ASCII Maze

[][ hi ...

trawling my archives ... (mag-tape2cd procedure)

i found something which some readers may find of passing interest

the snippet-items below were taken from:-

>
>From: gre...@otc.otca.oz.au (Greg McFarlane)
>Newsgroups: alt.sources
>Subject: VI SOLVES MAZE (commented macros)
>Date: 10 Feb 91 23:31:02 GMT
>
>A real working model. See it walk the maze in front of your very own eyes.
>To prove that you can do anything in vi, I wrote a couple of macros that
>allows vi to solve mazes. It will solve any maze produced by maze.c
>that was posted to alt.sources last month.
>
>After studying these macros, anyone who cannot write an emacs emulator
>in vi macros should just curl up and :q!.
>
>Coming soon to a newsgroup near you: "Vi macros solve Tower of Hanoi",
>and a repost of the original "Turing Machine implemented in Vi macros"
>
>Anyone who has a version of these macros for edlin or nroff, please post.
>


maze.c - John Tromp (tro...@piring.cwi.nl)

char*M,A,Z,E=40,J[40],T[40];main(C){for(*J=A=scanf(M="%d",&C);
--            E;             J[              E]             =T
[E   ]=  E)   printf("._");  for(;(A-=Z=!Z)  ||  (printf("\n|"
)    ,   A    =              39              ,C             --
)    ;   Z    ||    printf   (M   ))M[Z]=Z[A-(E   =A[J-Z])&&!C
&    A   ==             T[                                  A]
|6<<27<rand()||!C&!Z?J[T[E]=T[A]]=E,J[T[A]=A-Z]=A,"_.":" |"];}

[i got segmentation-fault with this][i486 gcc2.6 linux1.3.0]


maze.ansi.c - ant...@nntp-server.caltech.edu (Bill T. Cat)

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,"_.":" |"];}
     .
    /|\
     |
     `---- i had to change this to 27 (my rand.seed default is 1)


with an input value of 5 (rows)
produces:
._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
| ._| . . ._| | |_._._. . ._|_._._._._. ._|_. ._|_._. ._| | . ._|_. | . ._._. |
| ._|_| |_. | | | | ._._|_._|_._. . |_. | | | ._._| |_._._._| | ._. ._| . . |_|
|_._._._. | ._|_. ._._._. | | ._. |_._. . | ._._| |_. | . ._._._. |_. | |_|_| |
| | . |_._| . ._._._| ._._. ._._| | | |_| . | |_. . ._|_|_| ._._. |_._|_| . | |
|_._|_._._._|_._._._|_|_._._._|_._|_._._._|_._._._|_._._._._|_._._._._._._|_._|
|


and using this - maze.ansi40.c

char*M,A,Z,E=20,J[40],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    =              19              ,C             --
)    ;   Z    ||    printf   (T   ))T[Z]=Z[A-(E   =A[J-Z])&&!C
&    A   ==             M[                                  A]
|6<<27<rand()||!C&!Z?J[M[E]=M[A]]=E,J[M[A]=A-Z]=A,"_.":" |"];}

cc -o maze40 maze.ansi40.c
maze40
with an input value of 15
gave:
._._._._._._._._._._._._._._._._._._._
| |_. ._._| |_._._. |_. | |_. |_. ._._|
|_. | | | |_. . |_._._._._. ._|_. | ._|
| . | | |_. |_| | | ._._._. |_. ._. | |
|_| . ._._._._| | |_. ._| | |_._| | | |
|_. | ._|_. |_._._. |_| | | | | | |_. |
| | |_| ._. | | |_. . |_. | ._| | ._. |
|_._._. | ._._|_._._| |_. ._._. . . | |
|_._._. | ._._. ._._|_. ._._. | |_|_|_|
| | | ._|_|_._. | . |_. . ._|_| ._. ._|
|_. | . |_._. | . |_. ._| |_. | ._|_| |
|_._. |_. . . |_| ._|_| |_| | ._. ._| |
|_._. . | |_|_._| . ._._._| |_._| |_. |
| | . |_| ._._| | |_|_. | ._._. ._._| |
| |_| |_._._._._| | |_. |_. | |_|_. | |
|_._._|_._._._._._|_._._._._._._._._._|
|

watching vi solve the maze is quite entertaining
(the maze.vi.macros are binary)

[][ see yuz
[][ colin
-- 
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