This week’s summary is provided by Peter, originally posted to his
blog. I am providing it here with permission.
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’s 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’ll get to that). Sounds confusing? Don’t
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’s the basic idea, but we also need to ensure the candle
combinations are as non-repeating as possible. So let’s 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
Hmm… it 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’t check any algorithm/pseudocode/other solution though, I
wanted to roll my own) - after that the problem was pretty simple.