Questions

Hi. Has somebody an idear how to make this smaller. I need the max bearing difference between all enemys. My gut feeling says it should be easier, but i can't see it.

	private void calcBearingDiff()
	{
		if (myScans.isEmpty()) return;
		
		Comparator<ATarget> compi = new Comparator<ATarget>() 
		{
			@Override
			public int compare(ATarget a, ATarget b) 
			{
				double eAbsBearA = a.getBearing() + myRobot.getHeadingRadians();
				double eAbsBearB = b.getBearing() + myRobot.getHeadingRadians();
				return Double.compare(eAbsBearA, eAbsBearB);
			}
		};
		ArrayList<ATarget> enemys = new ArrayList<ATarget>(myScans.values());
		Collections.sort(enemys, compi);
		
		enemys.add(enemys.get(0));
		
		ATarget last = null;
		for (ATarget target : enemys) 
		{
			if (last != null) 
			{
				double eAbsBearingA = last.getBearing() + myRobot.getHeadingRadians();
				double eAbsBearingB = target.getBearing() + myRobot.getHeadingRadians();
				double diff = Utils.normalRelativeAngle(eAbsBearingA - eAbsBearingB);	

	                        // .... sort out the max
			}
			last = target;
		}
	}

Ah .. i'm looking for a different approach to the problem and not to make the code smaler.

Could make ATarget comparable

Thanks Chase. ATarget is a comparable but for other stuff. What i was looking for was something around another data structur. The above code snippet is way to much code for my taste. I thougth about to register the next ATarget on every ATarget and then simple ask the difference but it didn't work very well because i had to reorder the references to much. I'm looking for something that can handle ... update ATarget ... sort out the closest (bearing) ATarget ... and give the difference back ... I think i have to find a appropriate List/Tree structure for this.

Hmmm... what are you trying to use this for? I don't think I quite understood the question. Are you trying to get the biggest angle between any two in consecutive radial order? Or the one which is the closest to 180 degrees from it?

Area Picture

Maybe this picture can explain what i need. As you can see the white lines are the absBearing to every enemy and the transparent white area is the area from the two enemys who have the widest bearing to each other. I need this for my radar and my movement.
And yes i need the biggest angle between two in radial order (i guess :)). Nevermind the blue and green lines they are for something else.

I'm not sure, that i correctly understood your question and that its algorithm is completely correct, but idea must work, i think

double minAbsBearing = Integer.MAX_VALUE;
double minRelBearing = Integer.MAX_VALUE;
double maxRelBearing = Integer.MIN_VALUE;

for (ATarget target : enemys) {
    double absBearing = ...; // me.angleTo(target) in my case
    double relBearing = Utils.normalRelativeAngle(absBearing);
    if (relBearing < minRelBearing) {
        minAbsBearing = absBearing;
        minRelBearing = relBearing;
    } 
    if (relBearing > maxRelBearing) {
        maxRelBearing = relBearing;
    }
}

double widestArea = maxRelBearing - minRelBearing;
double startAngle = minAbsBearing;
double endAngle = Utils.normalAbsoluteAngle(minAbsBearing + widestArea);

You nailed it mate. Thanks. Looks like i was totaly looking at the wrong direction. But at least my gut was right :). Man this is even smaller than i hoped for.

You are welcome, but this code would not work:) Now I have no time to describe why (writing on english is difficult for me), but try to test different cases and you will easy understand the problem.

I also thought there might be some issues at certain angles with the single loop approach.

I think you need to take a <math>n^2</math> approach, ie, a loop within a loop. Otherwise you need to use a sorting technique (with nlogn), then a single loop. Here is pseudocode:

initialise min_sum = Inf
For each bot, b:
    me_b = find the angle from me to b
    initialise min_rel = Inf, max_rel = -Inf.
    for each bot c, excluding b:
        me_c = find the angle from me to c
        rel = the relative angle between me_b and me_c
        if rel < min_rel, min_rel = rel. if rel > max_rel, max_rel = rel.
    sum = max_rel - min_rel
    if sum < min_sum, min_sum = sum, min_ang = me_b + min_rel, max_ang = me_b + max_rel.

The other approach would be to sort all the angles into radial order, then go through looking for the largest space between two angles, and then use the opposite part of the circle. So in pseudocode:

for each bot, b:
    me_b = angle from me to b
    angles[i++] = me_b

sort_ascending(angles)

max_rel = abs(relative_angle(angles[0] - angles[angles.length - 1]))
min_angle = angles[0]
max_angle = angles[angles.length-1]
max_index = 0
for(i = 1; i < angles.length; i++)
    rel = abs(relative_angles(angles[i-1] - angles[i]));
    if rel > max_rel
        max_rel = rel
        min_angle = angles[i]
        max_angle = angles[i-1]
        max_index = i