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

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);
}

Algorithm for points inside a polyline inexplicably fails (AutoCAD)

I'm using the .NET API for Autocad, I have an algorithm (which I did not write) for determining if a point lies within a polygon (straight lines only).
I have been testing my command on the same 51 polygons repeatedly. 99% it will work perfectly. Every once in a while it will fail on 1 or more of the polygons, returning false for over 2000 points I am creating inside the bounding box of the polyline. I have seen it fail when the polyline isa simple rectangle and all of the points lie distributed in a grid within the polyline. It should have returned true over 2000 times in that case. It will never fail for just 1 of the points, it will fail all of them. I have confirmed that the points are being correctly created where I expect them to be and that the vertices of the polygon are where I expect them to be. When it fails, the last angle variable for the last point is at exactly double PI.
I am not doing any multi-threading. The only possibly 'funny' thing I am doing is COM Interop with Excel. This is happening after the transaction has been committed for the part with this algorithm, and I am sure I am cleaning up all my COM objects. I have not been able to reproduce the failure without the COM Interop part but I don't think I've tested it enough yet to have enough absence of evidence.
Any ideas what could be wrong?
bool IsInsidePolygon(Polyline polygon, Point3d pt)
{
int n = polygon.NumberOfVertices;
double angle = 0;
Point pt1, pt2;
for (int i = 0; i < n; i++)
{
pt1.X = polygon.GetPoint2dAt(i).X - pt.X;
pt1.Y = polygon.GetPoint2dAt(i).Y - pt.Y;
pt2.X = polygon.GetPoint2dAt((i + 1) % n).X - pt.X;
pt2.Y = polygon.GetPoint2dAt((i + 1) % n).Y - pt.Y;
angle += Angle2D(pt1.X, pt1.Y, pt2.X, pt2.Y);
}
if (Math.Abs(angle) < Math.PI)
return false;
else
return true;
}
public struct Point
{
public double X, Y;
};
public static double Angle2D(double x1, double y1, double x2, double y2)
{
double dtheta, theta1, theta2;
theta1 = Math.Atan2(y1, x1);
theta2 = Math.Atan2(y2, x2);
dtheta = theta2 - theta1;
while (dtheta > Math.PI)
dtheta -= (Math.PI * 2);
while (dtheta < -Math.PI)
dtheta += (Math.PI * 2);
return (dtheta);
}

Some ideas:
floating point comparison have to be done using a tolerence, this might cause kind of arbitrary results especially in case where the point lies on the polyline (same remark for point3d, they must be compared using a tolerence)
maybe the last point of your polyline is at the same location as the first one, in that case, the angle cannot be computed (perhaps this is why you get a double pi value for the last point). You should then test is first and last points are equals.
I'm not sure your algorithm works regardless if the polyline is clockwise or counterclockwise (I think yes)
you may convert your polyline into a region and rely on region point containment method

another method.
Make one "temporary" point outside the polygon (find the min X and Y and make a point with X-1 and Y-1).
Then make a line between your point and the new "temporary" point.
Check if this line crosses the polygon - use polyline.IntersectWith.
If number of cross points is odd - then your point is inside, if the number of crosses is even - the your point is not inside.
This works for me, hope it will helps you also.
If you find trouble with implementing this, i can send you an example code.
Regards,
Dobriyan Benov

I used some code from Kean Walmsley to convert 3d lines into 2d lines. But be aware that the following is not (always) true:
Point2d pt = lwp.GetPoint2dAt(i);
Point2d npt = new Point2d(lwp.GetPoint3dAt(i).X, lwp.GetPoint3dAt(i).Y);
pt == npt;
I encountered it using it on a polylines, with 3d vertices. I ended up using the npt.
http://through-the-interface.typepad.com/through_the_interface/2007/04/iterating_throu.html

