I created a test application that generates 10k random numbers in a range from 0 to 250 000. Then I calculated MAX and min values and noticed that the MAX value is always around 32k...
Do you have any idea how to extend the possible range? I need a range with MAX value around 250 000!
This is according to the definition of rand(), see:
http://cplusplus.com/reference/clibrary/cstdlib/rand/
http://cplusplus.com/reference/clibrary/cstdlib/RAND_MAX/
If you need larger random numbers, you can use an external library (for example http://www.boost.org/doc/libs/1_49_0/doc/html/boost_random.html) or calculate large random numbers out of multiple small random numbers by yourself.
But pay attention to the distribution you want to get. If you just sum up the small random numbers, the result will not be equally distributed.
If you just scale one small random number by a constant factor, there will be gaps between the possible values.
Taking the product of random numbers also doesn't work.
A possible solution is the following:
1) Take two random numbers a,b
2) Calculate a*(RAND_MAX+1)+b
So you get equally distributed random values up to (RAND_MAX+1)^2-1
Presumably, you also want an equal distribution over this extended
range. About the only way you can effectively do this is to generate a
sequence of smaller numbers, and scale them as if you were working in a
different base. For example, for 250000, you might 4 random numbers
in the range [0,10) and one in range [0,25), along the lines:
int
random250000()
{
return randomInt(10) + 10 * randomInt(10)
+ 100 * randomInt(10) + 1000 * randomInt(10)
+ 10000 * randomInt(25);
}
For this to work, your random number generator must be good; many
implementations of rand() aren't (or at least weren't—I've not
verified the situation recently). You'll also want to eliminate the
bias you get when you map RAND_MAX + 1 different values into 10 or
25 different values. Unless RAND_MAX + 1 is an exact multiple of
10 and 25 (e.g. is an exact multiple of 50), you'll need something
like:
int
randomInt( int upperLimit )
{
int const limit = (RAND_MAX + 1) - (RAND_MAX + 1) % upperLimit;
int result = rand();
while ( result >= limit ) {
result = rand();
return result % upperLimit;
}
(Attention when doing this: there are some machines where RAND_MAX + 1
will overflow; if portability is an issue, you'll need to take
additional precautions.)
All of this, of course, supposes a good quality generator, which is far
from a given.
You can just manipulate your number bitwise by generating smaller random numbers.
For instance, if you need a 32-bit random number:
int32 x = 0;
for (int i = 0; i < 4; ++i) { // 4 == 32/8
int8 tmp = 8bit_random_number_generator();
x <<= 8*i; x |= tmp;
}
If you don't need good randomness in your numbers, you can just use rand() & 0xff for the 8-bit random number generator. Otherwise, something better will be necessary.
Are you using short ints? If so, you will see 32,767 as your max number because anything larger will overflow the short int.
Scale your numbers up by N / RAND_MAX, where N is your desired maximum. If the numbers fit, you can do something like this:
unsigned long long int r = rand() * N / RAND_MAX;
Obviously if the initial part overflows you can't do this, but with N = 250000 you should be fine. RAND_MAX is 32K on many popular platforms.
More generally, to get a random number uniformly in the interval [A, B], use:
A + rand() * (B - A) / RAND_MAX;
Of course you should probably use the proper C++-style <random> library; search this site for many similar questions explaining how to use it.
Edit: In the hope of preventing an escalation of comments, here's yet another copy/paste of the Proper C++ solution for truly uniform distribution on an interval [A, B]:
#include <random>
typedef std::mt19937 rng_type;
typedef unsigned long int int_type; // anything you like
std::uniform_int_distribution<int_type> udist(A, B);
rng_type rng;
int main()
{
// seed rng first:
rng_type::result_type const seedval = get_seed();
rng.seed(seedval);
int_type random_number = udist(rng);
// use random_number
}
Don't forget to seend the RNG! If you store the seed value, you can replay the same random sequence later on.
Related
If 5 + rand() % 6 generates a random number from 5 to 10, how would I generate a decimal number? I tried assigning a float variable to ((4 + rand() % 8) / 100), but that does not do anything except assign the variable to zero.
