Generate random numbers uniformly over an entire range - c++

Question

I need to generate random numbers within a specified interval, [max;min].
Also, the random numbers should be uniformly distributed over the interval, not located to a particular point.
Currenly I am generating as:
for(int i=0; i<6; i++)
{
DWORD random = rand()%(max-min+1) + min;
}
From my tests, random numbers are generated around one point only.
Example
min = 3604607;
max = 7654607;
Random numbers generated:
3631594
3609293
3630000
3628441
3636376
3621404
From answers below: OK, RAND_MAX is 32767. I am on C++ Windows platform. Is there any other method to generate random numbers with a uniform distribution?

Answer 1

Why rand is a bad idea
Most of the answers you got here make use of the rand function and the modulus operator. That method may not generate numbers uniformly (it depends on the range and the value of RAND_MAX), and is therefore discouraged.
C++11 and generation over a range
With C++11 multiple other options have risen. One of which fits your requirements, for generating a random number in a range, pretty nicely: std::uniform_int_distribution. Here's an example:
#include <iostream>
#include <random>
int main()
{
const int range_from = 0;
const int range_to = 1000;
std::random_device rand_dev;
std::mt19937 generator(rand_dev());
std::uniform_int_distribution<int> distr(range_from, range_to);
std::cout << distr(generator) << '\n';
}
Try it online on Godbolt
And here's the running example.
Template function may help some:
template<typename T>
T random(T range_from, T range_to) {
std::random_device rand_dev;
std::mt19937 generator(rand_dev());
std::uniform_int_distribution<T> distr(range_from, range_to);
return distr(generator);
}
Other random generators
The <random> header offers innumerable other random number generators with different kind of distributions including Bernoulli, Poisson and normal.
How can I shuffle a container?
The standard provides std::shuffle, which can be used as follows:
#include <iostream>
#include <random>
#include <vector>
int main()
{
std::vector<int> vec = {4, 8, 15, 16, 23, 42};
std::random_device random_dev;
std::mt19937 generator(random_dev());
std::shuffle(vec.begin(), vec.end(), generator);
std::for_each(vec.begin(), vec.end(), [](auto i){std::cout << i << '\n';});
}
Try it online on Godbolt
The algorithm will reorder the elements randomly, with a linear complexity.
Boost.Random
Another alternative, in case you don't have access to a C++11+ compiler, is to use Boost.Random. Its interface is very similar to the C++11 one.

Answer 2

Warning: Do not use rand() for statistics, simulation, cryptography or anything serious.
It's good enough to make numbers look random for a typical human in a hurry, no more.
See Jefffrey's reply for better options, or this answer for crypto-secure random numbers.
Generally, the high bits show a better distribution than the low bits, so the recommended way to generate random numbers of a range for simple purposes is:
((double) rand() / (RAND_MAX+1)) * (max-min+1) + min
Note: make sure RAND_MAX+1 does not overflow (thanks Demi)!
The division generates a random number in the interval [0, 1); "stretch" this to the required range. Only when max-min+1 gets close to RAND_MAX you need a "BigRand()" function like posted by Mark Ransom.
This also avoids some slicing problems due to the modulo, which can worsen your numbers even more.
The built-in random number generator isn't guaranteed to have a the quality required for statistical simulations. It is OK for numbers to "look random" to a human, but for a serious application, you should take something better - or at least check its properties (uniform distribution is usually good, but values tend to correlate, and the sequence is deterministic). Knuth has an excellent (if hard-to-read) treatise on random number generators, and I recently found LFSR to be excellent and darn simple to implement, given its properties are OK for you.

