Explanation of Ruby code for building Trie data structures

Question

So I have this ruby code I grabbed from wikipedia and I modified a bit:

@trie = Hash.new()
def build(str)
    node = @trie
    str.each_char { |ch|
      cur = ch
      prev_node = node
      node = node[cur]
      if node == nil
        prev_node[cur] = Hash.new()
        node = prev_node[cur]
      end
    }
  end

build('dogs')

puts @trie.inspect

I first ran this on console irb, and each time I output node, it just keeps giving me an empty hash each time {}, but when I actually invoke that function build with parameter 'dogs' string, it actually does work, and outputs {"d"=>{"o"=>{"g"=>{"s"=>{}}}}}, which is totally correct.

This is probably more of a Ruby question than the actual question about how the algorithm works. I don't really have adequate Ruby knowledge to decipher what is going on there I guess.

Answer 1

You're probably getting lost inside that mess of code which takes an approach that seems a better fit for C++ than for Ruby. Here's the same thing in a more concise format that uses a special case Hash for storage:

class Trie < Hash
  def initialize
    # Ensure that this is not a special Hash by disallowing
    # initialization options.
    super
  end

  def build(string)
    string.chars.inject(self) do |h, char|
      h[char] ||= { }
    end
  end
end

It works exactly the same but doesn't have nearly the same mess with pointers and such:

trie = Trie.new
trie.build('dogs')
puts trie.inspect

Ruby's Enumerable module is full of amazingly useful methods like inject which is precisely what you want for a situation like this.

Answer 2

I think you are just using irb incorrectly. You should type the whole function in, then run it, and see if you get correct results. If it doesn't work, how about you post your entire IRB session here.

Also here is a simplified version of your code:

def build(str)
  node = @trie
  str.each_char do |ch|
    node = (node[ch] ||= {})
  end
  # not sure what the return value should be
end