Write a program that generates and prints a random maze using the algorithm of your choice. The maze should be different for multiple runs of the program. Height and width are given as command line arguments. Use | for vertical wall, - for horizontal wall and + for corner. The maze is bounded by walls and the entrances are marked by missing wall. The maze contains a treasure # which must be reachable from at least one entrance.
$ python2 random-maze.py 4 5
+-+-+
|#|
| |
+---+
import random,sys
H,V=map(int,sys.argv[1:])
H-=1
V-=1
b,h,v,p=' -|+'
M=H/2*h
n=random.randint(1,(H/2)*(V/2-1))
for i in range(V/2):
e=s=t='';N=v
for j in range(H/2):
if i and(random.randint(0,1)or j==0):s+=N+b;t+=v;N=v;M=M[1:]+p
else:s+=M[0]+h;t+=b;N=p;M=M[1:]+h
n-=1;t+=' #'[n==0]
if H&1:s+=s[-1];t+=b;e=h
print s+N+'\n'+t+v
if V&1:print t+v
print h.join(M)+e+h+p
This generates a maze with one entrance and a randomly placed treasure. The maze is a simple binary tree maze.
$ ./maze.py 15 15
--------------+
|
| | ----------+
| | |
| +-----+ | --+
| | | |
| --+ --+ +---+
| | | |
| --+-+ +---+ |
| | | |
| --+ +-+ --+ |
| | | |#|
| | | --+ --+-+
| | | | |
+-+-+---+-----+
eval ARGV[0]
z=[l="+"+"-"*@w+"+"]
@h.times{z<<"|"+" "*@w+"|"}
z[rand(@h)+1]="|#"
puts z<<l
Output
$> ruby maze.rb "@h=8;@w=8;"
+------+
| |
| |
| |
| |
|#
| |
+------+
Hey, no one said it had to be a good maze. OK, OK, I'll make a real one now.
#!/usr/bin/env python, for example, in their code. As I said I'll write a real solution too, this was just pointing out the poor quality of the question itself (and many others) and demonstrates we need to have better guidelines. And finally pointing to a question does not make the answer to the question the actual rules for the site. But fine, here's your new version...
\$\endgroup\$
I think this technically isn't a maze generator, but it creates a maze like result: https://gist.github.com/803450.
Some horrible code in there I know, and it only works less than half the time, and the result doesn't look quite right to do with walls sticking out from other walls. But its close enough that I can't be bothered fixing the rest.
Some example output:
→ ruby random-maze.rb 30 30
+----+-+-----------++-+----+
| + | ++ | |
++ + | ++ ++ + + ++ ++
| ++ ++ | | +---+ + |
| + | +| + +++ + + |
| + +| +| +-+ | + + |
| + + + + ++ |+ |
| + ++ + ++ + | + ++| |
| | | ++ + +----+ + +-+ | |
| + | +-+ |+ | | |
| +-+ +| ++ ++ + + | | |
| ++ + + | ++| + | ++ |
| + + + + +---++-+ +++ |
| + | +| + | ++ | |
| | +++ +| + ++ +--+ + |---+
|#+ | | | +++ + + |
++ | ++ +-+ ++ +--+ + + |
| ++ | + ++| + ++ |
| ++ +--------+ +| + + |
| | | +++ | + +-+
| | | +--+ |++ |+ | ++
| | | +--+ | | || | |
| | +-+ +-+ |+ |+ |
| | +---+ ++ + | |
| +-| + + ++ ++
| + ++ +---+ |
| ++ + +-+
| + ++
+-+ +-------------+---+
#include <stdlib.h>
#include <time.h>