artefacts during heightmap generation using plasma style fractal

I've spent a few hours today researching how random terrain generation tends to be done and after reading about the plasma fractal (midpoint displacement and diamond square algo's) I decided to try and have a go at implementing one. My result was actually not terriable, but I have these horrible square/line/grid type artefacts that I just can not seem to get rid of!
When rendered as a gray scale image my height map looks something like:
height map http://sphotos-d.ak.fbcdn.net/hphotos-ak-ash3/535816_10151739010123327_225111175_n.jpg
Obviously there is a fair amount of code involved in this but I will try to post what is only relevant. I've not not posted the code that turns it into a texture for example, but do not worry I have already tried just filling my height array with a smooth gradient and the texture comes out fine :)
I begin by setting the four corners of the map to random values between 0 and 1 and then start the recursive displacement algo:
public void GenerateTerrainLayer()
{
//set the four corners of the map to have random values
TerrainData[0, 0] = (float)RandomGenerator.NextDouble();
TerrainData[GenSize, 0] = (float)RandomGenerator.NextDouble();
TerrainData[0, GenSize] = (float)RandomGenerator.NextDouble();
TerrainData[GenSize, GenSize] = (float)RandomGenerator.NextDouble();
//begin midpoint displacement algorithm...
MidPointDisplace(new Vector2_I(0, 0), new Vector2_I(GenSize, 0), new Vector2_I(0, GenSize), new Vector2_I(GenSize, GenSize));
}
TerrainData is simply a 2D array of floats*. Vector2_I is just my own integer vector class. The last four functions are MidPointDisplace which is the recursive function, CalculateTerrainPointData which averages 2 data values and adds some noise, CalculateTerrainPointData2 which averages 4 data values and adds some noise and has a slightly higher scale value (its only used for center points) and finally my noise function which atm is just some random noise and not a real noise like perlin etc. They look like this:
private void MidPointDisplace(Vector2_I topleft, Vector2_I topright, Vector2_I bottomleft, Vector2_I bottomright)
{
//check size of square working on.. if its shorter than a certain amount stop the algo, we've done enough
if (topright.X - topleft.X < DisplacementMaxLOD)
{
return;
}
//calculate the positions of all the middle points for the square that has been passed to the function
Vector2_I MidLeft, MidRight, MidTop, MidBottom, Center;
MidLeft.X = topleft.X;
MidLeft.Y = topleft.Y + ((bottomleft.Y - topleft.Y) / 2);
MidRight.X = topright.X;
MidRight.Y = topright.Y + ((bottomright.Y - topright.Y) / 2);
MidTop.X = topleft.X + ((topright.X - topleft.X) / 2);
MidTop.Y = topleft.Y;
MidBottom.X = bottomleft.X + ((bottomright.X - bottomleft.X) / 2);
MidBottom.Y = bottomleft.Y;
Center.X = MidTop.X;
Center.Y = MidLeft.Y;
//collect the existing data from the corners of the area passed to algo
float TopLeftDat, TopRightDat, BottomLeftDat, BottomRightDat;
TopLeftDat = GetTerrainData(topleft.X, topleft.Y);
TopRightDat = GetTerrainData(topright.X, topright.Y);
BottomLeftDat = GetTerrainData(bottomleft.X, bottomleft.Y);
BottomRightDat = GetTerrainData(bottomright.X, bottomright.Y);
//and the center
//adverage data and insert for midpoints..
SetTerrainData(MidLeft.X, MidLeft.Y, CalculateTerrainPointData(TopLeftDat, BottomLeftDat, MidLeft.X, MidLeft.Y));
SetTerrainData(MidRight.X, MidRight.Y, CalculateTerrainPointData(TopRightDat, BottomRightDat, MidRight.X, MidRight.Y));
SetTerrainData(MidTop.X, MidTop.Y, CalculateTerrainPointData(TopLeftDat, TopRightDat, MidTop.X, MidTop.Y));
SetTerrainData(MidBottom.X, MidBottom.Y, CalculateTerrainPointData(BottomLeftDat, BottomRightDat, MidBottom.X, MidBottom.Y));
SetTerrainData(Center.X, Center.Y, CalculateTerrainPointData2(TopLeftDat, TopRightDat, BottomLeftDat, BottomRightDat, Center.X, Center.Y));
debug_displacement_iterations++;
//and recursively fire off new calls to the function to do the smaller squares
Rectangle NewTopLeft = new Rectangle(topleft.X, topleft.Y, Center.X - topleft.X, Center.Y - topleft.Y);
Rectangle NewTopRight = new Rectangle(Center.X, topright.Y, topright.X - Center.X, Center.Y - topright.Y);
Rectangle NewBottomLeft = new Rectangle(bottomleft.X, Center.Y, Center.X - bottomleft.X, bottomleft.Y - Center.Y);
Rectangle NewBottomRight = new Rectangle(Center.X , Center.Y, bottomright.X - Center.X, bottomright.Y - Center.Y);
MidPointDisplace(new Vector2_I(NewTopLeft.Left, NewTopLeft.Top), new Vector2_I(NewTopLeft.Right, NewTopLeft.Top), new Vector2_I(NewTopLeft.Left, NewTopLeft.Bottom), new Vector2_I(NewTopLeft.Right, NewTopLeft.Bottom));
MidPointDisplace(new Vector2_I(NewTopRight.Left, NewTopRight.Top), new Vector2_I(NewTopRight.Right, NewTopRight.Top), new Vector2_I(NewTopRight.Left, NewTopRight.Bottom), new Vector2_I(NewTopRight.Right, NewTopRight.Bottom));
MidPointDisplace(new Vector2_I(NewBottomLeft.Left, NewBottomLeft.Top), new Vector2_I(NewBottomLeft.Right, NewBottomLeft.Top), new Vector2_I(NewBottomLeft.Left, NewBottomLeft.Bottom), new Vector2_I(NewBottomLeft.Right, NewBottomLeft.Bottom));
MidPointDisplace(new Vector2_I(NewBottomRight.Left, NewBottomRight.Top), new Vector2_I(NewBottomRight.Right, NewBottomRight.Top), new Vector2_I(NewBottomRight.Left, NewBottomRight.Bottom), new Vector2_I(NewBottomRight.Right, NewBottomRight.Bottom));
}
//helper function to return a data value adveraged from two inputs, noise value added for randomness and result clamped to ensure a good value
private float CalculateTerrainPointData(float DataA, float DataB, int NoiseX, int NoiseY)
{
return MathHelper.Clamp(((DataA + DataB) / 2.0f) + NoiseFunction(NoiseX, NoiseY), 0.0f, 1.0f) * 1.0f;
}
//helper function to return a data value adveraged from four inputs, noise value added for randomness and result clamped to ensure a good value
private float CalculateTerrainPointData2(float DataA, float DataB, float DataC, float DataD, int NoiseX, int NoiseY)
{
return MathHelper.Clamp(((DataA + DataB + DataC + DataD) / 4.0f) + NoiseFunction(NoiseX, NoiseY), 0.0f, 1.0f) * 1.5f;
}
private float NoiseFunction(int x, int y)
{
return (float)(RandomGenerator.NextDouble() - 0.5) * 0.5f;
}
Ok thanks for taking the time to look - hopefully someone knows where this grid-like pattern is appearing from :)
*edit - accidently wrote ints, corrected to floats

