Odd Cycle Transversal - Tree decompositions of graphs and their eﬀect on algorithmic complexity

be colored using only the colors black and white. Thus, some vertexvonPb_j is colored red. The vertex v corresponds to a group assignment of some groupFi that satisfies Cbj. Asv is red, Lemma 2.17(3) implies thatVi is colored with the coloringµi that corresponds to this assignment. The construction then implies that our chosen assignment satisfies C_j. As this is true for every clause, this concludes the proof.

Observation 2.20. The vertices S

i≤tV_i form a feedback vertex set of G. Furthermore,pw(G)≤ pt+ 4

Proof. Observe that after removingS

i≤tV_i, all that is left are the gadgetsCb_j, which do not have any edges between each other. Each such gadget is a tree and henceS

i≤tV_i form a feedback vertex set ofG. If we place a searcher on each vertex of S

i≤tVi it is easy to see that each gadget Cbj can be searched with4 searchers. The pathwidth bound onGfollows using Proposition 2.2.

Lemma 2.21. Ifq-List Coloring can be solved in (q−)^fvs(G)·n^O(1) or(q−)^pw(G)·n^O(1) time for some >0, then SAT can be solved in (2−δ)ⁿ·n^O(1) time for someδ >0.

Proof. Let(q−)^fvs(G)·n^O(1)=O^∗(q^λfvs(G))time, whereλ= log_q(q−)<1. We choose a sufficiently largep such that δ⁰ =λ_p−1^p <1. Given an instanceφ of SAT, we construct a graph Gusing the construction above, and run the assumedq-List Coloring. Correctness follows from Lemmata 2.18 and 2.19. By Observation 2.20, the graph Ghas a feedback vertex set of sizepd_bp_logⁿ _qce. The choice ofp implies that

λpd n

bplogqce ≤λp n

(p−1) logq +p≤δ⁰ n

logq +p≤δ⁰⁰n,

for someδ⁰⁰<1. HenceSATcan be solved in time 2^δ⁰⁰ⁿ·n^O(1) =(2−δ)ⁿ·n^O(1), for someδ >0. By Observation 2.20, we also know that pw(G)≤pt+ 4. Thus, the feedback vertex set size and the pathwidth of the constructed graph just differs by4. This implies that q-List Coloringcannot be solved in (q−)^pw(G)·n^O(1) time.

Finally, observe that we can reduce q-List-Coloring to q-Coloring by adding a clique Q={q₁, . . . , q_c} onq vertices toGand making q_i adjacent tov when i /∈L(v). Any coloring ofQ must useq different colors, and without loss of generality q_i is colored with colori. Then one can complete the coloring if and only if one can properly colorGusing a color from L(v)for each v. We can add the clique Qto the feedback vertex set—this increases the size of the minimum feedback vertex set by q. Sinceq is a constant independent of the input, this yields Theorem 2.16.

2.6 Odd Cycle Transversal

An equivalent formulation of the Max Cut problem is to ask for a bipartite subgraph with the maximum number of edges, which is the same as asking for a set of edges ofminimumsize whose deletion makes the graph bipartite. We can also consider the vertex-deletion version of this problem.

An odd cycle transversal of a graph Gis a subset S ⊆ V(G) such that G\S is bipartite. In the Odd Cycle Transversalproblem, we are given a graph Gtogether with an integer kand asked whether Ghas an odd cycle transversal of size k.

Theorem 2.22. If Odd Cycle Transversal can be solved in(3−)^pw(G)·n^O(1) time for >0, then SAT can be solved in (2−δ)ⁿ·n^O(1) time for some δ >0.

ÂăÂăuÂă a1 a2 a3 v b1 b2 b3 b4

u a1 a2 a3 v

b1 b2 b3 b4

A(u, v) A(u, v)\ {u}

Figure 2.7: Reduction to Odd Cycle Transversal. The arrow A(u, v) from u to v with the passive odd cycle transversal shown in white (left) and the active odd cycle transversal ofA(u, v)\{u}

(right).

