I stumbled on the Factor language and got curious of stack based programming. Factor looks concise, uses a minimun of syntax and punctuation and offers an almost full-featured library for real world problem solving. At least enough for me to give it a try.
Before even getting started, I have a basic question on floating point implementation. How can I avoid this rounding issue when I want to output e.g. 8.12 in the Listener? I use a 32-bit Debian box.
IN: scratchpad 8.12
--- Data stack:
8.119999999999999
And why does it not show the same effect for other floats as well?
IN: scratchpad 8.23
--- Data stack:
8.23
It is because in binary floating point, the value 8.12 is not representable. It is an infinitely long decimal expansion. Similar to how the rational 1/3 is 0.333... as a decimal, 203/25 has an infinite number of digits in binary. So when you convert it back from binary to decimal for printing, you get rounding errors.
It's a known issue, see: https://github.com/slavapestov/factor/issues/1158. But it is only a cosmetic one and doesn't affect calculations.
Related
I came across this the other day and sure its not causing me any trouble but i'm just curious as to why it happens?
1.9.2p320 :001 > 0.39-0.09
=> 0.30000000000000004
This is because Ruby by default uses Double-precision floating-point format. You can read about issues related to it here. However here's a short and crisp answer:
Because internally, computers use a format (binary floating-point)
that cannot accurately represent a number like 0.1, 0.2 or 0.3 at all.
When the code is compiled or interpreted, your “0.1” is already
rounded to the nearest number in that format, which results in a small
rounding error even before the calculation happens.
Source: http://floating-point-gui.de/
Floating-point numbers cannot precisely represent all real numbers, and floating-point operations cannot precisely represent true arithmetic operations, this leads to many surprising situations.
I advise reading: https://en.wikipedia.org/wiki/Floating_point#Accuracy_problems
You may want to use BigDecimal to avoid such problems.
double r = 11.631;
double theta = 21.4;
In the debugger, these are shown as 11.631000000000000 and 21.399999618530273.
How can I avoid this?
These accuracy problems are due to the internal representation of floating point numbers and there's not much you can do to avoid it.
By the way, printing these values at run-time often still leads to the correct results, at least using modern C++ compilers. For most operations, this isn't much of an issue.
I liked Joel's explanation, which deals with a similar binary floating point precision issue in Excel 2007:
See how there's a lot of 0110 0110 0110 there at the end? That's because 0.1 has no exact representation in binary... it's a repeating binary number. It's sort of like how 1/3 has no representation in decimal. 1/3 is 0.33333333 and you have to keep writing 3's forever. If you lose patience, you get something inexact.
So you can imagine how, in decimal, if you tried to do 3*1/3, and you didn't have time to write 3's forever, the result you would get would be 0.99999999, not 1, and people would get angry with you for being wrong.
If you have a value like:
double theta = 21.4;
And you want to do:
if (theta == 21.4)
{
}
You have to be a bit clever, you will need to check if the value of theta is really close to 21.4, but not necessarily that value.
if (fabs(theta - 21.4) <= 1e-6)
{
}
This is partly platform-specific - and we don't know what platform you're using.
It's also partly a case of knowing what you actually want to see. The debugger is showing you - to some extent, anyway - the precise value stored in your variable. In my article on binary floating point numbers in .NET, there's a C# class which lets you see the absolutely exact number stored in a double. The online version isn't working at the moment - I'll try to put one up on another site.
Given that the debugger sees the "actual" value, it's got to make a judgement call about what to display - it could show you the value rounded to a few decimal places, or a more precise value. Some debuggers do a better job than others at reading developers' minds, but it's a fundamental problem with binary floating point numbers.
Use the fixed-point decimal type if you want stability at the limits of precision. There are overheads, and you must explicitly cast if you wish to convert to floating point. If you do convert to floating point you will reintroduce the instabilities that seem to bother you.
Alternately you can get over it and learn to work with the limited precision of floating point arithmetic. For example you can use rounding to make values converge, or you can use epsilon comparisons to describe a tolerance. "Epsilon" is a constant you set up that defines a tolerance. For example, you may choose to regard two values as being equal if they are within 0.0001 of each other.
It occurs to me that you could use operator overloading to make epsilon comparisons transparent. That would be very cool.
For mantissa-exponent representations EPSILON must be computed to remain within the representable precision. For a number N, Epsilon = N / 10E+14
System.Double.Epsilon is the smallest representable positive value for the Double type. It is too small for our purpose. Read Microsoft's advice on equality testing
I've come across this before (on my blog) - I think the surprise tends to be that the 'irrational' numbers are different.
By 'irrational' here I'm just referring to the fact that they can't be accurately represented in this format. Real irrational numbers (like π - pi) can't be accurately represented at all.
Most people are familiar with 1/3 not working in decimal: 0.3333333333333...
The odd thing is that 1.1 doesn't work in floats. People expect decimal values to work in floating point numbers because of how they think of them:
1.1 is 11 x 10^-1
When actually they're in base-2
1.1 is 154811237190861 x 2^-47
You can't avoid it, you just have to get used to the fact that some floats are 'irrational', in the same way that 1/3 is.
