I was just wondering what different strategies there are for division when dealing with big numbers. By big numbers, I mean ~50 digit numbers .
e.g.
9237639100273856744937827364095876289200667937278 / 8263744826271827396629934467882946252671
When both numbers are big, long division seems to lose its usefulness...
I thought one possibility is to count through multiplications of the divisor until you go over the dividend, but if it was the dividend in the example above divided by a small number, e.g. 4, then that's a huge amount of calculations to do.
So, is there simple, clean way to do this?
What language / platform do you use? This is most likely already solved, so you don't need to implement it from scratch. E.g. Haskell has the Integer type, Java the java.math.BigInteger class, .NET the System.Numerics.BigInteger structure, etc.
If your question is really a theoretical one, I suggest you read Knuth, The Art of Computer Programming, Volume 2, Section 4.3.1. What you are looking for is called "Algorithm D" there. Here is a C implementation of that algorithm along with a short explanation:
http://hackers-delight.org.ua/059.htm
Long division is not very complicated if you are working with binary representations of your numbers and probably the most efficient algorithm.
if you don't need very exact result, you can use logarithms and exponents.
Exponent is the function f(x)=e^x, where e is a mathmaticall constant equal to 2.71828182845...
Logarithm (marked by ln) is the inverse of the exponent.
Since ln(a/b)=ln(a)-ln(b), to calculate a/b you need to:
Calculate ln(a) and ln(b) [By library function, logarithm table or other methods]
substruct them: temp=ln(a)-lb(b)
calculate the exponent e^temp
Related
I'm trying to make a calculator using arbitrary-precision maths but I can't figure out how to handle negative exponents.
What is the most efficient way to preform an operation involving n**-x?
So far i've tried 1/n**x, the problem is that I have no way of knowing how many numbers will trail the decimal point and using integers for example defeats the purpose of making a calculator using arbitrary-precision as it would restrict the size of the allowed input numbers. I was wondering if there is any other way to do this.
I'm programming in C but any method for negative exponents works honestly.
If you need to support arbitrary-precision arithmetic with negative exponents, it sounds like you might want to consider storing your number as a fraction in simplest form with the numerator and denominator each storing arbitrary-precision integers. To implement something like x-n where x = a / b, you'd end up with the number bn / an. This way, you don't need to worry about decimal digits at all, which is a good thing because most real numbers don't have finite decimal representations.
I have a question about working on very big numbers. I'm trying to run RSA algorithm and lets's pretend i have 512 bit number d and 1024 bit number n. decrypted_word = crypted_word^d mod n, isn't it? But those d and n are very large numbers! Non of standard variable types can handle my 512 bit numbers. Everywhere is written, that rsa needs 512 bit prime number at last, but how actually can i perform any mathematical operations on such a number?
And one more think. I can't use extra libraries. I generate my prime numbers with java, using BigInteger, but on my system, i have only basic variable types and STRING256 is the biggest.
Suppose your maximal integer size is 64 bit. Strings are not that useful for doing math in most languages, so disregard string types. Now choose an integer of half that size, i.e. 32 bit. An array of these can be interpreted as digits of a number in base 232. With these, you can do long addition and multiplication, just like you are used to with base 10 and pen and paper. In each elementary step, you combine two 32-bit quantities, to produce both a 32-bit result and possibly some carry. If you do the elementary operation in 64-bit arithmetic, you'll have both of these as part of a single 64-bit variable, which you'll then have to split into the 32-bit result digit (via bit mask or simple truncating cast) and the remaining carry (via bit shift).
Division is harder. But if the divisor is known, then you may get away with doing a division by constant using multiplication instead. Consider an example: division by 7. The inverse of 7 is 1/7=0.142857…. So you can multiply by that to obtain the same result. Obviously we don't want to do any floating point math here. But you can also simply multiply by 14286 then omit the last six digits of the result. This will be exactly the right result if your dividend is small enough. How small? Well, you compute x/7 as x*14286/100000, so the error will be x*(14286/100000 - 1/7)=x/350000 so you are on the safe side as long as x<350000. As long as the modulus in your RSA setup is known, i.e. as long as the key pair remains the same, you can use this approach to do integer division, and can also use that to compute the remainder. Remember to use base 232 instead of base 10, though, and check how many digits you need for the inverse constant.
