I am trying to draw curves a specified distance away from a curve. Kind of like the sides of roads are always a set distance away from the yellow lines in the center. I have code to do this, however, when done using my code it ends up with overlap in one of the curves.
Here is my code:
for (int i = 0; i < curves.Count; i++)
{
for (float j = 0; j < 1; j += .001f)
{
Vector2 temp = Vector2.CatmullRom(curves[i].Points[0], curves[i].Points[1], curves[i].Points[2], curves[i].Points[3], j);
if (mid.Count != 0)
{
if (temp != mid[mid.Count - 1])
{
mid.Add(temp);
}
else Console.WriteLine(temp == mid[mid.Count - 1]);
}
else mid.Add(temp);
}
}
for (int i = 0; i < mid.Count; i++)
{
if (i == 0)
{
Vector2 slope = mid[i] - mid[i + 1];
double x = mid[i].X + (12 * Math.Cos(Math.Atan(-(slope.X / slope.Y)))),
y = mid[i].Y + (12 * Math.Sin(Math.Atan(-(slope.X / slope.Y))));
side1.Add(new Vector2((float)x, (float)y));
x = mid[i].X - (12 * Math.Cos(Math.Atan(-(slope.X / slope.Y))));
y = mid[i].Y - (12 * Math.Sin(Math.Atan(-(slope.X / slope.Y))));
side2.Add(new Vector2((float)x, (float)y));
}
else if (i < mid.Count - 1)
{
Vector2 slope1 = mid[i] - mid[i - 1],
slope2 = mid[i + 1] - mid[i];
Vector2 slope = (slope1 + slope2) / 2;
double x = mid[i].X + (12 * Math.Cos(Math.Atan(-(slope.X / slope.Y)))),
y = mid[i].Y + (12 * Math.Sin(Math.Atan(-(slope.X / slope.Y))));
side1.Add(new Vector2((float)x, (float)y));
x = mid[i].X - (12 * Math.Cos(Math.Atan(-(slope.X / slope.Y))));
y = mid[i].Y - (12 * Math.Sin(Math.Atan(-(slope.X / slope.Y))));
side2.Add(new Vector2((float)x, (float)y));
}
else
{
Vector2 slope = mid[i] - mid[i - 1];
double x = mid[i].X + (12 * Math.Cos(Math.Atan(-(slope.X / slope.Y)))),
y = mid[i].Y + (12 * Math.Sin(Math.Atan(-(slope.X / slope.Y))));
side1.Add(new Vector2((float)x, (float)y));
x = mid[i].X - (12 * Math.Cos(Math.Atan(-(slope.X / slope.Y))));
y = mid[i].Y - (12 * Math.Sin(Math.Atan(-(slope.X / slope.Y))));
side2.Add(new Vector2((float)x, (float)y));
}
}
And here is a photo of the result:
Curves with overlap
Any and all help would be appreciated.
Related
So I'm running iterations with this formula:
double x = 10 / 0.25 * ((0.0002 * x1 * (10 - 0.25 * x1)) + 0.00217 * x2 * (20 - 0.25 * x2)); With this process: Xn+1 = f(Xn).
And if you start from negative X you will eventually end up with (-/+) infinity, so after 6 iterations I'm supposed to get infinity, but what I got surprised me and I couldn't find anywhere what that is, I got "-?", I've tried comparing it to +/- infinity and tried to compare it to int numbers just to clarify what it is, but I cant get anything out of it, for example, I've tried if ("-?" > 1000) break;, and it doesn't outcome as "true". Neither am I getting any errors by comparing it to int/double, I need to stop iterations when I start going into infinity, how can I do that?
code:
public static double CalculateX1(double x1, double x2)
{
double x = 10 / 0.25 * ((0.0002 * x1 * (10 - 0.25 * x1)) + 0.00217 * x2 * (20 - 0.25 * x2));
return x;
}
public static double CalculateX2(double x2, double x1)
{
double y = 20 / 0.25 * ((0.00052 * x2 * (20 - 0.25 * x2)) + 0.0075 * x1 * (10 - 0.25 * x1));
return y;
}
static void Main(string[] args)
{
string writePath = #"C:\Users\evluc\Desktop\cord.txt";
double X = -5;
double Y = -5;
int pointer = 1;
double[,] coordinates = new double[10001, 2];
coordinates[0, 0] = X;
coordinates[0, 1] = Y;
for (int i = 0; i < 5000; i++)
{
//double XTemp = CalculateX1(X, Y);
//double YTemp = CalculateX2(Y, X);
//X = CalculateX1(coordinates[pointer - 1, 0], coordinates[pointer - 1, 1]);
//Y = CalculateX2(coordinates[pointer - 1, 1], coordinates[pointer - 1, 0]);
coordinates[pointer, 0] = CalculateX1(coordinates[pointer - 1, 0], coordinates[pointer - 1, 1]);
coordinates[pointer, 1] = CalculateX2(coordinates[pointer - 1, 1], coordinates[pointer - 1, 0]);
pointer++;
if (Math.Abs(coordinates[pointer, 0]) > 1000 || Math.Abs(coordinates[pointer, 1]) > 1000)
{
Console.WriteLine("infinity");
Console.ReadKey();
}
}
for (int i = 0; i < 5000; i++)
{
Console.WriteLine("X = " + coordinates[i, 0] + "," + "Y = " + coordinates[i, 1] + "; ");
}
}
I think whatever you use to display/inspect the value cannot print ∞.
