vertex vbi and one searcher on x0. Then one can search the gadgets Ccj one by one. In Gw it is sufficient to use2searchers for eachCcj, whereas in Gafter the edges have been replaced by multiple paths on three edges, we need4 searchers. This combined with Proposition 2.2 gives the desired upper bound on the pathwidth of the graph.
The construction, together with Lemmata 2.12, 2.13, 2.14 and 2.15 proves Theorem 2.11.
2.5 Graph Coloring
A q-coloring of G is a function µ : V(G) → [q]. A q-coloring µ of G is proper if for every edge uv ∈ E(G) we haveµ(u) 6= µ(v). In the q-Coloring problem we are given as input a graph G and the objective is to decide whetherGhas a proper q-coloring. In the List Coloring problem, every vertexv is given a listL(v)⊆[q] of admissible colors. A proper list coloring ofGis a function µ:V(G)→[q]such that µis a proper coloring ofGthat satisfiesµ(v)∈L(v) for everyv∈V(G).
In theq-List Coloringproblem we are given a graphG together with a listL(v)⊆[q]for every vertex v. The task is to determine whether there exists a proper list coloring ofG.
A feedback vertex set of a graph G is a set S ⊆ V(G) such that G\S is a forest; we denote by fvs(G) the size of the smallest such set. It is well-known that tw(G)≤fvs(G) + 1. Unlike the other sections, where we give lower bounds for algorithms parameterized by pw(G), the following theorem gives also a lower bound for algorithms parameterized by fvs(G). Such a lower bound follows very naturally from the construction we are doing here, but not from the constructions in the other sections. It would be interesting to explore whether it is possible to prove tight bounds parameterized by fvs(G) for the problems considered in the other sections.
Theorem 2.16. Letq be a fixed positive integer. If q-Coloringcan be solved in (q−)fvs(G)·nO(1) or(q−)pw(G)·nO(1) time for some >0, then SATcan be solved in (2−δ)n·nO(1) time for some δ >0.
Construction. We will show the result for List Coloring first, and then give a simple reduction that demonstrates that q-Coloringcan be solved in(q−)fvs(G)·nO(1) time if and only if q-List Coloring can.
Depending on and q we choose a parameter p. Now, given an instance φ to SAT we will construct a graphGwith a listL(v) for everyv, such thatGhas a proper list-coloring if and only if φis satisfiable. Throughout the construction we will call color1-red, color2-white and color 3-black.
We start by grouping the variables of φintot groupsF1, . . . , Ftof size at most blogqpc. Thus t = dblognqpce. We will call an assignment of truth values to the variables in a group Fi a group assignment. We will say that a group assignment satisfies a clauseCj of φif Cj contains at least one literal which is set to true by the group assignment. Notice thatCj can be satisfied by a group assignment of a groupFi, even though Cj also contains variables that are not inFi.
For each group Fi, we introduce a set Vi of p vertices v1i, . . . , vpi. The vertices in Vi get full lists, that is, they can be colored by any color in[q]. The coloring of the vertices inVi will encode the group assignment ofFi. There are qp ≥2|Fi| possible colorings ofVi. Thus, to each possible group assignment ofFi we attach a unique coloring ofVi. Notice that some colorings ofVi may not correspond to any group assignments ofFi.
For each clauseCj of φ, we introduce a gadget Cbj. The main part of Cbj is a long path Pbj that has one vertex for each group assignment that satisfiesCbj. Notice that there are at mosttqp possible group assignments, and thatq andp are constants independent of the input φ. The list of every
{red,2}
{red,2}
{red,2}
{red,4}
w02 vil
w2 w3 w4
{red,3}
{red,2,3,4}
w
v
w30 w40 {red,3}
{red,2}
{red,2} {red,4}
v
w {red,2,3,4}
w4
w3
w2
vli
Figure 2.5: Reduction to q-Coloring: the way the connector connects a vertex vi` with v for a particular “bad color” x∈[q]\ {µi(vi`)}. The left side shows the case x=red= 1, the right side x= 2 (q = 4).
vertex onPbj is{red,white,black}. We attach two vertices pstartj andpendj to the start and end ofPbj
respectively, and the two vertices are not counted as vertices of the path Pbj itself. The list ofpstartj is{white}. If|V(Pbj)|is even, then the list ofpendj is {white}, whereas if|V(Pbj)|is odd then the list of pendj is {black}. The intention is that to properly colorPbj, one needs to use the color red at least once, and that once is sufficient. The position of the red-colored vertex on the path Pbj encodes how the clauseCj is satisfied.
