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.NET Forum / Languages / Managed C++ / June 2007Random numbers
I would like to be able to create a random number generator that produces evenly distributed random numbers up to given number. For example, I would like to pick a random number less than 100000, or between 0 and 99999 (inclusive). Further, the I want the range to be a variable. Concretely, I would like to create the following method: unsigned long Random( unsigned long num ) { // return a uniformly distributed random number R in the range: 0 <= R <= (num-1) } I also would like this to be as executionally fast as possible. I have been trying to to this using rand() as my basis, but everything I try ends up with one problem or another (e.g., outside of range, not evenly distributed, execution too long). I'm using VS C++ 2005 Express Edition (managed /clr pure), and was hoping that there is a better alternative to rand( ) as the basis of random numbers, since this always generates an evenly distributed number from 0 to RAND_MAX (0x7fff), which is not easily adapted to the task I've described above... If there is a cheap third-party class I could buy to do this, I'd be interested in that too... Thanx in advance for responses! :) OK, I'm an idiot (feel free to throw stones....hehe). Here I am looking for some class in .NET that produces random numbers, and it never occurs to me to look for a class named 'Random'. Which, of course, exists; and, of course, does exactly what I want! :) >I would like to be able to create a random number generator that produces >evenly distributed random numbers up to given number. [quoted text clipped - 26 lines] > > Thanx in advance for responses! :) >I would like to be able to create a random number generator that produces >evenly distributed random numbers up to given number. A quick, though, not perfect, method is to compute a random number with rand() and then take the remainder of a division. e.g. rand() % 10 yields numbers in {0 ... 9}. if you want {1 ... 10} add 1. Of course, since the highest number rand() returns is RAND_MAX and that is not evenly divisible by 10, your numbers won't have an even distribution. Naively, I'd suggest in that case that you pick your own maximum such that it is evenly divisble by the size of the range in question and discard anything above it. If you need numbers larger than RAND_MAX you can put two 16 bit numbers together to make a 32 bit number. See the MAKELONG macro for example. Of course, this is the kind of problem that has no doubt been faced and solved thousands of times, so if you need a "good" random number generator I suggest that you search for an algorithm or make friends with a PhD in math. ;-) Regards, Will www.ivrforbeginners.com Hi William, Thanx! Yeah, I know of the method you speak. There are, in fact two basic methods I tried, but each fails my 'criteria' for one reason or another. For the purposes of the dicussion below, the range of numbers to find a random number in is 0 to (max-1). The method you mentioned, ala getting a big random number and then modulo'ing it with the 'max' value, does always return a random number in range and does span the whole range, but unless the 'max' is a integral divisor of RAND_MAX (which means it has to be a power of 2) the results will not be evenly distributed (ala, the lower numbers will be produced more than the upper numbers). The other basic method is to keep getting big random numbers until one falls in range, rejecting any out of range. As long as the big random numbers can range from 0 to outside 'max' (and are evely distributed themselves), this does indeed pick an evenly distributed random number in range, and does span the range. But it is execution-time VERY non-deterministic, and in fact can take a very long time (computer exectution-wise) to deal with some 'max' values. For example, trying to get a result of 0 or 1 (i.e., max = 2) and using big random numbers up to 64K will find one in range on average in only 1 in 32K tries, and thus on average would take many failed iterations to eventually get one in range. Thus I tried separating cases based on 'max', such as masking a big random number with 0xF for 'max's less than 16. This works, but one ends up with many many many case, and it gets so complex, the mechanism to figure out which method to use becomes intrusive. Thus, the solution I found was a cheat....sort of....hehe...it turns out there exists a Random class that has a Next(max) method that produces a non-negative random number less than 'max', where 'max' can be 32-bits in size (i.e., exactly what I'm looking for!!!)... I've impleneted it, but have yet to see if the MS class is indeed uniformly distributed. I will do tests to be sure, but I was VERY glad to find this class! [==P==] >>I would like to be able to create a random number generator that produces >>evenly distributed random numbers up to given number. [quoted text clipped - 26 lines] > Will > www.ivrforbeginners.com Hi, > Yeah, I know of the method you speak. There are, in fact two basic > methods I tried, but each fails my 'criteria' for one reason or [quoted text clipped - 15 lines] > random number in range, and does span the range. But it is > execution-time VERY non-deterministic Another one would be long nDiv = RAND_MAX / yourMax; long n = (rand() / nDiv) % yourMax; So if RAND_MAX == 100 and you want 0..9 then you get 0..9 as 0 and 10..19 as 1 and so on. You might get yourMax as result so I added the modulo operation to address that. > it turns > out there exists a Random class that has a Next(max) method that [quoted text clipped - 4 lines] > uniformly distributed. I will do tests to be sure, but I was VERY > glad to find this class! Im curious of your findings. -- SvenC >Another one would be > [quoted text clipped - 7 lines] >You might get yourMax as result so I added the modulo operation to address >that. That produces an uneven distribution a lot of the time. See the link to the C FAQ I posted in my reply to Peter for an approach that doesn't. (To see the problem in your approach, consider RAND_MAX = 100 and yourMax = 99. Then nDiv is 1, rand()/nDiv is 0..99, and 99%99 is 0. So you will have 0 appearing twice as often as any other number.)  