Which one executes faster: hitTestObject or Point.distance? - actionscript-3

I am trying to optimize my collision detection alghorithm written in AS3.
I am wondering if there is any improvements in performance if I use
Point.distance(pointObject1, pointObject2);
between two objects instead of using
object1.hitTestObject(object2);
My objects are all more or less convex so the border doesn't really matter.

Point.distance is much more faster (4 and more times!) if your test objects are complicated containers with several children inside. If you use simple Sprite's it'll be only near 25% difference in functions execution time.
It's so, because Point.distance is just counting a hypotenuse from Pythagoras' theorem. So, we've got only 2 subtractions, 1 addition and 3 involutions to count. Many modern processors have involution instruction, so it's fast. If we use hitTest, there are much more actions to be performed. And the number of these actions will grow with the complexity of hitTest'ing Sprite (because it's harder to count it's bounds).
I've just made some tests. The result confirmed that I was right.
var ar:Vector.<Sprite> = Vector.<Sprite>([]); //Sprites for hitTest
for(var i:int = 0; i < 100000; i++) {
var sp:Sprite = new Sprite(); //!The results will be other, is case of use a huge container with come objects here!
sp.graphics.drawCircle(0, 0, randomIntBetween(1, 200)); //add some shapes
sp.graphics.drawRect(0, 0, randomIntBetween(1, 200), randomIntBetween(1, 200));
sp.x = randomIntBetween(-800, 800);
sp.y = randomIntBetween(-800, 600);
sp.rotation = randomIntBetween(-360, 360); //rotate and scale in random way
sp.scaleX = sp.scaleY = Math.random();
ar.push(sp);
}
var tim:Number = new Date().time;
for each(var spr:Sprite in ar) {
ar[0].hitTestObject(spr);
}
tim = new Date().time - tim;
trace(tim);
var pn:Vector.<Point> = Vector.<Point>([]); //Points for Point.distance
for(i = 0; i < 100000; i++) {
var point:Point = new Point(randomIntBetween(-800, 800), randomIntBetween(-800, 800));
pn.push(point);
}
tim = new Date().time;
for each(var pnt:Point in pn) {
Point.distance(pn[0], pnt);
}
tim = new Date().time - tim;
trace(tim);

Actually, you can't compare those two with each other. If all your objects are single-pixel bitmaps the distance test would work fine. But I imagine that is not the case.
hitTestObject essentially checks the bounding rectangles of the objects, so it's pretty fast.
You could always check the distance to see if they're close enough before doing a pixel level check, because it's more expensive and you don't want to do it too much.
The distance check is great for when you set up circular bounds around an object and determining that two such circles do not intersect. If they do intersect then you have to do another check with hitTestObject to ensure the two object shapes actually overlap. So essentially you would use both, the distance check as a first pass check and hitTestObject as a follow up for accuracy.

Related

How to add Movie Clips of different sizes next to each other? AS3

Hello everyone so I am having some trouble with this I have Platforms that I add to the stage which are different sizes in Width. What I am trying to do in my for Loop is add more platforms on the right side of the current platforms x position on stage. I am having trouble because they are different sizes so they end up over lapping each other on this side-scroller game. I align the Platform MC's to the right of the registration like so:
here is the smaller size Movie clip:
I am doing this because I want to add different obstacles to each frame inside the Platform Movie Clip.
add Initial platform:
private function addInitPlatform():void
{
platforms = new mcPlatforms();
platforms.x = (stage.stageWidth / 2) - 380;
platforms.y = (stage.stageHeight / 2) + 175;
addChildAt(platforms, 1);
aPlatformArray.push(platforms);
}
Then add new platforms:
private function addPlatForms():void
{
//Loop trhough Platform Array
for (var i:int = 0; i < aPlatformArray.length; i++)
{
var currentPlat:mcPlatforms = aPlatformArray[i];
nOffSetX += currentPlat.width + 50;
//Add platforms
platforms = new mcPlatforms();
platforms.x = nOffSetX;
platforms.y = (stage.stageHeight / 2) + 175;
addChildAt(platforms, 1);
aPlatformArray.push(platforms);
break;
}
trace(aPlatformArray.length + " NPLATFORMS");
}
I am trying to get the current platform which is the last platform I added to the stage and get its width so i can add it at the end but it still is doing something weird and overlapping over time,
So I was wondering if anyone knows how I should go about solving this so whenever I add on a new platform Movie Clip to the stage it aligns on the right side of the last platform Movie clip added to the stage with some space in between like so:
Thank you in advance!
I'm guessing that you have bunch of different platforms in your library with Linkage names set. Not sure if you want them to be in random order or not, but anyways you probably want to pick them up from an array, so:
var aPlatformsArray:Array = [p1,p2,p3]; //Platform Mc's from library
var addedPlatforms:Array = new Array(); //Array where we store added platforms
The first method was to simply raise offset after adding each platform:
var offsetX:Number = 0; //Starting x for the first platform
for(var i:int=0; i<10; i++){
//Just picking random ones from the Linkage array:
var platform:MovieClip = new aPlatformsArray[Math.floor(Math.random() * aPlatformsArray.length)]();
platform.y = 200;
platform.x = offsetX;
offsetX = platform.x + platform.width + 50;
addChild(platform);
addedPlatforms.push(platform);
}
And actually you don't need to worry about the second method as it's technically slower.
But the main difference is that now we have an array where we pick platforms to the stage, and another array where we push added platforms. The second array is not needed at this point, but you'll probably need it later.
And for offset calculations we use x and width of the "previous" element +fixed gap.
And if you just needed fixed order for the platforms, you'd use similar condition for the loop as you already did, and pick platforms in correct order:
for(var i:int=0; i<aPlatformsArray.length; i++){
var platform:MovieClip = aPlatformsArray[i];
Let me know if this does the job or if I got something wrong.

