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Informal Problem Statement:

Given a string, e.g. $ACCABBAB$, we want to colour some letters red and some letters blue (and some not at all), such that reading only the red letters from left to right yields the same result as reading only the blue letters.

In the example we could colour them like this: $A\color{blue}{C}\color{red}{CAB}B\color{blue}{AB}$

Therefore, we say $CAB$ is a repeated subsequence of $ACCABBAB$. It is also a longest repeated subsequence (which is easy to check).

Can we compute longest repeated subsequences efficiently?

Formal Question:

Is it NP-hard to decide for a string and some $k$, whether a repeated subsequence of length $k$ exists in the string?

  • If so: Which problem can be reduced to this problem?
  • If not: What is an efficient algorithm? (obviously, this algorithm can then be used to compute a longest repeated subsequence)

Bonus Question:

Will their always be a repeated subsequence of length $n/2 - o(n)$ if the size of the alphabet is bounded by a constant?

(This is known to be true for binary alphabets.)

Edit 2: The negative answer to the Bonus Question is already known for alphabets of size at least $5$. In fact for alphabets of size $Σ$, there are strings with longest repeated subsequences of a length of merely $O(n · Σ^{-1/2})$. Random strings suffice to show this. The result already existed, but I overlooked it.

Edit: Note:

Some people mean "substring" when they say "subsequence". I don't. This is not the problem of finding a substrings twice.

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  • $\begingroup$ Sekti, thank you. You are right: my first comment was erroneous; I've now deleted it. On the other hand, my remaining comment is talking about non-contiguous subsequences. If $k$ is fixed, there is a way to solve your problem (with non-contiguous subsequences, mandated to be non-overlapping) in $O(n^{2k+2})$ time or so. Each d.p. subproblem keeps track of the indices of all of the red letters and all of the blue letters chosen so far. This is probably uninteresting, because it doesn't tell us what happens when $k$ is part of the input. $\endgroup$ – D.W. Jun 16 '14 at 22:36
  • $\begingroup$ @D.W. Why can't the formal question be answered efficiently with this modification of longest common subsequence? Perhaps I am missing something and someone can clarify for me. $\endgroup$ – Bryce Kille Apr 24 at 2:49
  • $\begingroup$ @BryceKille, I don't know; maybe it can. If you figure out how to do it I hope you'll write an answer! $\endgroup$ – D.W. Apr 24 at 4:39
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This can be solved in polynomial time by constructing a graph $G$ where each node represents a point $(i,j)$ in some repeated subsequence of $S$ such that $S[i]=S[j]$. Edge between nodes $u$ and $v$ means that $u$ can be continued by $v$ to form a repeated subsequence of length 2.

1. Find the nodes. This can be done in $O(n^2)$ time by building a sorted list of indices for each character $c$, and then enumerating the unique pairs. There are no more than $m=n^2$ nodes.

2. Find the edges. It takes $O(1)$ time to check if node $u$ can be continued by node $v$, so by considering all pairs $(u,v)$ this step takes $O(m^2)$ time.

3. Note that the longest path in $G$ may not be a valid repeated subsequence. Consider paths $ab$ and $bc$. If there exists an edge $ac$ then $abc$ is a valid repeated subsequence of length 3. Therefore it takes $O(m^4)$ time to find all repeated subsequences of length 3. In the general case it takes linear time to check whether two valid paths of length $n$ can be combined into a valid path of length $n+1$.

4. Iterate step 3 until no longer paths can be found.

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  • $\begingroup$ Hmm. Too quick. In step 3 the number of subsequences to consider gets bigger and bigger. So it's not polynomial. $\endgroup$ – noplogist Dec 11 '15 at 23:07
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    $\begingroup$ Welcome to CS.SE, and thank you for trying to solve this problem! I'm afraid I don't understand your algorithm. What is Step 3? All I see in "3." are some declarative statements/observations, but I don't see a procedural specification of what the algorithm is supposed to do. Also, I don't see what it means to iterate Step 3, or the rationale for your claim that $O(nnm^2)$ time suffices. Your subsequent comment makes it sound like you no longer believe your answer is correct. If so, it might be better to delete the answer, to avoid confusion. $\endgroup$ – D.W. Dec 11 '15 at 23:22
  • $\begingroup$ You can always undelete or post a new answer if you figure out the answer later. $\endgroup$ – D.W. Dec 11 '15 at 23:22
  • $\begingroup$ D.W., thanks. Input to step 3 is all repeated subsequences of length n, and output is all repeated subsequences of length n+1. I believe the algorithm is correct but that it is not a polynomial time algorithm. I have now marked the claims that I consider incorrect. $\endgroup$ – noplogist Dec 11 '15 at 23:40
  • $\begingroup$ Thank you. I understand. Unfortunately, with those revisions, I'm afraid this answer does not answer the question that was asked. The question was: Is this NP-hard? Is there an efficient algorithm? Showing that there exists an exponential-time algorithm does not help in answering either of those questions. Indeed, it's already trivial to see that there is an exponential-time algorithm for the problem, without invoking any fancy techniques. I do appreciate your attempt. $\endgroup$ – D.W. Dec 12 '15 at 2:11

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