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How do you get the 4 corners of a rectangle for any line that has a start coord and end coord in Java

I would like to know how to calculate the four corners of a rectangle.
I have various straight lines at different orientations. I would like
to use the start and end coordinates of each of these lines to determine a rectangle i.e using the addition or subtraction of diffx, diffy to the start and end points of the line dependent on the bearing of the line in whole circle quadrant. I am trying to do this in java.
My end goal is the have a rectangle (sleeve) to contain the line. The line would
appear as a centre-line to the rectangle or (rectangle buffer of the line).
Grateful for any help
Please see the slide on Assignment 2 on this link. It is a link to someone elses page and shows the image of what I am trying to do
http://blog.wijono.org/2015/07/simple-2d-tesselation-and-twist-with.html
link
link

Using points (x1, y1) and (x2, y2) for the start and end of the line.
Top left ( Math.min(x1, x2), Math.min(y1, y2) )
Top right: ( Math.max(x1, x2), Math.min(y1, y2) )
Bottom left: ( Math.min(x1, x2) , Math.max(y1, y2) )
Bottom right: ( Math.max(x1, x2) , Math.max(y1, y2) )

Assuming you intend the line to be the diagonal of the rectangle, and that line starts in x,y and ends in z,w with x < z.
Coordinates of the rectangle's vertices would be: (x,y), (x,w), (z,y), (z,w)

The other two folks covered the diagonal, here's the center-line method. I'm yoinking this from a shader I wrote, so forgive any type errors, I'll do my best to make them recognizable.
// Assume:
// Vec2 is a two-property float, with x and y
// Rect is a object that contains 4 Vec2s, representing each corner
// in counter-clockwise order.
function Rect getRectangle(Vec2 start, Vec2 end, float width) {
Vec2 parallel = end-start;
parallel = normalize(parallel);
Vec2 perpendicular = new Vec2(parallel.y,-parallel.x);
perpendicular *= width / 2;
return new Rect(start+perpendicular, start-perpendicular, end-perpendicular, end+perpendicular);
}

Given a line as two points and the amount of padding, this will return the four points of the bounding rectangle
public Point2D.Double[] padLine(Point2D lineStart, Point2D lineEnd, double padding)
{
//have the line start at the origin so as to be represented by one point
Point2D.Double line = new Point2D.Double(lineEnd.getX() - lineStart.getX(), lineEnd.getY() - lineStart.getY());
//find the unit vector perpendicular to the line
Point2D.Double perp = normalize(rotate90(line));
//create and fill in the array of the rectangle's coordinates
Point2D[] rect = new Point2D[4];
rect[0] = new Point2D.Double(lineStart.getX() + perp.x * padding, lineStart.getY() + perp.y * padding);
rect[1] = new Point2D.Double(lineStart.getX() - perp.x * padding, lineStart.getY() - perp.y * padding);
rect[2] = new Point2D.Double(lineEnd.getX() + perp.x * padding, lineEnd.getY() + perp.y * padding);
rect[3] = new Point2D.Double(lineEnd.getX() - perp.x * padding, lineEnd.getY() - perp.y * padding);
return rect;
}
//same directionality, but of length 1
public Point2D.Double normalize(Point2D p)
{
double length = Math.sqrt(p.getX() * p.getX() + p.getY() * p.getY());
return new Point2D.Double(p.getX() / length, p.getY() / length);
}
//rotates the given point by 90 degrees
public Point2D.Double rotate90(Point2D p)
{
return new Point2D.Double(p.getY(), -p.getX());
}
This just uses default java classes, java.awt.geom.Point2D.Double inside of java.awt.geom.Point2D

