How the heck does Ruby do this? Does Jörg or anyone else know what's happening behind the scenes?
Unfortunately I don't know C very well so bignum.c is of little help to me. I was just kind of curious it someone could explain (in plain English) the theory behind whatever miracle algorithm its using.
irb(main):001:0> 999**999
368063488259223267894700840060521865838338232037353204655959621437025609300472231530103873614505175218691345257589896391130393189447969771645832382192366076536631132001776175977932178658703660778465765811830827876982014124022948671975678131724958064427949902810498973271030787716781467419524180040734398996952930832508934116945966120176735120823151959779536852290090377452502236990839453416790640456116471139751546750048602189291028640970574762600185950226138244530187489211615864021135312077912018844630780307462205252807737757672094320692373101032517459518497524015120165166724189816766397247824175394802028228160027100623998873667435799073054618906855460488351426611310634023489044291860510352301912426608488807462312126590206830413782664554260411266378866626653755763627796569082931785645600816236891168141774993267488171702172191072731069216881668294625679492696148976999868715671440874206427212056717373099639711168901197440416590226524192782842896415414611688187391232048327738965820265934093108172054875188246591760877131657895633586576611857277011782497943522945011248430439201297015119468730712364007639373910811953430309476832453230123996750235710787086641070310288725389595138936784715274150426495416196669832679980253436807864187160054589045664027158817958549374490512399055448819148487049363674611664609890030088549591992466360050042566270348330911795487647045949301286614658650071299695652245266080672989921799342509291635330827874264789587306974472327718704306352445925996155619153783913237212716010410294999877569745287353422903443387562746452522860420416689019732913798073773281533570910205207767157128174184873357050830752777900041943256738499067821488421053870869022738698816059810579221002560882999884763252161747566893835178558961142349304466506402373556318707175710866983035313122068321102457824112014969387225476259342872866363550383840720010832906695360553556647545295849966279980830561242960013654529514995113584909050813015198928283202189194615501403435553060147713139766323195743324848047347575473228198492343231496580885057330510949058490527738662697480293583612233134502078182014347192522391449087738579081585795613547198599661273567662441490401862839817822686573112998663038868314974259766039340894024308383451039874674061160538242392803580758232755749310843694194787991556647907091849600704712003371103926967137408125713631396699343733288014254084819379380555174777020843568689927348949484201042595271932630685747613835385434424807024615161848223715989797178155169951121052285149157137697718850449708843330475301440373094611119631361702936342263219382793996895988331701890693689862459020775599439506870005130750427949747071390095256759203426671803377068109744629909769176319526837824364926844730545524646494321826241925107158040561607706364484910978348669388142016838792902926158979355432483611517588605967745393958061959024834251565197963477521095821435651996730128376734574843289089682710350244222290017891280419782767803785277960834729869249991658417000499998999
Simple: it does it the same way you do, ever since first grade. Except it doesn't compute in base 10, it computes in base 4 billion (and change).
Think about it: with our number system, we can only represent numbers from 0 to 9. So, how can we compute 6+7 without overflowing? Easy: we do actually overflow! We cannot represent the result of 6+7 as a number between 0 and 9, but we can overflow to the next place and represent it as two numbers between 0 and 9: 3×100 + 1×101. If you want to add two numbers, you add them digit-wise from the right and overflow ("carry") to the left. If you want to multiply two numbers, you have to multiply every digit of one number individually with the other number, then add up the intermediate results.
BigNum arithmetic (this is what this kind of arithmetic where the numbers are bigger than the native machine numbers is usually called) works basically the same way. Except that the base is not 10, and its not 2, either – it's the size of a native machine integer. So, on a 32 bit machine, it would be base 232 or 4 294 967 296.
Specifically, in Ruby Integer is actually an abstract class that is never instianted. Instead, it has two subclasses, Fixnum and Bignum, and numbers automagically migrate between them, depending on their size. In MRI and YARV, Fixnum can hold a 31 or 63 bit signed integer (one bit is used for tagging) depending on the native word size of the machine. In JRuby, a Fixnum can hold a full 64 bit signed integer, even on an 32 bit machine.
