Algorithm for drawing a 4-connected line - graphics

I'm looking for an algorithm (coded in Java would be nice, but anything clear enough to translate to Java is fine) to draw a 4-connected line. It seems that Bresenham's algorithm is the most widely used, but all the understandable implementations I've found are 8-connected. OpenCV's cvline function apparently has a 4-connected version, but the source code is, to me, as a mediocre and nearly C-illiterate programmer, impenetrable. Various other searches have turned up nothing.
Thanks for any help anyone can provide.

The following is a Bresenham-like algorithm that draws 4-connected lines. The code is in Python but I suppose can be understood easily even if you don't know the language.
def line(x0, y0, x1, y1, color):
dx = abs(x1 - x0) # distance to travel in X
dy = abs(y1 - y0) # distance to travel in Y
if x0 < x1:
ix = 1 # x will increase at each step
else:
ix = -1 # x will decrease at each step
if y0 < y1:
iy = 1 # y will increase at each step
else:
iy = -1 # y will decrease at each step
e = 0 # Current error
for i in range(dx + dy):
draw_pixel(x0, y0, color)
e1 = e + dy
e2 = e - dx
if abs(e1) < abs(e2):
# Error will be smaller moving on X
x0 += ix
e = e1
else:
# Error will be smaller moving on Y
y0 += iy
e = e2
The idea is that to draw a line you should increment X and Y with a ratio that matches DX/DY of the theoretic line. To do this I start with an error variable e initialized to 0 (we're on the line) and at each step I check if the error is lower if I only increment X or if I only increment Y (Bresenham check is to choose between changing only X or both X and Y).
The naive version for doing this check would be adding 1/dy or 1/dx, but multiplying all increments by dx*dy allows using only integer values and that improves both speed and accuracy and also avoids the need of special cases for dx==0 or dy==0 thus simplifying the logic.
Of course since we're looking for a proportion error, using a scaled increment doesn't affect the result.
Whatever is the line quadrant the two possibilities for the increment will always have a different sign effect on the error... so my arbitrary choice was to increment the error for an X step and decrement the error for an Y step.
The ix and iy variables are the real directions needed for the line (either +1 or -1) depending on whether the initial coordinates are lower or higher than the final coordinates.
The number of pixels to draw in a 4-connected line is obviously dx+dy, so I just do a loop for that many times to draw the line instead of checking if I got to the end point. Note that this algorithm draws all pixels except the last one; if you want also that final pixel then an extra draw_pixel call should be added after the end of the loop.
An example result of the above implementation can be seen in the following picture

For the Python-illiterate, here is a C version of 6502's code:
void drawLine(int x0, int y0, int x1, int y1) {
int dx = abs(x1 - x0);
int dy = abs(y1 - y0);
int sgnX = x0 < x1 ? 1 : -1;
int sgnY = y0 < y1 ? 1 : -1;
int e = 0;
for (int i=0; i < dx+dy; i++) {
drawPixel(x0, y0);
int e1 = e + dy;
int e2 = e - dx;
if (abs(e1) < abs(e2)) {
x0 += sgnX;
e = e1;
} else {
y0 += sgnY;
e = e2;
}
}
}

