Develop Reference

Efficient algorithm for finding which hexagon a point belongs to

I'm trying to find a more efficient way of determining which hexagon a point belongs to from the following:
an array of points - for the sake of argument, 10000 points.
an array of center points of hexagons, approximately 1000 hexagons.
every point will belong to exactly one hexagon, some (most) hexagons will be empty.
The hexagons form a perfect grid, with the point of one hexagon starting in the top left corner (it will overlap the edge of the total area).
My current solution works, but is rather slow n * (m log m) I think, where n=length(points) and m=length(hexagons).
I suspect I can do much better than this, one solution that comes to mind is to sort (just once) both the points and the hexagons by their distance to some arbitrary point (perhaps the middle, perhaps a corner) then iterate over the points and over a subset of the hexagons, starting from the first hexagon whose distance to this point is >= to the last hexagon matched. Similarly, we could stop looking at hexagons once the distance difference between the (point -> ref point) and (hexagon center -> ref point) is larger than the "radius" of the hexagon. In theory, since we know that every point will belong to a hexagon, I don't even have to consider this possibility.
My question is: Is there a Much better way of doing it than this? In terms of complexity, I think it's worst case becomes marginally better n * m but the average case should be very good, probably in the region of n * 20 (e.g., we only need to look at 20 hexagons per point). Below is my current inefficient solution for reference.
points.forEach((p) => {
p.hex = _.sortBy(hexes, (hex) => {
const xDist = Math.abs(hex.middle.x - p.x);
const yDist = Math.abs(hex.middle.y - p.y);
return Math.sqrt((xDist * xDist) + (yDist * yDist));
})[0];
});

For an arbitrary point, you can find the nearest hexagon center in two steps (assuming the same arrangement as that of Futurologist):
divide the abscissa by the horizontal spacing between the centers, and round to the nearest integer.
divide the ordinate by the half of the vertical spacing, and round to the nearest even or odd integer, depending on the parity found above.
consider this center and the six ones around it, and keep the closest to the target point.
This gives you the indexes of the tile, in constant time.

Just a suggestion: assume you have the centers of each regular hexagon from your regular hexagonal grid (if I have understood correctly, that's part of the information you have).
-----
/ \
- ----- -----------> x - axis
\ / \
----- -
/ \ /
- -----
\ / \
----- -
| \ /
| -----
|
|
V
y - axis
You can think that your coordinate system starts from the center of the hexagon in the upper left corner and the y coordinate axis runs vertically down, while the x axis runs from left to right horizontally. The centers of the hexagons from your regular hexagonal grid form an image of the regular square grid, where the integer vertices of the square grid are transformed into the centers of the polygons by simply multiplying the coordinates of points in the square grid by the 2 x 2 square matrix (a sheer metrix)
A = a*[ sqrt(3)/2 0;
1/2 1 ]
where a is a parameter of the hexagonal grid, the distance between the centers of two edge-adjacent hexagons. This provides a way to assign integer indices [m n] to the grid formed by the hexagonal centers. After that, if you are given a point with coordinates [x y] in the hexagonal grid, you can apply the inverse matrix of A
[u; v] = A^(-1)*[x; y]
where
A^(-1) = (2/(a*sqrt(3)))*[ 1 0 ;
-1/2 sqrt(3)/2 ]
([x; y] and [u; v] are column vectors) and then take m = floor(u) and n = floor(v) to determine the integer coordinates (also the indices) [m = floor(u), n = floor(v)] of the upper left corner of the square cell from the square grid (observe that we have chosen the coordinates for both grids to start from the upper left corner). Thus, your point [u, v] is in the square with vertices [m,n] [m+1, n] [m, n+1] [m+1, n+1]
which means that the original point [x y] is in one of the four hexagons whose centers have indices [m,n] [m+1, n] [m, n+1] [m+1, n+1]. So you can use that to check in which of the four hexagons the point [x y] is.
I hope this helps.

