Here’s the scenario:

• You have a book, and some of the words are underlined.

• I hand you another book.

• You have to figure out if it is possible to start with your book, erase some of the non-underlined words, and end up with my book (or at least, the same words in the same order as my book).

The straightforward “greedy” algorithm takes exponential time, and I’d like to do better.

It only takes linear time to check that the words of my book are a subsequence of the words in your book. It also only takes linear time to check that the underlined words in your book are a subsequence of the words in my book. Yet even with both of those conditions satisfied, I can’t see a good way to tackle the original problem.

## Looking for an algorithm to test for abridgement

**Moderators:** phlip, Moderators General, Prelates

### Looking for an algorithm to test for abridgement

wee free kings

### Re: Looking for an algorithm to test for abridgement

Here is a first attempt:

Name the words of the book as X_i for the ith word in the first book (which contains underlined words), and Y_j for the jth word in the second book. Let U(i) denote whether X_i is underlined.

Define A(i,j) to be a boolean function such that A(i,j) is true when X[i:] can be abridged to Y[j:] (in other words, consider all words after and including word X_i, and Y_j -- with 0 indexing).

So the answer to our question is given by A(0,0)

Then we can compute A(0,0) in time |X|*|Y| as follows:

A(|X|, |Y|) = True

A(|X|, z) = False, for z < |Y|

Other than these special cases, we have the following cases

A(i,j) = ...

A(i+1, j+1) if U(i) and X_i = Y_j

False if U(i) and X_i != Y_j

A(i+1, j) if X_i if not U(i) and X_i != Y_i

OR[A(i+1, j), A(i+1, j+1)] if not U(i) and X_i = Y_j

Name the words of the book as X_i for the ith word in the first book (which contains underlined words), and Y_j for the jth word in the second book. Let U(i) denote whether X_i is underlined.

Define A(i,j) to be a boolean function such that A(i,j) is true when X[i:] can be abridged to Y[j:] (in other words, consider all words after and including word X_i, and Y_j -- with 0 indexing).

So the answer to our question is given by A(0,0)

Then we can compute A(0,0) in time |X|*|Y| as follows:

A(|X|, |Y|) = True

A(|X|, z) = False, for z < |Y|

Other than these special cases, we have the following cases

A(i,j) = ...

A(i+1, j+1) if U(i) and X_i = Y_j

False if U(i) and X_i != Y_j

A(i+1, j) if X_i if not U(i) and X_i != Y_i

OR[A(i+1, j), A(i+1, j+1)] if not U(i) and X_i = Y_j

### Re: Looking for an algorithm to test for abridgement

**Spoiler:**

Edit: ninja'd

Zµ«VjÕ«ZµjÖZµ«VµjÕZµkVZÕ«VµjÖZµ«VjÕ«ZµjÖZÕ«VµjÕZµkVZÕ«VµjÖZµ«VjÕ«ZµjÖZÕ«VµjÕZµkVZÕ«ZµjÖZµ«VjÕ«ZµjÖZÕ«VµjÕZ

### Who is online

Users browsing this forum: No registered users and 6 guests