Longest common subsequence problem

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The longest common subsequence problem (LCS) is finding a longest sequence which is a subsequence of all sequences in a set of sequences (often just two). The problem is sometimes defined to be finding all longest common subsequences. It is a classic computer science problem, the basis of diff, and has applications in bioinformatics.

It should not be confused with the longest common substring problem (a substring is necessarily a contiguous part).

1 Complexity
2 Solution for two sequences
3 Relation to other problems
4 Dynamic programming solution
5 Example
6 Optimization
- 6.1 Reduce the problem set
- 6.2 Reduce the comparison time
  - 6.2.1 Reduce strings to hashes
  - 6.2.2 Reduce strings to identity integers
7 References
8 External links

[edit] Complexity

The problem is NP-hard for the general case of an arbitrary number of input sequences^[1]. When the number of sequences is constant, the problem is solvable in polynomial time by dynamic programming (see Solution below). Assume you have $N$ sequences of lengths $n_1, \dots, n_N$ . A naive search would be trying, for all possible lengths, for all the sequences, all possible subsequences, which numbers

$\sum_{L=0}^{\min_{i=1}^{N}{n_i}} \prod_{k=1}^{N}{n_k \choose L}$ .

A dynamic programming (DP) approach gives a solution in

$\Theta\left(N \prod_{k=1}^{N} n_k\right)$

There exist methods with lower complexity^[2], which often depend on the length of the LCS, the size of the alphabet, or both.

Notice that LCS is often defined to be finding all common subsequences of a maximum length. This problem inherently has higher complexity, as the number of such subsequences is exponential in the worst case^[3], even for only two input strings.

[edit] Solution for two sequences

Given the sequences $X_{1 \dots m}$ and $Y_{1 \dots n}$

$\textrm{LCS}\left(X_{1 \dots i},Y_{1 \dots j}\right) = \begin{cases} \emptyset & \mbox{ if } i = 0 \mbox{ or } j = 0 \\ \textrm{LCS}\left(X_{1 \dots i-1},Y_{1 \dots j-1}\right) + x_{i} & \mbox{ if } x_i = y_j \\ \max'\left(\textrm{LCS}\left(X_{1 \dots i},Y_{1 \dots j-1}\right),\textrm{LCS}\left(X_{1 \dots i-1},Y_{1 \dots j}\right)\right) & \mbox{ otherwise} \\ \end{cases}$

Here $+$ denotes concatenation, and $max'$ gives the longest sequence.

Since this problem has an optimal substructure property, it can be solved by dynamic programming. See ^[4] for an accessible explanation.

The rationale for this recurrence is that, if the last character of two sequences are equal, they must be part of the LCS. You could never get a larger LCS by matching $x m$ to $y j$ where $j < n$ , and vice versa. If they are not equal, check what gives the largest LCS of keeping $x m$ and $y n$ , and use that. If you want to find all the longest common subsequences, the LCS should be denoted as a set of sequences, and $max'$ should return both solutions if they are equally long.

[edit] Relation to other problems

For two strings $X_{1 \dots m}$ and $Y_{1 \dots n}$ , the length of the shortest common supersequence is^[2]

$\left|\textrm{SCS}(X,Y)\right| = n + m - \left|\textrm{LCS}(X,Y)\right|$

The edit distance when only insertion and deletion is allowed (no substitution), is

$d'(X,Y) = n + m - 2 \cdot \left|\textrm{LCS}(X,Y)\right|$

[edit] Dynamic programming solution

Wikibooks Algorithm implementation has a page on the topic of

Longest common subsequence

[edit] Computing the length of the LCS

The below function takes as input sequences X[1..m] and Y[1..n] computes the LCS between X[1..i] and Y[1..j] for all 1 ≤ i ≤ m and 1 ≤ j ≤ n, and stores it in C[i,j]. C[m,n] will contain the length of the LCS of X and Y.

function  LCSLength(X[1..m], Y[1..n])
    C = array(0..m, 0..n)
    for i := 0..m
       C[i,0] = 0
    for j := 1..n
       C[0,j] = 0
    for i := 1..m
        for j := 1..n
            if X[i] = Y[j]
                C[i,j] := C[i-1,j-1] + 1
            else:
                C[i,j] := max(C[i,j-1], C[i-1,j])
    return C[m,n]

