Consensus Patterns parameterized by input string length is W[1]-hard

TL;DR

This paper proves that the Consensus Patterns problem, which involves finding a common pattern in multiple strings with errors, is computationally hard (W[1]-hard) when parameterized by input string length.

Contribution

It establishes the W[1]-hardness of the Consensus Patterns problem based on input string length, highlighting its computational complexity.

Findings

Consensus Patterns problem is W[1]-hard when parameterized by string length.

The hardness result applies to the problem with errors allowed in pattern matching.

This work clarifies the computational limits of pattern extraction in strings.

Abstract

We consider the Consensus Patterns problem, where, given a set of input strings, one is asked to extract a long-enough pattern which appears (with some errors) in all strings. We prove that this problem is W[1]-hard when parameterized by the maximum length of input strings.

Full text

By reduction from Multi-Colored Clique. We are given a graph $G=(V,E)$, with a partition (coloring) $V=V_{1}\cup V_{2}\cup\ldots\cup V_{k}$, such that no edge has both endpoints of the same color. Assume that $|V_{h}|=n$ for all $h\in[k]$. Write $V_{h}=\{v_{h,1},v_{h,2},\ldots v_{h,n}\}$, i.e. each vertex has an index depending both on its color and its rank within its color. Let $m=|E|$. Multi-Colored Clique is W[1]-hard for parameter $k$ [2]. See Figure 1 for an example of the reduction.

We build an alphabet $\Sigma$ containing $V$ (i.e., one symbol per vertex) and two special characters \$$ and {\circ}$.

Define string {\mathcal{V}}_{i}=\v_{1,i}v_{2,i}\ldots v_{k,i}$. Let$e=(v_{h,i},v_{h^{\prime},i^{\prime}})$be the$j$th edge of$E$,$j\in[m]$. Define${\mathcal{E}}{j}$as the string starting with$$$, followed by$k+1$characters: all${\circ}$, except for two positions:${\mathcal{E}}{j}[k+h+1]=v_{h,i}$and${\mathcal{E}}{j}[h^{\prime}+2]=v{h^{\prime},i^{\prime}}$.

Let $N=m(k+2)+1$. The instance $\mathcal{I}$ of Consensus Patterns contains $N$ occurrences of strings ${\mathcal{V}}_{i}$, $i\in[n]$, and one occurrence of strings ${\mathcal{E}}_{j}$, $j\in[m]$. The target length is $\mathrm{m}=k+1$.

Note that due to the large value of $N$, any solution $S$ must have a minimal distance to the set of strings $\{{\mathcal{V}}_{i}\mid i\in[n]\}$. Otherwise, (if it is, say, at the minimum distance plus one), the distance to the whole instance $\mathcal{I}$ increases by at least $N$, which cannot be compensated by the remaining strings ${\mathcal{E}}_{j}$ (which have size $m(k+2)<N$). Hence we first enumerate the optimal solutions for the set $\{{\mathcal{V}}_{i}\mid i\in[n]\}$.

Consider now an optimal solution $S$ for $\mathcal{I}$. Let $\{i_{h}\mid h\in[k]\}$ be the set of indices as obtained from the lemma above. We show that the set of vertices $K=\{v_{h,i_{h}}\mid h\in[k]\}$ forms a clique of $G$ iff the distance is below a certain threshold. To this end, we compute the best possible alignment between $S$ and each string ${\mathcal{E}}_{j}$.

If $S[1]$ is aligned with ${\mathcal{E}}_{j}[1]$, then there is exactly one common character, namely S[1]={\mathcal{E}}_{j}[1]=\$$. Indeed, for all h\in[k]$,$S[h+1]\in V_{h}$and${\mathcal{E}}{j}[h+1]\in V{h-1}\cup{x}$, hence these two characters are different. The distance in this case is$k$.

If $S[1]$ is aligned with ${\mathcal{E}}_{j}[2]$, then first note that S[1]=\\neq x={\mathcal{E}}{j}[2]$. Consider index$h{\mathrm{s}}$. If$i_{\mathrm{s}}=i_{h_{\mathrm{s}}}$, then$S[{h_{\mathrm{s}}}]=v_{{h_{\mathrm{s}}},i_{h_{\mathrm{s}}}}={\mathcal{E}}{j}[{h{\mathrm{s}}}+1]$, otherwise$S[{h_{\mathrm{s}}}]\neq{\circ}={\mathcal{E}}{j}[{h{\mathrm{s}}}+1]$. Similarly for$h_{\mathrm{t}}$,$S[{h_{\mathrm{t}}}]={\mathcal{E}}{j}[{h{\mathrm{t}}}+1]$iff$i_{\mathrm{t}}=i_{h_{\mathrm{t}}}$. For other values of$h$(i.e.$h\in[k]\setminus{h_{\mathrm{s}},h_{\mathrm{t}}}$),$S[h]\neq{\circ}={\mathcal{E}}{j}[h+1]$. The distance is thus$k-1$iff$i{\mathrm{s}}=i_{h_{\mathrm{s}}}$and$i_{\mathrm{t}}=i_{h_{\mathrm{t}}}$, it is at least$k$ otherwise.

Overall, if $i_{\mathrm{s}}=i_{h_{\mathrm{s}}}$ and $i_{\mathrm{t}}=i_{h_{\mathrm{t}}}$ the optimal alignment has distance $k-1$, otherwise the optimal alignment has distance $k$.

∎

We can now conclude the proof. Let $S$ be an optimal solution of Consensus Pattern for instance $\mathcal{I}$ and $K$ its corresponding set of vertices. The distance from $S$ to the $N$ copies of strings ${\mathcal{V}}_{i}$ is $N(n-1)k$. The distance between $S$ and ${\mathcal{E}}_{j}$ is $k-1$ if both endpoints of $e_{j}$ are in $K$, and $k$ otherwise. $|E(K)|$ is the number of edges with both endpoints in $K$: the total distance from $S$ to strings ${\mathcal{E}}_{j}$ is thus $mk-|E(K)|$, and the total distance from $S$ to $\mathcal{I}$ is $N(n-1)k+mk-|E(K)|$. Overall, the optimal distance is at most $N(n-1)k+mk-\frac{k(k-1)}{2}$ if, and only if, $G$ contains a size-$k$ set of vertices $K$ with $|E(K)|\geq\frac{k(k-1)}{2}$, i.e. if $G$ contains a clique.