On the Complexity of Restoring Corrupted Colorings^††thanks: Work partially supported by the Emil Aaltonen Foundation (J.L).

Let $I=(G,W,c,r)$ be an instance of Precoloring Extension. To obtain, in polynomial time, an instance $I^{\prime}=(G^{\prime},r^{\prime},c^{\prime},k^{\prime})$ of Fix, we proceed as follows. First, let $G^{\prime}=G$, $r^{\prime}=r$, and set the number of recolorings $k^{\prime}=|V\setminus W|$. Then to each precolored vertex $w\in W$, we attach $(r^{\prime}-1)\cdot(k^{\prime}+1)$ pendant vertices, called $P_{w}$. Build $c^{\prime}$ from $c$ as follows. We color vertices in $P_{w}$ such that there are precisely $k^{\prime}+1$ vertices colored in every color $c\in([r^{\prime}]\setminus\{c(w)\})$. We retain the colors on the vertices in $W$, and color each uncolored vertex with color $1$. Observe that $k^{\prime}$ recolorings will not suffice to change the color of $w\in W$ as it has $k^{\prime}+1$ pendants colored in each color distinct from $c(w)$. Thus, it is easy to see $I=(G,W,c,r)$ is a YES-instance of Precoloring Extension iff $I^{\prime}=(G^{\prime},r^{\prime},c^{\prime},k^{\prime})$ is a YES-instance of Fix.

Let $V(G)=\{u_{1},u_{2},\ldots,u_{n}\}$. To construct the graph $G^{\prime}$, begin with $V(G^{\prime})=V(G)$ and $k^{\prime}=2k$. Add $k$ disjoint triangles $\{a_{1},b_{1},c_{1}\},\ldots\{a_{k},b_{k},c_{k}\}$, and color them such that $c(a_{j})=3$ and $c(b_{j})=c(c_{j})=2$, for $j\in[k]$. For each $i\in[n]$, add a disjoint triangle $C_{i}=\{u^{i}_{a},u^{i}_{b},u^{i}_{c}\}$. In each, color $c(u^{i}_{a})=1$, $c(u^{i}_{b})=2$, and $c(u^{i}_{c})=3$. Attach $2(k+1)$ pendant vertices to $u^{i}_{a}$ and color $k+1$ of them with color $2$, and $k+1$ of them with color $3$. For each $i$, add the edge $u_{i}u^{i}_{b}$. For every $i,j\in[n]$, if $j>i$ and $u_{i}u_{j}\in E(G)$, add the edge $u_{j}u^{i}_{c}$. Finally, for each $i\in[n]$, add $k+1$ disjoint triangles $\{t^{i}_{a,1},t^{i}_{b,1},t^{i}_{c,1}\},\ldots,\{t^{i}_{a,k+1},t^{i}_{b,k+1},t^{i}_{c,k+1}\}$. These $k+1$ disjoint triangles are colored such that $c(t^{i}_{a,j})=3$, $c(t^{i}_{b,j})=2$, and $c(t^{c}_{a,j})=1$, where $j\in[k+1]$. Add the edge $u_{i}t^{i}_{a,j}$, for $i\in[n]$ and $j\in[k+1]$. This completes the construction of $G^{\prime}$. An example is shown in Figure 1. Let us then prove $I=(G,k)$ is a YES-instance of Independent Set iff $I^{\prime}=(G^{\prime},r,c,k^{\prime})$ is a YES-instance of $r$-Swap.

Let $S=\{s_{1},\ldots,s_{k}\}\subseteq V(G)$ be an independent set. By construction, there are exactly $k$ conflicts in $G^{\prime}$ between $b_{j}$ and $c_{j}$, for $j\in[k]$. Swap $c_{j}$ with $u_{j}$, spending a total of $k$ swaps. By doing so, we fixed $k$ conflicts but also introduced $k$ new conflicts. In particular, the new conflicts are between $u_{i}$ and $u^{i}_{b}$, as they are both colored $2$. We swap $u^{i}_{b}$ with $u^{i}_{c}$ (colored $3$), and claim $G^{\prime}$ is properly colored. By construction, $u^{i}_{c}$ is adjacent to $u\in V(G)$ only if $u$ and $u_{i}$ are adjacent in $V(G)$. As $S$ is an independent set, each $u\in V(G)$ adjacent to $u^{i}_{c}$ is colored $1$. Thus, $c$ is a proper coloring for $G^{\prime}$, and we are done.

