Random integral numbers from uniform distribution

[jangaser]

I have to write a c++ program and have a problem only with the following part

for that purpose, you will need two random integral numbers from uniform distribution on the set: 1,2,3,4,5,6,7,8,9 (those two numbers have to be generated with equal probability)

I don't know math as good as c++, also I'm not allowed to use c++11. I've written the following simple code, but I doubt if it's correct?

srand(time(0));
int first_number = rand() % 10;
int second_number = rand() % 10;

[whiteflags]

Edit:
i'm dumb. brb.

[laserlight]

No, it is not correct. The first problem is due to the range: rand() % 10 would result in integers in the range [0, 10), but you want integers in the range [1, 10). This can be easily fixed by:

int first_number = rand() % 9 + 1;

But the above is still not likely to be correct in that while the two numbers would be generated with equal probability, the uniform distribution requirement might not be satisfied due to bias introduced when you used operator% in that way.

Read Prelude's article on using rand() for a simple solution, though one that has its own drawbacks. You could think of how you could avoid this drawback while maintaining a uniform distribution.

[Elysia]

The answer I came up with is: it's impossible. Give up. rand() in itself is so biased you're unlikely to get anything in the range [1, 10). What's worse is that rand is not deterministic. Change compilers, change operating system and everything fall into pieces again. You're not going to be able to pick two numbers that confirms to a uniform distribution unless you use a 3rd party library or C++11. Forget rand ever existed.

[Codeplug]

Not impossible. Code can be written to maintain the distribution provided by rand to a smaller domain - for a sufficiently large n anyways - Random function confusion

gg

[Elysia]

Color me surprised that something like that actually works... but it does. So I guess I'll have to retract that impossible statement.

[laserlight]

Color me surprised that something like that actually works... but it does. So I guess I'll have to retract that impossible statement.

This has been discussed here several times before though, as well as in my link to Prelude's article. The solution (or fixed variants thereof, if you read further in the thread) offered by Accelerated C++ is hardly the only solution.

[GReaper]

Why don't you implement a Mersenne Twister? It's really straightforward.

[Elysia]

This has been discussed here several times before though, as well as in my link to Prelude's article. The solution (or fixed variants thereof, if you read further in the thread) offered by Accelerated C++ is hardly the only solution.

The only one I remember from Prelude's article is to call rand until you get the number you want, which unsurprisingly does not always work since rand() is not a uniform distribution.

[laserlight]

The only one I remember from Prelude's article is to call rand until you get the number you want, which unsurprisingly does not always work since rand() is not a uniform distribution.

The distribution of rand() is not specified, but as far as I know it is always assumed to be a uniform distribution. If it is not a uniform distribution, then none of the fixes suggested will result in a uniform distribution.

The flaw in calling rand() until you get the number within the reduced range that you want is that with a sufficiently small range, this would be terribly inefficient, and it might even be plausible that a particular seed might result in a sequence that does not have any numbers within that range. Hence, I suggested that "you could think of how you could avoid this drawback while maintaining a uniform distribution".