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I need a function which would generate a random integer in a given range (including boundary values). I don't have unreasonable quality/randomness requirements; I have four requirements:
I need it to be fast. My project needs to generate millions (or sometimes even tens of millions) of random numbers and my current generator function has proven to be a bottleneck.
I need it to be reasonably uniform (use of rand() is perfectly fine).
the minimum-maximum ranges can be anything from <0, 1> to <-32727, 32727>.
it has to be seedable.
I currently have the following C++ code:
output = min + (rand() * (int)(max - min) / RAND_MAX)
The problem is that it is not really uniform - max is returned only when rand() = RAND_MAX (for Visual C++ it is 1/32727). This is a major issue for small ranges like <-1, 1>, where the last value is almost never returned.
So I grabbed pen and paper and came up with following formula (which builds on the (int)(n + 0.5) integer rounding trick):
But it still doesn't give me a uniform distribution. Repeated runs with 10000 samples give me ratio of 37:50:13 for values values -1, 0. 1.
Is there a better formula? (Or even whole pseudo-random number generator function?)
The simplest (and hence best) C++ (using the 2011 standard) answer is:
#include <random>
std::random_device rd; // Only used once to initialise (seed) engine
std::mt19937 rng(rd()); // Random-number engine used (Mersenne-Twister in this case)
std::uniform_int_distribution<int> uni(min,max); // Guaranteed unbiased
auto random_integer = uni(rng);
There isn't any need to reinvent the wheel, worry about bias, or worry about using time as the random seed.
A fast, somewhat better than yours, but still not properly uniform distributed solution is
output = min + (rand() % static_cast<int>(max - min + 1))
Except when the size of the range is a power of 2, this method produces biased non-uniform distributed numbers regardless the quality of rand(). For a comprehensive test of the quality of this method, please read this.
If your compiler supports C++0x and using it is an option for you, then the new standard <random> header is likely to meet your needs. It has a high quality uniform_int_distribution which will accept minimum and maximum bounds (inclusive as you need), and you can choose among various random number generators to plug into that distribution.
Here is code that generates a million random ints uniformly distributed in [-57, 365]. I've used the new std <chrono> facilities to time it as you mentioned performance is a major concern for you.
#include <iostream>
#include <random>
#include <chrono>
int main()
{
typedef std::chrono::high_resolution_clock Clock;
typedef std::chrono::duration<double> sec;
Clock::time_point t0 = Clock::now();
const int N = 10000000;
typedef std::minstd_rand G; // Select the engine
G g; // Construct the engine
typedef std::uniform_int_distribution<> D; // Select the distribution
D d(-57, 365); // Construct the distribution
int c = 0;
for (int i = 0; i < N; ++i)
c += d(g); // Generate a random number
Clock::time_point t1 = Clock::now();
std::cout << N/sec(t1-t0).count() << " random numbers per second.\n";
return c;
}
For me (2.8 GHz Intel Core i5) this prints out:
2.10268e+07 random numbers per second.
You can seed the generator by passing in an int to its constructor:
G g(seed);
If you later find that int doesn't cover the range you need for your distribution, this can be remedied by changing the uniform_int_distribution like so (e.g., to long long):
typedef std::uniform_int_distribution<long long> D;
If you later find that the minstd_rand isn't a high enough quality generator, that can also easily be swapped out. E.g.:
typedef std::mt19937 G; // Now using mersenne_twister_engine
Having separate control over the random number generator, and the random distribution can be quite liberating.
I've also computed (not shown) the first four "moments" of this distribution (using minstd_rand) and compared them to the theoretical values in an attempt to quantify the quality of the distribution:
min = -57
max = 365
mean = 154.131
x_mean = 154
var = 14931.9
x_var = 14910.7
skew = -0.00197375
x_skew = 0
kurtosis = -1.20129
x_kurtosis = -1.20001
(The x_ prefix refers to "expected".)
Let's split the problem into two parts:
Generate a random number n in the range 0 through (max-min).
