Recently we encountered an issue with math.log() . Since 243 is a perfect power of 3 , assumption that taking floor should be fine was wrong as it seems to have precision error on lower side.
So as a hack we started adding a small value before taking logarithm. Is there a way to configue math.log upfront or something similar that that we dont have to add EPS every time.
To clarify some of the comments Note we are not looking to round to nearest integer. Our goal is to keep the value exact or at times take floor. But if the precision error is on lower side floor screws up big time, that's what we are trying to avoid.
code:
import math
math.log(243, 3)
int(math.log(243, 3))
output:
4.999999999999999
4
code:
import math
EPS = 1e-09
math.log(243 + EPS, 3)
int(math.log(243 + EPS, 3))
output:
5.0000000000037454
5
Instead of trying to solve it might be easier to look at and just solve this iteratively, taking advantage of Python's integer type. This way you can avoid the float domain, and its associated precision loss, entirely.
Here's a rough attempt:
def ilog(a: int, p: int) -> tuple[int, bool]:
"""
find the largest b such that p ** b <= a
return tuple of (b, exact)
"""
if p == 1:
return a, True
b = 0
x = 1
while x < a:
x *= p
b += 1
if x == a:
return b, True
else:
return b - 1, False
There are plenty of opportunities for optimization if this is too slow (consider Newton's method, binary search...)
How about this? Is this what you are looking for?
import math
def ilog(a: int, p:int) -> int:
"""
find the largest b such that p ** b <= a
"""
float_log = math.log(a, p)
if (candidate := math.ceil(float_log))**p <= a:
return candidate
return int(float_log)
print(ilog(243, 3))
print(ilog(3**31, 3))
print(ilog(8,2))
Output:
5
31
3
You can use decimals and play with precision and rounding instead of floats in this case
Like this:
from decimal import Decimal, Context, ROUND_HALF_UP, ROUND_HALF_DOWN
ctx1 = Context(prec=20, rounding=ROUND_HALF_UP)
ctx2 = Context(prec=20, rounding=ROUND_HALF_DOWN)
ctx1.divide(Decimal(243).ln( ctx1) , Decimal(3).ln( ctx2))
Output:
Decimal('5')
First, the rounding works like the epsilon - the numerator is rounded up and denominator down. You always get a slightly higher answer
Second, you can adjust precision you need
However, fundamentally the problem is unsolvable.
I need to calculate the square root of some numbers, for example √9 = 3 and √2 = 1.4142. How can I do it in Python?
The inputs will probably be all positive integers, and relatively small (say less than a billion), but just in case they're not, is there anything that might break?
Related
Integer square root in python
How to find integer nth roots?
Is there a short-hand for nth root of x in Python?
Difference between **(1/2), math.sqrt and cmath.sqrt?
Why is math.sqrt() incorrect for large numbers?
Python sqrt limit for very large numbers?
Which is faster in Python: x**.5 or math.sqrt(x)?
Why does Python give the "wrong" answer for square root? (specific to Python 2)
calculating n-th roots using Python 3's decimal module
How can I take the square root of -1 using python? (focused on NumPy)
Arbitrary precision of square roots
Note: This is an attempt at a canonical question after a discussion on Meta about an existing question with the same title.
Option 1: math.sqrt()
The math module from the standard library has a sqrt function to calculate the square root of a number. It takes any type that can be converted to float (which includes int) as an argument and returns a float.
>>> import math
>>> math.sqrt(9)
3.0
Option 2: Fractional exponent
The power operator (**) or the built-in pow() function can also be used to calculate a square root. Mathematically speaking, the square root of a equals a to the power of 1/2.
The power operator requires numeric types and matches the conversion rules for binary arithmetic operators, so in this case it will return either a float or a complex number.
>>> 9 ** (1/2)
3.0
>>> 9 ** .5 # Same thing
3.0
>>> 2 ** .5
1.4142135623730951
(Note: in Python 2, 1/2 is truncated to 0, so you have to force floating point arithmetic with 1.0/2 or similar. See Why does Python give the "wrong" answer for square root?)
