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Quick Separation in Chordal and Split Graphs1
Pranabendu Misra2
Max-Planck Institute for Informatics, Saarbrucken, Germany. [email protected]
Fahad Panolan4
Indian Institute of Technology Hyderabad, Hyderabad, India. [email protected]
Ashutosh Rai6
Faculty of Mathematics and Physics, Charles University, Prague, Czech Republic.7
Saket Saurabh9
Institute of Mathematical Sciences, HBNI, India, UMI ReLax and University of Bergen, Norway.10
Roohani Sharma12
Institute of Mathematical Sciences, HBNI, India. [email protected]
Abstract14
In this paper we study two classical cut problems, namely Multicut and Multiway Cut on15
chordal graphs and split graphs. In the Multicut problem, the input is a graph G, a collection of `16
vertex pairs (si, ti), i ∈ [`], and a positive integer k and the goal is to decide if there exists a vertex17
subset S ⊆ V (G) \ {si, ti : i ∈ [`]} of size at most k such that for every vertex pair (si, ti), si and ti18
are in two different connected components of G− S. In Unrestricted Multicut, the solution S19
can possibly pick the vertices in the vertex pairs {(si, ti) : i ∈ [`]}. An important special case of the20
Multicut problem is the Multiway Cut problem, where instead of vertex pairs, we are given a set21
T of terminal vertices, and the goal is to separate every pair of distinct vertices in T × T . The fixed22
parameter tractability (FPT) of these problems was a long-standing open problem and has been23
resolved fairly recently. Multicut and Multiway Cut now admit algorithms with running times24
2O(k3)nO(1) and 2knO(1), respectively. However, the kernelization complexity of both these problems25
is not fully resolved: while Multicut cannot admit a polynomial kernel under reasonable complexity26
assumptions, it is a well known open problem to construct a polynomial kernel for Multiway Cut.27
Towards designing faster FPT algorithms and polynomial kernels for the above mentioned problems,28
we study them on chordal and split graphs. In particular we obtain the following results.29
1. Multicut on chordal graphs admits a polynomial kernel with O(k3`7) vertices. Multiway Cut30
on chordal graphs admits a polynomial kernel with O(k13) vertices.31
2. Multicut on chordal graphs can be solved in time min{O(2k · (k3 + `) · (n + m)), 2O(` log k) · (n +32
m) + `(n + m)}. Hence Multicut on chordal graphs parameterized by the number of terminals33
is in XP.34
3. Multicut on split graphs can be solved in time min{O(1.2738k +kn+`(n+m),O(2` ·`·(n+m))}.35
Unrestricted Multicut on split graphs can be solved in time O(4` · ` · (n + m)).36
2012 ACM Subject Classification Theory of computation → Parameterized complexity and exact37
algorithms38
Keywords and phrases chordal graphs, multicut, multiway cut, FPT, kernel39
Digital Object Identifier 10.4230/LIPIcs.CVIT.2016.2340
Full version of the MFCS 2020 submission.41
© John Q. Public and Joan R. Public;licensed under Creative Commons License CC-BY
42nd Conference on Very Important Topics (CVIT 2016).Editors: John Q. Open and Joan R. Access; Article No. 23; pp. 23:1–23:21
Leibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany
23:2 Quick Separation in Chordal and Split Graphs
1 Introduction42
Graph cuts and flows are a central topic in computer science and combinatorial optimization.43
A fundamental problem in this setting is Multicut, where the input is a graph G and a44
collection of ` terminal vertex pairs (si, ti), i ∈ [`], and the goal is to output a minimum45
sized vertex subset S ⊆ V (G) \ {si, ti : i ∈ [`]} such that for every vertex pair (si, ti), si46
and ti are in two different connected components of G− S. Another variant of the problem47
where a solution can possibly contain vertices from terminal pairs is called Unrestricted48
Multicut. Note that Unrestricted Multicut can be easily reduced to Multicut by49
adding a new terminal of degree one for each existing terminal and making it adjacent to the50
existing terminal. An important special case of Multicut is Multiway Cut, where we are51
given a set T of terminal vertices, and the goal is to separate every pair of distinct vertices52
in T × T . One can similarly define Unrestricted Multiway Cut, where the solution can53
possibly pick terminal vertices. When |T | = 2 this is the famous Min (s, t)-Cut problem54
which admits a classical polynomial time algorithm. However, Multiway Cut becomes55
NP-hard even for a set of three terminals [7].56
These problems appear in a number of applications, and they have been intensively57
studied over the past few decades, and several algorithmic tools and hardness results on them58
have been obtained in the field of approximation algorithms. These problems have played an59
important role in the development of the field of parameterized complexity. The first FPT60
algorithm for Multiway Cut parameterized by solution size k was given by Marx [17], who61
introduced the notion of important separators. This notion has since become an important62
algorithmic tool in the design of parameterized algorithms. Then, an algorithm of running63
time 4knO(1) was designed by Chen, Liu, and Lu [3], and later this was improved to 2knO(1)64
by Cygan et al. [5]. In fact, the algorithm of Cygan et al. [5] also gives a 4k−LPnO(1) time65
algorithm for Multiway Cut, where LP is the optimum LP solution for the problem. Marx66
and Razgon [18] and Bousquet et.al [1] independently proved that Multicut is FPT when67
parameterized by the solution size k. In particular, the algorithm of Marx and Razgon [18]68
developed the technique of randomized sampling of important separators which results in the69
best known algorithm for Multicut with running time 2O(k3)nO(1).70
These problems are very well studied from the perspective of kernelization as well. For71
Unrestricted Multiway Cut, a randomized polynomial kernel with O(k3) vertices72
was obtained by Kratsch and Wahlström using the technique of representative families on73
gammoids [16]. They also designed a randomized kernel for Multiway Cut and Multicut74
with O(k`+1) and O(kd√
2`e+1) vertices, respectively.1 Obtaining a polynomial kernel for75
Multiway Cut when the parameter is k alone remains a long standing open problem in76
the field of kernelization. This is also posed as an open problem in the recent book on77
kernelization [10]. However, for Multicut, it is known that if there exists a polynomial78
kernel with parameter k, then co-NP ⊆ NP/poly and the polynomial hierarchy collapses79
to the third level [4]. This effectively rules out a polynomial kernel for this problem when80
parameterized by k alone, while a polynomial kernel when parameterized by both k and ` is81
not ruled out for general graphs.82
Obtaining an algorithm faster than 2O(k3)nO(1) time for Multicut is an outstanding open83
problem, and one way forward towards this is to understand the complexity of Multicut on84
special classes of graphs. Let us recall some results obtained in this direction. Calinescu et al.85
1 This work is an extension of their work of randomized polynomial kernel for Odd Cycle Transversalwhich was awarded the EATCS-IPEC Nerode Prize 2018.
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:3
proved that Multicut is NP-hard even on bounded degree trees, whereas Unrestricted86
Multicut can be solved in polynomial time [6]. They also showed that even Unrestricted87
Multicut becomes NP-Complete on bounded degree graphs of tree-width two. On the88
other hand, Guo et al. proved that Unrestricted Multicut is NP-Complete on interval89
graphs, while Multicut is polynomial time solvable on interval graphs [13]. Papadopoulos90
proved that Multicut is polynomial time solvable on permutation graphs and co-bipartite91
graphs, but NP-Complete on split graphs (which is subclass of chordal graphs) [19]. For92
planar graphs, Dahlhaus et al. showed that Multiway Cut can be solved in time nO(`) [7].93
This running time is improved to 2O(`)nO(√`) and a matching lower bound under ETH is94
provided by Klein and Marx [15]. Note that this is not true in general graphs, as Multiway95
Cut is NP-hard even when ` = 3. Very recently, a polynomial kernel for Multiway Cut on96
planar graphs parameterized by k is obtained by Jansen et al. [14].97
In this paper, we study Multicut and Multiway Cut on chordal graphs and split98
graphs, and obtain new fast FPT algorithms and polynomial kernels for them. These problems99
are formally defined as follows.100
Multicut (MC)Input: A graph G = (V, E), a set of pairs of vertices T = {(si, ti) | i ∈ [`]} and
an integer k.Parameter: k, `
Question: Does there exist S ⊆ V (G) \ ∪i∈`{si, ti} such that |S| ≤ k and there isno path from si to ti for all i ∈ [`] in G− S?