Add min to that number
The first part is obviously the hardest. Let's assume that the return value of rand() is perfectly uniform. Using modulo will add bias
to the first (RAND_MAX + 1) % (max-min+1) numbers. So if we could magically change RAND_MAX to RAND_MAX - (RAND_MAX + 1) % (max-min+1), there would no longer be any bias.
It turns out that we can use this intuition if we are willing to allow pseudo-nondeterminism into the running time of our algorithm. Whenever rand() returns a number which is too large, we simply ask for another random number until we get one which is small enough.
The running time is now geometrically distributed, with expected value 1/p where p is the probability of getting a small enough number on the first try. Since RAND_MAX - (RAND_MAX + 1) % (max-min+1) is always less than (RAND_MAX + 1) / 2,
we know that p > 1/2, so the expected number of iterations will always be less than two
for any range. It should be possible to generate tens of millions of random numbers in less than a second on a standard CPU with this technique.
Although the above is technically correct, DSimon's answer is probably more useful in practice. You shouldn't implement this stuff yourself. I have seen a lot of implementations of rejection sampling and it is often very difficult to see if it's correct or not.
Use the Mersenne Twister. The Boost implementation is rather easy to use and is well tested in many real-world applications. I've used it myself in several academic projects, such as artificial intelligence and evolutionary algorithms.
Here's their example where they make a simple function to roll a six-sided die:
#include <boost/random/mersenne_twister.hpp>
#include <boost/random/uniform_int.hpp>
#include <boost/random/variate_generator.hpp>
boost::mt19937 gen;
int roll_die() {
boost::uniform_int<> dist(1, 6);
boost::variate_generator<boost::mt19937&, boost::uniform_int<> > die(gen, dist);
return die();
}
Oh, and here's some more pimping of this generator just in case you aren't convinced you should use it over the vastly inferior rand():
The Mersenne Twister is a "random
number" generator invented by Makoto
Matsumoto and Takuji Nishimura; their
website includes numerous
implementations of the algorithm.
Essentially, the Mersenne Twister is a
very large linear-feedback shift
register. The algorithm operates on a
19,937 bit seed, stored in an
624-element array of 32-bit unsigned
integers. The value 2^19937-1 is a
Mersenne prime; the technique for
manipulating the seed is based on an
older "twisting" algorithm -- hence
the name "Mersenne Twister".
An appealing aspect of the Mersenne
Twister is its use of binary
operations -- as opposed to
time-consuming multiplication -- for
generating numbers. The algorithm also
has a very long period, and good
granularity. It is both fast and
effective for non-cryptographic applications.
int RandU(int nMin, int nMax)
{
return nMin + (int)((double)rand() / (RAND_MAX+1) * (nMax-nMin+1));
}
This is a mapping of 32768 integers to (nMax-nMin+1) integers. The mapping will be quite good if (nMax-nMin+1) is small (as in your requirement). Note however that if (nMax-nMin+1) is large, the mapping won't work (For example - you can't map 32768 values to 30000 values with equal probability). If such ranges are needed - you should use a 32-bit or 64-bit random source, instead of the 15-bit rand(), or ignore rand() results which are out-of-range.
Assume min and max are integer values,
[ and ] means include this value,
( and ) means do not include this value,
using the above to get the right value using C++'s rand().
Reference:
For ()[] define, visit Interval (mathematics).
For the rand and srand function or RAND_MAX define,
visit std::rand.
[min, max]
int randNum = rand() % (max - min + 1) + min
(min, max]
int randNum = rand() % (max - min) + min + 1
[min, max)
int randNum = rand() % (max - min) + min
(min, max)
int randNum = rand() % (max - min - 1) + min + 1
Here is an unbiased version that generates numbers in [low, high]:
int r;
do {
r = rand();
} while (r < ((unsigned int)(RAND_MAX) + 1) % (high + 1 - low));
return r % (high + 1 - low) + low;
If your range is reasonably small, there is no reason to cache the right-hand side of the comparison in the do loop.
I recommend the Boost.Random library. It's super detailed and well-documented, lets you explicitly specify what distribution you want, and in non-cryptographic scenarios can actually outperform a typical C library rand implementation.
