I'm currently recreating a Civilization game in Processing. I'm planning to implement the feature in a which a given unit can see every possible move it can make with a given number of hexes it is allowed to move. All possible endpoints are marked with red circles. However, units cannot move through mountains or bodies of water. I'm trying to approach this by finding out every possible combination of moves I can make without the unit going into a mountain or body of water but I can't figure out how I can determine every combination.
There are 6 directions that any unit can go in, north-east, north, north-west, south-east, south, south-west. The max number of movements I'm assigning to any unit would probably go up to 6. Any higher and I'm afraid processing may become to slow every time I move a unit.
I'm trying to recreate this:
What I'm hoping the result will look like with two possible movements (without the black arrows):
Raw version of that image:
Here is the code I use to draw the hex grid. Immediately after drawing each individual hex, its center's x coords and y coords are stored in xHexes and yHexes respectively. Also, immediately after generating the type of tile (e.g. grass, beach), the type of tile is also stored in an array named hexTypes. Therefore, I can get the x and y coords and type of hex of any hex I want on the map just by referencing its index.
Code used to draw a single hexagon:
beginShape();
for (float a = PI/6; a < TWO_PI; a += TWO_PI/6) {
float vx = x + cos(a) * gs*2;
float vy = y + sin(a) * gs*2;
vertex(vx, vy);
}
x is the x coord for centre of hexagon
y is the y coord for centre of hexagon
gs = radius of hexagon
Code used to tesselate hex over the window creating a hex grid:
void redrawMap() {
float xChange = 1.7;
float yChange = 6;
for (int y = 0; y < ySize/hexSize; y++) {
for (int x = 0; x < xSize/hexSize; x++) {
if (x % 2 == 1) {
// if any part of this hexagon being formed will be visible on the window and not off the window.
if (x*hexSize*xChange <= width+2*hexSize && int(y*hexSize*yChange) <= height+3*hexSize) {
drawHex(x*hexSize*xChange, y*hexSize*yChange, hexSize);
}
// only record and allow player to react with it if the entire tile is visible on the window
if (x*hexSize*xChange < width && int(y*hexSize*yChange) < height) {
xHexes.add(int(x*hexSize*xChange));
yHexes.add(int(y*hexSize*yChange));
}
} else {
if (x*hexSize*xChange <= width+2*hexSize && int(y*hexSize*yChange) <= height+3*hexSize) {
drawHex(x*hexSize*xChange, y*hexSize*yChange+(hexSize*3), hexSize);
}
if (x*hexSize*xChange < width && int(y*hexSize*yChange+(hexSize*3)) < height) {
xHexes.add(int(x*hexSize*xChange));
yHexes.add(int(y*hexSize*yChange+(hexSize*3)));
}
}
}
}
}
hexSize is a user-specified size for each hexagon, determining the number of hexagons that will be on the screen.
This answer will help you get to this (green is plains, red is hills and blue is water, also please don't flame my terrible grid):
Note that there is no pathfinding in this solution, only some very simple "can I get there" math. I'll include the full code of the sketch at the end so you can reproduce what I did and test it yourself. One last thing: this answer doesn't use any advanced design pattern, but it assume that you're confortable with the basics and Object Oriented Programming. If I did something which you're not sure you understand, you can (and should) ask about it.
Also: this is a proof of concept, not a "copy and paste me" solution. I don't have your code, so it cannot be that second thing anyway, but as your question can be solved in a bazillion manners, this is only one which I deliberately made as simple and visual as possible so you can get the idea and run with it.
First, I strongly suggest that you make your tiles into objects. First because they need to carry a lot of information (what's on each tile, how hard they are to cross, maybe things like resources or yield... I don't know, but there will be a lot of stuff).