For the other direction, suppose $k^{\prime}=2k$ swaps suffice to obtain a proper vertex coloring from $c$. Again, each of the $k$ conflicts between $b_{j}$ and $c_{j}$, for $j\in[k]$, must be fixed. To resolve the conflicts, we must swap either $b_{j}$ or $c_{j}$ with a vertex colored $1$. Without loss, suppose we choose $c_{j}$ over $b_{j}$. Observe that for every $i\in[n]$, we cannot swap $u^{i}_{a}$ with any vertex as it has $k+1$ pendant vertices colored $2$, and $k+1$ pendant vertices colored $3$. Thus, there are two possibilities: either we swap $c_{j}$ with $u_{i}$, or with $t^{i}_{a,j}$, for some $i\in[n]$ and $j\in[k+1]$. If the swap occurs between $c_{j}$ and $t^{i}_{a,j}$, we must then swap $t^{i}_{b,j}$ with $u_{i}$ for some $i\in[n]$. This introduces a conflict between the chosen $u_{i}$ and its adjacent vertex $u^{i}_{b}$. After we fix this conflict, we have used a total of 3 swaps, totaling $3k>k^{\prime}$ swaps if we followed this strategy for each $c_{j}$. Thus, as $k^{\prime}=2k$ swaps suffice, and we are looking for an independent set of size exactly $k$, we must swap each of the vertices $c_{j}$ with some $u_{i}$. Clearly, two vertices $u_{i},u_{\ell}$, for $i,\ell\in[n]$, adjacent in $G$ cannot be used to fix the conflicts, for otherwise we would have a conflict between the vertices $u^{i}_{b}$ and $u^{\ell}_{b}$ (both colored $2$). We conclude that the vertices $u_{i}$ a vertex $c_{j}$ is swapped with form an independent set.

Assume the construction of an instance $I=(G^{\prime},r,c,k^{\prime})$ of $r$-Swap of Lemma 10. To prove the claim, it suffices to augment the construction to ensure the promise holds, i.e., that with a finite number of swaps we have that $G^{\prime}$ is properly $r$-colored and $\chi(G^{\prime})=r$.

A double star $S^{\prime}_{n}$ is the complete bipartite graph $K_{1,n}$ with each edge subdivided. Formally, $S^{\prime}_{n}=(\{s\}\cup\{q_{1},...,q_{n}\}\cup\{q^{\prime}_{1},...,q^{\prime}_{n}\},\{(s,q_{i}),(q_{i},q^{\prime}_{i})\mid i\in[n]\})$. To enforce the promise, add $n$ disjoint double stars $S^{\prime}_{k+1}$ to $G^{\prime}$. Each double star is colored such that the central vertex $s$ receives color 1, vertices $q_{i}$ adjacent to the central vertex color 2, and the remaining vertices $q^{\prime}_{i}$ color 3. Observe that after swapping the central vertex $s$ (colored 1) of a $S_{k+1}$ with (say) $b_{1}$ (colored 2), we require $k^{\prime}+1$ more swaps to fix the conflicts residing at that particular $S^{\prime}_{k+1}$. Nevertheless, given enough swaps, $k^{\prime}\cdot(k^{\prime}+1)$ to be precise, we can properly $r$-color $G^{\prime}$. Finally, to guarantee $\chi(G^{\prime})=r$, it suffices to add a disjoint properly $r$-colored clique.