Update: Leaving the below comment for posterity
I am now using the code provided here: https://www.redblobgames.com/grids/hexagons/
A really important note, is that your hexagon grid MUST start with the first hexagons mid point at (0, 0) - if it doesn't you get extremely odd results from this, which at first glance appeared as rounding errors (even after accounting for the expected offset). For me, it didn't matter where the first hexagon was positioned, so I just set it to be (0, 0) and it worked great.
Old solution
I'm still hoping for an optimal solution, but I ended up rolling my own which needs only check 6 hexagons per point, with a little overhead (approximately sqrt(m)) needed in addition.
With approximately 3000 points, and 768 hexagons (of which 310 were populated), it correctly assigned the point to the hexagon 100% of the time (as checked against a brute force approach) and took 29 milliseconds, compared to ~840 with brute force.
To start with, I store the hexagons in a map where the key is "${column},${row}". The columns technically overlap, so for the 0th row, the 0th column starts at -0.5 * hexWidth, and for row 1, the 0th column starts at 0px.
Next, I start from the position of the top left hexagon, item "0,0", which should also be at position 0, and increment y by either the height of the hexagon, or the edge length of the hexagon accordingly. When the y is > the points y, I've found the probable row, I then check the row above and below.
For the column within the row, I take the both the Math.floor and Math.ceil of x / hexWidth.
Doing this gives 6 hexagons to check, from this point the solution is identical to the solution in the question.
In theory, this could be used to just look up the correct hexagon, using the x/y position. However in practice, this didn't work for me about 5% of the time with off by 1 errors, likely a rounding problem.
Some other things I looked at:
As suggested by #jason-aller, https://www.redblobgames.com/grids/hexagons/#rounding. Unfortunately, this seems to assume some form of transformation on the hex grid (rotations) and is not easy to follow - continually referencing functions which have yet to be defined.
QuadTree (various implementations) unfortunately, this returned approximately 100 "potential matches" for each point - so the performance improvement was not good. I'm aware that insertion order changes how useful QuadTree is, I tried natural order, sorted by distance from top, left and shuffled, they all performed equally badly. It's likely that an optimal solution with QuadTree would involve populating the tree with the item closest to the mid point, then the items 1/2 from the mid point to each corner, recursively. Too much like hard work for me!

Placing points equidistantly along an Archimedean spiral

I have an Archimedean spiral determined by the parametric equations x = r t * cos(t) and y = r t * sin(t).
I need to place n points equidistantly along the spiral. The exact definition of equidistant doesn't matter too much - it only has to be approximate.
Using just r, t and n as parameters, how can I calculate the coordinates of each equidistant point?

You want to place points equidistantly corresponding to arc length. Arc length for Archimedean spiral (formula 4) is rather complex
s(t) = a/2 * (t * Sqrt(1 + t*t) + ln(t + Sqrt(1+t*t)))
and for exact positions one could use numerical methods, calculating t values for equidistant s1, s2, s3... arithmetical progression. It is possible though.
First approximation possible - calculate s(t) values for some sequence of t, then get intervals for needed s values and apply linear interpolation.
Second way - use Clackson scroll formula approximation, this approach looks very simple (perhaps inexact for small t values)
t = 2 * Pi * Sqrt(2 * s / a)
Checked: quite reliable result

Extract vertices from an ARC in generic way

I want to get all the vertices from an ARC. I have all the data (for ex : start point, end point, start angle, end angle, radius) which will used to draw an arc but my need is I have to generate all the vertices from the arc data.
I have already tried with one or two algorithm but I failed to get the exact vertices from an arc data.
I used Bresenham's algorithm but I failed.
Right now I am using below code but its not working ..
double theta = 2 * 3.1415926 / 100;
double c = Math.cos(theta);
double s = Math.sin(theta);
double t;
double x = ((ArcTo) element).getRadius();//we start at angle = 0
double y = 0;
for(int ii = 0; ii < 100; ii++) {
coordinates.add(new Coordinate(x + element.getCenterPoint().getX(), y + element.getCenterPoint().getY()));//output vertex
//apply the rotation matrix
t = x;
x = c * x - s * y;
y = s * t + c * y;
}
Please help me. Thank you.