The Basics
I organized my global variables like this:
// Debug
int unitTravelPoints = 30; // this is the number if "travel points" currently being tested, you can change it
// Golbals
float _tileSize = 60;
int _gridWidth = 10;
int _gridHeight = 20;
ArrayList<Tile> _tiles = new ArrayList<Tile>(); // all the tiles
ArrayList<Tile> _canTravel = new ArrayList<Tile>(); // tiles you can currently travel to
The basics being that I like to be able to change my grid size on the fly, but that's just a detail. What's next is to choose a coordinate system for the grid. I choose the simplest one as I didn't want to bust my brain on something complicated, but you may want to adapt this to another coordinate system. I choose the offset coordinate type of grid: my "every second row" is half a tile offset. So, instead of having this:
I have this:
The rest is just adjusting the spatial coordinates of the tiles so it doesn't look too bad, but their coordinates stays the same:
Notice how I consider that the spatial coordinates and the grid coordinates are two different things. I'll mostly use the spatial coordinates for the proximity checks, but that's because I'm lazy, because you could make a nice algorithm which do the same thing without the spatial coordinates and it would probably be less costly.
What about the travel points? Here's how I decided to work: your unit has a finite amount of "travel points". Here there's no unit, but instead a global variable unitTravelPoints which will do the same thing. I decided to work with this scale: one normal tile is worth 10 travel points. So:
Plains: 10 points
Hills: 15 points
Water: 1000 points (this is impassable terrain but without going into the details)
I'm not going to go into the details of drawing a grid, but that's mostly because your algorithm looks way better than mine on this front. On the other hand, I'll spend some time on explaining how I designed the Tiles.
The Tiles
We're entering OOP: they are Drawable. Drawable is a base class which contains some basic info which every visible thing should have: a position, and an isVisible setting which can be turned off. And a method to draw it, which I call Render() since draw() is already taken by Processing:
class Drawable {
PVector position;
boolean isVisible;
public Drawable() {
position = new PVector(0, 0);
isVisible = true;
}
public void Render() {
// If you forget to overshadow the Render() method you'll see this error message in your console
println("Error: A Drawable just defaulted to the catch-all Render(): '" + this.getClass() + "'.");
}
}
The Tile will be more sophisticated. It'll have more basic informations: row, column, is it currently selected (why not), a type like plains or hills or water, a bunch of neighboring tiles, a method to draw itself and a method to know if the unit can travel through it:
class Tile extends Drawable {
int row, column;
boolean selected = false;
TileType type;
ArrayList<Tile> neighbors = new ArrayList<Tile>();
Tile(int row, int column, TileType type) {
super(); // this calls the parent class' constructor
this.row = row;
this.column = column;
this.type = type;
// the hardcoded numbers are all cosmetics I included to make my grid looks less awful, nothing to see here
position.x = (_tileSize * 1.5) * (column + 1);
position.y = (_tileSize * 0.5) * (row + 1);
// this part checks if this is an offset row to adjust the spatial coordinates
if (row % 2 != 0) {
position.x += _tileSize * 0.75;
}
}
// this method looks recursive, but isn't. It doesn't call itself, but it calls it's twin from neighbors tiles
void FillCanTravelArrayList(int travelPoints, boolean originalTile) {
if (travelPoints >= type.travelCost) {
// if the unit has enough travel points, we add the tile to the "the unit can get there" list
if (!_canTravel.contains(this)) {
// well, only if it's not already in the list
_canTravel.add(this);
}
// then we check if the unit can go further
for (Tile t : neighbors) {
if (originalTile) {
t.FillCanTravelArrayList(travelPoints, false);
} else {
t.FillCanTravelArrayList(travelPoints - type.travelCost, false);
}
}
}
}
void Render() {
if (isVisible) {
// the type knows which colors to use, so we're letting the type draw the tile
type.Render(this);
}
}
}
The Tile Types
The TileType is a strange animal: it's a real class, but it's never used anywhere. That's because it's a common root for all tile types, which will inherit it's basics. The "City" tile may need very different variables than, say, the "Desert" tile. But both need to be able to draw themselves, and both need to be owned by the tiles.