smooth color interpolation along a "bresenham" line

I am trying to interpolate color along a line so that, given two points and their respective RGB values, I can draw a line with a smooth color gradient. Using Bresenham's Line Algorithm, I can now draw lines, but am not sure how to begin interpolating colors between the two end points. The following is part of the drawLine() function that works for all line whose slope are less than 1.
int x_start = p1.x, x_end = p2.x, y_start =p1.y, y_end = p2.y;
int dx = Math.abs(x_end-x_start), dy = Math.abs(y_end-y_start);
int x = x_start, y = y_start;
int step_x = x_start < x_end ? 1:-1;
int step_y = y_start < y_end ? 1:-1;
int rStart = (int)(255.0f * p1.c.r), rEnd = (int)(255.0f * p2.c.r);
int gStart = (int)(255.0f * p1.c.g), gEnd = (int)(255.0f * p2.c.g);
int bStart = (int)(255.0f * p1.c.b), bEnd = (int)(255.0f * p2.c.b);
int xCount = 0;
//for slope < 1
int p = 2*dy-dx;
int twoDy = 2*dy, twoDyMinusDx = 2*(dy-dx);
int xCount = 0;
// draw the first point
Point2D start = new Point2D(x, y, new ColorType(p1.c.r, p1.c.g, p1.c.b));
drawPoint(buff, start);
float pColor = xCount / Math.abs((x_end - x_start));
System.out.println(x_end + " " + x_start);
while(x != x_end){
x+= step_x;
xCount++;
if(p<0){
p+= twoDy;
}
else{
y += step_y;
p += twoDyMinusDx;
}
Point2D draw_line = new Point2D(x, y, new ColorType(p1.c.r*(1-pColor)+p2.c.r*pColor,p1.c.g*(1-pColor)+p2.c.g*pColor,p1.c.b*(1-pColor)+p2.c.b*pColor));
System.out.println(pColor);
drawPoint(buff,draw_line );
}
So what I'm thinking is that, just like drawing lines, I also need some sort of decision parameter p to determine when to change the RGB values. I am thinking of something along lines of as x increments, look at each rgb value and decide if I want to manipualte them or not.
I initialized rStart and rEnd(and so on for g and b) but have no idea where to start. any kind of help or suggestions would be greatly appreciated!
Edit: thanks #Compass for the great suggestion ! Now I've ran into another while trying to implementing that strategy, and I am almost certain it's an easy bug. I just can't see it right now. For some reason my pColor always return 0, I am not sure why. I ran some print statements to make sure xCount is indeed increasing, so I am not sure what else might've made this variable always 0.

I remember figuring this out way back when I was learning GUI! I'll explain the basic concepts for you.
Let's say we have two colors,
RGB(A,B,C)
and
RGB(X,Y,Z)
for simplicity.
If we know the position percentage-wise (we'll call this P, a float 0 for beginning, 1.0 at end) along the line, we can calculate what color should be there using the following:
Resultant Color = RGB(A*(1-P)+X*P,B*(1-P)+Y*P,C*(1-P)+Z*P)
In other words, you average out the individual RGB values along the line.

Actually you will be drawing the line in RGB space as well !
Bresenham lets you compute point coordinates from (X0, Y0) to (X1, Y1).
This is done by a loop on X or Y, with a linear interpolation on the other coordinate.
Just extend the algorithm to draw a line from (X0, Y0, R0, G0, B0) to (X1, Y1, R1, G1, B1), in the same loop on X or Y, with a linear interpolation on the other coordinates.

Circle-Rectangle collision side detection in libgdx

I have spent hours looking for the solution to this: I am developing a little top-down game with libgdx (maybe it matters what engine i am using). Now i have to implement the collision detection between my character (circle) and the wall (rectangle). I want the character to slide along the wall on collision, if sliding is possible.
Let me explain:
If i am moving 45 degrees right up i can collide with the down, the
left or the corner of a wall.
If i collide with the left i want to stop x-movement and move only up. If i leave the wall then i want to go on moving right up. The same
with the down side (stop y-movement)
If i collide with the Corner i want to stop movement (sliding not possible).
What i am doing actually is to check if the left line of the rectangle intersects my circle. Then i check intersection between the left line of wall and my circle and the bottom line of wall and my circle. Depending on which intersection occuret i set back x/y possition of my circle and set x/y Speed to 0. The Problem is, that most times not a collision bt an overlap occures. So the bottom check returns true, even if in reality the circle would only collide with the right. In this case both intersection test would return true and i would reset both speeds like on the Corner collision.
How can i solve this Problem? Is ther a better way to detect collision and collision side or corner?
I don't Need the exact Point of collision just the side of the rectangle.
Edit:
I have to say, that the rects aren't rotated just parallel to the x-axis.