Related

basic fractal coloring problems

I am trying to get more comfortable with the math behind fractal coloring and understanding the coloring algorithms much better. I am the following paper:
http://jussiharkonen.com/files/on_fractal_coloring_techniques%28lo-res%29.pdf
The paper gives specific parameters to each of the functions, however when I use the same, my results are not quite right. I have no idea what could be going on though.
I am using the iteration count coloring algorithm to start and using the following julia set:
c = 0.5 + 0.25i and p = 2
with the coloring algorithm:
The coloring function simply returns the number of
elements in the truncated orbit divided by 20
And the palette function:
I(u) = k(u − u0),
where k = 2.5 and u0 = 0, was used.
And with a palette being white at 0 and 1, and interpolating to black in-between.
and following this algorithm:
Set z0 to correspond to the position of the pixel in the complex plane.
Calculate the truncated orbit by iterating the formula zn = f(zn−1) starting
from z0 until either
• |zn| > M, or
• n = Nmax,
where Nmax is the maximum number of iterations.
Using the coloring and color index functions, map the resulting truncated
orbit to a color index value.
Determine an RGB color of the pixel by using the palette function
Using this my code looks like the following:
float izoom = pow(1.001, zoom );
vec2 z = focusPoint + (uv * 4.0 - 2.0) * 1.0 / izoom;
vec2 c = vec2(0.5f, 0.25f) ;
const float B = 2.0;
float l;
for( int i=0; i<100; i++ )
{
z = vec2( z.x*z.x - z.y*z.y, 2.0*z.x*z.y ) + c;
if( length(z)>10.0) break;
l++;
}
float ind = basicindex(l);
vec4 col = color(ind);
and have the following index and coloring functions:
float basicindex(float val){
return val / 20.0;
}
vec4 color(float index){
float r = 2.5 * index;
float g = r;
float b = g;
vec3 v = 0.5 - 0.5 * sin(3.14/2.0 + 3.14 * vec3(r, g, b));
return vec4(1.0 - v, 1.0) ;
}
The paper provides the following image:
https://imgur.com/YIZMhaa
While my code produces:
https://imgur.com/OrxdMsN
I get the correct results by using k = 1.0 instead of 2.5, however I would prefer to understand why my results are incorrect. When extending this to the smooth coloring algorithms, my results are still incorrect so I would like to figure this out first.
Let me know if this isn't the correct place for this kind of question and I can move it to the math stack exchange. I wasn't sure which place was more appropriate.
Your image is perfectly implemented for Figure 3.3 in the paper. The other image you posted uses a different routine.
Your figure seems to have that bit of perspective code there at top, but remove that and they should be the same.
If your objection is the color extremes you set that with the "0.5 - 0.5 * ..." part of your code. This makes the darkest black originally 0.5 when in the example image you're trying to duplicate the darkest black should be 1 and the lightest white should be 0.
You're making the whiteness equal to the distance from 0.5
If you ignore the fractal all together you are getting a bunch of values that can be normalized between 0 and 1 and you're coloring those in some particular ways. Clearly the image you are duplicating is linear between 0 and 1 so putting black as 0.5 cannot be correct.
o = {
length : 500,
width : 500,
c : [.5, .25], // c = x + iy will be [x, y]
maxIterate : 100,
canvas : null
}
function point(pos, color){
var c = 255 - Math.round((1 + Math.log(color)/Math.log(o.maxIterate)) * 255);
c = c.toString(16);
if (c.length == 1) c = '0'+c;
o.canvas.fillStyle="#"+c+c+c;
o.canvas.fillRect(pos[0], pos[1], 1, 1);
}
function conversion(x, y, R){
var m = R / o.width;
var x1 = m * (2 * x - o.width);
var y2 = m * (o.width - 2 * y);
return [x1, y2];
}
function f(z, c){
return [z[0]*z[0] - z[1] * z[1] + c[0], 2 * z[0] * z[1] + c[1]];
}
function abs(z){
return Math.sqrt(z[0]*z[0] + z[1]*z[1]);
}
function init(){
var R = (1 + Math.sqrt(1+4*abs(o.c))) / 2,
z, x, y, i;
o.canvas = document.getElementById('a').getContext("2d");
for (x = 0; x < o.width; x++){
for (y = 0; y < o.length; y++){
i = 0;
z = conversion(x, y, R);
while (i < o.maxIterate && abs(z) < R){
z = f(z, o.c);
if (abs(z) > R) break;
i++;
}
if (i) point([x, y], i / o.maxIterate);
}
}
}
init();
<canvas id="a" width="500" height="500"></canvas>
via: http://jsfiddle.net/3fnB6/29/

