[SUMMARY] Mix and Match (#186)

_This week's summary is provided by Peter, originally [posted to his  
blog][1]. I am providing it here with permission._

[1]: http://www.rubyrailways.com/ruby-quiz-mix-and-match/


In my interpretation, this is a simple combinatorial problem: say the  
number of recipients is `r` and candles per recipient is `c`, then you  
are looking for a (preferably non-repeating) random selection of `r`  
elements of `c`-combinations of the original set of candles. (In fact  
it痴 a bit more complicated than that: the `c`-combinations have to be  
recalculated from the remaining candles each time you give away a  
group of candles, so we値l get to that). Sounds confusing? Don稚  
worry, after the implementation everything will be clear!

So first, define a k-combination for a histogram (a Hash like candles  
above, where keys are elements and values are cardinalities):

     class Hash
       def comb(group_size)
         result = []
         inner_comb = lambda do |head,tail|
           tail[0..-(group_size-head.size)].each do |e|
             if (head.size >= group_size-1)
               tail.each {|t| result << head + [t]}
             else
               inner_comb[head + [e], tail[tail.index(e)+1..-1]]
             end
           end
         end
         inner_comb[[],self.inject([]) {|a,v| v[1].times{a << v[0]}; a}]
         result.uniq
       end

For example:

     candles = { :orange   => 2,
                 :vanilla  => 1,
                 :lavender => 1,
                 :garden => 1 }

     pp candles.comb(3)

     => [[:lavender, :garden, :orange],
         [:lavender, :garden, :vanilla],
         [:lavender, :orange, :orange],
         [:lavender, :orange, :vanilla],
         [:garden, :orange, :orange],
         [:garden, :orange, :vanilla],
         [:orange, :orange, :vanilla]]

So, for a set of candles, this method generates all possible 3- 
combinations of the candles. We can then pick one and assign it to one  
of the recipients. Then recalculate the above from the remaining  
candles, give it to the next recipient - and so on and so forth.  
That痴 the basic idea, but we also need to ensure the candle  
combinations are as non-repeating as possible. So let痴 define some  
further utility methods:

     class Hash
       def remove_set(set)
         set.each {|e| self[e] -= 1}
       end
     end

The above code adjusts the number of candles in the original hash once  
we give away some of them. So for example:

     candles = { :orange   => 2,
                 :vanilla  => 1,
                 :lavender => 1,
                 :garden => 1 }

     candles.remove_set([:orange,:orange,:lavender])
     p candles
     => {:lavender=>0, :garden=>1, :orange=>0, :vanilla=>1}

And some Array extensions:

     class Array
       def rand
         uniqs = self.select{|e| e.uniq.size == e.size}
       uniqs.empty? ? self[Kernel.rand(length)] :  
uniqs[Kernel.rand(uniqs.length)]
       end

       def unordered_include?(other)
         self.map{|e| e.map{|s| s.to_s}.sort}.include? other.map{|s|  
s.to_s}.sort
       end
     end

`Array#rand` is trying to pick a random non-repeating combination if  
there is one (e.g. `[:orange, :lavender, :garden]`) or, if there is no  
such combination, then just a random one (e.g.  
`[:orange, :orange, :garden]` - orange is repeating, but we have no  
other choice).

`Array#unordered_include?` is like normal `Array#include?`, but  
disregards the ordering of the elements. So for example:

     [[:lavender, :garden, :orange]].include?  
[:lavender, :orange, :garden] => false
     [[:lavender, :garden, :orange]].unordered_include?  
[:lavender, :orange, :garden] => true

Hmmit would have been much more effective to use a set here rather  
than the above CPU-sucker, but now I am lazy to change it. ;-)

OK, so finally for the solution:

     ERROR_STRING = "The number of recipients times the number of  
candles per recipient is more than the supply!"

     def mix_and_match(candles, recipients, candles_per_recipient)
       return ERROR_STRING if ((candles.values.inject{|a,v| a+v}) <  
(recipients.size * candles_per_recipient))
       candle_set = recipients.inject({}) do |a,v|
         tried = []
         tries = 0
         loop do
           random_pick = candles.comb(candles_per_recipient).rand
           tried << random_pick unless tried.unordered_include?  
random_pick
           break unless a.values.unordered_include? random_pick
           break if (tries+=1) > candles.values.size * 2
         end
         candles.remove_set(tried.last)
         a[v] = tried.last
         a
       end
       candle_set.merge({:extra => candles})
     end

So, in the inner loop we randomly pick a candles-per-recipient- 
combination of all the possible combinations; If no one has that combo  
yet, we assign it to the next recipient. If someone has it already, we  
try to find an unique combination (loop on), unless it is impossible.  
In this case we simply start giving out any combinations. Once we give  
away a set of candles, we remove them from the original set. Easy-peasy.

You can check out the source code here.

This was a great quiz, too bad that not many people took a stab at it  
(so far 1 except me ;-)). The hardest part for me was the  
implementation of the k-combination (and the result looks awful to me  
- I didn稚 check any algorithm/pseudocode/other solution though, I  
wanted to roll my own) - after that the problem was pretty simple.