You can find an explanation for circle/rectangle collision below, but please note that this type of collision might not be necessary for your needs. If, for example, you had a rectangle bounding box for your character the algorithm would be simpler and faster. Even if you are using a circle, it is probable that there is a simpler approach that is good enough for your purposes.
I though about writing the code for this, but it would take too long so here is only an explanation:
Here is a example movement of your character circle, with its last (previous) and current positions. Wall rectangle is displayed above it.
Here is that same movement, dotted lines represent the area the circle sweeps in this move. The sweep area is capsule shaped.
It would be difficult to calculate the collision of these two object, so we need to do this differently. If you look at the capsule on the previous image, you will see that it is simply the movement line extended in every direction by the radius of the circle. We can move that "extension" from the movement line to the wall rectangle. This way we get a rounded rectangle like on the image below.
The movement line will collide with this extended (rounded) rectangle if and only if the capsule collides with the wall rectangle, so they are somehow equivalent and interchangeable.
Since this collision calculation is still non-trivial and relatively expensive, you can first do a fast collision check between the extended wall rectangle (non-rounded this time) and the bounding rectangle of the movement line. You can see these rectangles on the image below - they are both dotted. This is a fast and easy calculation, and while you play the game there will probably NOT be an overlap with a specific wall rectangle >99% of the time and collision calculation will stop here.
If however there is an overlap, there is probably a collision of the character circle with wall rectangle, but it is not certain as will be demonstrated later.
Now you need to calculate the intersection between the movement line itself (not its bounding box) and the extended wall rectangle. You can probably find an algorithm how to do this online, search for line/rectangle intersection, or line/aabb intersection (aabb = Axis Aligned Bounding Box). The rectangle is axis-aligned and this makes the calculation simpler. The algorithm can give you intersection point or points since it is possible that there are two - in this case you choose the closest one to the starting point of the line. Below is an example of this intersection/collision.
When you get an intersection point, it should be easy to calculate on which part of the extended rectangle this intersection is located. You can see these parts on the image above, separated by red lines and marked with one or two letters (l - left, r - right, b - bottom, t - top, tl - top and left etc).
If the intersection is on parts l, r, b or t (the single letter ones, in the middle) then you are done. There is definitely a collision between character circle and wall rectangle, and you know on which side. In the example above, it is on the bottom side. You should probably use 4 variables called something like isLeftCollision, isRightCollision, isBottomCollsion and isTopCollision. In this case you would set isBottomCollision to true, while the other 3 would remain at false.
However, if the intersection is on the corner, on the two-letter sections, additional calculations are needed to determine if there is an actual collision between character circle and wall rectangle. Image below shows 3 such intersections on the corners, but there is an actual circle-rectangle collision on only 2 of them.
To determine if there is a collision, you need to find an intersection between the movement line and the circle centered in the closest corner of the original non-extended wall rectangle. The radius of this circle is equal to the radius of character circle. Again, you can google for line/circle intersection algorithm (maybe even libgdx has one), it isn't complex and shouldn't be hard to find.
There is no line/circle intersection (and no circle/rectangle collision) on bl part, and there are intersections/collisions on br and tr parts.
In the br case you set both isRightCollision, isBottomCollsion to true and in the tr case you set both isRightCollision and isTopCollision to true.
There is also one edge case you need to look out for, and you can see it on the image below.
This can happen if the movement of previous step ends in the corner of the the extended rectangle, but outside the radius of the inner rectangle corner (there was no collision).
To determine if this is the case, simply check if movement staring point is inside the extended rectangle.