class TileType {
// cosmetics
color fill = color(255, 255, 255);
color stroke = color(0);
float strokeWeight = 2;
// every tile has a "travelCost" variable, how much it cost to travel through it
int travelCost = 10;
// while I put this method here, it could have been contained in many other places
// I just though that it made sense here
void Render(Tile tile) {
fill(fill);
if (tile.selected) {
stroke(255);
} else {
stroke(stroke);
}
strokeWeight(strokeWeight);
DrawPolygon(tile.position.x, tile.position.y, _tileSize/2, 6);
textAlign(CENTER, CENTER);
fill(255);
text(tile.column + ", " + tile.row, tile.position.x, tile.position.y);
}
}
Each tile type can be custom, now, yet each tile is... just a tile, whatever it's content. Here are the TileType I used in this demonstration:
// each different tile type will adjust details like it's travel cost or fill color
class Plains extends TileType {
Plains() {
this.fill = color(0, 125, 0);
this.travelCost = 10;
}
}
class Water extends TileType {
// here I'm adding a random variable just to show that you can custom those types with whatever you need
int numberOfFishes = 10;
Water() {
this.fill = color(0, 0, 125);
this.travelCost = 1000;
}
}
class Hill extends TileType {
Hill() {
this.fill = color(125, 50, 50);
this.travelCost = 15;
}
}
Non-class methods
I added a mouseClicked() method so we can select a hex to check how far from it the unit can travel. In your game, you would have to make it so when you select a unit all these things fall into place, but as this is just a proof of concept the unit is imaginary and it's location is wherever you click.
void mouseClicked() {
// clearing the array which contains tiles where the unit can travel as we're changing those
_canTravel.clear();
for (Tile t : _tiles) {
// select the tile we're clicking on (and nothing else)
t.selected = IsPointInRadius(t.position, new PVector(mouseX, mouseY), _tileSize/2);
if (t.selected) {
// if a tile is selected, check how far the imaginary unit can travel
t.FillCanTravelArrayList(unitTravelPoints, true);
}
}
}
At last, I added 2 "helper methods" to make things easier:
// checks if a point is inside a circle's radius
boolean IsPointInRadius(PVector center, PVector point, float radius) {
// simple math, but with a twist: I'm not using the square root because it's costly
// we don't need to know the distance between the center and the point, so there's nothing lost here
return pow(center.x - point.x, 2) + pow(center.y - point.y, 2) <= pow(radius, 2);
}
// draw a polygon (I'm using it to draw hexagons, but any regular shape could be drawn)
void DrawPolygon(float x, float y, float radius, int npoints) {
float angle = TWO_PI / npoints;
beginShape();
for (float a = 0; a < TWO_PI; a += angle) {
float sx = x + cos(a) * radius;
float sy = y + sin(a) * radius;
vertex(sx, sy);
}
endShape(CLOSE);
}
How Travel is calculated
Behind the scenes, that's how the program knows where the unit can travel: in this example, the unit has 30 travel points. Plains cost 10, hills cost 15. If the unit has enough points left, the tile is marked as "can travel there". Every time a tile is in travel distance, we also check if the unit can get further from this tile, too.
Now, if you're still following me, you may ask: how do the tiles know which other tile is their neighbor? That's a great question. I suppose that an algorithm checking their coordinates would be the best way to handle this, but as this operation will happen only once when we create the map I decided to take the easy route and check which tiles were the close enough spatially:
void setup() {
// create the grid
for (int i=0; i<_gridWidth; i++) {
for (int j=0; j<_gridHeight; j++) {
int rand = (int)random(100);
if (rand < 20) {
_tiles.add(new Tile(j, i, new Water()));
} else if (rand < 50) {
_tiles.add(new Tile(j, i, new Hill()));
} else {
_tiles.add(new Tile(j, i, new Plains()));
}
}
}
// detect and save neighbor tiles for every Tile
for (Tile currentTile : _tiles) {
for (Tile t : _tiles) {
if (t != currentTile) {
if (IsPointInRadius(currentTile.position, t.position, _tileSize)) {
currentTile.neighbors.add(t);
}
}
}
}
}
Complete code for copy-pasting
Here's the whole thing in one place so you can easily copy and paste it into a Processing IDE and play around with it (also, it includes how I draw my awful grid):
