`========================================`

Maze FAQ

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

Sorry this is NOT an organized FAQ it's still useful though

------------------------------------------------------------

Oh, you're really going to love this one. It's an obfuscated C code

mazegenerator. Fun fun fun. Well, if you can figure it out, there's

youralgorithm. Fun fun fun.

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< <11

Pretty cute, no?

--

Brad Threatt | MISSING! Single white male Jesus look-alike in blue

| Members Only jacket. Answers to the name 'Dave Gillespie'.

Safe sex is | Last seen with roller-skating couple in Pasadena.

for wimps. | If found, please return to the cs10 lab immediately.

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

In <1992Mar5.210706.24104@wpi.WPI.EDU> rcarter@wpi.WPI.EDU (Randolph

Carter (nee. Joseph H. Allen)) writes:

>>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<<11
^^

This should be 28 if rand() returns a 32-bit quantity.

>>Pretty cute, no?

>No style at all.... :-)

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

>--

>/* rcarter@wpi.wpi.edu */ /* Amazing */ /* Joseph H.

Allen */

>int

a[1817];main(z,p,q,r){for(p=80;q+p-80;p-=2*a

`)for(z=9;z--;)q=3&(r=time(0)`

>+r*57)/7,q=q?q-1?q-2?1-p%79?-1:0:p%79-77?1:0:p<1659?79:0:p>158?-79:0,q?!a[p+q*

2

>]?a[p+=a[p+=q]=q]=q:0:0;for(;q++-1817;)printf(q%79?"%c":"%c\n","

#"[!a[q-1]]);}

Well, it doesn't produce a maze, but try this one...

int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;b-c;)f[b++]=a/5;for(;d=0,g=c*2;c

-=14,printf("%.4d",e+d/a),e=d%a)for(b=c;d+=f*a,f=d%--g,d/=g--,--b;d*=b);}

(I disclaim any credit for this!)

--

John Brownie

School of Mathematics and Statistics

University of Sydney

Internet: jhb@maths.su.oz.au

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

Excerpts from programming: 6-Mar-92 Re: Algorithm to create a m.. Mark

Howell@movies.enet. (5723)

Here's the single level maze algorithm, solver and printer.

/*

* MazeGen.c -- Mark Howell -- 8 May 1991

*

* Usage: MazeGen [width [height [seed]]]

*/

#include

#include

#include

#define WIDTH 39

#define HEIGHT 11

#define UP 0

#define RIGHT 1

#define DOWN 2

#define LEFT 3

#ifdef TRUE

#undef TRUE

#endif /* TRUE */

#define TRUE 1

#define cell_empty(a) (!(a)->up && !(a)->right && !(a)->down && !(a)->left)

typedef struct {

unsigned int up : 1;

unsigned int right : 1;

unsigned int down : 1;

unsigned int left : 1;

unsigned int path : 1;

unsigned int visited : 1;

} cell_t;

typedef cell_t *maze_t;

void CreateMaze (maze_t maze, int width, int height);

void SolveMaze (maze_t maze, int width, int height);

void PrintMaze (maze_t maze, int width, int height);

int main (int argc, char *argv [])

{

int width = WIDTH;

int height = HEIGHT;

maze_t maze;

if (argc >= 2)

width = atoi (argv [1]);

if (argc >= 3)

height = atoi (argv [2]);

if (argc >= 4)

srand (atoi (argv [3]));

else

srand ((int) time ((time_t *) NULL));

if (width <= 0 || height <= 0) {

(void) fprintf (stderr, "Illegal width or height value!\n");

exit (EXIT_FAILURE);

}

maze = (maze_t) calloc (width * height, sizeof (cell_t));

if (maze == NULL) {

(void) fprintf (stderr, "Cannot allocate memory!\n");

exit (EXIT_FAILURE);

}

CreateMaze (maze, width, height);

PrintMaze (maze, width, height);

(void) putchar ('\n');

SolveMaze (maze, width, height);

PrintMaze (maze, width, height);

free (maze);

exit (EXIT_SUCCESS);

return (0);

}/* main */

void CreateMaze (maze_t maze, int width, int height)

