Prefixes

The prefix S_n of S is defined as the first n characters of S.^[5] For example, the prefixes of S = (AGCA) are

S₀ = ()

S₁ = (A)

S₂ = (AG)

S₃ = (AGC)

S₄ = (AGCA).

Let LCS(X, Y) be a function that computes a longest subsequence common to X and Y. Such a function has two interesting properties.

First property

LCS(X^A,Y^A) = LCS(X,Y)^A, for all strings X, Y and all symbols A, where ^ denotes string concatenation. This allows one to simplify the LCS computation for two sequences ending in the same symbol. For example, LCS("BANANA","ATANA") = LCS("BANAN","ATAN")^"A", Continuing for the remaining common symbols, LCS("BANANA","ATANA") = LCS("BAN","AT")^"ANA".

Second property

If A and B are distinct symbols (A≠B), then LCS(X^A,Y^B) is one of the maximal-length strings in the set { LCS(X^A,Y), LCS(X,Y^B) }, for all strings X, Y.

For example, LCS("ABCDEFG","BCDGK") is the longest string among LCS("ABCDEFG","BCDG") and LCS("ABCDEF","BCDGK"); if both happened to be of equal length, one of them could be chosen arbitrarily.

To realize the property, distinguish two cases:

• If LCS("ABCDEFG","BCDGK") ends with a "G", then the final "K" cannot be in the LCS, hence LCS("ABCDEFG","BCDGK") = LCS("ABCDEFG","BCDG").
• If LCS("ABCDEFG","BCDGK") does not end with a "G", then the final "G" cannot be in the LCS, hence LCS("ABCDEFG","BCDGK") = LCS("ABCDEF","BCDGK").

LCS function defined

Let two sequences be defined as follows: ${\displaystyle X=(x_{1}x_{2}\cdots x_{m})}$ and ${\displaystyle Y=(y_{1}y_{2}\cdots y_{n})}$. The prefixes of ${\displaystyle X}$ are ${\displaystyle X_{0},X_{1},X_{2},\dots ,X_{m}}$; the prefixes of ${\displaystyle Y}$ are ${\displaystyle Y_{0},Y_{1},Y_{2},\dots ,Y_{n}}$. Let ${\displaystyle {\mathit {LCS}}(X_{i},Y_{j})}$ represent the set of longest common subsequence of prefixes ${\displaystyle X_{i}}$ and ${\displaystyle Y_{j}}$. This set of sequences is given by the following.

${\displaystyle {\mathit {LCS}}(X_{i},Y_{j})={\begin{cases}\epsilon &{\mbox{if }}i=0{\mbox{ or }}j=0\\{\mathit {LCS}}(X_{i-1},Y_{j-1}){\hat {}}x_{i}&{\mbox{if }}i,j>0{\mbox{ and }}x_{i}=y_{j}\\\operatorname {\max } \{{\mathit {LCS}}(X_{i},Y_{j-1}),{\mathit {LCS}}(X_{i-1},Y_{j})\}&{\mbox{if }}i,j>0{\mbox{ and }}x_{i}\neq y_{j}.\end{cases}}}$

To find the LCS of ${\displaystyle X_{i}}$ and ${\displaystyle Y_{j}}$, compare ${\displaystyle x_{i}}$ and ${\displaystyle y_{j}}$. If they are equal, then the sequence ${\displaystyle {\mathit {LCS}}(X_{i-1},Y_{j-1})}$ is extended by that element, ${\displaystyle x_{i}}$. If they are not equal, then the longest among the two sequences, ${\displaystyle {\mathit {LCS}}(X_{i},Y_{j-1})}$, and ${\displaystyle {\mathit {LCS}}(X_{i-1},Y_{j})}$, is retained. (If they are the same length, but not identical, then both are retained.) The base case, when either ${\displaystyle X_{i}}$ or ${\displaystyle Y_{i}}$ is empty, is the empty string, ${\displaystyle \epsilon }$.

Worked example

The longest subsequence common to R = (GAC), and C = (AGCAT) will be found. Because the LCS function uses a "zeroth" element, it is convenient to define zero prefixes that are empty for these sequences: R₀ = ε; and C₀ = ε. All the prefixes are placed in a table with C in the first row (making it a column header) and R in the first column (making it a row header).

This table is used to store the LCS sequence for each step of the calculation. The second column and second row have been filled in with ε, because when an empty sequence is compared with a non-empty sequence, the longest common subsequence is always an empty sequence.

LCS(R₁, C₁) is determined by comparing the first elements in each sequence. G and A are not the same, so this LCS gets (using the "second property") the longest of the two sequences, LCS(R₁, C₀) and LCS(R₀, C₁). According to the table, both of these are empty, so LCS(R₁, C₁) is also empty, as shown in the table below. The arrows indicate that the sequence comes from both the cell above, LCS(R₀, C₁) and the cell on the left, LCS(R₁, C₀).

