CODEWITHSUNDEEP
javaandroidxmlgame-engine
usrnvm43894
usrnvm43894

Reputation: 95

generate random location for buttons

I am making a simple game. I have 6 buttons and i want to shuffle them each time on different locations.

So i've made a simple method to achieve this.

    private void changeButtonPlace(ImageView button) {
    Random r = new Random();
    int newXloc = r.nextInt(getScreenWidth() - (2 * button.getLayoutParams().width));
    int newYloc = r.nextInt(getScreenHeight() - (3 * button.getLayoutParams().height));
    button.setX(newXloc);
    button.setY(newYloc);
}

It works pretty well, but sometimes the buttons override each other which means that it goes on the almost the same location. I want each button to be on a unique location and don't touch other buttons.

Any idea how i can achieve this? Thanks in advance!

Upvotes: 2

Views: 435

Answers (3)

2D3D
2D3D

Reputation: 353

You should take in account the width of the button , and before setting the location of the button make sure no button intersect with a simple if, if they do just randomize the intersecting buttons, should be easy to programm

Edit :

Lets make it a bit simpler an assume you have 3 buttons an assuming they are located on a 2d axis with these coordinates A : 1,0 B : 2,0 C : 5,0 button width : X=2 Y=2

As you can see in this example at first glance it seems as though no button intersects with each other yet if you add the width to B you understand that is real location is B [1,3] ,[-1,1] intersects with the real location of C [4,6],[-1,1] assuming the coordinates are the center of the button

first [,] represents X axis second [,] represents Y axis therefore your check will be something like so:

For each Button x
  For each button x
     calculate real coordinates of button x and y check intersection 
      if exists recalculate the coordinates and start the loop over

this solution is a bit expensive when talking about running time but once you get the idea you will be able to find something easier.

Upvotes: 0

Hoopla70
Hoopla70

Reputation: 111

What you are looking for is collision detection, and my answer will greatly simplified this process. I suggest searching for collision detection algorithms to learn more.

So, for super simple starts, we can compare the position, length, and height of 2 boxes. For my example, I am going to assume the origin of these to boxes are their upper left corner.

if((boxA.xPos + boxA.length ) < boxB.xPos || boxA.xPos > (boxB.xPos + boxB.length))

That will check if the two boxes are touching along the x-axis, and we can change the values for the y-axis as well

if((boxA.yPos + boxA.height ) < boxB.yPos || boxA.yPos > boxB.yPos + boxB.height)

Now, this is not a very efficient way of doing this. There are lots and lots of better ways to simplified this logic, and save on resources. But, it is quick and dirty, and probably good enough for a small application like your simple game involving only 6 buttons.

So, with these two equations, you can either nest them then run your collision code inside, or you can OR them together to one equation like this:

if(((boxA.yPos + boxA.height) < boxB.yPos || boxA.yPos > (boxB.yPos + boxB.height)) || ((boxA.xPos + boxA.length ) < boxB.xPos || boxA.xPos > (boxB.xPos + boxB.length)))

That is a lot to read for one line, and if you just starting out, I would suggest nesting them so you can better see the flow of logic through the equations. But, keep in mind for the future, if you ever need to squeeze those few extra bits of performance, OR them together to one if statement is alright place to start.

Upvotes: 3

duncan
duncan

Reputation: 1161

One way would be to make a grid and instead of a random location on the screen use a random point on the grid. That way you can check if the current grid location has a button on it already.

If you want them to be more scattered you could add each button to an array and check that the new doesn't touch the other buttons. Loop infinitely creating random locations and another loop to check they don't hit the other buttons in the array. Once a new location is found add the button and break out of the infinite loop.

Upvotes: 1

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