There is an alternative you might want to consider, to do modulo reduction more easily, perhaps even if n is variable. Instead of expressing your remainders as numbers 0 through n-1, you could also use 21024-n through 21024-1. So if your initial number is smaller than 21024-n, you add n to convert to this new encoding. The benefit of this is that you can do the reduction step without performing any division at all. 21024 is equivalent to 21024-n in this setup, so an elementary modulo reduction would start by splitting some number into its lower 1024 bits and its higher rest. The higher rest will be right-shifted by 1024 bits (which is just a change in your array indexing), then multiplied by 21024-n and finally added to the lower part. You'll have to do this until you can be sure that the result has no more than 1024 bits. How often that is depends on n, so for fixed n you can precompute that (and for large n I'd expect it to be two reduction steps after addition but hree steps after multiplication, but please double-check that) whereas for variable n you'll have to check at runtime. At the very end, you can go back to the usual representation: if the result is not smaller than n, subtract n. All of this should work as described if n>2512. If not, i.e. if the top bit of your modulus is zero, then you might have to make further adjustments. Haven't thought this through, since I only used this approach for fixed moduli close to a power of two so far.
Now for that exponentiation. I very much suggest you do the binary approach for that. When computing xd, you start with x, x2=x*x, x4=x2*x2, x8=…, i.e. you compute all power-of-two exponents. You also maintain some intermediate result, which you initialize to one. In every step, if the corresponding bit is set in the exponent d, then you multiply the corresponding power into that intermediate result. So let's say you have d=11. Then you'd compute 1*x1*x2*x8 because d=11=1+2+8=10112. That way, you'll need only about 1024 multiplications max if your exponent has 512 bits. Half of them for the powers-of-two exponentiation, the other to combine the right powers of two. Every single multiplication in all of this should be immediately followed by a modulo reduction, to keep memory requirements low.
Note that the speed of the above exponentiation process will, in this simple form, depend on how many bits in d are actually set. So this might open up a side channel attack which might give an attacker access to information about d. But if you are worried about side channel attacks, then you really should have an expert develop your implementation, because I guess there might be more of those that I didn't think about.
You may write some macros you may execute under Microsoft for functions like +, -, x, /, modulo, x power y which work generally for any integer of less than ten or hundred thousand digits (the practical --not theoretical-- limit being the internal memory of your CPU). Please note the logic is exactly the same as the one you got at elementary school.
E.g.: p= 1819181918953471 divider of (2^8091) - 1, q = ((2^8091) - 1)/p, mod(2^8043 ; q ) = 23322504995859448929764248735216052746508873363163717902048355336760940697615990871589728765508813434665732804031928045448582775940475126837880519641309018668592622533434745187004918392715442874493425444385093718605461240482371261514886704075186619878194235490396202667733422641436251739877125473437191453772352527250063213916768204844936898278633350886662141141963562157184401647467451404036455043333801666890925659608198009284637923691723589801130623143981948238440635691182121543342187092677259674911744400973454032209502359935457437167937310250876002326101738107930637025183950650821770087660200075266862075383130669519130999029920527656234911392421991471757068187747362854148720728923205534341236146499449910896530359729077300366804846439225483086901484209333236595803263313219725469715699546041162923522784170350104589716544529751439438021914727772620391262534105599688603950923321008883179433474898034318285889129115556541479670761040388075352934137326883287245821888999474421001155721566547813970496809555996313854631137490774297564881901877687628176106771918206945434350873509679638109887831932279470631097604018939855788990542627072626049281784152807097659485238838560958316888238137237548590528450890328780080286844038796325101488977988549639523988002825055286469740227842388538751870971691617543141658142313059934326924867846151749777575279310394296562191530602817014549464614253886843832645946866466362950484629554258855714401785472987727841040805816224413657036499959117701249028435191327757276644272944743479296268749828927565559951441945143269656866355210310482235520220580213533425016298993903615753714343456014577479225435915031225863551911605117029393085632947373872635330181718820669836830147312948966028682960518225213960218867207825417830016281036121959384707391718333892849665248512802926601676251199711698978725399048954325887410317060400620412797240129787158839164969382498537742579233544463501470239575760940937130926062252501116458281610468726777710383038372260777522143500312913040987942762244940009811450966646527814576364565964518092955053720983465333258335601691477534154940549197873199633313223848155047098569827560014018412679602636286195283270106917742919383395056306107175539370483171915774381614222806960872813575048014729965930007408532959309197608469115633821869206793759322044599554551057140046156235152048507130125695763956991351137040435703946195318000567664233417843805257728.