Construction. Given >0and an instanceφof SAT, we construct a graphGas follows. We choose an integer p based just on. Exactly how p is chosen will be discussed at the end of this section. We start by grouping the variables ofφintotgroups F₁, . . . , F_t of size at mosth=blog 3^pc.

Thust=d_{blog 3}ⁿpce. We will call an assignment of truth values to the variables in a group Fi agroup assignment. We will say that a group assignment satisfies a clause C_j of φ ifC_j contains at least one literal that is set to true by the group assignment. Notice that C_j can be satisfied by a group assignment of a groupFi even thoughCj also contains variables that are not inFi.

Now we describe an auxiliary gadget which will be very useful in our construction (see Figure 2.7).

For two vertices u and v by adding an arrow from u to v we will mean adding a path ua₁a₂a₃v on four edges starting in u and ending inv. Furthermore, we add four vertices b1, b2, b3 and b4

and edges ub₁,b₁a₁,a₁b₂,b₂a₂,a₂b₃,b₃a₃,a₃b₄,b₄v, andb₄v. Denote the resulting graphA(u, v).

None of the vertices in A(u, v) except foru and v will receive any further neighbours throughout the construction of G. The graph A(u, v) has the following properties, which are useful for our construction.

• The unique smallest odd cycle transversal ofA(u, v) is{a₁, a3}. We call this the passiveodd cycle transversal of the arrow.

• InA(u, v)\ {a₁, a₃},u and v are in different connected components.

• The set {a₂, v}is a smallest odd cycle transversal of A(u, v)\ {u}. We call this theactive odd cycle transversal of the arrow.

The intuition behind an arrow fromu to v is that ifu is put into the odd cycle transversal, then v can be put into the odd cycle transversal “for free.” When the active odd cycle transversal of the arrow is picked, we say the arrow is active, otherwise we say the arrow is passive.

To constructG, we maket·p paths, {P_i,j}for 1≤i≤t, 1≤j≤p(see Figure 2.8). Each path has3m(tp+ 1)vertices, and the vertices ofP_i,j are denoted byp^`_i,j for1≤`≤3m(tp+ 1). For a fixed i, the paths{P_i,j : 1≤j≤p} correspond to the setFi of variables. For every 1≤i≤t,1≤j ≤p and 1≤` <3m(tp+ 1)we add three verticesa^`_i,j, b^`_i,j andq^`_i,j adjacent to each other. We also add the edges a^`_i,jp^`_i,j andb^`_i,jp^`+1_i,j . One can think of the vertices of the paths {P_i,j}layed out as rows in a matrix, where for every fixed1 ≤`≤3m(tp+ 1)there is a column{p^`_i,j : 1≤i≤t,1 ≤j ≤p}.

We group the colums three by three. In particular, For every i ≤ t and 0 ≤ ` < m(tp+ 1) we define the sets P_i^` = {p^3`+1_i,j , p^3`+2_i,j , p^3`+3_i,j : 1 ≤ j ≤ p}, A^`_i = {a^3`+1_i,j , a^3`+2_i,j , a^3`+3_i,j : 1 ≤ j ≤ p}, B^`_i ={b^3`+1_i,j , b^3`+2_i,j , b^3`+3_i,j : 1≤j≤p} andQ^`_i ={q_i,j^3`+1, q_i,j^3`+2, q^3`+3_i,j : 1≤j ≤p}.

For everyi≤tand0≤` < m(tp+ 1)we make two new sets L^`_i andR^`_i of new vertices. BothL^`_i andR^`_i are independent sets of size 5p, and we add all the edges possible betweenL^`_i andR^`_i. From L^`_i we pick a special vertexλ^`_i and fromR^`_i we pick ρ^`_i. We make all the vertices inA^`_i adjacent to

2.6. ODD CYCLE TRANSVERSAL 37

Figure 2.8: Reduction toOdd Cycle Transversal. The path Pi,j with three different ways of removing a setZ and partitioning the remaining bipartite graph into classes L andR.

all vertices ofL^`_i, and we make all vertices inB_i^` adjacent to all vertices ofR_i^`. We make λ^`_i adjacent the distance along P_i,j between the vertex ofS onP_i,j and the vertex of S⁰ onP_i,j is divisible by3.

