Develop Reference

Rectangle packing algorithm with desired position?

I'd like to implement a variation of a rectangle packing algorithm in C#. In my case the rectangles have a width and height and a "desired" position in a 2D plane (on the screen). They must however not overlap. I want the algorithm to find the positions of the rectangles that minimizes the distances of their desired positions. I am aware that the order in which the rectangles are placed plays a role but I can't even find a performant algorithm for a fixed or random order. Anyone got an idea or references?
More formal definiton of the problem here

I implemented #tiliavirga's suggestion and it works quite well.
Some notes:
I made the repulsive force proportional to only the square root of the overlapping area because otherwise, the first few iterations had huge repulsive forces blowing the constellation apart. (On the other hand, it leads to quick termination which could be important, see below)
I reduced the attractive force over time towards 0, because otherwise, the alg oscillates in some cases, where overlapping rectangles are pushed away, then in the next iteration pulled together, then pushed away, and so on
The algorithm can take very long, depending on the parameters (1) how quickly the attractive force weakens, (2) how large the motion of the rectangles is in each iteration, and (3) the limit of the total overlapping area which can be tolerated, terminating the algorithm. In time-critical applications, e.g. in games where this computation is done every frame, these parameters should be adjusted to result in a quick termination with a not-so-optimal solution.
All in all, a good enough solution for me. Python code below:
DATA STRUCTURE:
class Rect(object):
def __init__(self, centerX, centerY, width, height):
self.centerX = centerX
self.centerY = centerY
self.desired_centerX = centerX
self.desired_centerY = centerY
self.left = centerX - width / 2
self.right = centerX + width / 2
self.bottom = centerY - height / 2
self.top = centerY + height / 2
self.width = width
self.height = height
def move(self, x, y):
self.centerX += x
self.centerY += y
self.left += x
self.right += x
self.bottom += y
self.top += y
UTILITY:
def normalize(vector):
length = np.linalg.norm(vector)
# define the normalization of the zero vector like this, because we need to move rectangles
# somewhere when they are perfectly centered on each other
if length == 0:
return np.random.rand(vector.shape[0])
else:
return vector / length
def isOverlapping(r1, r2):
#we define that a rects doesn't overlap with itself
if r1 is r2:
return False
if r1.left > r2.right or r1.right < r2.left or r1.bottom > r2.top or r1.top < r2.bottom:
return False
return True
def getOverlappingArea(r1, r2):
if not isOverlapping(r1, r2):
return 0
else:
return (min(r1.right, r2.right) - max(r1.left, r2.left)) * \
(min(r1.right, r2.right) - max(r1.left, r2.left))
#pointing from "r1" to "r2"
def getScaledPushingForce(r1, r2):
overlappingArea = getOverlappingArea(r1, r2)
if overlappingArea < 0:
raise ValueError("Something went wrong, negative overlapping area calculated!")
if overlappingArea == 0:
return np.array([0,0])
return np.sqrt(overlappingArea) * normalize( \
np.array([r2.centerX - r1.centerX, r2.centerY - r1.centerY]))
PARAMETERS:
# the strength of the pulling force towards the desired position decays to easy termination
# higher value = slower decay
# faster decay means faster termination but worse results
pullingForceHalfTime = 10
# the overlapping area which is considered to be small enough to stop the algorithm
# (recommended to assign according to the number and size of the rectangles)
acceptableOverlap = 2*len(rects)
# the scaling of the total forces, that moves the rectangles
# larger portions mean faster termination but possibly worse results
# (recommended 1/2<= forceScaling <= 1/20, the smaller pullingStrength is, the lower should forceScaling also be
# e.g. forceScaling = 1/20 * pullingStrength)
forceScaling = 1/10
ALGORITHM:
# calculates pulling and pushing forces and moves the rectangles a bit in the direction of the combination of these forces
# in every iteration. Stops when the overlapping area is sufficiently small
def unstack():
i = 1
#iterate until break
while True:
#pulling forces towards the desired position
#weakened over the course of the iteration (depending on d), since no overlapping is the stronger constraint
pulling_forces = [np.array([r.desired_centerX - r.centerX, r.desired_centerY - r.centerY]) * \
np.power(0.5, i/pullingForceHalfTime) for r in rects]
#pushing forces resulting from overlapping rectangles
#the directions of the forces for a pair of overlapping rectangles has the direction of the connecting vector
#between their centers and the magnitude is proportional to the are of the overlap
pushing_forces = [np.sum([getScaledPushingForce(r_, r) for r_ in rects], axis=0) for r in rects]
total_forces = np.sum([pulling_forces, pushing_forces], axis=0) * forceScaling
#move the rectangles by a portion of the total forces (smaller steps => more iterations but better results)
for j in range(len(rects)):
rects[j].move(total_forces[j][0], total_forces[j][1])
#stop iterating when the total overlapping area is sufficiently small
if np.sum(np.square([getOverlappingArea(r[0], r[1]) for r in itertools.combinations(rects, 2)])) <= acceptableOverlap:
break
i += 1
#print results
finalDistancesFromDesired = [np.array([r.desired_centerX - r.centerX, r.desired_centerY - r.centerY]) for r in rects]
print("Total distances to desired positions: " + str(np.sum(np.linalg.norm(finalDistancesFromDesired, axis = 1))))
and an example run through:
Example