This method can be generalized to nth root, though fractions that can't be exactly represented as a float (like 1/3 or any denominator that's not a power of 2) may cause some inaccuracy:
>>> 8 ** (1/3)
2.0
>>> 125 ** (1/3)
4.999999999999999
Edge cases
Negative and complex
Exponentiation works with negative numbers and complex numbers, though the results have some slight inaccuracy:
>>> (-25) ** .5 # Should be 5j
(3.061616997868383e-16+5j)
>>> 8j ** .5 # Should be 2+2j
(2.0000000000000004+2j)
Note the parentheses on -25! Otherwise it's parsed as -(25**.5) because exponentiation is more tightly binding than unary negation.
Meanwhile, math is only built for floats, so for x<0, math.sqrt(x) will raise ValueError: math domain error and for complex x, it'll raise TypeError: can't convert complex to float. Instead, you can use cmath.sqrt(x), which is more more accurate than exponentiation (and will likely be faster too):
>>> import cmath
>>> cmath.sqrt(-25)
5j
>>> cmath.sqrt(8j)
(2+2j)
Precision
Both options involve an implicit conversion to float, so floating point precision is a factor. For example:
>>> n = 10**30
>>> x = n**2
>>> root = x**.5
>>> n == root
False
>>> n - root # how far off are they?
0.0
>>> int(root) - n # how far off is the float from the int?
19884624838656
Very large numbers might not even fit in a float and you'll get OverflowError: int too large to convert to float. See Python sqrt limit for very large numbers?
Other types
Let's look at Decimal for example:
Exponentiation fails unless the exponent is also Decimal:
>>> decimal.Decimal('9') ** .5
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ** or pow(): 'decimal.Decimal' and 'float'
>>> decimal.Decimal('9') ** decimal.Decimal('.5')
Decimal('3.000000000000000000000000000')
Meanwhile, math and cmath will silently convert their arguments to float and complex respectively, which could mean loss of precision.
decimal also has its own .sqrt(). See also calculating n-th roots using Python 3's decimal module
SymPy
Depending on your goal, it might be a good idea to delay the calculation of square roots for as long as possible. SymPy might help.
SymPy is a Python library for symbolic mathematics.
import sympy
sympy.sqrt(2)
# => sqrt(2)
This doesn't seem very useful at first.
But sympy can give more information than floats or Decimals:
sympy.sqrt(8) / sympy.sqrt(27)
# => 2*sqrt(6)/9
Also, no precision is lost. (√2)² is still an integer:
s = sympy.sqrt(2)
s**2
# => 2
type(s**2)
#=> <class 'sympy.core.numbers.Integer'>
In comparison, floats and Decimals would return a number which is very close to 2 but not equal to 2:
(2**0.5)**2
# => 2.0000000000000004
from decimal import Decimal
(Decimal('2')**Decimal('0.5'))**Decimal('2')
# => Decimal('1.999999999999999999999999999')
Sympy also understands more complex examples like the Gaussian integral:
from sympy import Symbol, integrate, pi, sqrt, exp, oo
x = Symbol('x')
integrate(exp(-x**2), (x, -oo, oo))
# => sqrt(pi)
integrate(exp(-x**2), (x, -oo, oo)) == sqrt(pi)
# => True
Finally, if a decimal representation is desired, it's possible to ask for more digits than will ever be needed:
sympy.N(sympy.sqrt(2), 1_000_000)
# => 1.4142135623730950488016...........2044193016904841204
NumPy
>>> import numpy as np
>>> np.sqrt(25)
5.0
>>> np.sqrt([2, 3, 4])
array([1.41421356, 1.73205081, 2. ])
docs
Negative
For negative reals, it'll return nan, so np.emath.sqrt() is available for that case.
>>> a = np.array([4, -1, np.inf])
>>> np.sqrt(a)
<stdin>:1: RuntimeWarning: invalid value encountered in sqrt
array([ 2., nan, inf])
>>> np.emath.sqrt(a)
array([ 2.+0.j, 0.+1.j, inf+0.j])
Another option, of course, is to convert to complex first:
>>> a = a.astype(complex)
>>> np.sqrt(a)
array([ 2.+0.j, 0.+1.j, inf+0.j])
Newton's method
Most simple and accurate way to compute square root is Newton's method.