First some clarifications
I assume that by vertices you mean the pixels and ARC is standard 2D circular arc (not elliptic arc !!!) and your input data are:
int (x0,y0),(x1,y1) // star/end points on curve !!!
float a0,a1 // start end angles [rad]
int (xc,yc) // center of circle
int r // radius
Do not use Bresenham
because you would need to start from zero angle and compute all pixels until start point is hit. Then flip draw flag so you start filling the pixel from that point and stop on end point hit. Also you would need to handle the winding to match ARC direction.
You can use circle parametric equation
// just some arc data to test with
float r=25.0;
float a0= 45.0*M_PI/180.0;
float a1=270.0*M_PI/180.0;
int xc=100,x0=xc+floor(r*cos(a0)),x1=xc+floor(r*cos(a1));
int yc=100,y0=yc+floor(r*sin(a0)),y1=yc+floor(r*sin(a1));
// arc rasterize code
int x,y;
float a,da;
// here draw pixels x0,y0 and x1,y1 to avoid rounding holes ...
if (r) da=0.75/float(r); else da=0.1; // step slightly less then pixel to avoid holes
for (a=a0;;a+=da)
{
x=xc+int(floor(r*cos(a)));
y=yc+int(floor(r*sin(a)));
// here draw pixel x,y
if ((x==x1)&&(y==y1)) // stop if endpoint reach
if (fabs(a-a1)<da) // but ignore stop if not at end angle (full or empty circle arc)
break;
}
may be round instead of floor will have less pixel position error. If your endpoint does not match then this will loop infinitely. If you tweak a bit the end conditions you can avoid even that or recompute x1,y1 from a1 as I have ...
You can use equation (x-xc)^2+(y-yc)^2=r^2
you need to divide ARC to quadrants and handle each as separate arc looping through x or y and computing the other coordinate. Loop through coordinate that is changing more
so in blue areas loop y and in the red loop x. For example red area code can look like this:
int x,y;
for (x=_x0;;x++)
{
y=sqrt((r*r)-((x-xc)*(x-xc)));
// here draw pixel x,y
if (x==_x1) // stop if endpoint reach
break;
}
you need to compute (_x0,_y0),(_x1,_y1) start end points of cut part of ARC inside the quadrant and make _x0<=_x1.
The value for _x looped start/end point coordinate will be xc +/- sqrt(r) or x0 or x1
the value for _y looped start/end point coordinate will be yc +/- sqrt(r) or y0 or y1
The blue parts are done in analogically manner (just swap/replace x and y). This approach is a bit more complicated due to cutting but can be done solely on integers. sqrt can be speed up by LUT (limiting the max radius) and the ^2 can be also further optimized.
[Notes]
so if I recapitulate the parametric equation is the simplest to implement but slowest. Then is the sqrt approach which can be done as fast as Bresenham (and may be even faster with LUT) but need the code for cutting ARC to quadrants which need few ifs prior to rendering.
All codes are in C++ and can be further improved like avoiding some int/float conversions, pre-compute some values before loop, etc ...
The last goes the Bresenham but you need to change a few things inside and when you do not know what you are doing you can easily get lost. It also need to cut to octant's so the complexity of change is far bigger then sqrt approach

How to divide an ellipse to equal segments?

This calculates vertex coordinates on ellipse:
function calculateEllipse(a, b, angle)
{
var alpha = angle * (Math.PI / 180) ;
var sinalpha = Math.sin(alpha);
var cosalpha = Math.cos(alpha);
var X = a * cosalpha - b * sinalpha;
var Y = a * cosalpha + b * sinalpha;
}
But how can I calculate the "angle" to get equal or roughly equal circumference segments?