Drawing a parabolic arc between two points based on a known projectile angle

I am trying to draw an arc between two points that represents a projectile's path. The angle that the projectile leaves point A at is known, and the X/Y coordinates of both points are known.
I'm trying to figure out the math behind this, and how to draw it up in c#.
Here is my failed attempt, based on some path examples I found
var g = new StreamGeometry();
var xDistance = Math.Abs(pointA.X - pointB.X);
var yDistance = Math.Abs(pointA.Y - pointB.Y);
var angle = 60;
var radiusX = xDistance / angle;
var radiusY = yDistance / angle;
using (var gc = g.Open())
{
gc.BeginFigure(
startPoint: pointA,
isFilled: false,
isClosed: false);
gc.ArcTo(
point: pointB,
size: new Size(radiusX, radiusY),
rotationAngle: 0d,
isLargeArc: false,
sweepDirection: SweepDirection.Clockwise,
isStroked: true,
isSmoothJoin: false);
}
Any help would be greatly appreciated!
Edit #2 (added clarity): For this problem assume that physics play no role (no gravity, velocity, or any outside forces). The projectile is guaranteed to land at point B and move along a parabolic path. The vertex will be halfway between point A and point B on the horizontal axis. The angle that it launches at is the angle up from the ground (horizontal).
So Point A (Ax, Ay) is known.
Point B (Bx, By) is known.
The angle of departure is known.
The X half of the vertex is known (Vx = Abs(Ax - Bx)).
Does this really boil down to needing to figure out the Y coordinate of the vertex?

Following on from the comments, we need a quadratic Bezier curve. This is defined by 3 points, the start, end, and a control point:
It is defined by the following equation:
We therefore need to find P1 using the given conditions (note that the gravity strength is determined implicitly). For a 2D coordinate we need two constraints / boundary conditions. They are given by:
The tangent vector at P0:
We need to match the angle to the horizontal:
The apex of the curve must be directly below the control point P1:
Therefore the vertical coordinate is given by:
[Please let me know if you would like some example code for the above]
Now for how to add a quadratic Bezier; thankfully, once you have done the above, it is not too difficult
The following method creates the parabolic geometry for the simple symmetric case. The angle is measured in degrees counterclockwise from the horizontal.
public Geometry CreateParabola(double x1, double x2, double y, double angle)
{
var startPoint = new Point(x1, y);
var endPoint = new Point(x2, y);
var controlPoint = new Point(
0.5 * (x1 + x2),
y - 0.5 * (x2 - x1) * Math.Tan(angle * Math.PI / 180));
var geometry = new StreamGeometry();
using (var context = geometry.Open())
{
context.BeginFigure(startPoint, false, false);
context.QuadraticBezierTo(controlPoint, endPoint, true, false);
}
return geometry;
}

