Is there a function that returns the highest and lowest possible numeric values?
help(numeric) sends you to help(double) which has
Double-precision values:
All R platforms are required to work with values conforming tothe
IEC 60559 (also known as IEEE 754) standard. This basically works
with a precision of 53 bits, and represents to that precision a
range of absolute values from about 2e-308 to 2e+308. It also has
special values ‘NaN’ (many of them), plus and minus infinity and
plus and minus zero (although R acts as if these are the same).
There are also _denormal(ized)_ (or _subnormal_) numbers with
absolute values above or below the range given above but
represented to less precision.
See ‘.Machine’ for precise information on these limits. Note that
ultimately how double precision numbers are handled is down to the
CPU/FPU and compiler.
So you want to look at .Machine which on my 64-bit box has
$double.xmin
[1] 2.22507e-308
$double.xmax
[1] 1.79769e+308
help("numeric")
will ask you to do
help("double)
which will give the answer: range of absolute values from about 2e-308 to 2e+308.
Related
There is a lot of questions on rounding that i have looked at but tey all involve rounding a number to its nearest whole, or to a certain number of points. What i want to do is simply convert a string to a double without any added digits on the right of the decimal point. Here is my code and result as of now:
Convert the string 0.78240 to a double, which should be 0.78240 but instead is 0.78239999999999998 when i look at it in the debugger.
The string value is a QString and is converted to a double simply using the toDouble() function.
I don't understand how or where these extra numbers are coming from, but any help on converting from QString to double directly would be greatly appreciated!
The extra digits are there because you are converting a decimal real number to binary floating point.
Unlike real numbers, floating-point representations have infinite resolution and finite range, and also binary floating-point values do not exactly coincide with all (or even most) decimal real values.
The simple fact is that binary floating-point cannot exactly represent 0.7824010, your debugger is showing you all the available digits after round-tripping the binary value back to decimal.
It is not necessarily a problem, because the error is infinitesimally small compared to the magnitude of the value, and in any event the original 0.78240 value is no doubt some approximation of a real-world value - they are both approximations, just binary or decimal approximations.
The issue is normally dealt with at presentation rather then representation. For example, in this case, unlike your debugger which necessarily shows the full precision of the internal representation (you would not want it any other way in a debugger), the standard means of presenting such a value will limit itself to a small, or caller defined number of decimal places and this value presented to even 15 decimal places will be correctly presented as 0.782400000000000 (by default standard output methods will show just 0.7824).
Any double value presented at 15 significant decimal figures or fewer will display as expected, for a float this reduces to just 6 significant figures. I imagine your debugger is displaying more digits that can accurately be presented in an IEEE 754 64-bit FP (double) value because internally the x86 FPU uses an 80bit representation.
You are quite literally sweating the small stuff.
One place where this difference in representation does matter is in financial applications. For those, it is common to use decimal floating point and normally to many more significant figures than double can provide. However decimal floating-point is not normally implemented in hardware, so is much slower. Moreover decimal floating point is not directly supported in most programming languages, and requires library support. C# is an example of a language with built-in support for decimal floating-point; its decimal type is good for 28 significant figures.
Can I always trust that when I divide two IEEE 754 doubles that exactly represent integers, I get the exact value?
For example:
80.0/4.0
=> 20.0
The result is exactly 20.0.
Are there some potential values where the result is inaccurate like 19.99999999?
If both integers can be exactly represented by the floating point format (i.e. for doubles, they are less than 253), then yes you will get the exact value.
Floating point is not some crazy fuzzy random algorithm, it does have well-defined deterministic behaviour. Operations that are exact will always give exact answers, and operations that are not exact will round to the representable value (assuming you haven't done anything funny with your rounding modes): "problems" arise when (1) the inputs are not exactly representable (such as 0.1, or 1020), or (2) you're performing multiple operations, some of which may cause intermediate rounding.
According to Wikipedia,
[floating point] division is accomplished by subtracting the divisor's exponent from the dividend's exponent, and dividing the dividend's significand by the divisor's significand.
If one number evenly divides another, then the former's significand should evenly divide the latter's significand. This will give you an exact result.
I am often programming mathematical algorithms that assume a nondimensional parameter spans the continuous space from 0..1 inclusive. These algorithms could in theory benefit from maximum resolution over the parameter space and I've considered that it would be of use to expend the full 32 or 64 bits of precision over the parameter space, with none wasted for exponents or signs.
I imagine the methods would look similar to an unsigned integer divided by its maximum representable value. Does this exist already and if so where, if not, is there a compelling reason why?
Can't you simply do all calculations in integers from 0 to MAX_INT, keeping all the same formulas/algorithms/whatever and then use "unsigned integer divided by its maximum representable value" conversion as very final step before printing result to user (or otherwise outputting it - for example in intermediate logs)?
The representation doesn't make sense without algorithms. E.g. you could represent it as fixed point (i.e. 0..MAX_INT / MAX_INT) or floated point a mantissa and exponent (e.g. to have an ability to store a values like 1e-1000) or something custom (e.g. to have an ability to represent a number 1/π precisely). After it you have define algos to manipulate the numbers in such representations. So, in other words there is no silver bullet to cover all cases. Only you know your task and could choose the best solution.
Moreover, the continuous space is impossible to represent using computes, because the space has infinite number of elements, so it cannot be algorithmized.
If I run following line of code, I get DIVIDE BY ZERO error
1. System.out.println(5/0);
which is the expected behavior.
