CODEWITHSUNDEEP
c++11random
Josh C
Josh C

Reputation: 1111

Making a custom Distribution of Random Numbers

I am trying to make sense of the different distribution objects in c++11 and I am finding it overwhelming. I hope some of you can and will help.

This is why I am looking into all this:

I need a random number generator that I can adjust every time it is used so that it is more likely to produce the same number again. The second requirement I need to fill is that I need the random numbers generated to only be these numbers: {1, 2, 4, 8, 16, ..., 128}

Third and last requirement is that on certain occasions I need to skip one or more numbers from the above set.

My problem is that I don't understand the descriptions of various distribution objects. I, thus, cannot determine what tools I need to use to meet my above needs.

Can somebody tell me what tools I need and how I need to use them? The more clear, concise and detailed the response the better.

Upvotes: 2

Views: 854

Answers (2)

joshbooks
joshbooks

Reputation: 489

For skipping numbers I completely agree with Howard that manual checking is probably the way to go, but there might be a better way altering the probability of a given number being generated. Another way to do this would be to use a discrete_distribution object, which allows you to specify the probability of generating any given value, so for your example it would be something like

std::default_random_engine entropy;
std::array<double, 128> probs;
probs.fill(1.0);
std::discrete_distribution<int> choose(probs.begin(), probs.end());

then when you're in your loop, in addition to deciding whether or not to skip, you can increment one of those values by some amount to increase the odds of it coming up again, making sure to reinitialize the discrete distribution, like this:

int x;
double myValue = 0.2;//or whatever increment you want
for (something; something else; something else else)
{
  x = choose(entropy);
  if (skip(x))
    continue;//alternately you could set probs.at(x) = 0
  //only if you never want to generate it again
  probs.at(x) += myValue;
  choose = std::discrete_distribution<int>(probs.begin(), probs.end());
  output(x);
}

where skip and output are your functions to decide if x should be skipped and do whatever you want with the generated value respectively

Upvotes: 1

Howard Hinnant
Howard Hinnant

Reputation: 218770

Your range can be generated with a random number j in the range [0, 7], then you compute:

1 << j

to get your number. std::uniform_int_distribution<> would be handy for generating the value in [0, 7].

Additionally you could use a std::bernoulli_distribution (which returns a random bool) to decide if the next number is going to be the same as the last one, or if you should generate a new number. The std::bernoulli_distribution defaults to a 50/50 chance of true/false, but you can customize that distribution in the bernoulli_distribution constructor to anything you like (e.g. 80/20 or whatever).

If this isn't clear enough, just jump in with some code. Try coding it up, and if it isn't working, post what you have, and I'm sure somebody will help.

Oh, forgot about your 3rd requirement: For that just put your [0, 7] generation in a loop, and if you come up with a number you're supposed to skip, then iterate the loop, else break out of it.

Upvotes: 2

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