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Visualization of Noble Prime Numbers |
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When we draw noble prime numbers in a circle, we can see that all of them have nine corners. The proof of this is as fallows:
Theory of Finding The Right Position of Each Point in Circle 
     K : distance between every number thru p.
midx, midy: pointing the middle of the monitor in pixels. diz[] : A structure. For every number it keeps "distance", x and y coordinates. To locate correct position, it should be: x = midx + x y = midy - y The Algorithm 
C Source Code #include <graphics.h>
#include <stdlib.h>
#include <stdio.h>
#include <conio.h>
#include <math.h>
#define pi 3.141592653
typedef struct coor {
double k;
double x;
double y;
};
int init_graph(void) {
int gdriver = DETECT, gmode, errorcode;
initgraph(&gdriver, &gmode, "");
errorcode = graphresult();
if (errorcode != grOk) {
printf("\n2. Graphical Error: %s\n", grapherrormsg(errorcode));
printf("Press any key to exit:");
getch();
return 1;
}
return 0;
}
void main(void) {
randomize();
if (init_graph() == 1) exit(1);
int i, midx, midy;
long k, t;
double cevre, K, r=239.0, p=499.0; // we choose noble prime number p here
coor *diz;
midx = getmaxx() / 2;
midy = getmaxy() / 2;
cevre = 2*pi*r; // = 1501 pixel
K = cevre / p; // p must be lower than 1501
diz = (coor *)calloc(p,sizeof(coor));
diz[0].k = 0.0;
diz[0].x = midx*1.0;
diz[0].y = 0.0;
for(i=1;i<=p;i++) {
diz[i].k = i*K;
diz[i].x = r*sin(diz[i].k/r);
diz[i].y = sqrt(r*r - diz[i].x*diz[i].x);
if ((diz[i].x>0)&&(diz[i].y<diz[i+1].y)) diz[i].y=-diz[i].y;
else if ((diz[i].x>0)&&(diz[i].x<diz[i-1].x)) diz[i].y=-diz[i].y;
if ((diz[i].x<0)&&(diz[i].y<diz[i-1].y)) diz[i].y=-diz[i].y;
else if ((diz[i].x<0)&&(diz[i].x<diz[i-1].x)) diz[i].y=-diz[i].y;
}
circle(midx, midy, r);
t = 1; i = 1;
k = 10;
while (i<p) {
i++;
k = fmod(k,p);
line(midx+diz[t].x,midy-diz[t].y,midx+diz[k].x,midy-diz[k].y);
t = k;
k = k * 10;
}
getch();
closegraph();
}
Some Examples These examples show the nine corners of noble prime numbers. P = 193 
P = 337 
P = 499 
P = 701 
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