How to get a larger random number from the c function rand() - c

I am in a coding environment where I only have access to some most basic c functions. #include'ing other lib is not feasible.
In this environment, I can call rand() which gives me a random number between 0 to 32767 inclusively(I think) in my environment. Is the following code the right logic to get a larger random number that is evenly distributed like/as rand()?
rnum = rand() * (32767 + 1) + rand();

rnum = (rand() << 15) | rand() might be faster, but if you need good quality random numbers you should look for an external library. Built-in random functions are generally adequate only for the simplest applications.

static unsigned long next = 1;
int my_rand(void) {
next = next * 1103515245 + 12345;
return((unsigned)(next/65536) % (RAND_MAX+1));
}
void my_srand(unsigned int seed) {
next = seed;
}
on linux
#define RAND_MAX 2147483647
your environment RAND_MAX is probably 32767
reference: http://en.wikipedia.org/wiki/Linear_congruential_generator
if you are not memory constrained you can look also at http://en.wikipedia.org/wiki/Mersenne_twister the code is embeddable as like as the example above

Before using ANY random number generator for a non-trivial use, it should be tested extensively. Here is a link to one article on this subject.
A fun background on the weaknesses of even true random numbers is available in the various histories on the code-breakers at Bletchley Park in WWII. The Germans and Hitler for most of the war assumed their codes were unbreakable because of being encrypted with random numbers, whereas the British were breaking all of these codes completely within months because of various weaknesses in the German implementation of "randomness". Many codes were sufficiently "twisted", if not outright broken, within a few days or months, sufficiently to be of use even though not completely broken.

Besides the other excellent solutions proposed here, you can also do a power expansion in RAND_MAX, truncate at a user defined MY_RAND_MAX, and discard solutions which would cause uniformity to be destroyed.
int myrand(int MY_RAND_MAX)
{
int j = 0, num = 0;
// Generate digits for expansion in powers of RAND_MAX, up to acceptable range.
while (pow(RAND_MAX + 1, j) <= MY_RAND_MAX){
num = num + rand() * (int)pow(RAND_MAX + 1, j);
j++;
}
//compute remainder to determine truncation
int remainder = ((int)pow(RAND_MAX + 1, j)) % (MY_RAND_MAX + 1);
// If number is within accepted bounds, mod and return
if (num <= ( (int)pow(RAND_MAX + 1, j) - remainder ) ){
return (num % (MY_RAND_MAX + 1));
// Else, if number generated exceeds allowed bounds, rerun method.
}else if (num > ( (int)pow(RAND_MAX + 1, j) - remainder ) ){
return myrand(MY_RAND_MAX);
}else{
exit(-1);
}
}
You can check empirically that this method gives you a statistically uniform output in the given range.
I've done so for several trials with different ranges, each with a sample size of 100000, and got an agreement between the variance of the sample and the expected variance to at least 3 sig. fig.s each time.
p.s. I am not a coder, but a mathematician/physicist who has recently learnt to code, so any and all feedback on the quality of my code would be appreciated.

Related

Random integers in C, how bad is rand()%N compared to integer arithmetic? What are its flaws?