101
Multiway Cut (MWC)Input: A graph G = (V, E), T ⊆ V (G) and an integer k.Parameter: k
Question: Does there exist S ⊆ V (G) \ T such that |S| ≤ k and there is no pathfrom ti to tj for all ti, tj ∈ T , i 6= j, in G− S?
102
Our results and methods.103
Chordal graphs are a well studied sub-class of perfect graphs that contains several other104
graph classes such as split graphs, interval graphs, threshold graphs and block graphs. They105
are characterized by the property that every cycle of length 4 or more has a chord in it.106
Alternatively, they are the class of intersection graphs of a collection of sub-trees of a tree;107
and graphs that have a forest-decomposition where every bag induces a clique.108
Our first result is a polynomial kernel for Multicut on chordal graphs when parameterized109
by k and `. This is obtained by a sequence of reduction rules that are based on the structure110
of the clique-forest of the input chordal graph. First, we get rid of non-terminal simplicial111
vertices, which helps us bound the number of leaves and higher degree nodes in the clique-112
forest of G. One key step here is to ensure that each terminal vertex occurs in exactly one113
bag of the clique-forest decomposition. Then a marking procedure marks a bounded number114
of vertices in high degree bags, and deletes unmarked vertices to bound the size of high degree115
bags. After this, we only need to bound the lengths of degree 2 paths in the clique-forest and116
the number of vertices occurring in them. For bounding the first, we look at the end nodes of117
the degree 2 path which are either a bag containing a terminal or a high degree node in the118
forest, and since their sizes are bounded, it helps us mark bounded number of “interesting”119
degree 2 nodes on the path. Then we apply a reduction rule which deletes the “uninteresting”120
bags from the path while preserving the size of the min-cut between the interesting nodes.121
CVIT 2016
23:4 Quick Separation in Chordal and Split Graphs
Finally we do a similar marking procedure for vertices in the bags of degree 2, as we did for122
high degree nodes. Then, deleting unmarked vertices gives us a polynomial kernel for MC123
parameterized by k and `.124
I Theorem 1. MC admits a kernel with O(k3`7) vertices on chordal graphs.125
We extend this result to a polynomial kernel for MWC on chordal graphs, parameterized126
by k alone. One of the key reduction rules here, first described by [5], shows that the number127
of terminals in T can be reduced to 2k. Then, combined with a tighter analysis of our128
previous kernelization result, we prove the following.129
I Theorem 2. MWC admits a kernel with O(k13) vertices on chordal graphs.130
Next we present FPT algorithms for these problems on chordal graphs. These algorithms131
are based on two crucial ingredients. The first ingredient is a fact that there is a unique132
important ({v}, X)-separator in a chordal graph G of a fixed size k, for any v ∈ V (G) and133
X ⊂ V (G) \ {v} such that X induces a clique in G. This result uses the fact that minimal134
separators in a chordal graph are cliques and a lemma from [18] that bounds the number of135
important ({v},W )-separators inducing cliques, where W ⊂ V (G) and v ∈ V (G) \W . Our136
second ingredient is the design of a Pushing Lemma for MC on chordal graphs based on137
the clique-forest decomposition of the chordal graph. These two ingredients are combined138
with the structure of the graph, to yield fast FPT algorithms for MC on chordal graphs139
parameterized by k or k + `. This is formalized in the theorems below.140
I Theorem 3. MC on chordal graphs can be solved in O(2k · (k3 + `) · (n+m)) time.141
I Theorem 4. MC on chordal graphs can be solved in 2O(` log k) · ` · (n+m) + `(n+m) time.142
Finally, we turn to MC on split graphs. Split graphs are a sub-class of chordal graphs,143
where the vertex set can be partitioned into an independent set and a clique. It is known144
that the problem remains NP-hard on split graphs [19]. We design fast FPT algorithms for it145
parameterized by k or `. We also consider UMC on split graphs and design FPT algorithm146
for it parameterized by `. Let us note that, while UMC can be easily reduced to MC, the147
reduction does not produce a split graph. Hence, we need a slightly different algorithm for it.148
I Theorem 5. MC on split graphs can be solved in O(1.2738k + kn+ `(n+m)) time.149
I Theorem 6. MC on split graphs can be solved in O(2` · ` · (n+m)) time.150
I Theorem 7. UMC on split graphs can be solved in O(4` · ` · (n+m)) time.151
2 Preliminaries152
We use [n] to denote the set of first n positive integers {1, 2, 3, . . . n}. For a graph G, we153
denote the set of vertices of the graph by V (G) and the set of edges of the graph by E(G).154
We denote |V (G)| and |E(G)| by n and m respectively, where the graph is clear from context.155
We abbreviate an edge {u, v} as uv sometimes. For a set S ⊆ V (G), the subgraph of G156
induced by S is denoted by G[S] and it is defined as the subgraph of G with vertex set S157
and edge set {{u, v} ∈ E(G) : u, v ∈ S} and the subgraph obtained after deleting S (and158
the edges incident to the vertices in S) is denoted by G − S. For v ∈ V (G), we will use159
G− v to denote G− {v} for ease of notation. All vertices adjacent to a vertex v are called160
neighbours of v and the set of all such vertices is called the open neighbourhood of v, denoted161
by NG(v). For a set of vertices S ⊆ V (G), we define NG(S) = (∪v∈SN(v)) \ S. We define162
the closed neighbourhood of a vertex v in the graph G to be NG[v] := NG(v) ∪ {v} and163
closed neighbourhood of a set of vertices S ⊆ V (G) to be NG(S) := NG(S) ∪ S. We drop164
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:5
the subscript G when the graph is clear from the context. We say a vertex v is simplicial in165
G if N(v) forms a clique in G. For C ⊆ V (G), if G[C] is connected and N(C) = ∅, then we166
say that G[C] is a connected component of G.167
A path P in a graph G is a subgraph of G where V (P ) = {x1, x2, . . . , x`} ⊆ V (G)168
and E(P ) = {{x1, x2}, {x2, x3}, . . . , {x`−1, x`}} ⊆ E(G) for some ` ∈ [n]. We denote it by169
P := x1x2 . . . x`. The vertices x1 and x` are called endpoints of the path P and the remaining170
vertices in V (P ) are called internal vertices of P . The length of a path is the number of vertices171
in it. We also say that P is a path between x1 and x`. A cycle C in G is a subgraph of G where172
V (C) = {x1, x2, . . . , x`} ⊆ V (G) and E(C) = {{x1, x2}, {x2, x3}, . . . , {x`−1, x`}, {x`, x1}} ⊆173
E(G), i.e., it is a path with an additional edge between x1 and x`. Let P be a path in the174
graph G on at least three vertices. We say that {u, v} ∈ E(G) is a chord of P if u, v ∈ V (P )175