Informally, S and S⁰ are equal if they look identical when we superimpose P_i^` onto P_i^`⁰. To every group assignment of variablesFi, we designate a good subset of P_i^` for every `. We designate good subsets in such a way that good subsets corresponding to the same group assignment are equal.

Finally, for every clause C_j, 1 ≤ j ≤ m, we will introduce tp+ 1 cycles. That is, for every 0 ≤ r ≤ tp, we inroduce a cycle Cb_j^r. The cycle contains one vertex for every i ≤ t and group

assignment to F_i, and potentially one dummy vertex to make it have odd length. Going around the cycle counterclockwise we first encounter all the vertices corresponding to group assignments of F1, then all the vertices corresponding to group assignments ofF2, and so on. Fori≤tand every good subset S of P_i^rm+j that corresponds to a group assignment of F_i that satisfiesC_j we add an arrow from x^rm+j_i,S to the vertex on Cb_j^r that corresponds to the same group assignment ofF_i asS does. This concludes the construction ofG.

The intention behind the construction is that if φ is satisfiable, then a minimum odd cycle transversal ofGcan pick:

• One vertex from each triangle{a^`_i,j, b^`_i,j, q^`_i,j} for each1≤i≤t,1≤j ≤p, 1≤` <3m(tp+ 1).

There aretp(3m(tp+ 1)−1)such triangles in total.

• One vertex from {p^3`+1_i,j , p^3`+2_i,j , p^3`+3_i,j }for each1≤i≤t,1≤j≤p,0≤` < m(tp+ 1). There aretpm(tp+ 1)such triples.

• Two vertices from every arrow added,withoutcounting the starting point of the arrow. For each i≤t and0 ≤` < m(tp+ 1), there are2p3^p arrows ending in some cycle X_i,S^` . Hence there are 2p3^ptm(tp+ 1) such arrows. For every i≤ t and 0≤ ` < m(tp+ 1) there are 3^p arrows ending in the cycle Y_i^`. Hence there are3^ptm(tp+ 1)such arrows. For every clause C_j, there aretp+ 1 arrows added for every group assignment that satisfies that clause. Letµ be the sum over all clauses of the number of group assignments that satisfy that clause. The total number of arrows added is then µ(tp+ 1) + (2p+ 1)3^ptm(tp+ 1). Thus the odd cycle transversal can pick 2µ(tp+ 1) + 2(2p+ 1)3^ptm(tp+ 1)vertices from arrows.

• One vertexx^`_i,S for everyi≤tand0≤` < m(tp+ 1). There aretm(tp+ 1)choices for iand`.

We let the α be the value of the total budget, that is the sum of the items above.

Lemma 2.23. If φ is satisfiable, then Ghas an odd cycle transversal of size α.

Proof. Given a satisfying assignmentγ to φ, we construct an odd cycle transversalZ of Gof size α together with a partition of V(G)\Z intoL andR such that every edge ofG\Z goes between a vertex inL and a vertex inR. The assignment to φcorresponds to a group assignment of each Fi

for 1≤i≤t. For every1≤i≤tand0≤` < m(tp+ 1), we add toZ the good subset S ofP_i^` that corresponds to the group assignment ofF_i. Notice that for each fixedi, the sets picked fromP_i^` and P_i^`⁰ are equal for any`,`⁰. At this point we have picked one vertex from {p^3`+1_i,j , p^3`+2_i,j , p^3`+3_i,j } for each 1≤i≤t,1≤j≤p,0≤` < m(tp+ 1).

For every fixed1≤i≤t,1≤j ≤p, there are three cases. Ifp¹_i,j ∈Z, we putp²_i,j intoL andp³_i,j into R. Ifp²_i,j ∈Z we putp¹_i,j into R andp³_i,j intoL. If p³_i,j ∈Z we put p¹_i,j intoL andp²_i,j intoR.

