Longest Common Prefixes with k-Errors and Applications

Abstract

Although real-world text datasets, such as DNA sequences, are far from being uniformly random, string searching average-case algorithms perform significantly better than worst-case ones in most applications of interest. In this paper, we study the problem of computing the longest prefix of each suffix of a given string of length n that occurs elsewhere in the string with k-errors. This problem has already been studied under the Hamming distance model. Our first result is an improvement upon the state-of-the-art average-case time complexity for non-constant k and using only linear space under the Hamming distance model. Notably, we show that our technique can be extended to the edit distance model with the same time and space complexities. Specifically, our algorithms run in $\mathcal {O}(n\frac{(c\log n)^k}{k!})$ time on average, where $c>1$ is a constant, using $\mathcal {O}(n)$ space. Finally, we show that our technique is applicable to several algorithmic problems found in computational biology and elsewhere. The importance of our technique lies on the fact that it is the first one achieving this bound for non-constant k and using $\mathcal {O}(n)$ space.

A K-Combinations Generation Process

We generate all combinations of size K of the set [N] using the following folklore algorithm. We build a tree in a recursive way. Each node v has a label $\ell (v) \in [N]$, apart from the root which is labeled with 0. We say that a node v represents the set $r(v)=\{\ell (v)\} \cup \{\ell (u)|u\text { is an ancestor of }v\}\setminus \{0\}$. By construction, no two nodes will represent the same set. At the end of the process each of the $\mathcal {O}(\genfrac(){0.0pt}1{N}{K})$ leaves will represent a distinct K-combination. The procedure works as follows:

• 1. It takes as input a node v with satellite data $\ell (v)$ and |r(v)|.

• 2. It creates $N-\ell (v)-K+|r(v)|$ child nodes of v, labeled $\ell (v)+1$ to $N-K+|r(v)|+1$.

• 3. For each newly created node u, if $|r(u)|=|r(v)|+1<K$, the procedure recursively calls itself.

Initiating this procedure with an input node u, with $\ell (u)=0$ and $|r(u)|=0$, all K-combinations are generated in time $\mathcal {O}(\genfrac(){0.0pt}1{N}{K})$ if $K\le N/2$ as shown below. Note that each node of the tree is associated with a unique subset of [N] and neighbouring nodes only differ on whether they contain the largest element of their union.

To upper bound the time complexity, it suffices to bound the number of nodes in the tree with a single child, since the rest internal nodes are trivially upper bounded by the number of leaves. A node with label $N-i$ will have a single child only if it is the $(K-i)$th element added in a combination. This can happen $\mathcal {O}(\genfrac(){0.0pt}1{N-i-1}{K-i-1})$ times. Given our assumption that $K\le N/2$ and using Pascal’s identity $\genfrac(){0.0pt}1{m}{j}=\genfrac(){0.0pt}1{m-1}{j-1}+\genfrac(){0.0pt}1{m-1}{j}$, we bound the number of nodes with a single child as follows:

$$\begin{aligned}&\sum _{i=1}^{K-1}{\genfrac(){0.0pt}1{N-i-1}{K-i-1}}= 1+\sum _{i=1}^{K-2}{\genfrac(){0.0pt}1{N-i}{K-i-1}}-\sum _{i=1}^{K-2}{\genfrac(){0.0pt}1{N-i-1}{K-i-2}}\\&=1+\sum _{i=1}^{K-2}{\genfrac(){0.0pt}1{N-i}{K-i-1}}-\sum _{i=2}^{K-1}{\genfrac(){0.0pt}1{N-i}{K-i-1}}= 1+\genfrac(){0.0pt}1{N-1}{K-2}-1= \frac{K(K-1)}{N(N-K+1)}\genfrac(){0.0pt}1{N}{K}\le \genfrac(){0.0pt}1{N}{K}. \end{aligned}$$