Now I run the below line of code
2. System.out.println(5/0F);
here there is no DIVIDE BY ZERO error, rather it shows INFINITY
In the first line I am dividing two integers and in the second two real numbers.
Why does dividing by zero for integers gives DIVIDE BY ZERO error while in the case of real numbers it gives INFINITY
I am sure it is not specific to any programming language.
(EDIT: The question has been changed a bit - it specifically referred to Java at one point.)
The integer types in Java don't have representations of infinity, "not a number" values etc - whereas IEEE-754 floating point types such as float and double do. It's as simple as that, really. It's not really a "real" vs "integer" difference - for example, BigDecimal represents real numbers too, but it doesn't have a representation of infinity either.
EDIT: Just to be clear, this is language/platform specific, in that you could create your own language/platform which worked differently. However, the underlying CPUs typically work the same way - so you'll find that many, many languages behave this way.
EDIT: In terms of motivation, bear in mind that for the infinity case in particular, there are ways of getting to infinity without dividing by zero - such as dividing by a very, very small floating point number. In the case of integers, there's obviously nothing between zero and one.
Also bear in mind that the cases in which integers (or decimal floating point types) are used typically don't need to concept of infinity, or "not a number" results - whereas in scientific applications (where float/double are more typically useful), "infinity" (or at least, "a number which is too large to sensibly represent") is still a potentially valid result.
This is specific to one programming language or a family of languages. Not all languages allow integers and floats to be used in the same expression. Not all languages have both types (for example, ECMAScript implementations like JavaScript have no notion of an integer type externally). Not all languages have syntax like this to convert values inline.
However, there is an intrinsic difference between integer arithmetic and floating-point arithmetic. In integer arithmetic, you must define that division by zero is an error, because there are no values to represent the result. In floating-point arithmetic, specifically that defined in IEEE-754, there are additional values (combinations of sign bit, exponent and mantissa) for the mathematical concept of infinity and meta-concepts like NaN (not a number).
So we can assume that the / operator in this programming language is generic, that it performs integer division if both operands are of the language's integer type; and that it performs floating-point division if at least one of the operands is of a float type of the language, whereas the other operands would be implicitly converted to that float type for the purpose of the operation.
In real-number math, division of a number by a number close to zero is equivalent to multiplying the first number by a number whose absolute is very large (x / (1 / y) = x * y). So it is reasonable that the result of dividing by zero should be (defined as) infinity as the precision of the floating-point value would be exceeded.
Implementation details were to be found in that programming language's specification.
I have a method that deals with some geographic coordinates in .NET, and I have a struct that stores a coordinate pair such that if 256 is passed in for one of the coordinates, it becomes 0. However, in one particular instance a value of approximately 255.99999998 is calculated, and thus stored in the struct. When it's printed in ToString(), it becomes 256, which should not happen - 256 should be 0. I wouldn't mind if it printed 255.9999998 but the fact that it prints 256 when the debugger shows 255.99999998 is a problem. Having it both store and display 0 would be even better.
Specifically there's an issue with comparison. 255.99999998 is sufficiently close to 256 such that it should equal it. What should I do when comparing doubles? use some sort of epsilon value?
EDIT: Specifically, my problem is that I take a value, perform some calculations, then perform the opposite calculations on that number, and I need to get back the original value exactly.
This sounds like a problem with how the number is printed, not how it is stored. A double has about 15 significant figures, so it can tell 255.99999998 from 256 with precision to spare.
You could use the epsilon approach, but the epsilon is typically a fudge to get around the fact that floating-point arithmetic is lossy.
You might consider avoiding binary floating-points altogether and use a nice Rational class.
The calculation above was probably destined to be 256 if you were doing lossless arithmetic as you would get with a Rational type.
Rational types can go by the name of Ratio or Fraction class, and are fairly simple to write
Here's one example.
Here's another
Edit....
To understand your problem consider that when the decimal value 0.01 is converted to a binary representation it cannot be stored exactly in finite memory. The Hexidecimal representation for this value is 0.028F5C28F5C where the "28F5C" repeats infinitely. So even before doing any calculations, you loose exactness just by storing 0.01 in binary format.
Rational and Decimal classes are used to overcome this problem, albeit with a performance cost. Rational types avoid this problem by storing a numerator and a denominator to represent your value. Decimal type use a binary encoded decimal format, which can be lossy in division, but can store common decimal values exactly.
For your purpose I still suggest a Rational type.
You can choose format strings which should let you display as much of the number as you like.
The usual way to compare doubles for equality is to subtract them and see if the absolute value is less than some predefined epsilon, maybe 0.000001.
You have to decide yourself on a threshold under which two values are equal. This amounts to using so-called fixed point numbers (as opposed to floating point). Then, you have to perform the round up manually.
I would go with some unsigned type with known size (eg. uint32 or uint64 if they're available, I don't know .NET) and treat it as a fixed point number type mod 256.
Eg.
typedef uint32 fixed;
inline fixed to_fixed(double d)
{
return (fixed)(fmod(d, 256.) * (double)(1 << 24))
}
inline double to_double(fixed f)
{
return (double)f / (double)(1 << 24);
}
or something more elaborated to suit a rounding convention (to nearest, to lower, to higher, to odd, to even). The highest 8 bits of fixed hold the integer part, the 24 lower bits hold the fractional part. Absolute precision is 2^{-24}.
Note that adding and substracting such numbers naturally wraps around at 256. For multiplication, you should beware.