// Debug
int unitTravelPoints = 30; // this is the number if "travel points" currently being tested, you can change it
// Golbals
float _tileSize = 60;
int _gridWidth = 10;
int _gridHeight = 20;
ArrayList<Tile> _tiles = new ArrayList<Tile>();
ArrayList<Tile> _canTravel = new ArrayList<Tile>();
void settings() {
// this is how to make a window size's dynamic
size((int)(((_gridWidth+1) * 1.5) * _tileSize), (int)(((_gridHeight+1) * 0.5) * _tileSize));
}
void setup() {
// create the grid
for (int i=0; i<_gridWidth; i++) {
for (int j=0; j<_gridHeight; j++) {
int rand = (int)random(100);
if (rand < 20) {
_tiles.add(new Tile(j, i, new Water()));
} else if (rand < 50) {
_tiles.add(new Tile(j, i, new Hill()));
} else {
_tiles.add(new Tile(j, i, new Plains()));
}
}
}
// detect and save neighbor tiles for every Tile
for (Tile currentTile : _tiles) {
for (Tile t : _tiles) {
if (t != currentTile) {
if (IsPointInRadius(currentTile.position, t.position, _tileSize)) {
currentTile.neighbors.add(t);
}
}
}
}
}
void draw() {
background(0);
// show the tiles
for (Tile t : _tiles) {
t.Render();
}
// show how far you can go
for (Tile t : _canTravel) {
fill(0, 0, 0, 0);
if (t.selected) {
stroke(255);
} else {
stroke(0, 255, 0);
}
strokeWeight(5);
DrawPolygon(t.position.x, t.position.y, _tileSize/2, 6);
}
}
class Drawable {
PVector position;
boolean isVisible;
public Drawable() {
position = new PVector(0, 0);
isVisible = true;
}
public void Render() {
// If you forget to overshadow the Render() method you'll see this error message in your console
println("Error: A Drawable just defaulted to the catch-all Render(): '" + this.getClass() + "'.");
}
}
class Tile extends Drawable {
int row, column;
boolean selected = false;
TileType type;
ArrayList<Tile> neighbors = new ArrayList<Tile>();
Tile(int row, int column, TileType type) {
super(); // this calls the parent class' constructor
this.row = row;
this.column = column;
this.type = type;
// the hardcoded numbers are all cosmetics I included to make my grid looks less awful, nothing to see here
position.x = (_tileSize * 1.5) * (column + 1);
position.y = (_tileSize * 0.5) * (row + 1);
// this part checks if this is an offset row to adjust the spatial coordinates
if (row % 2 != 0) {
position.x += _tileSize * 0.75;
}
}
// this method looks recursive, but isn't. It doesn't call itself, but it calls it's twin from neighbors tiles
void FillCanTravelArrayList(int travelPoints, boolean originalTile) {
if (travelPoints >= type.travelCost) {
// if the unit has enough travel points, we add the tile to the "the unit can get there" list
if (!_canTravel.contains(this)) {
// well, only if it's not already in the list
_canTravel.add(this);
}
// then we check if the unit can go further
for (Tile t : neighbors) {
if (originalTile) {
t.FillCanTravelArrayList(travelPoints, false);
} else {
t.FillCanTravelArrayList(travelPoints - type.travelCost, false);
}
}
}
}
void Render() {
if (isVisible) {
// the type knows which colors to use, so we're letting the type draw the tile
type.Render(this);
}
}
}
class TileType {
// cosmetics
color fill = color(255, 255, 255);
color stroke = color(0);
float strokeWeight = 2;
// every tile has a "travelCost" variable, how much it cost to travel through it
int travelCost = 10;
// while I put this method here, it could have been contained in many other places
// I just though that it made sense here
void Render(Tile tile) {
fill(fill);
if (tile.selected) {
stroke(255);
} else {
stroke(stroke);
}
strokeWeight(strokeWeight);
DrawPolygon(tile.position.x, tile.position.y, _tileSize/2, 6);
textAlign(CENTER, CENTER);
fill(255);
text(tile.column + ", " + tile.row, tile.position.x, tile.position.y);
}
}
// each different tile type will adjust details like it's travel cost or fill color
class Plains extends TileType {
Plains() {
this.fill = color(0, 125, 0);
this.travelCost = 10;
}
}
class Water extends TileType {
// here I'm adding a random variable just to show that you can custom those types with whatever you need
int numberOfFishes = 10;
Water() {
this.fill = color(0, 0, 125);
this.travelCost = 1000;
}
}
class Hill extends TileType {
Hill() {
this.fill = color(125, 50, 50);
this.travelCost = 15;
}
}
void mouseClicked() {