[edit] Reading out an LCS

The following function backtracks the choices taken when computing the C table. If the last characters in the prefixes are equal, they must be in an LCS. If not, check what gave the largest LCS of keeping $x i$ and $y j$ , and make the same choice. Just choose one if they were equally long. Call the function with i=m and j=n.

function backTrack(C[0..m,0..n], X[1..m], Y[1..n], i, j)
    if i = 0 or j = 0
        return ""
    else if  X[i] = Y[j]
        return backTrack(C, X, Y, i-1, j-1) + X[i]
    else
        if C[i,j-1] > C[i-1,j]
            return backTrack(C, X, Y, i, j-1)
        else
            return backTrack(C, X, Y, i-1, j)

[edit] Reading out all LCSs

If choosing $x i$ and $y j$ would give an equally long result, read out both resulting subsequences. This is returned as a set by this function. Notice that this function is not polynomial, as it might branch in almost every step if the strings are similar.

function backTrackAll(C[0..m,0..n], X[1..m], Y[1..n], i, j)
    if i = 0 or j = 0
        return {}
    else if X[i] = Y[j]
        return {Z + X[i] for all Z in backTrackAll(C, X, Y, i-1, j-1)}
    else
        R := {}
        if C[i,j-1] ≥ C[i-1,j]
            R := R ∪ backTrackAll(C, X, Y, i, j-1)
        if C[i-1,j] ≥ C[i,j-1]
            R := R ∪ backTrackAll(C, X, Y, i-1, j)
        return R

[edit] Print the diff

This function will backtrack through the C matrix, and print the diff between the two sequences. Notice that you will get a different answer if you exchange ≥ and <, with > and ≤ below.

function printDiff(C[0..m,0..n], X[1..m], Y[1..n], i, j)
    if i > 0 and j > 0 and X[i] = Y[j]
        printDiff(C, X, Y, i-1, j-1)
        print "  " + X[i]
    else
        if j > 0 and (i = 0 or C[i,j-1] ≥ C[i-1,j])
            printDiff(C, X, Y, i, j-1)
            print "+ " + Y[j]
        else if i > 0 and (j == 0 or C[i,j-1] < C[i-1,j])
            printDiff(C, X, Y, i-1, j)
            print "- " + X[i]

[edit] Example

Let $X$ be "XMJYAUZ" and $Y$ be "MZJAWXU". The longest common subsequence between $X$ and $Y$ is "MJAU". The table C shown below, which is generated by the function LCSlength, shows the lengths of the longest common subsequences between prefixes of $X$ and $Y$ . The $i$ th row and $j$ th column shows the length of the LCS between $X 1.. i$ and $Y 1.. j$ .

     | 0 1 2 3 4 5 6 7
     |   M Z J A W X U
-----|-----------------
0    | 0 0 0 0 0 0 0 0
1  X | 0 0 0 0 0 0 1 1
2  M | 0 1 1 1 1 1 1 1
3  J | 0 1 1 2 2 2 2 2
4  Y | 0 1 1 2 2 2 2 2
5  A | 0 1 1 2 3 3 3 3
6  U | 0 1 1 2 3 3 3 4
7  Z | 0 1 2 2 3 3 3 4

The bold numbers show the path the function backTrack would follow from the bottom right to the top left corner, when reading out an LCS. If the current symbols in $X$ and $Y$ are equal, they are part of the LCS, and we go both up and left. If not, we go up or left, depending on which cell has a higher number. This corresponds to either taking the LCS between $X 1.. i - 1$ and $Y 1.. j$ , or $X 1.. i$ and $Y 1.. j - 1$

[edit] Optimization

Several optimizations can be made to the algorithm above to speed it up for real-world cases.