Correct use of cos() and sin() in Python

I'm trying to convert, just for fun, a code listing written in a language called Seed7 into Python. I've got it just about to beta phase, except for one part. This is the extract of the Seed7 listing:
x1 := flt(column) + 0.5;
y1 := flt(row) + 0.5;
angle := (course - 1.0) * 0.785398;
delta_x := cos(angle);
delta_y := -sin(angle);
inquad := TRUE;
blocked := FALSE;
number := 1;
while number <= distance do
y1 := y1 + delta_y;
x1 := x1 + delta_x;
row := trunc(y1);
column := trunc(x1);
if column < 1 or column > 8 or row < 1 or row > 8 then
inquad := FALSE;
number := distance;
else
if sect[row][column] <> 1 then (* Object blocking move *)
blocked := TRUE;
number := distance;
end if;
end if;
incr(number);
end while;
Which all makes sense, except for the fact that I don't understand how the functions cos() and sin() work in Seed7.
The manual says:
sin
const func float: sin (in float: x)
Compute the sine of x, where x is given in radians.
Returns:
the trigonometric sine of an angle.
but I can't make the equivalent in Python.
This problem is purely one caused by me not understanding Python properly (and not really being that great with maths either), so I come here to ask someone who does understand these things.
What code is required to make the above code work in Python? Help!!! :-)
Many thanks,
Joseph.
Edit: I think the problem is the incr() function. Basically, it is possible to warp in this game less than 1. From the help files:
writeln("Warp - One warp moves you the width of a quadrant. A warp of .5 will move you");
writeln("halfway through a quadrant. Moving diagonally across a quadrant to the next");
writeln("will require 1.414 warps. Warp 3 will move you 3 quadrants providing nothing");
writeln("in your present quadrant blocks your exit. Once you leave the quadrant that");
writeln("you were in, you will enter hyperspace; coming out of hyperspace will place you");
writeln("randomly in the new quadrant. Klingons in a given quadrant will fire at you");
writeln("whenever you leave, enter, or move within the quadrant. Entering a course or");
writeln("warp of zero can be used to return to the command mode.")
My code looks like this:
x1 = float(column) + 0.5
y1 = float(row) + 0.5
angle = (course - 1.0) * 0.785398
deltaX = math.cos(angle)
deltaY = -math.sin(angle)
inQuad = True
blocked = False
num = 1
while num <= distance:
y1 += deltaY
x1 += deltaX
row = int(round(y1))
column = int(round(x1))
if column < 0 or column > 7 or row < 0 or row > 7:
inQuad = False
num = distance
else:
if sect[row][column] != 1:
blocked = True
num = distance
num += 1
The thing is I'm using num+=1 at the end there, as opposed to incr, but I don't understand incr. As I have said, I've been a bit long out of the game, and certain things are really catching me out.
Any help in shining a light would be appreciated.
Joseph.
A guess:
If Seed7 expects radians, you may need to convert degrees to radians before applying sin or cos:
import math
math.sin(math.radians(1))

How to calculate a geometric cross field inside an arbitrary polygon?

I'm having troubles finding a way to calculate a "cross-field" inside an arbitrary polygon.
A Cross field, as defined by one paper is the smoothest field that is tangential to the domain boundary (in this case the polygon) I find it a lot in quad re-topology papers but surprisingly not even in Wikipedia I can find the definition of a Cross field.
I have images but since I'm new here the system said I need at least 10 reputation points to upload images.
Any ideas?
I think it could be something along the lines of an interpolation? given an inner point determine the distance to each edge and integrate or weight sum the tangent and perpendicular vector of every edge by the distance? (or any other factor in fact)
But other simpler approaches may exist?
Thanks in advance!
//I've come up with something like this (for the 3D case), very raw, educational purposes
float ditance2segment(Vector3D p, Vector3D p0, Vector3D p1){
Vector3D v = p1 - p0;
Vector3D w = p - p0;
float c1 = v.Dot(w);
if (c1 <= 0)
return (p - p1).Length();
float c2 = v.Dot(v);
if (c2 <= c1)
return (p - p1).Length();
float b = c1 / c2;
Vector3D pb = p0 + b*v;
return (p - pb).Length();
}
void CrossFieldInterpolation(List<Vector3D>& Contour, List<Vector3D>& ContourN, Vector3D p, Vector3D& crossU, Vector3D& crossV){
int N = Contour.Amount();
for (int i=0; i < N; i++){
Vector3D u = Contour[(i + 1) % N] - Contour[i];
Vector3D n = 0.5*(ContourN[(i + 1) % N] + ContourN[i]);
Vector3D v = -Vector3D::Cross(u,n); //perpendicular vector
u = Vector3D::Normalize(u);
n = Vector3D::Normalize(n);
v = Vector3D::Normalize(v);
float dist = ditance2segment(p, Contour[i], Contour[(i+1)%N]);
crossU += u / (1+dist); //to avoid infinity at points over the segment
crossV += v / (1+dist);
}
crossU = Vector3D::Normalize(crossU);
crossV = Vector3D::Normalize(crossV);
}
You can check the OpenSource Graphite software that I'm developping, it implements the "Periodic Global Parameterization" algorithm [1] that was developed in my research team. You may be also interested in the following research articles with algorithms that we developed more recently [2],[3]
Graphite website:
http://alice.loria.fr/software/graphite
How to use Periodic Global Parameterization:
http://alice.loria.fr/WIKI/index.php/Graphite/PGP
[1] http://alice.loria.fr/index.php/publications.html?Paper=TOG_pgp%402006
[2] http://alice.loria.fr/index.php/publications.html?Paper=DGF#2008
[3] http://alice.loria.fr/index.php/publications.html?redirect=0&Paper=DFD#2008&Author=vallet

