I'm creating a pop out action effect on a window after I click on some button in LibGdx.
May I know, Is there any methods available to scale the actor, in a way that keeping its center position at the center (scales in both direction of x and also both direction of y)?
addAction(sequence(scaleTo(1.3f,1.3f,0.2f),scaleTo(1f,1f,0.2f)));
This is the basic scaling method I used.
↑
O →
But I want something like this:
......↑
← O →
......↓
You need to set actor's origin to its center. First set the size, and then call:
actor.setOrigin(Align.center);
Note, that if your actor is a Table (and classes like Button, ImageButton extend Table), then you need to call:
actor.setTransform(true); // But keep in mind that has a performance impact "because {#link Batch#flush()} must be done before and after the transform"
I've been searching and having trouble doing this the right way.So my player has 2 arms which are joined in one mc. Now I'm trying to make the body(which is doesn't belong on the container mc) go between them, so it's like R arm first then body then L arm. The 2 arms rotate on mouse direction, i don't want the body too. Any ideas what way I should go?
EDIT:
please see my sample image
The checked thing is the result I'm looking for. But since the 2 arms is combined on one MC, I can't make the body appear on the middle between the 2 arms.
You should change your nesting model like this:
And then store your arms in an Array or a Vector, and change every arm's rotation separately.
It's the only way
In my cocos2dx game, I have a CCLayer that contains another CCLayer. The sublayer just cover part of the container layer. I 'think' I achieve this through:
this->setContentSize( CCSizeMake( 100, 200 ) );
however, the sublayer always capture touches even though it is outside its size and position area... Is it common?
I can filter through the touches position by comparing it inside the ccTouch** functions, but I think it is a hack, what is the proper way to set the sublayer to properly cover just the partial area of its parent?
The only thing i can think of straight away is making this inner layer a CCNODE and also extent it with CCTouchDelegate.
Now with this, when u register with the TouchDispatcher, you make sure it doesn't Swallowtouches(the boolean value given as the last parameter)...
This way when you receive a touch ... just see if it is within the boundary of this inner layer of urs and if it is not, send let the parent class use this touch.
Hope this helps.
I am playing with animation in AS3 and flex4, and I've come into a problem. My application is a game board (like a chess board), where each field is a border container added to some position.
Also I am adding a child element (shape), to this container on mouse click. What I want to achieve is to be able to move shapes smoothly from one field to another. But it appears that the shape goes behind the neighbor field this way http://screencast.com/t/iZ3DCdobs.
I believe this happens because shape is a child of specific border container, and to make it visible over every other container, I would need to use layers somehow....
I would be happy if anybody could suggest a solution
Yes you're right on that. You should add the movable objects to a different layer.
As there are no typical layers in AS, you could try to drop the fields in one sprite and any other objects to a different an than place them on each other, so that when you will move a object it won't go behind other objects.
If you place both sprites in the same position you will still have accurate x,y positions between movable objects and fields.
You have two options:
First one is to have different layers for your DisplayObjects: as an example, the bottom layer would hold all the boards, and the upper layer would hold all the pieces.
Second option is to manipulate the index of the objects with swapChildren(), swapChildrenAt(), and setChildIndex(). So to bring a MovieClip to the topmost front, you would do MovieClip(parent).setChildIndex(this, 0);
If the situation is that always the shape object gets hidden behind the next ( right side ) grid container, the I suggest you create your grid in reverse.
Suppose you are creating a chess grid. that is a 8x8 grid. Normally you would create your grid using 2 for loops, looping from 0 to 8 with say the x and y points starting at 0,0 for the first grid and going on till the end. What I suggest you to do is to create from 8,8 to 0,0.
Display objects in flash are stacked on top of each other based on their child index.
For example: If you create two objects. Rectangle and Circle as follows
var rect:Rectangle = new Rectangle();
this.addChild(rect);
var circ:Circle = new Circle();
this.addChild(circ);
The circle will always be on top of the rectangle in this scenario because the circle was added after the rectangle to the display list.
So if you reverse the order of creation of your grid, the right grid cell will be added to the display list first and so the grid cells to the left will always be on top of the right ones. Hence, the problem that you are facing will not occur.
I'm drawing rectangles at random positions on the stage, and I don't want them to overlap.
So for each rectangle, I need to find a blank area to place it.
I've thought about trying a random position, verify if it is free with
private function containsRect(r:Rectangle):Boolean {
var free:Boolean = true;
for (var i:int = 0; i < numChildren; i++)
free &&= getChildAt(i).getBounds(this).containsRect(r);
return free;
}
and in case it returns false, to try with another random position.
The problem is that if there is no free space, I'll be stuck trying random positions forever.
There is an elegant solution to this?