EDIT:
My question is: rand()%N is considered very bad, whereas the use of integer arithmetic is considered superior, but I cannot see the difference between the two.
People always mention:
low bits are not random in rand()%N,
rand()%N is very predictable,
you can use it for games but not for cryptography
Can someone explain if any of these points are the case here and how to see that?
The idea of the non-randomness of the lower bits is something that should make the PE of the two cases that I show differ, but it's not the case.
I guess many like me would always avoid using rand(), or rand()%N because we've been always taught that it is pretty bad. I was curious to see how "wrong" random integers generated with c rand()%N effectively are. This is also a follow up to Ryan Reich's answer in How to generate a random integer number from within a range.
The explanation there sounds very convincing, to be honest; nevertheless, I thought I’d give it a try. So, I compare the distributions in a VERY naive way. I run both random generators for different numbers of samples and domains. I didn't see the point of computing a density instead of histograms, so I just computed histograms and, just by looking, I would say they both look just as uniform. Regarding the other point that was raised, about the actual randomness (despite being uniformly distributed). I — again naively —compute the permutation entropy for these runs, which are the same for both sample sets, which tell us that there's no difference between both regarding the ordering of the occurrence.
So, for many purposes, it seems to me that rand()%N would be just fine, how can we see their flaws?
Here I show you a very simple, inefficient and not very elegant (but I think correct) way of computing these samples and get the histograms together with the permutation entropies.
I show plots for domains (0,i) with i in {5,10,25,50,100} for different number of samples:
There's not much to see in the code I guess, so I will leave both the C and the matlab code for replication purposes.
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
int main(int argc, char *argv[]){
unsigned long max = atoi(argv[2]);
int samples=atoi(argv[3]);
srand(time(NULL));
if(atoi(argv[1])==1){
for(int i=0;i<samples;++i)
printf("%ld\n",rand()%(max+1));
}else{
for(int i=0;i<samples;++i){
unsigned long
num_bins = (unsigned long) max + 1,
num_rand = (unsigned long) RAND_MAX + 1,
bin_size = num_rand / num_bins,
defect = num_rand % num_bins;
long x;
do {
x = rand();
}
while (num_rand - defect <= (unsigned long)x);
printf("%ld\n",x/bin_size);
}
}
return 0;
}
And here is the Matlab code to plot this and compute the PEs (the recursion for the permutations I took it from: https://www.mathworks.com/matlabcentral/answers/308255-how-to-generate-all-possible-permutations-without-using-the-function-perms-randperm):
system('gcc randomTest.c -o randomTest.exe;');
max = 100;
samples = max*10000;
trials = 200;
system(['./randomTest.exe 1 ' num2str(max) ' ' num2str(samples) ' > file1'])
system(['./randomTest.exe 2 ' num2str(max) ' ' num2str(samples) ' > file2'])
a1=load('file1');
a2=load('file2');
uni = figure(1);
title(['Samples: ' num2str(samples)])
subplot(1,3,1)
h1 = histogram(a1,max+1);
title('rand%(max+1)')
subplot(1,3,2)
h2 = histogram(a2,max+1);
title('Integer arithmetic')
as=[a1,a2];
ns=3:8;
H = nan(numel(ns),size(as,2));
for op=1:size(as,2)
x = as(:,op);
for n=ns
sequenceOcurrence = zeros(1,factorial(n));
sequences = myperms(1:n);
sequencesArrayIdx = sum(sequences.*10.^(size(sequences,2)-1:-1:0),2);
for i=1:numel(x)-n
[~,sequenceOrder] = sort(x(i:i+n-1));
out = sequenceOrder'*10.^(numel(sequenceOrder)-1:-1:0).';
sequenceOcurrence(sequencesArrayIdx == out) = sequenceOcurrence(sequencesArrayIdx == out) + 1;
end
chunks = length(x) - n + 1;
ps = sequenceOcurrence/chunks;
hh = sum(ps(logical(ps)).*log2(ps(logical(ps))));
H(n,op) = hh/log2(factorial(n));
end
end
subplot(1,3,3)
plot(ns,H(ns,:),'--*','linewidth',2)
ylabel('PE')
xlabel('Sequence length')
filename = ['all_' num2str(max) '_' num2str(samples) ];
export_fig(filename)
Due to the way modulo arithmetic works if N is significant compared to RAND_MAX doing %N will make it so you're considerably more likely to get some values than others. Imagine RAND_MAX is 12, and N is 9. If the distribution is good then the chances of getting one of 0, 1, or 2 is 0.5, and the chances of getting one of 3, 4, 5, 6, 7, 8 is 0.5. The result being that you're twice as likely to get a 0 instead of a 4. If N is an exact divider of RAND_MAX this distribution problem doesn't happen, and if N is very small compared to RAND_MAX the issue becomes less noticeable. RAND_MAX may not be a particularly large value (maybe 2^15 - 1), making this problem worse than you may expect. The alternative of doing (rand() * n) / (RAND_MAX + 1) also doesn't give an even distribution, however, it will be every mth value (for some m) that will be more likely to occur rather than the more likely values all being at the low end of the distribution.
If N is 75% of RAND_MAX then the values in the bottom third of your distribution are twice as likely as the values in the top two thirds (as this is where the extra values map to)
The quality of rand() will depend on the implementation of the system that you're on. I believe that some systems have had very poor implementation, OS Xs man pages declare rand obsolete. The Debian man page says the following:
The versions of rand() and srand() in the Linux C Library use the same
random number generator as random(3) and srandom(3), so the lower-order
bits should be as random as the higher-order bits. However, on older
rand() implementations, and on current implementations on different
systems, the lower-order bits are much less random than the higher-
order bits. Do not use this function in applications intended to be
portable when good randomness is needed. (Use random(3) instead.)
Both approaches have their pitfalls, and your graphs are little more than a pretty verification of the central limit theorem! For a sensible implementation of rand():
% N suffers from a "pigeon-holing" effect if 1u + RAND_MAX is not a multiple of N
/((RAND_MAX + 1u)/N) does not, in general, evenly distribute the return of rand across your range, due to integer truncation effects.
On balance, if N is small cf. RAND_MAX, I'd plump for % for its tractability. In any case test your generator to see it it has the appropriate statistical properties for your application.
rand() % N is considered extremely poor not because the distribution is bad, but because the randomness is poor-to-nonexistent. (If anything the distribution will be too good.)
If N is not small with respect to RAND_MAX, both
rand() % N
and
rand() / (RAND_MAX / N + 1)
will have more or less the same, poor distribution -- certain values will occur with significantly higher probability than others.
Looking at distribution histograms won't show you that for some implementations, rand() % N has a much, much worse problem -- to show that you'd have to perform some correlations with previous values. (For example, try taking rand() % 2, then subtracting from the previous value you got, and plotting a histogram of the differences. If the difference is never 0, you've got a problem.)
I would like to say that the implementations for which rand()'s low-order bits aren't random are simply buggy. I'd like to think that all those buggy implementations would have disappeared by now. I'd like to think that programmers shouldn't have to worry about calling rand()%N any more. But, unfortunately, my wishes don't change the fact that this seems to be one of those bugs that never get fixed, meaning that programmers do still have to worry.
See also the C FAQ list, question 13.16.