Now, for every4≤`≤3m(tp+ 1)such that p^`_i,j ∈/Z we putp^`_i,j into the same set out of {L, R} as p^`_i,j⁰ where1≤`⁰ ≤3 and`≡`⁰ mod 3.

For every 1 ≤i≤t, 0 ≤`≤ m(tp+ 1), we put L^`_i into L and R_i^` into R. For every triple of a, b, qof pairwise adjacent vertices such that a∈A^`_i,b∈B_i^`, andq ∈Q^`_i, we proceed as follows. The vertexahas a neighbour a⁰ inP_i^` and bhas a neighbour b⁰ inP_i^`. There is a j such that b⁰ is the successor of a⁰ onP_i,j. Thus, there are three cases;

• a⁰∈Z andb⁰ ∈L, we puta inR,q inL and binZ.

• a⁰∈R and b⁰∈Z, we put ainZ,q inRand b inL.

• a⁰∈L andb⁰ ∈R, we put ainR,q inZ and binL.

For every1≤i≤t, 0≤`≤m(tp+ 1), there are many arrows from vertices in P_i^` to vertices on cycles X_i,S^` for good subsets S of P_i^`. For each arrow, if its endpoint inP_i^` is inZ we add the active

2.6. ODD CYCLE TRANSVERSAL 39 odd cycle transversal of the arrow toZ, otherwise we add the passive odd cycle transversal of the arrow toZ. In either case, the remaining vertices on the arrow form a forest, and therefore we can insert the remaining vertices of the arrow intoL andR according to which sets out of {L, R, Z}u andv are in.

For every 1≤i≤t,0≤`≤m(tp+ 1), there is exactly one set S such that the cycle X_i,S^` only has passive arrows pointing into it. This is exactly the setS which corresponds to the restriction of γ toFi. Each cycle X_i,S^` 0 that has at least one arrow pointing into them already contains at least one vertex inZ—the endpoint of the active arrow pointing into the cycle. Thus we can partition the remaining vertices ofX_i,S^` 0 into L and R such that no edge has both endpoints in L or both endpoints inR. For the cycleX_i,S^` , we put x^`_i,S intoZ and partition the remaining vertices ofX_i,S^` intoL andR such that no edge has both endpoints inL or both endpoints in R. We add the active odd cycle transversal in the arrow fromx^`_i,S to the cycleY_i^` into Z. For all other good subsetsS⁰, we add the passive odd cycle transversal in the arrow from x^`_i,S to the cycle Y_i^` intoZ. Thus each cycleY_i^` contains one vertex inZ and the remaining vertices ofY_i^` can be distributed intoL andR.

For every arrow that goes from a vertex x^`_i,S into a cycle Cb_h^r, we add the active odd cycle transversal of the arrow toZ ifx^`_i,S ∈Z and add the passive odd cycle transversal toZ otherwise.

Again the remaining vertices on each arrow can easily be partitioned into L and R such that no edge has both endpoints in Lor both endpoints in R. This concludes the construction of Z. Since we have put the vertices intoZ in accordance to the budget described in the construction it follows that|Z| ≤α. All that remains to show, is that for each1≤h≤mand0≤r≤tp, the cycleCb_h^r has at least one active arrow pointing into it.

The cycle Cb_h^r corresponds to the clause Ch. The clause Ch is satisfied by γ and hence it is satisfied by the restriction ofγ to some groupF_i. This restriction is a group assignment ofF_i and hence it corresponds to a good subsetS of P_i^rm+h, which happens to be exactly Z∩P_i^rm+h. Thus x^rm+h_i,S ∈Z and since the restriction of γ toFi satisfiesCh, there is an arrow pointing fromx^rm+h_i,S and intoCb_h^r. Since this arrow is active, this concludes the proof.

Lemma 2.24. If G has an odd cycle transversal of size α, then φ is satisfiable.

