From: Ruby Quiz <james@...>
Date: 2007-05-24T21:18:00+09:00
Subject: [SUMMARY] Magic Squares (#124)

I was pleasantly surprised by the number of people that tackled the extra credit
this time around.  Essentially, there are different algorithms for building
magic squares depending on the size of the square.  There are in fact three
different algorithms:  one for odd sizes, one for doubly even (divisible by 4)
sizes, and another for singly even (divisible by 2 but not 4) sizes.

One solution that did handle all three cases came from David Tran.  Let's dive
right into how David constructs the squares:

	class MagicSquare
	
	  def initialize(size = 3)
	    raise "Error: size must greater than 2." if size < 3
	    @magic_square = if (size % 2 != 0)
	      OddMagicSquare.new(size)
	    elsif (size % 4 == 0)
	      DoublyEvenMagicSquare.new(size)
	    else
	      SinglyEvenMagicSquare.new(size)
	    end
	  end
	  
	  # ...

Here we see the initialization deferred to various classes based on the
requested square size.  These are just the conditions for the various algorithms
I mentioned earlier.

One point of interest is that this solution won't construct a magic square of
size one, though they are legal:

	+---+
	| 1 |
	+---+

Let's see some of the other methods in this class:

	  # ...
	  
	  def size
	    @magic_square.size
	  end
	
	  def [](i,j)
	    @magic_square[i,j]
	  end
	  
	  # ...

These two methods just delegate to the inner square object.  Nothing tricky
there.

Next we have the pretty printer:

	  # ...
	  
	  def to_s
	    digits = (size * size).to_s.size
	    divider = '+' + '-' * ((digits + 2) * size + (size - 1)) + "+\n"
	    (0...size).inject(divider) do |s, i|
	      (0...size).inject(s + "|") do |s, j|
	        s + " #{self[i,j].to_s.rjust(digits)} |"
	      end + "\n" + divider
	    end
	  end
	  
	  # ...

Most solutions included a routine pretty similar to this.  You first have to
find the width of the largest number and assemble a properly size border.  Then
you can iterate over the rows and cells printing them at the proper width and
with borders between.

David also included a method that verifies his work:

	  # ...
	  
	  def is_magic_square?
	    size = self.size
	    n = size * size
	
	    array = Array.new(n)
	    (0...size).each do |i|
	      (0...size).each do |j|
	        index = self[i,j] - 1
	        return false if (index < 0) || (index >= n) || array[index]
	        array[index] = true
	      end
	    end
	    return false unless array.all?
	
	    sum = size * (size * size + 1) / 2
	    (0...size).each do |i|
	      return false if sum != (0...size).inject(0) { |s,j| s + self[i,j] }
	      return false if sum != (0...size).inject(0) { |s,j| s + self[j,i] }
	    end
	    return false if sum != (0...size).inject(0) { |s,i| s + self[i,i] }
	    return false if sum != (0...size).inject(0) { |s,i|
	      s + self[i, size-1-i]
	    }
	    true
	  end
	  
	  # ...

This method begins by calculating a few sizes.  It then launches into verifying
that all the numbers in the expected range were used.  This code works by
filling an Array the length of all the numbers with nils.  It then walks all
numbers of the square, replacing that index with a true value.  Finally it
checks that they are all true.  The rest of the method does the standard magic
square validation by row, column, and diagonal.

We're now ready to examine the three individual algorithms:

	  # ...
	  
	  private
	  class OddMagicSquare
	    attr_reader :size
	
	    def initialize(size)
	      @size = size
	      n = @size * @size
	      @array = Array.new(n)
	      i, j = 0, @size/2
	      (1..n).each do |v|
	        @array[get_index(i,j)] = v
	        a, b = i-1, j+1
	        i, j = self[a,b] ? [i+1, j] : [a, b]
	      end
	    end
	
	    def [](i, j)
	      @array[get_index(i,j)]
	    end
	
	    private
	    def get_index(i, j)
	      (i % @size) * @size + (j % @size)
	    end
	  end
	  
	  # ...

The algorithm for odd size magic squares is pretty straightforward.  You begin
by placing a one in the top center of the of the square.  From there you just
count, placing each number you come to in the square above and to the right of
the last square you filled.  The board "wraps" for these movements, so moving
off the top brings you to the bottom and moving off the right side returns you
to the left.  If a normal move would take you to a filled square, you drop one
square instead.

The above is a Ruby implementation of this algorithm.  The values are all
calculated at the time of object construction and stored in an instance
variable.  They can then be accessed at any time via the [] method which uses
row major indexing.

	  # ...
	  
	  class DoublyEvenMagicSquare
	    attr_reader :size
	
	    def initialize(size)
	      @size = size
	    end
	
	    def [](i, j)
	      i, j = i % @size, j % @size
	      value = (i * @size) + j + 1
	      i, j = i % 4, j % 4
	      ((i == j) || (i + j == 3)) ? (@size*@size+1-value) : value
	    end
	  end
	  
	  # ...

Doubly even squares use an easy algorithm that can calculate a given value given
just the coordinates.  Because of that, no effort is made to pre-calculate the
values here and all the work is done in the []() method.

This algorithm divides the overall grid into two kinds of squares.  One type is
all squares that land on any diagonal created by subdividing the grid into four
by four subgrids.  All other squares make up the other type.  Once you know
which type of square you are dealing with, simple counting, from left to right
and top to bottom, will give you the value of the square.  Diagonal squares
count down from the highest number and the other squares count up from one.

We have one more algorithm to go:

	  class SinglyEvenMagicSquare
	    attr_reader :size
	
	    L = [4, 1, 2, 3]
	    U = [1, 4, 2, 3]
	    X = [1, 4, 3, 2]
	
	    def initialize(size)
	      @size = size
	      @odd_magic_square = MagicSquare.new(@size/2)
	    end
	
	    def [](i, j)
	      i, j = i % @size, j % @size
	      ii, jj = i / 2, j / 2
	      center = @size / 2 / 2
	      value = @odd_magic_square[ii, jj]
	      case
	        when ii < center then L
	        when ii == center then (jj == center) ? U : L
	        when ii == center+1 then (jj == center) ? L : U
	        else X
	      end [i%2*2 + j%2] + 4 * (value - 1)
	    end
	  end
	end
	
	# ...

The final algorithm is the trickiest.  You divide the grid into two by two
subgrids.  Each subgrid is assigned a letter:  L, U, or X.  The first (size - 2)
/ 4 + 1 rows are L's, there's one row of U's, and the rest of the rows are X's. 
You also swap the center U with the L just above it.  The letters describe the
order you fill subgrids.  You can see these orders defined in constants at the
top of David's class.  The only other element you need to know is the order to
fill in the subgrids.  That is determined by building a size / 2 magic square
using the odd pattern described early.  The order of those numbers dictate the
order the subgrids are filled in.

The final piece of the puzzle is the application code:

	# ...
	
	if __FILE__ == $0
	  puts MagicSquare.new(ARGV[0].to_i)
	end

This just builds and prints the correct square object from the choices we have
been examining.  Note that to_s() is called implicitly by puts().

My thanks to all the brave souls who went above and beyond the quiz requirements
to show us great solutions.

Tomorrow we will play with some fractal fun...