I want to write a program to convert hexadecimal numbers into their decimal forms without using a variable of fixed length to store the result because that would restrict the range of inputs that my program can work with.
Let's say I were to use a variable of type long long int to calculate, store and print the result. Doing so would limit the range of hexadecimal numbers that my program can handle to between 8000000000000001 and 7FFFFFFFFFFFFFFF. Anything outside this range would cause the variable to overflow.
I did write a program that calculates and stores the decimal result in a dynamically allocated string by performing carry and borrow operations but it runs much slower, even for numbers that are as big as 7FFFFFFFF!
Then I stumbled onto this site which could take numbers that are way outside the range of a 64 bit variable. I tried their converter with numbers much larger than 16^65 - 1 and still couldn't get it to overflow. It just kept on going and printing the result.
I figured that they must be using a much better algorithm for hex to decimal conversion, one that isn't limited to 64 bit values.
So far, Google's search results have only led me to algorithms that use some fixed-length variable for storing the result.
That's why I am here. I wanna know if such an algorithm exists and if it does, what is it?
Well, it sounds like you already did it when you wrote "a program that calculates and stores the decimal result in a dynamically allocated string by performing carry and borrow operations".
Converting from base 16 (hexadecimal) to base 10 means implementing multiplication and addition of numbers in a base 10x representation. Then for each hex digit d, you calculate result = result*16 + d. When you're done you have the same number in a 10-based representation that is easy to write out as a decimal string.
There could be any number of reasons why your string-based method was slow. If you provide it, I'm sure someone could comment.
The most important trick for making it reasonably fast, though, is to pick the right base to convert to and from. I would probably do the multiplication and addition in base 109, so that each digit will be as large as possible while still fitting into a 32-bit integer, and process 7 hex digits at a time, which is as many as I can while only multiplying by single digits.
For every 7 hex digts, I'd convert them to a number d, and then do result = result * (16^7) + d.
Then I can get the 9 decimal digits for each resulting digit in base 109.
This process is pretty easy, since you only have to multiply by single digits. I'm sure there are faster, more complicated ways that recursively break the number into equal-sized pieces.
Related
My application requires a fractional quantity multiplied by a monetary value.
For example, $65.50 × 0.55 hours = $36.025 (rounded to $36.03).
I know that floats should not be used to represent money, so I'm storing all of my monetary values as cents. $65.50 in the above equation is stored as 6550 (integer).
For the fractional coefficient, my issue is that 0.55 does not have a 32-bit float representation. In the use case above, 0.55 hours == 33 minutes, so 0.55 is an example of a specific value that my application will need to account for exactly. The floating point representation of 0.550000012 is insufficient, because the user will not understand where the additional 0.000000012 came from. I cannot simply call a rounding function on 0.550000012 because it will round to the whole number.
Multiplication solution
To solve this, my first idea was to store all quantities as integers and multiply × 1000. So 0.55 entered by the user would become 550 (integer) when stored. All calculations would happen without floats, and then simply divide by 1000 (integer division, not float) when presenting the result to the user.
I realize that this would permanently limit me to 3 decimal places of
precision. If I decide that 3 is adequate for the lifetime of my
application, does this approach make sense?
Are there potential rounding issues if I were to use integer division?
Is there a name for this process? EDIT: As indicated by #SergGr, this is fixed-point arithmetic.
Is there a better approach?
EDIT:
I should have clarified, this is not time-specific. It is for generic quantities like 1.256 pounds of flour, 1 sofa, or 0.25 hours (think invoices).
What I'm trying to replicate here is a more exact version of Postgres's extra_float_digits = 0 functionality, where if the user enters 0.55 (float32), the database stores 0.550000012 but when queried for the result returns 0.55 which appears to be exactly what the user typed.
I am willing to limit this application's precision to 3 decimal places (it's business, not scientific), so that's what made me consider the × 1000 approach.
I'm using the Go programming language, but I'm interested in generic cross-language solutions.
Another solution to store the result is using the rational form of the value. You can explain the number by two integer value which the number is equal p/q, such that both p and q are integers. Hence, you can have more precision for your numbers and do some math with the rational numbers in the format of two integers.
Note: This is an attempt to merge different comments into one coherent answer as was requested by Matt.
TL;DR
Yes, this approach makes sense but most probably is not the best choice
Yes, there are rounding issues but there inevitably will be some no matter what representation you use
What you suggest using is called Decimal fixed point numbers
I'd argue yes, there is a better approach and it is to use some standard or popular decimal floating point numbers library for your language (Go is not my native language so I can't recommend one)
In PostgreSQL it is better to use Numeric (something like Numeric(15,3) for example) rather than a combination of float4/float8 and extra_float_digits. Actually this is what the first item in the PostgreSQL doc on Floating-Point Types suggests:
If you require exact storage and calculations (such as for monetary amounts), use the numeric type instead.
