I have some math (in C++) which seems to be generating some very small, near zero, numbers (I suspect the trig function calls are my real problem), but I'd like to detect these cases so that I can study them in more detail.
I'm currently trying out the following, is it correct?
if ( std::abs(x) < DBL_MIN ) {
log_debug("detected small num, %Le, %Le", x, y);
}
Second, the nature of the mathematics is trigonometric in nature (aka using a lot of radian/degree conversions and sin/cos/tan calls, etc), what sort of transformations can I do to avoid mathematical errors?
Obviously for multiplications I can use a log transform - what else?
Contrary to widespread belief, DBL_MIN is not the smallest positive double value but the smallest positive normalized double value. Typically - for 64-bit ieee754 doubles - it's 2-1022, while the smallest positive double value is 2-1074. Therefore
I'm currently trying out the following, is it correct?
if ( std::abs(x) < DBL_MIN ) {
log_debug("detected small num, %Le, %Le", x, y);
}
may have an affirmative answer. The condition checks whether x is a denormalized (also called subnormal) number or ±0.0. Without knowing more about your specific situation, I cannot tell if that test is appropriate. Denormalized numbers can be legitimate results of calculations or the consequence of rounding where the correct result would be 0. It is also possible that rounding produces numbers of far greater magnitude than DBL_MIN when the mathematically correct result would be 0, so a much larger threshold could be sensible.
If x is a double, then one problem with this approach is that you can't distinguish between x being legitimately zero, and x being a positive value smaller than DBL_MIN. So this will work if you know x can never be legitimately zero, and you want to see when underflow occurs.
You could also try catching the SIGFPE signal, which will fire on a POSIX-compliant system any time there's a math error including floating-point underflow. See: http://en.wikipedia.org/wiki/SIGFPE
EDIT: To be clear, DBL_MIN is NOT the largest negative value that a double can hold, it is the smallest positive normalized value that a double can hold. So your approach is fine as long as the value can't be zero.
Another useful constant is DBL_EPSILON which is the smallest double value that can be added to 1.0 without getting 1.0 back. Note that this is a much larger value than DBL_MIN. But it may be useful to you since you're doing trigonometric functions that may tend toward 1 instead of tending toward 0.
Since you are using C++, the most idiomatic is to use std::numeric_limits from header <limits>.
For instance:
template <typename T>
bool is_close_to_zero(T x)
{
return std::abs(x) < std::numeric_limits<T>::epsilon();
}
The actual tolerance to be used heavily depends on your problem. Please complete your question with a concrete use case so that I can enhance my answer.
There is also std::numeric_limits<T>::min() and std::numeric_limits<T>::denorm_min() that may be useful. The first one is the smallest positive non-denormalized value of type T (equal to FLT/DBL/LDBL_MIN from <cfloat>), the second one is the smallest positive value of type T (no <cfloat> equivalent).
[You may find this document useful to read if you aren't at ease with floating point numbers representation.]
The first if check will actually only be true when your value is zero.
For your second question, you imply lots of conversions. Instead, pick one unit (deg or rad) and do all your computational operations in that unit. Then at the very end do a single conversion to the other value if you need to.
Related
I was solving an equation using double precision and I got -7.07649e-17 as a solution instead of 0.
I agree it's close enough that I can say it's equal but I've read that the machine epsilon for the C++ double type is 2^-52 which is larger than the value I get.
So why do I have an inferior value than the machine epsilon?
Why isn't the value rounded to zero?
It's not a big deal but when I do a logical test it appears that my value is not zero...
There are two different constants in this story. One is epsilon, which is a minimal value that when added to 1.0 produces a value different from 1.0. If you add a smaller value to 1.0 you will again get a 1.0, because there are physical limits to the representation of a number in a computer. But there are values that are less than epsilon and greater than zero. Smallest such number for a double you get with std::numeric_limits<double>::min.
For reference, you get epsilon with std::numeric_limits<double>::epsilon.
You are not guaranteed that rounding will take place at any particular time. The C++ standard permits the implementation to use additional precision pretty much anywhere it wants to and many real-world implementations do exactly that.
A common solution for the floating point precision problem is to define an epsilon value yourself and compare to that instead of zero.
e.g.
double epsilon = 0.00001;
if (abs(value) < epsilon) // treat value as 0 in your code
I was solving an equation using double precision and I got -7.07649e-17 as a solution instead of 0.
I agree it's close enough that I can say it's equal but I've read that the machine epsilon for the C++ double type is 2^-52 which is larger than the value I get.
So why do I have an inferior value than the machine epsilon?
Why isn't the value rounded to zero?
It's not a big deal but when I do a logical test it appears that my value is not zero...
