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In this section, some preliminaries concerning tree decompositions are presented.

Definition 2.11.

LetG= (V, E) be a graph. A pair (T,X) withX = (Xi)iV(T)is atree decompositionofGifT is a tree, XiV for every iV(T), and the following three properties are satisfied.

(T1) For everyvV, there is someiV(T) withvXi. (T2) For everyeE, there is someiV(T) witheXi.

(T3) For alli, j, hV(T), ifhis on the (unique)i,j-path inT, thenXiXjXh. Thewidthof a tree decomposition (T,X) is max

|Xi| −1: iV(T) . Thetree-widthtw(G) of a graphG is the smallest integertsuch thatGallows a tree decomposition of widtht.

Consider a tree decomposition (T,X) with X = (Xi)iV(T) of a graph G. To easily distinguish the vertices ofGfrom the vertices ofT, the vertices in V(T) are callednodes in the following. Furthermore, foriV(T), the setXiis referred to as thecluster ofiin (T,X). It is easy to show that (T3) is equivalent to the following condition, see also Section 2 in [Bod98]:

(T3’) For everyvV, the graph T[Iv] withIv={i∈V(T): vXi} is connected.

In the following, (T1), (T2), (T3), and (T3’) always refer to the properties defined here.

Every graphG= (V, E) has a tree decomposition. For example, letT = ({i},∅) andXi=V, then (T,X) where X consists only ofXiis a tree decomposition ofG. It follows that tw(G)n−1 for every graphG onn vertices. If a graphGallows a tree decomposition of width 0, thenE(G) =∅. Next, it is shown that every tree ˜T with at least two vertices has tree-width 1. Let ˜T = ( ˜V ,E) be a tree with˜ |V˜| ≥2. To construct a tree decomposition of ˜T, let T be the tree obtained from ˜T by subdividing each edge of ˜T once. For each eE, denote by˜ ie the vertex used to subdivide e. For eachvV˜, defineXv ={v}

and, for each e={v, w} ∈ E, define˜ Xie ={v, w}. Let X = (Xi)iV(T). Clearly, (T,X) satisfies (T1) and (T2). To see that (T3’) is satisfied, letvV˜, thenvXi if and only ifi=vori=ie for an edgee that is incident tovin ˜T. Hence, for each vV˜, the set{j∈V(T): vXj}={v} ∪ {ie: eE, v˜ ∈e}

induces a connected subgraph ofT, namely a star or an isolated vertex. Consequently, (T,X) is a tree decomposition of ˜T and tw( ˜T)≤1. As ˜T contains at least one edge, tw( ˜T) must be at least 1 due to (T2).

Next, a tree decomposition of the square grid is presented. Fix an integerk≥2 and let ˜G= ( ˜V ,E) be˜ thek×k grid. Recall that ˜V =(˜i,˜j): ˜i∈[k],˜j∈[k] . For eachi∈[k−1] and eachj∈[k] define

X(i−1)k+j :=(i,˜j)V˜: ˜jj ∪(i+ 1,˜j)V˜: ˜jj ,

see Figure2.3for a visualization. LetT be the path obtained fromPk(k−1) by removing the node 0 and letX = (Xh)h∈[k(k−1)]. For all ˜i,˜j∈[k],

i,˜j)Xhh∈(˜i−2)k+ ˜j, . . . ,i−1)k+ ˜j ∩[k(k−1)].

It follows that (T1) and (T3’) are satisfied. To see that (T2) is satisfied, leteE. If˜ e={(˜i,˜j),i+ 1,˜j)}

for some ˜i ∈[k−1] and ˜j ∈ [k], then eXh for h= (˜i−1)k+ ˜j. Otherwise,e ={(˜i,˜j),i,˜j+ 1)} for some ˜i∈[k] and ˜j ∈[k−1] and eXh for h= (˜i−1)k+ ˜j. Consequently, (T,X) satisfies (T2) and (T,X) is a tree decomposition of ˜Gof widthk. Hence, tw( ˜G)k. A matching lower bound is stated in Lemma 88 in [Bod98] or Exercise 21 in Chapter 12 of [Die12]. This yields the next proposition.

Proposition 2.12.

a) Each tree T on at least2 vertices satisfiestw(T) = 1.

(1,1)

(5,1) (1,5)

(5,5) X1

1 = (11)5 + 1

X2 2 = (11)5 + 2

X8

8 = (21)5 + 3

X16

16 = (41)5 + 1 X20

20 = (41)5 + 5 j

i

Figure 2.3:A few clusters of a tree decomposition of the 5×5 grid of width 5.

A tree decomposition (T,X) of a graph G is a path decomposition of G if T is a path. The tree decomposition of the square grid that was presented above is a path decomposition. The width of a path decomposition is defined as the width of a tree decomposition. Thepath-width of a graphGis the smallest integertsuch thatGallows a path decomposition of widtht and is denoted by pw(G). Clearly, all graphsGsatisfy pw(G)≥tw(G).

For planar graphs, the following bound on the tree-width is known. Its proof can be found in [Bod98].

There, Corollary 23 states that the path-width of a planar graph onnvertices is at mostO(√

n). Using that tw(G)≤pw(G) for all graphsGimplies the next proposition.

Proposition 2.13.

Every planar graphGon nvertices satisfiestw(G) =O(√ n).