Is it a good idea to try and use the entire rand() table?

I'm currently using this code to randomise some variables.
srand(time(NULL));
do{vkmhrand = rand ();vkmh = vkmhrand/78;}while(vkmh <= 0 || vkmh > vmax);
vmax won't exceed 415, which means that if I make it as simple as vkmh = rand() the majority of the randomized table will be discarded, plus I also get some nice decimals to make the results a little more interesting.
I know rand()'s not the best random function to use in existence. I'm just thinking in general, is it a good idea to try and use as much of the table as possible? Is it a waste of time? Or does it exacerbate any patterns that might already exist in the table?
Scaling the output of rand() effectively means in your case, that you can use more of its bits. Assuming their value is independent of each other (which is probably not true for all implementations of rand()), this is generally a good thing. However, you have to be aware, that those additional bits are not used to make the high order bits "more random" (less predictable) but only to give you more decimal digits, so your solution space has a finer resolution.
The usual way to generate floats in a certain range would be:
double vkmh = (double) rand() / RAND_MAX * (vmax-vmin) + vmin;
if you are fine with ints, the usual way would be:
unsigned int = rand() % (vmax - vmin + 1) + vmin;
Although this will result in a slightly skewed distribution if RAND_MAX isn't multiple of (vmax - vmin + 1) distributed values
Use this, I'm not vouching for it's randomness though
srand(time(NULL));
do {
vkmhrand = rand () % 415;
vkmh = vkmhrand/78;
}
while(vkmh <= 0 || vkmh > vmax);

About a criteria for random integers number generation (C)

