Add preview of 1984-2019
The 2020 winners are hinted it, but the source code is not there,
nor are the hints. Someone could look into the HTML and
see what was commented out, but that is not a problem because
WHO won is known and the name of their AWARD is known already.
Moreover a number of the winners have talked about what their
entirs do, so the commented out spoiler text is not a bit problem.
One will find HTML commented out with:
<!-- XXX remove when 2020 is ready
2020 links here
XXX remove when 2020 is ready -->
The "XXX remove when 2020 is ready" string is found at the
start and end comment blocks. Again, soneone could read
what was commented out, but that is OK since most of that
information is known or can be known. The following files
have such HTML comment blocks in them.
./winners.html
./years-spoiler.html
./years.html
In addition, ./all/summary.txt was modified to not
include the 2020 entries for the momemt.
The 2020 winner code AND write-ups are NOT part of this set.
p.s. The all/all.tar.bz2 does not contain the 2020 source code (yet?).
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@@ -0,0 +1,15 @@
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LDFLAGS+=-lm
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build: persano
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clean:
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clobber:
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-rm persano
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persano: persano.c
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${CC} ${CFLAGS} ${LDFLAGS} $< -o $@
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alt:
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data:
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@@ -0,0 +1,116 @@
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# Best of show
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Mauro Persano
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São Paulo
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Brazil
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mauro_persano@yahoo.com
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## Judges' Comments
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### To build
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cc -lm persano.c -o persano
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If your system is one of those that distinguish between text and binary
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streams, you must know what to do to switch the standard output into
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binary mode (e.g. adding a line to the source code or specifying a
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linker option).
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### Try
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persano 3 16 > knot3-16.gif
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Open the GIF file with a viewer capable of displaying animated GIFs and
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watch closely at the bottom of the picture (you may need to use
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"gifsicle" - a great tool to manipulate animated GIFs - to slow down the
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animation and to compress the GIFs generated by the entry).
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What needs to be done to fix the problem?
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Solving the author's exercises below may or may not help.
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At the time of judging, the approximation of π used in the program could
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not be found on mathworld.com.
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## Selected Author's Comments
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### NAME
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**knot**
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> writes animated GIF images of rendered torus knots to standard output.
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### SYNOPSIS
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knot p q [num frames]
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### DESCRIPTION
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A (p,q)-torus knot is a closed, non-intersecting curve on the surface of
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a torus in R3 that winds p times around one cycle of the torus and q
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times around the other. p and q should be coprime; otherwise, the
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resulting curve will _not_ be a knot (that is, imagining the resulting
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curve as a loop of rubber, there will be a way to bend, twist and
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stretch it, without cutting it [1], until it ends up looking just like a
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circle embedded in R3 [2]).
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Sample usage:
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knot 3 2 1 > foo.gif
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writes a non-animated GIF (a single frame) stream of a rendered trefoil
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knot to the file foo.gif.
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knot 5 2
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writes a GIF stream with a rendering of Solomon's Seal knot to standard
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output; since the number of frames is not specified, an animation with
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40 frames will be written.
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Please view the code with 8-character tabs!
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### OUTPUT FORMAT
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The program writes fully valid GIF 89 streams.
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It is suggested that the output be viewed with a program capable of
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displaying animated GIFs, such as almost any graphical HTTP browser.
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Due to code size constraints, full LZW compression was not implemented
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[3]. In fact, the output images are even somewhat expanded, thanks to
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GIF overhead.
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The GIF format supports only a limited number of bits per pixel, so the
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program uses ordered dithering to emulate full RGB output.
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### PORTABILITY
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The program was tested on Linux, with gcc 3.1; Cygwin, with gcc 3.3; and
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Solaris/Sparc, with Sun's C compiler.
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The only major portability issue should be reliance on ASCII.
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It should also be mentioned that the program won't work on systems that
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distinguish between text and binary streams; however, since hardly
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anyone uses such systems in this day and age, this is a minor issue. [4]
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### EXERCISES FOR THE READER
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1. (1 point) The surface is composed of many small polygons. Try to make it
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smoother by increasing the number of polygons.
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2. (2 points) Figure out how to change the radius of the tube cross section.
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3. (5 points) Locate the famous transcendental constant π in the code.
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### FOOTNOTES
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[1] You can call such a transformation an "ambient isotopy" if you want to
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impress people at parties.
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[2] Also known as the "unknot", which sounds slightly less impressive than
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"ambient isotopy".
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[3] Unfortunately (for the author), the convenient excuse that LZW compression
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was not implemented in order to avoid the Unisys patent can no longer be used.