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wpjo (willibrord oomen on academia.edu)
I have been testing the waters of competitive programming and I have already seen this statement mentioned a lot of times:
Print the result modulo 109 + 7
Now I can figure out that this is some way of preventing overflow of digits when dealing with very large numbers. But how and why does it work? I would be grateful if someone could explain the mathematical reasoning behind this.
Many contest questions ask you to compute some very, very large number (say, the number of permutations of an 150-element sequence containing some large number of duplicates). Many programming languages don't natively support arbitrary-precision arithmetic, so in the interest of fairness it makes sense for those contests not to ask you for the exact value. The challenge, then, is the following: how can the contest site know when you have the right answer given that you can't exactly compute it?
One initially appealing option would be to just ask for the answer modulo some large power of two (say, 232 or 264) so that competitors working in languages like C or C++ could just use uint32_t or uint64_ts to do all the computations, letting overflows occur normally, and then submit the results. However, this isn't particularly desirable. Suppose, for example, that the question is the following:
Compute 10,000!
This number is staggeringly huge and is way too big to fit into a 32-bit or 64-bit unsigned integer. However, if you just want to get the answer modulo 232 or 264, you could just use this program:
#include <stdio.h>
int main() {
puts("0");
}
The reason for this is that 10,000! is the product of at least 5,000 even numbers, so one of its factors is 25,000. Therefore, if you just want the answer modulo 232 or 264, you don't actually have to compute it at all. You can just say that the result is 0 mod 232 or 264.
The problem here is that working modulo 232 or 264 is troublesome if the resulting answer is cleanly divisible by either of those numbers. However, if we work modulo a large prime number, then this trick wouldn't work. As an example, the number 7,897,987 is prime. If you try to compute 10,000! mod 7,897,987, then you can't just say "the answer is 0" because none of the numbers multiplied together in 10,000! are divisors of 7,897,987. You'd actually have to do some work to figure out what this number is modulo that large prime. More generally, working modulo a large prime usually requires you to compute the actual answer modulo that large prime, rather than using number-theoretic tricks to skip all the work entirely.
So why work modulo 1,000,000,007? This number happens to be prime (so it's good to use as a modulus) and it's less than 231 - 1, the largest possible value you can fit in a signed 32-bit integer. The signedness is nice here because in some languages (like Java) there are no unsigned integer types and the default integer type is a 32-bit signed integer. This means that you can work modulo 1,000,000,007 without risking an integer overflow.
To summarize:
Working modulo a large prime makes it likely that if your program produces the correct output, it actually did some calculation and did so correctly.
Working modulo 1,000,000,007 allows a large number of languages to use their built-in integer types to store and calculate the result.
Hope this helps!
(I'm not sure whether I should post this problem on this site or on the math site. Please feel free to migrate this post if necessary.)
My problem at hand is that given a value of k I'd like to numerically compute a rational function of nonlinear polynomials in k which looks like the following: (sorry I don't know how to typeset equations here...)
where {a_0, ..., a_N; b_0, ..., b_N} are complex constants, {u_0, ..., u_N, v_0, ..., v_N} are real constants and i is the imaginary number. I learned from Numerical Recipes that there are whole bunch of ways to compute polynomials quickly, in the meanwhile keeping the rounding error small enough, if all coefficients were constant. But I do not think those ideas are useful in my case since the exponential prefactors also depend on k.
Currently I calculate it in a brute force way in C with complex.h (this is just a pseudo code):
double complex function(double k)
{
return (a_0+a_1*cexp(I*u_1*k)*k+a_2*cexp(I*u_2*k)*k*k+...)/(b_0+b_1*cexp(I*v_1*k)*k+v_2*cexp(I*v_2*k)*k*k+...);
}
However when the number of calls of function increases (because this is just a part of my real calculation), it is very slow and inaccurate (only 6 valid digits). I appreciate any comments and/or suggestions.
I trust that this isn't a homework assignment!
Normally the trick is to use a loop add the next coefficient to the running sum, and multiply by k. However, in your case, I think the "e" term in the coefficient is going to overwhelm any savings by factoring out k. You can still do it, but the savings will probably be small.
