Wow, these solutions are great fun to play with.  I think next week's quiz needs
to give me a little man icon and some controls!  Throw in some doors, some keys,
and little critters to chase me around and there's simply no chance at all I
would get a summary written next week.  Hmm, maybe it's not such a good idea.

Jokes aside, do run the solutions a few times each this week.  It's fun to see
what they build.  Then peek inside the code and read the comments.  Good stuff
in there.

Below, I want to look into Dominik Bathon's code.  It is a nice search and
lightning quick!  On my machine, it makes and solves quizzes faster than the
other solutions can just make them.  Even better, it uses a complex internal
representation (mainly for speed), yet still comes out with clean algorithms.  I
was quite impressed by that.

Let's get to the code.  Dominik starts off by defining a helper method in Hash:

	class Hash
	      # find the key for with the smallest value, delete it and return it
	      def delete_min_value
	            return nil if empty?
	            minkey=min=nil
	            each { |k, v|
	                  min, minkey=v, k if !min || v<min
	            }
	            delete(minkey)
	            minkey
	      end
	end
	
	# ...

The comment pretty much explains what's going on there.  Each pair of the Hash
is compared by value.  The pair with the lowest value is deleted and the key for
that value is returned.

On to the interesting parts.  Here's the start of the main class used by the
solution:

	# Maze represents the maze ;-)
	#
	# Cells/positions in the maze are represented by Numbers
	# (from 0 to w*h-1), each position corresponds to x/y coordinates,
	# you can convert between positions and coordinates by coord2pos
	# and pos2coord.
	#
	# The walls for each position are stored in the String @data. The walls
	# for position p are stored in the first two bits of @data[p], the
	# other bits are unused. If bit one is set then p has a north wall, if
	# bit two is set then p has a west wall.
	#
	# Maze#generate generates a (random) maze using the method described at
	# http://www.mazeworks.com/mazegen/mazetut/
	#
	# Maze#shortest_path uses Dijkstra's shortest path algorithm, so it can
	# not anly find shortest pathes in perfect mazes, but also in mazes
	# where different pathes between two position exist.

	class Maze
	      attr_reader :w, :h # width, height

	      def initialize(w, h)
	            @w, @h=[w, 1].max, [h, 1].max
	            @wh=@w*@h
	            @neighbors_cache={}
	            set_all_walls
	      end
	
	      # ...

I know that section is mostly a comment, but you'll want to read through it. 
It's interesting information and it introduces you to the internal format the
code uses.

After that, we see some readers defined and some simple initialization work. 
Set a width and height, ensuring they are both at least 1.  Nice use of max()
there.  Calculate width times height or the total number of cells, initialize a
cache and call set_all_walls().

That means we need some more code:

	      # ...
	
	      def set_all_walls
	            # set all bits
	            @data=3.chr * (@wh)
	            nil
	      end
	      def clear_all_walls
	            # all except outer border
	            @data=0.chr * (@wh)
	            # set north walls of row 0
	            w.times { |i| @data[i] |= 1 }
	            # set west walls of col 0
	            h.times { |i| @data[i*w] |= 2 }
	            nil
	      end

	      # ...

Okay, now we start to get tricky.  Remember the initial comment about using bits
for the walls.  We're only tracking two walls here, north and west.  Of course
cells can still have up to four walls, but your east wall is just your
neighbor's west wall and your south wall is the north wall for the cell below
you.

Now, what do you get if you turn two bits on?  3.  The set_all_walls() method
just translates that to a character and duplicates it for every cell.  That
gives us a String representing the entire maze with all the walls turned on.

That should make clear_all_walls() more obvious.  This time we want no walls so
we don't set any bits.  Translate 0 to a character and duplicate.  However, we
still need the edges of the maze.  All cells in the top row need a north wall
(set the 1 bit).  Then all the cells in the first column need a west wall (set
the 2 bit).  That makes up the rest of the method.

Ready for the next chunk?

	      # ...

	      # positions in path will be printed as "X"
	      def to_s(path=[])
	            ph={}
	            path.each { |i| ph[i]=true }
	            res=""
	            h.times { |y|
	                  w.times { |x|
	                        res << "+" << ( (@data[y*w+x] & 1 > 0) ? "---" :
	                                                                 "   " )
	                  }
	                  res << "+\n"
	                  w.times { |x|
	                        res << ((@data[y*w+x] & 2 > 0) ? "|" : " ")
	                        res << (ph[y*w+x] ? " X " : "   ")
	                  }
	                  res << "|\n"
	            }
	            res << ("+---"*w) << "+"
	      end
	      def inspect
	            "#<#{self.class.name} #{w}x#{h}>"
	      end

	      # ...

The to_s() method draws mazes.  The first two lines fill a Hash with the
solution path, if one is given.  The Hash is indexed identically as the maze
String and values can be true (if it's on the path) or the default nil, (when
it's not).