One way you can avoid this is to use a library that uses an alternative method of representing decimal numbers, such as BCD
If you are using Java and you need accuracy, use the BigDecimal class for floating point calculations. It is slower but safer.
Seems to me that 21.399999618530273 is the single precision (float) representation of 21.4. Looks like the debugger is casting down from double to float somewhere.
You cant avoid this as you're using floating point numbers with fixed quantity of bytes. There's simply no isomorphism possible between real numbers and its limited notation.
But most of the time you can simply ignore it. 21.4==21.4 would still be true because it is still the same numbers with the same error. But 21.4f==21.4 may not be true because the error for float and double are different.
If you need fixed precision, perhaps you should try fixed point numbers. Or even integers. I for example often use int(1000*x) for passing to debug pager.
Dangers of computer arithmetic
If it bothers you, you can customize the way some values are displayed during debug. Use it with care :-)
Enhancing Debugging with the Debugger Display Attributes
Refer to General Decimal Arithmetic
Also take note when comparing floats, see this answer for more information.
According to the javadoc
"If at least one of the operands to a numerical operator is of type double, then the
operation is carried out using 64-bit floating-point arithmetic, and the result of the
numerical operator is a value of type double. If the other operand is not a double, it is
first widened (§5.1.5) to type double by numeric promotion (§5.6)."
Here is the Source
I'm learning Ruby for fun, and for creating websites also (but that's irrelevant). While playing with it, i noticed something "weird"
When I compute 4.21 + 5 with irb, it answers 9.21 (weird, right?)
when I compute 4.23 + 5, it gives 9.23 (waw, that's definitely weird).
and when i type 4.22 + 5, it answers 9.21999... (w...wait! that's really weird).
Hence my question: what's going on?
I'd understand this behavior with division or really big numbers, but in this simple case....???
Does it mean that i can't develop an accounting app with Ruby?
Is there a patch or something to be applied? (to my brains, most likely)
You should read up on floating point representation in computers. This guide is a good place to start. Here's the short of it:
Because internally, computers use a
format (binary floating-point) that
cannot accurately represent a number
like 0.1, 0.2 or 0.3 at all.
When the code is compiled or
interpreted, your “0.1” is already
rounded to the nearest number in that
format, which results in a small
rounding error even before the
calculation happens.
By the way, I'm not sure why you think 4.21 + 5 = 9.21 or 4.23 + 5 = 9.23 is weird. If you think it's weird because one literal is an integer and one is a float, which ends up as a float response, that's how Ruby and some other languages handle differences in number types. It wouldn't be handy if Ruby dropped your float's precision and gave you just an integer back (9), so any integer literals are effectively changed to floats.
As for using floating-point numbers in a financial application, check out Developing financial application. The main takeaway is use the Decimal object vs. Float.
http://ruby-decimal.rubyforge.org/
This is the package you'd want to do reliable Floating Point maths in ruby. The Developint Financial Applications link doesn't reference it, but it makes the important point of using Decimal data types in your database.
Just noting this since it was non obvious for me :)
Why this code 7.30 - 7.20 in ruby returns 0.0999999999999996, not 0.10?
But if i'll write 7.30 - 7.16, for example, everything will be ok, i'll get 0.14.
What the problem, and how can i solve it?
What Every Computer Scientist Should Know About Floating-Point Arithmetic
The problem is that some numbers we can easily write in decimal don't have an exact representation in the particular floating point format implemented by current hardware. A casual way of stating this is that all the integers do, but not all of the fractions, because we normally store the fraction with a 2**e exponent. So, you have 3 choices:
Round off appropriately. The unrounded result is always really really close, so a rounded result is invariably "perfect". This is what Javascript does and lots of people don't even realize that JS does everything in floating point.
Use fixed point arithmetic. Ruby actually makes this really easy; it's one of the only languages that seamlessly shifts to Class Bignum from Fixnum as numbers get bigger.
Use a class that is designed to solve this problem, like BigDecimal
To look at the problem in more detail, we can try to represent your "7.3" in binary. The 7 part is easy, 111, but how do we do .3? 111.1 is 7.5, too big, 111.01 is 7.25, getting closer. Turns out, 111.010011 is the "next closest smaller number", 7.296875, and when we try to fill in the missing .003125 eventually we find out that it's just 111.010011001100110011... forever, not representable in our chosen encoding in a finite bit string.
The problem is that floating point is inaccurate. You can solve it by using Rational, BigDecimal or just plain integers (for example if you want to store money you can store the number of cents as an int instead of the number of dollars as a float).
BigDecimal can accurately store any number that has a finite number of digits in base 10 and rounds numbers that don't (so three thirds aren't one whole).
Rational can accurately store any rational number and can't store irrational numbers at all.
That is a common error from how float point numbers are represented in memory.
