Develop Reference

Java rotation of pixel array

I have tried to make an algorithm in java to rotate a 2-d pixel array(Not restricted to 90 degrees), the only problem i have with this is: the end result leaves me with dots/holes within the image.
Here is the code :
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
int xp = (int) (nx + Math.cos(rotation) * (x - width / 2) + Math
.cos(rotation + Math.PI / 2) * (y - height / 2));
int yp = (int) (ny + Math.sin(rotation) * (x - width / 2) + Math
.sin(rotation + Math.PI / 2) * (y - height / 2));
int pixel = pixels[x + y * width];
Main.pixels[xp + yp * Main.WIDTH] = pixel;
}
}
'Main.pixels' is an array connected to a canvas display, this is what is displayed onto the monitor.
'pixels' and the function itself, is within a sprite class. The sprite class grabs the pixels from a '.png' image at initialization of the program.
I've tried looking at the 'Rotation Matrix' solutions. But they are too complicated for me. I have noticed that when the image gets closer to a point of 45 degrees, the image is some-what stretched ? What is going wrong? And what is the correct code; that adds the pixels to a larger scale array(E.g. Main.pixels[]).
Needs to be java! and relative to the code format above. I am not looking for complex examples, simply because i will not understand(As said above). Simple and straight to the point, is what i am looking for.
How id like the question to be answered.
Your formula is wrong because ....
Do this and the effect will be...
Simplify this...
Id recommend...
Im sorry if im asking to much, but i have looked for an answer relative to this question, that i can understand and use. But to always either be given a rotation of 90 degrees, or an example from another programming language.

You are pushing the pixels forward, and not every pixel is hit by the discretized rotation map. You can get rid of the gaps by calculating the source of each pixel instead.
Instead of
for each pixel p in the source
pixel q = rotate(p, theta)
q.setColor(p.getColor())
try
for each pixel q in the image
pixel p = rotate(q, -theta)
q.setColor(p.getColor())
This will still have visual artifacts. You can improve on this by interpolating instead of rounding the coordinates of the source pixel p to integer values.
Edit: Your rotation formulas looked odd, but they appear ok after using trig identities like cos(r+pi/2) = -sin(r) and sin(r+pi/2)=cos(r). They should not be the cause of any stretching.

To avoid holes you can:
compute the source coordinate from destination
(just reverse the computation to your current state) it is the same as Douglas Zare answer
use bilinear or better filtering
use less then single pixel step
usually 0.75 pixel is enough for covering the holes but you need to use floats instead of ints which sometimes is not possible (due to performance and or missing implementation or other reasons)
Distortion
if your image get distorted then you do not have aspect ratio correctly applied so x-pixel size is different then y-pixel size. You need to add scale to one axis so it matches the device/transforms applied. Here few hints:
Is the source image and destination image separate (not in place)? so Main.pixels and pixels are not the same thing... otherwise you are overwriting some pixels before their usage which could be another cause of distortion.
Just have realized you have cos,cos and sin,sin in rotation formula which is non standard and may be you got the angle delta wrongly signed somewhere so
Just to be sure here an example of the bullet #1. (reverse) with standard rotation formula (C++):
float c=Math.cos(-rotation);
float s=Math.sin(-rotation);
int x0=Main.width/2;
int y0=Main.height/2;
int x1= width/2;
int y1= height/2;
for (int a=0,y=0; y < Main.height; y++)
for (int x=0; x < Main.width; x++,a++)
{
// coordinate inside dst image rotation center biased
int xp=x-x0;
int yp=y-y0;
// rotate inverse
int xx=int(float(float(xp)*c-float(yp)*s));
int yy=int(float(float(xp)*s+float(yp)*c));
// coordinate inside src image
xp=xx+x1;
yp=yy+y1;
if ((xp>=0)&&(xp<width)&&(yp>=0)&&(yp<height))
Main.pixels[a]=pixels[xp + yp*width]; // copy pixel
else Main.pixels[a]=0; // out of src range pixel is black
}

Two Shape objects appear to be intersecting at the wrong location?