The rest of that method does the drawing.  It walks row by row with h.times(),
down the maze drawing cells.  The first w.times() call handles the north walls. 
First it adds a "+", then it adds "---" if the 1 bit is set or "   " if it's 
not.  Next we need another "+" and a "\n".  Now the second w.times() block
handles the west wall and path.  First it checks to see if the 2 bit is set for
the current cell, outputting "|" if it is and " " if it's not.  Then the path is
checked.  If this cell is on the path, it's filled with " X " and if it's not,
the code adds a "   ".

The last two lines of the method are important.  They ensure a final "|" is
always added to the end of a row and a final "+---" is placed at the end each
column of the maze.  This handles the east and south borders of the maze, which
are not covered by the bits.

The other method, inspect(), just returns a class name, width and height.

	      # ...
	      
	      # maze positions are cell indices from 0 to w*h-1
	      # the following functions do conversions to and from coordinates
	      def coord2pos(x, y)
	            (y % h)*w+(x % w)
	      end
	      def pos2coord(p)
	            [p % w, (p/w) % h]
	      end

	      # ...

These convertors were explained in the initial comment and they are explained
again here.  No surprises there.

	      # returns valid neighbors to p, doesn't care about walls
	      def neighbors(p)
	            if ce=@neighbors_cache[p]; return ce; end
	            res=[p-w, p+w]
	            res << p-1 if p%w > 0
	            res << p+1 if p%w < w-1
	            @neighbors_cache[p] = res.find_all { |t| t>=0 && t<@wh }
	      end

This returns the indices of the up to four neighboring cells.  It caches this
lookup the first time it does it, since it will never change.  The first line
just uses the cache if it has already been figured.  The second line adds the
cell above and the cell below.  Note that these numbers are found by simple math
and could be outside the bounds of the maze.  The next two lines add the left
and right cells.  We're more careful with our math here, because a wrong answer
could look right:  The last cell of the first row is "left" of the first cell of
the second row, in our one dimensional String that holds the maze data.  The
final line, stores the indices to the cache and returns them, after using
find_all() to eliminate any bogus number that crept in.

	      # ...
      
	      def wall_between?(p1, p2)
	            p1, p2=[p1, p2].sort
	            if p2-p1==w # check north wall of p2
	                  @data[p2] & 1 > 0
	            elsif p2-p1==1 # check west wall of p2
	                  @data[p2] & 2 > 0
	            else
	                  false
	            end
	      end
	      def set_wall(p1, p2)
	            p1, p2=[p1, p2].sort
	            if p2-p1==w # set north wall of p2
	                  @data[p2] |= 1
	            elsif p2-p1==1 # set west wall of p2
	                  @data[p2] |= 2
	            end
	            nil
	      end
	      def unset_wall(p1, p2)
	            p1, p2=[p1, p2].sort
	            if p2-p1==w # unset north wall of p2
	                  @data[p2] &= ~1
	            elsif p2-p1==1 # unset west wall of p2
	                  @data[p2] &= ~2
	            end
	            nil
	      end
      
	      # ...

These three methods are all very similar.  Given two cells, the first checks if
there is a wall between them, the second sets the wall between them, and the
third unsets it.  The if's just figure out if we are talking about a north wall
or a west wall.  The rest is bit testing or setting.

On to maze generation:

	      # ...
	
	      # generate a (random) perfect maze
	      def generate(random=true)
	            set_all_walls
	            # (random) depth first search method
	            visited={0 => true}
	            stack=[0]
	            until stack.empty?
	                  n=neighbors(stack.last).reject { |p| visited[p] }
	                  if n.empty?
	                        stack.pop
	                  else
	                        # choose one unvisited neighbor
	                        np=n[random ? rand(n.size) : 0]
	                        unset_wall(stack.last, np)
	                        visited[np]=true
	                        # if all neighbors are visited then here is
	                        # nothing left to do
	                        stack.pop if n.size==1
	                        stack.push np
	                  end
	            end
	            self
	      end

	      # ...

This algorithm came out very clean, I think.  Not a bit operation in sight. 
First it turns all the walls on.  Then it sets up an Array for tracking visited
cells and another as a stack to drive the process.  While there is something on
the stack, the code looks at each not-yet-visited neighbor.  If there are no
neighbors in that set, the stack is popped and the routine moves on.  However,
if there are, one is chosen at random and the wall is knocked out between them. 
If that neighbor was the last unvisited one for this cell, the code pops the
current cell off the stack.  The neighbor cell is set to visited and pushed onto
the stack, moving the build process to that location for the next iteration.

