Question

Arising out of this question, I’m looking for an elegant (ruby) way to compute the word signature suggested in this answer.

The idea suggested is to sort the letters in the word, and also run length encode repeated letters. So, for example ‘mississippi’ first becomes ‘iiiimppssss’, and then could be further shortened by encoding as ‘4impp4s’.

I’m relatively new to ruby and though I could hack something together, I’m sure this is a one liner for somebody with more experience of ruby. I’d be interested to see people’s approaches and improve my ruby knowledge.

edit: to clarify, performance of computing the signature doesn’t much matter for my application. I’m looking to compute the signature so I can store it with each word in a large database of words (450K words), then query for words which have the same signature (i.e. all anagrams of a given word, that are actual english words). Hence the focus on space. The ‘elegant’ part is just to satisfy my curiosity.

Answer 1

I’m not much of a Ruby person either, but as I noted on the other comment this seems to work for the algorithm described.

s = 'mississippi' s.split('').sort.join.gsub(/(.)\1{2,}/) { |s| s.length.to_s + s[0,1] }  

Of course, you’ll want to make sure the word is lowercase, doesn’t contain numbers, etc.

As requested, I’ll try to explain the code. Please forgive me if I don’t get all of the Ruby or reg ex terminology correct, but here goes.

I think the split/sort/join part is pretty straightforward. The interesting part for me starts at the call to gsub. This will replace a substring that matches the regular expression with the return value from the block that follows it. The reg ex finds any character and creates a backreference. That’s the ‘(.)’ part. Then, we continue the matching process using the backreference ‘\1’ that evaluates to whatever character was found by the first part of the match. We want that character to be found a minimum of two more times for a total minimum number of occurrences of three. This is done using the quantifier ‘{2,}’.

If a match is found, the matching substring is then passed to the next block of code as an argument thanks to the ‘|s|’ part. Finally, we use the string equivalent of the matching substring’s length and append to it whatever character makes up that substring (they should all be the same) and return the concatenated value. The returned value replaces the original matching substring. The whole process continues until nothing is left to match since it’s a global substitution on the original string.

I apologize if that’s confusing. As is often the case, it’s easier for me to visualize the solution than to explain it clearly.