Notice that in most suggestions the initial random value that you have got from rand() function, which is typically from 0 to RAND_MAX, is simply wasted. You are creating only one random number out of it, while there is a sound procedure that can give you more.
Assume that you want [min,max] region of integer random numbers. We start from [0, max-min]
Take base b=max-min+1
Start from representing a number you got from rand() in base b.
That way you have got floor(log(b,RAND_MAX)) because each digit in base b, except possibly the last one, represents a random number in the range [0, max-min].
Of course the final shift to [min,max] is simple for each random number r+min.
int n = NUM_DIGIT-1;
while(n >= 0)
{
r[n] = res % b;
res -= r[n];
res /= b;
n--;
}
If NUM_DIGIT is the number of digit in base b that you can extract and that is
NUM_DIGIT = floor(log(b,RAND_MAX))
then the above is as a simple implementation of extracting NUM_DIGIT random numbers from 0 to b-1 out of one RAND_MAX random number providing b < RAND_MAX.
In answers to this question, rejection sampling was already addressed, but I wanted to suggest one optimization based on the fact that rand() % 2^something does not introduce any bias as already mentioned above.
The algorithm is really simple:
calculate the smallest power of 2 greater than the interval length
randomize one number in that "new" interval
return that number if it is less than the length of the original interval
reject otherwise
Here's my sample code:
int randInInterval(int min, int max) {
int intervalLen = max - min + 1;
//now calculate the smallest power of 2 that is >= than `intervalLen`
int ceilingPowerOf2 = pow(2, ceil(log2(intervalLen)));
int randomNumber = rand() % ceilingPowerOf2; //this is "as uniform as rand()"
if (randomNumber < intervalLen)
return min + randomNumber; //ok!
return randInInterval(min, max); //reject sample and try again
}
This works well especially for small intervals, because the power of 2 will be "nearer" to the real interval length, and so the number of misses will be smaller.
PS: Obviously avoiding the recursion would be more efficient (there isn't any need to calculate over and over the log ceiling...), but I thought it was more readable for this example.
The following is the idea presented by Walter. I wrote a self-contained C++ class that will generate a random integer in the closed interval [low, high]. It requires C++11.
#include <random>
// Returns random integer in closed range [low, high].
class UniformRandomInt {
std::random_device _rd{};
std::mt19937 _gen{_rd()};
std::uniform_int_distribution<int> _dist;
public:
UniformRandomInt() {
set(1, 10);
}
UniformRandomInt(int low, int high) {
set(low, high);
}
// Set the distribution parameters low and high.
void set(int low, int high) {
std::uniform_int_distribution<int>::param_type param(low, high);
_dist.param(param);
}
// Get random integer.
int get() {
return _dist(_gen);
}
};
Example usage:
UniformRandomInt ur;
ur.set(0, 9); // Get random int in closed range [0, 9].
int value = ur.get()
The formula for this is very simple, so try this expression,
int num = (int) rand() % (max - min) + min;
//Where rand() returns a random number between 0.0 and 1.0
The following expression should be unbiased if I am not mistaken:
std::floor( ( max - min + 1.0 ) * rand() ) + min;
I am assuming here that rand() gives you a random value in the range between 0.0 and 1.0 not including 1.0 and that max and min are integers with the condition that min < max.
This question already has answers here:
Random float number generation
(14 answers)
Closed 7 years ago.
The title pretty much says it all. I've looked online but I couldn't find anything for this language. I've seen the following:
((double) rand() / (RAND_MAX)) with no luck. I'd also like to avoid using external library's.
The value should be a float as I'm working out X,Y coordinates.
If you are using C++11, you can use the random header for this. You will need to create a generator, then define a distribution over this generator, and then you can use the generator and the distribution to get your results. You need to include random
#include <random>
Then define the generator and your distribution
std::default_random_engine generator;
std::uniform_real_distribution<double> distribution(-1,1); //doubles from -1 to 1
Then you can get random numbers like so
double random_number = distribution(generator);
If you need more information it is available here http://www.cplusplus.com/reference/random/
Maybe ((double) rand() / (RAND_MAX)) * 2 - 1.