double d = double.MinValue;
d *= 2;
Console.WriteLine($"{d}: IsInfinity: {double.IsNegativeInfinity(d)}");
-∞: IsInfinity: True
Stopping at infinity
Here's a loop that stops at infinity:
double d = 2;
var i = 1;
while(!double.IsInfinity(d))
{
d = i*d*d;
i = -i;
}
Console.WriteLine(d);
-∞
I am in need of a C# solution to some code i have in Javascript, which takes a array of latitude and longitude positions which forms an area or zone. and also an object X and Y position and returns the X and Y of the closest point from the object. This code works perfect in Javascript, however as i am now re-writing my work in C#, i am unable to find a workable solution which works, and does what this function does.
The Javascript function is below, which takes 2 arguments pXy, which is the X and Y position of your location and aXys, which is an array of X and Y positions forming the polylines of an area.
var getClosestPointOnLines = function(pXy, aXys) {
var minDist;
var fTo;
var fFrom;
var x;
var y;
var i;
var dist;
if (aXys.length > 1) {
for (var n = 1 ; n < aXys.length ; n++) {
if (aXys[n].x != aXys[n - 1].x) {
var a = (aXys[n].y - aXys[n - 1].y) / (aXys[n].x - aXys[n - 1].x);
var b = aXys[n].y - a * aXys[n].x;
dist = Math.abs(a * pXy.x + b - pXy.y) / Math.sqrt(a * a + 1);
}
else
dist = Math.abs(pXy.x - aXys[n].x)
// length^2 of line segment
var rl2 = Math.pow(aXys[n].y - aXys[n - 1].y, 2) + Math.pow(aXys[n].x - aXys[n - 1].x, 2);
// distance^2 of pt to end line segment
var ln2 = Math.pow(aXys[n].y - pXy.y, 2) + Math.pow(aXys[n].x - pXy.x, 2);
// distance^2 of pt to begin line segment
var lnm12 = Math.pow(aXys[n - 1].y - pXy.y, 2) + Math.pow(aXys[n - 1].x - pXy.x, 2);
// minimum distance^2 of pt to infinite line
var dist2 = Math.pow(dist, 2);
// calculated length^2 of line segment
var calcrl2 = ln2 - dist2 + lnm12 - dist2;
// redefine minimum distance to line segment (not infinite line) if necessary
if (calcrl2 > rl2)
dist = Math.sqrt(Math.min(ln2, lnm12));
if ((minDist == null) || (minDist > dist)) {
if (calcrl2 > rl2) {
if (lnm12 < ln2) {
fTo = 0;//nearer to previous point
fFrom = 1;
}
else {
fFrom = 0;//nearer to current point
fTo = 1;
}
}
else {
// perpendicular from point intersects line segment
fTo = ((Math.sqrt(lnm12 - dist2)) / Math.sqrt(rl2));
fFrom = ((Math.sqrt(ln2 - dist2)) / Math.sqrt(rl2));
}
minDist = dist;
i = n;
}
}
var dx = aXys[i - 1].x - aXys[i].x;
var dy = aXys[i - 1].y - aXys[i].y;
x = aXys[i - 1].x - (dx * fTo);
y = aXys[i - 1].y - (dy * fTo);
}
return { 'x': x, 'y': y, 'i': i, 'fTo': fTo, 'fFrom': fFrom };
}
How can i replicate the above in C#?
Thanks xdtTransform. I was not using Visual Studio so did not realize the Ctrl+Space. After installing it, i have managed to convert the code myself. I have included it here for anyone who will benefit from it in the future.