There are two different constants in this story. One is epsilon, which is a minimal value that when added to 1.0 produces a value different from 1.0. If you add a smaller value to 1.0 you will again get a 1.0, because there are physical limits to the representation of a number in a computer. But there are values that are less than epsilon and greater than zero. Smallest such number for a double you get with std::numeric_limits<double>::min.
For reference, you get epsilon with std::numeric_limits<double>::epsilon.
You are not guaranteed that rounding will take place at any particular time. The C++ standard permits the implementation to use additional precision pretty much anywhere it wants to and many real-world implementations do exactly that.
A common solution for the floating point precision problem is to define an epsilon value yourself and compare to that instead of zero.
e.g.
double epsilon = 0.00001;
if (abs(value) < epsilon) // treat value as 0 in your code
I am aware, that to compare two floating point values one needs to use some epsilon precision, as they are not exact. However, I wonder if there are edge cases, where I don't need that epsilon.
In particular, I would like to know if it is always safe to do something like this:
double foo(double x){
if (x < 0.0) return 0.0;
else return somethingelse(x); // somethingelse(x) != 0.0
}
int main(){
int x = -3.0;
if (foo(x) == 0.0) {
std::cout << "^- is this comparison ok?" << std::endl;
}
}
I know that there are better ways to write foo (e.g. returning a flag in addition), but I wonder if in general is it ok to assign 0.0 to a floating point variable and later compare it to 0.0.
Or more general, does the following comparison yield true always?
double x = 3.3;
double y = 3.3;
if (x == y) { std::cout << "is an epsilon required here?" << std::endl; }
When I tried it, it seems to work, but it might be that one should not rely on that.
Yes, in this example it is perfectly fine to check for == 0.0. This is not because 0.0 is special in any way, but because you only assign a value and compare it afterwards. You could also set it to 3.3 and compare for == 3.3, this would be fine too. You're storing a bit pattern, and comparing for that exact same bit pattern, as long as the values are not promoted to another type for doing the comparison.
However, calculation results that would mathematically equal zero would not always equal 0.0.
This Q/A has evolved to also include cases where different parts of the program are compiled by different compilers. The question does not mention this, my answer applies only when the same compiler is used for all relevant parts.
C++ 11 Standard,
§5.10 Equality operators
6 If both operands are of arithmetic or enumeration type, the usual
arithmetic conversions are performed on both operands; each of the
operators shall yield true if the specified relationship is true and
false if it is false.
The relationship is not defined further, so we have to use the common meaning of "equal".
§2.13.4 Floating literals
1 [...] If the scaled value is in the range of representable values
for its type, the result is the scaled value if representable, else
the larger or smaller representable value nearest the scaled value,
chosen in an implementation-defined manner. [...]
The compiler has to choose between exactly two values when converting a literal, when the value is not representable. If the same value is chosen for the same literal consistently, you are safe to compare values such as 3.3, because == means "equal".
Yes, if you return 0.0 you can compare it to 0.0; 0 is representable exactly as a floating-point value. If you return 3.3 you have to be a much more careful, since 3.3 is not exactly representable, so a conversion from double to float, for example, will produce a different value.
correction: 0 as a floating point value is not unique, but IEEE 754 defines the comparison 0.0==-0.0 to be true (any zero for that matter).
So with 0.0 this works - for every other number it does not. The literal 3.3 in one compilation unit (e.g. a library) and another (e.g. your application) might differ. The standard only requires the compiler to use the same rounding it would use at runtime - but different compilers / compiler settings might use different rounding.
It will work most of the time (for 0), but is very bad practice.
As long as you are using the same compiler with the same settings (e.g. one compilation unit) it will work because the literal 0.0 or 0.0f will translate to the same bit pattern every time. The representation of zero is not unique though. So if foo is declared in a library and your call to it in some application the same function might fail.
You can rescue this very case by using std::fpclassify to check whether the returned value represents a zero. For every finite (non-zero) value you will have to use an epsilon-comparison though unless you stay within one compilation unit and perform no operations on the values.
As written in both cases you are using identical constants in the same file fed to the same compiler. The string to float conversion the compiler uses should return the same bit pattern so these should not only be equal as in a plus or minus cases for zero thing but equal bit by bit.
Were you to have a constant which uses the operating systems C library to generate the bit pattern then have a string to f or something that can possibly use a different C library if the binary is transported to another computer than the one compiled on. You might have a problem.
Certainly if you compute 3.3 for one of the terms, runtime, and have the other 3.3 computed compile time again you can and will get failures on the equal comparisons. Some constants obviously are more likely to work than others.