How can I find the general form equation of a line from two points?

Given the input:
double x1,y1,x2,y2;
How can I find the general form equation (double a,b,c where ax + by + c = 0) ?
Note: I want to be able to do this computationally. So the equivalent for slope-intercept form would be something like:
double dx, dy;
double m, b;
dx = x2 - x1;
dy = y2 - y1;
m = dy/dx;
b = y1;
Obviously, this is very simple, but I haven't been able to find the solution for the general equation form (which is more useful since it can do vertical lines). I already looked in my linear algebra book and two books on computational geometry (both too advanced to explain this).
If you start from the equation y-y1 = (y2-y1)/(x2-x1) * (x-x1) (which is the equation of the line defined by two points), through some manipulation you can get (y1-y2) * x + (x2-x1) * y + (x1-x2)*y1 + (y2-y1)*x1 = 0, and you can recognize that:
a = y1-y2,
b = x2-x1,
c = (x1-x2)*y1 + (y2-y1)*x1.
Get the tangent by subtracting the two points (x2-x1, y2-y1). Normalize it and rotate by 90 degrees to get the normal vector (a,b). Take the dot product with one of the points to get the constant, c.
If you start from the equation of defining line from 2 points
(x - x1)/(x2 - x1) = (y - y1)/(y2 - y1)
you can end up with the next equation
x(y2 - y1) - y(x2 - x1) - x1*y2 + y1*x2 = 0
so the coefficients will be:
a = y2 - y1
b = -(x2 - x1) = x1 - x2
c = y1*x2 - x1*y2
My implementation of the algorithm in C
inline v3 LineEquationFrom2Points(v2 P1, v2 P2) {
v3 Result;
Result.A = P2.y - P1.y;
Result.B = -(P2.x - P1.x);
Result.C = P1.y * P2.x - P1.x * P2.y;
return(Result);
}
Shortcut steps:
"Problem : (4,5) (3,-7)"
Solve:
m=-12/1 then
12x-y= 48
"NOTE:m is a slope"
COPY THE NUMERATOR, AFFIX "X"
Positive fraction Negative sign on between.
(tip: simmilar sign = add + copy the sign)
1.Change the second set into opposite signs,
2.ADD y1 to y2 (means add or subtract them depending of the sign),
3.ADD x1 to x2 (also means add or subtract them depending of the sign),
4.Then Multiply 12 and 1 to any of the problem set.
After that "BOOM" Tada!, you have your answer
#include <stdio.h>
main()
{
int a,b,c;
char x,y;
a=5;
b=10;
c=15;
x=2;
y=3;
printf("the equation of line is %dx+%dy=%d" ,a,b,c);
}

How to test if a line segment intersects an axis-aligned rectange in 2D?