{

maze_t mp, maze_top;

char paths [4];

int visits, directions;

visits = width * height - 1;

mp = maze;

maze_top = mp + (width * height) - 1;

while (visits) {

directions = 0;

if ((mp - width) >= maze && cell_empty (mp - width))

paths [directions++] = UP;

if (mp < maze_top && ((mp - maze + 1) % width) && cell_empty (mp + 1))

paths [directions++] = RIGHT;

if ((mp + width) <= maze_top && cell_empty (mp + width))

paths [directions++] = DOWN;

if (mp > maze && ((mp - maze) % width) && cell_empty (mp - 1))

paths [directions++] = LEFT;

if (directions) {

visits--;

directions = ((unsigned) rand () % directions);

switch (paths [directions]) {

case UP:

mp->up = TRUE;

(mp -= width)->down = TRUE;

break;

case RIGHT:

mp->right = TRUE;

(++mp)->left = TRUE;

break;

case DOWN:

mp->down = TRUE;

(mp += width)->up = TRUE;

break;

case LEFT:

mp->left = TRUE;

(--mp)->right = TRUE;

break;

default:

break;

}

} else {

do {

if (++mp > maze_top)

mp = maze;

} while (cell_empty (mp));

}

}

}/* CreateMaze */

void SolveMaze (maze_t maze, int width, int height)

{

maze_t *stack, mp = maze;

int sp = 0;

stack = (maze_t *) calloc (width * height, sizeof (maze_t));

if (stack == NULL) {

(void) fprintf (stderr, "Cannot allocate memory!\n");

exit (EXIT_FAILURE);

}

(stack [sp++] = mp)->visited = TRUE;

while (mp != (maze + (width * height) - 1)) {

if (mp->up && !(mp - width)->visited)

stack [sp++] = mp - width;

if (mp->right && !(mp + 1)->visited)

stack [sp++] = mp + 1;

if (mp->down && !(mp + width)->visited)

stack [sp++] = mp + width;

if (mp->left && !(mp - 1)->visited)

stack [sp++] = mp - 1;

if (stack [sp - 1] == mp)

--sp;

(mp = stack [sp - 1])->visited = TRUE;

}

while (sp--)

if (stack [sp]->visited)

stack [sp]->path = TRUE;

free (stack);

}/* SolveMaze */

void PrintMaze (maze_t maze, int width, int height)

{

int w, h;

char *line, *lp;

line = (char *) calloc ((width + 1) * 2, sizeof (char));

if (line == NULL) {

(void) fprintf (stderr, "Cannot allocate memory!\n");

exit (EXIT_FAILURE);

}

maze->up = TRUE;

(maze + (width * height) - 1)->down = TRUE;

for (lp = line, w = 0; w < width; w++) {

*lp++ = '+';

if ((maze + w)->up)

*lp++ = ((maze + w)->path) ? '.' : ' ';

else

*lp++ = '-';

}

*lp++ = '+';

(void) puts (line);

for (h = 0; h < height; h++) {

for (lp = line, w = 0; w < width; w++) {

if ((maze + w)->left)

*lp++ = ((maze + w)->path && (maze + w - 1)->path) ? '.' : ' ';

else

*lp++ = '|';

*lp++ = ((maze + w)->path) ? '.' : ' ';

}

*lp++ = '|';

(void) puts (line);

for (lp = line, w = 0; w < width; w++) {

*lp++ = '+';

if ((maze + w)->down)

*lp++ = ((maze + w)->path && (h == height - 1 ||

(maze + w + width)->path)) ? '.' : ' ';

else

*lp++ = '-';

}

*lp++ = '+';

(void) puts (line);

maze += width;

}

free (line);

}/* PrintMaze */

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

Excerpts from programming: 6-Mar-92 Re: Algorithm to create a m.. "Jon

C. R. Bennett"@andr (4255)

gillies@m.cs.uiuc.edu (Don Gillies) writes:

> grid. Mark each square in the grid with a unique number. Make a list

what you want to do is make each grid in the maze into a set.

>

> rooms = n*n /* each spot in the grid is a unique room */

>

> repeat

> pick a random wall without replacement.