Moving Object to another Objects position

Hey everyone so I am having some trouble trying to get this to work correctly. I have a MC Object called character and another called "points". I have a container object called planetContainer I add the character to the planetContainer the character is rotating around the planets that are also added to the container. The main issue I am having is when the points power up is activated I want the points to move off the other planets and to the charactercenter position. It was working perfect but had to update some code and remove the Points out of the planetContainer and attach them to the planets instead. I know I might have to use localToGlobal but not too sure.
Here is how I setup the character:
private function newCounterClockWise():void
{
planetContainer.addChild(character);
character.rotation = (Math.atan2(character.y - planetHit.y, character.x - planetHit.x) * 180 / Math.PI);
}
How the points are added to the Planets:
private function addPoints():void
{
points = new mcPoints();
var planetPosition:Point = planetContainer.parent.localToGlobal(new Point(0, 0));
points.x = planetPosition.x;
points.y = planetPosition.y;
outerPlanets.addChild(points);
aPointsArray.push(points);
}
Now this is the main function that handles the points to move to the character but it is not working correctly. The points move but they move off the screen or cause the game to kinda tweak out and do different things. Also the "magnetHandler(); is in my EnterFRame Event:
private function magnetHandler():void
{
for (var i:int = 0; i < aPointsArray.length; i++)
{
var currentPoints:mcPoints = aPointsArray[i];
var characterPosition:Point = planetContainer.parent.globalToLocal(new Point(character.x, character.y));
if (currentPoints.hitTestObject(playScreen.mcPointsHit))
{
trace("POINTS MID STAGE");
currentPoints.x -= (currentPoints.x - characterPosition.x);
currentPoints.y -= (currentPoints.y - characterPosition.y);
//currentPoints.x = character.x;
//currentPoints.y = character.y;
//TweenMax.to(currentPoints, 0.5, {x:characterGlobalPosition.x, y:characterGlobalPosition.y , ease:Power1.easeInOut } );
}
}
}
Can anyone see what I am doing wrong?
It's a hard to understand your question fully (or to understand why you're putting things that relate to each other in separate containers), but likely this line is where it's falling down:
var characterPosition:Point = planetContainer.parent.globalToLocal(new Point(character.x, character.y));
What you want to do, is get the characters x/y coordinates in the currentPoints parent space. To do that, you would do something like this:
//first, find the global position of character:
var globalCharacterPoint:Point = character.localToGlobal(new Point());
//then, convert that to the currentPoints parent local space:
var localCharacterPoint:Point = currentPoints.parent.globalToLocal(globalCharacterPoint);
Also, in this code of yours:
points = new mcPoints();
var planetPosition:Point = planetContainer.parent.localToGlobal(new Point(0, 0));
points.x = planetPosition.x;
points.y = planetPosition.y;
You are getting the global space of the planetContainer's parent, which is probably NOT what you want. You likely want:
planetContainer.localToGlobal(new Point()); //this gives you the global location of the planet container's top left corner
And, since you're adding the points object to outerPlanets, you probably want to convert to its local space (unless it's positioned at 0,0 globally - then it doesn't especially matter).
var outerPoint:Point = outerPlanets.globalToLocal(planetPosition);
points.x = outerPoint.x;
points.y = outerPoint.y;
Needless to say, for games it's best to have everything in the global coordinate space unless it's truly encapsulated assets (like smoke on a rocket etc.)

