Randomly generated obstacles sometimes placed at the same position

Question

Below code is to generate the random positions for the obstacles. Obstacles are moving from right to left,so i am using its x coordinate to move it in left direction. When obstacles reaches left side of screen it again placed at some random position. But here problem is that sometimes the obstacles are placed at same position or they are too close.

 public void Randomize() 
 {
    int random = rand.Next(1,200);

    switch (random) 
    { 
        case 200:
            if (Texture.crabrect.X < 0)
              Texture.crabrect.X = rand.Next(1000,1500);
            break;
        case 12:
           if (Texture.samosarect.X < 0)
              Texture.samosarect.X = rand.Next(1000, 2000);
            break;
        case 10:
            if (Texture.mirchirect.X < 0)
              Texture.mirchirect.X = rand.Next(1800,3000);
            break;
        case 80:
            if (Texture.mushroomrect.X < 0)
              Texture.mushroomrect.X = rand.Next(1000, 2000);
            break;
        case 195:
            if (Texture.laddoorect.X < 0)
              Texture.laddoorect.X = rand.Next(1000, 2000);
            break;
        case 56:
            if (Texture.stonerect.X < 0)
              Texture.stonerect.X = rand.Next(1000, 2000);
            break;
        case 177:
            if (Texture.cactusrect.X < 0)
              Texture.cactusrect.X = rand.Next(1000, 2000);
            break;
    } 
 }

Then check if there is already another object at the spot you want to place the obstacle and try again if this is the case. — Nico Schertler, Oct 24 '13 at 15:15
@NicoSchertler you are right but if they do not placed at exact position ...like one is at 200 and another is at 210 then what to do?? — pravin, Oct 24 '13 at 15:31
Check if the textures' ranges (`position` to `position + width`) intersect. There are a few cases that you need to check (e.g. both `p1` and `p1 + w1` are less than `p2`. Then there is clearly no intersection). — Nico Schertler, Oct 24 '13 at 15:36

Cyral · Accepted Answer · 2013-10-25T13:46:53.633

Use the distance formula to see if the two objects are to close to each other.

Here is an example, using an Obstacle class to simplify things.

public void Randomize()
{
   int random = rand.Next(1,WIDTH);  
   thisObstacle = new Obstacle(); //Blah, make your obstacle.
   thisObstacle.rect = new Rectangle(random, Y,WIDTH, HEIGHT);

   foreach (Obstacle obstacle in obstacles)
   {
        //If less than 100 pixels distance, make the obstacle somewhere else
        if (GetDistance(thisObstacle.rect, obstacle.rect) < 100)
        {
             Randomize();
             return;
        }
   }
   //If we didn't get near an obstacle, place it
        //Do whatever you do
}

private static double GetDistance(Rectangle point1, Rectangle point2)
{
     //Get distance by using the pythagorean theorem
     double a = (double)(point2.X - point1.X);
     double b = (double)(point2.Y - point1.Y);
     return Math.Sqrt(a * a + b * b);
}

pinckerman · Answer 2 · 2013-10-25T11:09:49.353

2

Why don't you place your obstacles in progression?
I mean, you randomize the position of the first obstacle, then you add a default offset and then randomize another position for your obstacle. In this way you are sure that you won't have any obstacle placed at the same position, without checking previous obstacles.

edited Oct 25 '13 at 11:09

answered Oct 24 '13 at 22:32

pinckerman

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Randomly generated obstacles sometimes placed at the same position

2 Answers2