maze-faq

迷宫

        }
    }

    /*init hedges*/
    for(j=0;j<(y-1);j++){
        for(i=0;i<x;i++){
            hedge[x*j+i].x = i;
            hedge[x*j+i].y = j;
            hedge[x*j+i].pos=DOWN;
            hedge[x*j+i].on=1;
            hedge_list[x*j+i]= &hedge[x*j+i];
        }
    }
    k=x*(y-1);

    for(j=0;j<y;j++){
        for(i=0;i<(x-1);i++){
            hedge[k+(x-1)*j+i].x = i;
            hedge[k+(x-1)*j+i].y = j;
            hedge[k+(x-1)*j+i].pos=RIGHT;
            hedge[k+(x-1)*j+i].on=1;
            hedge_list[k+(x-1)*j+i]= &hedge[k+(x-1)*j+i];
        }
    }

    k=(x*(y-1))+((x-1)*y);

    /*randomize hedges*/
    for(i=(k-1);i>0;i--){
        htmp=hedge_list[i];
        j=random()%i;
        hedge_list[i]=hedge_list[j];
        hedge_list[j]=htmp;
    }
    fflush(stdout);

    l=k;

    /*create maze*/
    for(i=0;i<l;i++){
        j=hedge_list[i]->x;
        k=hedge_list[i]->y;
        a=find(&maze[x*k+j]);
        if(hedge_list[i]->pos==DOWN){
            b=find(&maze[x*(k+1)+j]);
        } else {
            b=find(&maze[x*k+j+1]);
        }
        if(a!=b){
            link(a,b);
            hedge_list[i]->on=0;
        }
    }

    printf("+");
    for(i=0;i<x;i++){
        printf("-+");
    }
    printf("\n");

    l=x*(y-1);

    for(j=0;j<y;j++){
        printf("|");
        for(i=0;i<(x-1);i++){
            if(hedge[l+(x-1)*j+i].on){
               printf(" |");
            } else {
               printf("  ");
            }
        }
        printf(" |\n|");
        if(j<(y-1)){
            for(i=0;i<x;i++){
               if(hedge[j*x+i].on){
                   printf("--");
               } else {
                   printf(" -");
               }
            }
            printf("\n");
        }

    }

    for(i=0;i<x;i++){
        printf("-+");
    }
    printf("\n");

}
========================================



Excerpts from programming: 9-Mar-92 Re: Algorithm to create a m.. Don
[e-mail] (609)

Here is another algorithm I just thought of -- how well does it work?

1.  create a GRAPH with n*n nodes.
2.  for each node in the graph create 4 edges, linking it to the
    node in the north, south, east, and west direction.
3.  Assign a random weight to every edge.
4.  Run a minimum spanning tree algorithm (take your choice
    from any algorithms textbook) to produce a graph.  A minimum
    spanning tree has a path from every node to every other node,
    and no cycles, hence, it corresponds to a perfect maze.

Don Gillies - [e-mail] - University of Illinois at Urbana-Champaign




========================================





Excerpts from programming: 8-Apr-92 re mazes [e-mail]. (5437)

Thanks for the messages.  FYI, here is the maze tutorial I used to send out.

Chris
--------------
This seems to come up every couple of months, doesn't it?  Maybe we
should make a FAQL for this group...

Anyway, maze generation--a topic near and dear to my heart!  I'm going
to describe the three most common methods.  Note that these will all
generate mazes without loops or rooms or anything like that.  Restricting
a maze to be a tree (in the graph-theoretical sense of the term) simplifies
things immensely.  Mazes with loops are much harder to generate automatically
in any nice way.  Perhaps the best way is start out generating a tree and
then knock a couple of extra walls (all the algorithms start out with all
walls present and then knock out walls as necessary).

Finally, note that all of these techniques are based on well-known graph
algorithms.  This isn't surprising since what is being generated is
really nothing more than a spanning tree.


Technique #1: Depth-first search

1. Pick a random starting location as the "current cell" and mark it
   as "visited".  Also, initialize a stack of cells to empty (the stack
   will be used for backtracking).  Initialize VISITED (the number of
   visited cells) to 1.
2. While VISITED < the total number of cells, do the following:
     If the current cell has any neighbors which haven't yet been visited,
       pick one at random.  Push the current cell on the stack and set the
       current cell to be the new cell, marking the new cell as visited.
       Knock out the wall between the two cells.  Increment VISITED.
     If all of the current cell's neighbors have already been visited, then
       backtrack.  Pop the previous cell off the stack and make it the current
       cell.