I identified 3 problems in your code. (2 of which are related)
You don't scale down the randomness in each step. There must be a reduction of the randomness in each step. Otherwise you get white(-ish) noise. You choose a factor (0.5-0.7 worked fine for my purposes) and multiply the reduction by alpha in each recursion and scale the generated random number by that factor.
You swapped the diamond and square step. First the diamonds, then the squares. The other way round is impossible (see next).
Your square step uses only points in one direction. This one probably causes the rectangular structures you are talking about. The squares must average the values to all four sides. This means that the square step depends on the point generated by the diamond step. And not only the diamond step of the rectangle you are currently looking at, also of the rectangles next to it. For values outside of the map, you can either wrap, use a fixed value or only average 3 values.

I see a problem in your CalculateTerrainPointData implementation: you're not scaling down the result of NoiseFunction with each iteration.
See this description of the Midpoint Displacement algorithm:
Start with a single horizontal line segment.
Repeat for a sufficiently large number of times:
Repeat over each line segment in the scene:
Find the midpoint of the line segment.
Displace the midpoint in Y by a random amount.
Reduce the range for random numbers.
A fast way to do it in your code without changing too much is by adding some scale parameter to MidPointDisplace (with default set to 1.0f) and CalculateTerrainPointData; use it in CalculateTerrainPointData to multiply result of NoiseFunction; and reduce it with each recursive call to MidPointDisplace(..., 0.5f * scale).
Not sure though if that is the only cause to your image looking wrong or there are other problems.

According to Wikipedia's summary of midpoint displacement, only the average for the center most point get noise added to it - try only adding noise via CalculateTerrainPointData2 & removing the noise in CalculateTerrainPointData.

