生成某些形式的两个值（x，y）的矢量的函数

Question

EDIT: I reformulate the question simply: How do I generate in C++ or Python, random points (x,y) following: a circular distribution, a square distribution, and a triangular distribution.

This is a simple code for square for example:

def generateSquare(min, max, size):
    data = []

    for i in range(size):
        x = randint(min, max)
        y = randint(min, max)
        data += [[x, y]]

    return data

Answer 1

First of all, instead of storing your coordinates in a vector, you would be better off using std::pair or a custom class:

struct Point
{
    int x;
    int y;
};

Then you just need to have a way of generating random points, such as

Point randomPoint(Point const & min, Point const & max)
{
    static std::mt19937 gen;
    std::uniform_int_distribution<> distribX(min.x, max.x);
    std::uniform_int_distribution<> distribY(min.y, max.y);

    return Point{distribX(gen), distribY(gen)};
}

You can then use this generation function to fill your vector, for instance with generate_n :

unsigned int const nbPoints = 100;

std::vector<Point> points;

std::generate_n(back_inserter(points), nbPoints, 
    std::bind(randomPoint, Point{0, 0}, Point{1000, 1000}));

Note that this will generate random points, so you are not guaranteed to end up with a square, a triangle, etc. If you want to generate a could, you could either use a non-uniform distribution (if you know what distribution your coordinates follow) to generate your numbers, or use rejection sampling to discard points that are not in the area you want them to be.

Generating a triangle boils down to drawing three random points.

To generate a square, you can draw two points, corresponding to two opposite corners of the square.

And so on... I don't think there is a "universal" solution that would work for any shapes.

Answer 2

As supplement to Luc Touraille's post.

For a square find two random points and let these two points be the two furhest apart corners of the square.

For a triangle find three random points and let the triangle be triangle these three points make.

For a circle find a random point as a center for the circle and another random point, and let the distance between the two be the radius of the circle.

A more general approach could be to find the center point of the figures and let the parameters (scale, rotation, etc.) be found by further randomly generated numbers. (I guess a bit like Rook suggests).

Answer 3

Your problem is underspecified.

There is no such thing as a "circular distribution" or "triangular distribution".

You probably meant: a uniform distribution in the shape of a circle, rectangle, triangle. There even is no uniquely specified triangle...

The key point is uniform .

Eg a standard normal distribution in 2D may appear to be somewhat circular, but it is not exactly the shape of a circle.

There is no random generator that directly produces a circle with uniform density; at least not that I know of. The most common way is just to generate a square, and reject those points that you do not want to have .

Eg generate (x,y) pairs on [0:1]x[0:1] and reject those with distance from .5,.5 larger than .5 - then you get the circle.

If you - as other users suggested - generate a radius and a distance, then the generated points will not be uniformly distributed on the circle.