but {u, v} /∈ E(P ). Similarly, for a cycle C on at least four vertices, {u, v} ∈ E(G) is a chord176
of C if u, v ∈ V (C) but {u, v} /∈ E(C). A path P or cycle C is chordless if it has no chords.177
Let us note that any chordless cycle has length at least 4. We also use P and C to denote178
the set of vertices or edges of the path P or cycle C respectively, when it is clear from the179
context. A walk W from u to v in G is a sequence W := v1v2 . . . vw of vertices of G such180
that v1 = u, vw = v, and for all i ∈ [w− 1], either vi = vi+1, or vivi+1 ∈ E(G). Observe that181
if there exists a walk from u to v in G, this implies that there is also a path from u to v in G.182
A set S ⊆ V (G) \ {u, v} is called a (u, v)-separators for u, v ∈ V (G), if there is no path183
from u to v in G− S. For X,Y ⊆ V (G), an (X,Y )-separator in G is a set S ⊆ V (G) such184
that there is no path from x to y in G − S for all x ∈ X, y ∈ Y . A set S ⊆ V (G) is a185
minimal separator of a graph G, if there exist u, v ∈ V (G) such that S is inclusion-wise186
minimal (u, v)-separator.187
I Definition 8. A forest-decomposition of a graph G is a pair (F, β), where F is a forest188
and and β : V (F )→ 2V (G) such that189
∪x∈V (F )β(x) = V (G),190
for every edge uv ∈ E(G) there exists x ∈ V (F ) such that {u, v} ⊆ β(x), and191
for every vertex v ∈ V (G) the subgraph of F induced by the set β−1(v) := {x | v ∈ β(x)}192
is connected.193
For x ∈ V (F ), we call β(x) the bag of x, and for the sake of clarity of presentation, we194
sometimes use x and β(x) interchangeably. We refer to the vertices in V (F ) as nodes. A195
tree-decomposition is a forest-decomposition where F is a tree. For two adjacent nodes x1196
and x2, β(x1) ∩ β(x2) is called adhesion of x1 and x2. For a path P = x1x2 . . . xp−1xp in F ,197
the set of adhesions on the path P refers to the set {β(xi) ∩ β(xi+1) : i ∈ [p− 1]}. We will198
state a simple property of adhesions which we will use repeatedly.199
I Lemma 9 (folklore). Let (F, β) be a forest-decomposition of G, and let P = x0x1x2 . . . xp200
be a path in F such that u ∈ β(x0), v ∈ β(xp) and {u, v}∩β(xi) = ∅ for all i ∈ {1, . . . , p−1}.201
Then for all Ai := β(xi−1) ∩ β(xi), there is no path from u to v in G−Ai.202
Chordal Graphs: A graph G is called chordal if it does not contain any chordless cycle203
of length at least four. It is well known that the set of chordal graphs is closed under the204
operation of taking induced subgraphs and contracting edges [12]. A clique-forest of G is205
a forest-decomposition of G where every bag is a maximal clique. We further insist that206
every bag of the clique-forest is distinct. The following lemma shows that the class of chordal207
graphs is exactly the class of graphs that have a clique forest.208
I Lemma 10 ( [12]). A graph G is a chordal graph if and only if G has a clique-forest.209
CVIT 2016
23:6 Quick Separation in Chordal and Split Graphs
It is also known that if G is chordal, then its clique-forest can be computed in O(m+ n)210
time [11]. Observe that since every bag is a maximal clique, not only the bags are distinct211
in the clique-forest (F, β) of G, but also for any x, y ∈ V (F ), we have that none of β(x)212
and β(y) is a subset of the other, i.e., β(x) * β(y) and β(y) * β(x). Also, given a forest213
F and a surjective function β : V (F ) → S where S ⊆ 2V , such that it satisfies property214
3 of Definition 8, we can associate a graph G with V (G) = ∪S∈SS and E(G) defined by215
uv ∈ E(G) if and only if there exists x ∈ V (F ) such that {u, v} ⊆ β(x). It is easy to see that216
in this case the graph G is chordal and that the bags of (F, β) correspond to the maximal217
cliques of G and we say that G is the chordal graph associated with the clique-forest (F, β).218
We need another property of clique-forests of chordal graphs, which says that deleting219
some adhesion is necessary to disconnect vertices that do not occur in the same bag.220
I Lemma 11. Let (F, β) be the clique-forest of a chordal graph G, and let P = x0x1x2 . . . xp221
be a path in F such that u ∈ β(x0), v ∈ β(xp) and {u, v}∩β(xi) = ∅ for all i ∈ {1, . . . , p−1}.222
Let Ai = β(xi) ∩ β(xi−1) be the adhesions on P for i ∈ [p]. Then for any (u, v)-separator S223
in G, there exists i ∈ [p] such that Ai ⊆ S.224
Proof. For the sake of contradiction, suppose that S does not contain any adhesion on P .225
This gives that Ai \ S is nonempty for all i ∈ [p]. Let vi be an arbitrary vertex in Ai \ S.226
Since Ai ∩ Ai+1 ⊆ β(xi), either vi = vi+1 or vivi+1 ∈ E(G) for all i ∈ [p− 1]. This means227
that W := uv1v2 . . . vrv is a walk in G− S, and this would imply there is a path from u to v228
in G− S, a contradiction. J229
3 A Polynomial Kernel for Multicut on Chordal Graphs230
In this section we will show that Multicut (MC) admits a polynomial kernel on chordal231
graphs parameterized by the solution size and the number of terminal pairs, that is we232
prove Theorem 1. Throughout this section, we will assume that the input graph G in an233
MC instance (G,T, k) is chordal, unless otherwise stated. We will use T ∗ to denote the set234
of all terminals, i.e., T ∗ :=⋃i∈`{si, ti}. We also associate a measure τ with the instance235
(G,T, k), where τ := |T ∗|. We present a series of reduction rules, which will be applied in236
order, assuming that while applying a reduction rule, none of the previous reduction rules237
apply to the current instance. In this section we will say that a reduction rule is correct,238
if in addition to the input and output instances being equivalent, the graph in the output239
instance is chordal. We first give a reduction rule which takes care of trivial instances and240
vertices which are in common neighborhood of some terminal pair.241
I Reduction Rule 1. Let (G,T, k) be an instance of MC. If there exist (si, ti) ∈ T such that242
siti ∈ E(G), say No. Let Si := N(si) ∩N(ti) for all i ∈ [`] and let S := ∪i∈[`]Si. Output243
(G− S, T, k − |S|).244
The correctness of the reduction rule follows from the fact that any solution must delete245
Si for all i ∈ [`]. After application of the reduction rule, we see that no two terminals are246
adjacent, so they do not appear together in any of the bags of the clique-forest of G. Now247
we present a rule which deletes simplicial vertices from the graph which are not terminals.248
I Reduction Rule 2. Let (G,T, k) be an instance of MC. If there exists v ∈ V (G) \T ∗ such249
that v is a simplicial vertex in G, delete v.250
I Lemma 12. Reduction Rule 2 is correct.251
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:7
Proof. Let G′ = G− v. As G′ is an induced subgraph of G, it is chordal, and any solution252
to (G,T, k) remains a solution to (G′, T, k). For the converse, let S ⊆ V (G′) be a solution253
to (G′, T, k). Suppose, for the sake of contradiction that S is not a solution to G. Then254
