Duplicate without overlapping?

Please see attached file.

Is there a way to duplicate a movie clip, filling up the screen with increasing speed, yet not have the movie clips overlap each other on the screen?

Currently I am using this actionscript:
//initialize counter
mycount = 0;
//function to create multiple movieclips from single movie clip
function createCircleMC() {
if (mycount<400) {
_root.circle.duplicateMovieClip(“circle”+mycount, _root.getNextHighestDepth());
randomX = 550Math.random();
randomY = 440
Math.random();
eval(“circle”+mycount)._x = randomX;
eval(“circle”+mycount)._y = randomY;
mycount++;
} else {
clearInterval(ID);
for(var i = 0;i<400;i++){
removeMovieClip(eval(“circle”+i));
}
}
}
//function to periodically call another function
ID = setInterval(createCircleMC, 25);

I appreciate any help!

nope, everytime you create a new movieClip, attach a movieClip, or duplicate a movieClip it will have to go to a new depth or replace whatever is already in that depth …

_root.getNextHighestDepth()

that codes tells the new movieClip what depth to go to

replace it with any interger(like 10) and see what happens … just replaces whatever was in depth level 10

another other way to make something fill up the screen and not “overlap” each other is if you come up with a drawing solution using Flash’s drawing commands

if you are using flash 8 you can setup a bitmap and draw into that bitmap which will retain one bitmap object in flash but will still look like you are overlaping the same object over and over

hope this helps

-optixburn

This code is not very good in my opinion but it might do what you want. By not very good I mean that you can’t control exactly what happens each time with a high level of precision.

Also it requires a lot of work to find an empty spot as the screen fills up. This extra work means that the speed at which the dots are placed slows down. To avoid slowly searching for empty places, you can turn down the number of times it will try before stopping. However, the lower you set the number of tries, the less screen is filled.

[FONT=Verdana][FONT=&quot][FONT=Verdana]Hope this helps.[/FONT]

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