A body movement subject only to the force of gravity (air resistance is ignored) can be evaluated with the following equations:
DistanceX(t) = dx0 + Vx0·t
DistanceY(t) = dy0 + Vy0·t - g/2·t^2
Where
g : gravity acceleration (9.8 m/s^2)
dx0 : initial position in the X axis
dy0 : initial position in the Y axis
Vy0 : initial X velocity component (muzzle speed)
Vy0 : initial Y velocity component (muzzle speed)
Well that doesn't seem very helpful, but lets investigate further. Your cannon has a muzzle speed V we can consider constant, so Vx0 and Vy0 can be written as:
Vx0 = V·cos(X)
Vy0 = V·sin(X)
Where X is the angle at which you are shooting. Ok, that seems interesting, we finally have an input that is useful to whoever is shooting the cannon: X. Lets go back to our equations and rewrite them:
DistanceX(t) = dx0 + V·cos(X)·t
DistanceY(t) = dy0 + V·sin(X)·t - g/2·t^2
And now, lets think through what we are trying to do. We want to figure out a way to hit a specific point P. Lets give it coordinates: (A, B). And in order to do that, the projectile has to reach that point in both projections at the same time. We'll call that time T. Ok, lets rewrite our equations again:
A = dx0 + V·cos(X)·T
B = dy0 + V·sin(X)·T - g/2·T^2
Lets get ourselves rid of some unnecessary constants here; if our cannon is located at (0, 0) our equations are now:
A = V·cos(X)·T [1]
B = V·sin(X)·T - g/2·T^2 [2]
From [1] we know that: T = A/(V·cos(X)), so we use that in [2]:
B = V·sin(X)·A/(V·cos(X)) - g/2·A^2/(V^2·cos^2(X))
Or
B = A·tan(X) - g/2·A^2/(V^2*cos^2(X))
And now some trigonometry will tell you that 1/cos^2 = 1 + tan^2 so:
B = A·tan(X) - g/2·A^2/V^2·(1+tan^2(X)) [3]
And now you have quadratic equation in tan(X) you can solve.
DISCLAIMER: typing math is kind of hard, I might have an error in there somewhere, but you should get the idea.
UPDATE The previous approach would allow you to solve the angle X that hits a target P given a muzzle speed V. Based on your comments, the angle X is given, so what you need to figure out is the muzzle speed that will make the projectile hit the target with the specified cannon angle. If it makes you more comfortable, don't think of V as muzzle speed, think of it as a form factor of the parabola you are trying to find.
Solve Vin [3]:
B = A·tan(X) - g/2·A^2/V^2·(1+tan^2(X))
This is a trivial quadratic equation, simply isolate V and take the square root. Obviously the negative root has no physical meaning but it will also do, you can take any of the two solutions. If there is no real number solution for V, it would mean that there is simply no possible shot (or parabola) that reaches P(angle X is too big; imagine you shoot straight up, you'll hit yourself, nothing else).
Now simply eliminate t in the parametrized equations:
x = V·cos(X)·t [4]
y = V·sin(X)·t - g/2·t^2 [5]
From [4] you have t = x/(V·cos(X)). Substitute in [5]:
y = tan(X)·x - g·x^2 /(2·V^2*cos^2(X))
And there is your parabola equation. Draw it and see your shot hit the mark.
I've given it a physical interpretation because I find it easier to follow, but you could change all the names I've written here to purely mathematical terms, it doesn't really matter, at the end of the day its all maths and the parabola is the same, any which way you want to think about it.