Personally, I'd go with something a little more robust than rand() (but that might be because I'm doing research in areas that require high-quality random numbers):
#include <random>
double randomDouble(double bottom, double top) {
std::uniform_real_distribution<double> dist(bottom, top);
std::mt19937 gen; // Mersenne Twister PRNG
return dist(gen);
}
This little function uses a Mersenne Twister, a common type of PRNG that has a period (number of random numbers before you get repetition) of 2^19937 - hence the name. Now just call it like this:
double foo = randomDouble(0.04, 0.11);
There are two basic ways to transform a number generator rand() that produces numbers in [0, RAND_MAX] range to make it produce numbers in range [a, b]:
n = rand() % (b - a + 1) + a;
n = rand() * (b - a + 1) / (RAND_MAX + 1) + a;
The second method is applicable to ranges with non-integer a and b.
Keep in mind though that aside from the already mentioned random number quality issues, this generator might not produce sufficiently dense coverage of the target range. The greater the value of RAND_MAX, the denser is the coverage.
I am using rand() to create random numbers. When I run my code:
int lowest=10000, highest=90000000000;
int range=(highest-lowest)+1;
for (int index=1; index<2; index++){
c = lowest+int(range*rand()/(RAND_MAX + 1.0));
C returns as both positive and negative numbers. Why is this, and how can I fix it?
It's because range * rand() overflowed and wrapped around.
This approach can't be "fixed", because you would just get RAND_MAX of the range possible values, leaving big holes that would never be chosen. (Or, if (INT_MAX - INT_MIN + 1) / (RAND_MAX + 1) is smaller than RAND_MAX, that's how many distinct results you can get)
Use the new C++11 random number generators.
If you can't use C++11, then use bitshifting and OR operators to combine multiple rand() results into a single number. If you got a number outside your range, just try again. Any method of mapping one range onto another will break most of the nice properties of the pseudo-random number, like uniform distribution.
In this StackOverflow question:
Generating random integer from a range
the accepted answer suggests the following formula for generating a random integer in between given min and max, with min and max being included into the range:
output = min + (rand() % (int)(max - min + 1))
But it also says that
This is still slightly biased towards lower numbers ... It's also
possible to extend it so that it removes the bias.
But it doesn't explain why it's biased towards lower numbers or how to remove the bias. So, the question is: is this the most optimal approach to generation of a random integer within a (signed) range while not relying on anything fancy, just rand() function, and in case if it is optimal, how to remove the bias?
EDIT:
I've just tested the while-loop algorithm suggested by #Joey against floating-point extrapolation:
static const double s_invRandMax = 1.0/((double)RAND_MAX + 1.0);
return min + (int)(((double)(max + 1 - min))*rand()*s_invRandMax);
to see how much uniformly "balls" are "falling" into and are being distributed among a number of "buckets", one test for the floating-point extrapolation and another for the while-loop algorithm. But results turned out to be varying depending on the number of "balls" (and "buckets") so I couldn't easily pick a winner. The working code can be found at this Ideone page. For example, with 10 buckets and 100 balls the maximum deviation from the ideal probability among buckets is less for the floating-point extrapolation than for the while-loop algorithm (0.04 and 0.05 respectively) but with 1000 balls, the maximum deviation of the while-loop algorithm is lesser (0.024 and 0.011), and with 10000 balls, the floating-point extrapolation is again doing better (0.0034 and 0.0053), and so on without much of consistency. Thinking of the possibility that none of the algorithms consistently produces uniform distribution better than that of the other algorithm, makes me lean towards the floating-point extrapolation since it appears to perform faster than the while-loop algorithm. So is it fine to choose the floating-point extrapolation algorithm or my testings/conclusions are not completely correct?
The problem is that you're doing a modulo operation. This would be no problem if RAND_MAX would be evenly divisible by your modulus, but usually that is not the case. As a very contrived example, assume RAND_MAX to be 11 and your modulus to be 3. You'll get the following possible random numbers and the following resulting remainders:
0 1 2 3 4 5 6 7 8 9 10
0 1 2 0 1 2 0 1 2 0 1
As you can see, 0 and 1 are slightly more probable than 2.
One option to solve this is rejection sampling: By disallowing the numbers 9 and 10 above you can cause the resulting distribution to be uniform again. The tricky part is figuring out how to do so efficiently. A very nice example (one that took me two days to understand why it works) can be found in Java's java.util.Random.nextInt(int) method.