The simplest operation is adding two numbers. And if you look at the implementation of + or rather bigadd_core in YARV's bignum.c, it's not too bad to follow. I can't read C either, but you can cleary see how it loops over the individual digits.
You could read the source for bignum.c...
At a very high level, without going into any implementation details, bignums are calculated "by hand" like you used to do in grade school. Now, there are certainly many optimizations that can be applied, but that's the gist of it.
I don't know of the implementation details so I'll cover how a basic Big Number implementation would work.
Basically instead of relying on CPU "integers" it will create it's own using multiple CPU integers. To store arbritrary precision, well lets say you have 2 bits. So the current integer is 11. You want to add one. In normal CPU integers, this would roll over to 00
But, for big number, instead of rolling over and keeping a "fixed" integer width, it would allocate another bit and simulate an addition so that the number becomes the correct 100.
Try looking up how binary math can be done on paper. It's very simple and is trivial to convert to an algorithm.
Beaconaut APICalc 2 just released on Jan.18, 2011, which is an arbitrary-precision integer calculator for bignum arithmetic, cryptography analysis and number theory research......
http://www.beaconaut.com/forums/default.aspx?g=posts&t=13
It uses the Bignum class
irb(main):001:0> (999**999).class
=> Bignum
Rdoc is available of course
Related
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 trying to seed a random number generator with the output of a hash. Currently I'm computing a SHA-1 hash, converting it to a giant integer, and feeding it to srand to initialize the RNG. This is so that I can get a predictable set of random numbers for an set of infinite cartesian coordinates (I'm hashing the coordinates).
I'm wondering whether Kernel::srand actually has a maximum value that it'll take, after which it truncates it in some way. The docs don't really make this obvious - they just say "a number".
I'll try to figure it out myself, but I'm assuming somebody out there has run into this already.
Knowing what programmers are like, it probably just calls libc's srand(). Either way, it's probably limited to 2^32-1, 2^31-1, 2^16-1, or 2^15-1.
There's also a danger that the value is clipped when cast from a biginteger to a C int/long, instead of only taking the low-order bits.
An easy test is to seed with 1 and take the first output. Then, seed with 2i+1 for i in [1..64] or so, take the first output of each, and compare. If you get a match for some i=n and all greater is, then it's probably doing arithmetic modulo 2n.
Note that the random number generator is almost certainly limited to 32 or 48 bits of entropy anyway, so there's little point seeding it with a huge value, and an attacker can reasonably easily predict future outputs given past outputs (and an "attacker" could simply be a player on a public nethack server).
EDIT: So I was wrong.
According to the docs for Kernel::rand(),
Ruby currently uses a modified Mersenne Twister with a period of 2**19937-1.
This means it's not just a call to libc's rand(). The Mersenne Twister is statistically superior (but not cryptographically secure). But anyway.
Testing using Kernel::srand(0); Kernel::sprintf("%x",Kernel::rand(2**32)) for various output sizes (2*16, 2*32, 2*36, 2*60, 2*64, 2*32+1, 2*35, 2*34+1), a few things are evident:
It figures out how many bits it needs (number of bits in max-1).
It generates output in groups of 32 bits, most-significant-bits-first, and drops the top bits (i.e. 0x[r0][r1][r2][r3][r4] with the top bits masked off).
If it's not less than max, it does some sort of retry. It's not obvious what this is from the output.
If it is less than max, it outputs the result.
I'm not sure why 2*32+1 and 2*64+1 are special (they produce the same output from Kernel::rand(2**1024) so probably have the exact same state) — I haven't found another collision.
The good news is that it doesn't simply clip to some arbitrary maximum (i.e. passing in huge numbers isn't equivalent to passing in 2**31-1), which is the most obvious thing that can go wrong. Kernel::srand() also returns the previous seed, which appears to be 128-bit, so it seems likely to be safe to pass in something large.
EDIT 2: Of course, there's no guarantee that the output will be reproducible between different Ruby versions (the docs merely say what it "currently uses"; apparently this was initially committed in 2002). Java has several portable deterministic PRNGs (SecureRandom.getInstance("SHA1PRNG","SUN"), albeit slow); I'm not aware of something similar for Ruby.