> if the numbers X and Y in the grid on both sides of the wall

> are different --

> delete the wall and use a recursive depth

> first search or brute-force loop to replace

> all Y in the grid with X's.

what you do here is instead

pick a wall

if the rooms on either side of the wall belong to differnent sets

delete the wall

union the two sets together.

the rest is the same

the brute force solution runs in O(n^2) this runs in O(n) (where n is the

number of grids) so if you had a 100 x 100 maze, this method takes 10,000

time steps, the brute force could take as many as 100,000,000 steps.

jon

p.s. below you will find some code to generate a maze this way

---------------------------------------------------

/* maze.c a maze generator

Jon Bennett

jcrb@cs.cmu.edu

*/

/* the maze is generated by making a list of all the internal hedges

and randomizing it, then going lineraly through the list, we take a

hedge and se if the maze squares adjacent to it are already connected

(with find) is not the we connect them (with link), this prevents us

from creating a maze with a cycle because we will not link two maze

squares that are already connect by some path */

#include

#define DOWN 1

#define RIGHT 2

struct maze_loc{

int rank;

int x,y;

struct maze_loc *ptr;

};

struct hedge_loc{

int x,y,pos,on;

};

struct maze_loc *maze;

struct hedge_loc *hedge;

struct hedge_loc **hedge_list;

void link(a,b)

struct maze_loc *a,*b;

{

if(a->rank == b->rank){

a->ptr=b;

b->rank++;

return;

}

if(a->rank > b->rank){

b->ptr=a;

return;

}

a->ptr=b;

}

struct maze_loc *find(a)

struct maze_loc *a;

{

if(a != a->ptr){

a->ptr = find(a->ptr);

}

return a->ptr;

}

main(argc,argv)

int argc;

char **argv;

{

int x,y,i,j,k,l,tmp;

struct maze_loc *a,*b;

struct hedge_loc *htmp;

if(argc!=3) exit(1);

srandom(time(0));

x=atoi(argv[1]);

y=atoi(argv[2]);

/*malloc the maze and hedges */

maze=(struct maze_loc *)malloc(sizeof(struct maze_loc)*x*y);

hedge=(struct hedge_loc *)malloc(sizeof(struct

hedge_loc)*((x*(y-1))+((x-1)*y)));

hedge_list=(struct hedge_loc **)malloc(sizeof(struct hedge_loc

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

/*init maze*/

for(j=0;j

maze[x*j+i].y = j;

maze[x*j+i].ptr = &maze[x*j+i];

maze[x*j+i].rank=0;

}

}

/*init hedges*/

for(j=0;j<(y-1);j++){

for(i=0;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

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;

j=random()%i;

hedge_list=hedge_list[j];

hedge_list[j]=htmp;

}

fflush(stdout);

l=k;

/*create maze*/

for(i=0;i

k=hedge_list->y;

a=find(&maze[x*k+j]);

if(hedge_list->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->on=0;

}

}

printf("+");

for(i=0;i

}

printf("\n");

l=x*(y-1);

for(j=0;j

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

printf("--");

} else {

printf(" -");

}

}

printf("\n");

}

}

for(i=0;i

}

printf("\n");

}

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

Excerpts from programming: 9-Mar-92 Re: Algorithm to create a m.. Don

Gillies@m.cs.uiuc.ed (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 - gillies@cs.uiuc.edu - University of Illinois at Urbana-Champaign

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

Excerpts from programming: 8-Apr-92 re mazes Chris_Okasaki@LOCH.MESS. (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 jallen@ic.sunysb.edu

* program cleaned and reorganized by mzraly@ldbvax.dnet.lotus.com

*

* 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

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 = 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 = hc;

else if (xx == 0 || xx == (x - 1))

maz = vc;

else if (maz == 1)

maz = fc;

else if (maz[i - x] != fc && maz[i - 1] == fc

&& (maz[i + x] == 0 || (i % x) == (y - 2)))

maz = vc;

else

maz = 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);

if (((i + 1) % x) == 0)

(void) putchar('\n');

}

} else {

(void) fprintf(stderr, "Couldn't make the maze\n");

}

exit(0);

}

#endif DEBUG

????

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);

if (((i + 1) % x) == 0)

(void) putchar('\n');

}

} else {

(void) fprintf(stderr, "Couldn't make the maze\n");

}

exit(0);

}

#endif DEBUG

??????