This algorithm is probably the simplest to implement, but it has a problem
in that there are many mazes which it cannot generate.  In particular, it will
continue generating one branch of the maze as long as there are unvisited cells
in the area instead of being able to give up and let the unvisited cells get
used by a different branch.  This can be partially remedied by allowing the
algorithm to randomly backtrack even when there are unvisited neighbors, but
this creates the possibility of (potentially large) dead spaces in the
maze.  For some applications, such as dungeon creation, this behavior
might be desirable.  A refinement which cuts down on the amount of dead space
is to vary the probability of backtracking as an increasing function on the
number of iterations since the last backtrack.



Technique #2: Prim's Algorithm
1. Maintain three sets of cells: IN, OUT, and FRONTIER.  Initially, choose
   one cell at random and place it in IN.  Place all of the cell's neighbors in
   FRONTIER and all remaining cells in OUT.
2. While FRONTIER is not empty do the following:  Remove one cell at random
   from FRONTIER and place it in IN.  If the cell has any neighbors in
   OUT, remove them from OUT and place them in FRONTIER.  The cell is
   guaranteed to have at least one neighbor in IN (otherwise it would
   not have been in FRONTIER); pick one such neighbor at random and connect
   it to the new cell (ie knock out a wall).

This algorithm works by growing a single tree (without concentrating
on any one particular branch like the previous algorithm).  An interesting
variation on this algorithm is to run two (or more) instances of it
at once, starting in different locations and generating non-interesecting
trees.  This can be useful, for instance, for generating multiple disjoint
areas on a single level of a dungeon (perhaps connected by secret doors
or perhaps requiring adventurers to blast through walls themselves).


Technique #3: Kruskal's Algorithm

This is perhaps the most advanced of the maze generation algorithms,
requiring some knowledge of the union-find algorithm.  It is also the
most fun to watch in progress!

Basically what the union-find algorithm does is give you a FAST
implementation of equivalence classes where you can do two things:
FIND which equivalence class a given object belongs to, or UNION
two equivalance classes into a single class.  Any moderately advanced
book on algorithms and data structures will have more details.
In this algorithm, the objects will be cells in the maze, and two
objects will be in the same equivalence class iff they are (perhaps
indirectly) connected.

1. Initialize the union-find structures so that every cell is in
   its own equivalence class.  Create a set containing all the interior
   walls of the maze (ie those walls which lie between neighboring cells).
   Set COUNT to the number of cells.  (COUNT is the number of connected
   components in the maze).
2. While COUNT > 1 do the following:
     Remove a wall from the set of walls at random.  If the two cells
     that this wall separates are already connected (test by doing a FIND
     on each), then do nothing; otherwise, connect the two cells (by
     UNIONing them and decrementing COUNT) and knock out the wall.



Note that none of these algorithms make any assumptions about the topology
of the maze.  They will work with 2-d or 3-d grids, toroids, hexagons,
whatever.  However, in the more highly connected topologies (such as 3-d
grids), the deficiencies of the first algorithm will become even more apparent
(it will tend to produce long, winding paths with very little branching).


Have fun with these!

Chris


================================
/*
 * maz.c - generate a maze
 *
 * algorithm posted to rec.games.programmer by [e-mail]
 * program cleaned and reorganized by [e-mail]
 *
 * don't make people pay for this, or I'll jump up and down and
 * yell and scream and embarass you in front of your friends...
 *
 * compile: cc -o maz -DDEBUG maz.c
 *
 */
#include <stdio.h>

static int      multiple = 57;  /* experiment with this? */
static int      offset = 1;     /* experiment with this? */

#ifdef __STDC__
int maze(char maz[], int y, int x, char vc, char hc, char fc);
void mazegen(int pos, char maz[], int y, int x, int rnd);
#else
int maze();
void mazegen();
#endif

