url_shorten.rb

def freq(letter)
 freq_dict = {"a" => 8.167,
  "b" => 1.492,
  "c" => 2.782,
  "d" => 4.253,
  "e" => 12.702,
  "f" => 2.228,
  "g" => 2.015,
  "h" => 6.094,
  "i" => 6.966,
  "j" => 0.153,
  "k" => 0.772,
  "l" => 4.025,
  "m" => 2.406,
  "n" => 6.749,
  "o" => 7.507,
  "p" => 1.929,
  "q" => 0.095,
  "r" => 5.987,
  "s" => 6.327,
  "t" => 9.056,
  "u" => 2.758,
  "v" => 0.978,
  "w" => 2.360,
  "x" => 0.150,
  "y" => 1.974,
  "z" => 0.074}
  freq_dict[letter] || 10 #hack so double-letters get a high score
end

 #shorten one piece of the url
def shorten(word,size)
  #leave intact if it's a number
  return word if word =~ /\A\d+\z/

   #pull out duplicates and put letters in an array
  # e.g., mississippi ==> ["m","i","ss","i","ss","i","pp","i"]
  letter_arr = word.scan(/((\w)\2*)/).collect{|a| a.first}

   #grab the indices so we can reassemble
  with_indices = []
  letter_arr[1..-2].each_with_index {|v,i| with_indices << [v,i]}

   #sort based on frequencies (giving double letters highest priority)
  with_indices.sort!{|a,b| freq(a[0]) <=> freq(b[0])}

   #prune the word down
  1.upto(2+with_indices.length-size) { with_indices.pop}

   #and reassemble the word
  with_indices.sort!{|a,b| a[1] <=> b[1]}
  letter_arr[0]+
    with_indices.collect {|v,index| v.length > 1 ? v[0..0] : v}.join +
    letter_arr.last
end

 #the main function just shortens all parts of the path after the host:
def shorten_url(url,host,size)
  to_shorten = url.split(host)[1]
  host + to_shorten.split("/").collect{|elem| shorten(elem,size)}.join("/")
end