How to test if a line segment intersects an axis-aligned rectange in 2D? The segment is defined with its two ends: p1, p2. The rectangle is defined with top-left and bottom-right points.
The original poster wanted to DETECT an intersection between a line segment and a polygon. There was no need to LOCATE the intersection, if there is one. If that's how you meant it, you can do less work than Liang-Barsky or Cohen-Sutherland:
Let the segment endpoints be p1=(x1 y1) and p2=(x2 y2).
Let the rectangle's corners be (xBL yBL) and (xTR yTR).
Then all you have to do is
A. Check if all four corners of the rectangle are on the same side of the line.
The implicit equation for a line through p1 and p2 is:
F(x y) = (y2-y1)*x + (x1-x2)*y + (x2*y1-x1*y2)
If F(x y) = 0, (x y) is ON the line.
If F(x y) > 0, (x y) is "above" the line.
If F(x y) < 0, (x y) is "below" the line.
Substitute all four corners into F(x y). If they're all negative or all positive, there is no intersection. If some are positive and some negative, go to step B.
B. Project the endpoint onto the x axis, and check if the segment's shadow intersects the polygon's shadow. Repeat on the y axis:
If (x1 > xTR and x2 > xTR), no intersection (line is to right of rectangle).
If (x1 < xBL and x2 < xBL), no intersection (line is to left of rectangle).
If (y1 > yTR and y2 > yTR), no intersection (line is above rectangle).
If (y1 < yBL and y2 < yBL), no intersection (line is below rectangle).
else, there is an intersection. Do Cohen-Sutherland or whatever code was mentioned in the other answers to your question.
You can, of course, do B first, then A.
Alejo
Wrote quite simple and working solution:
bool SegmentIntersectRectangle(double a_rectangleMinX,
double a_rectangleMinY,
double a_rectangleMaxX,
double a_rectangleMaxY,
double a_p1x,
double a_p1y,
double a_p2x,
double a_p2y)
{
// Find min and max X for the segment
double minX = a_p1x;
double maxX = a_p2x;
if(a_p1x > a_p2x)
{
minX = a_p2x;
maxX = a_p1x;
}
// Find the intersection of the segment's and rectangle's x-projections
if(maxX > a_rectangleMaxX)
{
maxX = a_rectangleMaxX;
}
if(minX < a_rectangleMinX)
{
minX = a_rectangleMinX;
}
if(minX > maxX) // If their projections do not intersect return false
{
return false;
}
// Find corresponding min and max Y for min and max X we found before
double minY = a_p1y;
double maxY = a_p2y;
double dx = a_p2x - a_p1x;
if(Math::Abs(dx) > 0.0000001)
{
double a = (a_p2y - a_p1y) / dx;
double b = a_p1y - a * a_p1x;
minY = a * minX + b;
maxY = a * maxX + b;
}
if(minY > maxY)
{
double tmp = maxY;
maxY = minY;
minY = tmp;
}
// Find the intersection of the segment's and rectangle's y-projections
if(maxY > a_rectangleMaxY)
{
maxY = a_rectangleMaxY;
}
if(minY < a_rectangleMinY)
{
minY = a_rectangleMinY;
}
if(minY > maxY) // If Y-projections do not intersect return false
{
return false;
}
return true;
}
Since your rectangle is aligned, Liang-Barsky might be a good solution. It is faster than Cohen-Sutherland, if speed is significant here.
Siggraph explanation
Another good description
And of course, Wikipedia
You could also create a rectangle out of the segment and test if the other rectangle collides with it, since it is just a series of comparisons. From pygame source:
def _rect_collide(a, b):
return a.x + a.w > b.x and b.x + b.w > a.x and \
a.y + a.h > b.y and b.y + b.h > a.y
Use the Cohen-Sutherland algorithm.
It's used for clipping but can be slightly tweaked for this task. It divides 2D space up into a tic-tac-toe board with your rectangle as the "center square".
then it checks to see which of the nine regions each of your line's two points are in.
If both points are left, right, top, or bottom, you trivially reject.
If either point is inside, you trivially accept.
