/* 
  How to tell whether a point is to the right or left side of a line ?

 http://stackoverflow.com/questions/1560492/how-to-tell-whether-a-point-is-to-the-right-or-left-side-of-a-line

  a, b = points
  line = ab
 pont to check = z

  position = sign((Bx - Ax) * (Y - Ay) - (By - Ay) * (X - Ax))
  It is 0 on the line, and +1 on one side, -1 on the other side.

*/

double CheckSide(double Zx, double Zy, double Ax, double Ay, double Bx, double By)
{
  return ((Bx - Ax) * (Zy - Ay) - (By - Ay) * (Zx - Ax));

}

/* 
c console program 
gcc t.c -Wall
./a.out

*/

# include <stdio.h>

// 3 points define triangle 
double Zax = -0.250000000000000;
double Zay = 0.433012701892219;
// left when y
double Zlx = -0.112538773749444;  
double Zly = 0.436719687479814 ;

double Zrx = -0.335875821657728;
double Zry = 0.316782798339332;

// points to test 
// = inside triangle 
double Zx = -0.209881783739630;
double Zy =   +0.4; 

// outside triangle 
double Zxo = -0.193503885412548  ;
double Zyo = 0.521747636163664;

double Zxo2 = -0.338750000000000;
double Zyo2 = +0.440690927838329; 





// ============ http://stackoverflow.com/questions/2049582/how-to-determine-a-point-in-a-2d-triangle
// In general, the simplest (and quite optimal) algorithm is checking on which side of the half-plane created by the edges the point is.
double sign (double  x1, double y1,  double x2, double y2, double x3, double y3)
{
    return (x1 - x3) * (y2 - y3) - (x2 - x3) * (y1 - y3);
}

int  PointInTriangle (double x, double y, double x1, double y1, double x2, double y2, double x3, double y3)
{
    double  b1, b2, b3;

    b1 = sign(x, y, x1, y1, x2, y2) < 0.0;
    b2 = sign(x, y, x2, y2, x3, y3) < 0.0;
    b3 = sign(x, y, x3, y3, x1, y1) < 0.0;

    return ((b1 == b2) && (b2 == b3));
}

int Describe_Position(double Zx, double Zy){
if (PointInTriangle( Zx, Zy, Zax, Zay, Zlx, Zly, Zrx, Zry))
  printf(" Z is inside \n");
  else printf(" Z is outside \n");

return 0;
}



// ======================================



int main(void){

Describe_Position(Zx, Zy);
Describe_Position(Zxo, Zyo);
Describe_Position(Zxo2, Zyo2);

return 0;
}

// gcc t.c -Wall
// ./a.out
# include <stdio.h>



// http://ncalculators.com/geometry/triangle-area-by-3-points.htm
double GiveTriangleArea(double xa, double ya, double xb, double yb, double xc, double yc)
{
return ((xb*ya-xa*yb)+(xc*yb-xb*yc)+(xa*yc-xc*ya))/2.0;
}


/*

wiki Curve_orientation
[http://mathoverflow.net/questions/44096/detecting-whether-directed-cycle-is-clockwise-or-counterclockwise]


The orientation of a triangle (clockwise/counterclockwise) is the sign of the determinant



matrix = { {1 , x1, y1}, {1 ,x2, y2} , {1,  x3, y3}}
 


where 
(x_1,y_1), (x_2,y_2), (x_3,y_3)$ 
are the Cartesian coordinates of the three vertices of the triangle.

:<math>\mathbf{O} = \begin{bmatrix}

1 & x_{A} & y_{A} \\
1 & x_{B} & y_{B} \\
1 & x_{C} & y_{C}\end{bmatrix}.</math>

A formula for its determinant may be obtained, e.g., using the method of [[cofactor expansion]]:
:<math>\begin{align}
\det(O) &= 1\begin{vmatrix}x_{B}&y_{B}\\x_{C}&y_{C}\end{vmatrix}
-x_{A}\begin{vmatrix}1&y_{B}\\1&y_{C}\end{vmatrix}
+y_{A}\begin{vmatrix}1&x_{B}\\1&x_{C}\end{vmatrix} \\
&= x_{B}y_{C}-y_{B}x_{C}-x_{A}y_{C}+x_{A}y_{B}+y_{A}x_{C}-y_{A}x_{B} \\
&= (x_{B}y_{C}+x_{A}y_{B}+y_{A}x_{C})-(y_{A}x_{B}+y_{B}x_{C}+x_{A}y_{C}).
\end{align}
</math>

If the determinant is negative, then the polygon is oriented clockwise.  If the determinant is positive, the polygon is oriented counterclockwise.  The determinant  is non-zero if points A, B, and C are non-[[collinear]].  In the above example, with points ordered A, B, C, etc., the determinant is negative, and therefore the polygon is clockwise.