We show that $3$-SAT cross-composes into $r$-Fix parameterized by the number of recolorings $k$. By choosing an appropriate polynomial equivalence relation $\mathcal{R}$, we can assume we are given a sequence $\varphi_{1},\varphi_{2},\ldots,\varphi_{t}$ of $3$-SAT instances with an equal number of variables, denoted by $n$, and an equal number of clauses, denoted by $m$.

Let us then proceed with an cross-composition algorithm that composes $t$ input instances $\varphi_{1}$,$\varphi_{2}$,$\ldots,\varphi_{t}$ which are equivalent under $\mathcal{R}$ into a single instance of $r$-Fix parameterized by the number of recolorings. Specifically, we construct an instance $(G,k)$ of $r$-Fix, where $G$ is a vertex-colored graph, and $k$ the number of recolorings. Our plan is to convert each $3$-SAT instance $\varphi_{1},\varphi_{2},\ldots,\varphi_{t}$ to an instance $\varphi^{\prime}_{1},\varphi^{\prime}_{2},\ldots,\varphi^{\prime}_{t}$ of $4$-SAT. For each resulting instance of $4$-SAT, we apply the standard reduction from $4$-SAT to $3$-Coloring (see e.g., [12]). Finally, the resulting graphs are connected by a spread gadget, which acts as an instance selector. Let us first describe the gadgets, and then the construction of the whole graph $G$. At the same time, we describe a 3-coloring $c:V(G)\to[3]$. We set $k=2\log_{2}(t)+2n+9m$. Our construction depends crucially on $k$, and its choice will become apparent later on.

To obtain $G$, we connect the described gadgets together as follows. For each $\varphi_{h}^{\prime}$, where $h\in[t]$, we add a vertex $w_{h}$. Make $w_{h}$ adjacent to each variable vertex, and set $c(w_{h})=3$. We give $w_{h}$ altogether $2(k+1)$ pendant vertices, and color them so that $k+1$ of them have color $1$, and $k+1$ of them have color $2$. This enforces $c(w_{h})=3$ if only $k$ recolorings are available. In the spread gadget, each leaf $\ell_{h}$ is made adjacent to both $u_{h}$ and $w_{h}$.

This completes the construction of the graph $G$. Recall $k=2\log_{2}(t)+2n+9m$, and output the instance $(G,k)$ of $r$-Fix. Let us then prove that $(G,k)$ is a YES-instance of $r$-Fix iff one of $\varphi_{h}$ is satisfiable for $h\in[t]$.

For the other direction, suppose $\varphi_{s}$ is satisfiable for some $s\in[t]$. Again, the initial vertex-coloring corresponds to a truth assignment $\tau=\{z_{1},z_{2},\ldots,z_{n}\}\in\{1\}^{n}$ setting each variable to true. Using at most $2n+9m$ recolorings, we turn the initial vertex-coloring to a vertex-coloring corresponding to $\tau^{\prime}=\{z^{\prime}_{1},z^{\prime}_{2},\ldots,z^{\prime}_{n}\}\in\{0,1\}^{n}$ such that $\varphi_{s}$ is satisfied under $\tau^{\prime}$. Indeed, observe that as $\varphi_{s}$ is satisfiable, $(P_{1})$ is not violated. Moreover, observe that $\tau^{\prime}$ satisfies $\varphi_{s}$ regardless of the truth value of $u_{s}$. Thus, we can freely let $c(u_{s})=1$. But now we introduce a conflict between $\ell_{s}$ (colored $1$) and its unique neighbor in the spread gadget. However, using precisely $2\log_{2}(t)$ recolorings, we propagate this conflict, and in particular the color $1$, up to the root $r$. At most $k=2\log_{2}(t)+2n+9m$ recolorings have been used, and no conflicts remain in $G$. Thus, if at least one of $\varphi_{1},\ldots,\varphi_{t}$ is a YES-instance of $3$-SAT, then $(G,k)$ is a YES-instance of $r$-Fix. This concludes the proof.