I am running a bunch of physical simulations in which I need random numbers. I'm using the standard rand() function in C++.
So it works like this: first I precalculate a bunch of probabilities that are of the form 1/(1+exp(a)), for a set of different a. They're of type double as returned by the exp function in the math library, and then things must happen with those probabilities, there are only two of them, so I generate a random number uniformly distributed between 0 and 1 and compared with those precalculated probabilities. To do that, I used:
double p = double(rand()%101)/100.0;
so I'm given random values between 0 and 1 both included. This didn't yield to correct physical results. I tried this:
double p = double(rand()%1000001)/1000000.0;
And this worked. I don't really understand why so I would like some criteria about how to do it. My intuition tells that if I do
double p = double(rand()%(N+1))/double(N);
with N big enough such that the smallest division (1/N) is much smaller than the smallest probability 1/1+exp(a) then I will be getting realistic random numbers.
I would like to understand why, though.
rand() returns a random number between 0 and RAND_MAX.
Therefore you need this:
double p = double(rand() % RAND_MAX) / double(RAND_MAX);
Also run this snippet and you will understand:
int i;
for (i = 1; i < 30; i++)
{
int rnd = rand();
double p0 = double(rnd % 101) / 100.0;
double p1 = double(rnd % 1000001) / 1000000.0;
printf ("%d\t%f\t%f\n", rnd, p0, p1);
}
for (i = 1; i < 30; i++)
{
int rnd = rand();
double p0 = double(rnd) / double(RAND_MAX);
printf ("%d\t%f\n", rnd, p0);
}
You have multiple problems.
rand() isn't very random at all. On almost all operating systems it returns badly distributed, horribly biased numbers. It's actually quite hard to find a good random number generator, but I can guarantee you that rand() will be among the worst you can find.
rand() % N gives a biased distribution. Think about the pigeonhole principle. Let's simplify it, assume that rand returns numbers [0,7) and your N is 6. 0 to 5 map to 0 to 5, 6 maps to 0 and 7 maps to 1, meaning that 0 and 1 are twice as likely to come out.
Converting the numbers to double before division does not remove the bias from 2, it just makes it less visible. The pigeonhole principle applies regardless of the conversions you do.
Converting a well-distributed random number from integer to float/double is harder than it looks. Simple division ignores the problems of how floating point math works.
I can't help you much with 1, you need to do research. Look around the net for random number libraries. If you want something very random and unpredictable you need to look for cryptographic random libraries. If you want a repeatable but good random number Mersenne Twister should probably be good enough. But you need to do the research here.
For 2 and 3 there are standard solutions. You are mapping a set from M elements to N elements and rand % N will only work iff N < M and N and M share prime factors. Since on most systems M will be a power of two it means that N also has to be a power of two. So assuming that M is a power of two the algorithm is: find the nearest power of 2 higher or equal to N, let's call it P. Generate randomness_source() % P. If the number is higher than N, throw it away and try again. This is the only safe way to do this. Cleverer people than you and me have spent years on this problem, there's no better way to remove the bias.
For 4, you can probably ignore the problem and just divide, in an absolute majority of cases this should be good enough. If you really want to study the problem, I've done some work on it and published the code on github. There I go through some basic principles of how floating point numbers work and how it relates to generating random numbers.
// produces pseudorandom bits. These are NOT crypto quality bits. Has the same underlying unpredictability as uncooked
// rand() output. It buffers rand() bits to produce a more convenient zero-to-the-argument range including negative
// arguments, corrects for the toward-zero bias of the modular construction I'd be using otherwise, eliminates the
// RAND_MAX range limitation, (use INT64_MAX instead) and effectively obscures biases and sequence telltales due to
// annoyingly bad rand libraries. It does not correct these biases; anyone tracking the arguments and outputs has
// enough information to reconstruct the rand() output and detect them. But it makes the relationships drastically more complicated.
// needs stdint, stdlib.
int64_t privaterandom(int64_t range, int reset){
static uint64_t state = 0;
int64_t retval;
if (reset != 0){
srand((unsigned int)range);
state = (uint64_t)range;
}
if (range == 0) return (0);
if (range < 0) return -privaterandom(-range, 0);
if (range > UINT64_MAX/0xFFFFFFFF){
retval = (privaterandom(range/0xFFFFFFFF, 0) * 0xFFFFFFFF); // order of operations matters
return (retval + privaterandom(0xFFFFFFFF, 0));
}
while (state < UINT64_MAX / 0xFF){
state *= RAND_MAX;
state += rand();
}
retval = (state % range);
// makes "pigeonhole" bias alternate unpredictably between toward-even and toward-odd
if ((state/range > (state - (retval) )/ range) && state % 2 == 0) retval++;
state /= range;
return retval;
}
int64_t Random(int64_t range){ return (privaterandom(range, 0));}
int64_t Random_Init(int64_t seed){return (privaterandom(seed, 1));}

