This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 6 years ago.
I'm solving some cryptographic problem,
I need the cube root of 4.157786362549383e+37, which gives me 3464341716380.1113
using
x = x ** (1. / 3)
I thought it was weird at first, so I did try:
x=1000
print(x)
x= pow(x,1/3)
print(x)
but got 9.99999998
I have even tried somewhere else. But i got the same result. Is there something wrong? How can I calculate the true cube root?
Due to floating-point arithmetic, this is hard to represent.
Using decimal somewhat resolves it but is still problematic in certain numbers, and allows rounding only to integrals. Try using a decimal like so:
>>> (1000 ** (Decimal(1)/3)).to_integral_exact()
Decimal('10')
This is normal when dealing with floating-point numbers on a computer. No decimal fraction can be represented precisely in binary, which deals in negative powers of 2, so you need to get used to getting very close approximations.
In this specific case, if you know your result is supposed to be an integer, simply use round().
Both answers are correct within the accuracy of the hardware's numerical representation. 1/3 is a repeating "decimal" in binary: 0.010101010101...
It can't be represented precisely.
If you want a "true" cube root, you need to implement an algorithm that handles the round-off problems and corner cases you find useful. Given the representation problem, you could certainly cover integer cubes. However, noting even simple cases, such as cube_root(1.728) => 1.2, would be problematic: neither decimal number converts precisely to binary.
As discussed in other answers, this is due to the limited precision of floating point numbers. It's just not possible to exactly represent values, unless your math is done symbolically. If you are fine with limited precision but just need more precision than the built in datatypes give you, I suggest an arbitrary precision arithmetic library, such as this one.
Related
Recently, I've been trying to implement the 15 tests for randomness described in NIST SP800-22. As a check of my function implementation, I have been running the examples that the NIST document provides for each of it's tests. Some of these tests require bit strings that are very long (up to a million bits). For example, on one of the examples, the input is "the first 100,000 bits of e." That brings up the question: how do I generate a bit representation of a float value that exceeds the precision available for floating point numbers in Python?
I have found articles converting integers to binary strings (the bin() function), and converting floating point fractions to binary (repeated division by 2 (slow!) and limited by floating point precision). I've considered constructing it iteratively in some way using $e=\sum_{n=0}^{\infty}\frac{2n+2}{(2n+1)!}$, calculating the next portion value, converting it to a binary representation, and somehow adding it to the cumulative representation (still thinking through how to do this). However, I've hit the same wall going down this path: the precision of the floating point values as I go farther out on this sum.
Does anyone have some suggestions on creating arbitrarily long bit strings from arbitrarily precise floating point values?
PS - Also, is there any way to get my Markdown math equation above to render properly here? :-)
I maintain the gmpy2 library and its supports arbitrary-precision binary arithmetic. Here is an example of generating the first 100 bits of e.
>>> import gmpy2
>>> gmpy2.get_context().precision=100
>>> gmpy2.exp(1).digits(2)[0]
'101011011111100001010100010110001010001010111011010010101001101010101
1111101110001010110001000000010'
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 6 years ago.
I'm trying to subtract to floating point numbers in python.
I've values
a = 1460356156116843.000000, b = 2301.93138123
When I try to print a-b, it's resulting the value 1460356156114541.000000 contrary to the actual value 1460356156114541.06861877.
What are the limitations in python while doing floating point arithmetic. Is there any way in python through which I can get the actual result of this subtraction?
Python has the same limitations for floating point arithmetic as all the other languages. You can use Decimal to get the accurate result:
from decimal import Decimal
a = Decimal('1460356156116843.000000')
b = Decimal('2301.93138123')
print a - b # 1460356156114541.06861877
Python uses IEEE 754 doubles for its floats. So you should treat anything after 15 significant figures or so as science fiction. And that's just for a freshly-initialised number. When you start doing operations with floats you can lose more precision, especially doing addition or subtraction between numbers that differ significantly in absolute magnitude.
OTOH, doing subtraction between numbers very close to each other in magnitude can lead to catastrophic cancellation.
If you are careful, you can reduce the impact of these problems, but you do need a good understanding of how floating-point arithmetic works, and well-behaved data.
Alternatively, you can work with a library that provides higher precision, eg Python's Decimal module. You still need to take care to avoid catastrophic cancellation and the other problems that lead to loss of significance, but at least you've got more significant digits to play with.
The Decimal module just provides basic arithmetic operations. If you need advanced mathematical functions like trig and exponential functions, take a look at the excellent 3rd-party arbitrary precision mathematics module mpmath. It can handle complex numbers, solve equations, and provides some calculus operations.
Using decimal is convenient. But for the sake of demonstration regarding the importance of keeping significant digits, let me throw in this example.
import sys
print(sys.maxsize)
9223372036854775807 # for 64 bit machine, the max integer number. But it can grow as needed.
So for the above case you can do the computation in two steps.
1460356156116842 - 2301 = 1460356156114541 # all integer digits preserved
1 - .93138123 = 0.06861877 # all significant float digits preserved.
So the answer would be adding the two. But if you do that you will lose all float digits. The 64bit is not big enough to keep all digits.
This question already has answers here:
Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 6 years ago.
From mathematics we know that the cosine of a 90 degree angle is 0 but Python says it's a bit more than that.
import math
math.cos(math.radians(90))
6.123233995736766e-17
What's the matter between Python and the number "0"?