GPS lap & Segment timer

I've been searching for a while but haven't found exactly what I'm looking for.
I'm working on an app that will go in a race car. It will give the driver the ability to press a button to mark a Start/Finish line. It will also have a button to allow a driver to set segment times.
Keep in mind a track can be an oval which I'm working on first. It could be a road course or it could be an auto cross where the start and finish line aren't the exact same location. They could be with 50 feet of each other or so but the car never crosses where it starts.
I have my gps data coming in and I convert the NMea messages to my classes and I store Lat, Lon, Speed, Course etc. In my research I've ran across this which is interesting. The GPS will be mounted outside the roof for better signal. It generates 10 hits per second. (Garmin Glo)
http://www.drdobbs.com/windows/gps-programming-net/184405690?pgno=1
It's old but it talks about UTM and the Cartesian coordinate system. So using the DecDeg2UTM, I convert Lat & Lon to X & coordinates as well.
I've also been trying to use the Intersect formula I found Here I took the intersect and tried to convert it to C# which I'll post at the end. However, feeding coordinates of an oval track, it doesn't seem to be working. Also, I'm not sure exactly what it's supposed to be doing. But the coordinates it returns when it does somethign like -35.xxx & 98.xxxx which out in an ocean somewhere 1000's of miles from where the track is.
I looking for answers to the following.
I assume I need to take the location recorded when a button is pressed for Start/Finish or Segment and calculate a line perpendicular to the direction the car in able to be able to do some sort of Line Intersection calculation. The Cartesian coordinates seems to calculate the bearing fairly well. But the question here is how do you get the "left and right coordinates". Also, keep in mind, an oval track may be 60 feet wide. But as mentioned an auto cross track may only be 20 ft wide and part of the track may be with 50 ft. Note I'm fine with indicating to set the points, the car needs to be going slow or stopped at the points to get an accurate coordinate. Some tracks they will have to be set while walking the track.
Based on this, should I be trying to use decimal lat lon or would utilizing the Cartesian coordinate system based on UTM be a more accurate method for what I'm trying to do?
Either one is there a .Net library or C based library with source code that has methods for making these calculations?
How can this be accurately handled. (Not that great with Math, links to code samples would help tremendously.)
Next, after I have the lines or whatever is needed for start/finish and segments, as I get GPS sign from the car racing, I need to figure out the most accurate way to tell when a car has crossed those segments. again if I'm lucky I'll get 10 hits per second but it will probably be lower. Then the vehicle speeds could vary significantly depending on the type of track and vehicle. So the GPS hit could be many feet "left or right" of a segment. Also, it could be many feet before or after a segment.
Again, if there is a GIS library out there I can feed coordinates and all this is calculated, that's would work as well as long as it's performant. If not again I'm trying to decide if it's best to break down coordinates to X Y or some geometry formulas for coordinates in decimal format. Mods, I assume there is hard data to support an answer of either way and this isn't responses aren't fully subjective to opinions.