Of course as written your 3.3 comparison is dead code and the compiler just removes it if optimizations are enabled.
You didnt specify the floating point format nor standard if any for that format you were interested in. Some formats have the +/- zero problem, some dont for example.
It is a common misconception that floating point values are "not exact". In fact each of them is perfectly exact (except, may be, some special cases as -0.0 or Inf) and equal to s·2e – (p – 1), where s, e, and p are significand, exponent, and precision correspondingly, each of them integer. E.g. in IEEE 754-2008 binary32 format (aka float32) p = 24 and 1 is represented as 0x800000·20 – 23. There are two things that are really not exact when you deal with floating point values:
Representation of a real value using a FP one. Obviously, not all real numbers can be represented using a given FP format, so they have to be somehow rounded. There are several rounding modes, but the most commonly used is the "Round to nearest, ties to even". If you always use the same rounding mode, which is almost certainly the case, the same real value is always represented with the same FP one. So you can be sure that if two real values are equal, their FP counterparts are exactly equal too (but not the reverse, obviously).
Operations with FP numbers are (mostly) inexact. So if you have some real-value function φ(ξ) implemented in the computer as a function of a FP argument f(x), and you want to compare its result with some "true" value y, you need to use some ε in comparison, because it is very hard (sometimes even impossible) to white a function giving exactly y. And the value of ε strongly depends on the nature of the FP operations involved, so in each particular case there may be different optimal value.
For more details see D. Goldberg. What Every Computer Scientist Should Know About Floating-Point Arithmetic, and J.-M. Muller et al. Handbook of Floating-Point Arithmetic. Both texts you can find in the Internet.
When comparing doubles for equality, we need to give a tolerance level, because floating-point computation might introduce errors. For example:
double x;
double y;
x = f();
y = g();
if (fabs(x-y)<epsilon) {
// they are equal!
} else {
// they are not!
}
However, if I simply assign a constant value, without any computation, do I still need to check the epsilon?
double x = 1;
double y = 1;
if (x==y) {
// they are equal!
} else {
// no they are not!
}
Is == comparison good enough? Or I need to do fabs(x-y)<epsilon again? Is it possible to introduce error in assigning? Am I too paranoid?
How about casting (double x = static_cast<double>(100))? Is that gonna introduce floating-point error as well?
I am using C++ on Linux, but if it differs by language, I would like to understand that as well.
Actually, it depends on the value and the implementation. The C++ standard (draft n3126) has this to say in 2.14.4 Floating literals:
If the scaled value is in the range of representable values for its type, the result is the scaled value if representable, else the larger or smaller representable value nearest the scaled value, chosen in an implementation-defined manner.
In other words, if the value is exactly representable (and 1 is, in IEEE754, as is 100 in your static cast), you get the value. Otherwise (such as with 0.1) you get an implementation-defined close match (a). Now I'd be very worried about an implementation that chose a different close match based on the same input token but it is possible.
(a) Actually, that paragraph can be read in two ways, either the implementation is free to choose either the closest higher or closest lower value regardless of which is actually the closest, or it must choose the closest to the desired value.
If the latter, it doesn't change this answer however since all you have to do is hardcode a floating point value exactly at the midpoint of two representable types and the implementation is once again free to choose either.
For example, it might alternate between the next higher and next lower for the same reason banker's rounding is applied - to reduce the cumulative errors.
No if you assign literals they should be the same :)
Also if you start with the same value and do the same operations, they should be the same.
Floating point values are non-exact, but the operations should produce consistent results :)
Both cases are ultimately subject to implementation defined representations.
Storage of floating point values and their representations take on may forms - load by address or constant? optimized out by fast math? what is the register width? is it stored in an SSE register? Many variations exist.
If you need precise behavior and portability, do not rely on this implementation defined behavior.
IEEE-754, which is a standard common implementations of floating point numbers abide to, requires floating-point operations to produce a result that is the nearest representable value to an infinitely-precise result. Thus the only imprecision that you will face is rounding after each operation you perform, as well as propagation of rounding errors from the operations performed earlier in the chain. Floats are not per se inexact. And by the way, epsilon can and should be computed, you can consult any numerics book on that.
Floating point numbers can represent integers precisely up to the length of their mantissa. So for example if you cast from an int to a double, it will always be exact, but for casting into into a float, it will no longer be exact for very large integers.
There is one major example of extensive usage of floating point numbers as a substitute for integers, it's the LUA scripting language, which has no integer built-in type, and floating-point numbers are used extensively for logic and flow control etc. The performance and storage penalty from using floating-point numbers turns out to be smaller than the penalty of resolving multiple types at run time and makes the implementation lighter. LUA has been extensively used not only on PC, but also on game consoles.
