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#### JeGX

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##### Decyphering the Business Card Raytracer
« on: September 26, 2013, 10:13:07 PM »
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I recently came across Paul Heckbert's business card raytracer. For those that have never heard of it: It is a very famous challenge in the Computer Graphics field that started on May 4th, 1984 via a post on comp.graphics by Paul Heckbert ( More about this in his article "A Minimal Ray Tracer" from the book Graphics Gems IV).

The goal was to produce the source code for a raytracer...that would fit on the back of a business card.

`    #include <stdlib.h>   // card > aek.ppm    #include <stdio.h>    #include <math.h>    typedef int i;typedef float f;struct v{    f x,y,z;v operator+(v r){return v(x+r.x    ,y+r.y,z+r.z);}v operator*(f r){return    v(x*r,y*r,z*r);}f operator%(v r){return    x*r.x+y*r.y+z*r.z;}v(){}v operator^(v r    ){return v(y*r.z-z*r.y,z*r.x-x*r.z,x*r.    y-y*r.x);}v(f a,f b,f c){x=a;y=b;z=c;}v    operator!(){return*this*(1/sqrt(*this%*    this));}};i G[]={247570,280596,280600,    249748,18578,18577,231184,16,16};f R(){    return(f)rand()/RAND_MAX;}i T(v o,v d,f    &t,v&n){t=1e9;i m=0;f p=-o.z/d.z;if(.01    <p)t=p,n=v(0,0,1),m=1;for(i k=19;k--;)    for(i j=9;j--;)if(G[j]&1<<k){v p=o+v(-k    ,0,-j-4);f b=p%d,c=p%p-1,q=b*b-c;if(q>0    ){f s=-b-sqrt(q);if(s<t&&s>.01)t=s,n=!(    p+d*t),m=2;}}return m;}v S(v o,v d){f t    ;v n;i m=T(o,d,t,n);if(!m)return v(.7,    .6,1)*pow(1-d.z,4);v h=o+d*t,l=!(v(9+R(    ),9+R(),16)+h*-1),r=d+n*(n%d*-2);f b=l%    n;if(b<0||T(h,l,t,n))b=0;f p=pow(l%r*(b    >0),99);if(m&1){h=h*.2;return((i)(ceil(    h.x)+ceil(h.y))&1?v(3,1,1):v(3,3,3))*(b    *.2+.1);}return v(p,p,p)+S(h,r)*.5;}i    main(){printf("P6 512 512 255 ");v g=!v    (-6,-16,0),a=!(v(0,0,1)^g)*.002,b=!(g^a    )*.002,c=(a+b)*-256+g;for(i y=512;y--;)    for(i x=512;x--;){v p(13,13,13);for(i r    =64;r--;){v t=a*(R()-.5)*99+b*(R()-.5)*    99;p=S(v(17,16,8)+t,!(t*-1+(a*(R()+x)+b    *(y+R())+c)*16))*3.5+p;}printf("%c%c%c"    ,(i)p.x,(i)p.y,(i)p.z);}}`