h,w,*m,y,x,z;d(t,b,l,r){int i=b-t,j=r-l;if(i>1&&j>1){i=(rand()%--i)|1;j=(rand()%--j)|1;z=rand()%4;x=rand()%i+t;x|=1;for(y=t;y<i+t;y++)if(y!=x||!z)m[y*w+j+l]=124;x=rand()%(b-i-t)+i+t;x|=1;for(y=t+i;y<b+1;y++)if(y!=x||!(z-1))m[y*w+j+l]=124;y=rand()%j+l;y|=1;for(x=l;x<j+l;x++)if(y!=x||!(z-2))m[(i+t)*w+x]=45;y=rand()%(r-j-l)+j+l;y|=1;for(x=l+j;x<r+1;x++)if(y!=x||!(z-3))m[(i+t)*w+x]=45;m[(i+t)*w+j+l]=43;m[(t-1)*w+l+j]=43;m[(b+1)*w+j+l]=43;m[(i+t)*w+l-1]=43;m[(i+t)*w+r+1]=43;d(t,t+i-1,l,l+j-1);d(t+i+1,b,l,l+j-1);d(t,t+i-1,l+j+1,r);d(t+i+1,b,l+j+1,r);}}main(int c,char**v){h=atoi(v[1]),w=atoi(v[2]),m=calloc(h*w,4);srand(time(0));while(y<h){while(x<w){m[y*h+x]=(!y||y==h-1)?(!x||x==w-1)?43:45:(!x||x==w-1)?124:32;x++;}y++;x=0;}d(1,h-2,1,w-2);z=rand()%(w-2);z|=1;m[z]=32;z=rand()%(w-2);z|=1;m[h*(w-2)+z]=35;}
To Test:
#include <stdio.h>//beginning
for(y=0;y<h;y++){for(x=0;x<w;x++){putchar(m[y*h+x]);}putchar('\n');}getchar();//end
3x3
+ + |#| +-+
7x8
+-+-- -+ | | + +-+--+ | | | +-+ -+ | | # | +-+-+--+
18x20
+-+-+ +---+---+-+--+ | | | | | | + + +-- +---+ +--+ | | | | + + +-+---+-- +-+ -+ | | | | +-+ +-+-+-+---+-+--+ | | | | | | | + + + +-+-- --+ | | | | | | | | | +-+-+ ----+ | | | | | | + + +-+-+-+-- --+ -+ | | | | | | + +-+-- +-- --+ | | | | | | | | | | #| | | +-+-+-+---+-----+--+
Here's a simple java solution:
import java.util.*;
public class MazeGen {
public static void main(String[]a){
int w,l;
Random rand=new Random(System.currentTimeMillis()%1000+System.nanoTime());
if(a.length==2){
w=Integer.parseInt(a[0]);
l=Integer.parseInt(a[1]);
}else{
System.out.println("No command line arguments, taking from STDIN.");
Scanner input=new Scanner(System.in);
w=input.nextInt();
l=input.nextInt();
input.close();
}
char[][]maze=new char[w][l];
for(int x=0;x<w;x++){
for(int y=0;y<l;y++){
maze[x][y]=' ';
}
}
for(int x=0;x<w;x++){
maze[x][0]=maze[x][l-1]='|';
}
for(int y=0;y<l;y++){
maze[0][y]=maze[w-1][y]='-';
}
maze[0][0]=maze[w-1][0]=maze[w-1][l-1]=maze[0][l-1]='+';
int dor=1+rand.nextInt(l-2);
maze[0][dor]=' ';
int tx=2+rand.nextInt(w-3),ty=1+rand.nextInt(l-2);
maze[tx][ty]='#';
if(ty<dor-1){
maze[tx][ty+1]='|';
if(tx==w-2){
maze[tx+1][ty+1]='+';
}
if(tx==1){
maze[0][ty+1]='+';
}
}
if(ty>dor+1){
maze[tx][ty-1]='|';
if(tx==w-2){
maze[tx+1][ty-1]='+';
}
if(tx==1){
maze[0][ty-1]='+';
}
}
if(ty==dor&&tx>3&&(maze[tx][ty+1]==' '||maze[tx][ty-1]==' ')){
maze[tx-1][ty]='-';
}
if(dor>5){
int z=2+rand.nextInt(dor-3);
int q=1+rand.nextInt(w-3);
for(int i=0;i<w;i++){
if(i==0||i==w-1){
maze[i][z]='+';
}else if(i!=q&&maze[i][z]==' '){
maze[i][z]='|';
}
}
}
if(l-dor>5){
int z=dor+2+rand.nextInt(l-dor-3);
int q=1+rand.nextInt(w-3);
for(int i=0;i<w;i++){
if(i==0||i==w-1){
maze[i][z]='+';
}else if(i!=q&&maze[i][z]==' '){
maze[i][z]='|';
}
}
}
for(char[]row:maze){
System.out.println(row);
}
}
}
3x3:
+ +
|#|
+-+
4x4:
+ -+
| #|
| |
+--+
4x5:
+-+ +
|#| |
| |
+---+
5x5:
+ --+
| |
| |
| |#|
+-+-+
5x8:
+ --+--+
| | |
| |
| # | |
+---+--+
8x15:
+---- ----+---+
| | |
| | |
| | |
| |
| #| | |
| | |
+---------+---+
*or is there two separate entrances? 2: You should probably specify that the exit must be reachable. \$\endgroup\$