If it is, after the initial rectangle overlap test (between extended wall rectangle and bounding rectangle of movement line), you should skip line/rectangle intersection test (because in this case there might not be any intersection AND still be a circle/rectangle collision), and also simply based on movement stating point determine which corner you are in, and then only check for line/circle intersection with that corner's circle. If there is intersection, there is a character circle/wall rectangle collision, otherwise not.
After all of this, the collision code should be simple:
// x, y - character coordinates
// r - character circle radius
// speedX, speedY - character speed
// intersectionX, intersectionY - intersection coordinates
// left, right, bottom, top - wall rect positions
// I strongly recomment using a const "EPSILON" value
// set it to something like 1e-5 or 1e-4
// floats can be tricky and you could find yourself on the inside of the wall
// or something similar if you don't use it :)
if (isLeftCollision) {
x = intersectionX - EPSILON;
if (speedX > 0) {
speedX = 0;
}
} else if (isRightCollision) {
x = intersectionX + EPSILON;
if (speedX < 0) {
speedX = 0;
}
}
if (isBottomCollision) {
y = intersectionY - EPSILON;
if (speedY > 0) {
speedY = 0;
}
} else if (isTopCollision) {
y = intersectionY + EPSILON;
if (speedY < 0) {
speedY = 0;
}
}
[Update]
Here is a simple and I believe efficient implementation of segment-aabb intersection that should be good enough for your purposes. It is a slightly modified Cohen-Sutherland algorithm. Also you can check out the second part of this answer.
public final class SegmentAabbIntersector {
private static final int INSIDE = 0x0000;
private static final int LEFT = 0x0001;
private static final int RIGHT = 0x0010;
private static final int BOTTOM = 0x0100;
private static final int TOP = 0x1000;
// Cohen–Sutherland clipping algorithm (adjusted for our needs)
public static boolean cohenSutherlandIntersection(float x1, float y1, float x2, float y2, Rectangle r, Vector2 intersection) {
int regionCode1 = calculateRegionCode(x1, y1, r);
int regionCode2 = calculateRegionCode(x2, y2, r);
float xMin = r.x;
float xMax = r.x + r.width;
float yMin = r.y;
float yMax = r.y + r.height;
while (true) {
if (regionCode1 == INSIDE) {
intersection.x = x1;
intersection.y = y1;
return true;
} else if ((regionCode1 & regionCode2) != 0) {
return false;
} else {
float x = 0.0f;
float y = 0.0f;
if ((regionCode1 & TOP) != 0) {
x = x1 + (x2 - x1) / (y2 - y1) * (yMax - y1);
y = yMax;
} else if ((regionCode1 & BOTTOM) != 0) {
x = x1 + (x2 - x1) / (y2 - y1) * (yMin - y1);
y = yMin;
} else if ((regionCode1 & RIGHT) != 0) {
y = y1 + (y2 - y1) / (x2 - x1) * (xMax - x1);
x = xMax;
} else if ((regionCode1 & LEFT) != 0) {
y = y1 + (y2 - y1) / (x2 - x1) * (xMin - x1);
x = xMin;
}
x1 = x;
y1 = y;
regionCode1 = calculateRegionCode(x1, y1, r);
}
}
}
private static int calculateRegionCode(double x, double y, Rectangle r) {
int code = INSIDE;
if (x < r.x) {
code |= LEFT;
} else if (x > r.x + r.width) {
code |= RIGHT;
}
if (y < r.y) {
code |= BOTTOM;
} else if (y > r.y + r.height) {
code |= TOP;
}
return code;
}
}
Here is some code example usage:
public final class Program {
public static void main(String[] args) {
float radius = 5.0f;
float x1 = -10.0f;
float y1 = -10.0f;
float x2 = 31.0f;
float y2 = 13.0f;
Rectangle r = new Rectangle(3.0f, 3.0f, 20.0f, 10.0f);
Rectangle expandedR = new Rectangle(r.x - radius, r.y - radius, r.width + 2.0f * radius, r.height + 2.0f * radius);
Vector2 intersection = new Vector2();
boolean isIntersection = SegmentAabbIntersector.cohenSutherlandIntersection(x1, y1, x2, y2, expandedR, intersection);
if (isIntersection) {
boolean isLeft = intersection.x < r.x;
boolean isRight = intersection.x > r.x + r.width;
boolean isBottom = intersection.y < r.y;
boolean isTop = intersection.y > r.y + r.height;
String message = String.format("Intersection point: %s; isLeft: %b; isRight: %b; isBottom: %b, isTop: %b",
intersection, isLeft, isRight, isBottom, isTop);
System.out.println(message);
}
long startTime = System.nanoTime();
int numCalls = 10000000;
for (int i = 0; i < numCalls; i++) {
SegmentAabbIntersector.cohenSutherlandIntersection(x1, y1, x2, y2, expandedR, intersection);
}
long endTime = System.nanoTime();
double durationMs = (endTime - startTime) / 1e6;
System.out.println(String.format("Duration of %d calls: %f ms", numCalls, durationMs));
}
}
This is the result I get from executing this:
Intersection point: [4.26087:-2.0]; isLeft: false; isRight: false; isBottom: true, isTop: false
Duration of 10000000 calls: 279,932343 ms
Please note that this is desktop performance, on an i5-2400 CPU. It will probably be much slower on Android devices, but I believe still more than sufficient.
I only tested this superficially, so if you find any errors, let me know.
If you use this algorithm, I believe you don't need special handling for that case where starting point is in the corner of the extended wall rectangle, since in this case you will get the intersection point at line start, and the collision detection procedure will continue to the next step (line-circle collision).