Don't simply use rand() because it's not random, just pseudo random!
unsigned int rand_interval(unsigned int min, unsigned int max)
{
int r;
const unsigned int range = 1 + max - min;
const unsigned int buckets = RAND_MAX / range;
const unsigned int limit = buckets * range;
/* Create equal size buckets all in a row, then fire randomly towards
* the buckets until you land in one of them. All buckets are equally
* likely. If you land off the end of the line of buckets, try again. */
do
{
r = rand();
} while (r >= limit);
return min + (r / buckets);
}
This is a fully functioning code. (Don't forget to seed rand with srand() in main!
Note> I can't take credit for it as I have also seen it somewhere online and I'm using it myself for some time.
This question already has answers here:
How to generate random number within range (-x,x)
(3 answers)
Closed 9 years ago.
Hi i wanna generate a random number between 2 values.
I have 2 variables.
This is the default value:
(MIN = MAX = 1)
Later this value can change!
I have use this:
rand()%(max-min)+min;
But i got debug about division for zero.
Any ideas?
Edit: With the default value the number generated must be 1.
In your initial case (max-min) is 0 and modulus by zero is undefined behavior. From the C++ draft standard section 5.6 Multiplicative operators says:
The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined. [...]
As for generating a random number between min and max you should use the random header and uniform_int_distrubution:
#include <iostream>
#include <random>
int main()
{
std::random_device rd;
std::mt19937 e2(rd());
int min = 1, max = 1 ;
std::uniform_int_distribution<int> dist(min,max);
for (int n = 0; n < 10; ++n) {
std::cout << dist(e2) << ", " ;
}
std::cout << std::endl ;
}
If for some reason C++11 is not an option then the C FAQ gives us the proper formula when using rand in the section How can I get random integers in a certain range? which indicates to generate random numbers in the range [M, N] you use the following:
M + rand() / (RAND_MAX / (N - M + 1) + 1)
The range [min, max] contains max - min + 1 numbers! So your code should have been like this:
rand() % (max - min + 1) + min;
Don't forget to check the value of RAND_MAX though, your range shouldn't get close to that. In fact, you want it to be much smaller than that to avoid too much bias.
I wrote a function that takes integers. It won't crash if the user types for example, -5, but it will convert it into positive =-(
int getRandoms(int size, int upper, int lower)
{
int randInt = 0;
randInt = 1 + rand() % (upper -lower + 1);
return randInt;
}
What should I change in the function in order to build random negative integers?
The user inputs the range.
There are two answers to this, if you are using C++11 then you should be using uniform_int_distribtion, it is preferable for many reasons for example Why do people say there is modulo bias when using a random number generator? is one example and rand() Considered Harmful presentation gives a more complete set of reasons. Here is an example which generates random integers from -5 to 5:
#include <iostream>
#include <random>
int main()
{
std::random_device rd;
std::mt19937 e2(rd());
std::uniform_int_distribution<int> dist(-5, 5);
for (int n = 0; n < 10; ++n) {
std::cout << dist(e2) << ", " ;
}
std::cout << std::endl ;
}
If C++11 is not an option then the method described in C FAQ in How can I get random integers in a certain range? is the way to go. Using that method you would do the following to generate random integers from [M, N]:
M + rand() / (RAND_MAX / (N - M + 1) + 1)
For a number in the closed range [lower,upper], you want:
return lower + rand() % (upper - lower + 1); // NOT 1 + ...
This will work for positive or negative values, as long as upper is greater than or equal to lower.
Your version returns numbers from a range of the same size, but starting from 1 rather than lower.
You could also use Boost.Random, if you don't mind the dependency. http://www.boost.org/doc/libs/1_54_0/doc/html/boost_random.html
You want to start by computing the range of the numbers, so (for example) -10 to +5 is a range of 15.