That covers creation.  Now we need a solver:

	      # ...
	
	      # central part of Dijkstra's shortest path algorithm:
	      # returns a hash that associates each reachable (from start)
	      # position p, with the previous position on the shortest path
	      # from start to p and the length of that path.
	      # example: if the shortest path from 0 to 2 is [0, 1, 2], then
	      # prev[2]==[1, 2], prev[1]==[0, 1] and prev[0]==[nil, 0].
	      # so you can get all shortest paths from start to each reachable
	      # position out of the returned hash.
	      # if stop_at!=nil the method stops when the previous cell on the
	      # shortest path from start to stop_at is found.
	      def build_prev_hash(start, stop_at=nil)
	            prev={start=>[nil, 0]} # hash to be returned
	            return prev if stop_at==start
	            # positions which we have seen, but we are not yet sure about
	            # the shortest path to them (the value is length of the path,
	            # for delete_min_value):
	            active={start=>0}
	            until active.empty?
	                  # get the position with the shortest path from the
	                  # active list
	                  cur=active.delete_min_value
	                  return prev if cur==stop_at
	                  newlength=prev[cur][1]+1 # path to cur length + 1
	                  # for all reachable neighbors of cur, check if we found
	                  # a shorter path to them
	                  neighbors(cur).each { |n|
	                        # ignore unreachable
	                        next if wall_between?(cur, n)
	                        if old=prev[n] # was n already visited
	                              # if we found a longer path, ignore it
	                              next if newlength>=old[1]
	                        end
	                        # (re)add new position to active list
	                        active[n]=newlength
	                        # set new prev and length
	                        prev[n]=[cur, newlength]
	                  }
	            end
	            prev
	      end

	      # ...

I really don't think I need to launch into too deep an explanation here as the
comments guide you right through it.  The short story is that this method
branches out from a starting cell, walking to each neighbor and always counting
its steps.  While doing this, it is building the Hash described in the first
comment, which points to the cell that came before on the shortest path.  Using
that Hash, returned by this method, you can easily construct the shortest path
to any cell the algorithm visited.  Handy stuff!  Let's see how it gets put to
use:

	      # ...
      
	      def shortest_path(from, to)
	            prev=build_prev_hash(from, to)
	            if prev[to]
	                  # path found, build it by following the prev hash from
	                  # "to" to "from"
	                  path=[to]
	                  path.unshift(to) while to=prev[to][0]
	                  path
	            else
	                  nil
	            end
	      end
      
	      # ...

Given a starting and ending cell, this returns just what the name implies.  It
builds the magic Hash we just looked at on the first line, then just walks the
path in reverse until it reaches the start (nil in the Hash).  Again, clean and
simple.  Nice coding Dominik.

Let's look at the other search the code provides:

	      # ...

	      # finds the longest shortest path in this maze, only works if
	      # there is at least one position that can only reach one
	      # neighbor, because we search only starting at those positions.
	      def longest_shortest_path
	            startp=endp=nil
	            max=-1
	            @wh.times { |p|
	                  # if current p can only reach 1 neighbor
	                  if neighbors(p).reject { |n|
	                           wall_between?(p, n)
	                     }.size==1
	                        prev=build_prev_hash(p)
	                        # search longest path from p
	                        tend, tmax=nil, -1
	                        prev.each { |k, v|
	                              if v[1]>tmax
	                                    tend=k
	                                    tmax=v[1]
	                              end
	                        }
	                        if tmax>max
	                              max=tmax
	                              startp, endp=p, tend
	                        end
	                  end
	            }
	            if startp # path found
	                  shortest_path(startp, endp)
	            else
	                  nil
	            end
	      end
	end

	# ...

This method walks the maze, looking for cells that are dead-ends.  From each of
those, it builds the path Hash and checks the lengths of each path found.  In
the end, it will return the longest path it saw.

Just a little more code is needed for human interface:

	# ...

	if $0 == __FILE__
	      ARGV.shift if search_longest=ARGV[0]=="-l"
	      w, h, from, to=ARGV
	      m=Maze.new(w.to_i, h.to_i)
	      m.generate
	      puts "Maze:", m.to_s
	      if from=~/(\d+),(\d+)/
	            p1=m.coord2pos($1.to_i, $2.to_i)
	      else
	            p1=rand(m.w*m.h)
	      end
	      if to=~/(\d+),(\d+)/
	            p2=m.coord2pos($1.to_i, $2.to_i)
	      else
	            p2=rand(m.w*m.h)
	      end

	      path=m.shortest_path(p1, p2)
	      puts "\nShortest path from #{m.pos2coord(p1).inspect} to " \
	      "#{m.pos2coord(p2).inspect}:", m.to_s(path)

	      if search_longest
	            path=m.longest_shortest_path
	            puts "\nLongest shortest path (from " \
	            "#{m.pos2coord(path[0]).inspect} to " \
	            "#{m.pos2coord(path[-1]).inspect}:",
	            m.to_s(path)
	      end
	end

This is just option parsing and display.  The code checks for a special first
"-l" option, which just sets a flag to add the long search.

The next chunk reads a width and height then builds and displays a maze of the
indicated size.   The code next reads from and to cells for a solution search,
if they where provided.  Random coordinates are used when from or to cells are
absent.  Note the use of the coord2pos() convertor in here.

Finally, the shortest path is displayed.  The longer search is also added, if
requested.  Dominik uses an unusual Ruby idiom here, "string" "string".  Ruby
will concatenate these, even without the + between them.  (I didn't know this!) 
However, the rumor is that this feature may vanish in a future version of Ruby,
so it's probably not a good habit to get into.

My thanks to those who braved the mazes this week.  Really interesting (and
fun!) solutions were given by all.

Tomorrow's quiz is a little client and server fun, care of Pat Eyler's
children...