From: Carl Porth <badcarl@...>
Date: 2007-04-09T07:12:32+09:00
Subject: Re: Getting to 100 (#119)

After going back and reading the current solutions, I like Ken Bloom's
each_partition method.  It's much cleaner than my combinations method.


On Apr 8, 2:59 pm, "Carl Porth" <badc...@gmail.com> wrote:
> here is my first pass:
>
> class Array
>   def combinations(n)
>     case n
>     when 0: []
>     when 1: self.map { |e| [e] }
>     when size: [self]
>     else
>       (0..(size - n)).to_a.inject([]) do |mem,i|
>         mem += self[(i+1)..size].combinations(n-1).map do |rest|
>           [self[i],*rest]
>         end
>       end
>     end
>   end
> end
>
> equations = 0
> separator = "************************"
>
> (1..8).to_a.combinations(3).each do |partitions|
>   3.times do |n|
>     equation = "123456789"
>
>     partitions.reverse.each_with_index do |partition,index|
>       equation = equation.insert(partition, (index == n ? ' + ' : ' -
> '))
>     end
>
>     result = eval(equation)
>     equation << " = #{result}"
>
>     if result == 100
>       equation = "#{separator}\n#{equation}\n#{separator}"
>     end
>
>     puts equation
>
>     equations += 1
>   end
> end
>
> puts "#{equations} possible equations tested"