Reputation: 94
I have 'n' number of polygons as like below.
<Polygon Points="544,245,544,175,568,175,568,175" Stroke="Black" StrokeThickness="1" />
<Polygon Points="2,223,96,223,96,153,96,153" Stroke="Black" StrokeThickness="1" />
<Polygon Points="350,315,350,333,306,333,306,395,306,395" Stroke="Black" StrokeThickness="1" />
<Polygon Points="164,53,160,53,160,51,160,55,160,55" Stroke="Black" StrokeThickness="1" />
<Polygon Points="264,63,264,58,264,68,264,63,267,63,267,60,267,66,267,63,270,63,270,58,270,68,270,68" Stroke="Black" StrokeThickness="1" />
<Polygon Points="8,63,444,63,444,168,444,168" Stroke="Black" StrokeThickness="1" />
<Polygon Points="212,169,212,93,285,93,285,63,285,63" Stroke="Black" StrokeThickness="1" />
<Polygon Points="26,93,127,93,127,148,29,148,29,148" Stroke="Black" StrokeThickness="1" />
<Polygon Points="152,116,152,132,212,132,212,132" Stroke="Black" StrokeThickness="1" />
<Polygon Points="121,316,121,333,70,333,70,366,70,366" Stroke="Black" StrokeThickness="1" />
<Polygon Points="464,395,488,395,488,284,527,284,527,284" Stroke="Black" StrokeThickness="1" />
<Polygon Points="168,63,168,67,180,59,180,67,168,59,168,59" Stroke="Black" StrokeThickness="1" />
<Polygon Points="173,62,173,56,165,56,165,51,175,51,175,61,175,61" Stroke="Black" StrokeThickness="1" />
<Polygon Points="3,285,121,285,121,316,211,316,211,304,211,304" Stroke="Black" StrokeThickness="1" />
Please help me to identify what are the triangles in these polygons
I have tried to identify the vertices like below..
Polygon polygon = new Polygon();
polygon.Points = new System.Windows.Media.PointCollection()
{
new Point(446,134),
new Point(442,134),
new Point(444,140),
new Point(444,140),
};
List<double> verticesPoints = new List<double>();
for (int i = 0; i < polygon.Points.Count - 1; i++)
{
var point1 = polygon.Points[i];
var point2 = polygon.Points[i + 1];
//calculate delta x and delta y between the two points
var deltaX = Math.Pow((point2.X - point1.X), 2);
var deltaY = Math.Pow((point2.Y - point1.Y), 2);
//pythagras theorem for distance
var distance = Math.Sqrt(deltaY + deltaX);
//distance is zero..then same point
if (distance != 0)
{
verticesPoints.Add(distance);
}
}
///Here is the code to calculate angle and consider the triangle
///three vertices then it might be triangle.
if (verticesPoints.Count == 3)
{
///use The Law of Cosines
///cos(C) = a2 + b2 − c2 /2ab
///cos(A) = b2 + c2 − a2 /bc
///cos(B) = c2 + a2 − b2 /ca
var a = ((Math.Pow(verticesPoints[1], 2)) + (Math.Pow(verticesPoints[2], 2)) - (Math.Pow(verticesPoints[0], 2)))
/ (2 * verticesPoints[1] * verticesPoints[2]);
var b = ((Math.Pow(verticesPoints[0], 2)) + (Math.Pow(verticesPoints[2], 2)) - (Math.Pow(verticesPoints[1], 2)))
/ (2 * verticesPoints[0] * verticesPoints[2]);
var c = ((Math.Pow(verticesPoints[0], 2)) + (Math.Pow(verticesPoints[1], 2)) - (Math.Pow(verticesPoints[2], 2)))
/ (2 * verticesPoints[0] * verticesPoints[1]);
///Inverse of cos
var radians1 = Math.Acos(a);
///Convert radian to degree
double degrees1 = (radians1 * 180.0) / Math.PI;
///Inverse of cos
var radians2 = Math.Acos(b);
//Convert radian to degree
double degrees2 = (radians2 * 180.0) / Math.PI;
///Inverse of cos
var radians3 = Math.Acos(c);
///Convert radian to degree
double degrees3 = (radians3 * 180.0) / Math.PI;
var totalDegrees = degrees1 + degrees2 + degrees3;
if (totalDegrees == 180)
{
// Consider triangle
}
}
But above code is not working for <Polygon Points="446,134,442,134,444,140,444,140" Stroke="Black" StrokeThickness="1" />
it is giving only two vertices but it is a triangle and some scenario getting 3 vertices but totalDegrees is not as 180
Upvotes: 2
Views: 389
Reputation: 94
I have solved the above problem using "AForge.NET"
Polygon polygon = new Polygon();
polygon.Points = new PointCollection()
{
new Point(446,134),
new Point(442,134),
new Point(444,140),
new Point(444,140),
};
SimpleShapeChecker shapeChecker = new SimpleShapeChecker();
List<IntPoint> edgePoints = new List<IntPoint>();
List<IntPoint> corners;
for (int i = 0; i <= polygon.Points.Count - 1; i++)
{
edgePoints.Add(new IntPoint((int)polygon.Points[i].X, (int)polygon.Points[i].Y));
}
shapeChecker.MinAcceptableDistortion = 0.2f;
shapeChecker.LengthError = 0;
shapeChecker.AngleError = 5;
shapeChecker.RelativeDistortionLimit = 0;
if (shapeChecker.IsTriangle(edgePoints, out corners))
{
//shape is triangle
}
Need to add below namespace
using AForge;
using AForge.Math.Geometry;
Reference :http://aforgenet.com/articles/shape_checker/
Upvotes: 0
Reputation: 2013
This code here iterates through the points and calculates the gradient between each of them. If the gradient is the same for two sequential points they must be the same line, so the noOfPoints
is not incremented otherwise it is incremented.
The first gradient is stored in firstGradient
in order to check if the gradient connecting the last and first point is the same as the gradient between the first and second point.
Polygon polygon = new Polygon();
polygon.Points = new System.Windows.Media.PointCollection()
{
new Point(446,134),
new Point(442,134),
new Point(444,140),
new Point(444,140),
};
List<double> verticesPoints = new List<double>();
double? firstGradient = null;
double? gradient = null;
double? newGradient = null;
int noOfSides = 1;
for (int i = 0; i < polygon.Points.Count - 1; i++)
{
var point1 = polygon.Points[i];
var point2 = polygon.Points[i + 1];
if(point1 == point2) { continue;}
//calculate delta x and delta y between the two points
var deltaX = point2.X - point1.X;
var deltaY = point2.Y - point1.Y;
//calculate gradient
newGradient = (deltaY / deltaX);
if (i == 0)
{
firstGradient = newGradient;
}
if ((gradient != newGradient) && (i != polygon.Points.Count - 2))
{
noOfSides++;
}
else if (i == polygon.Points.Count - 2)
{
if ((gradient != newGradient) && (firstGradient != newGradient)) //This now checks the gradient between the last and first point.
{
point1 = polygon.Points[i+1];
point2 = polygon.Points[0];
if (point1 == point2) { continue; }
//calculate delta x and delta y between the two points
deltaX = point2.X - point1.X;
deltaY = point2.Y - point1.Y;
//calculate gradient
newGradient = (deltaY / deltaX);
if(newGradient != firstGradient)
{
noOfSides++;
}
}
gradient = newGradient;
}
Upvotes: 1