/*
 * maze() : generate a random maze of size (y by x) in maz, using vc as the
 * vertical character, hc as the horizontal character, and fc as the floor
 * character
 *
 * maz is an array that should already have its memory allocated - you could
 * malloc a char string if you like.
 */
#ifdef __STDC__
int maze(char maz[], int y, int x, char vc, char hc, char fc)
#else
int maze(maz, y, x, vc, hc, fc)
char            maz[], vc, hc, fc;
int             y, x;
#endif
{
   int             i, yy, xx;
   int             max = (y * x);
   int             rnd = time(0L);

   /* For now, return error on even parameters */
   /* Alternative is to decrement evens by one */
   /* But really that should be handled by caller */
   if (!(y & 1) | !(x & 1))
      return (1);

   /* I never assume... */
   for (i = 0; i < max; ++i)
      maz[i] = 0;

   (void) mazegen((x + 1), maz, y, x, rnd);

   /* Now replace the 1's and 0's with appropriate chars */
   for (yy = 0; yy < y; ++yy) {
      for (xx = 0; xx < x; ++xx) {
         i = (yy * x) + xx;

         if (yy == 0 || yy == (y - 1))
            maz[i] = hc;
         else if (xx == 0 || xx == (x - 1))
            maz[i] = vc;
         else if (maz[i] == 1)
            maz[i] = fc;
         else if (maz[i - x] != fc && maz[i - 1] == fc
                 && (maz[i + x] == 0 || (i % x) == (y - 2)))
            maz[i] = vc;
         else
            maz[i] = hc;       /* for now... */
      }
   }
   return (0);
}


/*
 * mazegen : do the recursive maze generation
 *
 */
#ifdef __STDC__
void mazegen(int pos, char maz[], int y, int x, int rnd)
#else
void
mazegen(pos, maz, y, x, rnd)
   int             pos, y, x, rnd;
   char            maz[];
#endif
{
   int             d, i, j;

   maz[pos] = 1;
   while (d = (pos <= x * 2 ? 0 : (maz[pos - x - x] ? 0 : 1))
          | (pos >= x * (y - 2) ? 0 : (maz[pos + x + x] ? 0 : 2))
          | (pos % x == x - 2 ? 0 : (maz[pos + 2] ? 0 : 4))
          | (pos % x == 1 ? 0 : (maz[pos - 2] ? 0 : 8))) {

      do {
         rnd = (rnd * multiple + offset);
         i = 3 & (rnd / d);
      } while (!(d & (1 << i)));

      switch (i) {
      case 0:
         j = -x;
         break;
      case 1:
         j = x;
         break;
      case 2:
         j = 1;
         break;
      case 3:
         j = -1;
         break;
      default:
         break;
      }

      maz[pos + j] = 1;

      mazegen(pos + 2 * j, maz, y, x, rnd);
   }

   return;
}
#ifdef DEBUG
#define MAXY 24
#define MAXX 80

#ifdef __STDC__
main(int argc, char *argv[])
#else
main(argc, argv)
   int             argc;
   char           *argv[];
#endif
{
   extern int      optind;
   extern char    *optarg;
   int             x = 79;
   int             y = 23;
   char            hor = '-';
   char            ver = '|';
   char            flo = ' ';
   char            maz[MAXY * MAXX];
   int             i;

   while ((i = getopt(argc, argv, "h:v:f:y:x:m:o:")) != EOF)
      switch (i) {
      case 'h':
         hor = *optarg;
         break;
      case 'v':
         ver = *optarg;
         break;
      case 'f':
         flo = *optarg;
         break;
      case 'y':
         y = atoi(optarg);
         break;
      case 'x':
         x = atoi(optarg);
         break;
      case 'm':
         multiple = atoi(optarg);
         break;
      case 'o':
         offset = atoi(optarg);
         break;
      case '?':
      default:
         (void) fprintf(stderr, "usage: maz [xyhvfmo]\n");
         break;
      }

   if (maze(maz, y, x, ver, hor, flo) == 0) {
      for (i = 0; i < (x * y); ++i) {
         (void) putchar(maz[i]);
         if (((i + 1) % x) == 0)
            (void) putchar('\n');
      }
   } else {
      (void) fprintf(stderr, "Couldn't make the maze\n");
   }

   exit(0);
}
#endif DEBUG