So from what Jozi's said in the OP's comments, what's needed isn't how to subdivide an ellipse into equal segments (which would require a whole bunch of horrible integrals), it's to construct an ellipse from line segments of roughly equal length.
There are a whole pile of ways to do that, but I think the best suited for the OP's purposes would be the concentric circle method, listed on the page as 'the draftman's method'. If you don't mind installing the Mathematica player, there's a neat lil' app here which illustrates it interactively.
The problem with those methods is that the segment lengths are only roughly equal at low eccentricities. If you're dealing in extreme eccentricities, things get a lot more complicated. The simplest solution I can think of is to linearly approximate the length of a line segment within each quadrant, and then solve for the positions of the endpoints in that quadrant exactly.
In detail: this is an ellipse quadrant with parameters a = 5, b = 1:
And this is a plot of the length of the arc subtended by an infinitesimal change in the angle, at each angle:
The x axis is the angle, in radians, and the y axis is the length of the arc that would be subtended by a change in angle of 1 radian. The formula, which can be derived using the equations in the Wikipedia article I just linked, is y = Sqrt(a^2 Sin^2(x) + b^2 Cos^2(x)). The important thing to note though is that the integral of this function - the area under this curve - is the length of the arc in the whole quadrant.
Now, we can approximate it by a straight line:
which has gradient m = (a-b) / (Pi/2) and y intercept c = b. Using simple geometry, we can deduce that the area under the red curve is A = (a+b)*Pi/4.
Using this knowledge, and the knowledge that the area under the curve is the total length of the curve, the problem of constructing an approximation to the ellipse reduces to finding say a midpoint-rule quadrature (other quadratures would work too, but this is the simplest) of the red line such that each rectangle has equal area.
Converting that sentence to an equation, and representing the position of a rectangle in a quadrature by it's left hand boundary x and its width w, we get that:
(v*m)*w^2 + (m*x+c)*w - A/k == 0
where k is the number of pieces we want to use to approximate the quadrant, and v is a weighting function I'll come to shortly. This can be used to construct the quadrature by first setting x0 = 0 and solving for w0, which is then used to set x1 = w0 and solve for w1. Then set x2 = w1, etc etc until you've got all k left-hand boundary points. The k+1th boundary point is obviously Pi/2.
The weighting function v effectively represents where the rectangle crosses the red line. A constant v = 0.5 is equivalent to it crossing in the middle, and gets you this with 10 points:
but you can play around with it to see what better balances the points. Ideally it should stay in the range [0, 1] and the sum of the values you use should be k/2.
If you want an even better approximation without messing around with weighting functions, you could try least-squares fitting a line rather than just fitting it to the endpoints, or you could try fitting a cubic polynomial to the blue curve instead of a linear polynomial. It'll entail solving quartics but if you've a maths package on hand that shouldn't be a problem.

Too long for a comment, so I suppose this has to be an answer ...
Here's a mathematically simple approach to forming a first order approximation. Pick one quadrant. You can generate the data for the other quadrants by reflection in the X and Y axis. Calculate (x,y) for the angle = 0 degrees, 1 degree, ... 90 degrees. Now you want the little lengths joining consecutive points. If (x_n, y_n) are the coordinates at angle = n, then Pythagoras tells us the distance D between points (x_n, y_n) and (x_n+1, y_n+1) is D = sqrt((x_n+1 - x_n)^2 + (y_n+1 - y_n)^2). Use this formula to produce a table of cumulative distances around the ellipse for angles from 0 degrees to 90 degrees. This is the inverse of the function you seek. Of course, you don't have to pick a stepsize of 1 degree; you could use any angle which exactly divides 90 degrees.
If you want to find the angle which corresponds to a perimeter step size of x, find the largest angle n in your table producing a partial perimeter less than or equal to x. The partial perimeter of angle n+1 will be larger than x. Use linear interpolation to find the fractional angle which corresponds to x.
All we are doing is approximating the ellipse with straight line segments and using them instead of the original curve; its a first order approximation. You could do somewhat better by using Simpson's rule or similar instead of linear interpolation.
Yes, you have to calculate the table in advance. But once you have the table, the calculations are easy. If you don't need too much accuracy, this is pretty simple both mathematically and coding-wise.

How to calculate depth coordinate of the 3D rectangle?

I have 3D rectangle, as shown in the image.
Here I know depth distance and x and y coordinates of the one end. Based on these two values I would like calculate coordinates at the other end.
For clear view, I have attached a screen.

If you don't know the relation between 2D and 3D (i.e. the projection formulas used) then you can't apply the depth.
That said, if you make the following assumptions:
the projection type is orthographic
a depth line is projected as a 45 degrees line
the length of a 45 degrees line is the same as if it was a normal line
... then you could calculate it with Pythagoras' theorem as follows:
The red lines are equal (in case of a 45 degree line), so:
x1 = 100 + 50 * (1 / sqrt(2))
y1 = 50 - 50 * (1 / sqrt(2))