there exist (si, ti) ∈ T such that G has a path between si and ti in G− S. Let us look at255
a shortest (hence chordless) path P between si and ti in G− S. As there is no such path256
between si and ti in G′ − S, P must pass through v. Let u1 and u2 be the neighbours of257
v on P . Then u1u2 ∈ E(G) as v is simplicial, and so u1u2 is a chord of P , a contradiction.258
This shows that S is a solution to (G,T, k) and finishes the proof of the lemma. J259
Now we can show that each leaf bag in the clique-forest of G must contain a terminal.260
I Lemma 13. Let (G,T, k) be an instance of MC after applying Reduction Rule 2, and let261
(F, β) be a clique-forest of G. Then for each leaf node x ∈ V (F ), β(x) ∩ T ∗ 6= ∅.262
Proof. Let us suppose, for the sake of contradiction, that there exists a leaf x ∈ V (F ),263
such that β(x) ∩ T ∗ = ∅. Let NF (x) = {x1}. Since all the bags are distinct and also264
maximal cliques, we must have a vertex v ∈ V (G) \ T ∗ such that v ∈ β(x) \ β(x1). But then265
N [v] = β(x) which is a clique and hence v is simplicial in G. It is a contradiction to that266
fact that Reduction Rule 2 does not apply and proves the statement of the lemma. J267
Now we apply another reduction rule to make sure that all the terminals are part of268
exactly one bag in the clique-forest of G.269
I Reduction Rule 3. Let (G,T, k) be an instance of MC. Let G′ be obtained from G by270
making k + 1 copies of each t ∈ T ∗ and introducing a new terminal vertex t′ to G′ which271
forms a clique with the copies of t. More formally, V (G′) = (V (G) \ T ∗)⋃
(∪t∈T∗Ct))⋃T ′∗,272
where Ct = {v1t , . . . , v
k+1t }, T ′∗ = ∪t∈T∗{t′}, V (G) ∩ (T ′∗ ∪ (∪t∈T∗Ct)) = ∅, NG′(vjt ) =273
NG(t) ∪ Ct ∪ {t′}, and NG′(t′) = Ct for all t ∈ T ′∗ and j ∈ [k + 1]. (G′, T ′, k) is the new274
instance, where T ′ = ∪(si,ti)∈T {(s′i, t′i)}.275
I Lemma 14. Reduction Rule 3 is correct.276
Proof. We first show that G′ is chordal. Let us assume it is not, then it contains a277
chordless cycle C := v1v2v3 . . . vqv1 where q ≥ 4. Since NG′(t′) is a clique for all t ∈ T ′∗,278
C does not contain t′ for all t′ ∈ T ′∗. Now we look at some Ci. Since for all vxt , vyt ∈ Ci,279
NG′ [vxt ] = NG′ [vyt ] = NG(t) ∪ Ct ∪ {t′}, if there are two vertices in C from Ct, then that280
would give rise to a chord in C. So we have that C contains at most one vertex from every281
Ct. For a fixed t, let this vertex be vjt . Now, since NG′(vjt ) = NG(t) ∪ Ct ∪ {t′}, and t′ /∈ C282
and C ∩ Ct = {vjt }, we have that the neighbours of vjt on C in G′ are also neighbours of283
t in G, and the non-neighbours of vjt on C in G′ are either new vertices or they are also284
non-neighbours of t in G. This enables us to replace each occurrence of a unique vertex from285
some Ct with t and get a chordless cycle in G, which is a contradiction.286
Let S be a solution for (G,T, k). We show that S is also a solution for (G′, T ′, k). Suppose287
that is not the case, then there exist (s′i, t′i) ∈ T ′ such that there is a path from s′i to t′i288
in G′ − S. Let P be one shortest path from s′i to t′i in G′ − S. Since Csiforms a clique289
and all vertices from Csi have the same closed neighbourhood, we have that exactly one290
vertex from Csiand Cti respectively appears on P . Let P := s′iv
xsiv1 . . . vpv
ytit′i. Since291
NG′(vjsi) = NG(si) ∪ Csi
∪ {s′i} for all j ∈ [k + 1], and vxsiis the only vertex from Csi
, we292
have that v1 ∈ NG(si). Similarly, we can argue that vp ∈ NG(ti). This means that the path293
P ∗ := siv1 . . . vpti exists in G− S between si and ti, which is a contradiction.294
For the converse, let S′ be a minimal solution for (G′, T ′, k). We claim that S′ \ (∪t∈T∗Ct)295
is a solution to (G,T, k). Suppose not, and let P := siv1 . . . vpti be a path from si to ti in296
CVIT 2016
23:8 Quick Separation in Chordal and Split Graphs
G− S′ for some (si, ti) ∈ T . Since |Csi | = |Cti | = k + 1, we have that Csi , Cti * S′. So we297
have some x, y ∈ [k + 1] such that vxsi, vyti /∈ S
′. Since NG′(vjsi) = NG(si) ∪ Csi
∪ {s′i} for298
all j ∈ [k + 1], it is easy to see that P ′ := s′ivxsiv1 . . . vpv
ytit′j is a path in G′ − S′, which is a299
contradction. J300
Observe that Reduction Rule 3 preserves the number of terminals. Also, if reduction301
rules 1 and 2 do not apply before the application of Reduction Rule 3, then they do not302
apply after the application of Reduction Rule 3 as well. This happens because we do not303
add any simplicial vertices, and also do not introduce any edge between terminal pairs or304
vertex in the common neighbourhood of terminal pairs.305
I Lemma 15. Let (G,T, k) be an instance of MC obtained after applying Reduction Rule 3.306
For each t ∈ T ∗, t belongs to exactly one bag of the clique-forest of G, and |N(t)| = k + 1.307
Proof. Let (G,T, k) be an instance obtained after applying Reduction Rule 3. Clearly, for308
each t ∈ T ∗, |N(t)| = k + 1 due to Reduction Rule 3. To see that a terminal t ∈ T ∗ belongs309
to exactly one bag of the clique-forest (F, β) of G, let us suppose that it is not the case and310
let x1, x2 ∈ V (F ) be two adjacent nodes such that t ∈ β(x1) ∩ β(x2). We know that all the311
bags in the clique-forest are distinct, and none is a subset of the other. This means that there312
exist u1 ∈ β(x1) \ β(x2) and u2 ∈ β(x2) \ β(x1) and u1u2 /∈ E(G), as no bag contains both313
u1 and u2 because of the connectivity property of the clique-forest (F, β). But we also know314
that β(x1) and β(x2) form a clique in G and hence u1, u2 ∈ NG(t). This is a contradiction315
to the fact that the neighbourhood of a terminal forms a clique after applying Reduction316
Rule 3, and proves the lemma. J317
I Lemma 16. Let (G,T, k) be an instance obtained after applying Reduction Rule 3 and let318
(F, β) be the clique forest of G. Let V (F ) = F1 ∪ F2 ∪ F≥3, where F1 is the set of leaves of319
F , F2 is the set of nodes of F with degree exactly 2 and F≥3 is the set of nodes of degree at320
least 3. Let FT be the set of nodes that contain terminals. Then F1 ⊆ FT , |FT |, |F≥3| ≤ τ ,321
and |⋃x∈FT
β(x)| ≤ (k + 2)τ .322
Proof. We know from Lemma 13 that after applying Reduction Rule 2, each leaf bag contains323
a terminal. This gives us F1 ⊆ FT . We also know that because of Reduction Rule 3 and324
Lemma 15, each terminal occurs in exactly one bag of (F, β). This gives us |FT | ≤ τ . As the325
number of vertices with degree at least three in a forest is at most the number of leaves, we326
also get |F≥3| ≤ τ as desired. We know that each bag in FT contains a terminal and is a327
clique in G, and hence⋃x∈FT
β(x) ⊆ NG[T ]. From Lemma 15, we get that |N(t)| = k + 1328
for all t ∈ T ∗. This gives |⋃x∈FT
β(x)| ≤ (k + 2)τ . J329
So far, we have bounded the number of leaf bags, the number of bags with degree at330
least three in the clique-forest of G, and also the number of vertices appearing in the bags331