I am trying to create a Shape with the centre of the ship being in the middle.
one.x and one.z is the X and Z positions of the ship. The ship size is about 100 on the X, and 50 on the Z.
Shape my = new Rectangle(
(int) one.x - disToLeft, // upper-left corner X
(int) one.z - disToTop, // upper-left corner Y
disToLeft + disToRight, // width
disToTop + disToBottom // height
);
I'm then rotating the Shape, to of course be facing the correct way. This appears to work:
int rectWidth = (disToLeft + disToRight);
int rectHeight = (disToTop + disToBottom);
AffineTransform tr = new AffineTransform();
// rotating in central axis
tr.rotate(
Math.toRadians(one.rotation),
x + (disToLeft + disToRight) / 2,
z + (disToTop + disToBottom) / 2
);
my = tr.createTransformedShape(my);
I am then doing the exact same thing with another Shape, and testing for intersection. This works.
However, it feels like the Shape is the incorrect dimensions. Or something. My ship is colliding very far out to one side (outside where it graphically exists), but through the other side, I can almost go right through the ship before any collision is detected!
Basically the Shapes are simply intersecting at the wrong location. And I cannot work out why. Either the shape, the location, or the rotation must be wrong.
int disToLeft = 100;
int disToRight = 100;
int disToTop = 150;
int disToBottom = 100;
These are the distance from the centre to the left, right, top, and bottom sides.
I am using Z instead of Y because my game is in a 3D world and the sea level is pretty much the same (hence I don't need to worry about Y axis collision!).
Update:
After doing a lot of testing, I have discovered that the top of the rectangle is in the middle! I have done a lot of messing around, but without being able to graphically see the squares, it's been very hard to test.
This means that the box is on the side of the ship, like this:
Obviously when the ship on the left rotates to what it's like in this picture, a collision is detected.

It seems that your rotation is wrong. From my understanding of math it should be
tr.rotate(Math.toRadians(one.rotation), x + (disToRight - disToLeft) /2, z + (disToBottom - disToTop) /2);
Note the signs and the order of the variables
Edit:
Let's take apart the formula:
Your Rectangle is defined like this:
x-coordinate (x): one.x-disToLeft
y-coordinate (y): one.z-disToTop
width: disToLeft+disToRight
height: disToTop+disToBottom
The centre of the Rectangle (where you are rotating) is therefore:
(x+width/2)
(y+height/2)
if you replace x, width, y and height with the declarations above you get
(one.x-disToLeft + (disToLeft+disToRight)/2)
(one.z-disToTop + (disToTop+disToBottom)/2)
This is already the point you need, but it can be simplyfied:
one.x- disToLeft + (disToLeft+disToRight)/2
is equal to
one.x-(2*disToLeft/2)+(disToLeft/2)+(disToRight/2)
is equal to
one.x-(distoLeft/2) + (disToRight/2)
is equal to
one.x+(disToRight-disToLeft)/2
The other coordinate works exactly the same.

Change coordinates to bottom left corner

How can I change coordinates to bottom left corner?
I know that's in Java the coordinates begin from Top=Left corner, but I'm asking if can someone help me how can I change it to begin (0,0) coordinates from Bottom-Left corner?

I think it's too late, but for people like me (new to android development). The above answers are correct but here is a more detailed one
If you get the coordinate with respect to top left as (a,b),
then the coordinates with respect to the bottom left are simply
(a, h-b) where h is the height of the view.
Example:
float x = getXcoordinatesonTouch();
float y = getYcoordinatesonTouch();
//should return height
float h = getHeightoftheView();
float transformY = h - y;
//"x" should be as it is
//Now you can show "x" and "transformY"

getHeight() will get you the size height. so (0, getHeight()) will give you the left-bottom point. But take into consideration the height of the object you want to place. So you may want to use
(0, getHeight() - heightOfObject)

Use the value (x, HEIGHT - y).

Isometric Tile Selection

I'm not all that good with Maths, so i was hoping some of you guys could help?
I'm trying to make a function to convert mouse coordiantes into a particular tile in an isometric view.
It won't let me post images for a stupid reason, so ill just link the image:
Link
All of the algorithms i have seen so far work with the X & Y axes going diagonal, my game is currently set up like this, and i would like to keep it so.
Is there an algorithm so that if the mouse was at the red dot, it would return the coordinates of the tile that it is sitting on? (6,2)
Thanks in advance!

There is a good start : http://www.java-gaming.org/index.php?topic=23656.0
Enjoy :)
EDIT
Full-trusted "DrDobb's" website, full article on this : http://www.drdobbs.com/parallel/designing-isometric-game-environments/184410055

<0;4>
x <0;3> <1;4>
<0;2> <1;3> <2;4>
<0;1> <1;2> <2;3> <3;4>
<0;0> <1;1> <2;2> <3;3> <4;4>
<1;0> <2;1> <3;2> <4;3>
<2;0> <3;1> <4;2>
y <3;0> <4;1>
<4;0>
I rendered the tiles like above.
the sollution is VERY simple!
first thing:
my Tile width and height are both = 32 this means that in isometric view, the width = 32 and height = 16! Mapheight in this case is 5 (max. Y value)
y_iso & x_iso == 0 when y_mouse=MapHeight/tilewidth/2 and x_mouse = 0
when x_mouse +=1, y_iso -=1
so first of all I calculate the "per-pixel transformation"
TileY = ((y_mouse*2)-((MapHeight*tilewidth)/2)+x_mouse/2;
TileX = x_mouse-TileY;
to find the tile coordinates I just devide both by tilewidth
TileY = TileY/32; TileX = TileX/32;
DONE!! never had any problems!