You have a number which you want to compute its square root (num) and you have a guess of its square root (estimate). Estimate can be any number bigger than 0, but a number that makes sense shortens the recursive call depth significantly.
new_estimate = (estimate + num/estimate) / 2
This line computes a more accurate estimate with those 2 parameters. You can pass new_estimate value to the function and compute another new_estimate which is more accurate than the previous one or you can make a recursive function definition like this.
def newtons_method(num, estimate):
# Computing a new_estimate
new_estimate = (estimate + num/estimate) / 2
print(new_estimate)
# Base Case: Comparing our estimate with built-in functions value
if new_estimate == math.sqrt(num):
return True
else:
return newtons_method(num, new_estimate)
For example we need to find 30's square root. We know that the result is between 5 and 6.
newtons_method(30,5)
number is 30 and estimate is 5. The result from each recursive calls are:
5.5
5.477272727272727
5.4772255752546215
5.477225575051661
The last result is the most accurate computation of the square root of number. It is the same value as the built-in function math.sqrt().
This answer was originally posted by gunesevitan, but is now deleted.
Python's fractions module and its class, Fraction, implement arithmetic with rational numbers. The Fraction class doesn't implement a square root operation, because most square roots are irrational numbers. However, it can be used to approximate a square root with arbitrary accuracy, because a Fraction's numerator and denominator are arbitrary-precision integers.
The following method takes a positive number x and a number of iterations, and returns upper and lower bounds for the square root of x.
from fractions import Fraction
def sqrt(x, n):
x = x if isinstance(x, Fraction) else Fraction(x)
upper = x + 1
for i in range(0, n):
upper = (upper + x/upper) / 2
lower = x / upper
if lower > upper:
raise ValueError("Sanity check failed")
return (lower, upper)
See the reference below for details on this operation's implementation. It also shows how to implement other operations with upper and lower bounds (although there is apparently at least one error with the log operation there).
Daumas, M., Lester, D., Muñoz, C., "Verified Real Number Calculations: A Library for Interval Arithmetic", arXiv:0708.3721 [cs.MS], 2007.
Alternatively, using Python's math.isqrt, we can calculate a square root to arbitrary precision:
Square root of i within 1/2n of the correct value, where i is an integer:Fraction(math.isqrt(i * 2**(n*2)), 2**n).
Square root of i within 1/10n of the correct value, where i is an integer:Fraction(math.isqrt(i * 10**(n*2)), 10**n).
Square root of x within 1/2n of the correct value, where x is a multiple of 1/2n:Fraction(math.isqrt(x * 2**(n)), 2**n).
Square root of x within 1/10n of the correct value, where x is a multiple of 1/10n:Fraction(math.isqrt(x * 10**(n)), 10**n).
In the foregoing, i or x must be 0 or greater.
Binary search
Disclaimer: this is for a more specialised use-case. This method might not be practical in all circumstances.
Benefits:
can find integer values (i.e. which integer is the root?)
no need to convert to float, so better precision (can be done that well too)
I personally implemented this one for a crypto CTF challenge (RSA cube root attack),where I needed a precise integer value.
The general idea can be extended to any other root.
def int_squareroot(d: int) -> tuple[int, bool]:
"""Try calculating integer squareroot and return if it's exact"""
left, right = 1, (d+1)//2
while left<right-1:
x = (left+right)//2
if x**2 > d:
left, right = left, x
else:
left, right = x, right
return left, left**2==d
EDIT:
As #wjandrea have also pointed out, **this example code can NOT compute **. This is a side-effect of the fact that it does not convert anything into floats, so no precision is lost. If the root is an integer, you get that back. If it's not, you get the biggest number whose square is smaller than your number. I updated the code so that it also returns a bool indicating if the value is correct or not, and also fixed an issue causing it to loop infinitely (also pointed out by #wjandrea). This implementation of the general method still works kindof weird for smaller numbers, but above 10 I had no problems with.
Overcoming the issues and limits of this method/implementation:
For smaller numbers, you can just use all the other methods from other answers. They generally use floats, which might be a loss of precision, but for small integers that should mean no problem at all. All of those methods that use floats have the same (or nearly the same) limit from this.
If you still want to use this method and get float results, it should be trivial to convert this to use floats too. Note that that will reintroduce precision loss, this method's unique benefit over the others, and in that case you can also just use any of the other answers. I think the newton's method version converges a bit faster, but I'm not sure.
