Maze solving: Difference between revisions

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For a maze generated by this task, write a function that finds (and displays) the shortest path between two cells. Note that because these mazes are generated by the Depth-first search algorithm, they contain no circular paths, and a simple depth-first tree search can be used.

D

Basic define <lang d>import std.stdio, std.random ;

immutable string[] Tiles =

   [" ","╞","╥","╔","╡","═", "╗","╦","╨","╚","║","╠","╝","╩","╣","╬"] ;

immutable string[] Paths =

   [" "," "," ","┌"," ","─", "┐"," "," ","└","│"," ","┘"," "," "," "] ;

alias uint Room ; enum Open { EMPTY = 0U , EAST = 1, SOUTH = 2, WEST = 4, NORTH = 8,

           EPATH = 16, SPATH = 32, WPATH = 64, NPATH = 128,
           ENTRY = 256, EXIT = 512, Visited = 1024 }

struct Where {

   immutable int x, y ;
   Where opBinary(string op)(Where rhs) if (op=="+") {
       return Where(x + rhs.x, y + rhs.y) ;
   }
   bool bounded(int w, int h) {
       return (x >= 0 && y >= 0 && x < w && y < h) ;
   }
   bool bounded(Room[][] r) {
       return (x >= 0 && y >= 0 && r.length > 0 && y < r.length && x < r[0].length ) ;
   }

} struct Door { Open opening = Open.EMPTY ; Where where ; alias where this ; }

immutable Open[Open] Opposite ; immutable Where[Open] NeibrPos ;

static this() { // workarround for Associative Array init.

   Opposite = [Open.EAST:Open.WEST, Open.SOUTH:Open.NORTH,
               Open.WEST:Open.EAST, Open.NORTH:Open.SOUTH] ;
   NeibrPos = [Open.EAST:Where( 1,0), Open.SOUTH:Where(0, 1),
               Open.WEST:Where(-1,0), Open.NORTH:Where(0,-1)] ;

}

const AsPath = true ;

void connectTo(ref Room[][] r, Door d, bool asPath = false) {

       auto s/*hift*/ = asPath ? 4 : 0 ;
       r[d.y][d.x] |= (d.opening << s) ;
       auto there = d.where + NeibrPos[d.opening] ;
       if(there.bounded(r))
           r[there.y][there.x] |= (Opposite[d.opening] << s) ;

} bool connected(ref Room[][] r, Door d) {

       auto there = d.where + NeibrPos[d.opening] ;
       return (r[    d.y][    d.x] & d.opening) && there.bounded(r) &&
              (r[there.y][there.x] & Opposite[d.opening]) ;

}

void printMaze(Room[][] maze, bool showPath = false) {

   foreach(row ; maze) {
       foreach(cell;row)
           if(showPath)
               write(Paths[0xf & (cell >> 4)]) ;
           else
               write(Tiles[0xf & (cell)]) ;
       writeln() ;
   }

} </lang> Generate Maze: <lang d>Room[][] genMaze(uint w, uint h, ref Door entry = Door.init, ref Door exit = Door.init) {

   Room[][] r = new Room[][](h, w) ; // default value = 0 means un-visited empty room

   bool validDoor(Door d) {
       if(d.where.bounded(w,h))
           switch(d.opening) {
               case Open.EAST : return (d.x == w - 1) ;
               case Open.WEST : return (d.x == 0) ;
               case Open.NORTH: return (d.y == 0) ;
               case Open.SOUTH: return (d.y == h - 1) ;
           }
       return false ;
   }

   Door makeDoor() { // create door at random edge
       switch([Open.EAST, Open.WEST, Open.NORTH, Open.SOUTH][uniform(0,4)]) {
           case Open.EAST : return Door(Open.EAST , Where(w-1, uniform(0,h))) ;
           case Open.WEST : return Door(Open.WEST , Where(  0, uniform(0,h))) ;
           case Open.NORTH: return Door(Open.NORTH, Where(uniform(0,w),   0)) ;
           case Open.SOUTH: return Door(Open.SOUTH, Where(uniform(0,w), h-1)) ;
       }
   }

   Open[] neibrDir = NeibrPos.keys ;

