VI SOLVES MAZE (and lies about it)

[Ivan Batinic]

	First of all, the maze program does infact compile on a sun/3
under SunOS 3.5;  if one manually cb's the source, one sees the pgm:


char*M,A,Z,E=40,J[80],T[3];

main(C){
	for(M=J+E,*J=A=scanf("%d",&C); --E;J[E] = M[E] = E)
		printf("._"); 
	for(;(A -= Z = !Z) || (printf("\n|"), A = 39 ,C--); Z || printf (T))
		T[Z] = Z[A - (E = A[J - Z]) && !C & A == M[A] | 6 << 11 <
			 rand()|| !C &!Z ? J[M[E] = M[A]] = E,
			 J[M[A] = A - Z] = A, "_." : " |"];
}
	


	The above is pure 'C'onfusion to hide the simplicity of the
pgm.  If you run it, and give it an integer input, it prints n times:

._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
|_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._|
|


	But the real joke that seems to be escaping some of you is the
original form of the source:


    char*M,A,Z,E=40,J[80],T[3];main(C){for(M=J+E,*J=A=scanf("%d",&
    C)            ;--            E;J             [E            ]=M
    [E   ]=  E)   printf("._");  for(;(A-=Z=!Z)  ||  (printf("\n|"
    )    ,   A    =              39              ,C             --
    )    ;   Z    ||    printf   (T   ))T[Z]=Z[A-(E   =A[J-Z])&&!C
    &    A   ==             M[                                  A]
    |6<<11<rand()||!C&!Z?J[M[E]=M[A]]=E,J[M[A]=A-Z]=A,"_.":" |"];}


Which forms a maze if you look far away and squint your eyes.
Therefore, the poster was not lying, it is a "Maze C-Program" in the 
truest possible way:

	a) It attempts to print a maze
	b) It looks like a maze in source
	c) Parsing the source is like solving a maze

A true Maze program in form, structure and functionality -- Congrats 
to the author.  I don't know who started this 'thread' on a source
newsgroup, but do I win a prize?

	Ivan
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