Consider a graphG= (V, E), a tree decomposition (T,X) withX = (Xi)iV(T), and a graphHG. A tree decomposition forHcan be easily obtained from (T,X) by deleting all vertices that are not inH. More precisely, foriV(T), let ˜Xi=XiV(H). Then, it is easy to check that (T,X˜) with ˜X = ( ˜Xi)iV(T)

is a tree decomposition ofH. The tree decomposition (T,X˜) is called theinduced tree decomposition ofH with respect to (T,X). Observing that the width of (T,X˜) is at most the width of (T,X) yields the next proposition.

Proposition 2.14.

LetGbe a graph and let(T,X)be a tree decomposition ofGof widtht−1. For every subgraphHG, the induced tree decomposition ofH with respect to(T,X)is a tree decomposition of H of width at mostt−1.

Furthermore,tw(H)≤tw(G).

Similarly to the construction for subgraphs, a tree decomposition of a minor can be constructed, see also Lemma 12.3.3 and Proposition 12.3.6 in [Die12]. The following proposition is obtained.

Proposition 2.15.

For all graphsH andG such thatGcontainsH as a minor, tw(H)≤tw(G).

Often, when constructing a cut in a tree ˜T, the vertex set of ˜T is partitioned by removing all edges incident to a vertexvV( ˜T) and considering the vertex sets of the resulting components. Then, a cut in ˜T of width at most ∆( ˜T) is obtained when combining these vertex sets in an arbitrary way. This idea can be generalized by considering clusters of a tree decomposition, as done in the next lemma. It uses the following notation: Consider a graphG= (V, E) and a tree decomposition (T,X) ofG. For each nodei inT define

EG(i) =

eE:eXi6=∅ and eG(i) =|EG(i)|,

where Xi denotes the cluster of i in (T,X). Observe that, when t−1 denotes the width of (T,X), theneG(i)≤ |Xi|∆(G)≤t∆(G) for everyiV(T). We say that two subgraphsH1GandH2G are disjoint parts of G if V(H1)∩V(H2) = ∅ and there is no edge e = {v, w} in G withvV(H1) andwV(H2). Note that, ifGis not connected, then two distinct components ofGare disjoint parts ofG, but the subgraphHi fori∈ {1,2}in the definition of disjoint parts does not have to be connected.

The next lemma says that, if the vertices inXior the edges inEG(i) are removed for someiV(T), then the graph Gsplits into several disjoint parts. So, these disjoint parts can be combined in an arbitrary way to obtain a cut inGof width at mosteG(i)≤t∆(G). The lemma is a widely known fact about tree decompositions, similar statements are Fact 10.13 and Fact 10.14 in [KT06] or Corollary 1.8 in [Ree97].

Lemma 2.16.

LetG= (V, E)be an arbitrary graph and let(T,X)be a tree decomposition ofGwithX = (Xj)jV(T). Fix an arbitrary nodeiV(T), letk:= degT(i), and denote byi1, i2, . . . , ik the neighbors ofiinT. For`∈[k], letV`T be the node set of the component ofTithat contains i` and define V` :=S

jV`TXj\Xi. a) Removing the vertices inXi from GdecomposesGintok disjoint parts, which areG[V1], . . . , G[Vk].

b) Removing the edges inEG(i)fromGdecomposesGintok+|Xi|disjoint parts, which are ({v},∅) for every vXi andG[V`] for every`∈[k].

Proof. LetG= (V, E), (T,X) withX = (Xj)jV(T),i,k,i1, . . . , ik, as well asV`T andV`for each`∈[k]

be as in the statement.

a) For eachvV letIv:={j∈V(T): vXj}, which is the same as in (T3’). Due to (T1), it follows that

[

`∈[k]

V` =

 [

jV(T)\{i}

Xj

\Xi = V \Xi.

Consider a vertex vV. IfvXi, thenv /V` for every`∈[k]. Otherwise,v /Xi andIvV`T for a unique `∈[k] asT[Iv] is connected by (T3’), nonempty by (T1), and does not contain the node i. So vV` impliesv /V`0 for all `0 6= ` and therefore the setsV1, . . . , Vk are a partition ofV \Xi.

It remains to show that, for all distinct `1, `2 ∈[k], there is no edge{v1, v2} ∈E withv1V`1

andv2V`2. Assume, for a contradiction, that there is such an edge inE. Property (T2) says that there is a nodej withv1Xj andv2Xj. So,jIv1Iv2. Alsoj6=ibecausev1/Xi. As argued before,Iv1V`T1 andIv2V`T2, because v1/Xi andv2/ Xi. As`16=`2, the treesT[V`T1] andT[V`T2] are different components ofTi. Hence,V`T1V`T2 =∅ and thereforeIv1Iv2=∅, but this contradicts that jIv1Iv2.

b) Clearly, every vertexvXi is an isolated vertex inGEG(i). So, it suffices to show thatGXi decomposes into the kdisjoint partsG[V1], G[V2], . . . , G[Vk], which is equivalent to Parta). 2

The next proposition says that low tree-width implies that a graph does not have many edges.

Proposition 2.17 (Fact 1.10 in [Ree97]).

Every graphG onnvertices satisfies|E(G)| ≤ntw(G). For a proof see Fact 1.10 in [Ree97].