I started with a class to hold the Latitude and Longitude details of the points.
namespace Classes
{
public class AppGeoPoint
{
public double X { get; set; }
public double Y { get; set; }
}
}
And the function converted
public static Classes.AppGeoPoint getClosestPointOnLines(Classes.AppGeoPoint pXy, Classes.AppGeoPoint[] aXys)
{
double? minDist = null;
double fTo = 0.0;
double fFrom;
double x = 0.0;
double y = 0.0;
int i = 0;
double dist;
if (aXys.Length > 1)
{
for (var n = 1; n < aXys.Length; n++)
{
if (aXys[n].X != aXys[n - 1].X)
{
var a = (aXys[n].Y - aXys[n - 1].Y) / (aXys[n].X - aXys[n - 1].X);
var b = aXys[n].Y - a * aXys[n].X;
dist = Math.Abs(a * pXy.X + b - pXy.Y) / Math.Sqrt(a * a + 1);
}
else
dist = Math.Abs(pXy.X - aXys[n].X);
// length^2 of line segment
double rl2 = Math.Pow(aXys[n].Y - aXys[n - 1].Y, 2) + Math.Pow(aXys[n].X - aXys[n - 1].X, 2);
// distance^2 of pt to end line segment
double ln2 = Math.Pow(aXys[n].Y - pXy.Y, 2) + Math.Pow(aXys[n].X - pXy.X, 2);
// distance^2 of pt to begin line segment
double lnm12 = Math.Pow(aXys[n - 1].Y - pXy.Y, 2) + Math.Pow(aXys[n - 1].X - pXy.X, 2);
// minimum distance^2 of pt to infinite line
double dist2 = Math.Pow(dist, 2);
// calculated length^2 of line segment
double calcrl2 = ln2 - dist2 + lnm12 - dist2;
// redefine minimum distance to line segment (not infinite line) if necessary
if (calcrl2 > rl2)
dist = Math.Sqrt(Math.Min(ln2, lnm12));
if ((minDist == null) || (minDist > dist))
{
if (calcrl2 > rl2)
{
if (lnm12 < ln2)
{
fTo = 0;//nearer to previous point
fFrom = 1;
}
else
{
fFrom = 0;//nearer to current point
fTo = 1;
}
}
else
{
// perpendicular from point intersects line segment
fTo = ((Math.Sqrt(lnm12 - dist2)) / Math.Sqrt(rl2));
fFrom = ((Math.Sqrt(ln2 - dist2)) / Math.Sqrt(rl2));
}
minDist = dist;
i = n;
}
}
var dx = aXys[i - 1].X - aXys[i].X;
var dy = aXys[i - 1].Y - aXys[i].Y;
x = aXys[i - 1].X - (dx * fTo);
y = aXys[i - 1].Y - (dy * fTo);
}
return new Classes.AppGeoPoint { X = x, Y = y };
}
Then finally, to build an array list and check for the closest point
Classes.AppGeoPoint YourLocation= new Classes.AppGeoPoint { X = 50.83737, Y = -1.07428 };
Classes.AppGeoPoint[] AreaCheck = new[] {
new Classes.AppGeoPoint { X = 50.847550000000005, Y = -1.0863200000000002 },
new Classes.AppGeoPoint { X = 50.83975, Y = -1.0859800000000002 },
new Classes.AppGeoPoint { X = 50.83845, Y = -1.06487 },
new Classes.AppGeoPoint { X = 50.84723, Y = -1.0645200000000001 }
};
Classes.AppGeoPoint ReturnVal = getClosestPointOnLines(YourLocation, AreaCheck);
Console.WriteLine("X " + ReturnVal.X);
Console.WriteLine("Y " + ReturnVal.Y);
ReturnVal.X and ReturnVal.Y will return the closest points latitude and longitude.
Hopefully, this may come in helpful to others.
I purchased the book Numerical Methods, Algorithms and Tools in C# by Waldemar Dos Passos.
On page 463, there is the method:
public static double FactorialLn(int n)
{
if(n < 0)
{
throw new Exception("Input value must be > 0");
}
else
{
return GammaLn(n+1.0);
}
}
I found the namespace where the function GammaLn resides: MicrosoftResearch.Infer.Maths
but Visual Studio 2010 does not recognize it and I was unable to find it in the .NET reference.
Please help me get the program to compile.
Thank you in advance.
I finally found the dource code of the GammaLn method itself here: http://seungwon.tistory.com/9
One does need nothing else when one possesses the very real thing, as Andrey correctly pointed out, but until then ...
I reproduce the method here for future reference, in case the site above stops from functioning.