Repeat after me:
Computers cannot process real numbers.
Python uses double precision IEEE floats, which round to 53 binary digits of precision and have limits on range. Since π/2 is an irrational number, the computer rounds it to the nearest representable number (or to a close representable number — some operations have exact rounding, some have error greater than 1/2 ULP).
Therefore, you never asked the computer to compute cos(π/2), you really asked it to compute cos(π/2+ε), where ε is the roundoff error for computing π/2. The result is then rounded again.
Why does Excel (or another program) show the correct result?
Possibility 1: The program does symbolic computations, not numeric ones. This applies to programs like Mathematica and Maxima, not Excel.
Possibility 2: The program is hiding the data (most likely). Excel will only show you the digits you ask for, e.g.,
>>> '%.10f' % math.cos(math.radians(90))
'0.0000000000'
Python has a finely tuned function for printing out floats so that they survive a round trip to text and back. This means that Python prints more digits by default than, for example, printf.
Possibility 3: The program you are using had two round-off errors that canceled.
As Dietrich points out, the included Math package uses numerical approximations to calculate trig functions - pi has some level of precision represented through a float. But there are a bunch of good python packages for doing symbolic calculations too - Sympy is an easy way do more precise calculations, if you'd like.
consider:
import math
math.cos( 3*math.pi/2 )
# Outputs -1.8369701987210297e-16
as apposed to
import sympy
sympy.cos( 3*sympy.pi/2 )
# Outputs 0
There aren't a lot of cases where this makes a difference, and sympy is considerably slower. I tested how many cosine calculations my computer could do in five seconds with math, and with sympy, and the it did 38 times more calculations with math. It depends what you're looking for.
Why the result of these two expressions should be different ?
The same thing happens in gcc and python. what is happening in here ? Is there any way to prevent it ?
Floating point numbers have limited precision. If you add a small number (3) to a large number (1e20), the result often is the same as the large number. That is the case here, hence
(3 + 1e20) - 1e20 = 1e20 - 1e20 = 0
The precision of double is roughly 15 decimal digits, floats have about 7 decimal digits of precision.
Although it's related to timestamps, the article “Don't store that in a float” gives a rough overview of the pitfalls you can get when using floating point arithmetics, most importantly:
This real example demonstrates a few things:
Any time you add or subtract floats of widely varying magnitudes you need to watch for loss of precision
Sometimes using ‘double’ instead of ‘float’ is the correct solution, but often a more stable algorithm is more important
In your second case you're adding 10²⁰ to 3, which is a widely varying magnitude. Due to the limited precision of doubles (14 digits approx, 7 for four byte floats (single precision)), the 3 will just get lost in the result. If you however first subtract 10²⁰ from itself, you get a zero, which added to 3 does not change the result at all.
These slight difference in operation ordering can become important in certain calculations and is a thing one should always bear in mind when dealing with floating point numbers on IEEE platforms. A simulation which ran fine for hours suddenly breaking without any reason or only when something specific happens can easily be caused by floating point arithmetics.
This question already has answers here:
Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 8 years ago.
Straight to the chase I want to create a function which calculates the length of a step.
This is what I have:
def humanstep(angle):
global human
human[1]+=math.sin(math.radians(angle))*0.06
human[2]+=math.cos(math.radians(angle))*0.06
So if the angle is 90 then the x value (human[1]) should equal 0.06 and the y value equal to 0.
Instead the conversion between radians and degrees is not quite perfect and these values are returned.
[5.99999999999999, 3.6739403974420643e-16]
Is there anyway to fix this?
This is representation error due to how floating point arithmetic works. See the following page from the Python documentation: Floating Point Arithmetic: Issues and Limitations.
FTA:
Note that this is in the very nature of binary floating-point: this is not a bug in Python, and it is not a bug in your code either. You’ll see the same kind of thing in all languages that support your hardware’s floating-point arithmetic (although some languages may not display the difference by default, or in all output modes).
For further reading, see the following pages:
The Perils of Floating Point
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Exactly how accurate do you want? The above is accurate to 15dp.
If you want accurate results, you are doing it correctly.
If you want mathematically exact results like [6, 0], use a symbolic math library such as sympy
Notice that these are very different goals.
You should read up on floating-point numbers, as these calculations are naturally imperfect and some numbers cannot be represented accurately using Python's floating point numbers. (Obviously, a fixed number of bits cannot represent the infinitely many real numbers.
The short answer is no. You can round if you want.
As the others have said, it's floating point error. You can use the Decimal module, which can give you arbitrary precision math
If you want to avoid the representation issues inherent in floating-point numbers, you can use a Decimal, but you will need to implement your own trigonometric functions. This will get you arbitrary precision but it will be rather slow.
As everyone has said, this is to do with binary representation issues - 0.06 is not a "round" figure in binary (just like 1/3 is not in decimal). It has nothing to do with the trig functions. If you drop them out and just do:
human[1]+=0.06
human[2]+=0.0
and look at the results you will see the same.
However, in Python 2.7 up the representation of floating point numbers has been improved. It will now display the shortest decimal number that gives that binary number - in this case I think it would display the answer you expect (only running 2.5 here so I can't quickly test). See this article for more information.