Generating a uniform distribution of INTEGERS in C

I've written a C function that I think selects integers from a uniform distribution with range [rangeLow, rangeHigh], inclusive. This isn't homework--I'm just using this in some embedded systems tinkering that I'm doing for fun.
In my test cases, this code appears to produce an appropriate distribution. I'm not feeling fully confident that the implementation is correct, though.
Could someone do a sanity check and let me know if I've done anything wrong here?
//uniform_distribution returns an INTEGER in [rangeLow, rangeHigh], inclusive.
int uniform_distribution(int rangeLow, int rangeHigh)
{
int myRand = (int)rand();
int range = rangeHigh - rangeLow + 1; //+1 makes it [rangeLow, rangeHigh], inclusive.
int myRand_scaled = (myRand % range) + rangeLow;
return myRand_scaled;
}
//note: make sure rand() was already initialized using srand()
P.S. I searched for other questions like this. However, it was hard to filter out the small subset of questions that discuss random integers instead of random floating-point numbers.
Let's assume that rand() generates a uniformly-distributed value I in the range [0..RAND_MAX],
and you want to generate a uniformly-distributed value O in the range [L,H].
Suppose I in is the range [0..32767] and O is in the range [0..2].
According to your suggested method, O= I%3. Note that in the given range, there are 10923 numbers for which I%3=0, 10923 number for which I%3=1, but only 10922 number for which I%3=2. Hence your method will not map a value from I into O uniformly.
As another example, suppose O is in the range [0..32766].
According to your suggested method, O=I%32767. Now you'll get O=0 for both I=0 and I=32767. Hence 0 is twice as likely than any other value - your method is again nonuniform.
The suggest way to generate a uniform mapping is as follow:
Calculate the number of bits that are needed to store a random value in the range [L,H]:
unsigned int nRange = (unsigned int)H - (unsigned int)L + 1;
unsigned int nRangeBits= (unsigned int)ceil(log((double(nRange) / log(2.));
Generate nRangeBits random bits
this can be easily implemented by shifting-right the result of rand()
Ensure that the generated number is not greater than H-L.
If it is - repeat step 2.
Now you can map the generated number into O just by adding a L.
On some implementations, rand() did not provide good randomness on its lower order bits, so the modulus operator would not provide very random results. If you find that to be the case, you could try this instead:
int uniform_distribution(int rangeLow, int rangeHigh) {
double myRand = rand()/(1.0 + RAND_MAX);
int range = rangeHigh - rangeLow + 1;
int myRand_scaled = (myRand * range) + rangeLow;
return myRand_scaled;
}
Using rand() this way will produce a bias as noted by Lior. But, the technique is fine if you can find a uniform number generator to calculate myRand. One possible candidate would be drand48(). This will greatly reduce the amount of bias to something that would be very difficult to detect.
However, if you need something cryptographically secure, you should use an algorithm outlined in Lior's answer, assuming your rand() is itself cryptographically secure (the default one is probably not, so you would need to find one). Below is a simplified implementation of what Lior described. Instead of counting bits, we assume the range falls within RAND_MAX, and compute a suitable multiple. Worst case, the algorithm ends up calling the random number generator twice on average per request for a number in the range.
int uniform_distribution_secure(int rangeLow, int rangeHigh) {
int range = rangeHigh - rangeLow + 1;
int secureMax = RAND_MAX - RAND_MAX % range;
int x;
do x = secure_rand(); while (x >= secureMax);
return rangeLow + x % range;
}
I think it is known that rand() is not very good. It just depends on how good of "random" data you need.
http://www.azillionmonkeys.com/qed/random.html
http://www.linuxquestions.org/questions/programming-9/generating-random-numbers-in-c-378358/
http://forums.indiegamer.com/showthread.php?9460-Using-C-rand%28%29-isn-t-as-bad-as-previously-thought
I suppose you could write a test then calculate the chi-squared value to see how good your uniform generator is:
http://en.wikipedia.org/wiki/Pearson%27s_chi-squared_test
Depending on your use (don't use this for your online poker shuffler), you might consider a LFSR
http://en.wikipedia.org/wiki/Linear_feedback_shift_register
It may be faster, if you just want some psuedo-random output. Also, supposedly they can be uniform, although I haven't studied the math enough to back up that claim.
A version which corrects the distribution errors (noted by Lior),
involves the high-bits returned by rand() and
only uses integer math (if that's desirable):
int uniform_distribution(int rangeLow, int rangeHigh)
{
int range = rangeHigh - rangeLow + 1; //+1 makes it [rangeLow, rangeHigh], inclusive.
int copies=RAND_MAX/range; // we can fit n-copies of [0...range-1] into RAND_MAX
// Use rejection sampling to avoid distribution errors
int limit=range*copies;
int myRand=-1;
while( myRand<0 || myRand>=limit){
myRand=rand();
}
return myRand/copies+rangeLow; // note that this involves the high-bits
}
//note: make sure rand() was already initialized using srand()
This should work well provided that range is much smaller than RAND_MAX, otherwise
you'll be back to the problem that rand() isn't a good random number generator in terms of its low-bits.