In the rare remaining cases you can do the math to intersect with whichever sides of the rectangle are possible to intersect with, based on which regions they're in.
Or just use/copy the code already in the Java method
java.awt.geom.Rectangle2D.intersectsLine(double x1, double y1, double x2, double y2)
Here is the method after being converted to static for convenience:
/**
* Code copied from {#link java.awt.geom.Rectangle2D#intersectsLine(double, double, double, double)}
*/
public class RectangleLineIntersectTest {
private static final int OUT_LEFT = 1;
private static final int OUT_TOP = 2;
private static final int OUT_RIGHT = 4;
private static final int OUT_BOTTOM = 8;
private static int outcode(double pX, double pY, double rectX, double rectY, double rectWidth, double rectHeight) {
int out = 0;
if (rectWidth <= 0) {
out |= OUT_LEFT | OUT_RIGHT;
} else if (pX < rectX) {
out |= OUT_LEFT;
} else if (pX > rectX + rectWidth) {
out |= OUT_RIGHT;
}
if (rectHeight <= 0) {
out |= OUT_TOP | OUT_BOTTOM;
} else if (pY < rectY) {
out |= OUT_TOP;
} else if (pY > rectY + rectHeight) {
out |= OUT_BOTTOM;
}
return out;
}
public static boolean intersectsLine(double lineX1, double lineY1, double lineX2, double lineY2, double rectX, double rectY, double rectWidth, double rectHeight) {
int out1, out2;
if ((out2 = outcode(lineX2, lineY2, rectX, rectY, rectWidth, rectHeight)) == 0) {
return true;
}
while ((out1 = outcode(lineX1, lineY1, rectX, rectY, rectWidth, rectHeight)) != 0) {
if ((out1 & out2) != 0) {
return false;
}
if ((out1 & (OUT_LEFT | OUT_RIGHT)) != 0) {
double x = rectX;
if ((out1 & OUT_RIGHT) != 0) {
x += rectWidth;
}
lineY1 = lineY1 + (x - lineX1) * (lineY2 - lineY1) / (lineX2 - lineX1);
lineX1 = x;
} else {
double y = rectY;
if ((out1 & OUT_BOTTOM) != 0) {
y += rectHeight;
}
lineX1 = lineX1 + (y - lineY1) * (lineX2 - lineX1) / (lineY2 - lineY1);
lineY1 = y;
}
}
return true;
}
}
A quick Google search popped up a page with C++ code for testing the intersection.
Basically it tests the intersection between the line, and every border or the rectangle.
Rectangle and line intersection code
Here's a javascript version of #metamal's answer
var isRectangleIntersectedByLine = function (
a_rectangleMinX,
a_rectangleMinY,
a_rectangleMaxX,
a_rectangleMaxY,
a_p1x,
a_p1y,
a_p2x,
a_p2y) {
// Find min and max X for the segment
var minX = a_p1x
var maxX = a_p2x
if (a_p1x > a_p2x) {
minX = a_p2x
maxX = a_p1x
}
// Find the intersection of the segment's and rectangle's x-projections
if (maxX > a_rectangleMaxX)
maxX = a_rectangleMaxX
if (minX < a_rectangleMinX)
minX = a_rectangleMinX
// If their projections do not intersect return false
if (minX > maxX)
return false
// Find corresponding min and max Y for min and max X we found before
var minY = a_p1y
var maxY = a_p2y
var dx = a_p2x - a_p1x
if (Math.abs(dx) > 0.0000001) {
var a = (a_p2y - a_p1y) / dx
var b = a_p1y - a * a_p1x
minY = a * minX + b
maxY = a * maxX + b
}
if (minY > maxY) {
var tmp = maxY
maxY = minY
minY = tmp
}
// Find the intersection of the segment's and rectangle's y-projections
if(maxY > a_rectangleMaxY)
maxY = a_rectangleMaxY
if (minY < a_rectangleMinY)
minY = a_rectangleMinY
// If Y-projections do not intersect return false
if(minY > maxY)
return false
return true
}
I did a little napkin solution..
Next find m and c and hence the equation y = mx + c
y = (Point2.Y - Point1.Y) / (Point2.X - Point1.X)
Substitute P1 co-ordinates to now find c
Now for a rectangle vertex, put the X value in the line equation, get the Y value and see if the Y value lies in the rectangle bounds shown below
(you can find the constant values X1, X2, Y1, Y2 for the rectangle such that)
X1 <= x <= X2 &
Y1 <= y <= Y2
