From: Mark Hubbart <discordantus@...>
Date: 2004-11-10T08:08:49+09:00
Subject: Re: Iterating trough hash

On Wed, 10 Nov 2004 06:51:50 +0900, Brian Schr�der
<ruby@brian-schroeder.de> wrote:
> On Wed, 10 Nov 2004 06:41:08 +0900
> 
> 
> Michael DeHaan <michael.dehaan@gmail.com> wrote:
> 
> > Hmm, that's not a good bug there, so I'll pull the "history" thing out
> > (I was trying to be too clever)...
> >
> > module Enumerable
> >  def each_permutation
> >    each do |item1|
> >      each do |item2|
> >          yield [item1,item2] unless (item1 === item2)
> >      end
> >    end
> >  end
> > end
> >
> >
> 
> Hello Michael,
> 
> I think, that this is not each_permutation, but each_ordered_pair_of_elements_in_the_set_except_for_equal_elements.
> 
> each_permutation would be all shuffles of the original enumerable.
> 
> And it is not what the OP wanted, as he wanted all unordered pairs. So maybe

oops. I'll try to correct myself now...

The method I defined should be called each_combination_pair; they
weren't permutations at all.

So, I submit a revised method:

module Enumerable
  def combinations(min=0, max = nil)
    size = 0
    com = [[]]
    each do |item1|
      size += 1
      com += com.map{|item2| item2.dup << item1 }
    end
    com.select{|n| (min..max||size) === n.size}
  end
end

Then:

h.combinations(2,2).each do |(person1, age1), (person2, age2)|
  puts "#{person1} shakes hands with #{person2}." +
    "Together they are #{age1+age2} years old."
end

I'm sure #combinations could be optimized to not bother generating
combinations of the wrong size, but it's already questionable whether
the method is short to justify replacing the original code, which is
probably more efficient :)

cheers,
Mark