*rounding errors*. Consider the following C snippet

double pi = 3.1415926535897932384626433832795; double val = sqrt(pi) * sqrt(pi); printf("pi = %f\n", pi); printf("val = %f\n", val);which yields

pi = 3.141593 val = 3.141593Which is what we expected!

`val`

is the same as pi (the square is the inverse of the square root as we all know). However, the following C snippet shows the problem we run into:
if (val == pi) { printf ("pi == val\n"); } else { printf ("pi != val\n"); }which yields

pi != valSo two apparently equal values are not equal any more! Where is the problem? It is in the precision of

`printf()`

, of course. We modify the first snippet so the `printf()`

calls print the floating point values to more significant places
printf("pi = %.16f\n", pi); printf("val = %.16f\n", val);and now we get

pi = 3.1415926535897931 val = 3.1415926535897927So how do we resolve this? Instead of comparing floating point values directly, we calculate their difference and make sure it is below a small value, for example

`1e-9`

:
if (abs(val - pi) < 1e-9) { printf ("pi == val\n"); } else { printf ("pi != val\n"); }which now gives us

pi == valI guess most programmers are aware about the problems with floating point numbers and I have also known about the problem for a long time. What made me stumble, however, was that I trusted the printed value - which was inaccurate. As far as I know, there is no generic solution to the problem with floating point inaccuracies. You will get away with comparing whether the difference between two floating point values is smaller than a given value for comparing. In Ruby, you could hack the built in

`Float`

class with the following:
class Float def ==(value) self - value < 0.01 end endThis allows us to compare floating point values “fuzzily”:

>> Math.sqrt(2) ** 2 == 2.0 =>However, I dofalse>> require 'floathack' => true >> Math.sqrt(2) ** 2 == 2.0 =>true

**not**think that this is a good idea! Overwriting the

`==`

operator of the `Float`

as above breaks its *transitivity*, i.e. we get the following (in-)equalities 1.99 == 2.00 == 2.01 but 1.99 != 2.01 Another idea would be only to compare the first

*n*numbers after the point, e.g. consider

*3.1415926535897931*to be

*3.141*for

*n = 2*. However, this does not help us if a floating point error gives us two values just next to a “boundary”, i.e. a result of

*1.999*would not be considered correct if

*2.001*was expected and we only cared about the result being “sufficiently close” to the expected value. So be careful when using floating point values are checking two of them for equality. Trouble will strike and most probably find some way to trick you into wasting time. An extensive, academic - and arguably dry - discussion of the implications of floating point arithmetics can be found in What Every Computer Scientist Should Know About Floating-Point Arithmetic, by David Goldberg.

## 5 comments:

The python decimal module lets you mess around with arbitrary precision numbers. There is even a (super slow) implementation of the standard math functions for decimal numbers called dmath.

cshesse, thank you for your comment.

Yes, arbitrary-precision arithmetics can solve floating point related problems in some cases (e.g. correct financial calculations).

However, you do not profit from the native - and fast - floating point operations of modern processors.

Additionally, some problems are inherent to the representation of something infinte - like sqrt(2) - in a computer system which can only store finitely long objects:

>>> import decimal

>>> Decimal(2).sqrt()

Decimal("1.414213562373095048801688724")

>>> a = Decimal(2).sqrt()

>>> a * a

Decimal("1.999999999999999999999999999")

I guess you can find an example where arbitrary precision numbers lead to inaccurate results for any number of significant places.

This is usually not such a big deal in practice (i.e. scientific computing).

We use floating point numbers in numerical algorithms, and usually the algorithmic error (i.e. the error obtained when doing computation over the real numbers) is much larger than floating point error.

I.e., use RK4/5 to solve a differential equation, with dt=0.01. RK4/5 is O(dt^6)=1e-12. Floating point error tends to be of order 1e-17; i.e. 5 orders of magnitude smaller.

In hard problems (modelling global climate, parts of plasma physics or the stock market), the situation is even worse since we don't even know the right equation to solve.

So except for trivial problems, I'd say not to worry about floating point errors.

"This is usually not such a big deal in practice (i.e. scientific computing)."

In computational geometry and solid modeling, floating point problems can be a very big deal.

Excellent article. Very interesting to read. I really love to read such a nice article. Thanks! keep rocking. wärmerückgewinnung industrie

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