// clearing the array which contains tiles where the unit can travel as we're changing those
_canTravel.clear();
for (Tile t : _tiles) {
// select the tile we're clicking on (and nothing else)
t.selected = IsPointInRadius(t.position, new PVector(mouseX, mouseY), _tileSize/2);
if (t.selected) {
// if a tile is selected, check how far the imaginary unit can travel
t.FillCanTravelArrayList(unitTravelPoints, true);
}
}
}
// checks if a point is inside a circle's radius
boolean IsPointInRadius(PVector center, PVector point, float radius) {
// simple math, but with a twist: I'm not using the square root because it's costly
// we don't need to know the distance between the center and the point, so there's nothing lost here
return pow(center.x - point.x, 2) + pow(center.y - point.y, 2) <= pow(radius, 2);
}
// draw a polygon (I'm using it to draw hexagons, but any regular shape could be drawn)
void DrawPolygon(float x, float y, float radius, int npoints) {
float angle = TWO_PI / npoints;
beginShape();
for (float a = 0; a < TWO_PI; a += angle) {
float sx = x + cos(a) * radius;
float sy = y + sin(a) * radius;
vertex(sx, sy);
}
endShape(CLOSE);
}
Hope it'll help. Have fun!
You will have to use similar algorithms we use on pathfinding. you create a stack or queue that will hold a class storing the position of the hex and the number of moves left from that point, initially you insert your starting position with the number of moves you have and mark that hex as done ( to not re-use a position you have already been on ), then you pop an element, and you insert every neighbor of that hex with a number of moves -1. when you insert the hexes with zero moves, those are your endpoints. And before inserting any hex check if it's not already done.
I hope I was clear, your question was a bit vague but I tried to give you an idea of how these solutions are usually done, also I think your question fits more into algorithms rather then processing
Best of luck
I'm using the method of dividing the x and y coordinates by the z value, then rendering it just like you would a 2D game/thing to create 3D looking thing. The problem is when the z value is less then 0 it flips the quadrants for the coordinate and looks very weird. At the moment there is only movement if someone could help me fix the negative z value thing and show me how to rotate. I'm not using and matrices only vectors that take in x,y and z for the maths if that helps. I'm making this using Java with no extra libraries.
Thanks for the help.
I used the perspective matrix and multiplied it by my vector but it didn't work here is my code there might be something wrong with it. I just turned the vector into a 1 by 3 matrix and then did this.
public Matrix multiply(Matrix matrix)
{
Matrix result = new Matrix(getWidth(),getHeight());
for(int y = 0; y < getHeight()-1; y++)
{
for(int x = 0; x < getWidth()-1; x++)
{
float sum = 0.0f;
for(int e = 0; e < this.getWidth()-1; e++)
{
sum += this.matrix[e + y * getWidth()] * matrix.matrix[x + e * matrix.getWidth()];
}
result.matrix[x + y * getWidth()] = sum;
}
}
return result;
}
Just guessing here, but it sounds like you are trying to do a projection transform: You are modeling 3D objects (things with X, Y, and Z coordinates) and you want to project them onto a 2D window (i.e., the screen).
The meaning of Z in the naive projection transform is the distance between the point and a plane parallel to the screen, that passes through your eyeball. If you have points with -Z, those represent points that are behind your head.
Sounds like you need to translate the Z coordinates so that Z=0 is the plane of the screen, or a plane parallel to and behind the screen. (In other words, Add a constant to all of your Zs, so that none of them is negative.)
http://en.wikipedia.org/wiki/3D_projection
I'm trying to get my head around this- and I've literally been looking for a whole day!
I think I understand the main concepts behind it, but I'm struggling to figure out the math I need to create the axis on which to project my shapes on to?
So if I have a rectange I find out each of the points and then use these to find the side of the shape edge = v(n) - v(n-1) and go through all sides.
But I don't know how to then create the separating axis.
