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Why is math.sqrt() incorrect for large numbers?

Why does the math module return the wrong result?
First test
A = 12345678917
print 'A =',A
B = sqrt(A**2)
print 'B =',int(B)
Result
A = 12345678917
B = 12345678917
Here, the result is correct.
Second test
A = 123456758365483459347856
print 'A =',A
B = sqrt(A**2)
print 'B =',int(B)
Result
A = 123456758365483459347856
B = 123456758365483467538432
Here the result is incorrect.
Why is that the case?

Because math.sqrt(..) first casts the number to a floating point and floating points have a limited mantissa: it can only represent part of the number correctly. So float(A**2) is not equal to A**2. Next it calculates the math.sqrt which is also approximately correct.
Most functions working with floating points will never be fully correct to their integer counterparts. Floating point calculations are almost inherently approximative.
If one calculates A**2 one gets:
>>> 12345678917**2
152415787921658292889L
Now if one converts it to a float(..), one gets:
>>> float(12345678917**2)
1.5241578792165828e+20
But if you now ask whether the two are equal:
>>> float(12345678917**2) == 12345678917**2
False
So information has been lost while converting it to a float.
You can read more about how floats work and why these are approximative in the Wikipedia article about IEEE-754, the formal definition on how floating points work.

The documentation for the math module states "It provides access to the mathematical functions defined by the C standard." It also states "Except when explicitly noted otherwise, all return values are floats."
Those together mean that the parameter to the square root function is a float value. In most systems that means a floating point value that fits into 8 bytes, which is called "double" in the C language. Your code converts your integer value into such a value before calculating the square root, then returns such a value.
However, the 8-byte floating point value can store at most 15 to 17 significant decimal digits. That is what you are getting in your results.
If you want better precision in your square roots, use a function that is guaranteed to give full precision for an integer argument. Just do a web search and you will find several. Those usually do a variation of the Newton-Raphson method to iterate and eventually end at the correct answer. Be aware that this is significantly slower that the math module's sqrt function.
Here is a routine that I modified from the internet. I can't cite the source right now. This version also works for non-integer arguments but just returns the integer part of the square root.
def isqrt(x):
"""Return the integer part of the square root of x, even for very
large values."""
if x < 0:
raise ValueError('square root not defined for negative numbers')
n = int(x)
if n == 0:
return 0
a, b = divmod(n.bit_length(), 2)
x = (1 << (a+b)) - 1
while True:
y = (x + n//x) // 2
if y >= x:
return x
x = y

If you want to calculate sqrt of really large numbers and you need exact results, you can use sympy:
import sympy
num = sympy.Integer(123456758365483459347856)
print(int(num) == int(sympy.sqrt(num**2)))

The way floating-point numbers are stored in memory makes calculations with them prone to slight errors that can nevertheless be significant when exact results are needed. As mentioned in one of the comments, the decimal library can help you here:
>>> A = Decimal(12345678917)
>>> A
Decimal('123456758365483459347856')
>>> B = A.sqrt()**2
>>> B
Decimal('123456758365483459347856.0000')
>>> A == B
True
>>> int(B)
123456758365483459347856
I use version 3.6, which has no hardcoded limit on the size of integers. I don't know if, in 2.7, casting B as an int would cause overflow, but decimal is incredibly useful regardless.

Comparing two floats in python [duplicate]

It's well known that comparing floats for equality is a little fiddly due to rounding and precision issues.
For example: Comparing Floating Point Numbers, 2012 Edition
What is the recommended way to deal with this in Python?
Is a standard library function for this somewhere?

Python 3.5 adds the math.isclose and cmath.isclose functions as described in PEP 485.
If you're using an earlier version of Python, the equivalent function is given in the documentation.
def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):
return abs(a-b) <= max(rel_tol * max(abs(a), abs(b)), abs_tol)
rel_tol is a relative tolerance, it is multiplied by the greater of the magnitudes of the two arguments; as the values get larger, so does the allowed difference between them while still considering them equal.
abs_tol is an absolute tolerance that is applied as-is in all cases. If the difference is less than either of those tolerances, the values are considered equal.

