Determine if string can be split up into words

[ArmedMonkey]

I need to write an algorithm using dynamic programming to take an input in the form of a string without spaces or punctuation and return

a) if it can be split up into a sequence of words. "themailer" -> true; "thdfapple" -> false

b) the sequence produced. "the", "mailer"

In the dynamic programming approach, you have to find a subproblem... I've tried several with no avail... i feel like I'm closer right now with saying the subproblem is the longest word adjacent to the current position i'm testing

My general approach is to fill a table that is n x n long where n is the length of the string, the horizontal axis is the START of the substring and the vertical axis is the END of the substring (that i am testing)

Can anyone offer some insight / help on doing this? I realize this is highly esoteric and probably not explained very well... Sorry

[Antaris Veteran]

Hmm, that looks to be quite a challenge. You'd need some sort of dictionary to compare valid words to, but it might get a little interesting when you encounter words which could exist within other words, e.g. "ten" is a valid word, but so is "tenth". Unless you have a defined set of words and your work backwards (longest -> shortest) and break the string down that way? Never done anything like this before, so it would be interesting to see how you do it...

[ArmedMonkey]

well, i have a dictionary lookup. that's not the problem. And yes, words can occur in the beginning, end or middle of other words.

[Argote]

If you want to do this with DYNAMIC PROGRAMMING you basically have to keep pointers of where your current words start/end and try to find a matching item. If the last word doesn't fit anywhere then try to move the previous one and so on.

[AdamLC]

  On 03/03/2010 at 20:38, Argote said:

If you want to do this with DYNAMIC PROGRAMMING you basically have to keep pointers of where your current words start/end and try to find a matching item. If the last word doesn't fit anywhere then try to move the previous one and so on.

That sounds like a good approach! Rather you than me though! Sounds very tricky indeed :rolleyes:

[ArmedMonkey]

The constraint is that it has to be done in O(n^2) time. Not quite sure what you mean Argote, but if your algorithm has to go back, then you might end up with n^n time because for every letter, you can have n mistakes.

I'm glad you seem to be familiar with dynamic programming though. I think I have a solution; perhaps you can verify that what I have runs in O(n^2) and is "dynamic programming"

pseudo python... the check to see if something is a word is considered to take unit time (that's just how it is.)

for start in range(0, length):
     for end in range(start+1, length):
         isWord = #check the current substring based on start and end.
         if isWord: table[start][end] = substring length

This should take n^2 time, according to a source, but it takes a little less since it only covers half of a table

due to the fact that end >= start is a constraint. This basically goes through all the substrings:

[a]bcd

[ab]cd

[abc]d

acd

a[bc]d (etc)

and when it hits one that is a word, it puts the length of that word into that cell (so to get the word, you trace the table UP that many rows)

I then propose to do a DEPTH FIRST SEARCH from the end of this table back, (as if it were a graph).

If I can get to the start state, then I can split the string up (and will have actually built it by then).

Depth First Search is O(|V| + |E|)... |V| = n, and |E| seems to be constrained to be n(n-1)=n^2-n in this case.

You have n nodes, and each node can have n-1 edges since it's basically an undirected graph.

so I have O(n^2 + n^2 -n) = O(n^2) runtime, I THINK....

please feel free to correct me or comment on it. I'm not sure if this is actually dynamic programming since I have to do a multipass (the DFS after I fill in the table)

This is my test string, by the way.... I hammered it out by combining interesting cases... it covers things such as choices like "than" or "thank" or "an"... "mail" or "mailer" or "email"

  thankillthemailer
t 00000000000000000
h 00000000000000000
a 00000000000000000
n 40000000000000000
k 50000000000000000
i 00000000000000000
l 00000000000000000
l 00004300000000000
t 00000000000000000
h 00000000000000000
e 00000000320000000
m 00000000430000000
a 00000000000000000
i 00000000000000000
l 00000000005400000
e 00000000000000000
r 00000000000600000

so the columns represent the START and the rows represent the END from my algorithm, so if you look at table['k','l'], you get a 4, which means that the 'l' is the end of a 4 letter word. (yes, i realize there are two l's, but it's easier to explain that way. i'm obviously talking about the second l)

[Xilo]

You seem to even to account for multiple results:

than kill the mailer

thank ill the mailer

Would you output both of those? Are you accounting that in the O() equation (might not matter thought...)

[ArmedMonkey]

I suppose I could, but that is what is making me wonder if the algorithm is to run sufficiently fast.

[Dustin Fineout]

There is a much less complex and more elegant solution.

You propose to finish with a DFS working backwards to locate all solutions (paths through the directed graph back to the start). This itself is very close to the more elegant solution I will attempt to describe. We are indeed going to be starting at the end of the string rather than the beginning.

Let us call a string that has been successfully split into words a "sentence". Consider that every sentence must end with a word. Working from the end of the string forward, we will construct all valid sentences recursively.

Rather than iterating through all end and then all start positions, let us consider our end position static - we simply always end with the end of our "current" string as we are not interested in words ending before the end of our current string because those solutions don't provide a valid word at the end of the sentence. We will thus iterate only the start position, working backwards from end-1 to 0. If we ever get to start = 0 without finding a valid word, then there aren't any valid sentences because there are no valid words that the sentence can end with. We don't need to iterate again with a new end position, that would be wasted computation.

As we work backwards by start position, each time we encounter a valid word we will recurse with the substring preceding the word, and then prepend each result from the recursion to our "current" valid word to generate an array of our valid solution sentences. We don't recurse if start is 0 as we have reached the beginning of the string and can simply add the valid word to our array of solutions and return.

In pseudo-code (I have a running solution in PHP that I can generalize and post if there is interest, though it is much better left as an exercise for the reader)

 function find_sentences(string) {
     sentences = [];
     end = length(string) - 1;
     for (start = end; start >= 0; start--) {
         sublength = end - start + 1;
         substring = substring(string, start, sublength);
         if (is_word(substring)) {
             if (start > 0) {
                 next = substring(string, 0, start);
                 recurse = find_sentences(next);
                 foreach (recurse as result) {
                     sentences[] = result + ' ' + substring;
                 }
             } else {
                 sentences[] = substring;
             }
         }
     }
     return sentences;
 }

Enjoy! Any questions, I'll be happy to answer as best I can.

Also, thanks to ArmedMonkey - when I originally began on a solution for a similar problem, it was your solution that 'tipped' me off to the solution I described above.

[ViZioN]

Whilst I don't have anything more constructive to say to contribute to the topic itself, nice first post, Dustin :)