Find the simplest fraction for a given float or interval
Project description
The simplefractions package finds the simplest fraction that converts to a given float, or more generally, the simplest fraction that lies within a given interval.
Definition. Given fractions x = a/b and y = c/d (written in lowest terms with positive denominators), say that x is simpler than y if abs(a) <= abs(c), b <= d, and at least one of these two inequalities is strict.
For example, 22/7 is simpler than 23/8, but neither of 3/8 and 4/7 is simpler than the other.
Then it's a theorem that given any subinterval I of the real line that contains at least one fraction, that interval contains a unique simplest fraction. That is, there's a fraction a/b in I such that a/b is simpler (in the above sense) than all other fractions in I. As a consequence, for any given finite Python float x, there's a unique simplest fraction that rounds to that float.
The simplefractions package provides two functions:
simplest_from_floatreturns, for a given floatx, the unique simplest fraction with the property thatfloat(simplest_from_float(x)) == x.simplest_in_intervalreturns the unique simplest fraction in a given (open or closed, bounded or unbounded) nonempty interval.
Example usage
Start by importing the functions from the module:
>>> from simplefractions import *
The simplest_from_float function takes a single finite float x and
produces a Fraction object that recovers that float:
>>> simplest_from_float(0.25)
Fraction(1, 4)
>>> simplest_from_float(0.33)
Fraction(33, 100)
No matter what x is given, the invariant float(simplest_from_float(x)) == x
will always be true.
>>> x = 0.7429667872099244
>>> simplest_from_float(x)
Fraction(88650459, 119319545)
>>> float(simplest_from_float(x))
0.7429667872099244
>>> float(simplest_from_float(x)) == x
True
If the float x was constructed by dividing two small integers, then
more often than not, simplest_from_float will recover those integers:
>>> x = 231 / 199
>>> x
1.1608040201005025
>>> simplest_from_float(x)
Fraction(231, 199)
More precisely, if x was constructed by dividing two relatively prime
integers smaller than or equal to 67114657 in absolute value,
simplest_from_float will recover those integers.
>>> simplest_from_float(64841043 / 66055498)
Fraction(64841043, 66055498)
But 67114657 is the best we can do here:
>>> simplest_from_float(67114658 / 67114657)
Fraction(67114657, 67114656)
In larger cases, simplest_from_float might discover a simpler fraction
that gives the same float:
>>> x = 818421477165 / 1580973145504
>>> simplest_from_float(x)
Fraction(5171, 9989)
>>> 818421477165 / 1580973145504 == 5171 / 9989
True
Note that simplest_from_float does not magically fix floating-point
inaccuracies. For example:
>>> x = 1.1 + 2.2
>>> simplest_from_float(x)
Fraction(675539944105597, 204709073971393)
You might have expected Fraction(33, 10) here, but when converted to float,
that gives a value very close to, but not exactly equal to, x. In contrast,
the return value of simplest_from_float(x) will always produce exactly x
when converted to float.
To fix this, you might want to ask for the simplest float that lies within
some small error bound of x - for example, within 5 ulps (units in the
last place) in either direction. simplest_from_float can't do that, but
simplest_in_interval can! For example
>>> from math import ulp
>>> x = 1.1 + 2.2
>>> simplest_in_interval(x - 5*ulp(x), x + 5*ulp(x))
Fraction(33, 10)
Alternatively, you might ask for the simplest fraction approximating x
with a relative error of at most 0.000001:
>>> relerr = 1e-6
>>> simplest_in_interval(x - relerr*x, x + relerr*x)
Fraction(33, 10)
Here are some more examples of simplest_in_interval at work. The inputs
to simplest_in_interval can be floats, integers, or Fraction objects.
>>> simplest_in_interval(3.14, 3.15)
Fraction(22, 7)
By default, simplest_in_interval assumes that you're specifying an
open interval:
>>> simplest_in_interval(3, 4)
Fraction(7, 2)
Keyword arguments include_left and include_right allow you to specify
that one or both endpoints should be included in the interval:
>>> simplest_in_interval(3, 4, include_left=True, include_right=True)
Fraction(3, 1)
The left and right endpoints of the interval are also both optional, alowing a semi-infinite or infinite interval to be specified:
>>> simplest_in_interval(right=4) # simplest in (-inf, 4)
Fraction(0, 1)
>>> simplest_in_interval(left=4, include_left=True) # simplest in [4, inf)
Fraction(4, 1)
>>> simplest_in_interval() # simplest in (-inf, inf)
Fraction(0, 1)
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