/// <summary>
/// http://seungwon.tistory.com/9
/// GammaLn函数
/// </summary>
/// <param name="x"></param>
/// <returns></returns>
public static double GammaLn(double x)
{
double result = 0.0;
double d1 = -5.772156649015328605195174e-1;
double[,] p1 = {{4.945235359296727046734888e0},{ 2.018112620856775083915565e2},
{2.290838373831346393026739e3},{ 1.131967205903380828685045e4},
{2.855724635671635335736389e4},{ 3.848496228443793359990269e4},
{2.637748787624195437963534e4},{ 7.225813979700288197698961e3}};
double[,] q1 = {{6.748212550303777196073036e1},{ 1.113332393857199323513008e3},
{7.738757056935398733233834e3},{ 2.763987074403340708898585e4},
{5.499310206226157329794414e4},{ 6.161122180066002127833352e4},
{3.635127591501940507276287e4},{ 8.785536302431013170870835e3}};
double d2 = 4.227843350984671393993777e-1;
double[,] p2 = {{4.974607845568932035012064e0},{ 5.424138599891070494101986e2},
{1.550693864978364947665077e4},{ 1.847932904445632425417223e5},
{1.088204769468828767498470e6},{ 3.338152967987029735917223e6},
{5.106661678927352456275255e6},{ 3.074109054850539556250927e6}};
double[,] q2 = {{1.830328399370592604055942e2},{ 7.765049321445005871323047e3},
{1.331903827966074194402448e5},{ 1.136705821321969608938755e6},
{5.267964117437946917577538e6},{ 1.346701454311101692290052e7},
{1.782736530353274213975932e7},{ 9.533095591844353613395747e6}};
double d4 = 1.791759469228055000094023e0;
double[,] p4 = {{1.474502166059939948905062e4},{ 2.426813369486704502836312e6},
{1.214755574045093227939592e8},{ 2.663432449630976949898078e9},
{2.940378956634553899906876e10},{ 1.702665737765398868392998e11},
{4.926125793377430887588120e11},{5.606251856223951465078242e11}};
double[,] q4 = {{2.690530175870899333379843e3},{ 6.393885654300092398984238e5},
{4.135599930241388052042842e7},{ 1.120872109616147941376570e9},
{1.488613728678813811542398e10},{1.016803586272438228077304e11},
{3.417476345507377132798597e11},{ 4.463158187419713286462081e11}};
double[,] c = {{-1.910444077728e-03},{ 8.4171387781295e-04},
{-5.952379913043012e-04},{ 7.93650793500350248e-04},
{-2.777777777777681622553e-03},{ 8.333333333333333331554247e-02},
{ 5.7083835261e-03}};
double eps = 2.2204e-016;
if (x <= 0)
{
//报错！
}
else
{
double xden = 0.0;
double xnum = 0.0;
result = x;
if (x > 0 && x <= eps)
{
result = -Math.Log(x);
}
else if ((x > eps) && (x <= 0.5))
{
double y = x;
xden = 1;
xnum = 0;
for (int i = 0; i < 8; i++)
{
xnum = xnum * y + p1[i, 0];
xden = xden * y + q1[i, 0];
}
result = -Math.Log(y) + (y * (d1 + y * (xnum / xden)));
}
else if ((x > 0.5) && (x <= 0.6796875))
{
double xm1 = (x - 0.5) - 0.5;
xden = 1;
xnum = 0;
for (int i = 0; i < 8; i++)
{
xnum = xnum * xm1 + p2[i, 0];
xden = xden * xm1 + q2[i, 0];
}
result = -Math.Log(x) + xm1 * (d2 + xm1 * (xnum / xden));
}
else if ((x > 0.6796875) && (x <= 1.5))
{
double xm1 = (x - 0.5) - 0.5;
xden = 1;
xnum = 0;
for (int i = 0; i < 8; i++)
{
xnum = xnum * xm1 + p1[i, 0];
xden = xden * xm1 + q1[i, 0];
}
result = xm1 * (d1 + xm1 * (xnum / xden));
}
else if ((x > 1.5) && (x <= 4))
{
double xm2 = x - 2;
xden = 1;
xnum = 0;
for (int i = 0; i < 8; i++)
{
xnum = xnum * xm2 + p2[i, 0];
xden = xden * xm2 + q2[i, 0];
}
result = xm2 * (d2 + xm2 * (xnum / xden));
}
else if ((x > 4) && (x <= 12))
{
double xm4 = x - 4;
xden = -1;
xnum = 0;
for (int i = 0; i < 8; i++)
{
xnum = xnum * xm4 + p4[i, 0];
xden = xden * xm4 + q4[i, 0];
}
result = d4 + xm4 * (xnum / xden);
}
else if (x > 12)
{
double y = x;
double r = c[6, 0];// 等于：double r = repmat(c[6, 0], 1)[0,0];
double ysq = y * y;
for (int i = 0; i < 6; i++)
{
r = r / ysq + c[i, 0];
}
r = r / y;
double corr = Math.Log(y);
double spi = 0.9189385332046727417803297;
result = r + spi - 0.5 * corr + y * (corr - 1);
}
}
return result;
}
Can you guys help me converting this C# code to Objective-C?