Here is the C# code I came up with from the Script page above. I'm starting to feel UTM and the Cartesian Coordinate system would be better for accuracy and performance. But again I'm open to evidence to the contrary if it exists.
Thanks
P.S. Note GeoCoordinate is from the .Net System.Device.Location assemble. GpsData is just a class I use to convert NMEA messages into Lat, Lon, Course, NumSats, DateTime etc.
The degree Radian methods are extensions as as follows.
public static double DegreeToRadians(this double angle)
{
return Math.PI * angle / 180.0;
}
public static double RadianToDegree(this double angle)
{
return angle * (180.0 / Math.PI);
}
}
public static GeoCoordinate CalculateIntersection(GpsData p1, double brng1, GpsData p2, double brng2)
{
// see http://williams.best.vwh.net/avform.htm#Intersection
// Not sure I need to use Cosine
double _p1LatRadians = p1.Latitude.DegreeToRadians();
double _p1LonToRadians = p1.Longitude.DegreeToRadians();
double _p2LatToRadians = p2.Latitude.DegreeToRadians();
double _p2LonToRadians = p2.Longitude.DegreeToRadians();
double _brng1ToRadians = brng1.DegreeToRadians();
double _brng2ToRadians = brng2.DegreeToRadians();
double _deltaLat = _p2LatToRadians - _p1LatRadians;
double _deltaLon = _p2LonToRadians - _p1LonToRadians;
var _var1 = 2 * Math.Asin(Math.Sqrt(Math.Sin(_deltaLat / 2) * Math.Sin(_deltaLat / 2)
+ Math.Cos(_p1LatRadians) * Math.Cos(_p2LatToRadians) * Math.Sin(_deltaLon / 2) * Math.Sin(_deltaLon / 2)));
if (_var1 == 0) return null;
// initial/final bearings between points
var _finalBrng = Math.Acos((Math.Sin(_p2LatToRadians) - Math.Sin(_p1LatRadians) * Math.Cos(_var1)) / (Math.Sin(_var1) * Math.Cos(_p1LatRadians)));
//if (isNaN(θa)) θa = 0; // protect against rounding
var θb = Math.Acos((Math.Sin(_p1LatRadians) - Math.Sin(_p2LatToRadians) * Math.Cos(_var1)) / (Math.Sin(_var1) * Math.Cos(_p2LatToRadians)));
var θ12 = Math.Sin(_p2LonToRadians - _p1LonToRadians) > 0 ? _finalBrng : 2 * Math.PI - _finalBrng;
var θ21 = Math.Sin(_p2LonToRadians - _p1LonToRadians) > 0 ? 2 * Math.PI - θb : θb;
var α1 = (_brng1ToRadians - θ12 + Math.PI) % (2 * Math.PI) - Math.PI; // angle 2-1-3
var α2 = (θ21 - _brng2ToRadians + Math.PI) % (2 * Math.PI) - Math.PI; // angle 1-2-3
if (Math.Sin(α1) == 0 && Math.Sin(α2) == 0) return null; // infinite intersections
if (Math.Sin(α1) * Math.Sin(α2) < 0) return null; // ambiguous intersection
α1 = Math.Abs(α1);
α2 = Math.Abs(α2);
// ... Ed Williams takes abs of α1/α2, but seems to break calculation?
var α3 = Math.Acos(-Math.Cos(α1) * Math.Cos(α2) + Math.Sin(α1) * Math.Sin(α2) * Math.Cos(_var1));
var δ13 = Math.Atan2(Math.Sin(_var1) * Math.Sin(α1) * Math.Sin(α2), Math.Cos(α2) + Math.Cos(α1) * Math.Cos(α3));
var _finalLatRadians = Math.Asin(Math.Sin(_p1LatRadians) * Math.Cos(δ13) + Math.Cos(_p1LatRadians) * Math.Sin(δ13) * Math.Cos(_brng1ToRadians));
var _lonBearing = Math.Atan2(Math.Sin(_brng1ToRadians) * Math.Sin(δ13) * Math.Cos(_p1LatRadians), Math.Cos(δ13) - Math.Sin(_p1LatRadians) * Math.Sin(_finalLatRadians));
var _finalLon = _p1LonToRadians + _lonBearing;
var _returnLat = _finalLatRadians.RadianToDegree();
var _latToDegree = _finalLon.RadianToDegree();
var _returnLon = ( _latToDegree + 540) % 360 - 180;
return new GeoCoordinate(_returnLat, _returnLon);
//return new LatLon(φ3.toDegrees(), (λ3.toDegrees() + 540) % 360 - 180); // normalise to −180..+180°
}