Something as simple as the following may be good enough:
return abs(f1 - f2) <= allowed_error

I would agree that Gareth's answer is probably most appropriate as a lightweight function/solution.
But I thought it would be helpful to note that if you are using NumPy or are considering it, there is a packaged function for this.
numpy.isclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False)
A little disclaimer though: installing NumPy can be a non-trivial experience depending on your platform.

Use Python's decimal module, which provides the Decimal class.
From the comments:
It is worth noting that if you're
doing math-heavy work and you don't
absolutely need the precision from
decimal, this can really bog things
down. Floats are way, way faster to
deal with, but imprecise. Decimals are
extremely precise but slow.

The common wisdom that floating-point numbers cannot be compared for equality is inaccurate. Floating-point numbers are no different from integers: If you evaluate "a == b", you will get true if they are identical numbers and false otherwise (with the understanding that two NaNs are of course not identical numbers).
The actual problem is this: If I have done some calculations and am not sure the two numbers I have to compare are exactly correct, then what? This problem is the same for floating-point as it is for integers. If you evaluate the integer expression "7/3*3", it will not compare equal to "7*3/3".
So suppose we asked "How do I compare integers for equality?" in such a situation. There is no single answer; what you should do depends on the specific situation, notably what sort of errors you have and what you want to achieve.
Here are some possible choices.
If you want to get a "true" result if the mathematically exact numbers would be equal, then you might try to use the properties of the calculations you perform to prove that you get the same errors in the two numbers. If that is feasible, and you compare two numbers that result from expressions that would give equal numbers if computed exactly, then you will get "true" from the comparison. Another approach is that you might analyze the properties of the calculations and prove that the error never exceeds a certain amount, perhaps an absolute amount or an amount relative to one of the inputs or one of the outputs. In that case, you can ask whether the two calculated numbers differ by at most that amount, and return "true" if they are within the interval. If you cannot prove an error bound, you might guess and hope for the best. One way of guessing is to evaluate many random samples and see what sort of distribution you get in the results.
Of course, since we only set the requirement that you get "true" if the mathematically exact results are equal, we left open the possibility that you get "true" even if they are unequal. (In fact, we can satisfy the requirement by always returning "true". This makes the calculation simple but is generally undesirable, so I will discuss improving the situation below.)
If you want to get a "false" result if the mathematically exact numbers would be unequal, you need to prove that your evaluation of the numbers yields different numbers if the mathematically exact numbers would be unequal. This may be impossible for practical purposes in many common situations. So let us consider an alternative.
A useful requirement might be that we get a "false" result if the mathematically exact numbers differ by more than a certain amount. For example, perhaps we are going to calculate where a ball thrown in a computer game traveled, and we want to know whether it struck a bat. In this case, we certainly want to get "true" if the ball strikes the bat, and we want to get "false" if the ball is far from the bat, and we can accept an incorrect "true" answer if the ball in a mathematically exact simulation missed the bat but is within a millimeter of hitting the bat. In that case, we need to prove (or guess/estimate) that our calculation of the ball's position and the bat's position have a combined error of at most one millimeter (for all positions of interest). This would allow us to always return "false" if the ball and bat are more than a millimeter apart, to return "true" if they touch, and to return "true" if they are close enough to be acceptable.
So, how you decide what to return when comparing floating-point numbers depends very much on your specific situation.
As to how you go about proving error bounds for calculations, that can be a complicated subject. Any floating-point implementation using the IEEE 754 standard in round-to-nearest mode returns the floating-point number nearest to the exact result for any basic operation (notably multiplication, division, addition, subtraction, square root). (In case of tie, round so the low bit is even.) (Be particularly careful about square root and division; your language implementation might use methods that do not conform to IEEE 754 for those.) Because of this requirement, we know the error in a single result is at most 1/2 of the value of the least significant bit. (If it were more, the rounding would have gone to a different number that is within 1/2 the value.)
Going on from there gets substantially more complicated; the next step is performing an operation where one of the inputs already has some error. For simple expressions, these errors can be followed through the calculations to reach a bound on the final error. In practice, this is only done in a few situations, such as working on a high-quality mathematics library. And, of course, you need precise control over exactly which operations are performed. High-level languages often give the compiler a lot of slack, so you might not know in which order operations are performed.
There is much more that could be (and is) written about this topic, but I have to stop there. In summary, the answer is: There is no library routine for this comparison because there is no single solution that fits most needs that is worth putting into a library routine. (If comparing with a relative or absolute error interval suffices for you, you can do it simply without a library routine.)