Let $(G=(V,E),W,c)$ be an instance of $r$-PrExt, where $G$ is an $n$-vertex bipartite planar graph, $W\subseteq V$ a set of precolored vertices, and $c:W\to\{0,1,2\}$ a precoloring of the vertices in $W$. By Theorem 15, we may assume without loss that each (precolored) vertex in $W$ has degree 1. Our construction crucially depends on this fact. Let $r=3$ be a fixed color bound, and let $h=|V\setminus W|$. We will construct a graph $H$ along with its vertex-coloring $c_{H}$ such that the precoloring $c$ can be extended to a valid $r$-coloring of $G$ iff at most $h$ swaps are needed to transform $c_{H}$ to an optimal proper vertex-coloring of $H$, i.e., $B(c_{H})\leq h$. To enforce the promise of the problem, it shall hold for $c_{H}$ that (i) it uses precisely $\chi(H)$ colors, and that (ii) by using a finite number of swaps $c_{H}$ can be transformed into a proper coloring of $H$.

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  We retain the coloring on the vertices of $W$, that is, $c_{H}(w)=c(w)$, for every $w\in W$.

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  If $x\in X$ has one or more neighbors colored with color $i$ (observe it cannot have distinctly colored neighbors), we set $c_{H}(x)=i$. If all neighbors of $x$ are uncolored, we set $c_{H}(x)=0$ if $x\in A$. Otherwise, $x\in B$, so we set $c_{H}(x)=1$. For each $x$, we also add two vertices $x_{1}$ and $x_{2}$ along with the edges $xx_{1}$ and $xx_{2}$. We color $c_{H}(x_{1})=(i+1)\bmod 3$ and $c_{H}(x_{2})=(i+2)\bmod 3$.

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  For each precolored vertex $w\in W$ we add $2(h+1)$ new vertices $s_{w,1},\ldots,s_{w,h+1}$ and $t_{w,1},\ldots,t_{w,h+1}$. These will be made pendant vertices of $w$ by adding the altogether $2(h+1)$ edges $ws_{w,\ell}$ and $wt_{w,\ell}$ where $\ell\in[h+1]$. They receive a color as follows, where $c(w)$ denotes the color of $w$:

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    if $w\in A\wedge c(w)\neq 0$, then $c_{H}(s_{w,\ell})=0$ and $c_{H}(t_{w,\ell})=f$, where $f\in C\setminus\{0,c(w)\}$;

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    if $w\in B\wedge c(w)\neq 1$, then $c_{H}(s_{w,\ell})=1$ and $c_{H}(t_{w,\ell})=g$, where $g\in C\setminus\{1,c(w)\}$; and

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    in all other cases $c_{H}(s_{w,\ell})=(i+1)\bmod 3$, and $c_{H}(t_{w,\ell})=(i+2)\bmod 3$.

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  For every precolored vertex $w\in A\wedge c(w)\neq 0$, we add $h$ new vertices $s^{\prime}_{w,1},\ldots,s^{\prime}_{w,h}$ with the edges $s_{w,j}s^{\prime}_{w,j}$, and set $c_{H}(s^{\prime}_{w,j})=c(w)$, where $j\in[h]$.

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  For every precolored vertex $w\in B\wedge c(w)\neq 1$, we add $h$ new vertices $s^{\prime}_{w,1},\ldots,s^{\prime}_{w,h}$ with the edges $s_{w,j}s^{\prime}_{w,j}$, and set $c_{H}(s^{\prime}_{w,j})=c(w)$, where $j\in[h]$.

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  Finally, consider an arbitrary precolored vertex $w\in W$, and its set of $h+1$ pendant vertices $t_{w,j}$. We choose an arbitrary vertex among the $t_{w,j}$ vertices, and call it $v$. Then, we add two vertices $r$ and $r^{\prime}$ along with the edges $vr$, $vr^{\prime}$, and $rr^{\prime}$. These vertices are colored such that $c_{H}(r)=(c_{H}(v)+1)\bmod 3$ and $c_{H}(r^{\prime})=(c_{H}(v)+2)\bmod 3$.