List:

# Postprocessing

• Two types of edge detection
• Pseudo-3D projections
• Star field generator
• Random dot stereograms (aka Magic Eye)
• Motion blur for animations
• Interlacing
• Embossing
• Antialiasing
• Palette emulation to allow color cycling on true-color displays
• True color emulation that provides dithering on 256-color display

# dense image

Dense image

• downscaling with gamma correction
• path finding
• aliasing 
• changing algorithm ( representation function) from discrete to continous, like from level set method of escape time to continous ( DEM )
  "the denser the area, the more heavy the anti-aliasing have to be in order to make it look good."  knighty
  "the details are smaller than pixel spacing, so all that remains is the bands of colour shift from period-doubling of features making it even denser"  claude

path finding

Algorithms

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

/*
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));

}


## Testing if point is inside triangle

/*
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;
}


## Orientation and area of the triangle

Orientation and area of the triangle : how to do it ?

// 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.

:$\mathbf{O} = \begin{bmatrix} 1 & x_{A} & y_{A} \\ 1 & x_{B} & y_{B} \\ 1 & x_{C} & y_{C}\end{bmatrix}.$

A formula for its determinant may be obtained, e.g., using the method of [[cofactor expansion]]:
:\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}

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;
}


## Testing if point is inside polygon

• stackoverflow
• description by W Muła
• point_in_polygon by Patrick Glauner
/*

gcc p.c -Wall
./a.out

----------- git --------------------
cd existing_folder
git init
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-0.001, zzy);
CheckPoint(LENGTH, zzx, zzy, zzx+0.001, zzy);

return 0;
}


# test external tangency of 2 circles

/*
distance between 2 points
z1 = x1 + y1*I
z2 = x2 + y2*I
en.wikipedia.org/wiki/Distance#Geometry

*/

double GiveDistance(int x1, int y1, int x2, int y2){
return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}

/*
mutually and externally tangent circles
mathworld.wolfram.com/TangentCircles.html
Two circles are mutually and externally tangent if distance between their centers is equal to the sum of their radii

*/

double TestTangency(int x1, int y1, int r1, int x2, int y2, int r2){
double distance;

distance = GiveDistance(x1, y1, x2, y2);
return ( distance - (r1+r2));
// return should be zero
}