math.isclose() has been added to Python 3.5 for that (source code). Here is a port of it to Python 2. It's difference from one-liner of Mark Ransom is that it can handle "inf" and "-inf" properly.
def isclose(a, b, rel_tol=1e-09, abs_tol=0.0):
'''
Python 2 implementation of Python 3.5 math.isclose()
https://github.com/python/cpython/blob/v3.5.10/Modules/mathmodule.c#L1993
'''
# sanity check on the inputs
if rel_tol < 0 or abs_tol < 0:
raise ValueError("tolerances must be non-negative")
# short circuit exact equality -- needed to catch two infinities of
# the same sign. And perhaps speeds things up a bit sometimes.
if a == b:
return True
# This catches the case of two infinities of opposite sign, or
# one infinity and one finite number. Two infinities of opposite
# sign would otherwise have an infinite relative tolerance.
# Two infinities of the same sign are caught by the equality check
# above.
if math.isinf(a) or math.isinf(b):
return False
# now do the regular computation
# this is essentially the "weak" test from the Boost library
diff = math.fabs(b - a)
result = (((diff <= math.fabs(rel_tol * b)) or
(diff <= math.fabs(rel_tol * a))) or
(diff <= abs_tol))
return result

I'm not aware of anything in the Python standard library (or elsewhere) that implements Dawson's AlmostEqual2sComplement function. If that's the sort of behaviour you want, you'll have to implement it yourself. (In which case, rather than using Dawson's clever bitwise hacks you'd probably do better to use more conventional tests of the form if abs(a-b) <= eps1*(abs(a)+abs(b)) + eps2 or similar. To get Dawson-like behaviour you might say something like if abs(a-b) <= eps*max(EPS,abs(a),abs(b)) for some small fixed EPS; this isn't exactly the same as Dawson, but it's similar in spirit.

If you want to use it in testing/TDD context, I'd say this is a standard way:
from nose.tools import assert_almost_equals
assert_almost_equals(x, y, places=7) # The default is 7

In terms of absolute error, you can just check
if abs(a - b) <= error:
print("Almost equal")
Some information of why float act weird in Python:
Python 3 Tutorial 03 - if-else, logical operators and top beginner mistakes
You can also use math.isclose for relative errors.

This is useful for the case where you want to make sure two numbers are the same 'up to precision', and there isn't any need to specify the tolerance:
Find minimum precision of the two numbers
Round both of them to minimum precision and compare
def isclose(a, b):
astr = str(a)
aprec = len(astr.split('.')[1]) if '.' in astr else 0
bstr = str(b)
bprec = len(bstr.split('.')[1]) if '.' in bstr else 0
prec = min(aprec, bprec)
return round(a, prec) == round(b, prec)
As written, it only works for numbers without the 'e' in their string representation (meaning 0.9999999999995e-4 < number <= 0.9999999999995e11)
Example:
>>> isclose(10.0, 10.049)
True
>>> isclose(10.0, 10.05)
False

For some of the cases where you can affect the source number representation, you can represent them as fractions instead of floats, using integer numerator and denominator. That way you can have exact comparisons.
See Fraction from fractions module for details.