LCS(R₁, C₂) is determined by comparing G and G. They match, so G is appended to the upper left sequence, LCS(R₀, C₁), which is (ε), giving (εG), which is (G).

For LCS(R₁, C₃), G and C do not match. The sequence above is empty; the one to the left contains one element, G. Selecting the longest of these, LCS(R₁, C₃) is (G). The arrow points to the left, since that is the longest of the two sequences.

LCS(R₁, C₄), likewise, is (G).

LCS(R₁, C₅), likewise, is (G).

"G" Row Completed

	ε	A	G	C	A	T
ε	ε	ε	ε	ε	ε	ε
G	ε	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$ε	${\displaystyle {\overset {\nwarrow }{\ }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)
A	ε
C	ε

For LCS(R₂, C₁), A is compared with A. The two elements match, so A is appended to ε, giving (A).

For LCS(R₂, C₂), A and G do not match, so the longest of LCS(R₁, C₂), which is (G), and LCS(R₂, C₁), which is (A), is used. In this case, they each contain one element, so this LCS is given two subsequences: (A) and (G).

For LCS(R₂, C₃), A does not match C. LCS(R₂, C₂) contains sequences (A) and (G); LCS(R₁, C₃) is (G), which is already contained in LCS(R₂, C₂). The result is that LCS(R₂, C₃) also contains the two subsequences, (A) and (G).

For LCS(R₂, C₄), A matches A, which is appended to the upper left cell, giving (GA).

For LCS(R₂, C₅), A does not match T. Comparing the two sequences, (GA) and (G), the longest is (GA), so LCS(R₂, C₅) is (GA).

"G" & "A" Rows Completed

	ε	A	G	C	A	T
ε	ε	ε	ε	ε	ε	ε
G	ε	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$ε	${\displaystyle {\overset {\nwarrow }{\ }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)
A	ε	${\displaystyle {\overset {\nwarrow }{\ }}}$(A)	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$(A) & (G)	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$(A) & (G)	${\displaystyle {\overset {\nwarrow }{\ }}}$(GA)	${\displaystyle {\overset {\ }{\leftarrow }}}$(GA)
C	ε

For LCS(R₃, C₁), C and A do not match, so LCS(R₃, C₁) gets the longest of the two sequences, (A).

For LCS(R₃, C₂), C and G do not match. Both LCS(R₃, C₁) and LCS(R₂, C₂) have one element. The result is that LCS(R₃, C₂) contains the two subsequences, (A) and (G).

For LCS(R₃, C₃), C and C match, so C is appended to LCS(R₂, C₂), which contains the two subsequences, (A) and (G), giving (AC) and (GC).

For LCS(R₃, C₄), C and A do not match. Combining LCS(R₃, C₃), which contains (AC) and (GC), and LCS(R₂, C₄), which contains (GA), gives a total of three sequences: (AC), (GC), and (GA).

Finally, for LCS(R₃, C₅), C and T do not match. The result is that LCS(R₃, C₅) also contains the three sequences, (AC), (GC), and (GA).

Completed LCS Table

	ε	A	G	C	A	T
ε	ε	ε	ε	ε	ε	ε
G	ε	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$ε	${\displaystyle {\overset {\nwarrow }{\ }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)	${\displaystyle {\overset {\ }{\leftarrow }}}$(G)
A	ε	${\displaystyle {\overset {\nwarrow }{\ }}}$(A)	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$(A) & (G)	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$(A) & (G)	${\displaystyle {\overset {\nwarrow }{\ }}}$(GA)	${\displaystyle {\overset {\ }{\leftarrow }}}$(GA)
C	ε	${\displaystyle {\overset {\ \uparrow }{\ }}}$(A)	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$(A) & (G)	${\displaystyle {\overset {\nwarrow }{\ }}}$(AC) & (GC)	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$(AC) & (GC) & (GA)	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$(AC) & (GC) & (GA)

The final result is that the last cell contains all the longest subsequences common to (AGCAT) and (GAC); these are (AC), (GC), and (GA). The table also shows the longest common subsequences for every possible pair of prefixes. For example, for (AGC) and (GA), the longest common subsequence are (A) and (G).

Traceback approach

Calculating the LCS of a row of the LCS table requires only the solutions to the current row and the previous row. Still, for long sequences, these sequences can get numerous and long, requiring a lot of storage space. Storage space can be saved by saving not the actual subsequences, but the length of the subsequence and the direction of the arrows, as in the table below.