The reason why Java's algorithm is a little tricky is that they avoid slow operations like multiplication and division for the check. If you don't care too much you can also do it the naïve way:
int n = (int)(max - min + 1);
int remainder = RAND_MAX % n;
int x, output;
do {
x = rand();
output = x % n;
} while (x >= RAND_MAX - remainder);
return min + output;
EDIT: Corrected a fencepost error in above code, now it works as it should. I also created a little sample program (C#; taking a uniform PRNG for numbers between 0 and 15 and constructing a PRNG for numbers between 0 and 6 from it via various ways):
using System;
class Rand {
static Random r = new Random();
static int Rand16() {
return r.Next(16);
}
static int Rand7Naive() {
return Rand16() % 7;
}
static int Rand7Float() {
return (int)(Rand16() / 16.0 * 7);
}
// corrected
static int Rand7RejectionNaive() {
int n = 7, remainder = 16 % n, x, output;
do {
x = Rand16();
output = x % n;
} while (x >= 16 - remainder);
return output;
}
// adapted to fit the constraints of this example
static int Rand7RejectionJava() {
int n = 7, x, output;
do {
x = Rand16();
output = x % n;
} while (x - output + 6 > 15);
return output;
}
static void Test(Func<int> rand, string name) {
var buckets = new int[7];
for (int i = 0; i < 10000000; i++) buckets[rand()]++;
Console.WriteLine(name);
for (int i = 0; i < 7; i++) Console.WriteLine("{0}\t{1}", i, buckets[i]);
}
static void Main() {
Test(Rand7Naive, "Rand7Naive");
Test(Rand7Float, "Rand7Float");
Test(Rand7RejectionNaive, "Rand7RejectionNaive");
}
}
The result is as follows (pasted into Excel and added conditional coloring of cells so that differences are more apparent):
Now that I fixed my mistake in above rejection sampling it works as it should (before it would bias 0). As you can see, the float method isn't perfect at all, it just distributes the biased numbers differently.
The problem occurs when the number of outputs from the random number generator (RAND_MAX+1) is not evenly divisible by the desired range (max-min+1). Since there will be a consistent mapping from a random number to an output, some outputs will be mapped to more random numbers than others. This is regardless of how the mapping is done - you can use modulo, division, conversion to floating point, whatever voodoo you can come up with, the basic problem remains.
The magnitude of the problem is very small, and undemanding applications can generally get away with ignoring it. The smaller the range and the larger RAND_MAX is, the less pronounced the effect will be.
I took your example program and tweaked it a bit. First I created a special version of rand that only has a range of 0-255, to better demonstrate the effect. I made a few tweaks to rangeRandomAlg2. Finally I changed the number of "balls" to 1000000 to improve the consistency. You can see the results here: http://ideone.com/4P4HY
Notice that the floating-point version produces two tightly grouped probabilities, near either 0.101 or 0.097, nothing in between. This is the bias in action.
I think calling this "Java's algorithm" is a bit misleading - I'm sure it's much older than Java.
int rangeRandomAlg2 (int min, int max)
{
int n = max - min + 1;
int remainder = RAND_MAX % n;
int x;
do
{
x = rand();
} while (x >= RAND_MAX - remainder);
return min + x % n;
}
It's easy to see why this algorithm produces a biased sample. Suppose your rand() function returns uniform integers from the set {0, 1, 2, 3, 4}. If I want to use this to generate a random bit 0 or 1, I would say rand() % 2. The set {0, 2, 4} gives me 0, and the set {1, 3} gives me 1 -- so clearly I sample 0 with 60% and 1 with 40% likelihood, not uniform at all!
To fix this you have to either make sure that your desired range divides the range of the random number generator, or otherwise discard the result whenever the random number generator returns a number that's larger than the largest possible multiple of the target range.
In the above example, the target range is 2, the largest multiple that fits into the random generation range is 4, so we discard any sample that is not in the set {0, 1, 2, 3} and roll again.
By far the easiest solution is std::uniform_int_distribution<int>(min, max).
You have touched on two points involving a random integer algorithm: Is it optimal, and is it unbiased?
Optimal
There are many ways to define an "optimal" algorithm. Here we look at "optimal" algorithms in terms of the number of random bits it uses on average. In this sense, rand is a poor method to use for randomly generated numbers because, among other problems with rand(), it need not necessarily produce random bits (because RAND_MAX is not exactly specified). Instead, we will assume we have a "true" random generator that can produce unbiased and independent random bits.
