There is a variable first_variable which is always a mod of some number, mod_value.
In every step first_variable is multiplied with some number second_variable.
And the range of all three variables is from 1 to 10^18.
For that I build a formula,
first_variable = ((first_variable%mod_value)*(second_variable%mod_value))%mod_value
But this gives a wrong answer,
For example, If first_variable and second_variable is (10^18)-1 and mod_value = 10^18
Please suggest me method, so that first_variable will always give right answer.
Seems you are using a runtime where arithmetic is implemented using 64-bit integers. You can check this using multipliers like 2^32: if their product is 0, my guess is true. In that case, you should switch to an arbitrary long arithmetic implementation, or at least one that is much longer than the current one. E.g. Python supports integers up to 2^1016 (256^127), same for Erlang.
I've seen in comments you use C++. If so, look for GMP library and analogs. Or, if 128 bits is enough, modern GCC support it through own library.
This is basically overflows, so you should either use different value for mod_value (up to 10^9) or limit the range for first value and second value.
Your number is O(10^36) which is O(2^108) which cannot fit in any primitive data type in languages like java or C++. Use BigInt in C++ or Java or use numpy in python to get over it.
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 want to know the exact method used to generate random numbers in gcc compiler of linux.
I know that the Linear Congruental Generator is used to generate random numbers in gcc which has general formula:
X(n+1) = (a* X(n) +c) mod m
and I came to know that the general formula used, has these constant values as given in wikipedia :
http://en.wikipedia.org/wiki/Linear_congruential_generator
which are m=2^3, a =1103515245 and 12345
But the results obtained by putting these constants do not match with the result obtained by rand() function in gcc.
Can someone please help me where i am wrong, or is there something which i dont know.
Do the numbers match this:
http://www.mathstat.dal.ca/~selinger/random/
Otherwise, the source code is here:
http://sourceware.org/git/?p=glibc.git;a=blob;f=stdlib/rand.c;h=92916e833f7fc94ac16a2bd047c33f8a6ef6ec49;hb=HEAD
which leads to here:
http://sourceware.org/git/?p=glibc.git;a=blob;f=stdlib/random.c;h=ff6bdd2b5d5a8f7633a914282f4c6ab1991df0cf;hb=HEAD
There looks like a call to DES::SetKey(unsigned long long int); in the stdlib.h don't know if this is anything to do with the Random Generator function but you never know as there is a much faster version of the DES encryption/decryption algorithm which was the one used once used in the old version of GNUPG which actually used DES as the encryption/decryption system of choice so this is what they might be using to generate random numbers.
This may not be a programming question but it's a problem that arised recently at work. Some background: big C development with special interest in performance.
I've a set of integers and want to test the membership of another given integer. I would love to implement an algorithm that can check it with a minimal set of algebraic functions, using only a integer to represent the whole space of integers contained in the first set.
I've tried a composite Cantor pairing function for instance, but with a 30 element set it seems too complicated, and focusing in performance it makes no sense. I played with some operations, like XORing and negating, but it gives me low estimations on membership. Then I tried with successions of additions and finally got lost.
Any ideas?
For sets of unsigned long of size 30, the following is one fairly obvious way to do it:
store each set as a sorted array, 30 * sizeof(unsigned long) bytes per set.
to look up an integer, do a few steps of a binary search, followed by a linear search (profile in order to figure out how many steps of binary search is best - my wild guess is 2 steps, but you might find out different, and of course if you test bsearch and it's fast enough, you can just use it).
So the next question is why you want a big-maths solution, which will tell me what's wrong with this solution other than "it is insufficiently pleasing".
I suspect that any big-math solution will be slower than this. A single arithmetic operation on an N-digit number takes at least linear time in N. A single number to represent a set can't be very much smaller than the elements of the set laid end to end with a separator in between. So even a linear search in the set is about as fast as a single arithmetic operation on a big number. With the possible exception of a Goedel representation, which could do it in one division once you've found the nth prime number, any clever mathematical representation of sets is going to take multiple arithmetic operations to establish membership.
Note also that there are two different reasons you might care about the performance of "look up an integer in a set":
You are looking up lots of different integers in a single set, in which case you might be able to go faster by constructing a custom lookup function for that data. Of course in C that means you need either (a) a simple virtual machine to execute that "function", or (b) runtime code generation, or (c) to know the set at compile time. None of which is necessarily easy.
You are looking up the same integer in lots of different sets (to get a sequence of all the sets it belongs to), in which case you might benefit from a combined representation of all the sets you care about, rather than considering each set separately.
I suppose that very occasionally, you might be looking up lots of different integers, each in a different set, and so neither of the reasons applies. If this is one of them, you can ignore that stuff.
One good start is to try Bloom Filters.
Basically, it's a probabilistic data structure that gives you no false negative, but some false positive. So when an integer matches a bloom filter, you then have to check if it really matches the set, but it's a big speedup by reducing a lot the number of sets to check.
if i'd understood your correctly, python example:
>>> a=[1,2,3,4,5,6,7,8,9,0]
>>>
>>>
>>> len_a = len(a)
>>> b = [1]
>>> if len(set(a) - set(b)) < len_a:
... print 'this integer exists in set'
...
this integer exists in set
>>>
math base: http://en.wikipedia.org/wiki/Euler_diagram
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
How do I quickly generate a random prime number, that is for sure 1024 bit long?
Generate 1024 random bits. Use a random source that is strong enough for your intended purpose.
Set the highest and lowest bits to 1. This makes sure there are no leading zeros (the prime candidate is big enough) and it is not an even number (definitely not prime).
Test for primality. If it's not a prime, go back to 1.
Alternatively, use a library function that generates primes for you.
Use a library function, such as OpenSSL. There's no need to write this yourself.
Example: http://ardoino.com/7-maths-openssl-primes-random/
The above link doesn't work so you can use this archive link.
1024 is a lot.
Are you sure a probabilistic prime won't do?
Probabilistic prime generator is part of JDK
You do not specify a context/language/platform.. if you'd like to use unix/linux-like system and shell, you might consider a solution involving OpenSSL version >= 1.0.0:
$ openssl prime -generate -bits 1024
140750877582727333214379261853877378646889234118675380673028200387281415297520423589261211081966230040412916644372766351028035798201654335110081318739796178745233127842988596480299276295476504358587725867882394416543075082108266054273016211760684113070285409887820598314292803190900634009988950624354964653677
If you got the same result, something is very wrong with the universe.
Add -hex option if you like hexadecimal system.
To trade memory for speed you could just generate them and store them in a list and then randomly pick one.
Edit:
Naturally you can't generate them all so the best you could achieve is pseudo randomness at a high memory cost. Also this isn't good if you want it for security.
In PARI/GP:
randomprime([2^1023,2^1024])
If you'd like to do this in 'library mode'
#include <pari/pari.h>
// ...
randomprime(mkvec2(int2u(1023), int2u(1024)))