SignatureDoug Harrison Visual C++ MVP >> Another one would be >> [quoted text clipped - 14 lines] > is 0. So you will have 0 appearing twice as often as any other > number.) Yes, figured that out after thinking again about my post. I'll have a look at the FAQ. -- SvenC >The other basic method is to keep getting big random numbers until one falls >in range, rejecting any out of range. As long as the big random numbers can [quoted text clipped - 6 lines] >1 in 32K tries, and thus on average would take many failed iterations to >eventually get one in range. You don't need to do that much iteration, or for max = 2 with odd RAND_MAX, any at all. See: http://c-faq.com/lib/randrange.html SignatureDoug Harrison Visual C++ MVP Keep in mind, my desire is to produce random rumber is a ranges GREATER than RAND_MAX, and that the range be sometihng I can establish on the fly (i.e., it's a variable). I did experiements on the innate Random class, and it works just fine. It can be seeded (I use current 'time'), and can range from 0 to ANY number up to 31 bits (the return value is signed). 31-bits gets me ranges up to 2 billion or so, which is (more than) just fine for my purposes! : ) All that is required is the following: Random^ R = gcnew Random( /* int seed if desired*/ ) ; int r = R->Next( max ) ; // 0 <= r < max [==P==] >>The other basic method is to keep getting big random numbers until one >>falls [quoted text clipped - 17 lines] > > http://c-faq.com/lib/randrange.html > <snipped> > [quoted text clipped - 8 lines] > 1 in 32K tries, and thus on average would take many failed iterations to > eventually get one in range. I would say this method is not guaranteed to ever finish. > ... > > The other basic method is to keep getting big random numbers until one falls > > in range, rejecting any out of range. As long as the big random numbers can [quoted text clipped - 8 lines] > > I would say this method is not guaranteed to ever finish. If the numbers are evenly distributed, then it's guaranteed that each number within the range will occur *eventually*, isn't it? Put it this way - if it's a PRNG, it will have a cycle eventually. If it cycles without ever hitting a particular number, the distribution cannot be said to be even. Therefore so long as at least one number within the "target" range is within the range of the PRNG, it will finish at some point. If it's a truly random number generator that makes things harder... SignatureJon Skeet - <skeet@pobox.com> http://www.pobox.com/~skeet Blog: http://www.msmvps.com/jon.skeet If replying to the group, please do not mail me too >> <snipped> >> [quoted text clipped - 12 lines] > > > ... Actually, if the generator of the big random numbers is a truly evenly distributed one, then it is guaranteed to EVENTUALLY (i.e., in FINTE time) produce a value in any sub-range it spans. This would include a range from 0 to any value smaller than its max. Hence, it is GUARANTEED to finish in finite time, but like I said, in NON_DETERMINISTIC finite time. It could happen the first attempt, and there is no number of attempts for which it will definitiely happen within. But as the number of attempts increase, the probability of continued failure goes to zero (viewed as a probability over the entire history, not individual attempt probability, which is assumed in this example to be fixed). Specifically, the probability it will happen in the first N attempts is: p = 1 - (1 - (1/32K)) ^ N That is, the compliment of the probability it WON'T happen (which is the compliment of it happening) in N attempts (which is the product of the probabilities of the individual attempts failing, hence the exponent of N). Note that as N gets larger the second term goes to zero, so the probability of succes goes to 1... [==P==] > Hi William, > [quoted text clipped - 23 lines] > 64K will find one in range on average in only 1 in 32K tries, and thus on > average would take many failed iterations to eventually get one in range. Combine these two ideas. If you desire a random number [0, N): Let M = RAND_MAX / N (integral division) Now, if rand() returns either a number less than N * M, take the modulus, resulting in an even distribution if rand() returns >= N * M, try again Now the probability of needing to retry is maximum at N = (RAND_MAX+1)/2 for odd RAND_MAX and is never greater than 0.5 for any N. Cache the computed value of N*M for efficiency. > Thus I tried separating cases based on 'max', such as masking a big random > number with 0xF for 'max's less than 16. This works, but one ends up with [quoted text clipped - 43 lines] >> Will >> www.ivrforbeginners.com >I would like to be able to create a random number generator that >produces evenly distributed random numbers up to given number. Computer-generated random numbers tend to not be very good, unless you REALLY know what you're doing. rand() in the C standard library tends to be barely usable-- on a lot of systems, it only has 16-bit accuracy. You really should stick with finding a better random number generator and use it -- the Mersenne Twister from http://www.math.sci.hiroshima-u.ac.jp/~m-mat/MT/emt.html has gotten a bunch of good reviews, and is quite free (BSD license, which basically says "feel free to use it, even in commercial apps." That site has C code available, and allows you to get a random # up to a range. Nathan Mates -- <*> Nathan Mates - personal webpage http://www.visi.com/~nathan/ # Programmer at Pandemic Studios -- http://www.pandemicstudios.com/ # NOT speaking for Pandemic Studios. "Care not what the neighbors # think. What are the facts, and to how many decimal places?" -R.A. Heinlein
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