I suppose you determine the collision by calculating the distance of the circles center with the lines.
We can simplify the case and tell that the circle colliding with the corner if both distances are equal and smaller than the radius. The equality should have a tolerance of course.
More - may be not necessary- realistic approach would be to consider x,y speed and factor it in the equality check.

how to save values of variables and prevent them from changing when doing recursion? also recursion assistance?

import gpdraw.*;
public class Y2K {
// Attributes
SketchPad pad;
DrawingTool pen;
// Constructor
public Y2K() {
pad = new SketchPad(600, 600, 50);
pen = new DrawingTool(pad);
// Back the pen up so the Y is drawn in the middle of the screen
pen.up();
pen.setDirection(270);
pen.forward(150);
pen.down();
pen.setDirection(90);
}
public void drawY(int level, double length) {
// Base case: Draw an Y
if (level == 0) {
//pen.setDirection(90);
pen.forward(length);
pen.turnRight(60);
pen.forward(length);
pen.backward(length);
pen.turnLeft(120);
pen.forward(length);
pen.backward(length);
}
// Recursive case: Draw an L at each midpoint
// of the current L's segments
else {
//Drawing the bottom "leg" of our Y shape
pen.forward(length / 2);
double xpos1 = pen.getXPos();
double ypos1 = pen.getYPos();
double direction1 = pen.getDirection();
pen.turnRight(90);
drawY(level - 1, length / 2.0);
pen.up();
pen.move(xpos1, ypos1);
pen.setDirection(direction1);
pen.down();
pen.forward(length / 2);
double xpos2 = pen.getXPos();
double ypos2 = pen.getYPos();
double direction2 = pen.getDirection();
//Drawing upper Right Leg
pen.turnRight(60);
pen.forward(length / 2); //going to the midpoint
double xpos3 = pen.getXPos();
double ypos3 = pen.getYPos();
double direction3 = pen.getDirection();
pen.turnLeft(90);
drawY(level - 1, length / 2.0);
pen.up();
pen.move(xpos3, ypos3);
pen.setDirection(direction3);
pen.down();
pen.forward(length / 2);
//drawing upper left leg
pen.up();
pen.move(xpos1, ypos1);
pen.setDirection(direction1);
pen.down();
pen.forward(length / 2);
pen.turnLeft(60);
pen.forward(length / 2);
double xpos4 = pen.getXPos();
double ypos4 = pen.getYPos();
double direction4 = pen.getDirection();
pen.turnLeft(90);
drawY(level - 1, length / 2.0);
pen.up();
pen.move(xpos4, ypos4);
pen.setDirection(direction4);
pen.down();
pen.forward(length / 2);
pen.forward(length / 2);
}
}
public static void main(String[] args) {
Y2K fractal = new Y2K();
// Draw Y with given level and side length
fractal.drawY(8, 200);
}
}
output:
one certain leg of the triangle is too long, and that makes the output slightly off. maybe its because the code went (length/2) too far? lets debug this.
otherwise it is completely fine, the recursion is great, and its exactly what i wanted to do