*/

double IsTriangleCounterclockwise(double xa, double ya, double xb, double yb, double xc, double yc)
{return  ((xb*yc + xa*yb +ya*xc) - (ya*xb +yb*xc + xa*yc)); }



int DescribeTriangle(double xa, double ya, double xb, double yb, double xc, double yc)
{
 double t = IsTriangleCounterclockwise( xa,  ya, xb,  yb,  xc,  yc);
 double a = GiveTriangleArea( xa,  ya, xb,  yb,  xc,  yc);
 if (t>0)  printf("this triangle is oriented counterclockwise,     determinent = %f ; area = %f\n", t,a);
 if (t<0)  printf("this triangle is oriented clockwise,            determinent = %f; area = %f\n", t,a);
 if (t==0) printf("this triangle is degenerate: colinear or identical points, determinent = %f; area = %f\n", t,a);

 return 0;
}





int main()
{
 // clockwise oriented triangles 
 DescribeTriangle(-94,   0,  92,  68, 400, 180); // https://www-sop.inria.fr/prisme/fiches/Arithmetique/index.html.en
 DescribeTriangle(4.0, 1.0, 0.0, 9.0, 8.0, 3.0); // clockwise orientation https://people.sc.fsu.edu/~jburkardt/datasets/triangles/tex5.txt
 
 //  counterclockwise oriented triangles
 DescribeTriangle(-50.00, 0.00, 50.00,  0.00, 0.00,  0.02); // a "cap" triangle. This example has an area of 1.
 DescribeTriangle(0.0,  0.0, 3.0,  0.0, 0.0,  4.0); // a right triangle. This example has an area of (?? 3 ??)  =  6
 DescribeTriangle(4.0, 1.0, 8.0, 3.0, 0.0, 9.0);  //      https://people.sc.fsu.edu/~jburkardt/datasets/triangles/tex1.txt
 DescribeTriangle(-0.5, 0.0,  0.5,  0.0, 0.0,  0.866025403784439); // an equilateral triangle. This triangle has an area of sqrt(3)/4.

 // degenerate triangles 
 DescribeTriangle(1.0, 0.0, 2.0, 2.0, 3.0, 4.0); // This triangle is degenerate: 3 colinear points. https://people.sc.fsu.edu/~jburkardt/datasets/triangles/tex6.txt
 DescribeTriangle(4.0, 1.0, 0.0, 9.0, 4.0, 1.0); //2 identical points 
 DescribeTriangle(2.0, 3.0, 2.0, 3.0, 2.0, 3.0); // 3 identical points

 return 0; 
}

/*

gcc p.c -Wall
./a.out

----------- git --------------------
cd existing_folder
git init
git remote add origin git@gitlab.com:adammajewski/PointInPolygonTest_c.git
git add .
git commit
git push -u origin master


*/


#include <stdio.h>

#define LENGTH 6



/*

Argument	Meaning
nvert	Number of vertices in the polygon. Whether to repeat the first vertex at the end is discussed below.
vertx, verty	Arrays containing the x- and y-coordinates of the polygon's vertices.
testx, testy	X- and y-coordinate of the test point.

https://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html
PNPOLY - Point Inclusion in Polygon Test
W. Randolph Franklin (WRF)

	

I run a semi-infinite ray horizontally (increasing x, fixed y) out from the test point, 
and count how many edges it crosses. 
At each crossing, the ray switches between inside and outside. 
This is called the Jordan curve theorem.
The case of the ray going thru a vertex is handled correctly via a careful selection of inequalities. 
Don't mess with this code unless you're familiar with the idea of Simulation of Simplicity. 
This pretends to shift the ray infinitesimally down so that it either clearly intersects, or clearly doesn't touch. 
Since this is merely a conceptual, infinitesimal, shift, it never creates an intersection that didn't exist before, 
and never destroys an intersection that clearly existed before.

The ray is tested against each edge thus:

Is the point in the half-plane to the left of the extended edge? and
Is the point's Y coordinate within the edge's Y-range?
Handling endpoints here is tricky.


I run a semi-infinite ray horizontally (increasing x, fixed y) out from the test point, 
and count how many edges it crosses. At each crossing, 
the ray switches between inside and outside. This is called the Jordan curve theorem.
The variable c is switching from 0 to 1 and 1 to 0 each time the horizontal ray crosses any edge. 
So basically it's keeping track of whether the number of edges crossed are even or odd. 
0 means even and 1 means odd.

*/


int pnpoly(int nvert, double *vertx, double *verty, double testx, double testy)
{
  int i, j, c = 0;
  for (i = 0, j = nvert-1; i < nvert; j = i++) {
    if ( ((verty[i]>testy) != (verty[j]>testy)) &&
	 (testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
       c = !c;
  }

  return c;
}


void CheckPoint(int nvert, double *vertx, double *verty, double testx, double testy){

int flag;

flag =  pnpoly(nvert, vertx, verty, testx, testy);

 switch(flag){
   case 0  : printf("outside\n"); break;
   case 1  : printf("inside\n"); break;
   default : printf(" ??? \n");
 }
}


int main (){



// values from http://stackoverflow.com/questions/217578/how-can-i-determine-whether-a-2d-point-is-within-a-polygon
// number from 0 to (LENGTH-1)
double zzx[LENGTH] = { 13.5,  6.0, 13.5, 42.5, 39.5, 42.5};
double zzy[LENGTH] = {100.0, 70.5, 41.5, 56.5, 69.5, 84.5};
       

CheckPoint(LENGTH, zzx, zzy, zzx[4]-0.001, zzy[4]);
CheckPoint(LENGTH, zzx, zzy, zzx[4]+0.001, zzy[4]);

return 0;
}