Proof. Let Z be an odd cycle transversal of G of size α. Since G\Z is bipartite, the vertices of G\Z can be partitioned into L andR such that every edge ofG\Z has one endpoint in L and the other inR. Given Z,L andR, we construct a satisfying assignment to φ. Every arrow inGmust contain at least two vertices inZ, not counting the startpoint of the arrow. Let Z~ be a subset of Z containing two vertices from each arrow, but no arrow start point. Observe that no two arrows have the same endpoint, and therefore |Z~|is exactly two times the number of arrows in G. Let Z⁰ =Z\Z.~

We argue that for any1≤i≤tand0≤` < m(tp+1)we have|Z⁰∩(L^`_i∪R^`_i∪A^`_i∪B_i^`∪Q^`_i∪P_i^`)| ≥ 4p. Observe that no vertices in L^`_i, R_i^`,A^`_i,B_i^`, Q^`_i or P_i^` are endpoints of arrows, and hence they do not contain any vertices ofZ. Suppose for contradiction that~ |Z⁰∩(L^`_i∪R^`_i∪A^`_i∪B_i^`∪Q^`_i∪P_i^`)|<4p.

Then there is a vertex inλ∈L^`_i\Z⁰, and a vertexρ ∈R^`_i \Z⁰. Without loss of generality, λ∈L andρ∈R. Furthermore, there is a1≤j ≤p such that

|Z⁰∩ {p^3`+1_i,j , p^3`+2_i,j , p^3`+3_i,j , a^3`+1_i,j , a^3`+2_i,j , a^3`+3_i,j , b^3`+1_i,j , b^3`+2_i,j , b^3`+3_i,j , q_i,j^3`+1, q^3`+2_i,j , q^3`+3_i,j }|<4.

Since{a^3`+1_i,j , b^3`+1_i,j , q_i,j^3`+1}, {a^3`+2_i,j , b^3`+2_i,j , q^3`+2_i,j }and{a^3`+3_i,j , b^3`+3_i,j , q_i,j^3`+3}form triangles and must contain a vertex fromZ⁰ each, it follows that each of these triangles contain exactly one vertex from

Z⁰, and thatZ⁰∩ {p^3`+1_i,j , p^3`+2_i,j , p^3`+3_i,j }=∅. Sinceλ∈L andρ∈R,λis adjacent to all vertices of A^`_i,j and ρ is adjacent to all vertices ofB^`_i,j, it follows that A^`_i,j\Z⁰ ⊆R and B_i,j^` \Z⁰⊆L.

Hence, there are two cases to consider either (1) {p^3`+1_i,j , p^3`+3_i,j } ⊆ L and p^3`+2_i,j ∈ R or (2) {p^3`+1_i,j , p^3`+3_i,j } ⊆ R and p^3`+2_i,j ∈ L. In the first case, observe that either a^3`+2_i,j ∈ R or b^3`+2_i,j ∈ L and hence either a^3`+2_i,j p^3`+2_i,j or b^3`+2_i,j p^3`+3_i,j have both endpoints in the same set out of {L, R}, a contradiction. The second case is similar, either a^3`+1_i,j ∈R orb^3`+1_i,j ∈Land hence either a^3`+1_i,j p^3`+1_i,j or b^3`+1_i,j p^3`+2_i,j have both endpoints in the same set out of{L, R}, a contradiction. We conclude that

|Z⁰∩(L^`_i ∪R^`_i∪A^`_i ∪B^`_i ∪Q^`_i ∪P_i^`)| ≥4p.

For any 1≤i≤tand0≤` < m(tp+ 1),Y_i^` is an odd cycle soY_i^` contains a vertex inZ. If Y_i^` contains no vertices ofZ⁰, then it contains a vertex fromZ~ and there is an active arrow pointing into Y_i^`. The starting point of this arrow is a vertex x^`_i,S for some good subsetS ofP_i^`. Since the arrow is active and x^`_i,S is not the endpoint of any arrow, we know that x^`_i,S∈Z⁰. Hence for any1≤i≤t and 0≤` < m(tp+ 1), we have that either there is a good subsetS ofP_i^` such thatx^`_i,S∈Z⁰ or at least one vertex of Y_i^` is in Z⁰.