It's quite easy actually once you get your head wrapped around it. All you do is find out where your mouse is relative to the map and then reverse to how you are drawing the tiles.
I draw my map in the double "for" loop like this:
For x coord: x * (TileWidth / 2) - (y * (TileWidth / 2))
For y coord: x * (TileHeight / 2) + (y * (TileHeight / 2))
So my x goes from top left to bottom right and my y goes from top right to bottom left. Mind though, like for the first tile the world coord will be 0,0 but the top pixel starts at x=0 + (tilewidth / 2) so we have to compensate for that when we are looking to find which tile the mouse is over. (or we could do that for the whole world itself by giving it a offset).
Now first we have to find the mouse position in relation to the world since you probably want a moving camera. My camera's centre starts as 0,0 so i have to compensate the mouse by half the screen width like so:
mouseWorldPosX = mouse.x + cam.x - (screen.width / 2)
mouseWorldPosY = mouse.y + cam.y - (screen.height / 2)
This is all we need to calculate the mouse position back to tile position.
For X:
tileX = (mouseWorldPosX + (2 * mouseWorldPosY) - (tileWidth / 2)) / tileWidth
As you can see we divide the whole thing by the tilewidth since we multiplied it in the draw method. The (tileWidth / 2) is just there to compensate for the offset i mentioned earlier.
For Y:
tileY = (mouseWorldPosX - (2 * mouseWorldPosY) - (tileHeight / 2) / -tileWidth
It's practically the same but the other way around. We subtract the Y world position since the Y axis runs the other way around. This time we compensate the offset for the height of the tile and we divide the whole thing by negative tilewidth, again since it runs the other way.
I hope this helps below is a working example of a method i looked up, it returns a vector with the tile coordinates:
public Vector2 MouseTilePosition(Camera cam, GraphicsDevice device)
{
float mPosX = newMouseState.X + (cam.Position.X - (device.Viewport.Width / 2));
float mPosY = newMouseState.Y + (cam.Position.Y - (device.Viewport.Height / 2));
float posx = (mPosX + (2 * mPosY) - (Map.TileWidth / 2)) / Map.TileWidth;
float posy = (mPosX - (2 * mPosY) - (Map.TileHeight / 2)) / -Map.TileWidth;
return new Vector2((int)posx, (int)posy);
}

Detect if monitor is rotated

I have an application in JavaSE and i want my application always starts at center of screen. If two monitors are plugged, the right one should be used. So i wrote a code like this:
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
if(ge.getScreenDevices().length == 2) {
int w_1 = ge.getScreenDevices()[0].getDisplayMode().getWidth();
int h_1 = ge.getScreenDevices()[0].getDisplayMode().getHeight();
int w_2 = ge.getScreenDevices()[1].getDisplayMode().getWidth();
int h_2 = ge.getScreenDevices()[1].getDisplayMode().getHeight();
int x = w_1 + w_2 / 2 - getWidth() / 2;
int y = h_2 / 2 - getHeight() / 2;
setLocation(x, y);
}
Unfortunately if monitor is rotated at 90°, width and height should be flipped. Is there any way to detect such rotation?

You don't need to know whether the second monitor is in portrait mode. Just find the bounds of the screen in device coordinates and use the center. (If it is in portrait mode, then height>width, but that isn't an important piece of information.)
Your formula to determine the center point of the second device is wrong. You are assuming that the coordinates of the second screen runs from (w_1,0) to (w_1 + w_2, h_2), but that isn't necessarily true. You need to find the GraphicsConfiguration object of the second screen and call GraphicsConfiguration.getBounds() on it. You can then calculate the center point of that rectangle.
If you want to know which device is on the left or the right (or top or bottom), you can compare the x (or y) values of their bounding rectangles. Note that the x or y values may be negative.

You should consider if the height is bigger than the width(portrait). I haven't heard anyone using portrait monitors yet, though.

Here is the code that works fine in most cases (from answer of Enwired).
GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
if(ge.getScreenDevices().length == 2) {
int x = (int)ge.getScreenDevices()[1].getDefaultConfiguration().getBounds().getCenterX() - frame.getWidth() / 2;
int y = (int)ge.getScreenDevices()[1].getDefaultConfiguration().getBounds().getCenterY() - frame.getHeight() / 2;
setLocation(x, y);
}
The only problem is that device index is not always 0 - left, 1 - right.