Maze solving
You are encouraged to solve this task according to the task description, using any language you may know.
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.
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 'a12' to the bottom-righ exit 'p1':
: (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
Tcl
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 {} {
set visited [lrepeat $x [lrepeat $y 0]] set queue {0 0 {}} while 1 { if {[llength $queue] == 0} { error "cannot reach finish" } set queue [lassign $queue cx cy path] lset visited $cx $cy 1 lappend path $cx $cy if {$cx == $x-1 && $cy == $y-1} break foreach {dx dy} {0 1 1 0 0 -1 -1 0} { if { [set nx [expr {$cx + $dx}]] >= 0 && $nx < $x && [set ny [expr {$cy + $dy}]] >= 0 && $ny < $y && !idx($visited, $nx, $ny) && ($dx && idx($verti, min($cx,$nx), $cy) || $dy && idx($horiz, $cx, min($cy,$ny))) } then { lappend queue $nx $ny $path } } } 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 $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 | +---+ +---+---+ +---+---+---+ + + + | > > > ^ | | > +---+---+---+---+---+---+---+---+---+---+---+