
e/log((6 + (-4/3)*e))
(3**3*5*e*catalan**2)/(2*7**2)
sqrt(((-13) + 1*e + 22*catalan))
log(((-6) + 24*e + 4*catalan)/e)
exp(catalan*((-1/5) + (8/15)*e))
catalan*(6 + (-6)*e + 15*catalan)
sqrt((5 + 26*e + (-3)*catalan))/e
e*sqrt(((-27) + 2*e + 25*catalan))
log(((-1) + (-11)*e + 59*catalan))
((3/20) + (21/20)*e + (3/20)*catalan)
...
The numerical values are roughly as close to `\pi` as permitted by the
specified tolerance:
>>> e/log(6-4*e/3)
3.14157719846001
>>> 135*e*catalan**2/98
3.14166950419369
>>> sqrt(e-13+22*catalan)
3.14158000062992
>>> log(24*e-6+4*catalan)-1
3.14158791577159
**Symbolic processing**
The output formula can be evaluated as a Python expression.
Note however that if fractions (like '2/3') are present in
the formula, Python's :func:`~mpmath.eval()` may erroneously perform
integer division. Note also that the output is not necessarily
in the algebraically simplest form::
>>> identify(sqrt(2))
'(sqrt(8)/2)'
As a solution to both problems, consider using SymPy's
:func:`~mpmath.sympify` to convert the formula into a symbolic expression.
SymPy can be used to pretty-print or further simplify the formula
symbolically::
>>> from sympy import sympify # doctest: +SKIP
>>> sympify(identify(sqrt(2))) # doctest: +SKIP
2**(1/2)
Sometimes :func:`~mpmath.identify` can simplify an expression further than
a symbolic algorithm::
>>> from sympy import simplify # doctest: +SKIP
>>> x = sympify('-1/(-3/2+(1/2)*5**(1/2))*(3/2-1/2*5**(1/2))**(1/2)') # doctest: +SKIP
>>> x # doctest: +SKIP
(3/2 - 5**(1/2)/2)**(-1/2)
>>> x = simplify(x) # doctest: +SKIP
>>> x # doctest: +SKIP
2/(6 - 2*5**(1/2))**(1/2)
>>> mp.dps = 30 # doctest: +SKIP
>>> x = sympify(identify(x.evalf(30))) # doctest: +SKIP
>>> x # doctest: +SKIP
1/2 + 5**(1/2)/2
(In fact, this functionality is available directly in SymPy as the
function :func:`~mpmath.nsimplify`, which is essentially a wrapper for
:func:`~mpmath.identify`.)
**Miscellaneous issues and limitations**
The input `x` must be a real number. All base constants must be
positive real numbers and must not be rationals or rational linear
combinations of each other.
The worst-case computation time grows quickly with the number of
base constants. Already with 3 or 4 base constants,
:func:`~mpmath.identify` may require several seconds to finish. To search
for relations among a large number of constants, you should
consider using :func:`~mpmath.pslq` directly.
The extended transformations are applied to x, not the constants
separately. As a result, ``identify`` will for example be able to
recognize ``exp(2*pi+3)`` with ``pi`` given as a base constant, but
not ``2*exp(pi)+3``. It will be able to recognize the latter if
``exp(pi)`` is given explicitly as a base constant.
"""
solutions = []
def addsolution(s):
if verbose: print("Found: ", s)
solutions.append(s)
x = ctx.mpf(x)
# Further along, x will be assumed positive
if x == 0:
if full: return ['0']
else: return '0'
if x < 0:
sol = ctx.identify(-x, constants, tol, maxcoeff, full, verbose)
if sol is None:
return sol
if full:
return ["-(%s)"%s for s in sol]
else:
return "-(%s)" % sol
if tol:
tol = ctx.mpf(tol)
else:
tol = ctx.eps**0.7
M = maxcoeff
if constants:
if isinstance(constants, dict):
constants = [(ctx.mpf(v), name) for (name, v) in sorted(constants.items())]
else:
namespace = dict((name, getattr(ctx,name)) for name in dir(ctx))
constants = [(eval(p, namespace), p) for p in constants]
else:
constants = []
# We always want to find at least rational terms
if 1 not in [value for (name, value) in constants]:
constants = [(ctx.mpf(1), '1')] + constants
# PSLQ with simple algebraic and functional transformations
for ft, ftn, red in transforms:
for c, cn in constants:
if red and cn == '1':
continue
t = ft(ctx,x,c)
# Prevent exponential transforms from wreaking havoc
if abs(t) > M**2 or abs(t) < tol:
continue
# Linear combination of base constants
r = ctx.pslq([t] + [a[0] for a in constants], tol, M)
s = None
if r is not None and max(abs(uw) for uw in r) <= M and r[0]:
s = pslqstring(r, constants)
# Quadratic algebraic numbers
else:
q = ctx.pslq([ctx.one, t, t**2], tol, M)
if q is not None and len(q) == 3 and q[2]:
aa, bb, cc = q
if max(abs(aa),abs(bb),abs(cc)) <= M:
s = quadraticstring(ctx,t,aa,bb,cc)
if s:
if cn == '1' and ('/$c' in ftn):
s = ftn.replace('$y', s).replace('/$c', '')
else:
s = ftn.replace('$y', s).replace('$c', cn)
addsolution(s)
if not full: return solutions[0]
if verbose:
print(".")
# Check for a direct multiplicative formula
if x != 1:
# Allow fractional powers of fractions
ilogs = [2,3,5,7]
# Watch out for existing fractional powers of fractions
logs = []
for a, s in constants:
if not sum(bool(ctx.findpoly(ctx.ln(a)/ctx.ln(i),1)) for i in ilogs):
logs.append((ctx.ln(a), s))
logs = [(ctx.ln(i),str(i)) for i in ilogs] + logs
r = ctx.pslq([ctx.ln(x)] + [a[0] for a in logs], tol, M)
if r is not None and max(abs(uw) for uw in r) <= M and r[0]:
addsolution(prodstring(r, logs))
if not full: return solutions[0]
if full:
return sorted(solutions, key=len)
else:
return None
IdentificationMethods.pslq = pslq
IdentificationMethods.findpoly = findpoly
IdentificationMethods.identify = identify
if __name__ == '__main__':
import doctest
doctest.testmod()