For larger numbers, where loss of precision with floats come into play, this method can give results closer to the actual answer (depending on how big is the input). If you want to work with non-integers in this range, you can use other types, for example fixed precision numbers in this method too.
Edit 2, on other answers:
Currently, and afaik, the only other answer that has similar or better precision for large numbers than this implementation is the one that suggest SymPy, by Eric Duminil. That version is also easier to use, and work for any kind of number, the only downside is that it requires SymPy. My implementation is free from any huge dependencies if that is what you are looking for.
Arbitrary precision square root
This variation uses string manipulations to convert a string which represents a decimal floating-point number to an int, calls math.isqrt to do the actual square root extraction, and then formats the result as a decimal string. math.isqrt rounds down, so all produced digits are correct.
The input string, num, must use plain float format: 'e' notation is not supported. The num string can be a plain integer, and leading zeroes are ignored.
The digits argument specifies the number of decimal places in the result string, i.e., the number of digits after the decimal point.
from math import isqrt
def str_sqrt(num, digits):
""" Arbitrary precision square root
num arg must be a string
Return a string with `digits` after
the decimal point
Written by PM 2Ring 2022.01.26
"""
int_part , _, frac_part = num.partition('.')
num = int_part + frac_part
# Determine the required precision
width = 2 * digits - len(frac_part)
# Truncate or pad with zeroes
num = num[:width] if width < 0 else num + '0' * width
s = str(isqrt(int(num)))
if digits:
# Pad, if necessary
s = '0' * (1 + digits - len(s)) + s
s = f"{s[:-digits]}.{s[-digits:]}"
return s
Test
print(str_sqrt("2.0", 30))
Output
1.414213562373095048801688724209
For small numbers of digits, it's faster to use decimal.Decimal.sqrt. Around 32 digits or so, str_sqrt is roughly the same speed as Decimal.sqrt. But at 128 digits, str_sqrt is 2.2× faster than Decimal.sqrt, at 512 digits, it's 4.3× faster, at 8192 digits, it's 7.4× faster.
Here's a live version running on the SageMathCell server.
find square-root of a number
while True:
num = int(input("Enter a number:\n>>"))
for i in range(2, num):
if num % i == 0:
if i*i == num:
print("Square root of", num, "==>", i)
break
else:
kd = (num**0.5) # (num**(1/2))
print("Square root of", num, "==>", kd)
OUTPUT:-
Enter a number: 24
Square root of 24 ==> 4.898979485566356
Enter a number: 36
Square root of 36 ==> 6
Enter a number: 49
Square root of 49 ==> 7
✔ Output 💡 CLICK BELOW & SEE ✔
I have to write a function that uses another function, but the other function has to return integers which get fairly innacurate with large numbers.
My code:
import math
def reduce(n, d):
m = min(n, d)
for i in range(m, 1, -1):
if n%i==0 and d%i==0:
n = n//i
d = d//i
return (n, d)
def almost_square(n, d):
f = n/d
c = math.ceil(f)
n*=c
return reduce(n, d)
def destiny(n, d):
b = n/d
fraction = n, d
while not b.is_integer():
breuk = almost_square(fraction[0], fraction[1])
b = fraction[0]/fraction[1]
return int(b)
What the functions are supposed to do:
reduce: just simplifying the fraction, so 2/4 becomes 1/2 for example
almost_square: multiplying the fraction with the rounded up integer of the fraction
destiny: applying almost square on a fraction until it returns an integer.
The thing is, my uni works with a program that tries 50 test cases for each function and you only completed the exercise when every function works for all 50 test cases, and they expect the function 'reduce' to return a tuple of integers, but making integers of the numbers there makes my function 'destiny' innacurate, or at least I think so.
So out of the 50 test cases, all 50 work on the function reduce, all 50 work on the function almost_square, but 5 fail for the function destiny which are:
destiny(10, 6), my output: 1484710602474311424, expected output: 1484710602474311520
destiny(17, 13), my output: 59832260230817688435680083968, expected output: 59832260230817687634146563200
destiny(10, 3), my output: 1484710602474311424, expected output: 1484710602474311520
destiny(15, 9), my output: 1484710602474311424, expected output: 1484710602474311520
destiny(11, 5), my output: 494764640798827343035498496, expected output: 494764640798827359861461484
Anything that could fix this?