As you're constantly drawing Y's, I'd recommend you create a method that draws a Y given certain parameters (e.g. length, angle of separation between the two branches of the Y, rotation, etc.). This will make your code much more readable and easier to understand.
As for moving to the center, just think of the Y on a coordinate plane. Based upon the rotation of the Y, and its starting point you can calculate the center point.
Just break it up into its x and y components.
Given this information, we can solve for a and for b.
a = length * sin(θ)
b = length * cos(θ)
Then add this to your x and y to calculate the center point of the Y.
As for keeping the constant length, you know the level. At the first level, level == 1. But the length of this next level should be length * (2^level). In this case, length/2 (as length would be -1).
In pseudo code terms:
public void drawY(int level, double length)
{
//Drawing the bottom "leg" of our Y shape
Move Forward length/2
Save our position
Save our direction
Turn to the right 90 degrees
Recursion (call drawY())
revert to original location
revert to original direction
move forward length/2 (to go to center point of Y)
save our new position
save our new direction
//Drawing upper Right Leg
Turn 60 to the right
Move Forward length/2 //going to the midpoint
save our new position (don't forget the center point)
save our new direction (don't forget the center point direction)
Turn 90 to the left
Recursion (call drawY())
return to our saved position (not center one)
return to our saved direction (not center one)
move forward length/2
//drawing upper left leg
return to center point
return to center direction
turn left 60
move forward length/2
save position (you can overwrite the center one now
save direction (you can overwrite)
turn left 90
Recursion (call drawY())
return to position
return to direction
move forward length/2
}

quaternion to angle

Alright, so this is how I am doing it:
float xrot = 0;
float yrot = 0;
float zrot = 0;
Quaternion q = new Quaternion().fromRotationMatrix(player.model.getRotation());
if (q.getW() > 1) {
q.normalizeLocal();
}
float angle = (float) (2 * Math.acos(q.getW()));
double s = Math.sqrt(1-q.getW()*q.getW());
// test to avoid divide by zero, s is always positive due to sqrt
// if s close to zero then direction of axis not important
if (s < 0.001) {
// if it is important that axis is normalised then replace with x=1; y=z=0;
xrot = q.getXf();
yrot = q.getYf();
zrot = q.getZf();
// z = q.getZ();
} else {
xrot = (float) (q.getXf() / s); // normalise axis
yrot = (float) (q.getYf() / s);
zrot = (float) (q.getZf() / s);
}
But it doesn't seem to work when I try to put it into use:
player.model.addTranslation(xrot * player.speed, 0, zrot * player.speed);
AddTranslation takes 3 numbers to move my model by than many spaces (x, y, z), but hen I give it the numbers above it doesn't move the model in the direction it has been rotated (on the XZ plane)
Why isn't this working?
Edit: new code, though it's about 45 degrees off now.
Vector3 move = new Vector3();
move = player.model.getRotation().applyPost(new Vector3(1,0,0), move);
move.multiplyLocal(-player.speed);
player.model.addTranslation(move);

xrot, yrot, and zrot define the axis of the rotation specified by the quaternion. I don't think you want to be using them in your addTranslation() call...in general, that won't have anything to do with the direction of motion.
What I mean by that is: your 3-D object -- let's say for the sake of argument that it's an airplane -- will have a certain preferred direction of motion in its original coordinate
system. So if the original orientation has the center of mass at the origin, and the
propeller somewhere along the +X axis, the plane wants to fly in the +X direction.
Now you introduce some coordinate transformation that rotates the airplane into some other orientation. That rotation is described by a rotation matrix, or equivalently, by a
quaternion. Which way does the plane want to move after the rotation?
You could find
that by taking a unit vector in the +X direction: (1.0, 0.0, 0.0), then applying the
rotation matrix or quaternion to that vector to transform it into the new coordinate
system. (Then scale it by the speed, as you're doing above.) The X, Y, and Z components
of the transformed, scaled vector give the desired incremental motion along each axis. That transformed vector is generally not going to be the rotation axis of the quaternion, and I think that's probably your problem.