The above arguments, together with the budget constraints, imply that for every1≤i≤tand 0≤` < m(tp+1), we have|Z⁰∩(L^`_i∪R^`_i∪A^`_i∪B_i^`∪Q^`_i∪P_i^`)|= 4pand that|Z⁰∩S

{x^`_i,S}∪V(Y_i^`)|= 1, where the union is taken over all good subsets S of P_i^`. It followsZ⁰∩P_i^` is a good subset ofP_i^`. Let S=Z⁰∩P_i^`. The cycleX_i,S^` has odd length, and hence it must contain some vertex fromZ. On the other hand, all the arrows pointing into X_i,S^` are passive, soX_i,S^` cannot contain any vertices from Z. Thus~ X_i,S^` contains a vertex fromZ⁰, and by the budget constraints this must bex^`_i,S.

Now, consider three consecutive vertices p^`_i,j, p^`+1_i,j , p^`+2_i,j for some 1 ≤ i ≤ t, 1 ≤ j ≤ p, 1 ≤` ≤ 3m(tp+ 1)−2. We prove that at least one of them has to be in Z. Suppose not. We know that neither λ^b`/3c_i , ρ^b`/3c_i , λ^b`/3c+1_i nor ρ^b`/3c+1_i are in Z. Thus, without loss of generality {λ^b`/3c_i , λ^b`/3c+1_i } ⊆Land{ρ^b`/3c_i , ρ^b`/3c+1_i } ⊆R. There are two cases. Either p^`_i,j ∈R andp^`+1_i,j ∈L orp^`+1_i,j ∈Landp^`+3_i,j ∈R. In the first case, we obtain a contradiction since eithera^`_i,j ∈R orb^`_i,j ∈L.

In the second case, we get a contradiction since eithera^`+1_i,j ∈R or b^`+1_i,j ∈L. Hence for any three consecutive vertices on P_i,j, at least one of them is inZ. Since the budget constraints ensure that there are at most|V(Pi,j)|/3vertices in Pi,j∩Z, it follows from the pigeon hole principle that there is an 0 ≤r ≤tp such that for any 1 ≤i≤t and1 ≤h≤m and1 ≤h⁰ ≤m, the setP_i^rm+h∩Z equals P_i^rm+h⁰∩Z. Here equality is in the sense of equality of good subsets of P_i^`.

For every1≤i≤t, P_i^rm+1∩Z is a good subset ofP_i^rm+1. IfP_i^rm+1∩Z corresponds to a group assignment ofF_i, then we set the variables inF_ito this assignment. Otherwise we set all the variables inFi to false. We need to argue that every clauseCh is satisfied by this assignment. Consider the cycleCb_h^r. Since it is an odd cycle, it must contain a vertex from Z, the budget constraints and the discussion above implies that this vertex is fromZ~. Hence there must be an active arrow pointing intoCb_h^r. The starting point of this active arrow is a vertexx^mr+h_i,S for some iand good subset S of P_i^mr+h. The set S corresponds to a group assignment of Fi that satisfies C_h. Since the arrow is activex^mr+h_i,S ∈Z⁰, and by the discussion above we have thatP_i^mr+h∩Z⁰=S. Now,S=P_i^mr+h∩Z⁰ andS is equal toP_i^mr+1∩Z⁰ and hence the assignment to the variables ofFi satisfies Ch. Since this holds for all clauses, this concludes the proof.

Lemma 2.25. pw(G)≤t(p+ 1) + 10p3^p.

Proof. We show how to search the graph using at most t(p+ 1) + 10p3^p searchers. The strategy consists of m(tp+ 1) rounds numbered from round 0 to round m(tp+ 1)−1. Each round has t

2.7. PARTITION INTO TRIANGLES 41

In document Tree decompositions of graphs and their eﬀect on algorithmic complexity (Pldal 35-41)