In 1976, D. E. Knuth and A. C. Yao showed that any algorithm that produces random integers with a given probability, using only random bits, can be represented as a binary tree, where random bits indicate which way to traverse the tree and each leaf (endpoint) corresponds to an outcome. (Knuth and Yao, "The complexity of nonuniform random number generation", in Algorithms and Complexity, 1976.) They also gave bounds on the number of bits a given algorithm will need on average for this task. In this case, an optimal algorithm to generate integers in [0, n) uniformly, will need at least log2(n) and at most log2(n) + 2 bits on average.
There are many examples of optimal algorithms in this sense. See the following answer of mine:
How to generate a random integer in the range [0,n] from a stream of random bits without wasting bits?
Unbiased
However, any optimal integer generator that is also unbiased will, in general, run forever in the worst case, as also shown by Knuth and Yao. Going back to the binary tree, each one of the n outcomes labels leaves in the binary tree so that each integer in [0, n) can occur with probability 1/n. But if 1/n has a non-terminating binary expansion (which will be the case if n is not a power of 2), this binary tree will necessarily either—
Have an "infinite" depth, or
include "rejection" leaves at the end of the tree,
And in either case, the algorithm won't run in constant time and will run forever in the worst case. (On the other hand, when n is a power of 2, the optimal binary tree will have a finite depth and no rejection nodes.)
And for general n, there is no way to "fix" this worst case time complexity without introducing bias. For instance, modulo reductions (including the min + (rand() % (int)(max - min + 1)) in your question) are equivalent to a binary tree in which rejection leaves are replaced with labeled outcomes — but since there are more possible outcomes than rejection leaves, only some of the outcomes can take the place of the rejection leaves, introducing bias. The same kind of binary tree — and the same kind of bias — results if you stop rejecting after a set number of iterations. (However, this bias may be negligible depending on the application. There are also security aspects to random integer generation, which are too complicated to discuss in this answer.)
Without loss of generality, the problem of generating random integers on [a, b] can be reduced to the problem of generating random integers on [0, s). The state of the art for generating random integers on a bounded range from a uniform PRNG is represented by the following recent publication:
Daniel Lemire,"Fast Random Integer Generation in an Interval." ACM Trans. Model. Comput. Simul. 29, 1, Article 3 (January 2019) (ArXiv draft)
Lemire shows that his algorithm provides unbiased results, and motivated by the growing popularity of very fast high-quality PRNGs such as Melissa O'Neill's PCG generators, shows how to the results can be computed fast, avoiding slow division operations almost all of the time.
An exemplary ISO-C implementation of his algorithm is shown in randint() below. Here I demonstrate it in conjunction with George Marsaglia's older KISS64 PRNG. For performance reasons, the required 64×64→128 bit unsigned multiplication is typically best implemented via machine-specific intrinsics or inline assembly that map directly to appropriate hardware instructions.
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
/* PRNG state */
typedef struct Prng_T *Prng_T;
/* Returns uniformly distributed integers in [0, 2**64-1] */
uint64_t random64 (Prng_T);
/* Multiplies two 64-bit factors into a 128-bit product */
void umul64wide (uint64_t, uint64_t, uint64_t *, uint64_t *);
/* Generate in bias-free manner a random integer in [0, s) with Lemire's fast
algorithm that uses integer division only rarely. s must be in [0, 2**64-1].
Daniel Lemire, "Fast Random Integer Generation in an Interval," ACM Trans.
Model. Comput. Simul. 29, 1, Article 3 (January 2019)
*/
uint64_t randint (Prng_T prng, uint64_t s)
{
uint64_t x, h, l, t;
x = random64 (prng);
umul64wide (x, s, &h, &l);
if (l < s) {
t = (0 - s) % s;
while (l < t) {
x = random64 (prng);
umul64wide (x, s, &h, &l);
}
}
return h;
}
#define X86_INLINE_ASM (0)
/* Multiply two 64-bit unsigned integers into a 128 bit unsined product. Return
the least significant 64 bist of the product to the location pointed to by
lo, and the most signfiicant 64 bits of the product to the location pointed
to by hi.