The theorem is not difficult to understand: If you can find a line for which all points of shape A are on the one side, and all points of shape B are on the other (dot product positive or negative), that line is separating the shapes.
What do you want to do? Find separating lines for arbitrary shapes?
I would recommend to have a look at projective geometry, as the edge for two vertices of a polygon extended to infinity can be represented by the cross product of the two vertices (x, y, 1). For convex polygons you can simply create lines for all edges and then take the dot product of all vertices of your other polygon to check on which side they are. If for one edge all points are outside, that edge is a separating line.
Also keep in mind that by keeping the line normalized you get the distance of a point to the line using dot product. The sign identifies the side on which the point lies.
If your problem is more difficult, please explain it in more detail. Maybe you can use some sort of clipping to solve it fast.
Example: projective geometry
public double[] e2p(double x, double y) {
return new double[] { x, y, 1 };
}
// standard vector maths
public double[] getCrossProduct(double[] u, double[] v) {
return new double[] { u[1] * v[2] - u[2] * v[1],
u[2] * v[0] - u[0] * v[2], u[0] * v[1] - u[1] * v[0] };
}
public double getDotProduct(double[] u, double[] v) {
return u[0] * v[0] + u[1] * v[1] + u[2] * v[2];
}
// collision check
public boolean isCollision(List<Point2D> coordsA, List<Point2D> coordsB) {
return !(isSeparate(pointsA, pointsB) || isSeparate(pointsB, pointsA));
}
// the following implementation expects the convex polygon's vertices to be in counter clockwise order
private boolean isSeparate(List<Point2D> coordsA, List<Point2D> coordsB) {
edges: for (int i = 0; i < coordsA.size(); i++) {
double[] u = e2p(coordsA.get(i).getX(), coordsA.get(i).getY());
int ni = i + 1 < coordsA.size() ? i + 1 : 0;
double[] v = e2p(coordsA.get(ni).getX(), coordsA.get(ni).getY());
double[] pedge = getCrossProduct(u, v);
for (Point2D p : coordsB) {
double d = getDotProduct(pedge, e2p(p.getX(), p.getY()));
if (d > -0.001) {
continue edges;
}
}
return true;
}
return false;
}
The separating axis is one of the sides. You can find the signs of the vertexes of the shape itself when plugged in the equation of this side:
(X - Xn).(Y - Yn-1) - (X - Xn-1).(Y - Yn) = 0
Check that the vertices of the other shape yield opposite signs.
I have a circle drawn, and I want to make it so I can have more slices than four. I can easily do four quadrants because I just check if the mouse in in the circle and inside a box.
This is how I am checking if the point is in the circle.
if( Math.sqrt((xx-x)*(xx-x) + (yy-y)*(yy-y)) <= radius)
{
return true;
}
else
{
return false;
}
How can I modify this if the circle is divided into more than 4 regions?
For radial slices (circular sectors), you have a couple of alternatives:
Use Math.atan2 to calculate the 4-quadrant angle of the line from the circle center to the point. Compare to the slice angles to determine the slice index.
For a particular slice, you can test which side of each slice edge the point falls. Classify the point accordingly. This is more complicated to calculate but probably faster (for a single slice) than calling Math.atan2.
The following sample code calculates the slice index for a particular point:
int sliceIndex(double xx, double yy, double x, double y, int nSlices) {
double angle = Math.atan2(yy - y, xx - x);
double sliceAngle = 2.0 * Math.PI / nSlices;
return (int) (angle / sliceAngle);
}
The above code makes the following assumptions:
slices are all the same (angular) width
slices are indexed counter-clockwise
slice 0 starts at the +x axis
slices own their right edge but not their left edge
You can adjust the calculations if these assumptions do not apply. (For instance, you can subtract the start angle from angle to eliminate assumption 3.)
First we can check that the point is within the circle as you did. But I woudln't combine this with a check for which quadrant (is that why you have radius/2 ?)
if( (xx-x)*(xx-x) + (yy-y)*(yy-y) > radius*radius)
return false;
Now we can look to see which region the point is in by using the atan2 function. atan2 is like Arctan except the Arctangent function always returns a value between -pi/2 and pi/2 (-90 and +90 degrees). We need the actual angle in polar coordinate fashion. Now assuming that (x,y) is the center of your circle and we are interested in the location of the point (xx,yy) we have
double theta = Math.atan2(yy-y,xx-x);
//theta is now in the range -Math.PI to Math.PI
if(theta<0)
theta = Math.PI - theta;
//Now theta is in the range [0, 2*pi]
//Use this value to determine which slice of the circle the point resides in.