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.
I have an array of numbers that potentially have up to 8 decimal places and I need to find the smallest common number I can multiply them by so that they are all whole numbers. I need this so all the original numbers can all be multiplied out to the same scale and be processed by a sealed system that will only deal with whole numbers, then I can retrieve the results and divide them by the common multiplier to get my relative results.
Currently we do a few checks on the numbers and multiply by 100 or 1,000,000, but the processing done by the *sealed system can get quite expensive when dealing with large numbers so multiplying everything by a million just for the sake of it isn’t really a great option. As an approximation lets say that the sealed algorithm gets 10 times more expensive every time you multiply by a factor of 10.
What is the most efficient algorithm, that will also give the best possible result, to accomplish what I need and is there a mathematical name and/or formula for what I’m need?
*The sealed system isn’t really sealed. I own/maintain the source code for it but its 100,000 odd lines of proprietary magic and it has been thoroughly bug and performance tested, altering it to deal with floats is not an option for many reasons. It is a system that creates a grid of X by Y cells, then rects that are X by Y are dropped into the grid, “proprietary magic” occurs and results are spat out – obviously this is an extremely simplified version of reality, but it’s a good enough approximation.
So far there are quiet a few good answers and I wondered how I should go about choosing the ‘correct’ one. To begin with I figured the only fair way was to create each solution and performance test it, but I later realised that pure speed wasn’t the only relevant factor – an more accurate solution is also very relevant. I wrote the performance tests anyway, but currently the I’m choosing the correct answer based on speed as well accuracy using a ‘gut feel’ formula.
My performance tests process 1000 different sets of 100 randomly generated numbers.
Each algorithm is tested using the same set of random numbers.
Algorithms are written in .Net 3.5 (although thus far would be 2.0 compatible)
I tried pretty hard to make the tests as fair as possible.
Greg – Multiply by large number
and then divide by GCD – 63
milliseconds
Andy – String Parsing
– 199 milliseconds
Eric – Decimal.GetBits – 160 milliseconds
Eric – Binary search – 32
milliseconds
Ima – sorry I couldn’t
figure out a how to implement your
solution easily in .Net (I didn’t
want to spend too long on it)
Bill – I figure your answer was pretty
close to Greg’s so didn’t implement
it. I’m sure it’d be a smidge faster
but potentially less accurate.
So Greg’s Multiply by large number and then divide by GCD” solution was the second fastest algorithm and it gave the most accurate results so for now I’m calling it correct.
I really wanted the Decimal.GetBits solution to be the fastest, but it was very slow, I’m unsure if this is due to the conversion of a Double to a Decimal or the Bit masking and shifting. There should be a
similar usable solution for a straight Double using the BitConverter.GetBytes and some knowledge contained here: http://blogs.msdn.com/bclteam/archive/2007/05/29/bcl-refresher-floating-point-types-the-good-the-bad-and-the-ugly-inbar-gazit-matthew-greig.aspx but my eyes just kept glazing over every time I read that article and I eventually ran out of time to try to implement a solution.
I’m always open to other solutions if anyone can think of something better.
I'd multiply by something sufficiently large (100,000,000 for 8 decimal places), then divide by the GCD of the resulting numbers. You'll end up with a pile of smallest integers that you can feed to the other algorithm. After getting the result, reverse the process to recover your original range.
Multiple all the numbers by 10
until you have integers.
Divide
by 2,3,5,7 while you still have all
integers.
I think that covers all cases.
2.1 * 10/7 -> 3
0.008 * 10^3/2^3 -> 1
That's assuming your multiplier can be a rational fraction.
If you want to find some integer N so that N*x is also an exact integer for a set of floats x in a given set are all integers, then you have a basically unsolvable problem. Suppose x = the smallest positive float your type can represent, say it's 10^-30. If you multiply all your numbers by 10^30, and then try to represent them in binary (otherwise, why are you even trying so hard to make them ints?), then you'll lose basically all the information of the other numbers due to overflow.