*/
void umul64wide (uint64_t a, uint64_t b, uint64_t *hi, uint64_t *lo)
{
#if X86_INLINE_ASM
uint64_t l, h;
__asm__ (
"movq %2, %%rax;\n\t" // rax = a
"mulq %3;\n\t" // rdx:rax = a * b
"movq %%rax, %0;\n\t" // l = (a * b)<31:0>
"movq %%rdx, %1;\n\t" // h = (a * b)<63:32>
: "=r"(l), "=r"(h)
: "r"(a), "r"(b)
: "%rax", "%rdx");
*lo = l;
*hi = h;
#else // X86_INLINE_ASM
uint64_t a_lo = (uint64_t)(uint32_t)a;
uint64_t a_hi = a >> 32;
uint64_t b_lo = (uint64_t)(uint32_t)b;
uint64_t b_hi = b >> 32;
uint64_t p0 = a_lo * b_lo;
uint64_t p1 = a_lo * b_hi;
uint64_t p2 = a_hi * b_lo;
uint64_t p3 = a_hi * b_hi;
uint32_t cy = (uint32_t)(((p0 >> 32) + (uint32_t)p1 + (uint32_t)p2) >> 32);
*lo = p0 + (p1 << 32) + (p2 << 32);
*hi = p3 + (p1 >> 32) + (p2 >> 32) + cy;
#endif // X86_INLINE_ASM
}
/* George Marsaglia's KISS64 generator, posted to comp.lang.c on 28 Feb 2009
https://groups.google.com/forum/#!original/comp.lang.c/qFv18ql_WlU/IK8KGZZFJx4J
*/
struct Prng_T {
uint64_t x, c, y, z, t;
};
struct Prng_T kiss64 = {1234567890987654321ULL, 123456123456123456ULL,
362436362436362436ULL, 1066149217761810ULL, 0ULL};
/* KISS64 state equations */
#define MWC64 (kiss64->t = (kiss64->x << 58) + kiss64->c, \
kiss64->c = (kiss64->x >> 6), kiss64->x += kiss64->t, \
kiss64->c += (kiss64->x < kiss64->t), kiss64->x)
#define XSH64 (kiss64->y ^= (kiss64->y << 13), kiss64->y ^= (kiss64->y >> 17), \
kiss64->y ^= (kiss64->y << 43))
#define CNG64 (kiss64->z = 6906969069ULL * kiss64->z + 1234567ULL)
#define KISS64 (MWC64 + XSH64 + CNG64)
uint64_t random64 (Prng_T kiss64)
{
return KISS64;
}
int main (void)
{
int i;
Prng_T state = &kiss64;
for (i = 0; i < 1000; i++) {
printf ("%llu\n", randint (state, 10));
}
return EXIT_SUCCESS;
}
If you really want to get a perfect generator assuming rand() function that you have is perfect, you need to apply the method explained bellow.
We will create a random number, r, from 0 to max-min=b-1, which is then easy to move to the range that you want, just take r+min
We will create a random number where b < RAND_MAX, but the procedure can be easily adopted to have a random number for any base
PROCEDURE:
Take a random number r in its original RAND_MAX size without any truncation
Display this number in base b
Take first m=floor(log_b(RAND_MAX)) digits of this number for m random numbers from 0 to b-1
Shift each by min (i.e. r+min) to get them into the range (min,max) as you wanted
Since log_b(RAND_MAX) is not necessarily an integer, the last digit in the representation is wasted.
The original approach of just using mod (%) is mistaken exactly by
(log_b(RAND_MAX) - floor(log_b(RAND_MAX)))/ceil(log_b(RAND_MAX))
which you might agree is not that much, but if you insist on being precise, that is the procedure.
My output is 20 random 1's, not between 10 and 1, can anyone explain why this is happening?
#include <iostream>
#include <ctime>
#include <cstdlib>
using namespace std;
int main()
{
srand((unsigned)time(0));
int random_integer;
int lowest=1, highest=10;
int range=(highest-lowest)+1;
for(int index=0; index<20; index++){
random_integer = lowest+int(range*rand()/(RAND_MAX + 1.0));
cout << random_integer << endl;
}
}
output:
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Because, on your platform, RAND_MAX == INT_MAX.
The expression range*rand() can never take on a value greater than INT_MAX. If the mathematical expression is greater than INT_MAX, then integer overflow reduces it to a number between INT_MIN and INT_MAX. Dividing that by RAND_MAX will always yield zero.
Try this expression:
random_integer = lowest+int(range*(rand()/(RAND_MAX + 1.0)))
It's much easier to use the <random> library correctly than rand (assuming you're familiar enough with C++ that the syntax doesn't throw you).