that contain terminals (which includes leaf bags). Next we will bound the number of vertices332
appearing in the bags with degree at least three. For that, we need to define some notions333
first.334
We have established that after applying the aforementioned reduction rules, every terminal335
appears in exactly one bag of the clique-forest (F, β) of G. This enables us to define the336
notion of clique paths between terminals. For (si, ti) ∈ T , let xsiand xti be the unique and337
distinct nodes in V (F ) that contain si and ti respectively. Now, we look at the unique path338
between xsi and xti in F and call it ΠG(si, ti). We will drop the subscript G if the graph is339
clear from context.340
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:9
Now, for each pair (si, ti) ∈ T such that Π(si, ti) is non-empty, for each bag β(x) for341
x ∈ Π(si, ti) of degree at least 3 that does not contain a terminal, we want to mark at most342
2k + 2 vertices in β(x). Let Π(si, ti) := x1x2 . . . xd be a nonempty path where xsi= x1 and343
xti = xd. Let xp ∈ Π(si, ti), p ∈ {2, . . . , d− 1} be an internal node of P with degree at least344
3 that does not contain a terminal. We define two orderings ≤(si,ti) and ≤(ti,si) on vertices345
of β(xp) as follows. For u, v ∈ β(x), u ≤(si,ti) v if and only if, for all q ≥ p, p, q ∈ [d], if346
u ∈ β(xq) then v ∈ β(xq). Similarly, for defining ≤(ti,si), we say that u ≤(tj ,si) v if and347
only if, for all q ≤ p, p, q ∈ [d], if u ∈ β(xq) then v ∈ β(xq). In other words, the ordering348
represents how far along Π(si, ti) the vertices of β(xp) go, ranking the ones that go the349
farthest on either side as the highest.350
Now we describe the marking procedure. For each bag x ∈ F≥3 \ FT , for each (si, ti) ∈ T351
for which x is an internal vertex of Π(si, ti), we mark k + 1 vertices which are largest in the352
ordering ≤(si,ti) and call the set Mx(si, ti). We also mark k + 1 vertices which are largest353
in the ordering ≤(ti,si) and call that set Mx(ti, si). Let the set of all marked vertices inside354
a bag β(x), such that x ∈ F≥3 \ FT be M(x) :=⋃
(si,ti)∈T (Mx(si, ti) ∪Mx(ti, si)) and let355
M :=(∪x∈F≥3\FT
M(x))∪(∪t∈T∗ β(xt)
).356
Now we are ready to give the next reduction rule which will help us bound the size of357
bags in F≥3.358
I Reduction Rule 4. Let (G,T, k) be an instance of MC where G is chordal graph and let359
(F, β) be the clique-forest of G. If there exists a node x ∈ F≥3 \ FT such that β(x) \M is360
nonempty, then delete an arbitrary vertex v ∈ β(x) \M from G.361
I Lemma 17. Reduction Rule 4 is correct.362
Proof. Let G′ := G− v. As G′ is an induced subgraph of G, G′ is chordal and any solution363
for (G,T, k) is a solution for (G′, T, k). For the other direction, let S be a solution for364
(G′, T, k). We claim that S is also a solution for (G,T, k). Suppose that it is not the case,365
then there exists (si, ti) ∈ T such that there is a path between si and ti in G − S. Also,366
this path must pass through v, as otherwise it would also exist in G′ − S. Let the shortest367
path between si and ti in G−S be P := siv1v2 . . . vp−1vpvp+1 . . . vqti, which has length q+ 1368
and let vp := v for some p ∈ [q]. We claim that a path P ′ := siv1v2 . . . vp−1v′pv′′pvp+1 . . . vqti369
exists in G′ − S between si and ti, where v′p, v′′p ∈ M , which will be a contradiction to S370
being a solution for (G′, T, k).371
To see that, let us first observe that since P is a shortest path between si and ti in372
G− S, V (P ) ⊆⋃x∈ΠG(si,ti) β(x). If ΠG(si, ti) does not contain any node of degree at least373
3, then v /∈ ∪x∈V (ΠG(si,ti))β(x), and hence v /∈ P which is a contradiction. Clearly, v is374
also not in the bags containing si or ti because then by definition of M , v would not be375
deleted. So v occurs in a bag of degree at least 3, that is an internal vertex of ΠG(si, ti)376
and does not contain a terminal. Let xv ∈ F≥3 \ FT be a node such that v ∈ β(xv).377
Clearly, v /∈Mxvas otherwise v would not be deleted. We know that the marking procedure378
marked k + 1 vertices in β(xv) as Mx(si, ti). Let v′′p be an arbitrary vertex in Mx(si, ti) \ S.379
Similarly, let v′p be an arbitrary vertex in Mx(ti, si) \S. Now, we want to show that the path380
P ′ := siv1v2 . . . vp−1v′pv′′pvp+1 . . . vqti exists in G′ − S. We first observe that V (P ′) ⊆ V (G′).381
Now, all we need to show is that the edges vp−1v′p, v′pv′′p , v
′′pvp+1 ∈ E(G). As G′ is an induced382
subgraph of G, this would mean that these edges also exist in G′, proving the claim.383
Clearly, v′pv′′p ∈ E(G) as v′p, v′′p ∈ β(xv) which induces a clique in G. Let ΠG(si, ti) :=384
xsix1x2 . . . xsxti where xr = xv for some r ∈ [s]. Since v /∈ xti , we have that there exists385
α ∈ {r, r+1, . . . s} such that v, vp+1 ∈ β(xα). Now, since v′′p ∈Mxv (si, ti) and v /∈Mxv (si, ti),386
we have that v′′p ≥(si,ti) v. Which means that for all xα′ , α′ ∈ {r, r + 1, . . . s}, if v ∈ β(xα′)387
CVIT 2016
23:10 Quick Separation in Chordal and Split Graphs
then v′′p ∈ β(xα′). This would mean that v′′p ∈ β(xα) and v′′pvp+1 ∈ E(G) as β(xα) induces388
a clique in G. Similarly we can show that vpv′p ∈ E(G), which finishes the proof of the389
lemma. J390
I Lemma 18. Let (G,T, k) be an instance of MC after applying Reduction Rule 4 exhaus-391
tively, and let (F, β) be the clique-forest of G. Let F≥3 be set of nodes of F with degree at392
least 3. Then, |β(x)| = O(k`τ) for all x ∈ F≥3 and |⋃x∈F≥3
β(x)| = O(k`τ2).393
Proof. We know from Lemma 16 that |F≥3| ≤ τ . So all we need to show that for all x ∈ F≥3,394
|β(x)| = O(k`τ). For a bag β(xt) containing a terminal t, we know from Lemma 15 that395
|β(xt)| ≤ k + 2. We already know due to Lemma 16 that | ∪t∈T β(xt)| ≤ (k + 2)τ Now, for396
a bag β(x) such that x ∈ F≥3 \ FT , we want to give a bound for Mx. We mark at most397
2k + 2 vertices for each (si, ti) ∈ T . That gives us |Mx| ≤ `(2k + 2). Now, combining that398
with |F≥3| ≤ τ , we get that | ∪x∈F≥3 M(x)| ≤ `τ(2k + 2). Now, if for some x ∈ F≥3 \ FT ,399
|β(x)| > (2k+2)`τ+(k+2)τ , then β(x)\M is non-empty and Reduction Rule 4 applies. This400
gives us |β(x)| ≤ (2k + 2)`τ + (k + 2)τ = O(k`τ) for all x ∈ F≥3 and proves the lemma. J401
Now we have bounded the number of vertices in the graph which are part of degree 1402
and degree at least 3 bags of the clique-forest of G, and also the number of vertices which403
appear in any bag that contains a terminal. What remains to be bounded is the number of404
degree 2 nodes and the number of vertices of G that appears in the bags corresponding to405
the degree 2 nodes. For that, first we will bound the length of a path which consists of only406
degree 2 nodes. To that end, we first describe a marking procedure, that marks a bounded407
number of degree 2 nodes.408