You can compute numbers in that range with code like this:
int rand_lim(int limit) {
/* return a random number in the range [0..limit)
*/
int divisor = RAND_MAX/limit;
int retval;
do {
retval = rand() / divisor;
} while (retval == limit);
return retval;
}
Having done that, getting the numbers to the correct range is pretty trivial: add the lower bound to each number you get.
Note that C++11 has added both random number generator and distribution classes that can take care of most of this for you.
If you do attempt to do this on your own, when you reduce numbers to a range, you pretty much need to use a loop as I've shown above to avoid skew. Essentially any attempt at just using division or remainder on its own almost inevitably introduces skew into the result (i.e., some results will happen more often than others).
You only need to sum to the lower-bound of the range [lbound, ubound]:
int rangesize = ubound - lbound + 1;
int myradnom = (rand() % rangesize) + lbound;
I need a function which would generate a random integer in a given range (including boundary values). I don't have unreasonable quality/randomness requirements; I have four requirements:
I need it to be fast. My project needs to generate millions (or sometimes even tens of millions) of random numbers and my current generator function has proven to be a bottleneck.
I need it to be reasonably uniform (use of rand() is perfectly fine).
the minimum-maximum ranges can be anything from <0, 1> to <-32727, 32727>.
it has to be seedable.
I currently have the following C++ code:
output = min + (rand() * (int)(max - min) / RAND_MAX)
The problem is that it is not really uniform - max is returned only when rand() = RAND_MAX (for Visual C++ it is 1/32727). This is a major issue for small ranges like <-1, 1>, where the last value is almost never returned.
So I grabbed pen and paper and came up with following formula (which builds on the (int)(n + 0.5) integer rounding trick):
But it still doesn't give me a uniform distribution. Repeated runs with 10000 samples give me ratio of 37:50:13 for values values -1, 0. 1.
Is there a better formula? (Or even whole pseudo-random number generator function?)
The simplest (and hence best) C++ (using the 2011 standard) answer is:
#include <random>
std::random_device rd; // Only used once to initialise (seed) engine
std::mt19937 rng(rd()); // Random-number engine used (Mersenne-Twister in this case)
std::uniform_int_distribution<int> uni(min,max); // Guaranteed unbiased
auto random_integer = uni(rng);
There isn't any need to reinvent the wheel, worry about bias, or worry about using time as the random seed.
A fast, somewhat better than yours, but still not properly uniform distributed solution is
output = min + (rand() % static_cast<int>(max - min + 1))
Except when the size of the range is a power of 2, this method produces biased non-uniform distributed numbers regardless the quality of rand(). For a comprehensive test of the quality of this method, please read this.
If your compiler supports C++0x and using it is an option for you, then the new standard <random> header is likely to meet your needs. It has a high quality uniform_int_distribution which will accept minimum and maximum bounds (inclusive as you need), and you can choose among various random number generators to plug into that distribution.
Here is code that generates a million random ints uniformly distributed in [-57, 365]. I've used the new std <chrono> facilities to time it as you mentioned performance is a major concern for you.
#include <iostream>
#include <random>
#include <chrono>
int main()
{
typedef std::chrono::high_resolution_clock Clock;
typedef std::chrono::duration<double> sec;
Clock::time_point t0 = Clock::now();
const int N = 10000000;
typedef std::minstd_rand G; // Select the engine
G g; // Construct the engine
typedef std::uniform_int_distribution<> D; // Select the distribution
D d(-57, 365); // Construct the distribution
int c = 0;
for (int i = 0; i < N; ++i)
c += d(g); // Generate a random number
Clock::time_point t1 = Clock::now();
std::cout << N/sec(t1-t0).count() << " random numbers per second.\n";
return c;
}
For me (2.8 GHz Intel Core i5) this prints out:
2.10268e+07 random numbers per second.
You can seed the generator by passing in an int to its constructor:
G g(seed);
If you later find that int doesn't cover the range you need for your distribution, this can be remedied by changing the uniform_int_distribution like so (e.g., to long long):
typedef std::uniform_int_distribution<long long> D;
If you later find that the minstd_rand isn't a high enough quality generator, that can also easily be swapped out. E.g.:
typedef std::mt19937 G; // Now using mersenne_twister_engine
Having separate control over the random number generator, and the random distribution can be quite liberating.