If the Y value satisfies the above condition and lies between (Point1.Y, Point2.Y) - we have an intersection.
Try every vertex if this one fails to make the cut.
I was looking at a similar problem and here's what I came up with. I was first comparing the edges and realized something. If the midpoint of an edge that fell within the opposite axis of the first box is within half the length of that edge of the outer points on the first in the same axis, then there is an intersection of that side somewhere.
But that was thinking 1 dimensionally and required looking at each side of the second box to figure out.
It suddenly occurred to me that if you find the 'midpoint' of the second box and compare the coordinates of the midpoint to see if they fall within 1/2 length of a side (of the second box) of the outer dimensions of the first, then there is an intersection somewhere.
i.e. box 1 is bounded by x1,y1 to x2,y2
box 2 is bounded by a1,b1 to a2,b2
the width and height of box 2 is:
w2 = a2 - a1 (half of that is w2/2)
h2 = b2 - b1 (half of that is h2/2)
the midpoints of box 2 are:
am = a1 + w2/2
bm = b1 + h2/2
So now you just check if
(x1 - w2/2) < am < (x2 + w2/2) and (y1 - h2/2) < bm < (y2 + h2/2)
then the two overlap somewhere.
If you want to check also for edges intersecting to count as 'overlap' then
change the < to <=
Of course you could just as easily compare the other way around (checking midpoints of box1 to be within 1/2 length of the outer dimenions of box 2)
And even more simplification - shift the midpoint by your half lengths and it's identical to the origin point of that box. Which means you can now check just that point for falling within your bounding range and by shifting the plain up and to the left, the lower corner is now the lower corner of the first box. Much less math:
(x1 - w2) < a1 < x2
&&
(y1 - h2) < b1 < y2
[overlap exists]
or non-substituted:
( (x1-(a2-a1)) < a1 < x2 ) && ( (y1-(b2-b1)) < b1 < y2 ) [overlap exists]
( (x1-(a2-a1)) <= a1 <= x2 ) && ( (y1-(b2-b1)) <= b1 <= y2 ) [overlap or intersect exists]
coding example in PHP (I'm using an object model that has methods for things like getLeft(), getRight(), getTop(), getBottom() to get the outer coordinates of a polygon and also has a getWidth() and getHeight() - depending on what parameters were fed it, it will calculate and cache the unknowns - i.e. I can create a polygon with x1,y1 and ... w,h or x2,y2 and it can calculate the others)
I use 'n' to designate the 'new' item being checked for overlap ($nItem is an instance of my polygon object) - the items to be tested again [this is a bin/sort knapsack program] are in an array consisting of more instances of the (same) polygon object.
public function checkForOverlaps(BinPack_Polygon $nItem) {
// grab some local variables for the stuff re-used over and over in loop
$nX = $nItem->getLeft();
$nY = $nItem->getTop();
$nW = $nItem->getWidth();
$nH = $nItem->getHeight();
// loop through the stored polygons checking for overlaps
foreach($this->packed as $_i => $pI) {
if(((($pI->getLeft() - $nW) < $nX) && ($nX < $pI->getRight())) &&
((($pI->getTop() - $nH) < $nY) && ($nY < $pI->getBottom()))) {
return false;
}
}
return true;
}
Some sample code for my solution (in php):
// returns 'true' on overlap checking against an array of similar objects in $this->packed
public function checkForOverlaps(BinPack_Polygon $nItem) {
$nX = $nItem->getLeft();
$nY = $nItem->getTop();
$nW = $nItem->getWidth();
$nH = $nItem->getHeight();
// loop through the stored polygons checking for overlaps
foreach($this->packed as $_i => $pI) {
if(((($pI->getLeft() - $nW) < $nX) && ($nX < $pI->getRight())) && ((($pI->getTop() - $nH) < $nY) && ($nY < $pI->getBottom()))) {
return true;
}
}
return false;
}

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