Is u_i a constant? Depending on how many times you need to run this formula, maybe you could premultiply u_i * k (unless k changes each run). It's been so many decades since I took a Numerical Analysis course that I have only vague recollections of the tricks of the trade. Let's see... is e^(i*u_i*k) the same as (e^(i*u_i))^k? I don't remember the rules on imaginary numbers, or whether you'll save anything since you've got a real^real (assuming k is real) anyway (internally done using e^power).
If you're getting only 6 digits, that suggests that your math, and maybe your library, is working in single precision (32 bit) reals. Check your library and check your declarations that you are using at least double precision (64 bit) reals everywhere.
I am attempting to generate QR codes on an extremely limited embedded platform. Everything in the specification seems fairly straightforward except for generating the error correction codewords. I have looked at a bunch of existing implementations, and they all try to implement a bunch of polynomial math that goes straight over my head, particularly with regards to the Galois fields. The most straightforward way I can see, both in mathematical complexity and in memory requirements is a circuit concept that is laid out in the spec itself:
With their description, I am fairly confident I could implement this with the exception of the parts labeled GF(256) addition and GF(256) Multiplication.
They offer this help:
The polynomial arithmetic for QR Code shall be calculated using bit-wise modulo 2 arithmetic and byte-wise
modulo 100011101 arithmetic. This is a Galois field of 2^8
with 100011101 representing the field's prime modulus
polynomial x^8+x^4+x^3+x^2+1.
which is all pretty much greek to me.
So my question is this: What is the easiest way to perform addition and multiplication in this kind of Galois field arithmetic? Assume both input numbers are 8 bits wide, and my output needs to be 8 bits wide also. Several implementations precalculate, or hardcode in two lookup tables to help with this, but I am not sure how those are calculated, or how I would use them in this situation. I would rather not take the 512 byte memory hit for the two tables, but it really depends on what the alternative is. I really just need help understanding how to do a single multiplication and addition operation in this circuit.
In practice only one table is needed. That would be for the GP(256) multiply. Note that all arithmetic is carry-less, meaning that there is no carry-propagation.
Addition and subtraction without carry is equivalent to an xor.
So in GF(256), a + b and a - b are both equivalent to a xor b.
GF(256) multiplication is also carry-less, and can be done using carry-less multiplication in a similar way with carry-less addition/subtraction. This can be done efficiently with hardware support via say Intel's CLMUL instruction set.
However, the hard part, is reducing the modulo 100011101. In normal integer division, you do it using a series of compare/subtract steps. In GF(256), you do it in a nearly identical manner using a series of compare/xor steps.
In fact, it's bad enough where it's still faster to just precompute all 256 x 256 multiplies and put them into a 65536-entry look-up table.
page 3 of the following pdf has a pretty good reference on GF256 arithmetic:
http://www.eecs.harvard.edu/~michaelm/CS222/eccnotes.pdf
(I'm following up on the pointer to zxing in the first answer, since I'm the author.)
The answer about addition is exactly right; that's why working in this field is convenient on a computer.
See http://code.google.com/p/zxing/source/browse/trunk/core/src/com/google/zxing/common/reedsolomon/GenericGF.java
Yes multiplication works, and is for GF256. a * b is really the same as exp(log(a) + log(b)). And because GF256 has only 256 elements, there are only 255 unique powers of "x", and same for log. So these are easy to put in a lookup table. The tables would "wrap around" at 256, so that is why you see the "% size". "/ size" is slightly harder to explain in a sentence -- it's because really 1-255 "wrap around", not 0-255. So it's not quite just a simple modulus that's needed.
The final piece perhaps is how you reduce modulo an irreducible polynomial. The irreducibly polynomial is x^8 plus some lower-power terms, right -- call it I(x) = x^8 + R(x). And the polynomial is congruent to 0 in the field, by definition; I(x) == 0. So x^8 == -R(x). And, conveniently, addition and subtraction are the same, so x^8 == -R(x) == R(x).
The only time we need to reduce higher-power polynomials is when constructing the exponents table. You just keep multiplying by x (which is a shift left) until it gets too big -- gets an x^8 term. But x^8 is the same as R(x). So you take out the x^8 and add in R(x). R(x) merely has powers up to x^7 so it's all in a byte still, all in GF(256). And you know how to add in this field.
Helps?