Faster Space-Efficient STR-IC-LCS Computation

TL;DR

The paper addresses the STR-IC-LCS problem, seeking the longest common subsequence of $A$ and $B$ that contains a given pattern $P$ as a substring. It introduces three algorithms with progressively better time-space trade-offs: Algorithm I computes the length in $O(n^2)$ time using $O((\ell+1)(n-\ell+1))$ space, Algorithm II accelerates to $O(nr/\log r + n(n-\ell+1))$ time while maintaining the same space bound by pruning candidates, and Algorithm III further refines the approach using $\ell'$ with $O(nr/\log r + n(\ell'+1))$ time and $O((\ell'+1)(n-\ell'+1))$ space. All methods build on Nakatsu et al.'s sparse LCS framework and incorporate minimal intervals for $P$ (via Das et al. for Algorithm II) to achieve subquadratic memory and, in many cases, subquadratic time, while ensuring practical reconstruction of the STR-IC-LCS. The results advance constrained LCS computation, with implications for sequence analysis where pattern constraints are essential, and establish near-optimal time bounds under standard hardness assumptions like SETH.

Abstract

One of the most fundamental method for comparing two given strings $A$ and $B$ is the longest common subsequence (LCS), where the task is to find (the length) of an LCS of $A$ and $B$. In this paper, we deal with the STR-IC-LCS problem which is one of the constrained LCS problems proposed by Chen and Chao [J. Comb. Optim, 2011]. A string $Z$ is said to be an STR-IC-LCS of three given strings $A$, $B$, and $P$, if $Z$ is a longest string satisfying that (1) $Z$ includes $P$ as a substring and (2) $Z$ is a common subsequence of $A$ and $B$. We present three efficient algorithms for this problem: First, we begin with a space-efficient solution which computes the length of an STR-IC-LCS in $O(n^2)$ time and $O((\ell+1)(n-\ell+1))$ space, where $\ell$ is the length of an LCS of $A$ and $B$ of length $n$. When $\ell = O(1)$ or $n-\ell = O(1)$, then this algorithm uses only linear $O(n)$ space. Second, we present a faster algorithm that works in $O(nr/\log{r}+n(n-\ell+1))$ time, where $r$ is the length of $P$, while retaining the $O((\ell+1)(n-\ell+1))$ space efficiency. Third, we give an alternative algorithm that runs in $O(nr/\log{r}+n(n-\ell'+1))$ time with $O((\ell'+1)(n-\ell'+1))$ space, where $\ell'$ denotes the STR-IC-LCS length for input strings $A$, $B$, and $P$.

Figures (8)

• Figure 1: Let $A = \mathtt{bcdababcb}$, $B = \mathtt{cbacbaaba}$, and $P = \mathtt{abb}$. The length of an STR-IC-LCS of these strings is 6. One of such strings can be obtained by minimal intervals $[4..7]$ over $A$ and $[6..8]$ over $B$ because $\mathsf{lcs}(\mathtt{bca},\mathtt{cbacb}) = 2$, $|P| = 3$, and $\mathsf{lcs}(\mathtt{cb},\mathtt{c}) = 1$.
• Figure 2: The LCS-table $f_A$ which is defined by Nakatsu et al. of $A = \mathtt{bcdababcb}$. This figure also illustrates the table $f_B$ of $B = \mathtt{cbacbaaba}$.
• Figure 3: A sparse table $F_A$ of $f_A$ for $A = \mathtt{bcdababcb}$ and $B = \mathtt{cbacbaaba}$ does not give $\mathsf{lcs}(A[1..i], B[1..j])$ for some $(i, j)$.
• Figure 4: Due to Observation \ref{['obs:lcs']}, $f_A(3,7)$ gives the fact that $\mathsf{lcs}(A[1..7],B[1..4]) = 3$. However, $F_A(3,7) = \mathsf{undefined}$. Then we can obtain the fact that $\mathsf{lcs}(A[1..7],B[1..4]) = 3$ by using $F_B$. Namely, $F_B(3,4)$ gives the LCS value.
• Figure 5: This figure shows an illustration for the proof of Lemma \ref{['lem:recover-lcs']} (and Lemma \ref{['lem:visible-lcs']}). The length $s$ of an LCS of $A[1..i]$ and $B[1..j]$ cannot be obtained over $F_A$ because $F_A(s,i) = \mathsf{undefined}$ (the highlighted cell). However, the length can be obtained by $F_B(s,j)$ over $F_B$. The existence of $F_B(s+m,j_{s+m})$ from an LCS path guarantees the fact that $F_B(s,j) \neq \mathsf{undefined}$.

Theorems & Definitions (16)

• Theorem 1
• Lemma 1
• Lemma 2
• proof
• proof : Proof of Lemma \ref{['lem:recover-lcs']}
• Lemma 3
• proof
• Theorem 2
• Lemma 4
• proof