Let Q := x1x2 . . . xq be a path in F such that x1, xq ∈ F≥3 ∪ FT and xi /∈ F≥3 ∪ FT for409
all i ∈ {2, 3, . . . , q − 1}. That is, Q is a path in F with all internal nodes having degree 2410
and not containing any terminal, while the first and the last nodes either have degree at411
least 3 or they contain a terminal. Now, we mark some nodes in Q as D(Q) ⊆ V (Q). For412
that, let B1 := β(x1) and let Bq = β(xq). Suppose xi and xi+1, i ∈ [q − 1] are such that413
B1 ∩ (β(xi) \ β(xi+1)) 6= ∅. In such a case, we add xi and xi+1 to D(Q). Similarly, if xj and414
xj−1, j ∈ {2, 3, . . . , q} are such that Bq ∩ (β(xj) \ β(xj−1)) 6= ∅, then we add xj and xj−1 to415
D(Q).416
B Observation 1. Let Q := x1x2 . . . xq be a path in F such that x1, xq ∈ F≥3 ∪ FT and xi /∈417
F≥3 ∪FT for all i ∈ {2, 3, . . . , q− 1}. Let B1 := β(x1) and Bq := β(xq). Let Q′ := y1y2 . . . yr418
be a a subpath of Q such that y1, yr ∈ D(Q) but yi /∈ D(Q) for all i ∈ {2, 3, . . . , r− 1}. Then419
B1 ∩ β(y1) = B1 ∩ β(y2) = . . . = B1 ∩ β(yr) and Bq ∩ β(y1) = Bq ∩ β(y2) = . . . = Bq ∩ β(yr).420
We next state a lemma, which shows that it is necessary and sufficient to pick one421
adhesion in the solution for every terminal pair, and that any minimal solution can be looked422
at as a collections of adhesions of the clique-forest, at most one of which comes from any423
degree 2 path in the clique-forest.424
I Lemma 19. Let S be a minimal solution to an instance (G,T, k) of MC where G is425
a chordal graph and let (F, β) be the clique-forest of G. Then, there exist Si ⊆ S for all426
(si, ti) ∈ T , such that427
1. Si is an adhesion on Π(si, ti),428
2. S =⋃
(si,ti)∈T Si, and429
3. if Q := x1x2 . . . xq is a path in F such that xz /∈ F≥3 ∪ FT for all z ∈ {2, 3, . . . , q − 1},430
then at most one adhesion from the path Q is picked as Si for some pair(s) (si, ti) ∈ T .431
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:11
Proof. The first part of the lemma is obvious from Lemma 11, as it is necessary to pick one432
adhesion on Π(si, ti) to disconnect the pair (si, ti). For the second part, we observe that it433
suffices to delete one adhesion from Π(si, ti) to disconnect si from ti due to Lemma 9.434
For the third part, suppose that an adhesion SQ is picked from Q in S. Then observe435
that, by Lemma 9, SQ separates every path in G that contains a vertex in β(x1) and a vertex436
in β(xq). Hence, for the purpose of separating such paths in G, it is sufficient to include at437
most one adhesion in Q. Finally observe that, as every internal vertex xz of Q has degree438
2 in F and β(xz) doesn’t contain any terminals and every bag of the forest decomposition439
(F, β) induces a clique in G, any induced path in G between two terminals si and ti that are440
separated by an adhesion from Q, must contain a vertex in β(x1) and a vertex in β(xq), i.e.441
Q is a sub-path of Π(si, ti). Therefore, any adhesion from Q (in S) is sufficient to separate442
si and ti in G. Hence, if S contains more than one adhesion from Q, then it is safe to drop443
all but one of them from S. J444
Observe that the adhesions picked by Lemma 19 for a minimal solution for different pairs445
might not be distinct or disjoint. Also that the collection of adhesions might not be unique446
for a given minimal solution. Now we are ready to give the reduction rule which decreases447
the number of degree 2 nodes. We give the reduction rule in terms of the clique-forest of G,448
but as clique-forests are in one-to-one correspondence with chordal graphs, it can also be449
viewed as a well defined operation on the graph G.450
I Reduction Rule 5. Let (G,T, k) be an instance of MC where G is a chordal graph and451
(F, β) be the clique-forest of G. Let Q := x1x2 . . . xq be a path in F such that x1, xq ∈ F≥3∪FT452
and xγ /∈ F≥3 ∪ FT for all γ ∈ {2, 3, . . . , q − 1}. Let Q′ := y1, . . . yr be a subpath of Q of453
length at least 3 such that y1, yr ∈ D(Q) but yα /∈ D(Q) for all α ∈ {2, 3, . . . , r − 1}. Let454
W = ∪x∈Q′β(x). Consider an auxiliary graph G∗ with vertex set W ∪ {s, t} where s and455
t are new vertices such that NG∗(s) = β(y1), NG∗(t) = β(yr) and G∗[W ] = G[W ]. Let456
the size of a minimum vertex cut between s and t in G∗ be c. Let z := c − |β(y1) ∩ β(yr)|457
and let U = {u1, . . . , uz} be a set of new vertices such that U ∩ V (G) = ∅. To get a new458
clique-forest (F ′, β′), delete yα for each α ∈ {2, 3, . . . , q − 1}, make y1 and yr adjacent in F ′459
while preserving all other adjacencies of F , and put β′(y1) = β(y1) ∪ U , β′(yr) = β(yr) ∪ U460
and β′(x) = β(x) for all x /∈ V (Q′). Let G′ be the chordal graph corresponding to (F ′, β′).461
Output (G′, T, k).462
We first show that the reduction rule is well-defined.463
I Lemma 20. Reduction Rule 5 is well defined. That is, z ≥ 0, F ′ is a forest, and (F ′, β′)464
satisfies property 3 of Definition 8.465
Proof. First, we show that z ≥ 0. For that, we observe that β(y1) ∩ β(yr) ⊆ N(s) ∩N(t)466
and hence c ≥ |β(y1) ∩ β(yr)|. It is easy to see that F ′ is a forest, as it can be obtained by467
contracting all but one edges of the path Q′ in F . To show that (F ′, β′) satisfies property 3 of468
Definition 8, we first observe that F ′[β′−1(v)] is connected for all v ∈ V (G′) \ V (G), as they469
occur only in adjacent bags β′(y1) and β′(y2). For v ∈ V (G′) such that v ∈ β(y1) ∩ β(yr),470
F ′[β′−1(v)] remains connected because F [β−1(v)] was connected, and y1y2 ∈ E(F ′). For any471
other v ∈ V (G′), if F [β−1(v)] did not intersect Q′, then F ′[β′−1(v)] remains connected, as it472
is identical to F [β−1(v)]. The only remaining case is F [β−1(v)]∩Q 6= ∅ and v /∈ β(y1)∩β(yr).473
In this case, F [β−1(v)] contains either y1 or yr but not both, and a proper subpath of Q′.474
Without loss of generality, let y1 ∈ F [β−1(v)] and let the subpath of Q′ in F [β−1(v)] be475
Q′′. Then we have that F ′[β′−1(v)] = (F [β−1(v)] \Q′′) ∪ {y1}, which can be obtained from476
CVIT 2016
23:12 Quick Separation in Chordal and Split Graphs
contracting the edges of Q′′ in F [β−1(v)] and is connected. Hence, (F ′, β′) satisfies property477
3 of definition 8. J478
This shows that G′ is a chordal graph and (G′, T, k) is a valid instance of MC. Now we479
prove a lemma that relates the adhesions of the input and output instances of the reduction480
rule.481
I Lemma 21. Let (G′, T, k) be obtained by applying Reduction Rule 5 on (G,T, k). Let482
A∗ := β′(y1)∩β′(yr), and let A be set of adhesions on the path Q′. Let (si, ti) ∈ T . Then for483
any adhesion A = β(z1) ∩ β(z2) on ΠG(si, ti) such that A /∈ A, A = β′(z1) ∩ β′(z2) is also484
an adhesion on ΠG′(si, ti). Similarly, for any adhesion A′ = β′(z1) ∩ β′(z2) on ΠG′(si, ti)485
such that A′ 6= A∗, A′ = β(z1) ∩ β(z2) is also an adhesion on ΠG(si, ti).486
Proof. From the construction of (F ′, β′), we can look at F ′ as the forest obtained from487
