From: Sander Land <sander.land@...>
Date: 2006-08-14T05:34:21+09:00
Subject: Re: [Solution] [QUIZ] Pen and Paper (#90)

My solution also uses the Warnsdorff heuristic (moving to the square
with the least amount of possible moves left).
It works very well for sizes under 100 or so, gets worse after that
and starts taking hundreds of tries at about n=175.

I also included an option to see how long tours are starting from each
square (give the script any second parameter to show this).
This shows how well (or poor) the algorithm works for that size, but
it's obviously quite slow for larger sizes as it runs the algorithm
for every square.


Pastie link: http://pastie.caboo.se/8427
Code:

module Enumerable
  def x(enum)  # Cartesian product
    map{|a| enum.map{|b| [a,b].flatten }}.inject([]) {|a,b| a+b}
  end
  def **(n) #  Cartesian power
    if n == 1
      self
    else
      self.x(self**(n-1))
    end
  end
end

module OpenStructable
  def method_missing(method,*args)
    (class<<self;self;end).send(:attr_accessor, method.to_s.sub('=','') )
    send(method,*args)
  end
end

class Vertex < Array
  include OpenStructable
  def initialize(name)
    self.name = name
  end
end

class Graph
  def initialize
    @vertices = Hash.new{|h,k| h[k] = Vertex.new(k) }
  end

  def add_edge(from,to)
    @vertices[from] << @vertices[to]
  end

  def warnsdorff_tour(start)
    @vertices.each_value{|v|
      v.score = v.size
      v.done = false
    }
    curr_node = @vertices[start]
    tour = []
    while curr_node
      tour << curr_node
      curr_node.done = true
      curr_node.each{|v| v.score -= 1}
      curr_node = curr_node.reject{|v| v.done}.sort_by{|v| v.score}.first
    end
    tour.map{|t| t.name}
  end

end

n = (ARGV[0] || 5).to_i
graph = Graph.new

dirs = [2,-2]**2 + [[0,3],[3,0],[0,-3],[-3,0]]

valid = 0...n

(valid**2).each {|x,y|
  dirs.each{|dx,dy|
    to_x, to_y =  x + dx, y + dy
    graph.add_edge([x,y] , [to_x,to_y]) if valid.include?(to_x) &&
valid.include?(to_y)
  }
}
grid = Array.new(n){Array.new(n)}

# This shows how long a tour starting from each square is
if ARGV[1]
  full_count = 0
  (valid**2).each{|x,y|
    grid[y][x] =  graph.warnsdorff_tour([x,y]).length
    full_count += 1 if grid[y][x] == n*n
  }
  puts "#{full_count} / #{n*n} = #{'%.2f' %
[100*full_count.to_f/(n*n)]} % of the possible starting positions give
a full tour."
else

solution = []
try = 0
([0,n-1]**2 + [0].x(1..n-2) + (1..n-2)**2).each{|sx,sy|  # for
starting square, try corners first, then one edge, then inside
  try += 1
  solution = graph.warnsdorff_tour([sx,sy])
  break if solution.length == n*n
}

if solution.length != n*n
  puts "Failed to find tour."
  exit!
end
puts "Found tour in #{try} tries."

solution.each_with_index{|(x,y),i| grid[y][x] = i+1 }
end

max_len = (n * n).to_s.length

sep = ('+' + '-' * (max_len+2) ) * n + '+'
puts sep, grid.map{|row| '| ' + row.map{|n| n.to_s.center(max_len)
}.join(' | ') + ' |' }.join("\n" + sep + "\n"), sep