For every vertex v on Pbj, we proceed as follows. The vertex v corresponds to some group assignment toFi that satisfies the clauseCj. This assignment in turn corresponds to a coloring of the vertices ofVi. Let this coloring be µi. We build a connector whose role is to enforce thatv can be red only if coloringµi appears onVi. To build the connector, for each vertexvi`∈Vi and color x ∈[q]\ {µi(vi`)} we do the following to enforce that if v is red, thenvi` cannothave colorx (see Figure 2.5).
• If x is red, then we introduce one vertex wy for every color y except for red. We make wy
adjacent to v`i and the list ofwy is{red, y}. Then we introduce a vertexw that is adjacent to v and to all verticeswy. The list ofw is all of[q].
• Ifx is not red, we introduce two verticeswy andwy0 for each colory except for red. We make wy adjacent to v`i and wy0 adjacent to wy. The list of wy is {x,red} while the list of wy0 is {y,red}. Finally, we introduce a vertexwadjacent tov and tow0y for ally. The list ofwis all of [q].
Notice that in the above construction we have reused the namesw,wy andw0y for many different vertices: in each connector, there is a separate vertex w for each vertex v`i ∈ Vi and color x ∈ [q]\ {µi(v`i)}. Building a connector for each vertex v on Pbj concludes the construction of the clause gadgetCbj, and creating one such gadget for each clause concludes the construction ofG(see Figure 2.6). The following lemma, summarizes the most important properties of the connector:
Lemma 2.17. Consider the connector corresponding to a vertex v on Pbj and a coloring µi of Vi.
2.5. GRAPH COLORING 33
v1p v11
Pbj
pstartj pendj
V1 Vt
vt1 vtp
Figure 2.6: Reduction toq-Coloring. The tgroups of vertices V1, . . ., Vt represent thetgroups of variables F1,. . .,Ft (each of size dlogqpe). Each vertex of the clause pathPbj is connected to one groupVi via a connector (multiple vertices on the path can be connected to the same group).
1. Any coloring on Vi and any color c∈ {white,black} on v can be extended to the rest of the connector.
2. Coloring µi on Vi and any color c∈ {red,white,black} on v can be extended to the rest of the connector.
3. In any coloring of the connector, if v is red, then µi appears on Vi.
Proof. 1. For each vertexv`i ∈Vi and color x∈[q]\ {µi(v`i)}we do the following.
• Ifx is red, then in the construction ofCbj we introduced a vertexwy with list {y,red} for every color y 6= red adjacent to v`i, and a vertex w with list [q] adjacent to wy for every y 6= red.
If v`i is colored red, then we color each vertex wy with y and w with red. Notice thatw is adjacent tov, butv is colored either white or black, so it is safe to colorw red. If, on the other hand,v`i is not colored red, we can colorwy red for everyy. Then all the neighbours of whave been colored with red, except for v which has been colored white or black. Thus it is safe to color wwith the color out of black and white which was not used to colorv.
• Ifx is not red, then in the construction ofCbj we introduced two verticeswy and w0y for each colory except for red, and also introduced a vertexw. The vertices wy are adjacent tovi` and for everyy6=red, the vertex wy0 is adjacent to wy. Finally,w is adjacent to all the verticeswy0 and tov. For everyy the list of wy is{x,red}while the list of wy0 is{y,red}. The list ofw is [q]. Ifv`i is colored withx, then we letwy take color red andw0y take color yfor every y6=red.
We colorw with red. In the case thatvi` is colored with a color different fromx, we let wy be colored withx andw0y be colored red for everyy6=red. Finally, all the neighours of wexcept forv have been colored red, while v is colored with either black or white. According to the color ofv, we can either colorw black or white.