Answer 3

I'd like to complement Shoe's and peterchen's excellent answers with a short overview of the state of the art in 2015:
Some good choices
randutils
The randutils library (presentation) is an interesting novelty, offering a simple interface and (declared) robust random capabilities. It has the disadvantages that it adds a dependence on your project and, being new, it has not been extensively tested. Anyway, being free (MIT License) and header-only, I think it's worth a try.
Minimal sample: a die roll
#include <iostream>
#include "randutils.hpp"
int main() {
randutils::mt19937_rng rng;
std::cout << rng.uniform(1,6) << "\n";
}
Even if one is not interested in the library, the website provides many interesting articles about the theme of random number generation in general and the C++ library in particular.
Boost.Random
Boost.Random (documentation) is the library which inspired C++11's <random>, with whom it shares much of the interface. While theoretically also being an external dependency, Boost has by now a status of "quasi-standard" library, and its Random module could be regarded as the classical choice for good-quality random number generation. It features two advantages with respect to the C++11 solution:
it is more portable, just needing compiler support for C++03
its random_device uses system-specific methods to offer seeding of good quality
The only small flaw is that the module offering random_device is not header-only, one has to compile and link boost_random.
Minimal sample: a die roll
#include <iostream>
#include <boost/random.hpp>
#include <boost/nondet_random.hpp>
int main() {
boost::random::random_device rand_dev;
boost::random::mt19937 generator(rand_dev());
boost::random::uniform_int_distribution<> distr(1, 6);
std::cout << distr(generator) << '\n';
}
While the minimal sample does its work well, real programs should use a pair of improvements:
make mt19937 a thread_local: the generator is quite plump (more than 2 KB) and is better not allocated on the stack
seed mt19937 with more than one integer: the Mersenne Twister has a big state and can take benefit of more entropy during initialization
Some not-so-good choices
The C++11 library
While being the most idiomatic solution, the <random> library does not offer much in exchange for the complexity of its interface even for the basic needs. The flaw is in std::random_device: the Standard does not mandate any minimal quality for its output (as long as entropy() returns 0) and, as of 2015, MinGW (not the most used compiler, but hardly an esoteric choice) will always print 4 on the minimal sample.
Minimal sample: a die roll
#include <iostream>
#include <random>
int main() {
std::random_device rand_dev;
std::mt19937 generator(rand_dev());
std::uniform_int_distribution<int> distr(1, 6);
std::cout << distr(generator) << '\n';
}
If the implementation is not rotten, this solution should be equivalent to the Boost one, and the same suggestions apply.
Godot's solution
Minimal sample: a die roll
#include <iostream>
#include <random>
int main() {
std::cout << std::randint(1,6);
}
This is a simple, effective and neat solution. The only defect is it will take a while to compile – about two years, providing C++17 is released on time and the experimental randint function is approved into the new Standard. Maybe by that time also the guarantees on the seeding quality will improve.
The worse-is-better solution
Minimal sample: a die roll
#include <cstdlib>
#include <ctime>
#include <iostream>
int main() {
std::srand(std::time(nullptr));
std::cout << (std::rand() % 6 + 1);
}
The old C solution is considered harmful, and for good reasons (see the other answers here or this detailed analysis). Still, it has its advantages: is is simple, portable, fast and honest, in the sense it is known that the random numbers one gets are hardly decent, and therefore one is not tempted to use them for serious purposes.
The accounting troll solution
Minimal sample: a die roll
#include <iostream>
int main() {
std::cout << 9; // http://dilbert.com/strip/2001-10-25
}
While 9 is a somewhat unusual outcome for a regular die roll, one has to admire the excellent combination of good qualities in this solution, which manages to be the fastest, simplest, most cache-friendly and most portable one. By substituting 9 with 4, one gets a perfect generator for any kind of Dungeons & Dragons die, while still avoiding the symbol-laden values 1, 2 and 3. The only small flaw is that, because of the bad temper of Dilbert's accounting trolls, this program actually engenders undefined behavior.

Answer 4

If RAND_MAX is 32767, you can double the number of bits easily.
int BigRand()
{
assert(INT_MAX/(RAND_MAX+1) > RAND_MAX);
return rand() * (RAND_MAX+1) + rand();
}

Answer 5

If you are concerned about randomness and not about speed, you should use a secure random number generation method. There are several ways to do this... The easiest one being to use OpenSSL's Random Number Generator.
You can also write your own using an encryption algorithm (like AES). By picking a seed and an IV and then continuously re-encrypting the output of the encryption function. Using OpenSSL is easier, but less manly.

Answer 6

If you are able to, use Boost. I have had good luck with their random library.
uniform_int should do what you want.

Answer 7

You should look at RAND_MAX for your particular compiler/environment.
I think you would see these results if rand() is producing a random 16-bit number. (you seem to be assuming it will be a 32-bit number).
I can't promise this is the answer, but please post your value of RAND_MAX, and a little more detail on your environment.

Answer 8

This should provide a uniform distribution over the range [low, high) without using floats, as long as the overall range is less than RAND_MAX.
uint32_t rand_range_low(uint32_t low, uint32_t high)
{
uint32_t val;
// only for 0 < range <= RAND_MAX
assert(low < high);
assert(high - low <= RAND_MAX);
uint32_t range = high-low;
uint32_t scale = RAND_MAX/range;
do {
val = rand();
} while (val >= scale * range); // since scale is truncated, pick a new val until it's lower than scale*range
return val/scale + low;
}
and for values greater than RAND_MAX you want something like
uint32_t rand_range(uint32_t low, uint32_t high)
{
assert(high>low);
uint32_t val;
uint32_t range = high-low;
if (range < RAND_MAX)
return rand_range_low(low, high);
uint32_t scale = range/RAND_MAX;
do {
val = rand() + rand_range(0, scale) * RAND_MAX; // scale the initial range in RAND_MAX steps, then add an offset to get a uniform interval
} while (val >= range);
return val + low;
}
This is roughly how std::uniform_int_distribution does things.

Answer 9

Check what RAND_MAX is on your system -- I'm guessing it is only 16 bits, and your range is too big for it.
Beyond that see this discussion on: Generating Random Integers within a Desired Range and the notes on using (or not) the C rand() function.