I liked Sesquipedal's suggestion, but with modification (a special use case when both values are 0 returns False). In my case, I was on Python 2.7 and just used a simple function:
if f1 ==0 and f2 == 0:
return True
else:
return abs(f1-f2) < tol*max(abs(f1),abs(f2))

If you want to do it in a testing or TDD context using the pytest package, here's how:
import pytest
PRECISION = 1e-3
def assert_almost_equal():
obtained_value = 99.99
expected_value = 100.00
assert obtained_value == pytest.approx(expected_value, PRECISION)

I found the following comparison helpful:
str(f1) == str(f2)

To compare up to a given decimal without atol/rtol:
def almost_equal(a, b, decimal=6):
return '{0:.{1}f}'.format(a, decimal) == '{0:.{1}f}'.format(b, decimal)
print(almost_equal(0.0, 0.0001, decimal=5)) # False
print(almost_equal(0.0, 0.0001, decimal=4)) # True

This maybe is a bit ugly hack, but it works pretty well when you don't need more than the default float precision (about 11 decimals).
The round_to function uses the format method from the built-in str class to round up the float to a string that represents the float with the number of decimals needed, and then applies the eval built-in function to the rounded float string to get back to the float numeric type.
The is_close function just applies a simple conditional to the rounded up float.
def round_to(float_num, prec):
return eval("'{:." + str(int(prec)) + "f}'.format(" + str(float_num) + ")")
def is_close(float_a, float_b, prec):
if round_to(float_a, prec) == round_to(float_b, prec):
return True
return False
>>>a = 10.0
10.0
>>>b = 10.0001
10.0001
>>>print is_close(a, b, prec=3)
True
>>>print is_close(a, b, prec=4)
False
Update:
As suggested by #stepehjfox, a cleaner way to build a rount_to function avoiding "eval" is using nested formatting:
def round_to(float_num, prec):
return '{:.{precision}f}'.format(float_num, precision=prec)
Following the same idea, the code can be even simpler using the great new f-strings (Python 3.6+):
def round_to(float_num, prec):
return f'{float_num:.{prec}f}'
So, we could even wrap it up all in one simple and clean 'is_close' function:
def is_close(a, b, prec):
return f'{a:.{prec}f}' == f'{b:.{prec}f}'

If you want to compare floats, the options above are great, but in my case, I ended up using Enum's, since I only had few valid floats my use case was accepting.
from enum import Enum
class HolidayMultipliers(Enum):
EMPLOYED_LESS_THAN_YEAR = 2.0
EMPLOYED_MORE_THAN_YEAR = 2.5
Then running:
testable_value = 2.0
HolidayMultipliers(testable_value)
If the float is valid, it's fine, but otherwise it will just throw an ValueError.

Use == is a simple good way, if you don't care about tolerance precisely.
# Python 3.8.5
>>> 1.0000000000001 == 1
False
>>> 1.00000000000001 == 1
True
But watch out for 0:
>>> 0 == 0.00000000000000000000000000000000000000000001
False
The 0 is always the zero.
Use math.isclose if you want to control the tolerance.
The default a == b is equivalent to math.isclose(a, b, rel_tol=1e-16, abs_tol=0).
If you still want to use == with a self-defined tolerance:
>>> class MyFloat(float):
def __eq__(self, another):
return math.isclose(self, another, rel_tol=0, abs_tol=0.001)
>>> a == MyFloat(0)
>>> a
0.0
>>> a == 0.001
True
So far, I didn't find anywhere to config it globally for float. Besides, mock is also not working for float.__eq__.

Converting An "Infinite" Float To An Int [duplicate]

This question already has answers here:
Integer square root in python
(14 answers)
Closed 8 years ago.
I'm trying to check if a number is a perfect square. However, i am dealing with extraordinarily large numbers so python thinks its infinity for some reason. it gets up to 1.1 X 10^154 before the code returns "Inf". Is there anyway to get around this? Here is the code, the lst variable just holds a bunch of really really really really really big numbers
import math
from decimal import Decimal
def main():
for i in lst:
root = math.sqrt(Decimal(i))
print(root)
if int(root + 0.5) ** 2 == i:
print(str(i) + " True")

Replace math.sqrt(Decimal(i)) with Decimal(i).sqrt() to prevent your Decimals decaying into floats