[edit] Reduce the problem set

The C matrix in the naive algorithm grows quadratically with the lengths of the sequences. For two 100-item sequences, a 10,000-item matrix would be needed, and 10,000 comparisons would need to be done. In most real-world cases, especially source code diffs and patches, the beginnings and ends of files rarely change, and almost certainly not both at the same time. If only a few items have changed in the middle of the sequence, the beginning and end can be eliminated. This reduces not only the memory requirements for the matrix, but also the number of comparisons that must be done.

function  LCS(X[1..m], Y[1..n])
    m_start = 1
    m_end = m
    n_start = 1
    n_end = n
    trim off the matching items at the beginning
    while m_start ≤ m_end and n_start ≤ n_end and X[m_start] = Y[n_start]
        m_start := m_start + 1
        n_start := n_start + 1
    trim off the matching items at the end
    while m_start ≤ m_end and n_start ≤ n_end and X[m_end] = Y[n_end]
        m_end := m_end - 1
        n_end := m_end - 1
    C = array(m_start-1..m_end, n_start-1..n_end)
    only loop over the items that have changed
    for i := m_start..m_end
        for j := n_start..n_end
            the algorithm continues as before ...

In the best case scenario, a sequence with no changes, this optimization would completely eliminate the need for the C matrix. In the worst case scenario, a change to the very first and last items in the sequence, only two additional comparisons are performed.

[edit] Reduce the comparison time

Most of the time taken by the naive algorithm is spent performing comparisons between items in the sequences. For textual sequences such as source code, you want to view lines as the sequence elements instead of single characters. This can mean comparisons of relatively long strings for each step in the algorithm. Two optimizations can be made that can help to reduce the time these comparisons consume.

[edit] Reduce strings to hashes

A hash function or checksum can be used to reduce the size of the strings in the sequences. That is, for source code where the average line is 60 or more characters long, the hash or checksum for that line might be only 8 to 40 characters long. Additionally, the randomized nature of hashes and checksums would guarantee that comparisons would short-circuit faster, as lines of source code will rarely be changed at the beginning.

There are three primary drawbacks to this optimization. First, an amount of time needs to be spent beforehand to precompute the hashes for the two sequences. Second, additional memory needs to be allocated for the new hashed sequences. However, in comparison to the naive algorithm used here, both of these drawbacks are relatively minimal.

The third drawback is that of collisions. Since the checksum or hash is not guaranteed to be unique, there is a small chance that two different items could be reduced to the same hash. This is unlikely in source code, but it is possible. A cryptographic hash would therefore be far better suited for this optimization, as its entropy is going to be significantly greater than that of a simple checksum. However, the setup and computational requirements of a cryptographic hash may not be worth it for small sequence lengths.

[edit] Reduce strings to identity integers

In source code, there will often be lines that are exactly the same. This could include an empty line between two functions, or a trailing brace, or closing comment, and so on. The algorithm does not need to distinguish between specific instances of these similar lines. That is, one blank line is the same as any other blank line. Therefore, the previous optimization can be taken one step further by reducing each line to an identity integer. String buckets, such as Python dict dictionary objects, Perl hashes, or C++ STL map objects, are well-suited for this optimization.

1. Create an empty bucket.
2. For each item in the sequence, look to see if it is already present in the bucket.
3. If not, add the item to the bucket and tag it with a new identity integer.
4. If so, get the identity integer from the bucket.
5. Replace the item with its identity integer.
6. Repeat.

This method has advantages over the previous optimization. First, it does not suffer from the hash collision problem, as language-specific bucket objects are almost always coded to handle the case of two different items with the same hash. More importantly, each item is reduced from a string to a unique integer, meaning that each comparison is now an integer comparison. There is overhead to the identity allocation, but in terms of the original naive algorithm, this optimization is almost always worth it.

[edit] References

^ David Maier (1978). "The Complexity of Some Problems on Subsequences and Supersequences". J. ACM 25: 322–336. DOI:10.1145/322063.322075.
^ ^a ^b L. Bergroth and H. Hakonen and T. Raita (2000). "A Survey of Longest Common Subsequence Algorithms". SPIRE 00: 39–48. DOI:10.1109/SPIRE.2000.878178.
^ Ronald I. Greenberg (2003-08-06). "Bounds on the Number of Longest Common Subsequences". v2.
^ Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest and Clifford Stein (2001). "15.4", Introduction to Algorithms, second edition, MIT Press and McGraw-Hill, 350-355. ISBN 0-262-53196-8.

[edit] External links

Retrieved from "http://en.wikipedia.org../../../l/o/n/Longest_common_subsequence_problem.html"

Categories: Algorithms on strings | Combinatorics | Dynamic programming | Polynomial-time problems | NP-complete problems