Now, many compilers have an optional switch that disables IEEE-754 compatibility. Then compromises are made. Denormalized numbers (very very small numbers where the exponent has reached smallest possible value) are often treated as zero, and approximations in implementation of power, logarithm, sqrt, and 1/(x^2) can be made, but addition/subtraction, comparison and multiplication should retain their properties for numbers which can be exactly represented.
The easy answer: For constants == is ok.
There are two exceptions which you should be aware of:
First exception:
0.0 == -0.0
There is a negative zero which compares equal for the IEEE 754 standard. This means
1/INFINITY == 1/-INFINITY which breaks f(x) == f(y) => x == y
Second exception:
NaN != NaN
This is a special caveat of NotaNumber which allows to find out if a number is a NaN
on systems which do not have a test function available (Yes, that happens).
Logically speaking, given the nature of floating point values, the maximum and minimum representable values of a float are positive and negative infinity, respectively.
Why, then, are FLT_MAX and FLT_MIN not set to them? I understand that this is "just how the standard called for". But then, what use could FLT_MAX or FLT_MIN have as they currently lie in the middle of the representable numeric range of float? Other numeric limits have some utility because they make guarantees about comparisons (e.g. "No INT can test greater than INT_MAX"). Without that kind of guarantee, what use are these float limits at all?
A motivating example for C++:
#include <vector>
#include <limits>
template<typename T>
T find_min(const std::vector<T> &vec)
{
T result = std::numeric_limits<T>::max();
for (std::vector<T>::const_iterator p = vec.start() ; p != vec.end() ; ++p)
if (*p < result) result = *p;
return result;
}
This code works fine if T is an integral type, but not if it is a floating point type. This is annoying. (Yes yes, the standard library provides min_element, but that is not the point. The point is the pattern.)
The purpose of FLT_MIN/MAX is to tell you what the smallest and largest representable floating-point numbers are. Infinity isn't a number; it's a limit.
what use could FLT_MAX or FLT_MIN have as they currently lie in the middle of the representable numeric range of float?
They do not lie in the middle of the representable range. There is no positive float value x which you can add to FLT_MAX and get a representable number. You will get +INF. Which, as previously stated, is not a number.
This code works fine if T is an integral type, but not if it is a floating point type. This is annoying. (Yes yes, the standard library provides min_element, but that is not the point. The point is the pattern.)
And how doesn't it "work fine?" It gives you the smallest value. The only situation where it doesn't "work fine" is if the table contains only +INF. And even in that case, it returns an actual number, not an error-code. Which is probably the better option anyway.
FLT_MAX is defined in section 5.2.4.2.2(9) as
maximum representable finite floating-point number
Positive infinity is not finite.
FLT_MIN is defined in section 5.2.4.2.2(10) as
minimum normalized positive floating-point number
Negative infinity is neither normalized nor positive.
Unlike integer types, floating-point types are (almost?) universally symmetric about zero, and I think the C floating-point model requires this.
On two's-complement systems (i.e., almost all modern systems), INT_MIN is -INT_MAX-1; on other systems, it may be -INT_MAX. (Quibble: a two's-complement system can have INT_MIN equal to -INT_MAX if the lowest representable value is treated as a trap representation.) So INT_MIN conveys information that INT_MAX by itself doesn't.
And a macro for the smallest positive value would not be particularly useful; that's just 1.
In floating-point, on the other hand, the negative value with the greatest magnitude is just -FLT_MAX (or -DBL_MAX, or -LDBL_MAX).
As for why they're not Infinity, there's already a way to represent infinite values (at least in C99): the macro INFINITY. That might cause problems for some C++ applications, but these were defined for C, which doesn't have things like std::numeric_limits<T>::max().
Furthermore, not all floating-point systems have representations for infinity (or NaN).
If FLT_MAX were INFINITY (on systems that support it), then there would probably need to be another macro for the largest representable real value.
I would say the broken pattern you're seeing is only an artifact of poor naming in C, whereas in C++ with numeric_limits and templates, it's an actual semantic flaw that breaks template code that wants to handle both integer and floating point values. Of course you can write a little bit of extra code to test if you have an integer or floating point type (e.g. if ((T)1/2) /* floating point */ else /* integer */) and the problem goes away.
As for why somebody would care about the values FLT_MIN and FLT_MAX give you, they're useful for avoiding underflow and overflow. For example, suppose I need to compute sqrt(x²-1). This is well-defined for any floating point x greater than or equal to 1, but performing the squaring, subtraction, and square root could easily overflow and render the result meaningless when x is large. One might want to test whether x > FLT_MAX/x and handle this case some other way (such as simply returning x :-).