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[4] ;)
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@@ -0,0 +1,138 @@
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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#include <math.h>
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#define H 20
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#define X for
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#define F(v,b) X(k = 0; k < 3; k++) v[k] = b;
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typedef double R, V[3], U[4], (*J)(R);
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typedef void Z;
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typedef int _;
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V o, d, v;
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R (*q)[H][6], C[4*H][H][6], sr = 15, br = 25, A, s, t;
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_ P, Q, g, S, k, e = 256, (*u)(_) = putchar, D[64];
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J f = cos, Y = sin;
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char B[256];
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Z
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T(_ *p, _ k, _ n)
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{
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_ g;
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*p = n;
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X(g = 0; k && g < 4; g++)
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T(p + (9 & (g * 4|g))*(1<<k - 1), k - 1,
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n + (1<<6-2 * k)*(4 - g&3));
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}
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Z
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O(_ c)
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{
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B[++S] = c,
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(S > e-2 || c==129) && (*B = S, S = fwrite(B, S+1, 1, stdout) - 1);
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}
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Z N
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(R*v){R w=sqrt
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(*v**v +1[v]*v[1]
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+v[2]*2[v]);F(v,v[k]
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/w)}Z K(_ j){V
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n,p ;R*b=(*
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q) [j],w=f
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(A),x=
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Y(A);
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F(p,
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br*o[
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k]+sr
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* (n[ k] =f(t)*k[v]+
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Y(t)*(v[(k+1)%3]* d[( k+2)%3]-v[(k+2)%
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3]*d[(k+1)%3])))*b++=p[1 ]+e/ 2,*b++=-*p*x+
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p[2]*w-e*e,*b++=(*p*w+p[2]*x)+e/2;F( b,fabs(*n
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*!!k+n[1]*( k<2)+n[2])/sqrt(2+ !(k-1)))
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}Z M(R*a,_ s,Z(*m)(_)){_ i;R d=
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pow(25/log (36),4)/s/ 377;X(
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*a=i=0;i<s ;i++,*a+=d )m(i);
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}Z G(_ i ){R w=-Y (P*s),
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x=f(P*s), y=Y(Q*s ),z=f
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(Q*s);F( v,k[o] =(k<2
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)*(x+3)*( k?y:z) -!(k-2 )*w)N(
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v);F(d,P* (k?k- 1?x:w*y:w*z )+Q*(k?
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k-1?0:*o:- o[1]) )N(d),w=*d**v+d[1]
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*v[1]+d[2] *v[2 ];F(v,k[v]-w
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*d[k])N(v),q =&C[
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i],M(&t,H,K); }Z W
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(char*s){*s&&( W(s
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+1),u(*s-98*(*s>
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97)));}Z E(_ z){R
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*r,*a,*b;U*q,l,x,d
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,I[256];_ i,j,m,c,y,
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w,h,Y=sizeof l;fputc(
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46,stderr),M(&s,4*H,G),
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W("ibcbcbbbbb,"),g=130,
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O(e/2);X(y=0;y<e;y++){X(i
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=e;i;)*(I[--i])=0;X(;i<4
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*H*H;i++)X(*l=k=0;k-5;k++
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,a=b)if(b=C[(i/H+((k^k/
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2)&1))%(4*H)][(i+(k/2&1
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))%H],k&&y<*a^y<*b){X(
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h=0;h<4;h++)h[x]=a[h
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+2]+(b[h+2]-a[h+2])
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*(y-*a)/(*b-*a),h?
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*l&&(d[h]=(l[h]
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-h [x])/w):(w=1+
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fabs (*l-(*x=(_)*x)
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) ); X(q=I+(_)*x;
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*l&& w--;q+=2*(
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*x<* l)-1)X(**q
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>C[i /H][i%H][1
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]&&( memcpy(*q,
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x,Y),* *q=C[i/H][
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i%H][1 ]),h=1;h<
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4;h++)x [h]+=h[d];
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memcpy(l ,x,Y);}X(
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i=0;i<e;i++ ,O(*r?c:
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127),g=g<e-1 ?g+1:(O(e/
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2),130))X(r=&I[ i][3],c=0,j
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=2;j+1;j--,r--)k=3|!( j-1)*4,m=*r*k+
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(63*( *r*k-(_)(*r*k))>D[(y*8&56)+(i&7)
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]), c<<=k/4+2,c|=(m|-(m>k))&
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k;} O(129);u(0);}_
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main(_ c, char **v)
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{
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if (c < 3)
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return fprintf(stderr, "Usage: %s P Q [F]\n", *v), 1;
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P = atoi(v[1]),
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Q = atoi(v[2]),
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T(D, 3, 0);
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W("bb\346cbcba98FIG");
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X(; k < 384; k++)
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u(k / 3<<("gdb"[k%3] - 97) | 31 | (k % 3!=1)<<5);
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W("bbbce0.2EPACSTEN\x0b\xff!"),
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M(&A, c > 3 ? atoi(v[3]) : 40, E),
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u(59);
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return 0;
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}
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/* vi:set ts=8: */
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