From: Colin Bartlett <colinb2r@...>
Date: 2009-05-19T00:49:18+09:00
Subject: Re: permute each element of a ragged array?

Rick DeNatale:

> On Mon, May 18, 2009 at 10:35 AM, Mark Thomas <mark@thomaszone.com> wrote:
>> Not sure why the test must be as written, because the order doesn't
>> fall in a logical pattern (as far as I can tell).

> Exactly, I don't see a general way to see a pattern when the
> assertions are expressed as indexed elements equaling specific values.

Given the clarification, I think there is a pattern:
the element of the first set is varying slowest in the examples,
with the empty set being returned last,
but I assume that that is not absolutely necessary?

To summarise, using the example given, we have:
* one non-recursive solution which varies the element of the first set slowest;
two recursive (therefore more elegant) solutions:
* one recursive solution (RDN) which varies the element of the first
set fastest;
* one recursive solution (MT) which varies the element of the first set slowest
  (returns array of strings rather than array of arrays, but easily modified?);

All three produce the same results (albeit not in the same order) for:
  sets = [
          %w( android hero ),
          %w( insane clown posse ),
          %w( phenomenauts ),
         ]

But using:
  sets = [ %w( android ),
           %w( insane clown ) ]

ruby 1.8.6 (2007-09-24 patchlevel 111) [i386-mswin32]

[["android"], ["insane", "clown"]]

non-recursive
["android", "insane"]
["android", "clown"]
["android"]
["insane"]
["clown"]
[]

RDN recursive
["android", "insane"]
["insane"]
["android", "clown"]
["clown"]
["android"]
[]

MT recursive
"android insaneclown"
"android"
"insaneclown"
""