F by contracting all the edges of Q′ except one. This means that for any (si, ti) ∈ T488
and x ∈ V (F ′) \ {y2, . . . , yr−1}, x ∈ ΠG(si, ti) if and only if x ∈ ΠG′(si, ti). Moreover, if489
x ∈ V (F ′) \ {y1, yr} then β(x) = β′(x), else x ∈ {y1, yr} and β′(x) = β(x) ∪ U . Hence,490
for any pair z1, z2 ∈ V (F ′) such that at least one of them is not in {y1, yr}, we have491
β(z1) ∩ β(z2) = β′(z1) ∩ β′(z2).492
Now, to prove the first part of the lemma, let A = β(z1) ∩ β(z2) be an adhesion on493
ΠG(si, ti) (for the graph G) such that A /∈ A. Since we have that β(x) = β′(x) for all494
x ∈ V (F ) \Q′, if {z1, z2} ∩Q′ = ∅, A = β′(z1) ∩ β′(z2) remains an adhesion on ΠG′(si, ti)495
(for the graph G′). Now we look at the case when z1 ∈ V (F ) \ Q′ and z2 = y1. Since496
β(y1) ⊆ β′(y1) and U ∩β(z1) = ∅ (U is the new set of vertices added, i.e., U = V (G′)\V (G)),497
we have that β′(z1) ∩ β′(z2) = β(z1) ∩ β(z2). The case when z2 ∈ V (F ) \ Q′ and z1 = yr498
is similar. A similar argument holds for the second part of the lemma, where we consider499
adhesions on ΠG′(si, ti) (for the graph G′) excluding the adhesion A∗ = β′(y1) ∩ β′(yr). We500
obtain that each of them is an adhesion on ΠG(si, ti) (for the graph G). J501
I Lemma 22. Reduction Rule 5 is correct.502
Proof. We have already shown that G′ is chordal. Now we will show that there exists a503
solution for (G,T, k) if and only if there exists a solution for (G′, T, k). Let us recall that504
we applied the reduction rule on the subpath Q′ = (y1, . . . , yr) of the path Q = (x1, . . . , xq)505
in the forest F . In the forward direction, let S be a minimal solution for (G,T, k). Let506
A := {Aa} be the set of adhesions of the path Q′ where Aa := β(ya)∩β(ya+1) for a ∈ [r− 1]507
and let B1 := β(x1) and Bq := β(xq). We want to show that there exists a solution for508
(G′, T, k). For that, we look at a collection of adhesions Si,j as described in Lemma 19 such509
that S =⋃{ti,tj}∈(T
2) Si,j . Now we look at two possible cases.510
Case 1. The first case is when there does not exist any Aa ∈ A such that Aa = Si for511
some (ti, tj) ∈ T . In this case we claim that S is also a solution for (G′, T, k). First we512
observe that none of the vertices of β(y1) and β(yr) are deleted by the reduction rule. So,513
Si ⊆ V (G′) for all (si, ti) ∈ T and we have that S ⊆ V (G′). Also, for any (si, ti ∈ T , we514
know from Lemma 21 that Si remains an adhesion on ΠG′(si, ti), so deleting S disconnects515
si from ti, because Si ⊆ S. Hence S is a solution to (G′, T, k).516
Case 2. The second case is when there exists some Aa ∈ A such that Aa = Si for517
some (ti, tj(∈ T . Because of Lemma 19, we know that there exists unique such Aa. Let518
A∗ := β′(y1) ∩ β′(yr). We claim that S′ := (S \Aa) ∪A∗ is a solution to (G′, T, k). We see519
that |S′| ≤ |S|, because |Aa| ≥ c = |A∗|. Suppose that Aa separates some pair (si, ti) ∈ T520
in G, i.e. Aa is an adhesion on ΠG(si, ti). This implies that Q′ ⊆ ΠG(si, ti) and Aa is an521
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:13
adhesion on Q′. In G′, we have that that y1, yr ∈ ΠG′(si, ti), and hence A∗ is an adhesion522
which will disconnect si from ti in G′. Therefore deleting S′ in G′ disconnects si from ti.523
Next consider a pair (si, ti) ∈ T such that Aa 6= Si. By Lemma 19, we know that Si524
is not an adhesion on the path Q′ (for the graph G). Then, by Lemma 21, we know that525
Si is still an adhesion on ΠG′(si, ti) (for the graph G′). So to show that deleting S′ in G′526
disconnects si from ti, it is sufficient to show that Si ⊆ S′. Let v ∈ Si be an arbitrary vertex.527
We will show that v ∈ S′, which will finish the proof of forward direction.528
We first look at the case when v ∈ B1 ∪ Bq. Let v ∈ B1. If v /∈ Aa, then we have that529
v ∈ S′ and we are done. If v ∈ Aa, then v ∈ β(y1) from the connectivity property of the clique-530
forest. We also know from Observation 1 that B1 ∩ β(y1) = B1 ∩ β(y2) = . . . = B1 ∩ β(yr).531
That is, v ∈ β(y1) ∩ β(yr). By definition of S′, β(y1) ∩ β(yr) ⊆ A∗ ⊆ S′, and hence v ∈ S′.532
The case when v ∈ Bq is similar. Next we look at the case when v /∈ B1 ∪ Bq. Then we533
claim that v /∈ Aa and hence v ∈ S′. Suppose this is not the case and v ∈ Aa. We know534
that v ∈ Si,j and Si,j is not an adhesion on Q. Hence v occurs in the bag β(x) for some535
x /∈ V (Q). However, as v is in an adhesion Aa on Q′, it must be that v ∈ B1 ∪Bq to satisfy536
the connectivity property of the clique-forest (which is a tree-decomposition of G). This is a537
contradiction to our assumption that v /∈ B1 ∪Bq.538
In the reverse direction, let S′ be a minimal solution to (G′, T, k). We consider a539
collection of adhesions S′i such that S′ = ∪(ti,tj)∈TS′i as described in Lemma 19. Recall that540
A∗ = β′(y1)∩β′(yr). We have two cases. The first case is when S′i = A∗ for some (ti, ti) ∈ T .541
Due to Lemma 9 and Lemma 11, it is sufficient to delete an adhesion on Q′ to disconnect si542
from ti. Let C be a min-cut of size c in the auxiliary graph G∗, as described in the Reduction543
Rule 5. Observe that, by construction of G∗ and Lemma 11, C is a minimum-size adhesion on544
path Q′ (for the graph G). We claim that S := (S′ \A∗)∪C is a solution to (G,T, k). To see545
that, let us first observe that |S| = |S′| since A∗ ⊆ S′, A∗∩(β(y1)∪β(yr)) = C∩(β(y1)∪β(yr))546
and |C| = |A∗| = |β′(y1) ∩ β′(yr)| = |β(y1) ∩ β(yr)| + |U |. Now, if for a pair of terminals547
(si, ti) ∈ T , if S′i = A∗, then observe that, in G, Q′ ⊆ ΠG(si, ti), and hence deleting C in548
G disconnects si and ti. If S′i 6= A, then S′i ∩ U = ∅, S′i ⊆ S and S′i is still an adhesion on549
ΠG(si, ti) by Lemma 21. Hence deleting S′i would disconnect si from ti in G. Therefore, in550
this case, S is a solution to (G,T, k).551
The other case is when A∗ 6= S′i for any (si, tj) ∈ T . We claim that S′ itself is a solution552
to (G,T, k). To see that, let us first observe that U ∩A = ∅ for all adhesions A in G′ such553
that A 6= β(y1)∩β(yr). So we have that S′ ⊆ V (G). Also, S′i is also an adhesion on ΠG(si, ti)554
for all {si, ti} ∈(T2)due to Lemma 21, and deleting it would disconnect si from ti in G. This555
finishes the proof of the lemma. J556
I Lemma 23. Let (G,T, k) be an instance obtained after exhaustively applying Reduction557
Rule 5 and (F, β) be the clique-forest of G. Then |V (F )| = O(k`τ2).558
Proof. From Lemma 16 and Lemma 18, we know that for all x ∈ VT ∪ V≥3, |β(x)| = O(k`τ).559
Let Q := x1x2 . . . xq be a path in F such that x1, xq ∈ F≥3 ∪ FT and xi /∈ F≥3 ∪ FT for all560
i ∈ {2, 3, . . . , q − 1}. Since the for every vertex in β(x1) ∪ β(x2), at most two nodes in Q are561
marked, and we have that |β(x1) ∪ β(x2)| = O(k`τ) due to Lemma 18, we mark at most562
O(k`τ) nodes on Q. If there are any unmarked nodes on Q, then Reduction Rule 5 would563