   // depth-first search algorithm to generate maze
   void visit(Where here) {
       r[here.y][here.x] |= Open.Visited ;
       randomShuffle(neibrDir) ;
       foreach(dir; neibrDir) {
           auto there = here + NeibrPos[dir] ;
           if(there.bounded(w,h))                              // bounded?
               if((r[there.y][there.x] & Open.Visited) == 0) { // un-visited?
                   connectTo(r, Door(dir, here)) ;
                   visit(there) ;
               }
       }
   }

   // make entry & exit doors if not provided or invalid
   while(entry.opening == Open.EMPTY || entry.where == exit.where || !validDoor(entry))
       entry = makeDoor() ;
   while( exit.opening == Open.EMPTY || entry.where == exit.where || !validDoor(exit))
       exit  = makeDoor() ;
   r[entry.y][entry.x] |= (entry.opening | Open.ENTRY) ;
   r[ exit.y][ exit.x] |= ( exit.opening | Open.EXIT) ;

   // generate maze starting from entry
   visit(entry.where)  ;

   return r ;

}</lang> Solver: <lang d>bool solveMaze(ref Room[][] r, Door entry, Door exit ) {

   Open[] neibrDir = NeibrPos.keys ;

   foreach(row ; r) // clear old path, if any
       foreach(cell;row)
           cell &= ~(Open.EPATH|Open.WPATH|Open.SPATH|Open.NPATH) ;

   bool trace(Door here) {
       if(here.where == exit.where) { // reach exit
           r.connectTo(exit, AsPath) ;
           return true ;
       }
       foreach(dir ; neibrDir) {
           //  here.opening is direction to enter the room@here
           //  _dir_ is direction to exit the room@here, which is opposite to there.opening
           if( (dir != here.opening) && ((dir & r[here.y][here.x]) != 0)) { // path exist?
               auto there = Door(Opposite[dir], here.where + NeibrPos[dir]) ;
               if( there.bounded(r))
                   if(trace(there)) { // reach exit, use stack to trace back the path
                       r.connectTo(Door(dir, here.where), AsPath) ;
                       return true ;
                   }
           }
       }
       return false ; // dead end
   }

   auto success = trace(entry) ;
   if(success)
       r.connectTo(entry, AsPath) ;
   return success ;

}</lang> Sample run: <lang d>void main() {

   Door entry, exit;
   auto maze = genMaze(40, 12, entry, exit) ;

   printMaze(maze) ;
   if(solveMaze(maze, entry, exit))
       printMaze(maze, AsPath) ;
   else
       writeln("No solution!?") ;

}</lang> Output:)

╔═╗╞╗╔╡╔══╗╔╦╗╔═╗╞╗╔═══╦══╗╞╗╔═╗╔╦╗╚╗╞═╗
╚╗╚═╝╠╗║╔╗╚╝║║║╞╩═╝╚╗╔═╝╔╗╠═╝╠╡╠╝╨╚╗╚═╦╝
╔╝╔══╝╚╩╝╚╡╔╝║╠════╗║╚═╗║║║╔╗║╔╝╔══╩═╡╚╗
╠╗║╔╗╔═╗╥╔╗║╥║╨╔╗╔╗║║╞╗║╨╚╝║╚╝║╥║╞╦╗╔╦╡║
║╚╝╨╚╣╔╩╣║╚╝║╚╗║╚╝╚╣╚╦╝╚╗╔═╝╞╗║║╠╗║║║╚╗║
╚═══╗╨╚╗║║╔╡╚╗╚╣╥╔╗╚╗╚╗╥║╚═╗╔╝║║║╚╝╨║╥║║
╔═══╝╔═╣║╚╩═╗║╔╝╚╝╚═╝╔╝║╚═╗║╚╗║╠╝╔══╝║╚╝
╚═╗╔═╝╞╝╚═╗╔╝║╚╗╔═╗╔╗║╞╬╗╥║╚╗║║╠╡╚══╗╚═╗
╔═╝║╔╗╔═╗╔╝╚╗╚╦╝║╥╚╝╚╝╔╝╚╣╚╗║║║╨╔══╗╚╗╔╣
╩══╝║╚╝╥╠╝╞╗║╥║╔╝╠╦╗╔╗╚═╗╚╗║║║╚═╝╔╡║╔╩╝╨
╔╡╔╦╝╔═╣╚═╗║║╚╝║╞╝║║╨╠══╝╥║╠╝╠╗╞═╣╔╝╚══╗
╚═╝╚═╝╞╝╞═╩╝╚══╩══╝╚═╝╞══╩╝╚═╝╚══╝╚════╝