Use BigDecimal if you need exact results.
result=BigDecimal.new("7.3")-BigDecimal("7.2")
puts "%2.2f" % result
It is interesting to note that a number that has few decimals in one base may typically have a very large number of decimals in another. For instance, it takes an infinite number of decimals to express 1/3 (=0.3333...) in the base 10, but only one decimal in the base 3. Similarly, it takes many decimals to express the number 1/10 (=0.1) in the base 2.
Since you are doing floating point math then the number returned is what your computer uses for precision.
If you want a closer answer, to a set precision, just multiple the float by that (such as by 100), convert it to an int, do the math, then divide.
There are other solutions, but I find this to be the simplest since rounding always seems a bit iffy to me.
This has been asked before here, you may want to look for some of the answers given before, such as this one:
Dealing with accuracy problems in floating-point numbers
As I started coding my first app I used NSNumber for money values without thinking twice. Then I thought that maybe c types were enough to deal with my values. Yet, I was advised in the iPhone SDK forum to use NSDecimalNumber, because of its excellent rounding capabilities.
Not being a mathematician by temperament, I thought that the mantissa/exponent paradigm might be overkill; still, googlin' around, I realised that most talks about money/currency in cocoa were referred to NSDecimalNumber.
Notice that the app I am working on is going to be internationalised, so the option of counting the amount in cents is not really viable, for the monetary structure depends greatly on the locale used.
I am 90% sure that I need to go with NSDecimalNumber, but since I found no unambiguous answer on the web (something like: "if you deal with money, use NSDecimalNumber!") I thought I'd ask here. Maybe the answer is obvious to most, but I want to be sure before starting a massive re-factoring of my app.
Convince me :)
Marcus Zarra has a pretty clear stance on this: "If you are dealing with currency at all, then you should be using NSDecimalNumber." His article inspired me to look into NSDecimalNumber, and I've been very impressed with it. IEEE floating point errors when dealing with base-10 math have been irritating me for a while (1 * (0.5 - 0.4 - 0.1) = -0.00000000000000002776) and NSDecimalNumber does away with them.
NSDecimalNumber doesn't just add another few digits of binary floating point precision, it actually does base-10 math. This gets rid of the errors like the one shown in the example above.
Now, I'm writing a symbolic math application, so my desire for 30+ decimal digit precision and no weird floating point errors might be an exception, but I think it's worth looking at. The operations are a little more awkward than simple var = 1 + 2 style math, but they're still manageable. If you're worried about allocating all sorts of instances during your math operations, NSDecimal is the C struct equivalent of NSDecimalNumber and there are C functions for doing the exact same math operations with it. In my experience, these are plenty fast for all but the most demanding applications (3,344,593 additions/s, 254,017 divisions/s on a MacBook Air, 281,555 additions/s, 12,027 divisions/s on an iPhone).
As an added bonus, NSDecimalNumber's descriptionWithLocale: method provides a string with a localized version of the number, including the correct decimal separator. The same goes in reverse for its initWithString:locale: method.
Yes. You have to use
NSDecimalNumber and
not double or float when you deal with currency on iOS.
Why is that??
Because we don't want to get things like $9.9999999998 instead of $10
How that happens??
Floats and doubles are approximations. They always comes with a rounding error. The format computers use to store decimals cause this rouding error.
If you need more details read
http://floating-point-gui.de/
According to apple docs,
NSDecimalNumber is an immutable subclass of NSNumber, provides an object-oriented wrapper for doing base-10 arithmetic. An instance can represent any number that can be expressed as mantissa x 10^exponent where mantissa is a decimal integer up to 38 digits long, and exponent is an integer from –128 through 127.wrapper for doing base-10 arithmetic.
So NSDecimalNumber is recommonded for deal with currency.
(Adapted from my comment on the other answer.)
Yes, you should. An integral number of pennies works only as long as you don't need to represent, say, half a cent. If that happens, you could change it to count half-cents, but what if you then need to represent a quarter-cent, or an eighth of a cent?
The only proper solution is NSDecimalNumber (or something like it), which puts off the problem to 10^-128¢ (i.e.,
0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000001¢).
(Another way would be arbitrary-precision arithmetic, but that requires a separate library, such as the GNU MP Bignum library. GMP is under the LGPL. I've never used that library and don't know exactly how it works, so I couldn't say how well it would work for you.)
[Edit: Apparently, at least one person—Brad Larson—thinks I'm talking about binary floating-point somewhere in this answer. I'm not.]
I've found it convenient to use an integer to represent the number of cents and then divide by 100 for presentation. Avoids the whole issue.
A better question is, when should you not use NSDecimalNumber to deal with money. The short answer to that question is, when you can't tolerate the performance overhead of NSDecimalNumber and you don't care about small rounding errors because you're never dealing with more than a few digits of precision. The even shorter answer is, you should always use NSDecimalNumber when dealing with money.
VISA, MasterCards and others are using integer values while passing amounts. It's up to sender and reciever to parse amouts correctly according to currency exponent (divide or multiply by 10^num, where num - is an exponent of the currency). Note that different currencies have different exponents. Usually it's 2 (hence we divide and multiply by 100), but some currencies have exponent = 0 (VND,etc), or = 3.