Storing length, rather than sequences

	ε	A	G	C	A	T
ε	0	0	0	0	0	0
G	0	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$0	${\displaystyle {\overset {\nwarrow }{\ }}}$1	${\displaystyle {\overset {\ }{\leftarrow }}}$1	${\displaystyle {\overset {\ }{\leftarrow }}}$1	${\displaystyle {\overset {\ }{\leftarrow }}}$1
A	0	${\displaystyle {\overset {\nwarrow }{\ }}}$1	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$1	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$1	${\displaystyle {\overset {\nwarrow }{\ }}}$2	${\displaystyle {\overset {\ }{\leftarrow }}}$2
C	0	${\displaystyle {\overset {\ \uparrow }{\ }}}$1	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$1	${\displaystyle {\overset {\nwarrow }{\ }}}$2	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$2	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$2

The actual subsequences are deduced in a "traceback" procedure that follows the arrows backwards, starting from the last cell in the table. When the length decreases, the sequences must have had a common element. Several paths are possible when two arrows are shown in a cell. Below is the table for such an analysis, with numbers colored in cells where the length is about to decrease. The bold numbers trace out the sequence, (GA).^[6]

Traceback example

	ε	A	G	C	A	T
ε	0	0	0	0	0	0
G	0	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$0	${\displaystyle {\overset {\nwarrow }{\ }}}$1	${\displaystyle {\overset {\ }{\leftarrow }}}$1	${\displaystyle {\overset {\ }{\leftarrow }}}$1	${\displaystyle {\overset {\ }{\leftarrow }}}$1
A	0	${\displaystyle {\overset {\nwarrow }{\ }}}$1	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$1	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$1	${\displaystyle {\overset {\nwarrow }{\ }}}$2	${\displaystyle {\overset {\ }{\leftarrow }}}$2
C	0	${\displaystyle {\overset {\ \uparrow }{\ }}}$1	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$1	${\displaystyle {\overset {\nwarrow }{\ }}}$2	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$2	${\displaystyle {\overset {\ \ \uparrow }{\leftarrow }}}$2

Computing the length of the LCS

The function below 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.^[7]

function LCSLength(X[1..m], Y[1..n])
    C = array(0..m, 0..n)
    for i := 0..m
        C[i,0] = 0
    for j := 0..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]

Alternatively, memoization could be used.

Reading out a 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 ${\displaystyle x_{i}}$ and ${\displaystyle y_{j}}$, 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 ""
    if  X[i] = Y[j]
        return backtrack(C, X, Y, i-1, j-1) + X[i]
    if C[i,j-1] > C[i-1,j]
        return backtrack(C, X, Y, i, j-1)
    return backtrack(C, X, Y, i-1, j)

Reading out all LCSs

If choosing ${\displaystyle x_{i}}$ and ${\displaystyle y_{j}}$ 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 {""}
    if X[i] = Y[j]
        return {Z + X[i] for all Z in backtrackAll(C, X, Y, i-1, j-1)}
    R := {}
    if C[i,j-1] ≥ C[i-1,j]
        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

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]
    else
        print ""

Example

Let ${\displaystyle X}$ be “XMJYAUZ” and ${\displaystyle Y}$ be “MZJAWXU”. The longest common subsequence between ${\displaystyle X}$ and ${\displaystyle Y}$ 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 ${\displaystyle X}$ and ${\displaystyle Y}$. The ${\displaystyle i}$th row and ${\displaystyle j}$th column shows the length of the LCS between ${\displaystyle X_{1..i}}$ and ${\displaystyle Y_{1..j}}$.

The highlighted 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 ${\displaystyle X}$ and ${\displaystyle Y}$ are equal, they are part of the LCS, and we go both up and left (shown in bold). If not, we go up or left, depending on which cell has a higher number. This corresponds to either taking the LCS between ${\displaystyle X_{1..i-1}}$ and ${\displaystyle Y_{1..j}}$, or ${\displaystyle X_{1..i}}$ and ${\displaystyle Y_{1..j-1}}$.

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

In the best-case scenario, a sequence with no changes, this optimization would 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.

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.

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 benefits may not be worth the setup and computational requirements of a cryptographic hash for small sequence lengths.

Reduce the required space

If only the length of the LCS is required, the matrix can be reduced to a ${\displaystyle 2\times \min(n,m)}$ matrix, or to a ${\displaystyle \min(m,n)+1}$ vector as the dynamic programming approach requires only the current and previous columns of the matrix. Hirschberg's algorithm allows the construction of the optimal sequence itself in the same quadratic time and linear space bounds.^[8]

Reduce cache misses

Chowdhury and Ramachandran devised a quadratic-time linear-space algorithm^[9][10] for finding the LCS length along with an optimal sequence which runs faster than Hirschberg's algorithm in practice due to its superior cache performance.^[9] The algorithm has an asymptotically optimal cache complexity under the Ideal cache model.^[11] Interestingly, the algorithm itself is cache-oblivious^[11] meaning that it does not make any choices based on the cache parameters (e.g., cache size and cache line size) of the machine.

Further optimized algorithms

Several algorithms exist that run faster than the presented dynamic programming approach. One of them is Hunt–Szymanski algorithm, which typically runs in ${\displaystyle O((n+r)\log(n))}$ time (for ${\displaystyle n>m}$), where ${\displaystyle r}$ is the number of matches between the two sequences.^[12] For problems with a bounded alphabet size, the Method of Four Russians can be used to reduce the running time of the dynamic programming algorithm by a logarithmic factor.^[13]