Some more details on how non-integer numbers can be stored
First of all there is a fundamental fact that there are infinitely many numbers in the range [0;1] so you obviously can't store every number there in any finite data structure. It means you have to make some compromises: no matter what way you choose, there will be some numbers you can't store exactly so you'll have to round.
Another important point is that people are used to 10-based system and in that system only results of division by numbers in a form of 2^a*5^b can be represented using a finite number of digits. For every other rational number even if you somehow store it in the exact form, you will have to do some truncation and rounding at the formatting for human usage stage.
Potentially there are infinitely many ways to store numbers. In practice only a few are widely used:
floating point numbers with two major branches of binary (this is what most today's hardware natively implements and what is support by most of the languages as float or double) and decimal. This is the format that store mantissa and exponent (can be negative), so the number is mantissa * base^exponent (I omit sign and just say it is logically a part of the mantissa although in practice it is usually stored separately). Binary vs. decimal is specified by the base. For example 0.5 will be stored in binary as a pair (1,-1) i.e. 1*2^-1 and in decimal as a pair (5,-1) i.e. 5*10^-1. Theoretically you can use any other base as well but in practice only 2 and 10 make sense as the bases.
fixed point numbers with the same division in binary and decimal. The idea is the same as in floating point numbers but some fixed exponent is used for all the numbers. What you suggests is actually a decimal fixed point number with the exponent fixed at -3. I've seen a usage of binary fixed-point numbers on some embedded hardware where there is no built-in support of floating point numbers, because binary fixed-point numbers can be implemented with reasonable efficiency using integer arithmetic. As for decimal fixed-point numbers, in practice they are not much easier to implement that decimal floating-point numbers but provide much less flexibility.
rational numbers format i.e. the value is stored as a pair of (p, q) which represents p/q (and usually q>0 so sign stored in p and either p=0, q=1 for 0 or gcd(p,q) = 1 for every other number). Usually this requires some big integer arithmetic to be useful in the first place (here is a Go example of math.big.Rat). Actually this might be an useful format for some problems and people often forget about this possibility, probably because it is often not a part of a standard library. Another obvious drawback is that as I said people are not used to think in rational numbers (can you easily compare which is greater 123/456 or 213/789?) so you'll have to convert the final results to some other form. Another drawback is that if you have a long chain of computations, internal numbers (p and q) might easily become very big values so computations will be slow. Still it may be useful to store intermediate results of calculations.
In practical terms there is also a division into arbitrary length and fixed length representations. For example:
IEEE 754 float or double are fixed length floating-point binary representations,
Go math.big.Float is an arbitrary length floating-point binary representations
.Net decimal is a fixed length floating-point decimal representations
Java BigDecimal is an arbitrary length floating-point decimal representations
In practical terms I'd says that the best solution for your problem is some big enough fixed length floating point decimal representations (like .Net decimal). An arbitrary length implementation would also work. If you have to make an implementation from scratch, than your idea of a fixed length fixed point decimal representation might be OK because it is the easiest thing to implement yourself (a bit easier than the previous alternatives) but it may become a burden at some point.
As mentioned in the comments, it would be best to use some builtin Decimal module in your language to handle exact arithmetic. However, since you haven't specified a language, we cannot be certain that your language may even have such a module. If it does not, here is how to go about doing so.
Consider using Binary Coded Decimal to store your values. The way it works is by restricting the values that can be stored per byte to 0 through 9 (inclusive), "wasting" the rest. You can encode a decimal representation of a number byte by byte that way. For example, 613 would become
6 -> 0000 0110
1 -> 0000 0001
3 -> 0000 0011
613 -> 0000 0110 0000 0001 0000 0011
Where each grouping of 4 digits above is a "nibble" of a byte. In practice, a packed variant is used, where two decimal digits are packed into a byte (one per nibble) to be less "wasteful". You can then implement a few methods to do your basic addition, subtract, multiplication, etc. Just iterate over an array of bytes, and perform your classic grade school addition / multiplication algorithms (keep in mind for the packed variant that you may need to pad a zero to get an even number of nibbles). You just need to keep a variable to store where the decimal point is, and remember to carry where necessary to preserve the encoding.
I'm working on a problem out of Cracking The Coding Interview which requires that I swap odd and even bits in an integer with as few instructions as possible (e.g bit 0 and 1 are swapped, bits 2 and 3 are swapped, etc.)