Calculate a specific curve with specific rotation, c#

I am trying to code for a game I am working on a specific curve with a specific rotation. I am not a great mathematician... At all... Tried searching for solutions for a few hours, but I'm affraid I do not find any solution.
So, a small picture to illustrate first:
This is an eighth of a circle, radius of 9, beggining is (0,0)
The end is now at about 6.364, -2.636. But I need this same curve, with a 45° direction at the end, but ending at aexactly 6.0,-3.0.
Could any of you show me how to do this? I need to be able to calculate precisly any point on this curve & its exact length. I would suppose using some kind of eliptical math could be a solution? I admit my math class are reaaaly far now and have now good clue for now...
Thank for any possible help

I think I found a quadratic curve which sastisfies your requirement:
f(x) = -1/12 x^2 + 9
Copy the following into https://www.desmos.com/calculator to see it:
-\frac{1}{12}x^2+9
f'(x) would be -1/6x, so when x=6, the derivative would be -1, which corresponds to a -45° inclination. There are probably infinite curves that satisfy your requirement but if my calculus isn't too rusty this is one of them.
I tried to fit an ellipse with foci starting at y=6 here and starting at y=9 here to your points but the slope doesn't look like 45°.
Also starting at any height k, here doesn't seem to work.

I don't think you've fully understood the question I asked in the comments about the "inclination" angle. So I will give a general case solution, where you have an explicit tangent vector for the end of the curve. (You can calculate this using the inclination angle; if we clarify what you mean by it then I will be happy to edit with a formula to calculate the tangent vector if necessary)
Let's draw a diagram of how the setup can look:
(Not 100% accurate)
A and B are your fixed points. T is the unit tangent vector. r and C are the radius and center of the arc we need to calculate.
The angle θ is given by the angle between BA and T minus π/2 radians (90 degrees). We can calculate it using the dot product:
The (signed) distance from the center of AB to C is given by:
Note that this is negative for the case on the right, and positive for the left. The radius is given by:
(You can simplify by substituting and using a cosine addition rule, but I prefer to keep things in terms of variables in the diagram). To obtain the point C, we need the perpendicular vector to AB (call it n):
Now that we have the radius and center of the circular arc, we still need to determine which direction we are moving in, i.e. whether we are moving clockwise or anti-clockwise when going from A to B. This is a simple test, using the cross-product:
If this is negative, then T is as in the diagram, and we need to move clockwise, and vice versa. The length of the arc l, and the angular displacement γ when we move by a distance x along the arc:
Nearly there! Just one more step - we need to work out how to rotate the point A by angle γ around point C, to get the point we want (call it D):
(Adapted from this Wikipedia page)
Now for some code, in case the above was confusing (it probably was!):
public Vector2 getPointOnArc(Vector2 A, Vector2 B, Vector2 T, double x)
{
// calculate preliminaries
Vector2 BA = B - A;
double d = BA.Length();
double theta = Math.Acos(Vector2.DotProduct(BA, T) / d) - Math.PI * 0.5;
// calculate radius
double r = d / (2.0 * Math.Cos(theta));
// calculate center
Vector2 n = new Vector2(BA.y, -BA.x);
Vector2 C = 0.5 * (A + B + n * Math.Tan(theta));
// calculate displacement angle from point A
double l = (Math.PI - 2.0 * theta) * r;
double gamma = (2.0 * Math.PI * x) / l;
// sign change as discussed
double cross = T.x * BA.y - T.y * BA.x;
if (cross < 0.0) gamma = -gamma;
// finally return the point we want
Vector2 disp = A - C;
double c_g = Math.Cos(gamma), s_g = Math.Sin(gamma);
return new Vector2(disp.X * c_g + disp.Y * s_g + C.X,
disp.Y * c_g - disp.X * s_g + C.Y);
}

Approximating an ellipse with a polygon

I am working with geographic information, and recently I needed to draw an ellipse. For compatibility with the OGC convention, I cannot use the ellipse as it is; instead, I use an approximation of the ellipse using a polygon, by taking a polygon which is contained by the ellipse and using arbitrarily many points.
The process I used to generate the ellipse for a given number of point N is the following (using C# and a fictional Polygon class):
Polygon CreateEllipsePolygon(Coordinate center, double radiusX, double radiusY, int numberOfPoints)
{
Polygon result = new Polygon();
for (int i=0;i<numberOfPoints;i++)
{
double percentDone = ((double)i)/((double)numberOfPoints);
double currentEllipseAngle = percentDone * 2 * Math.PI;
Point newPoint = CalculatePointOnEllipseForAngle(currentEllipseAngle, center, radiusX, radiusY);
result.Add(newPoint);
}
return result;
}
This has served me quite while so far, but I've noticed a problem with it: if my ellipse is 'stocky', that is, radiusX is much larger than radiusY, the number of points on the top part of the ellipse is the same as the number of points on the left part of the ellipse.
That is a wasteful use of points! Adding a point on the upper part of the ellipse would hardly affect the precision of my polygon approximation, but adding a point to the left part of the ellipse can have a major effect.
What I'd really like, is a better algorithm to approximate the ellipse with a polygon. What I need from this algorithm:
It must accept the number of points as a parameter; it's OK to accept the number of points in every quadrant (I could iteratively add points in the 'problematic' places, but I need good control on how many points I'm using)
It must be bounded by the ellipse
It must contain the points straight above, straight below, straight to the left and straight to the right of the ellipse's center
Its area should be as close as possible to the area of the ellipse, with preference to optimal for the given number of points of course (See Jaan's answer - appearantly this solution is already optimal)
The minimal internal angle in the polygon is maximal
What I've had in mind is finding a polygon in which the angle between every two lines is always the same - but not only I couldn't find out how to produce such a polygon, I'm not even sure one exists, even if I remove the restrictions!
Does anybody have an idea about how I can find such a polygon?