So here are two suggestions:
If you have control over all the related code, find another
approach. For example, if you have some function that takes only
int's, but you have floats, and you want to stuff your floats into
the function, just re-write or overload this function to accept
floats as well.
If you don't have control over the part of your system that requires
int's, then choose a precision to which you care about, accept that
you will simply have to lose some information sometimes (but it will
always be "small" in some sense), and then just multiply all your
float's by that constant, and round to the nearest integer.
By the way, if you're dealing with fractions, rather than float's, then it's a different game. If you have a bunch of fractions a/b, c/d, e/f; and you want a least common multiplier N such that N*(each fraction) = an integer, then N = abc / gcd(a,b,c); and gcd(a,b,c) = gcd(a, gcd(b, c)). You can use Euclid's algorithm to find the gcd of any two numbers.
Greg: Nice solution but won't calculating a GCD that's common in an array of 100+ numbers get a bit expensive? And how would you go about that? Its easy to do GCD for two numbers but for 100 it becomes more complex (I think).
Evil Andy: I'm programing in .Net and the solution you pose is pretty much a match for what we do now. I didn't want to include it in my original question cause I was hoping for some outside the box (or my box anyway) thinking and I didn't want to taint peoples answers with a potential solution. While I don't have any solid performance statistics (because I haven't had any other method to compare it against) I know the string parsing would be relatively expensive and I figured a purely mathematical solution could potentially be more efficient.
To be fair the current string parsing solution is in production and there have been no complaints about its performance yet (its even in production in a separate system in a VB6 format and no complaints there either). It's just that it doesn't feel right, I guess it offends my programing sensibilities - but it may well be the best solution.
That said I'm still open to any other solutions, purely mathematical or otherwise.
What language are you programming in? Something like
myNumber.ToString().Substring(myNumber.ToString().IndexOf(".")+1).Length
would give you the number of decimal places for a double in C#. You could run each number through that and find the largest number of decimal places(x), then multiply each number by 10 to the power of x.
Edit: Out of curiosity, what is this sealed system which you can pass only integers to?
In a loop get mantissa and exponent of each number as integers. You can use frexp for exponent, but I think bit mask will be required for mantissa. Find minimal exponent. Find most significant digits in mantissa (loop through bits looking for last "1") - or simply use predefined number of significant digits.
Your multiple is then something like 2^(numberOfDigits-minMantissa). "Something like" because I don't remember biases/offsets/ranges, but I think idea is clear enough.
So basically you want to determine the number of digits after the decimal point for each number.
This would be rather easier if you had the binary representation of the number. Are the numbers being converted from rationals or scientific notation earlier in your program? If so, you could skip the earlier conversion and have a much easier time. Otherwise you might want to pass each number to a function in an external DLL written in C, where you could work with the floating point representation directly. Or you could cast the numbers to decimal and do some work with Decimal.GetBits.
The fastest approach I can think of in-place and following your conditions would be to find the smallest necessary power-of-ten (or 2, or whatever) as suggested before. But instead of doing it in a loop, save some computation by doing binary search on the possible powers. Assuming a maximum of 8, something like:
int NumDecimals( double d )
{
// make d positive for clarity; it won't change the result
if( d<0 ) d=-d;
// now do binary search on the possible numbers of post-decimal digits to
// determine the actual number as quickly as possible:
if( NeedsMore( d, 10e4 ) )
{
// more than 4 decimals
if( NeedsMore( d, 10e6 ) )
{
// > 6 decimal places
if( NeedsMore( d, 10e7 ) ) return 10e8;
return 10e7;
}
else
{
// <= 6 decimal places
if( NeedsMore( d, 10e5 ) ) return 10e6;
return 10e5;
}
}
else
{
// <= 4 decimal places
// etc...
}
}
bool NeedsMore( double d, double e )
{
// check whether the representation of D has more decimal points than the
// power of 10 represented in e.
return (d*e - Math.Floor( d*e )) > 0;
}
PS: you wouldn't be passing security prices to an option pricing engine would you? It has exactly the flavor...