AS3: How to refer an object by his properties

Well, I'm doing a checkers game and I need to refer a piece by its position (x and y, both) and remove it from the screen (no problem with this).
I've been traying combinations with "this." but nothing.
How would you do that?
this.x and this.y are functional from the scope of your checkers pieces object; however, if you're accessing a piece outside of their scope, you must use a piece's instance name. Although not optimal, you could loop through children DisplayObjects.
// create a collection of your checker pieces
var checkers:Array = [];
// create a checker piece, whatever your DisplayObject class is.
var checker:Checker;
checkers.push(checker);
// add it to the stage, probably your game board
addChild(checker);
checker.x = 100;
checker.y = 100;
// loop through the children (from your game board)
for (var i:uint = 0; i < numChildren; i++)
{
var checker:DisplayObject = getChildAt(i);
trace(checker.x);
trace(checker.y);
}
Using coordinates to reference a piece may not be optimal for game play. You might want to consider a row / column or approach it from how your game board works.
If this is not clear, you should specify some code or expand your question with more detail.

How to avoid create new bitmapData object in loop in AS3?

I want to store the bitmap data from _sampleTile in array, but I was wondering how to increase the performance. If I do it like this:
var _sampleTile:BitmapData;
var _arrayLenght:int = _tileClipArray.length;
for(var i:int = 0; i < _arrayLenght; ++i){
_sampleTile = new BitmapData(65, 65, false);
_sampleTile.draw(_tileClipArray[int(i)]);
_tileBitmapDataArray[i] = _sampleTile;
}
Then it would do too much constructing job in the loop, right? But if I do as bellow:
var _sampleTile:BitmapData = new BitmapData(65, 65, false);
var _arrayLenght:int = _tileClipArray.length;
for(var i:int = 0; i < _arrayLenght; ++i){
_sampleTile.fillRect(_sourceRectangle, 0x00FFFFFF);
_sampleTile.draw(_tileClipArray[int(i)]);
_tileBitmapDataArray[i] = _sampleTile.clone();
}
The .clone() returns a new BitmapData object so basically the result is the same, right?
In the second example if we replace the _sampleTile.clone() with _sampleTile - is it somehow possible to not store in array a reference to _sampleTile, but get the actual bitmapData from the _simpleTile?
No, you need to create a new BitmapData each iteration... either with clone() or new.
I see a couple alternatives though:
Make your creation asynchronous. Do just a few each frame, till you finish the whole batch.
Create a big BitmapData, draw all tiles in there and use references for the position of each tile. If the tiles are always the same, then you could eventually save the final BitmapData + positions and load them instead of creating them each time you run the application.

How to get unrotated display object width/height of a rotated display object?