I've also computed (not shown) the first four "moments" of this distribution (using minstd_rand) and compared them to the theoretical values in an attempt to quantify the quality of the distribution:
min = -57
max = 365
mean = 154.131
x_mean = 154
var = 14931.9
x_var = 14910.7
skew = -0.00197375
x_skew = 0
kurtosis = -1.20129
x_kurtosis = -1.20001
(The x_ prefix refers to "expected".)
Let's split the problem into two parts:
Generate a random number n in the range 0 through (max-min).
Add min to that number
The first part is obviously the hardest. Let's assume that the return value of rand() is perfectly uniform. Using modulo will add bias
to the first (RAND_MAX + 1) % (max-min+1) numbers. So if we could magically change RAND_MAX to RAND_MAX - (RAND_MAX + 1) % (max-min+1), there would no longer be any bias.
It turns out that we can use this intuition if we are willing to allow pseudo-nondeterminism into the running time of our algorithm. Whenever rand() returns a number which is too large, we simply ask for another random number until we get one which is small enough.
The running time is now geometrically distributed, with expected value 1/p where p is the probability of getting a small enough number on the first try. Since RAND_MAX - (RAND_MAX + 1) % (max-min+1) is always less than (RAND_MAX + 1) / 2,
we know that p > 1/2, so the expected number of iterations will always be less than two
for any range. It should be possible to generate tens of millions of random numbers in less than a second on a standard CPU with this technique.
Although the above is technically correct, DSimon's answer is probably more useful in practice. You shouldn't implement this stuff yourself. I have seen a lot of implementations of rejection sampling and it is often very difficult to see if it's correct or not.
Use the Mersenne Twister. The Boost implementation is rather easy to use and is well tested in many real-world applications. I've used it myself in several academic projects, such as artificial intelligence and evolutionary algorithms.
Here's their example where they make a simple function to roll a six-sided die:
#include <boost/random/mersenne_twister.hpp>
#include <boost/random/uniform_int.hpp>
#include <boost/random/variate_generator.hpp>
boost::mt19937 gen;
int roll_die() {
boost::uniform_int<> dist(1, 6);
boost::variate_generator<boost::mt19937&, boost::uniform_int<> > die(gen, dist);
return die();
}
Oh, and here's some more pimping of this generator just in case you aren't convinced you should use it over the vastly inferior rand():
The Mersenne Twister is a "random
number" generator invented by Makoto
Matsumoto and Takuji Nishimura; their
website includes numerous
implementations of the algorithm.
Essentially, the Mersenne Twister is a
very large linear-feedback shift
register. The algorithm operates on a
19,937 bit seed, stored in an
624-element array of 32-bit unsigned
integers. The value 2^19937-1 is a
Mersenne prime; the technique for
manipulating the seed is based on an
older "twisting" algorithm -- hence
the name "Mersenne Twister".
An appealing aspect of the Mersenne
Twister is its use of binary
operations -- as opposed to
time-consuming multiplication -- for
generating numbers. The algorithm also
has a very long period, and good
granularity. It is both fast and
effective for non-cryptographic applications.
int RandU(int nMin, int nMax)
{
return nMin + (int)((double)rand() / (RAND_MAX+1) * (nMax-nMin+1));
}
This is a mapping of 32768 integers to (nMax-nMin+1) integers. The mapping will be quite good if (nMax-nMin+1) is small (as in your requirement). Note however that if (nMax-nMin+1) is large, the mapping won't work (For example - you can't map 32768 values to 30000 values with equal probability). If such ranges are needed - you should use a 32-bit or 64-bit random source, instead of the 15-bit rand(), or ignore rand() results which are out-of-range.
Assume min and max are integer values,
[ and ] means include this value,
( and ) means do not include this value,
using the above to get the right value using C++'s rand().
Reference:
For ()[] define, visit Interval (mathematics).
For the rand and srand function or RAND_MAX define,
visit std::rand.
