Number identification¶
Most function in mpmath are concerned with producing approximations from exact mathematical formulas. It is also useful to consider the inverse problem: given only a decimal approximation for a number, such as 0.7320508075688772935274463, is it possible to find an exact formula?
Subject to certain restrictions, such “reverse engineering” is indeed possible thanks to the existence of integer relation algorithms. Mpmath implements the PSLQ algorithm (developed by H. Ferguson), which is one such algorithm.
Automated number recognition based on PSLQ is not a silver bullet. Any occurring transcendental constants (\(\pi\), \(e\), etc) must be guessed by the user, and the relation between those constants in the formula must be linear (such as \(x = 3 \pi + 4 e\)). More complex formulas can be found by combining PSLQ with functional transformations; however, this is only feasible to a limited extent since the computation time grows exponentially with the number of operations that need to be combined.
The number identification facilities in mpmath are inspired by the Inverse Symbolic Calculator (ISC). The ISC is more powerful than mpmath, as it uses a lookup table of millions of precomputed constants (thereby mitigating the problem with exponential complexity).
Constant recognition¶
identify()
¶

mpmath.
identify
(ctx, x, constants=[], tol=None, maxcoeff=1000, full=False, verbose=False)¶ 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. Withfull=True
, a list of matching formulas is returned.As a simple example,
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)'
identify()
can identify simple algebraic numbers and simple combinations of given base constants, as well as certain basic transformations thereof. More specifically,identify()
looks for the following: Fractions
 Quadratic algebraic numbers
 Rational linear combinations of the base constants
 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
 Products of fractional powers of the base constants and small integers
Base constants can be given as a list of strings representing mpmath expressions (
identify()
willeval
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,
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,
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 andidentify()
more correctly returnsNone
:>>> 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 userdefined constants, can be identified if they are provided explicitly:
>>> identify(3*pi2*e, ['pi', 'e']) '(3*pi + (2)*e)'
Here is an example using a dict of constants. Note that the constants need not be “atomic”;
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*pi4*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*pi4*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=1e2) '(22/7)' >>> identify(pi, tol=1e3) '(355/113)' >>> identify(pi, tol=1e10) '(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=1e5, full=True): ... print(p) ... 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(64*e/3) 3.14157719846001 >>> 135*e*catalan**2/98 3.14166950419369 >>> sqrt(e13+22*catalan) 3.14158000062992 >>> log(24*e6+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
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
sympify()
to convert the formula into a symbolic expression. SymPy can be used to prettyprint or further simplify the formula symbolically:>>> from sympy import sympify >>> sympify(identify(sqrt(2))) 2**(1/2)
Sometimes
identify()
can simplify an expression further than a symbolic algorithm:>>> from sympy import simplify >>> x = sympify('1/(3/2+(1/2)*5**(1/2))*(3/21/2*5**(1/2))**(1/2)') >>> x (3/2  5**(1/2)/2)**(1/2) >>> x = simplify(x) >>> x 2/(6  2*5**(1/2))**(1/2) >>> mp.dps = 30 >>> x = sympify(identify(x.evalf(30))) >>> x 1/2 + 5**(1/2)/2
(In fact, this functionality is available directly in SymPy as the function
nsimplify()
, which is essentially a wrapper foridentify()
.)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 worstcase computation time grows quickly with the number of base constants. Already with 3 or 4 base constants,
identify()
may require several seconds to finish. To search for relations among a large number of constants, you should consider usingpslq()
directly.The extended transformations are applied to x, not the constants separately. As a result,
identify
will for example be able to recognizeexp(2*pi+3)
withpi
given as a base constant, but not2*exp(pi)+3
. It will be able to recognize the latter ifexp(pi)
is given explicitly as a base constant.
Algebraic identification¶
findpoly()
¶

mpmath.
findpoly
(ctx, x, n=1, **kwargs)¶ 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,findpoly()
returnsNone
.findpoly()
works by successively callingpslq()
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 tofindpoly()
are forwarded verbatim topslq()
. In particular, you can specify a tolerance for \(P(x)\) withtol
and a maximum permitted coefficient size withmaxcoeff
.For large values of \(n\), it is recommended to run
findpoly()
at high precision; preferably 50 digits or more.Examples
By default (degree \(n = 1\)),
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
andpolyroots
:>>> nprint(polyval(findpoly(phi, 2), phi), 1) 2.0e16 >>> 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,
findpoly()
will usually find the correct minimal polynomial of a given algebraic number.Nonalgebraic numbers
If
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:
 Increasing the permitted degree
 Allowing larger coefficients
 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=1e7) [4, 22, 29, 2]
It is unknown whether Euler’s constant is transcendental (or even irrational). We can use
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=1e100, maxsteps=1000) >>>
Note that the high precision and strict tolerance is necessary for such highdegree runs, since otherwise unwanted lowaccuracy 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.
Integer relations (PSLQ)¶
pslq()
¶

mpmath.
pslq
(ctx, x, tol=None, maxcoeff=1000, maxsteps=100, verbose=False)¶ 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\[c_1 x_1 + c_2 x_2 + ... + c_n x_n < \mathrm{tol}\]and such that \(\max c_k < \mathrm{maxcoeff}\). If no such vector exists,
pslq()
returnsNone
. 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,
\[\frac{\pi}{4} = 4 \operatorname{acot} 5  \operatorname{acot} 239\]There are actually infinitely many formulas of this type. Two others are
\[ \begin{align}\begin{aligned}\frac{\pi}{4} = \operatorname{acot} 1\\\frac{\pi}{4} = 12 \operatorname{acot} 49 + 32 \operatorname{acot} 57 + 5 \operatorname{acot} 239 + 12 \operatorname{acot} 110443\end{aligned}\end{align} \]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 Machinlike 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 fixedpoint instead of floatingpoint arithmetic, since this is significantly (about 7x) faster.