My answers assumes an odd numbered spirals (I inferred it from "The number zero represents the center of the spiral"). Sorry for my beginners ruby (are there some standard min(x,y)/max(x,y,) functions?) The approach is to work out a standard equation for the value in any cell (I ended up with two, for the top left and bottom right) and then to iterate through each cell and calculate the value. The algorithm is stateless. class SpiralMaker def make_spiral(size) # only allow odd numbered squares (as zero is centre) if (size.modulo(2) == 0) exit(1) end #step along row (1..size).each do |y| # step down columns (1..size).each do |x| # are we in top left or bottom right half of spiral? if (y+x <= size) # top left - calculate value sn = size - (2 * (min(x,y) - 1)) val = (sn*sn) - (3*sn) + 2 - y + x else # bottom right - calculate value sn = size - (2 * (size - max(x,y))) val = (sn*sn) - sn + y - x end # Print value STDOUT.printf "%03d ", val end # Next line STDOUT.print "\n" end end def min(a,b) (a <= b) ? a : b end def max(a,b) (a >= b) ? a : b end end maker = SpiralMaker.new maker.make_spiral 21

on 2007-01-19 16:31

on 2007-09-25 22:38

I think the standard idiom is to use the min/max functions of an array: a = 5 b = 10 [a, b].max #=> 10 You can also give max a block, similar to sort: a = "Hello" b = "Hi" [a, b].max {|x, y| x.length <=> y.length} Or you can write a method so it works more like sort_by (the interface, not the implementation): class Array def max_by &blk max {|a, b| blk.call(a) <=> blk.call(b)} end end a = "Hello" b = "Hi" [a, b].max {|x| x.length} And then, in Ruby 1.9, you should be able to do this (using max_by from above): a = "Hello" b = "Hi" [a, b].max(&:length) # Not sure if the syntax is 100% And if you still want your max() function: def min(*args) args.min end Everything also applies to minimums using the min function. Dan

on 2007-09-25 22:44

This one works for even and odd spirals (obvious really) class SpiralMaker def make_spiral(square_size) # allow for even numbered squares by missing off the last row and column size = square_size if (square_size.modulo(2) == 0) square_size = square_size+1 end #step along row (1..size).each do |y| # step down columns (1..size).each do |x| # are we in top left or bottom right half of spiral? if (y+x <= square_size) # top left - calculate value sn = square_size - (2 * (min(x,y) - 1)) val = (sn*sn) - (3*sn) + 2 - y + x else # bottom right - calculate value sn = square_size - (2 * (square_size - max(x,y))) val = (sn*sn) - sn + y - x end # Print value STDOUT.printf "%03d ", val end # Next line STDOUT.print "\n" end end def min(a,b) (a <= b) ? a : b end def max(a,b) (a >= b) ? a : b end end maker = SpiralMaker.new maker.make_spiral 3

on 2007-09-25 22:44

On Jan 14, 2007, at 1:46 PM, Tom Ayerst wrote: > My answers assumes an odd numbered spirals (I inferred it from "The > number zero represents the center of the spiral"). But the quiz example is an even spiral. ;) James Edward Gray II