finding a polygon in which the angle between every two lines is
always the same
Yes, it is possible. We want to find such points of (the first) ellipse quadrant, that angles of tangents in these points form equidistant (the same angle difference) sequence. It is not hard to find that tangent in point
x=a*Cos(fi)
y=b*Sin(Fi)
derivatives
dx=-a*Sin(Fi), dy=b*Cos(Fi)
y'=dy/dx=-b/a*Cos(Fi)/Sin(Fi)=-b/a*Ctg(Fi)
Derivative y' describes tangent, this tangent has angular coefficient
k=b/a*Cotangent(Fi)=Tg(Theta)
Fi = ArcCotangent(a/b*Tg(Theta)) = Pi/2-ArcTan(a/b*Tg(Theta))
due to relation for complementary angles
where Fi varies from 0 to Pi/2, and Theta - from Pi/2 to 0.
So code for finding N + 1 points (including extremal ones) per quadrant may look like (this is Delphi code producing attached picture)
for i := 0 to N - 1 do begin
Theta := Pi/2 * i / N;
Fi := Pi/2 - ArcTan(Tan(Theta) * a/b);
x := CenterX + Round(a * Cos(Fi));
y := CenterY + Round(b * Sin(Fi));
end;
// I've removed Nth point calculation, that involves indefinite Tan(Pi/2)
// It would better to assign known value 0 to Fi in this point
Sketch for perfect-angle polygon:

One way to achieve adaptive discretisations for closed contours (like ellipses) is to run the Ramer–Douglas–Peucker algorithm in reverse:
1. Start with a coarse description of the contour C, in this case 4
points located at the left, right, top and bottom of the ellipse.
2. Push the initial 4 edges onto a queue Q.
while (N < Nmax && Q not empty)
3. Pop an edge [pi,pj] <- Q, where pi,pj are the endpoints.
4. Project a midpoint pk onto the contour C. (I expect that
simply bisecting the theta endpoint values will suffice
for an ellipse).
5. Calculate distance D between point pk and edge [pi,pj].
if (D > TOL)
6. Replace edge [pi,pj] with sub-edges [pi,pk], [pk,pj].
7. Push new edges onto Q.
8. N = N+1
endif
endwhile
This algorithm iteratively refines an initial discretisation of the contour C, clustering points in areas of high curvature. It terminates when, either (i) a user defined error tolerance TOL is satisfied, or (ii) the maximum allowable number of points Nmax is used.
I'm sure that it's possible to find an alternative that's optimised specifically for the case of an ellipse, but the generality of this method is, I think, pretty handy.

I assume that in the OP's question, CalculatePointOnEllipseForAngle returns a point whose coordinates are as follows.
newPoint.x = radiusX*cos(currentEllipseAngle) + center.x
newPoint.y = radiusY*sin(currentEllipseAngle) + center.y
Then, if the goal is to minimize the difference of the areas of the ellipse and the inscribed polygon (i.e., to find an inscribed polygon with maximal area), the OP's original solution is already an optimal one. See Ivan Niven, "Maxima and Minima Without Calculus", Theorem 7.3b. (There are infinitely many optimal solutions: one can get another polygon with the same area by adding an arbitrary constant to currentEllipseAngle in the formulae above; these are the only optimal solutions. The proof idea is quite simple: first one proves that these are the optimal solutions in case of a circle, i.e. if radiusX=radiusY; secondly one observes that under a linear transformation that transforms a circle into our ellipse, e.g. a transformation of multiplying the x-coordinate by some constant, all areas are multiplied by a constant and therefore a maximal-area inscribed polygon of the circle is transformed into a maximal-area inscribed polygon of the ellipse.)
One may also regard other goals, as suggested in the other posts: e.g. maximizing the minimal angle of the polygon or minimizing the Hausdorff distance between the boundaries of the polygon and ellipse. (E.g. the Ramer-Douglas-Peucker algorithm is a heuristic to approximately solve the latter problem. Instead of approximating a polygonal curve, as in the usual Ramer-Douglas-Peucker implementation, we approximate an ellipse, but it is possible to devise a formula for finding on an ellipse arc the farthest point from a line segment.) With respect to these goals, the OP's solution would usually not be optimal and I don't know if finding an exact solution formula is feasible at all. But the OP's solution is not as bad as the OP's picture shows: it seems that the OP's picture has not been produced using this algorithm, as it has less points in the more sharply curved parts of the ellipse than this algorithm produces.