       ┌──┐┌┐      ┌───┐     ┌─┐   └┐   
     ┌┐│  └┘│      └┐┌─┘     │ │    └─┐ 
  ┌──┘└┘   ┌┘       │└─┐   ┌┐│┌┘      └┐
┌┐│      ┌┐│        │  │   │└┘│     ┌┐ │
│└┘      │└┘        └┐ └┐┌─┘  │     │└┐│
└───┐    │           └┐ │└─┐  │     │ ││
┌───┘    └──┐        ┌┘ └─┐│  │  ┌──┘ └┘
└─┐        ┌┘   ┌─┐┌┐│    │└┐ │  └──┐   
┌─┘        └┐   │ └┘└┘    └┐│ │ ┌──┐└┐  
┘           │  ┌┘          ││ └─┘  │┌┘  
            │  │           └┘     ┌┘└──┐
            └──┘                  └────┘

PicoLisp

<lang PicoLisp>(de shortestPath (Goal This Maze)

  (let (Path NIL  Best NIL  Dir " > ")
     (recur (This Path Dir)
        (when (and This (not (: mark)))
           (push 'Path (cons This Dir))
           (if (== Goal This)
              (unless (and Best (>= (length Path) (length Best)))
                 (setq Best Path) )
              (=: mark T)
              (recurse (: west) Path " > ")
              (recurse (: east) Path " < ")
              (recurse (: south) Path " \^ ")
              (recurse (: north) Path " v ")
              (=: mark NIL) ) ) )
     (disp Maze 0
        '((Fld) (if (asoq Fld Best) (cdr @) "   ")) ) ) )</lang>

Using the maze produced in Maze generation#PicoLisp, this finds the shortest path from the top-left cell 'a8' to the bottom-right exit 'k1':

: (shortestPath 'a8 'k1 (maze 11 8))
   +   +---+---+---+---+---+---+---+---+---+---+
 8 | >   >   v | >   v |                       |
   +   +   +   +   +   +   +---+   +---+---+   +
 7 |   |   | >   ^ | v |   |       |       |   |
   +---+   +---+---+   +   +   +---+   +   +   +
 6 |   |       |     v |   |           |   |   |
   +   +---+   +---+   +---+---+---+   +   +---+
 5 |       |       | >   >   >   v |   |       |
   +---+   +---+   +---+---+---+   +---+---+   +
 4 |   |       |       |       | v | >   >   v |
   +   +---+   +---+   +---+   +   +   +---+   +
 3 |       |       |   |       | v | ^   < | v |
   +   +---+---+   +   +   +   +   +---+   +   +
 2 |       |       |   |   |   | v | >   ^ | v |
   +   +   +   +---+   +   +---+   +   +---+   +
 1 |   |               |         >   ^ |     >
   +---+---+---+---+---+---+---+---+---+---+---+
     a   b   c   d   e   f   g   h   i   j   k

Python

python 3

def Dijkstra(Graph, source):

   
       +   +---+---+
       | 0   1   2 |
       +---+   +   +
       | 3   4 | 5  
       +---+---+---+

       >>> graph = (        # or ones on the diagonal
       ...     (0,1,0,0,0,0,),
       ...     (1,0,1,0,1,0,),
       ...     (0,1,0,0,0,1,),
       ...     (0,0,0,0,1,0,),
       ...     (0,1,0,1,0,0,),
       ...     (0,0,1,0,0,0,),
       ... )
       ...
       >>> Dijkstra(graph, 0)
       ([0, 1, 2, 3, 2, 3], [1e+140, 0, 1, 4, 1, 2])
       >>> display_solution([1e+140, 0, 1, 4, 1, 2])
       5<2<1<0
   
