From: Jim Haungs <jhaungs@...>
Date: 2004-07-12T01:17:26+09:00
Subject: Re: A little algorithmic help requested...

Here's an iterator approach. 

The maximum group size and the splitting criterion can be
parameterized.

def groups(anArray, split)
	list = []
	anArray.each do | n |
		if list.empty? or split[list.last, n]
			list << n
		else
			yield list
			list = [n]
			end
		end
	yield list
end

def ranges(anArray, maxGroupSize, split)
	groups(anArray, split) do | group |
		if group.size > maxGroupSize then
			yield group
		else
			group.each {|g| yield [g]}
			end
		end
end

test = [1,2,3,4,6,7,8,11,12,15,16,17,18] 

maxGroupSize = 2
criterion = proc{|x,y| x.succ == y}
ranges(test, maxGroupSize, criterion) { | range | p range }
	
[1, 2, 3, 4]
[6, 7, 8]
[11]
[12]
[15, 16, 17, 18]

On Sun, 11 Jul 2004 16:10:34 +0900, Hal Fulton
<hal9000@hypermetrics.com> wrote:

>Here's a problem my tired brain is having trouble with.
>
>Given a sorted array of integers, convert them into as many
>ranges as possible (ranges of three or more).
>
>Example:
>[1,2,3,4,6,7,8,11,12,15,16,17] ==> [1..4,6..8,11,12,15..17]
>
>How would *you* do this?
>
>
>Thanks,
>Hal
>
>