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...

on 2007-05-24 14:19