I suggest you switch to polar coordinates:
Ellipse in polar coord is:
x(t) = XRadius * cos(t)
y(t) = YRadius * sin(t)
for 0 <= t <= 2*pi
The problems arise when Xradius >> YRadius (or Yradius >> Yradius)
Instead of using numberOfPoints you can use an array of angles obviously not all identical.
I.e. with 36 points and dividing equally you get angle = 2*pi*n / 36 radiants for each sector.
When you get around n = 0 (or 36) or n = 18 in a "neighborhood" of these 2 values the approx method doesn't works well cause the ellipse sector is significantly different from the triangle used to approximate it. You can decrease the sector size around this points thus increasing precision. Instead of just increasing the number of points that would also increase segments in other unneeded areas. The sequence of angles should become something like (in degrees ):
angles_array = [5,10,10,10,10.....,5,5,....10,10,...5]
The first 5 deg. sequence is for t = 0 the second for t = pi, and again the last is around 2*pi.

Here is an iterative algorithm I've used.
I didn't look for theoretically-optimal solution, but it works quit well for me.
Notice that this algorithm gets as an input the maximal error of the prime of the polygon agains the ellipse, and not the number of points as you wish.
public static class EllipsePolygonCreator
{
#region Public static methods
public static IEnumerable<Coordinate> CreateEllipsePoints(
double maxAngleErrorRadians,
double width,
double height)
{
IEnumerable<double> thetas = CreateEllipseThetas(maxAngleErrorRadians, width, height);
return thetas.Select(theta => GetPointOnEllipse(theta, width, height));
}
#endregion
#region Private methods
private static IEnumerable<double> CreateEllipseThetas(
double maxAngleErrorRadians,
double width,
double height)
{
double firstQuarterStart = 0;
double firstQuarterEnd = Math.PI / 2;
double startPrimeAngle = Math.PI / 2;
double endPrimeAngle = 0;
double[] thetasFirstQuarter = RecursiveCreateEllipsePoints(
firstQuarterStart,
firstQuarterEnd,
maxAngleErrorRadians,
width / height,
startPrimeAngle,
endPrimeAngle).ToArray();
double[] thetasSecondQuarter = new double[thetasFirstQuarter.Length];
for (int i = 0; i < thetasFirstQuarter.Length; ++i)
{
thetasSecondQuarter[i] = Math.PI - thetasFirstQuarter[thetasFirstQuarter.Length - i - 1];
}
IEnumerable<double> thetasFirstHalf = thetasFirstQuarter.Concat(thetasSecondQuarter);
IEnumerable<double> thetasSecondHalf = thetasFirstHalf.Select(theta => theta + Math.PI);
IEnumerable<double> thetas = thetasFirstHalf.Concat(thetasSecondHalf);
return thetas;
}
private static IEnumerable<double> RecursiveCreateEllipsePoints(
double startTheta,
double endTheta,
double maxAngleError,
double widthHeightRatio,
double startPrimeAngle,
double endPrimeAngle)
{
double yDelta = Math.Sin(endTheta) - Math.Sin(startTheta);
double xDelta = Math.Cos(startTheta) - Math.Cos(endTheta);
double averageAngle = Math.Atan2(yDelta, xDelta * widthHeightRatio);
if (Math.Abs(averageAngle - startPrimeAngle) < maxAngleError &&
Math.Abs(averageAngle - endPrimeAngle) < maxAngleError)
{
return new double[] { endTheta };
}
double middleTheta = (startTheta + endTheta) / 2;
double middlePrimeAngle = GetPrimeAngle(middleTheta, widthHeightRatio);
IEnumerable<double> firstPoints = RecursiveCreateEllipsePoints(
startTheta,
middleTheta,
maxAngleError,
widthHeightRatio,
startPrimeAngle,
middlePrimeAngle);
IEnumerable<double> lastPoints = RecursiveCreateEllipsePoints(
middleTheta,
endTheta,
maxAngleError,
widthHeightRatio,
middlePrimeAngle,
endPrimeAngle);
return firstPoints.Concat(lastPoints);
}
private static double GetPrimeAngle(double theta, double widthHeightRatio)
{
return Math.Atan(1 / (Math.Tan(theta) * widthHeightRatio)); // Prime of an ellipse
}
private static Coordinate GetPointOnEllipse(double theta, double width, double height)
{
double x = width * Math.Cos(theta);
double y = height * Math.Sin(theta);
return new Coordinate(x, y);
}
#endregion
}