This should be right, if not, lemme know. I tried to shorten in a bit. Supposedly this can be modified to do ovals but I've not tried it.
Code:void circle(int x, int y, int r) { r++; int target = 0; int a = r; int b = 0; int i; while(a >= b) { b = sqrt(r * r - a * a); SWAP(target, b); while(b < target) { plot(x + a, y + b); plot(x - a, y + b); plot(x - a, y - b); plot(x + a, y - b); plot(x + b, y + a); plot(x - b, y + a); plot(x - b, y - a); plot(x + b, y - a); b++; } a--; } }
I've tried it several times and it seems fineOriginally Posted by joed
Thanks
P.S. Do you also know how to make ovals? It would be helpful too.
Yes, you goto www.mathworld.com and look up the relevent parametric equations.
Then you put that into the code we've shown you how to do.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
What I really wanted wasnīt an oval but an ellipse.
I tried that site and I found the formula but I don't have the knowladge to make any sort of ellipses.
Just replace the circle formula in your code to the elipse formula. Is there any specific part of the equation you don't understand?
Code:#include <stdio.h> void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){ puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9 /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i] ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][ t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}
Circle
x = r * cos( theta );
y = r * sin( theta );
Ellipse
x = r1 * cos( theta );
y = r2 * sin( theta );
A circle is just an ellipse with r1 == r2
See, you latched onto some quick hack which did what you want without having any clue about what it really meant.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
