An Algorithmic Bridge Between Hamming and Levenshtein Distances

The edit distance between strings classically assigns unit cost to every character insertion, deletion, and substitution, whereas the Hamming distance only allows substitutions. In many real-life scenarios, insertions and deletions (abbreviated indels) appear frequently but significantly less so than substitutions. To model this, we consider substitutions being cheaper than indels, with cost $1\mathrm{/}a$ for a parameter $a\ge 1$. This basic variant, denoted $E{D}_a$, bridges classical edit distance ($a=1$) with Hamming distance ($a\to \mathrm{\infty }$), leading to interesting algorithmic challenges: Does the time complexity of computing $E{D}_a$ interpolate between that of Hamming distance (linear time) and edit distance (quadratic time)? What about approximating $E{D}_a$? We first present a simple deterministic exact algorithm for $E{D}_a$ and further prove that it is near-optimal assuming the Orthogonal Vectors Conjecture. Our main result is a randomized algorithm computing a $(1+\epsilon )$-approximation of $E{D}_a(X,Y)$, given strings $X,Y$ of total length $n$ and a bound $k\ge E{D}_a(X,Y)$. For simplicity, let us focus on $k\ge 1$ and a constant $\epsilon >0$; then, our algorithm takes $\tilde{O}(n\mathrm{/}a+a{k}^3)$ time. Unless $a=\tilde{O}(1)$ and for small enough $k$, this running time is sublinear in $n$. We also consider a very natural version that asks to find a $(k_I,k_S)$-alignment – an alignment with at most $k_I$ indels and $k_S$ substitutions. In this setting, we give an exact algorithm and, more importantly, an $\tilde{O}(n{k}_I\mathrm{/}{k}_S+k_S\cdot k_I^3)$-time $(1,1+\epsilon )$-bicriteria approximation algorithm. The latter solution is based on the techniques we develop for $E{D}_a$ for $a=\mathrm{\Theta }(k_S\mathrm{/}{k}_I)$. These bounds are in stark contrast to unit-cost edit distance, where state-of-the-art algorithms are far from achieving $(1+\epsilon )$-approximation in sublinear time, even for a favorable choice of $k$.