#include <random>
#include <iostream>
int main() {
std::random_device r;
std::seed_seq seed{r(), r(), r(), r(), r(), r(), r(), r()};
std::mt19937 eng(seed);
std::uniform_int_distribution<> dist(1, 10);
for(int i = 0; i < 20; ++i)
std::cout << dist(eng) << " ";
}
random_integer = (rand() % 10) + 1
That should give you a pseudo-random number between 1 & 10.
A somewhat late answer, but it should provide some additional
information if the quality of the generation is important. (Not all
applications need this—a slight bias is often not a problem.)
First, of course, the problem in the original code is the fact that
range * rand() has precedence over the following division, and is done
using integer arithmetic. Depending on RAND_MAX, this can easily
result in overflow, with implementation defined results; on all
implementations that I know, if it does result in overflow (because
RAND_MAX > INT_MAX / range, the actual results will almost certainly
be smaller than RAND_MAX + 1.0, and the division will result in a
value less than 1.0. There are several ways of avoiding this: the
simplest and most reliable is simply rand() % range + lowest.
Note that this supposes that rand() is of reasonable quality. Many
earlier implementations weren't, and I've seen at least one where
rand() % 6 + 1 to simulate a dice throw alternated odd and even. The
only correct solution here is to get a better implementation of
rand(); it has lead to people trying alternative solutions, such as
(range * (rand() / (RAND_MAX + 1.0))) + lowest. This masks the
problem, but it won't change a bad generator into a good one.
A second issue, if the quality of the generation is important, is
that when generating random integers, you're discretizing: if you're
simulating the throw of a die, for example, you have six possible
values, which you want to occur with equal probability. The random
generator will generate RAND_MAX + 1 different values, with equal
probability. If RAND_MAX + 1 is not a multiple of 6, there's no
possible way of distributing the values equaly amont the 6 desired
values. Imagine the simple case where RAND_MAX + 1 is 10. Using the
% method above, the values 1–4 are twice as likely as the the
values 5 and 6. If you use the more complicated formula 1 + int(6 *
(rand() / (RAND_MAX + 1.0))) (in the case where RAND_MAX + 1 == 10,
it turns out that 3 and 6 are only half as likely as the other values.
Mathematically, there's simply no way of distributing 10 different
values into 6 slots with an equal number of elements in each slot.
Of course, RAND_MAX will always be considerably larger than 10, and
the bias introduced will be considerably less; if the range is
significantly less than RAND_MAX, it could be acceptable. If it's
not, however, the usual procedure is something like:
int limit = (RAND_MAX + 1LL) - (RAND_MAX + 1LL) % range;
// 1LL will prevent overflow on most machines.
int result = rand();
while ( result >= limit ) {
result = rand();
}
return result % range + lowest;
(There are several ways of determining the values to throw out. This
happens to be the one I use, but I remember Andy Koenig using something
completely different—but which resulted in the same values being
thrown out in the end.)
Note that most of the time, you won't enter the loop; the worst case is
when range is (RAND_MAX + 1) / 2 + 1, in which case, you'll still
average just under one time through the loop.
Note that these comments only apply when you need a fixed number of
discrete results. For the (other) common case of generating a random
floating point number in the range of [0,1), rand() / (RAND_MAX +
1.0) is about as good as you're going to get.
Visual studio 2008 has no trouble with that program at all and happily generates a swathe of random numbers.
What I would be careful of is the /(RAND_MAX +1.0) as this will likely fall foul of integer problems and end up with a big fat zero.
Cast to double before dividing and then cast back to int afterwards
I suggest you replace rand()/(RAND_MAX + 1.0) with range*double(rand())/(RAND_MAX + 1.0)). Since my solution seems to give headaches ...
possible combinations of arguments:
range*rand() is an integer and overflows.
double(range*rand()) overflows before you convert it to double.
range*double(rand()) is not overflowing and yields expected results.
My original post had two braces but they did not change anything (results are the same).
(rand() % highest) + lowest + 1
Probably "10 * rand()" is smaller than "RAND_MAX + 1.0", so the value of your calculation is 0.
You are generating a random number (ie (range*rand()/(RAND_MAX + 1.0))) whose value is between -1 and 1 (]-1,1[) and then casting it to an integer. The integer value of such number is always 0 so you end up with the lower + 0
EDIT: added the formula to make my answer clearer
What about using a condition to check if the last number is the same as the current one? If the condition is met then generate another random number. This solution works but it will take more time though.