Answer 10

Using a Mersenne Twister engine (C++11):
#include <random>
// Returns a random integer within the range [min, max]
int generateRandomInt(const int min, const int max) {
static bool is_seeded = false;
static std::mt19937 generator;
// Seed once
if (!is_seeded) {
std::random_device rd;
generator.seed(rd());
is_seeded = true;
}
// Use a Mersenne Twister engine to pick a random number
// within the given range
std::uniform_int_distribution<int> distribution(min, max);
return distribution(generator);
}

Answer 11

If you want numbers to be uniformly distributed over the range, you should break your range up into a number of equal sections that represent the number of points you need. Then get a random number with a min/max for each section.
As another note, you should probably not use rand() as it's not very good at actually generating random numbers. I don't know what platform you're running on, but there is probably a better function you can call like random().

Answer 12

This is not the code, but this logic may help you.
static double rnd(void)
{
return (1.0 / (RAND_MAX + 1.0) * ((double)(rand())));
}
static void InitBetterRnd(unsigned int seed)
{
register int i;
srand(seed);
for(i = 0; i < POOLSIZE; i++)
{
pool[i] = rnd();
}
}
// This function returns a number between 0 and 1
static double rnd0_1(void)
{
static int i = POOLSIZE - 1;
double r;
i = (int)(POOLSIZE*pool[i]);
r = pool[i];
pool[i] = rnd();
return (r);
}

Answer 13

The solution given by man 3 rand for a number between 1 and 10 inclusive is:
j = 1 + (int) (10.0 * (rand() / (RAND_MAX + 1.0)));
In your case, it would be:
j = min + (int) ((max-min+1) * (rand() / (RAND_MAX + 1.0)));
Of course, this is not perfect randomness or uniformity as some other messages are pointing out, but this is enough for most cases.

Answer 14

This is the solution I came up with:
#include "<stdlib.h>"
int32_t RandomRange(int32_t min, int32_t max) {
return (rand() * (max - min + 1) / (RAND_MAX + 1)) + min;
}
This is a bucket solution, conceptually similar to the solutions that use rand() / RAND_MAX to get a floating point range between 0-1 and then round that into a bucket. However, it uses purely integer math, and takes advantage of integer division flooring to round down the value to the nearest bucket.
It makes a few assumptions. First, it assumes that RAND_MAX * (max - min + 1) will always fit within an int32_t. If RAND_MAX is 32767 and 32 bit int calculations are used, the the maximum range you can have is 32767. If your implementation has a much larger RAND_MAX, you can overcome this by using a larger integer (like int64_t) for the calculation. Secondly, if int64_t is used but RAND_MAX is still 32767, at ranges greater than RAND_MAX you will start to get "holes" in the possible output numbers. This is probably the biggest issue with any solution derived from scaling rand().
Testing over a huge number of iterations nevertheless shows this method to be very uniform for small ranges. However, it is possible (and likely) that mathematically this has some small bias and possibly develops issues when the range approaches RAND_MAX. Test it for yourself and decide if it meets your needs.

Answer 15

By their nature, a small sample of random numbers doesn't have to be uniformly distributed. They're random, after all. I agree that if a random number generator is generating numbers that consistently appear to be grouped, then there is probably something wrong with it.
But keep in mind that randomness isn't necessarily uniform.

Answer 16

A solution
((double) rand() / (RAND_MAX+1)) * (max-min+1) + min
Warning: Don't forget due to stretching and possible precision errors (even if RAND_MAX were large enough), you'll only be able to generate evenly distributed "bins" and not all numbers in [min,max].
A solution using Bigrand
Warning: Note that this doubles the bits, but still won't be able to generate all numbers in your range in general, i.e., it is not necessarily true that BigRand() will generate all numbers between in its range.
Information: Your approach (modulo) is "fine" as long as the range of rand() exceeds your interval range and rand() is "uniform". The error for at most the first max - min numbers is 1/(RAND_MAX +1).
Also, I suggest to switch to the new random package in C++11 too, which offers better and more varieties of implementations than rand().

Answer 17

Minimal implementation with C++11:
#include <random>
int randrange (int min, int max) {
static std::random_device rd; // Static in case init is costly
return std::uniform_int_distribution {min, max} (rd);
}

Answer 18

This is an old thread, but I just stumbled on it now. Here's my take:
As others rightly pointed out, MSBs tend to be more randomly distributed than LSBs in most RNGs. That implies that taking the modulo (%) of rand() is doomed: e.g. the extremely frequent random(2) would only return the single LSB... which is extremely badly distributed in most RNGs.
On the other hand, if you need your random(N) to be very fast (as I do: I'm in HPC and in heavily randomized GAs in particular), the modulo is cool for its speed.
Both of the above concerns can be addressed by (1) computing the (fast) modulo... of (2) rand()'s reversed bits.

Answer 19

Of course, the following code won't give you random numbers, but pseudo random numbers.
Use the following code
#define QUICK_RAND(m,n) m + ( std::rand() % ( (n) - (m) + 1 ) )
For example:
int myRand = QUICK_RAND(10, 20);
You must call
srand(time(0)); // Initialize random number generator.
Otherwise the numbers won't be near random.

Answer 20

I just found this on the Internet. This should work:
DWORD random = ((min) + rand()/(RAND_MAX + 1.0) * ((max) - (min) + 1));