2. We can assume that v is red, otherwise we are done by the previous statement. For each vertex v`i ∈Vi and color x∈[q]\ {µi(v`i)}, we do the following.
• Ifx is red, then in the construction ofCbj we introduced a vertexwy with list {y,red} for every color y 6= red adjacent to v`i, and a vertex w with list [q] adjacent to wy for every y 6= red.
Sincevi`0 is not colored red byµi, we can color wy red for everyy. Then all the neighbours of w includingv have been colored with red and it is safe to colorw with white.
• Ifx is not red, then in the construction ofCbj we introduced two verticeswy andwy0 for each colory except for red, and also introduced a vertex w. The verticeswy are adjacent to vi` and for everyy6=red the vertexw0y is adjacent towy. Finally, wis adjacent to all the vertices w0y and to v. For every y, the list ofwy is{x,red} while the list ofw0y is{y,red}. The list ofwis [q]. Since µi colors v`i with a color different from x, we let wy be colored with x andw0y be colored red for everyy6=red. Finally, all the neighours ofwincludingv have been colored red so it is safe to color wwhite.
3. Suppose for contradiction thatv is red, but some vertexvi`∈Vi has been colored with a color x 6= µi(vi`). There are two cases. If x is red, then in the construction we introduced vertices wy adjacent tovi` for every colory 6=red. Also we introduced a vertexw adjacent tov and towy for each y6=red. The list of wy is {red, y}and hence wy must have been colored y for everyy 6=red.
But then w is adjacent to v which is colored red, and to wy which is colored y for everyy 6=red.
Thus vertexwhas all colors in its neighborhood, a contradiction. In the case whenxis not red, then in the construction we introduced two vertices wy and w0y for each y6=red. Eachwy was adjacent to vi` and had {x,red} as its list. Since vi` is colored x, all the wy vertices must be colored red.
For every y6=red, we have that wy0 is adjacent to wy and has {red, y} as its list. Hence for every y 6=red, the vertexw0y is colored withy. But, in the construction we also introduced a vertex w adjacent to v and tow0y for eachy6=red. Thus again, vertexw has all colors in its neighbourhood, a contradiction.
Lemma 2.18. If φ is satisfiable, then Ghas a proper list-coloring.
Proof. Starting from a satisfying assignment of φ, we construct a coloring γ of G. The assignment to φcorresponds to a group assignment to each group Fi. Each group assignment corresponds to a coloring ofVi. For everyi, we letγ color the vertices of Vi using the coloring corresponding to the group assignment of Fi.
Now we show how to complete this coloring to a proper coloring ofG. Since the gadgetsCbj are pairwise disjoint, and there are no edges going between them, it is sufficient to show that we can complete the coloring for every gadget Cbj. Consider the clause Cj. The clause contains a literal that is set to true, and this literal belongs to a variable in some group Fi. The group assignment of Fi satisfies the clause Cj. Thus, there is a vertex v onPbj that corresponds to this assignment.
We set γ(v) as red (that is, γ colorsv red),pstartj is colored white and pendj is colored with its only admissible color, namely black if |V(Pbj)|is even and white if |V(Pbj)|is odd. The remaining vertices of Pbj are colored alternatingly white or black. By Lemma 2.17(2), the coloring can be extended to every vertex of the connector between Vi andv: the coloring appearing onVi is the coloring µi corresponding to the group assignment Fi. For every other vertexu on Pbj, the color ofuis black or white, thus Lemma 2.17(1) ensures that the coloring can be extended to any connector onu.
As this procedure can be repeated to color the gadgetCbj for every clause Cj, we can complete γ to a proper list-coloring of G.
Lemma 2.19. If G has a proper list-coloring γ, then φ is satisfiable.
Proof. Givenγ, we construct an assignment to the variables ofφ as follows. For every groupFi of variables, ifγ colorsVi with a coloring that corresponds to a group assignment ofFi, then we set this assignment for the variables inFi. Otherwise, we set all the variables inFi to false. We need to argue that this assignment satisfies all the clauses of φ.
2.6. ODD CYCLE TRANSVERSAL 35