//For example:
int numSlices = 8;
int whichSlice = 0;
double sliceSize = Math.PI*2 / numSlices;
double sliceStart;
for(int i=1; i<=numSlices; i++) {
sliceStart = i*sliceSize;
if(theta < sliceStart) {
whichSlice = i;
break;
}
}
//whichSlice should now be a number from 1 to 8 representing which part of the circle
// the point is in, where the slices are numbered 1 to numSlices starting with
// the right middle (positive x-axis if the center is (0,0).
It is more a trig problem Try something like this.
int numberOfSlices=8;
double angleInDegrees=(Math.toDegrees(Math.atan2(xx-x ,yy-y)));
long slice= Math.round(( numberOfSlices*angleInDegrees )/360 );
I have created a polygon with 6 vertices. Lets call this one, outside polygon. Inside the outside polygon I created smaller polygons. I want to flip all of it vertically one point at the time.
I know the vertices of the outside polygon and I have an ArrayList<Polygon> for the inner polygons. I was able to flip the outside polygon. but how do I flipped the inner polygons keeping their relative positions in the new one? I know the center of the outside polygon and the flipped version.
correction: I needed to flip horizontal.
I flipped the outer polygon (triangle shape), and I was able to move the inner polygons. but the distance is incorrect. this is a picture of what I have done,
(https://docs.google.com/drawings/d/1cPYJqxTWVu5gSHFQyHxHWSTysNzxJvNuJIwsgCQInfc/edit) https://docs.google.com/drawings/d/1cPYJqxTWVu5gSHFQyHxHWSTysNzxJvNuJIwsgCQInfc/edit
I tried this:
for (Polygon p : polygonList) {
Polygon tempP = new Polygon(p.xpoints, p.ypoints, p.npoints);
firstPointinPolygon = new Point(p.xpoints[0], p.ypoints[0]);
// find frist point in the polygon
float adjacent = (float) firstPointinPolygon.getX() - 400;
float opposite = (float) firstPointinPolygon.getY() - 400;
float hypotenuse = (float) Math.sqrt(opposite * opposite + adjacent * adjacent);
float cosine = adjacent / hypotenuse;
float sine = opposite / hypotenuse;
float endX = 400 * cosine;
float endY = 400 * sine;
float endXDelta =400-endX;
float endYDelta=400-endY;
Polygon pM = move(tempP, endX, endY);
polygonListMirror.add(pM);
tempP = new Polygon();
}
public Polygon move(Polygon p, double xMove, double yMove) {
// Change the values of the points for the Polygon
for (int i = 0; i < p.xpoints.length; i++) {
p.xpoints[i] += xMove;
p.ypoints[i] += yMove;
}
return p;
}
But did not get the result, I expected. What am I doing wrong? The end result should be like the picture in this link:
(https://docs.google.com/drawings/d/1vYdWkCelWW1_NUypNhtmckBYfEMzCf6bMVtoB-AyPkw/edit) https://docs.google.com/drawings/d/1vYdWkCelWW1_NUypNhtmckBYfEMzCf6bMVtoB-AyPkw/edit
I think something like this will do it:
Polygon outerPolygon, oldOuterPolygon;
ArrayList<Polygon> innerPolygons;
// set up objects
for (Polygon polygon: innerPolygons)
{
for (int i = 0; i < polygon.ypoints.length; i++)
{
polygon.ypoints[i] = center(outerPolygon) - polygon.ypoints[i] + center(oldOuterPolygon);
}
}
If you just to flip it vertically where it stands, such that the y-coordinate of top-most and bottom-most points just switch around, center for both should be the same (thus you can just say 2*center).
I'm pretty sure you can replace center(outerPolygon) and center(oldOuterPolygon) with any point from the applicable Polygon, as long as both use the same point.