need id's to delete to get a hexagrid

Programming Questions

clankill3r

need id's to delete to get a hexagrid

in Programming Questions • 8 months ago

I'm trying to get a hexagon grid, not in processing but in ICE, a node based programming system ( https://vimeo.com/19028797 ).

for example, i have this grid:

[ ][ ][ ][ ][ ]

Then i need to keep the dark ones:

[|][ ][|][ ][|]

[ ][|][ ][|][ ]

[|][ ][|][ ][|]

[ ][|][ ][|][ ]

To place a hexagon shape over it.

Since i'm better in processing (and i like processing) i thought lets try it here first (with ICE back in my head).

For ICE the easiest will be to delete the id's of the points i don't want, so in processing i try to create an array of the id's i have to remove.

There are 4 cases,

xSteps odd and ySteps odd

xSteps odd and ySteps even

xSteps even and ySteps odd

xSteps even and ySteps even

I solved the first 2, i only have problems with the last 2 (they are quite similair).

for xSteps - ySteps 4 by 5 it will be like this:

0 2 5 7 8 10 13 15 16 18 - 21 23 24 26 29 31...

hope someone can help

import java.util.*;
// try different odds and even for x and y
// make sure z works with more then 1 layer
int xSteps = 5;
int ySteps = 5;
int zSteps = 3;
int maxPoints = xSteps*ySteps*zSteps;
float spacing = 20;
ArrayList<Point> points = new ArrayList<Point>();
void setup() {
size(800, 800, OPENGL);
smooth();
createGrid();
ArrayList<Integer> toRemoveIndexes = getToRemoveIndexes();
println(toRemoveIndexes.size());
for (int i = toRemoveIndexes.size()-1; i >= 0; i--) {
println(toRemoveIndexes.get(i));
int r = toRemoveIndexes.get(i);
points.remove(r);
}
println(points.size());
}
ArrayList<Integer> getToRemoveIndexes() {
ArrayList<Integer> indexes = new ArrayList<Integer>();
if (odd(xSteps) && odd(ySteps)) {
for (int i = 0; i < maxPoints; i++) {
// i += 2 would also work instead of modullo
// but this is probably easier in ICE
if (i%2==0) indexes.add(i);
}
}
else if (odd(xSteps) && even(ySteps)) {
int offset = 0;
for (int i = 0; i < maxPoints; i++) {
if (i % (xSteps * ySteps) == 0) {
if (offset == 1) offset = 0;
else offset = 1;
}
if (i%2==0) indexes.add(i+offset);
}
}
// wrong
else if (even(xSteps) && odd(ySteps)) {
for (int i = 0; i < maxPoints; i++) {
int offset = 0;
if ( i % xSteps == 0) {
if (offset == 1) offset = 0;
else offset = 1;
}
if (i%2==0) indexes.add(i+offset);
}
}
// todo
else if (even(xSteps) && even(ySteps)) {
}
return indexes;
}
boolean odd(int number) {
return !(number % 2 == 0);
}
boolean even(int number) {
return number % 2 == 0;
}
void draw() {
background(0);
noStroke();
lights();
translate(width/2, height/2, 400);
rotateX(map(mouseY, 0, height, -PI, PI));
rotateY(map(mouseX, 0, width, -PI, PI));
translate(-(spacing*xSteps/2), -(spacing*ySteps/2));
for (Point p : points) {
pushMatrix();
translate(p.x, p.y, p.z);
box(5);
popMatrix();
}
}
void createGrid() {
// clear
points = new ArrayList<Point>();
int id = 0;
for (int z = 0 ; z < zSteps; z++) {
for (int y = 0 ; y < ySteps; y++) {
for (int x = 0 ; x < xSteps; x++) {
points.add(new Point(id, x*spacing, y*spacing, z*spacing*2));
id++;
}
}
}
}
class Point {
float x, y, z;
int id;
Point(int id, float x, float y, float z) {
this.id = id;
this.x = x;
this.y = y;
this.z = z;
}
}

Replies(1)

Chrisir

Re: need id's to delete to get a hexagrid

8 months ago

maybe this helps you, it paints a chess board :

see
http://www.openprocessing.org/sketch/31784

// Board outputs are here.
// Delivers the simple Board (when figures are 3D)
// OR
// the Board with the 2D-Figures painted on the Board (when figures are 2D).
// ========================================
void CheckeredBoard() {
color colLocal=0;
stroke(col1);
iBox=0;
for (int i = 0; i < 8; i = i+1) {
for (int j = 0; j < 8; j = j+1) {
// % is modulo, meaning rest of division
if (i%2 == 0) {
if (j%2 == 0) {
fill ( col1 );
colLocal=col1;
}
else
{
fill ( col2 );
colLocal=col2;
}
}
else {
if (j%2 == 0) {
fill ( col2 );
colLocal=col2;
}
else
{
fill ( col1 );
colLocal=col1;
}
} // if
pushMatrix();
PVector MyPVector = new PVector( 0.0, 0.0, 0.0 );
MyPVector = GetValuePVectorBoard (i, j);
box[iBox].moveTo(MyPVector.x, MyPVector.y+11, MyPVector.z);
// wenn die Figuren als Bilder dargestellt werden:
if (showFiguresAsWhat==showFiguresAsImages) {
if (Board [i][j] != " ") {
box[iBox].setTexture (GetFileName(i, j), 16 ) ;
}
}
else
{
if (colLocal == col1) {
box[iBox].setTexture ( "burloak.jpg", 16 );
}
else {
box[iBox].setTexture ( "BURLOAL.JPG", 16 );
}
}
box[iBox].fill(colLocal, 63); // ALL_SIDES
box[iBox].fill(colLocal); // ALL_SIDES
box[iBox].tagNo = (i+1)*10 + (j+1);
box[iBox].draw();
iBox++;
popMatrix();
} // for
} // for
} // function

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