I think that you need to take a look at the BigFloat module, e.g.:
import bigfloat as bf
b = bf.BigFloat('1e1000', bf.precision(21))
print bf.sqrt(b)
Prints BigFloat.exact('9.9999993810013282e+499', precision=53)

#casevh has the right answer -- use a library that can do math on arbitrarily large integers. Since you're looking for squares, you presumably are working with integers, and one could argue that using floating point types (including decimal.Decimal) is, in some sense, inelegant.
You definitely shouldn't use Python's float type; it has limited precision (about 16 decimal places). If you do use decimal.Decimal, be careful to specify the precision (which will depend on how big your numbers are).
Since Python has a big integer type, one can write a reasonably simple algorithm to check for squareness; see my implementation of such an algorithm, along with illustrations of problems with float, and how you could use decimal.Decimal, below.
import math
import decimal
def makendigit(n):
"""Return an arbitraryish n-digit number"""
return sum((j%9+1)*10**i for i,j in enumerate(range(n)))
x=makendigit(30)
# it looks like float will work...
print 'math.sqrt(x*x) - x: %.17g' % (math.sqrt(x*x) - x)
# ...but actually they won't
print 'math.sqrt(x*x+1) - x: %.17g' % (math.sqrt(x*x+1) - x)
# by default Decimal won't be sufficient...
print 'decimal.Decimal(x*x).sqrt() - x:',decimal.Decimal(x*x).sqrt() - x
# ...you need to specify the precision
print 'decimal.Decimal(x*x).sqrt(decimal.Context(prec=30)) - x:',decimal.Decimal(x*x).sqrt(decimal.Context(prec=100)) - x
def issquare_decimal(y,prec=1000):
x=decimal.Decimal(y).sqrt(decimal.Context(prec=prec))
return x==x.to_integral_value()
print 'issquare_decimal(x*x):',issquare_decimal(x*x)
print 'issquare_decimal(x*x+1):',issquare_decimal(x*x+1)
# you can check for "squareness" without going to floating point.
# one option is a bisection search; this Newton's method approach
# should be faster.
# For "industrial use" you should use gmpy2 or some similar "big
# integer" library.
def isqrt(y):
"""Find largest integer <= sqrt(y)"""
if not isinstance(y,(int,long)):
raise ValueError('arg must be an integer')
if y<0:
raise ValueError('arg must be positive')
if y in (0,1):
return y
x0=y//2
while True:
# newton's rule
x1= (x0**2+y)//2//x0
# we don't always get converge to x0=x1, e.g., for y=3
if abs(x1-x0)<=1:
# nearly converged; find biggest
# integer satisfying our condition
x=max(x0,x1)
if x**2>y:
while x**2>y:
x-=1
else:
while (x+1)**2<=y:
x+=1
return x
x0=x1
def issquare(y):
"""Return true if non-negative integer y is a perfect square"""
return y==isqrt(y)**2
print 'isqrt(x*x)-x:',isqrt(x*x)-x
print 'issquare(x*x):',issquare(x*x)
print 'issquare(x*x+1):',issquare(x*x+1)

math.sqrt() converts the argument to a Python float which has a maximum value around 10^308.
You should probably look at using the gmpy2 library. gmpy2 provide very fast multiple precision arithmetic.
If you want to check for arbitrary powers, the function gmpy2.is_power() will return True if a number is a perfect power. It may be a cube or fifth power so you will need to check for power you are interested in.
>>> gmpy2.is_power(456789**372)
True
You can use gmpy2.isqrt_rem() to check if it is an exact square.
>>> gmpy2.isqrt_rem(9)
(mpz(3), mpz(0))
>>> gmpy2.isqrt_rem(10)
(mpz(3), mpz(1))
You can use gmpy2.iroot_rem() to check for arbitrary powers.
>>> gmpy2.iroot_rem(13**7 + 1, 7)
(mpz(13), mpz(1))