   # Graph[u][v] is the weight from u to v (however 0 means infinity)
   infinity = 1e140
   undefined = 1e140
   n = len(graph)
   dist = [infinity]*n   # Unknown distance function from source to v
   previous = [infinity]*n # Previous node in optimal path from source
   dist[source] = 0        # Distance from source to source
   Q = list(range(n)) # All nodes in the graph are unoptimized - thus are in Q
   while Q:           # The main loop
       Q.sort(key=lambda n:dist[n], reverse=True) # (Not exactly a heap!)
                                   # improvement, install heapq from library
       u = Q.pop()                 # vertex in Q with smallest dist[]
       if dist[u] == infinity:
           break # all remaining vertices are inaccessible from source
       for v in range(n):               # each neighbor v of u
           if Graph[u][v] and (v in Q): # where v has not yet been visited
               alt = dist[u] + Graph[u][v]
               if alt < dist[v]:       # Relax (u,v,a)
                   dist[v] = alt
                   previous[v] = u
   return dist,previous

def display_solution(predecessor):

   cell = len(predecessor)-1
   while cell:
       print(cell,end='<')
       cell = predecessor[cell]
   print(0)

</lang>

Tcl

Works with: Tcl version 8.6

This script assumes that the contents of the generation task have already been sourced. Note that the algorithm implemented here does not assume that the maze is free of circuits, and in the case that there are multiple routes, it finds the shortest one because it is a breadth-first search (by virtue of the queue variable being used as a queue). <lang tcl>oo::define maze {

   method solve {} {

### Initialization of visited matrix and location/path queue set visited [lrepeat $x [lrepeat $y 0]] set queue {0 0 {}}

### Loop to do the searching ### while 1 { # Check for running out of path; an error in maze construction if {[llength $queue] == 0} { error "cannot reach finish" } # Visit the next square from the queue set queue [lassign $queue cx cy path] if {[lindex $visited $cx $cy]} continue lset visited $cx $cy 1 lappend path $cx $cy # Check for reaching the goal if {$cx == $x-1 && $cy == $y-1} break # Add the square in each direction to the queue if a move there is legal foreach {dx dy} {0 1 1 0 0 -1 -1 0} { set nx [expr {$cx + $dx}]; set ny [expr {$cy + $dy}] if { $nx >= 0 && $nx < $x && $ny >= 0 && $ny < $y && ($dx && idx($verti, min($cx,$nx), $cy) || $dy && idx($horiz, $cx, min($cy,$ny))) } then { lappend queue $nx $ny $path } } }

### Loop to set up the path rendering ### # (-2,-2) is just a marker that isn't next to the maze at all, so # guaranteeing the use of the last 'else' clause foreach {cx cy} $path {nx ny} [concat [lrange $path 2 end] -2 -2] { if {$nx-$cx == 1} { lset content $cx $cy "v" } elseif {$nx-$cx == -1} { lset content $cx $cy "^" } elseif {$ny-$cy == -1} { lset content $cx $cy "<" } else { lset content $cx $cy ">" } }

### Return the path ### return $path

}

Do the solution (we ignore the returned path here...)

m solve

Print it out

puts [m view]</lang> Example output:

+   +---+---+---+---+---+---+---+---+---+---+
| v     |                                   |
+   +---+   +---+---+---+---+---+---+---+   +
| v |       | >   v | >   v |   |           |
+   +   +---+   +   +   +   +   +   +---+   +
| v     | >   ^ | v | ^ | v |   |       |   |
+   +---+   +---+   +   +   +   +---+   +---+
| v | >   ^ | v   < | ^ | v |       |   |   |
+   +   +---+   +---+   +   +   +---+   +   +
| >   ^ | v   < | >   ^ | v |       |       |
+---+---+   +---+   +---+   +---+   +---+---+
| v   <   < | >   ^ | v   < | >   >   >   v |
+   +---+---+   +---+   +---+   +---+---+   +
| >   v |     ^   < | >   >   ^ |       | v |
+---+   +---+---+   +---+---+---+   +   +   +
|     >   >   >   ^ |               |     >  
+---+---+---+---+---+---+---+---+---+---+---+