* aliasing <ref>[https://www.overclockersclub.com/reviews/serious_statistics_aliasing/ serious_statistics_aliasing by Guest_Jim]</ref>
* aliasing <ref>[https://www.overclockersclub.com/reviews/serious_statistics_aliasing/ serious_statistics_aliasing by Guest_Jim]</ref>
* changing algorithm ( representation function) from discrete to continous, like from level set method of escape time to continous ( DEM )
* changing algorithm <ref>[https://www.scaler.com/topics/c/if-else-statement-in-c/]</ref> ( 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<ref>[https://fractalforums.org/fractal-mathematics-and-new-theories/28/newton-raphson-zooming/ fractalforums.org : newton-raphson-zooming]</ref>
"the denser the area, the more heavy the anti-aliasing have to be in order to make it look good." knighty<ref>[https://fractalforums.org/fractal-mathematics-and-new-theories/28/newton-raphson-zooming/ fractalforums.org : newton-raphson-zooming]</ref>
changing algorithm [25] ( 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[26]
"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[27]
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.*/doubleCheckSide(doubleZx,doubleZy,doubleAx,doubleAy,doubleBx,doubleBy){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 doubleZax=-0.250000000000000;doubleZay=0.433012701892219;// left when ydoubleZlx=-0.112538773749444;doubleZly=0.436719687479814;doubleZrx=-0.335875821657728;doubleZry=0.316782798339332;// points to test // = inside triangle doubleZx=-0.209881783739630;doubleZy=+0.4;// outside triangle doubleZxo=-0.193503885412548;doubleZyo=0.521747636163664;doubleZxo2=-0.338750000000000;doubleZyo2=+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.doublesign(doublex1,doubley1,doublex2,doubley2,doublex3,doubley3){return(x1-x3)*(y2-y3)-(x2-x3)*(y1-y3);}intPointInTriangle(doublex,doubley,doublex1,doubley1,doublex2,doubley2,doublex3,doubley3){doubleb1,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));}intDescribe_Position(doubleZx,doubleZy){if(PointInTriangle(Zx,Zy,Zax,Zay,Zlx,Zly,Zrx,Zry))printf(" Z is inside \n");elseprintf(" Z is outside \n");return0;}// ======================================intmain(void){Describe_Position(Zx,Zy);Describe_Position(Zxo,Zyo);Describe_Position(Zxo2,Zyo2);return0;}
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.htmdoubleGiveTriangleArea(doublexa,doubleya,doublexb,doubleyb,doublexc,doubleyc){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 determinantmatrix = { {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.*/doubleIsTriangleCounterclockwise(doublexa,doubleya,doublexb,doubleyb,doublexc,doubleyc){return((xb*yc+xa*yb+ya*xc)-(ya*xb+yb*xc+xa*yc));}intDescribeTriangle(doublexa,doubleya,doublexb,doubleyb,doublexc,doubleyc){doublet=IsTriangleCounterclockwise(xa,ya,xb,yb,xc,yc);doublea=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);return0;}intmain(){// clockwise oriented triangles DescribeTriangle(-94,0,92,68,400,180);// https://www-sop.inria.fr/prisme/fiches/Arithmetique/index.html.enDescribeTriangle(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 trianglesDescribeTriangle(-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 ??) = 6DescribeTriangle(4.0,1.0,8.0,3.0,0.0,9.0);// https://people.sc.fsu.edu/~jburkardt/datasets/triangles/tex1.txtDescribeTriangle(-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.txtDescribeTriangle(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 pointsreturn0;}
/*gcc p.c -Wall./a.out----------- git --------------------cd existing_foldergit initgit remote add origin git@gitlab.com:adammajewski/PointInPolygonTest_c.gitgit add .git commitgit push -u origin master*/#include<stdio.h>#define LENGTH 6/*Argument Meaningnvert 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.htmlPNPOLY - Point Inclusion in Polygon TestW. 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? andIs 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.*/intpnpoly(intnvert,double*vertx,double*verty,doubletestx,doubletesty){inti,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;}returnc;}voidCheckPoint(intnvert,double*vertx,double*verty,doubletestx,doubletesty){intflag;flag=pnpoly(nvert,vertx,verty,testx,testy);switch(flag){case0:printf("outside\n");break;case1:printf("inside\n");break;default:printf(" ??? \n");}}intmain(){// 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)doublezzx[LENGTH]={13.5,6.0,13.5,42.5,39.5,42.5};doublezzy[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]);return0;}
/* distance between 2 points z1 = x1 + y1*I z2 = x2 + y2*I en.wikipedia.org/wiki/Distance#Geometry*/doubleGiveDistance(intx1,inty1,intx2,inty2){returnsqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}/*mutually and externally tangent circlesmathworld.wolfram.com/TangentCircles.htmlTwo circles are mutually and externally tangent if distance between their centers is equal to the sum of their radii*/doubleTestTangency(intx1,inty1,intr1,intx2,inty2,intr2){doubledistance;distance=GiveDistance(x1,y1,x2,y2);return(distance-(r1+r2));// return should be zero}