Prevent Rounding to Zero in Python

I have a program meant to approximate pi using the Chudnovsky Algorithm, but a term in my equation that is very small keeps being rounded to zero.
Here is the algorithm:
import math
from decimal import *
getcontext().prec = 100
pi = Decimal(0.0)
C = Decimal(12/(math.sqrt(640320**3)))
k = 0
x = Decimal(0.0)
result = Decimal(0.0)
sign = 1
while k<10:
r = Decimal(math.factorial(6*k)/((math.factorial(k)**3)*math.factorial(3*k)))
s = Decimal((13591409+545140134*k)/((640320**3)**k))
x += Decimal(sign*r*s)
sign = sign*(-1)
k += 1
result = Decimal(C*x)
pi = Decimal(1/result)
print Decimal(pi)
The equations may be clearer without the "decimal" terms.
import math
pi = 0.0
C = 12/(math.sqrt(640320**3))
k = 0
x = 0.0
result = 0.0
sign = 1
while k<10:
r = math.factorial(6*k)/((math.factorial(k)**3)*math.factorial(3*k))
s = (13591409+545140134*k)/((640320**3)**k)
x += sign*r*s
sign = sign*(-1)
k += 1
result = C*x
pi = 1/result
print pi
The issue is with the "s" variable. For k>0, it always comes to zero. e.g. at k=1, s should equal about 2.1e-9, but instead it is just zero. Because of this all of my terms after the first =0. How do I get python to calculate the exact value of s instead of rounding it down to 0?

Try:
s = Decimal((13591409+545140134*k)) / Decimal(((640320**3)**k))
The arithmetic you're doing is native python - by allowing the Decimal object to perform your division, you should eliminate your error.
You can do the same, then, when computing r.

A couple of comments.
If you are using Python 2.x, the / returns an integer result. If you want a Decimal result, you convert at least one side to Decimal first.
math.sqrt() only return ~16 digits of precision. Since your value for C will only be accurate to ~16 digits, your final result will only be accurate to 16 digits.

If you're doing maths in Python 2.x, you should probably be putting this line into every module:
from __future__ import division
This changes the meaning of the division operator so that it will return a floating point number if needed to give a (closer to) precise answer. The historical behaviour is for x / y to return an int if both x and y are ints, which usually forces the answer to be rounded down.
Returning a float if necessary is generally regarded as a better way to handle division in a language like Python where duck typing is encouraged, since you can just worry about the value of your numbers rather than getting different behaviour for different types.
In Python 3 this is in fact the default, but since old programs relied on the historical behaviour of the division operator it was felt the change was too backwards-incompatible to be made in Python 2. This is why you have to explicitly turn it on with the __future__ import. I would recommend always adding that import in any module that might be doing any mathematics (or just any module at all, if you can be bothered). You'll almost never be upset that it's there, but not having it there has been the cause of a number of obscure bugs I've had to chase.

I feel that the problem with 's' is that all terms are integers, thus you are doing integer maths. A very simple workaround, would be to use 3.0 in the denominator. It only takes one float in the calculation to get a float returned.

In what contexts do programming languages make real use of an Infinity value?

So in Ruby there is a trick to specify infinity:
1.0/0
=> Infinity
I believe in Python you can do something like this
float('inf')
These are just examples though, I'm sure most languages have infinity in some capacity. When would you actually use this construct in the real world? Why would using it in a range be better than just using a boolean expression? For instance
(0..1.0/0).include?(number) == (number >= 0) # True for all values of number
=> true
To summarize, what I'm looking for is a real world reason to use Infinity.
EDIT: I'm looking for real world code. It's all well and good to say this is when you "could" use it, when have people actually used it.

Dijkstra's Algorithm typically assigns infinity as the initial edge weights in a graph. This doesn't have to be "infinity", just some arbitrarily constant but in java I typically use Double.Infinity. I assume ruby could be used similarly.

Off the top of the head, it can be useful as an initial value when searching for a minimum value.
For example:
min = float('inf')
for x in somelist:
if x<min:
min=x
Which I prefer to setting min initially to the first value of somelist
Of course, in Python, you should just use the min() built-in function in most cases.