It is one of the simplest logics, got it from a blog. in this logic you can limit the random numbers with that given modulus(%) operator inside the for loop, its just a copy and paste from that blog, but any way check it out:
// random numbers generation in C++ using builtin functions
#include <iostream>
using namespace std;
#include <iomanip>
using std::setw;
#include <cstdlib> // contains function prototype for rand
int main()
{
// loop 20 times
for ( int counter = 1; counter <= 20; counter++ ) {
// pick random number from 1 to 6 and output it
cout << setw( 10 ) << ( 1 + rand() % 6 );
// if counter divisible by 5, begin new line of output
if ( counter % 5 == 0 )
cout << endl;
}
return 0; // indicates successful termination
} // end main
- See more at: http://www.programmingtunes.com/generation-of-random-numbers-c/#sthash.BTZoT5ot.dpuf
I need your help and please give me some advice. From programming pearls I know that to generate random 30 bit integer we should write it like this:
RAND_MAX*rand()+rand()
But what could I do for generating not 30, but 64 bit random integer instead? I think that is very inefficient method if I multiply two 30 bit integers and then multiply again 4 bit integer, so what kind of method should I use?
I am using now popcount_1 different method for 64 bit one and I would like to test it on random integers(I am also measuring the time which each one takes to accomplish the task)
First, I have my doubts about the solution you post for a 30 bit
integer. RAND_MAX itself could be a 31 bit value, and RAND_MAX *
rand() + rand() is likely to overflow, producing undefined behavior
(and in practice, negative values).
If you need a value larger than the guaranteed minimum of RAND_MAX, or
for that matter, anything that isn't significantly smaller than
RAND_MAX, the only solution will be to use successive calls to
rand(), and combine the values, but you need to do this carefully, and
validate the results. (Most implementations of rand() use linear
congruent generators, which while adequate for some tasks, aren't
particularly good in this case.) Anyway, something like:
unsigned
rand256()
{
static unsigned const limit = RAND_MAX - RAND_MAX % 256;
unsigned result = rand();
while ( result >= limit ) {
result = rand();
}
return result % 256;
}
unsigned long long
rand64bits()
{
unsigned long long results = 0ULL;
for ( int count = 8; count > 0; -- count ) {
results = 256U * results + rand256();
}
return results;
}
(The code in rand256 is designed to eliminate the otherwise
unavoidable bias you get when mapping RAND_MAX values to 256 values.)
This could be a solution, without multiplication:
r30 = RAND_MAX*rand()+rand()
s30 = RAND_MAX*rand()+rand()
t4 = rand() & 0xf
res = (r30 << 34) + (s30 << 4) + t4
If boost is an option, you could use boost random.
A random 64 bit int is essentially 64 random bits interpreted as an int.
Fill a byte array of length 8 with random bytes (see here for how) and interpret these as an int (see here for how).
A generic solution:
template <unsigned long long I> struct log2 {
static const int result = 1 + log2<I/2>::result;
};
template <> struct log2<1> {
static const int result = 0;
};
template <typename UINT> UINT genrand() {
UINT result = 0;
int bits = std::numeric_limits<UINT>::digits;
int rand_bits = log2<RAND_MAX>::result;
while (bits > 0) {
int r = rand();
while (r >= (1<<rand_bits)) r = rand(); // Retry if too big.
result <<= rand_bits;
result += r;
bits -= rand_bits;
}
return result;
}
Use: unsigned long long R = genrand<unsigned long long>();.
The bits counter keeps track of the number of bits still needed.
'Returns a pseudo-random integral value between 0 and RAND_MAX (0 and RAND_MAX included).' - http://en.cppreference.com/w/cpp/numeric/random/rand
So you should use RAND_MAX + 1 (it's like generating a number digit by digit and then converting it to base 10) instead of RAND_MAX.
This way you can generate numbers with one, two, three etc. digits in base RAND_MAX + 1(possibly with leading zeroes) and convert them to base 10 and get arbitrarily large numbers.
Everything that you obtain larger than your desired MAX_VALUE can be discarded and you still get 1/(MAX_VALUE + 1) probability for obtaining each number.
Note, that this method might take a while, especially if your desired MAX_VALUE is a lot less than the maximum value that can be obtained before discarding the numbers that are not desired, as the expected number of steps to obtain a random number in [0, MAX_VALUE] with this algorithm is: (MAX_OBTAINABLE_VALUE + 1)/(MAX_VALUE + 1)