Java collision detection not functioning as expected

Question

With the following code, collisions are detected, but sides are incorrectly registered.

public int checkBoxes(int aX, int aY, int aWidth, int aHeight, int bX, int bY, int bWidth, int bHeight){

    /*
     * Returns int
     * 0 - No collisions
     * 1 - Top
     * 2 - Left
     * 3 - Bottom
     * 4 - Right
     */

    Vector2f aMin = new Vector2f(aX, aY);
    Vector2f aMax = new Vector2f(aX + aWidth, aY + aHeight);

    Vector2f bMin = new Vector2f(bX, bY);
    Vector2f bMax = new Vector2f(bX + bWidth, bY + bHeight);

    float left = bMin.x - aMax.x;
    float right = bMax.x - aMin.x;
    float top = bMin.y - aMax.y;
    float bottom = bMax.y - aMin.y;

    if(left > 0) return 0;
    if(right < 0) return 0;
    if(top > 0) return 0;
    if(bottom < 0) return 0;

    int returnCode = 0;

    if (Math.abs(left) < right)
    {
        returnCode = 2;
    } else {
        returnCode = 4;
    }

    if (Math.abs(top) < bottom)
    {
        returnCode = 1;
    } else {
        returnCode = 3;
    }

    return returnCode;
}

When A is colliding with the top, left, or right of shape B, the number 3 is returned, and when colliding with the bottom, the number 1 is returned. I don't really know what's causing this. What is wrong with my code?

auselen · Answer 1 · 2012-12-22T00:47:20.173

2

With these if blocks, you'll always get 1 or 3 because second if block will always execute independent of what you've set in the first one.

if (Math.abs(left) < right)
{
    returnCode = 2;
} else {
    returnCode = 4;
}

if (Math.abs(top) < bottom)
{
    returnCode = 1;
} else {
    returnCode = 3;
}

edited Dec 22 '12 at 00:47

answered Dec 22 '12 at 00:15

auselen

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Ah, so do I need to reverse the order, or modify the code in some way? – Jack Wilsdon Dec 22 '12 at 00:16
It now acts differently, the top, right and bottom being 4, and the left being 2. I'm starting to think it's something wrong with my maths. – Jack Wilsdon Dec 22 '12 at 00:21
1

I actually didn't think about your collision algorithm. However I don't understand it anyway. What would happen if they are overlapping? One inside another? how you can say it with your 4 state? – auselen Dec 22 '12 at 00:26
To be perfectly honest, I don't know. I was provided this code in ActionScript format, without an explanation. I have been trying to understand and run it, but to no avail. Hopefully someone can explain it. – Jack Wilsdon Dec 22 '12 at 00:32
Sorry i think it is my suggestion that's wrong. Now with this one it will be always 4 or 2 :) - removing from the answer. – auselen Dec 22 '12 at 00:45

score 1 · Accepted Answer · answered Dec 22 '12 at 15:53

The problem is you are checking for one side, but when you check for left by example and the bottom is also collided you are neglecting that one. I tested the code here: http://wonderfl.net/c/i90L

What I did was first getting the distances for the X and Y of the sides. And then checking which distance was the biggest, multiplied by the size of the rectangle itself, because that side will always be the good edge on a square.

    Vector2f returnCode = new Vector2f(0, 0);

    returnCode.x = (Math.abs(left) - right) * aWidth;
    returnCode.y = (Math.abs(top) - bottom) * aHeight;

    int temp = 0;

    if(returnCode.x > 0){
        //Hits left
        temp = 2;
    }else{
        //Hits right
        temp = 4;
    }

    if(returnCode.y > 0){
        //Hits top
        if(returnCode.y > Math.abs(returnCode.x)){
            temp = 1;
        }
    }else{
        //Hits bottom
        if(returnCode.y < -Math.abs(returnCode.x)){
            temp = 3;
        }
    }

    return temp;

Works perfectly with a switch of left and right, thankyou so much! :D — Jack Wilsdon, Dec 22 '12 at 16:23

Java collision detection not functioning as expected

2 Answers2