There seems to be an implied "Why does this functionality even exist?" in your question. And the reason is that Ruby and Python are just giving access to the full range of values that one can specify in floating point form as specified by IEEE.
This page seems to describe it well:
http://steve.hollasch.net/cgindex/coding/ieeefloat.html
As a result, you can also have NaN (Not-a-number) values and -0.0, while you may not immediately have real-world uses for those either.

In some physics calculations you can normalize irregularities (ie, infinite numbers) of the same order with each other, canceling them both and allowing a approximate result to come through.
When you deal with limits, calculations like (infinity / infinity) -> approaching a finite a number could be achieved. It's useful for the language to have the ability to overwrite the regular divide-by-zero error.

Use Infinity and -Infinity when implementing a mathematical algorithm calls for it.
In Ruby, Infinity and -Infinity have nice comparative properties so that -Infinity < x < Infinity for any real number x. For example, Math.log(0) returns -Infinity, extending to 0 the property that x > y implies that Math.log(x) > Math.log(y). Also, Infinity * x is Infinity if x > 0, -Infinity if x < 0, and 'NaN' (not a number; that is, undefined) if x is 0.
For example, I use the following bit of code in part of the calculation of some log likelihood ratios. I explicitly reference -Infinity to define a value even if k is 0 or n AND x is 0 or 1.
Infinity = 1.0/0.0
def Similarity.log_l(k, n, x)
unless x == 0 or x == 1
k * Math.log(x.to_f) + (n-k) * Math.log(1.0-x)
end
-Infinity
end
end

Alpha-beta pruning

I use it to specify the mass and inertia of a static object in physics simulations. Static objects are essentially unaffected by gravity and other simulation forces.

In Ruby infinity can be used to implement lazy lists. Say i want N numbers starting at 200 which get successively larger by 100 units each time:
Inf = 1.0 / 0.0
(200..Inf).step(100).take(N)
More info here: http://banisterfiend.wordpress.com/2009/10/02/wtf-infinite-ranges-in-ruby/

I've used it for cases where you want to define ranges of preferences / allowed.
For example in 37signals apps you have like a limit to project number
Infinity = 1 / 0.0
FREE = 0..1
BASIC = 0..5
PREMIUM = 0..Infinity
then you can do checks like
if PREMIUM.include? current_user.projects.count
# do something
end

I used it for representing camera focus distance and to my surprise in Python:
>>> float("inf") is float("inf")
False
>>> float("inf") == float("inf")
True
I wonder why is that.

I've used it in the minimax algorithm. When I'm generating new moves, if the min player wins on that node then the value of the node is -∞. Conversely, if the max player wins then the value of that node is +∞.
Also, if you're generating nodes/game states and then trying out several heuristics you can set all the node values to -∞/+∞ which ever makes sense and then when you're running a heuristic its easy to set the node value:
node_val = -∞
node_val = max(heuristic1(node), node_val)
node_val = max(heuristic2(node), node_val)
node_val = max(heuristic2(node), node_val)

I've used it in a DSL similar to Rails' has_one and has_many:
has 0..1 :author
has 0..INFINITY :tags
This makes it easy to express concepts like Kleene star and plus in your DSL.

I use it when I have a Range object where one or both ends need to be open

I've used symbolic values for positive and negative infinity in dealing with range comparisons to eliminate corner cases that would otherwise require special handling:
Given two ranges A=[a,b) and C=[c,d) do they intersect, is one greater than the other, or does one contain the other?
A > C iff a >= d
A < C iff b <= c
etc...
If you have values for positive and negative infinity that respectively compare greater than and less than all other values, you don't need to do any special handling for open-ended ranges. Since floats and doubles already implement these values, you might as well use them instead of trying to find the largest/smallest values on your platform. With integers, it's more difficult to use "infinity" since it's not supported by hardware.