I don't have a clue about C#/Visual Studio!
public static class BezierSpline
{
public static void GetCurveControlPoints(Point[] knots,
out Point[] firstControlPoints, out Point[] secondControlPoints)
{
int n = knots.Length - 1;
// Calculate first Bezier control points
// Right hand side vector
double[] rhs = new double[n];
// Set right hand side X values
for (int i = 1; i < n - 1; ++i)
rhs[i] = 4 * knots[i].X + 2 * knots[i + 1].X;
rhs[0] = knots[0].X + 2 * knots[1].X;
rhs[n - 1] = (8 * knots[n - 1].X + knots[n].X) / 2.0;
// Get first control points X-values
double[] x = GetFirstControlPoints(rhs);
// Set right hand side Y values
for (int i = 1; i < n - 1; ++i)
rhs[i] = 4 * knots[i].Y + 2 * knots[i + 1].Y;
rhs[0] = knots[0].Y + 2 * knots[1].Y;
rhs[n - 1] = (8 * knots[n - 1].Y + knots[n].Y) / 2.0;
// Get first control points Y-values
double[] y = GetFirstControlPoints(rhs);
// Fill output arrays.
firstControlPoints = new Point[n];
secondControlPoints = new Point[n];
for (int i = 0; i < n; ++i)
{
// First control point
firstControlPoints[i] = new Point(x[i], y[i]);
// Second control point
if (i < n - 1)
secondControlPoints[i] = new Point(2 * knots
[i + 1].X - x[i + 1], 2 *
knots[i + 1].Y - y[i + 1]);
else
secondControlPoints[i] = new Point((knots
[n].X + x[n - 1]) / 2,
(knots[n].Y + y[n - 1]) / 2);
}
}
private static double[] GetFirstControlPoints(double[] rhs)
{
int n = rhs.Length;
double[] x = new double[n]; // Solution vector.
double[] tmp = new double[n]; // Temp workspace.
double b = 2.0;
x[0] = rhs[0] / b;
for (int i = 1; i < n; i++) // Decomposition and forward substitution.
{
tmp[i] = 1 / b;
b = (i < n - 1 ? 4.0 : 3.5) - tmp[i];
x[i] = (rhs[i] - x[i - 1]) / b;
}
for (int i = 1; i < n; i++)
x[n - i - 1] -= tmp[n - i] * x[n - i]; // Backsubstitution.
return x;
}
}
thanks.
double[] tmp = new double[n];
tmp is an array of length n. Each value is not initialized explicitly, but it is implicitly set to the default value of the double type, which is 0. So tmp is an n length array of zeros. {0,0,0,0,0, ... 0}
How to draw triangle wave (symmetrical) using ZedGraph?
alt text http://img101.imageshack.us/img101/8482/okr20troj.jpg
Preferably with option to adjust period and amplitude.
//Edit: the function has to be [related/based on]? x (x-axis).
Something like this:
for (x = 0; x <= 10; x += .005)
{
if (Math.Sin(x * (2 * Math.PI / period)) >= 0)
y = amplitude;
else
y = -amplitude;
originalList.Add(x, y);
}
double amplitude = 1.7;
double period = 2;
PointPairList ppl = new PointPairList();
double y=0;
for (double x = 0; x <= 10; x += .005)
{
double p = (x % (period)) / period ;
if (p >= 0 && p <= 0.25)
y = 4 * p * amplitude;
if (p > 0.25 && p < 0.5)
y = amplitude - (p - 0.25) * 4 * amplitude;
if(p>0.5 && p<=0.75)
y = - 4 * (p-0.5) * amplitude;
if(p>0.75 && p<=1)
y = - (amplitude - (p - 0.75) * 4 * amplitude);
ppl.Add(x,y);
}
var line = zg1.MasterPane[0].AddCurve("", ppl, Color.Blue);
line.Symbol.IsVisible = false;
zg1.AxisChange();
zg1.Refresh();