From: "Eric I." <rubytraining@...>
Date: 2007-01-15T00:40:09+09:00
Subject: Re: Number Spiral (#109) [solution]

Here's my solution to the quiz.  I used a recursive solution.  An
odd-sized spiral is the smaller-by-one even-sized spiral with a number
added to the end of each row, and a new row across the bottom.  And an
even-sized spiral is the smaller-by-one odd-sized spiral with a number
added to the beginning of each row and a new row across the top.

I decided to make my solution use relatively few lines of code.
However in doing that I added some inefficiencies, where a value might
be calculated multiple times in a loop/iterator when it would have been
better to calculate it once before the loop/iterator, and use that
stored value in the loop/iterator.

With respect to not building the solution in an array and then
displaying the array, I read that to mean not creating a
two-dimensional array (i.e., array of arrays) in which to build the
entire spiral.  I assemble each line of output in an array before
displaying that line, but each line is displayed before any subsequent
lines are calculated.  The solution could be adapted to avoid even the
one-dimensional array.

Eric
----------------
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================

def odd_spiral(size, row)
  if row < size - 1 : even_spiral(size - 1, row) << (size - 1)**2 + row
  else (0...size).collect { |n| size**2 - 1 - n }
  end
end

def even_spiral(size, row)
  if row == 0 : (0...size).collect { |n| size**2 - size + n }
  else odd_spiral(size - 1, row - 1).unshift(size**2 - size - row)
  end
end

size = (ARGV[0] || 8).to_i
(0...size).each do |row|
  puts ((size % 2 == 0 ? even_spiral(size, row) : odd_spiral(size,
row)).
          map { |n| n.to_s.rjust((size**2 - 1).to_s.length) }.join("
"))
end