create a random number less than a max given value

What i would love to do is to create a function that takes a parameter that is the limit of which number the random generation should create. I have experienced that some generators that just repeat the number generated over and over again.
How can I make a generator that doesn't return the same number consecutively. Can someone please help me to achieve my goal?
int randomGen(int max)
{
int n;
return n;
}
The simplest way to get uniformly distributed results from rand is something like this:
int limited_rand(int limit)
{
int r, d = RAND_MAX / limit;
limit *= d;
do { r = rand(); } while (r >= limit);
return r / d;
}
The result will be in the range 0 to limit-1, and each will occur with equal probability as long as the values 0 through RAND_MAX all had equal probability with the original rand function.
Other methods such as modular arithmetic or dividing without the loop I used introduce bias. Methods that go through floating point intermediates do not avoid this problem. Getting good random floating point numbers from rand is at least as difficult. Using my function for integers (or an improvement of it) is a good place to start if you want random floats.
Edit: Here's an explanation of what I mean by bias. Suppose RAND_MAX is 7 and limit is 5. Suppose (if this is a good rand function) that the outputs 0, 1, 2, ..., 7 are all equally likely. Taking rand()%5 would map 0, 1, 2, 3, and 4 to themselves, but map 5, 6, and 7 to 0, 1, and 2. This means the values 0, 1, and 2 are twice as likely to pop up as the values 3 and 4. A similar phenomenon happens if you try to rescale and divide, for instance using rand()*(double)limit/(RAND_MAX+1) Here, 0 and 1 map to 0, 2 and 3 map to 1, 4 maps to 2, 5 and 6 map to 3, and 7 maps to 4.
These effects are somewhat mitigated by the magnitude of RAND_MAX, but they can come back if limit is large. By the way, as others have said, with linear congruence PRNGs (the typical implementation of rand), the low bits tend to behave very badly, so using modular arithmetic when limit is a power of 2 may avoid the bias problem I described (since limit usually divides RAND_MAX+1 evenly in this case), but you run into a different problem in its place.
How about this:
int randomGen(int limit)
{
return rand() % limit;
}
/* ... */
int main()
{
srand(time(NULL));
printf("%d", randomGen(2041));
return 0;
}
Any pseudo-random generator will repeat the values over and over again with some period. C only has rand(), if you use that you should definitively initialize the random seed with srand(). But probably your platform has better than that.
On POSIX systems there is a whole family of functions that you should find under the man drand48 page. They have a well defined period and quality. You probably find what you need, there.
Without explicit knowledge of the random generator of your platform, do not do rand() % max. The low-order bytes of simple random number generators are usually not random at all.
Use instead (returns a number between min inclusive and max non-inclusive):
int randomIntegerInRange(int min, int max)
{
double tmp = (double)rand() / (RAND_MAX - 1.);
return min + (int)floor(tmp * (max - min));
}
Update: The solution above is biased (see comments for explanation), and will likely not produce uniform results. I do not delete it since it is a non natural example of what not to do. Please use rejection methods as recommended elsewhere in this thread.

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