If I create a rectangle with 100px width and 100px height and then rotate it, the size of the element's "box" will have increased.
With 45 rotation, the size becomes about 143x143 (from 100x100).
Doing sometimes like cos(angleRad) * currentWidth seems to work for 45 rotation, but for other bigger angles it doesn't.
At the moment I am doing this:
var currentRotation = object.rotation;
object.rotation = 0;
var normalizedWidth = object.width;
var normalizedHeight = object.height;
object.rotation = currentRotation;
Surely, there must be a better and more efficient way. How should I get the "normalized" width and height of a displayobject, aka the size when it has not been rotated?
The best approach would probably be to use the code posted in the question - i.e. to unrotate the object, check its width, and then re-rotate it. Here's why.
First, simplicity. It's obvious what's being done, and why it works. Anyone coming along later should have no trouble understanding it.
Second, accuracy. Out of curiosity I coded up all three suggestions currently in this thread, and I was not really surprised to find that for an arbitrarily scaled object, they give three slightly different answers. The reason for this, in a nutshell, is that Flash's rendering internals are heavily optimized, and among other things, width and height are not stored internally as floats. They're stored as "twips" (twentieths of a pixel) on the ground that further accuracy is visually irrelevant.
Anyway, if the three methods give different answers, which is the most accurate? For my money, the most correct answer is what Flash thinks the width of the object is when it's unrotated, which is what the simple method gives us. Also, this method is the only one that always give answers rounded to the nearest 1/20, which I surmise (though I'm guessing) to mean it's probably equal to the value being stored internally, as opposed to being a calculated value.
Finally, speed. I assume this will surprise you, but when I coded the three methods up, the simple approach was the fastest by a small margin. (Don't read too much into that - they were all very close, and if you tweak my code, a different method might edge into the lead. The point is they're very comparable.)
You probably expected the simple method to be slower on the grounds that changing an object's rotation would cause lots of other things to be recalculated, incurring overhead. But all that really happens immediately when you change the rotation is that the object's transform matrix gets some new values. Flash doesn't really do much with that matrix until it's next time to draw the object on the screen. As for what math occurs when you then read the object's width/height, it's difficult to say. But it's worth noting that whatever math takes place in the simple method is done by the Player's heavily optimized internals, rather than being done in AS3 like the algebraic method.
Anyway I invite you to try out the sample code, and I think you'll find that the simple straightforward method is, at the least, no slower than any other. That plus simplicity makes it the one I'd go with.
Here's the code I used:
// init
var clip:MovieClip = new MovieClip();
clip.graphics.lineStyle( 10 );
clip.graphics.moveTo( 12.345, 37.123 ); // arbitrary
clip.graphics.lineTo( 45.678, 29.456 ); // arbitrary
clip.scaleX = .87; // arbitrary
clip.scaleY = 1.12; // arbitrary
clip.rotation = 47.123; // arbitrary
// run the test
var iterations:int = 1000000;
test( method1, iterations );
test( method2, iterations );
test( method3, iterations );
function test( fcn:Function, iter:int ) {
var t0:uint = getTimer();
for (var i:int=0; i<iter; i++) {
fcn( clip, i==0 );
}
trace(["Elapsed time", getTimer()-t0]);
}
// the "simple" method
function method1( m:MovieClip, traceSize:Boolean ) {
var rot:Number = m.rotation;
m.rotation = 0;
var w:Number = m.width;
var h:Number = m.height;
m.rotation = rot;
if (traceSize) { trace([ "method 1", w, h ]); }
}
// the "algebraic" method
function method2( m:MovieClip, traceSize:Boolean ) {
var r:Number = m.rotation * Math.PI/180;
var c:Number = Math.abs( Math.cos( r ) );
var s:Number = Math.abs( Math.sin( r ) );
var denominator:Number = (c*c - s*s); // an optimization
var w:Number = (m.width * c - m.height * s) / denominator;
var h:Number = (m.height * c - m.width * s) / denominator;
if (traceSize) { trace([ "method 2", w, h ]); }
}
// the "getBounds" method
function method3( m:MovieClip, traceSize:Boolean ) {
var r:Rectangle = m.getBounds(m);
var w:Number = r.width*m.scaleX;
var h:Number = r.height*m.scaleY;
if (traceSize) { trace([ "method 3", w, h ]); }
}
And my output:
method 1,37.7,19.75
Elapsed time,1416
method 2,37.74191378925391,19.608455916982187
Elapsed time,1703
method 3,37.7145,19.768000000000004
Elapsed time,1589
Surprising, eh? But there's an important lesson here about Flash development. I hereby christen Fen's Law of Flash Laziness:
Whenever possible, avoid tricky math by getting the renderer to do it for you.
It not only gets you done quicker, in my experience it usually results in a performance win anyway. Happy optimizing!
Here's the algorithmic approach, and its derivation.
First, let's do the opposite problem: Given a rectangle of unrotated width w, unrotated height h, and rotation r, what is the rotated width and height?
wr = abs(sin(r)) * h + abs(cos(r)) * w
hr = abs(sin(r)) * w + abs(cos(r)) * h
Now, try the problem as given: Given a rectangle of rotated width wr, rotated height hr, and rotation r, what is the unrotated width and height?
We need to solve the above equations for h and w. Let c represent abs(cos(r)) and s represent abs(sin(r)). If my rusty algebra skills still work, then the above equations can be solved with:
w = (wr * c - hr * s) / (c2 - s2)
h = (hr * c - wr * s) / (c2 - s2)
You should get the bounds of your square in your object's coordinate space (which means no rotations).
e.g.
var b:Sprite = new Sprite();
b.graphics.lineStyle(0.1);
b.graphics.drawRect(0,0,100,100);
b.rotation = 10;
trace('global coordinate bounds: ' + b.getBounds(this));//prints global coordinate bounds: (x=-17.35, y=0, w=115.85, h=115.85);
trace('local coordinate bounds: ' + b.getBounds(b));//prints local coordinate bounds: (x=0, y=0, w=100, h=100)
HTH,
George
Chip's answer in code:
// convert degrees to radians
var r:Number = this.rotation * Math.PI/180;
// cos, c in the equation
var c:Number = Math.abs(Math.cos(r));
// sin, s in the equation
var s:Number = Math.abs(Math.sin(r));
// get the unrotated width
var w:Number = (this.width * c - this.height * s) / (Math.pow(c, 2) - Math.pow(s, 2));