I ran across this because I'm looking for an "infinite" value to set for a maximum, if a given value doesn't exist, in an attempt to create a binary tree. (Because I'm selecting based on a range of values, and not just a single value, I quickly realized that even a hash won't work in my situation.)
Since I expect all numbers involved to be positive, the minimum is easy: 0. Since I don't know what to expect for a maximum, though, I would like the upper bound to be Infinity of some sort. This way, I won't have to figure out what "maximum" I should compare things to.
Since this is a project I'm working on at work, it's technically a "Real world problem". It may be kindof rare, but like a lot of abstractions, it's convenient when you need it!
Also, to those who say that this (and other examples) are contrived, I would point out that all abstractions are somewhat contrived; that doesn't mean they are useful when you contrive them.

When working in a problem domain where trig is used (especially tangent) infinity is an answer that can come up. Trig ends up being used heavily in graphics applications, games, and geospatial applications, plus the obvious math applications.

I'm sure there are other ways to do this, but you could use Infinity to check for reasonable inputs in a String-to-Float conversion. In Java, at least, the Float.isNaN() static method will return false for numbers with infinite magnitude, indicating they are valid numbers, even though your program might want to classify them as invalid. Checking against the Float.POSITIVE_INFINITY and Float.NEGATIVE_INFINITY constants solves that problem. For example:
// Some sample values to test our code with
String stringValues[] = {
"-999999999999999999999999999999999999999999999",
"12345",
"999999999999999999999999999999999999999999999"
};
// Loop through each string representation
for (String stringValue : stringValues) {
// Convert the string representation to a Float representation
Float floatValue = Float.parseFloat(stringValue);
System.out.println("String representation: " + stringValue);
System.out.println("Result of isNaN: " + floatValue.isNaN());
// Check the result for positive infinity, negative infinity, and
// "normal" float numbers (within the defined range for Float values).
if (floatValue == Float.POSITIVE_INFINITY) {
System.out.println("That number is too big.");
} else if (floatValue == Float.NEGATIVE_INFINITY) {
System.out.println("That number is too small.");
} else {
System.out.println("That number is jussssst right.");
}
}
Sample Output:
String representation: -999999999999999999999999999999999999999999999
Result of isNaN: false
That number is too small.
String representation: 12345
Result of isNaN: false
That number is jussssst right.
String representation: 999999999999999999999999999999999999999999999
Result of isNaN: false
That number is too big.

It is used quite extensively in graphics. For example, any pixel in a 3D image that is not part of an actual object is marked as infinitely far away. So that it can later be replaced with a background image.

I'm using a network library where you can specify the maximum number of reconnection attempts. Since I want mine to reconnect forever:
my_connection = ConnectionLibrary(max_connection_attempts = float('inf'))
In my opinion, it's more clear than the typical "set to -1 to retry forever" style, since it's literally saying "retry until the number of connection attempts is greater than infinity".

Some programmers use Infinity or NaNs to show a variable has never been initialized or assigned in the program.

If you want the largest number from an input but they might use very large negatives. If I enter -13543124321.431 it still works out as the largest number since it's bigger than -inf.
enter code here
initial_value = float('-inf')
while True:
try:
x = input('gimmee a number or type the word, stop ')
except KeyboardInterrupt:
print("we done - by yo command")
break
if x == "stop":
print("we done")
break
try:
x = float(x)
except ValueError:
print('not a number')
continue
if x > initial_value: initial_value = x
print("The largest number is: " + str(initial_value))

You can to use:
import decimal
decimal.Decimal("Infinity")
or:
from decimal import *
Decimal("Infinity")

For sorting
I've seen it used as a sort value, to say "always sort these items to the bottom".

To specify a non-existent maximum
If you're dealing with numbers, nil represents an unknown quantity, and should be preferred to 0 for that case. Similarly, Infinity represents an unbounded quantity, and should be preferred to (arbitrarily_large_number) in that case.
I think it can make the code cleaner. For example, I'm using Float::INFINITY in a Ruby gem for exactly that: the user can specify a maximum string length for a message, or they can specify :all. In that case, I represent the maximum length as Float::INFINITY, so that later when I check "is this message longer than the maximum length?" the answer will always be false, without needing a special case.