Let S be the area of the stage. Let A be the area of the smallest rectangle we want to draw. Let N = S/A
One possible deterministic approach:
When you draw a rectangle on an empty stage, this divides the stage into at most 4 regions that can fit your next rectangle. When you draw your next rectangle, one or two regions are split into at most 4 sub-regions (each) that can fit a rectangle, etc. You will never create more than N regions, where S is the area of your stage, and A is the area of your smallest rectangle. Keep a list of regions (unsorted is fine), each represented by its four corner points, and each labeled with its area, and use weighted-by-area reservoir sampling with a reservoir size of 1 to select a region with probability proportional to its area in at most one pass through the list. Then place a rectangle at a random location in that region. (Select a random point from bottom left portion of the region that allows you to draw a rectangle with that point as its bottom left corner without hitting the top or right wall.)
If you are not starting from a blank stage then just build your list of available regions in O(N) (by re-drawing all the existing rectangles on a blank stage in any order, for example) before searching for your first point to draw a new rectangle.
Note: You can change your reservoir size to k to select the next k rectangles all in one step.
Note 2: You could alternatively store available regions in a tree with each edge weight equaling the sum of areas of the regions in the sub-tree over the area of the stage. Then to select a region in O(logN) we recursively select the root with probability area of root region / S, or each subtree with probability edge weight / S. Updating weights when re-balancing the tree will be annoying, though.
Runtime: O(N)
Space: O(N)
One possible randomized approach:
Select a point at random on the stage. If you can draw one or more rectangles that contain the point (not just one that has the point as its bottom left corner), then return a randomly positioned rectangle that contains the point. It is possible to position the rectangle without bias with some subtleties, but I will leave this to you.
At worst there is one space exactly big enough for our rectangle and the rest of the stage is filled. So this approach succeeds with probability > 1/N, or fails with probability < 1-1/N. Repeat N times. We now fail with probability < (1-1/N)^N < 1/e. By fail we mean that there is a space for our rectangle, but we did not find it. By succeed we mean we found a space if one existed. To achieve a reasonable probability of success we repeat either Nlog(N) times for 1/N probability of failure, or N² times for 1/e^N probability of failure.
Summary: Try random points until we find a space, stopping after NlogN (or N²) tries, in which case we can be confident that no space exists.
Runtime: O(NlogN) for high probability of success, O(N²) for very high probability of success
Space: O(1)
You can simplify things with a transformation. If you're looking for a valid place to put your LxH rectangle, you can instead grow all of the previous rectangles L units to the right, and H units down, and then search for a single point that doesn't intersect any of those. This point will be the lower-right corner of a valid place to put your new rectangle.
Next apply a scan-line sweep algorithm to find areas not covered by any rectangle. If you want a uniform distribution, you should choose a random y-coordinate (assuming you sweep down) weighted by free area distribution. Then choose a random x-coordinate uniformly from the open segments in the scan line you've selected.
I'm not sure how elegant this would be, but you could set up a maximum number of attempts. Maybe 100?
Sure you might still have some space available, but you could trigger the "finish" event anyway. It would be like when tween libraries snap an object to the destination point just because it's "close enough".
HTH
One possible check you could make to determine if there was enough space, would be to check how much area the current set of rectangels are taking up. If the amount of area left over is less than the area of the new rectangle then you can immediately give up and bail out. I don't know what information you have available to you, or whether the rectangles are being laid down in a regular pattern but if so you may be able to vary the check to see if there is obviously not enough space available.
This may not be the most appropriate method for you, but it was the first thing that popped into my head!
Assuming you define the dimensions of the rectangle before trying to draw it, I think something like this might work:
Establish a grid of possible centre points across the stage for the candidate rectangle. So for a 6x4 rectangle your first point would be at (3, 2), then (3 + 6 * x, 2 + 4 * y). If you can draw a rectangle between the four adjacent points then a possible space exists.
for (x = 0, x < stage.size / rect.width - 1, x++)
for (y = 0, y < stage.size / rect.height - 1, y++)
if can_draw_rectangle_at([x,y], [x+rect.width, y+rect.height])
return true;
This doesn't tell you where you can draw it (although it should be possible to build a list of the possible drawing areas), just that you can.
I think that the only efficient way to do this with what you have is to maintain a 2D boolean array of open locations. Have the array of sufficient size such that the drawing positions still appear random.
When you draw a new rectangle, zero out the corresponding rectangular piece of the array. Then checking for a free area is constant^H^H^H^H^H^H^H time. Oops, that means a lookup is O(nm) time, where n is the length, m is the width. There must be a range based solution, argh.
Edit2: Apparently the answer is here but in my opinion this might be a bit much to implement on Actionscript, especially if you are not keen on the geometry.
Here's the algorithm I'd use
Put down N number of random points, where N is the number of rectangles you want
iteratively increase the dimensions of rectangles created at each point N until they touch another rectangle.
You can constrain the way that the initial points are put down if you want to have a minimum allowable rectangle size.
If you want all the space covered with rectangles, you can then incrementally add random points to the remaining "free" space until there is no area left uncovered.