There is some floating point arithmetic in that code, which can slightly throw off the results, and apparently it did. Forget about float, don't use any "floats, but larger" libraries either, integer arithmetic is the way to go.
For example,
f = n/d
c = math.ceil(f)
n*=c
This code looks like it computes n * ⌈n / d⌉, but it only approximately computes that because it uses floating point arithmetic, requiring values to be rounded to the nearest float (for example, int(float(1484710602474311520)) is 1484710602474311424). It should be implemented using integer arithmetic, for example like this:
n *= (n + d - 1) // d
The destiny function should not use floating point division either, and it does not need to. The "is b an integer" test can be stated equivalently as "does d divide n", which can be implemented with integer arithmetic.
Also for that reduce function you can use math.gcd, or implement gcd yourself, the implementation that you have now is very slow.
With those changes, I get the right results for the test cases that you mentioned. I could show the code, but since it is an assignment, you should probably write the code yourself. Asking this question at all is already risky.
Integers don't get inaccurate with large numbers. Floating point numbers do. And you are using floating point numbers.
Rewrite your algorithm to only use integers.
I'm trying to compare two coordinates. The idea is that two coordinates are "the same" if they have equal values up to 5 decimals.
That means:
Decimal(45.00001) =/= Decimal(45.00002)
Decimal(45.000001) == Decimal(45.000002)
because the first 5 decimals are the same
Is there a built in way to do that or do I need to do a string splitting and comparison?
Here are some things you can do with Decimal:
from decimal import Decimal, localcontext
ax = Decimal(45.00001)
ay = Decimal(45.00002)
bx = Decimal(45.000001)
by = Decimal(45.000002)
# 1. using a local context with given precision
with localcontext() as ctx:
ctx.prec = 7
a_tst = ax.normalize() == ay.normalize()
b_tst = bx.normalize() == by.normalize()
>>> a_tst, b_tst
(False, True)
# 2. using the Python built-in round()
# (which also works on native floats and ints
k = 5
a_tst = round(ax, k) == round(ay, k)
b_tst = round(bx, k) == round(by, k)
>>> a_tst, b_tst
(False, True)
Note: the localcontext() version is cumbersome, because prec refers to the full precision (all digits), not just the number of digits after the decimal point. I'd recommend going with round(x, k).
Why did I say that round works on floats and ints? You can use round(x, k) with negative k:
>>> round(54321, -2)
54300
BTW, I'm not sure why you are using Decimal. If you are dealing with lots of coordinates, you may want to use numpy. When using numpy, the canonical way to compare values (single scalars or vectors or tensors of any compatible dimension) is with np.allclose(x, y). You can customize the precision of the comparison, for example, in your case, you could say:
import numpy as np
prec = 0.5e-5
a_tst = np.allclose(float(ax), float(ay), rtol=0, atol=prec)
b_tst = np.allclose(float(bx), float(by), rtol=0, atol=prec)
>>> a_tst, b_tst
(False, True)
In python, suppose the code is:
import.math
a = math.sqrt(2.0)
if a * a == 2.0:
x = 2
else:
x = 1
This is a variant of "Floating Point Numbers are Approximations -- Not Exact".
Mathematically speaking, you are correct that sqrt(2) * sqrt(2) == 2. But sqrt(2) can not be exactly represented as a native datatype (read: floating point number). (Heck, the sqrt(2) is actually guaranteed to be an infinite decimal!). It can get really close, but not exact:
>>> import math
>>> math.sqrt(2)
1.4142135623730951
>>> math.sqrt(2) * math.sqrt(2)
2.0000000000000004
Note the result is, in fact, not exactly 2.
If you want the x = 2 branch to execute, you will need to use an epsilon value of "is the result close enough?":
epsilon = 1e-6 # 0.000001
if abs(2.0 - a*a) < epsilon:
x = 2
else:
x = 1
Numbers with decimals are stored as floating point numbers and they can only be an approximation to the real number in some cases.
So your comparison needs to be not "are these two numbers exactly equal (==)" but "are they sufficiently close as to be considered equal".
Fortunately, in the math library, there's a function to do that conveniently. Using isClose(), you can compare with a defined tolerance. The function isn't too complicated, you could do it yourself.
math.isclose(a*a, 2, abs_tol=0.0001)
>> True