[min, max]
int randNum = rand() % (max - min + 1) + min
(min, max]
int randNum = rand() % (max - min) + min + 1
[min, max)
int randNum = rand() % (max - min) + min
(min, max)
int randNum = rand() % (max - min - 1) + min + 1
Here is an unbiased version that generates numbers in [low, high]:
int r;
do {
r = rand();
} while (r < ((unsigned int)(RAND_MAX) + 1) % (high + 1 - low));
return r % (high + 1 - low) + low;
If your range is reasonably small, there is no reason to cache the right-hand side of the comparison in the do loop.
I recommend the Boost.Random library. It's super detailed and well-documented, lets you explicitly specify what distribution you want, and in non-cryptographic scenarios can actually outperform a typical C library rand implementation.
Notice that in most suggestions the initial random value that you have got from rand() function, which is typically from 0 to RAND_MAX, is simply wasted. You are creating only one random number out of it, while there is a sound procedure that can give you more.
Assume that you want [min,max] region of integer random numbers. We start from [0, max-min]
Take base b=max-min+1
Start from representing a number you got from rand() in base b.
That way you have got floor(log(b,RAND_MAX)) because each digit in base b, except possibly the last one, represents a random number in the range [0, max-min].
Of course the final shift to [min,max] is simple for each random number r+min.
int n = NUM_DIGIT-1;
while(n >= 0)
{
r[n] = res % b;
res -= r[n];
res /= b;
n--;
}
If NUM_DIGIT is the number of digit in base b that you can extract and that is
NUM_DIGIT = floor(log(b,RAND_MAX))
then the above is as a simple implementation of extracting NUM_DIGIT random numbers from 0 to b-1 out of one RAND_MAX random number providing b < RAND_MAX.
In answers to this question, rejection sampling was already addressed, but I wanted to suggest one optimization based on the fact that rand() % 2^something does not introduce any bias as already mentioned above.
The algorithm is really simple:
calculate the smallest power of 2 greater than the interval length
randomize one number in that "new" interval
return that number if it is less than the length of the original interval
reject otherwise
Here's my sample code:
int randInInterval(int min, int max) {
int intervalLen = max - min + 1;
//now calculate the smallest power of 2 that is >= than `intervalLen`
int ceilingPowerOf2 = pow(2, ceil(log2(intervalLen)));
int randomNumber = rand() % ceilingPowerOf2; //this is "as uniform as rand()"
if (randomNumber < intervalLen)
return min + randomNumber; //ok!
return randInInterval(min, max); //reject sample and try again
}
This works well especially for small intervals, because the power of 2 will be "nearer" to the real interval length, and so the number of misses will be smaller.
PS: Obviously avoiding the recursion would be more efficient (there isn't any need to calculate over and over the log ceiling...), but I thought it was more readable for this example.
The following is the idea presented by Walter. I wrote a self-contained C++ class that will generate a random integer in the closed interval [low, high]. It requires C++11.
#include <random>
// Returns random integer in closed range [low, high].
class UniformRandomInt {
std::random_device _rd{};
std::mt19937 _gen{_rd()};
std::uniform_int_distribution<int> _dist;
public:
UniformRandomInt() {
set(1, 10);
}
UniformRandomInt(int low, int high) {
set(low, high);
}
// Set the distribution parameters low and high.
void set(int low, int high) {
std::uniform_int_distribution<int>::param_type param(low, high);
_dist.param(param);
}
// Get random integer.
int get() {
return _dist(_gen);
}
};
Example usage:
UniformRandomInt ur;
ur.set(0, 9); // Get random int in closed range [0, 9].
int value = ur.get()
The formula for this is very simple, so try this expression,
int num = (int) rand() % (max - min) + min;
//Where rand() returns a random number between 0.0 and 1.0
The following expression should be unbiased if I am not mistaken:
std::floor( ( max - min + 1.0 ) * rand() ) + min;
I am assuming here that rand() gives you a random value in the range between 0.0 and 1.0 not including 1.0 and that max and min are integers with the condition that min < max.