Logarithmic Spiral - Is Point on Spiral (cartesian coordinates

Lets Say I have a 3d Cartesian grid. Lets also assume that there are one or more log spirals emanating from the origin on the horizontal plane.
If I then have a point in the grid I want to test if that point is in one of the spirals. I acutally want to test if it within a certain range of the spirals but determining if it is on the point is a good start.
So I guess the question has a couple parts.
How to generate the arms from parameters (direction, tightness)
How to tell if a point in the grid is in one of the spiral arms
Any ideas? I have been googling all day and don't feel I am any closer to a solution than when I started.
Here is a bit more information that might help:
I don't actually need to render the spirals. I want to set the pitch and rotation and then pass a point to a method that can tell me if the point I passed is within the spiral (within a given range of any point on the spiral). Based on the value returned (true or false) my program will make a decision on whether or not something exists at the point in space.
How to parametrically define the log spirals (pitch and rotation and ??)
Test if a point (x, y, z) is withing a given range of any point on the spiral.
Note: Both of the above would be just on the horizontal plane

These are two functions defining an anti-clockwise spiral:
PolarPlot[{
Exp[(t + 10)/100],
Exp[t/100]},
{t, 0, 100 Pi}]
Output:
These are two functions defining a clockwise spiral:
PolarPlot[{
- Exp[(t + 10)/100],
- Exp[t/100]},
{t, 0, 100 Pi}]
Output:
Cartesian coordinates
The conversion Cartesian <-> Polar is
(1) Ro = Sqrt[x^2+y^2]
t = ArcTan[y/x]
(2) x = Ro Cos[t]
y = Ro Sin[t]
So, If you have a point in Cartesian Coords (x,y) you transform it to your equivalent polar coordinates using (1). Then you use the forula for the spiral function (any of the four mentinoned above the plots, or similar ones) putting in there the value for t, and obtaining Ro. The last step is to compare this Ro with the one we got from the coordinates converion. If they are equal, the point is on the spiral.
Edit Answering your comment
For a Log spiral is almost the same, but with multiple spirals you need to take care of the logs not going to negative values. That's why I used exponentials ...
Example:
PolarPlot[{
Log[t],
If[t > 3, Log[ t - 2], 0],
If[t > 5, Log[ t - 4], 0]
}, {t, 1, 10}]
Output:

Not sure this is what you want, but you can reverse the log function (or "any" other for that matter).
Say you have ln A = B, to get A from B you do e^B = A.
So you get your point and pass it as B, you'll get A. Then you just need to check if that A (with a certain +- range) is in the values you first passed on to ln to generate the spiral.
I think this might work...

Unfortunately, you will need to know some mathematics notation anyway - this is a good read about the logarithmic sprial.
http://en.wikipedia.org/wiki/Logarithmic_spiral
we will only need the top 4 equations.
For your question 1
- to control the tightness, you tune the parameter 'a' as in the wiki page.
- to control the direction, you offset theta by a certain amount.
For your question 2
In floating point arithmetic, you will never get absolute precision, which mean there will be no point falling exactly on the sprial. On the screen, however, you will know which pixel get rendered, and you can test whether you are hitting a point that is rendered.
To render a curve, you usually render it as a sequence of line segments, short enough so that overall it looks like a curve. If you want to know whether a point lies within certain distance from the spiral, you can render the curve (on a off-screen buffer if you wish) by having thicker lines.

here a C++ code drawing any spiral passing where the mouse here
(sorry for my English)
int cx = pWin->vue.right / 2;
int cy = pWin->vue.bottom / 2;
double theta_mouse = atan2((double)(pWin->y_mouse - cy),(double)(pWin->x_mouse - cx));
double square_d_mouse = (double)(pWin->y_mouse - cy)*(double)(pWin->y_mouse - cy)+
(double)(pWin->x_mouse - cx)*(double)(pWin->x_mouse - cx);
double d_mouse = sqrt(square_d_mouse);
double theta_t = log( d_mouse / 3.0 ) / log( 1.19 );
int x = cx + (3 * cos(theta_mouse));
int y = cy + (3 * sin(theta_mouse));
MoveToEx(hdc,x,y,NULL);
for(double theta=0.0;theta < PI2*5.0;theta+=0.1)
{
double d = pow( 1.19 , theta ) * 3.0;
x = cx + (d * cos(theta-theta_t+theta_mouse));
y = cy + (d * sin(theta-theta_t+theta_mouse));
LineTo(hdc,x,y);
}
Ok now the parameter of spiral is 1.19 (slope) and 3.0 (radius at center)
Just compare the points where theta is a mutiple of 2 PI = PI2 = 6,283185307179586476925286766559
if any points is near of a non rotated spiral like
x = cx + (d * cos(theta));
y = cy + (d * sin(theta));
then your mouse is ON the spiral... I searched this tonight and i googled your past question