The author's solution revolves around using a mask to grab, in one number, the odd bits, and in another num the even bits, and then shifting them off by 1.
I get her solution, but I don't understand how she grabbed the even/odd bits. She creates two bit masks --both in hex -- for a 32 bit integer. The two are: 0xaaaaaaaa and 0x55555555. I understand she's essentially creating the equivalent of 1010101010... for a 32 bit integer in hexadecimal and then ANDing it with the original num to grab the even/odd bits respectively.
What I don't understand is why she used hex? Why not just code in 10101010101010101010101010101010? Did she use hex to reduce verbosity? And when should you use one over the other?
It's to reduce verbosity. Binary 10101010101010101010101010101010, hexadecimal 0xaaaaaaaa, and decimal 2863311530 all represent exactly the same value; they just use different bases to do so. The only reason to use one or another is for perceived readability.
Most people would clearly not want to use decimal here; it looks like an arbitrary value.
The binary is clear: alternating 1s and 0s, but with so many, it's not obvious that this is a 32-bit value, or that there isn't an adjacent pair of 1s or 0s hiding in the middle somewhere.
The hexadecimal version takes advantage of chunking. Assuming you recognize that 0x0a == 0b1010, you can mentally picture the 8 groups of 1010 in the assumed value.
Another possibility would be octal 25252525252, since... well, maybe not. You can see that something is alternating, but unless you use octal a lot, it's not clear what that alternating pattern in binary is.
I am using IEEE fixed point package in VHDL.
It works well, but I now facing a problem concerning their string representation in a test bench : I would like to dump them in a text file.
I have found that it is indeed possible to directly write ufixed or sfixed using :
write(buf, to_string(x)); --where x is either sfixed or ufixed (and buf : line)
But then I get values like 11110001.10101 (for sfixed q8.5 representation).
So my question : how to convert back these fixed point numbers to reals (and then to string) ?
The variable needs to be split into two std-logic-vector parts, the integer part can be converted to a string using standard conversion, but for the fraction part the string conversion is a bit different. For the integer part you need to use a loop and divide by 10 and convert the modulo remainder into ascii character, building up from the lower digit to the higher digit. For the fractional part it also need a loop but one needs to multiply by 10 take the floor and isolate this digit to get the corresponding character, then that integer is used to be substracted to the fraction number, etc. This is the concept, worked in MATLAB to test and making a vhdl version I will share soon. I was surprised not to find such useful function anywhere. Of course fixed-point format can vary Q(N,M) N and M can have all sorts of values, while for floating point, it is standardized.
This line:
NSLog(#"DBL_MAX: %f", DBL_MAX);
prints this very large value:
17976931348623157081452742373170435679807056752584499659891747680315726078002853876058955863276687817154045895351438246423432132688946418276846754670353751698604991057655128207624549009038932894407586850845513394230458323690322294816580855933212334827479
However, when I test a double value like this:
double test = 9999999999999999.0;
NSLog(#"test: %f", test);
I get this unexpected result:
test: 10000000000000000.000000
This appears to be the maximum number of digits that produce the expected result:
double test = 999999999999999.0;
NSLog(#"test: %f", test);
test: 999999999999999.000000
How can I work with higher positive fractions?
Thanks.
Unfortunately, I can't answer the question directly, as I don't understand what you mean by "How can I work with higher positive fractions?".
However, I can shed some light on what a floating-point number is ans what it isn't.
A floating-point number consists of:
A sign (plus or minus)
An exponent
A value (known as the "mantissa").
These are combined using a clever encoding typically into 32, 64, 80, or 128 bits. In addition, some special encodings are used to represent +-infinity, Not a Number (NaN), and +-Zero.
As the mantissa has a limited number of bits, your value can only have this number of significant bits. A really small floating-point number can represent values in the 10^-308 and and large one 10^308. However, any number can only have about 16 decimal digits.
On other words, the print-out if DBL_MAX does not corresponds the amount of information stored in the variable. For example, there is no way to represent the same number but with a ...7480 instead of ...7479 at the end.
So back to the question, in order to tell how to represent your values, you must describe what kind of values you want to represent. Are they really fractions (i.e. one integer divided by another integer), in that case you might want to represent this using two integers. If you want to represent really large values, you might want to use packages like http://gmplib.org
Floating point in C# doesn't produce accurate results all the time. There are numbers that cannot be represented in double, floats or decimals. You can improve your accuracy by using "decimal" instead of "double", but it still doesn't ensure that all numbers will be represented exactly.
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...