apply, so we have that after the exhaustive application of the rule, |V (Q)| = O(k`τ). But564
since |FT ∪ F≥3| = O(τ), there can be O(τ) many such paths, and hence the total number565
of nodes on such paths is O(k`τ2). This, combined with |FT ∪ F≥3| = O(τ) proves the566
lemma. J567
CVIT 2016
23:14 Quick Separation in Chordal and Split Graphs
Now, we are ready to give the final reduction rule which would bound the size of the568
graph. For that, we make use of the marking procedure defined for Reduction Rule 4 once569
again. Let F2 = V (F ) \ (F≥3 ∪ FT ) where (F, β) is the clique-forest of G.570
For each pair (si, ti) ∈ T such that Π(si, ti) is non-empty, for each bag x ∈ F2 that571
is an internal node of Π(si, ti), we want to mark at most 2k + 2 vertices in β(x). Let572
Π(si, ti) := x1x2 . . . xd be a nonempty path where xti = x1 and xtj = xd. Let xp ∈573
Π(si, ti), p ∈ {2, 3, d− 1} be an internal node of Π(si, ti) such that xp ∈ F2. We define two574
orderings ≤(si,ti) and ≤(ti,si) on β(x) for all x ∈ F2 as before. That is, for u, v ∈ β(x),575
u ≤(si,ti) v if and only if, for all q ≥ p, p, q ∈ [d], if u ∈ β(xq) then v ∈ β(xq). Similarly,576
for ≤(ti,si), we say that u ≤(ti,si) v if and only if, for all q ≤ p, p, q ∈ [d], if u ∈ β(xq) then577
v ∈ β(xq).578
Now we describe the marking procedure. For each bag x ∈ F2, for each pair of terminals579
(si, ti) ∈ T for which x is an internal vertex of Π(si, ti), we mark k + 1 vertices which are580
largest in the ordering ≤(si,ti) and call the set Zx(si, ti). We also mark k + 1 vertices which581
are largest in the ordering ≤(ti,si) and call that set Zx(ti, si). Let the set of all marked582
vertices inside a bag β(x), such that x ∈ F2 be Z(x) :=⋃
(si,ti)∈T (Zx(si, ti) ∪ Zx(ti, si)) and583
let Z :=(∪x∈F2 Z(x)
)∪(∪x∈FT∪F≥3 β(x)
).584
I Reduction Rule 6. Let (G,T, k) be an instance of MC and let (F, β) be the clique-forest585
of G. If there exists a node x ∈ F2 such that β(x) \ Z is nonempty, then delete an arbitrary586
vertex v ∈ β(x) \ Z from G.587
The proof of correctness of Reduction Rule 6 is exactly the same as proof of Lemma 17.588
Now we are ready to prove the final lemma that bounds the size of the instance.589
I Lemma 24. Let (G,T, k) be an instance of MC after exhaustive application of Reduction590
Rule 6. Then |V (G)| = O(k3`3τ4).591
Proof. Let (F, β) be the clique-forest of G. We already know due to Lemmas 16, 18, and592
23 that |V (F )| = O(k`τ2) and |β(x)| = O(k`τ) for all x ∈ FT ∪ F≥3. We also know that593
|F2| = O(k`τ2). So if we can show |β(x)| = O(k2`2τ2) for all x ∈ F2, this would prove594
the lemma. For that, we want to show that |Z| = O(k2`2τ2). This would mean that595
|β(x)| = O(k2`2τ2) for all x ∈ F2, as otherwise Reduction Rule 6 would apply.596
Now, for a bag β(x) such that x ∈ F2, we want to give a bound for Zx. We mark at most597
2k + 2 vertices for each pair of terminals (si, ti) ∈ T . That gives us |Zx| ≤ (2k + 2)`. Now,598
combining that with |F2| = O(k`τ2), we get that | ∪x∈F2 Z(x)| = O(k2`2τ2). We already599
know that | ∪x∈FT∪F≥3 β(x)| = O(k`τ2), so this gives us |Z| = O(k2`2τ2) as desired. J600
Proof of Theorem 1. Given an instance (G,T, k), we first find the clique-forest (F, β) of G601
and then apply reduction rules 1-6 in order exhaustively. It is easy to see that the reduction602
rules can be applied in polynomial time and we have argued that they are correct. From603
Lemma 24, we know that if none of the reduction rules apply, then the number of vertices in604
G is O(k3`7), since τ = O(`). J605
4 Kernel for Multiway Cut on Chordal Graphs606
In this section we will give a polynomial kernel for Multiway Cut (MWC) on chordal607
graphs parameterized by k alone, that is, we prove Theorem 2. The following preprocessing608
decreases the number of terminals to a linear function of the solution size.609
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:15
I Lemma 25 ( [5]). Given an instance (G′, T ′, k′) of Multiway Cut, in polynomial time610
we can arrive at an equivalent instance (G,T, k) such that T ⊆ T ′, k ≤ k′, |T | ≤ 2k and G611
is obtained from G′ by performing one of the following two operations iteratively.612
1. Taking an induced subgraph, and613
2. contracting an edge.614
I Reduction Rule 7. Apply Lemma 25 to get an instance (G,T, k) such that |T | ≤ 2k.615
The correctness follows from Lemma 25 and the fact that chordal graphs are closed under616
taking induced subgraphs and contracting edges.617
Proof of Theorem 2. Due to Reduction Rule 7, we can assume that for the MWC instance618
(G,T, k), |T | ≤ 2k. Now, given an instance (G,T, k) of MWC, we can obtain an equivalent619
instance (G′, T ′, k′) of MC, by putting G′ := G, k′ := k, and T ′ := {(ti, tj) | ti, tj ∈ T, i < j}.620
Clearly, since |T | ≤ 2k and⋃
(si,ti)∈T ′{si, ti} = T , for the MC instance (G′, T ′, k), we have621
` = O(k2) and τ ≤ 2k. Now, we apply reduction rules 1-6 on the MC instance (G′, T ′, k′).622
Let the output instance, on which none of the reduction rules apply, be (G′′, T ′′, k′′). Due to623
Lemma 24, we have that |V (G′′)| = O(k3`3τ4). Let T f :=⋃
(si,ti)∈T ′′{si, ti}. Observe that624
none of the reduction rules alter the set of terminals, except Reduction Rule 3, which also625
preserves one-to-one correspondence between the terminals (and between pairs of terminals).626
Because of this, there is a one-to-one correspondence between the pairs of terminals of the627
starting instance (G′, T ′, k′) and the kernelized instance (G′′, T ′′, k′′). This means that for628
all {ti, tj} ∈(T f
2), we have that (ti, tj) ∈ T ′′. This implies that (G′′, T f , k′′) is an equivalent629
instance to (G′, T ′, k′) and hence to (G,T, k). Since τ = O(k) and ` = O(k2), this gives a630
kernel for MWC on chordal graphs with O(k13) vertices. J631
5 FPT Algorithms for Multicut on Chordal Graphs632
In this section, we design FPT algorithms for Multicut (MC) on chordal graphs. In633
particular, we prove Theorems 3 and 4. The most crucial ingredient for the proofs in this634
section is the fact that in any graph (not necessarily chordal), amongst all the important635
({v},W )-separators that induce a clique, where W ⊂ V (G) and v ∈ V (G) \W , there is636
at most one important separator of a fixed size k [18]. Below we state a classical result637
concerning chordal graphs that we use to design our algorithms.638
I Proposition 26 ( [9]). Every minimal separator in a chordal graph is a clique.639
I Corollary 27. Let G be a chordal graph, and let X ⊆ V (G) such that G[X] is a clique.640
Then every minimal ({v}, X)-separator in G is a clique.641
Proof. Consider a chordal graph G′ obtained from G by introducing a new vertex x that642
is adjacent to all the vertices of X. Then by applying Proposition 26 to the minimal643
(v, x)-separators in G′, we obtain the corollary. J644
Let G be a graph and X,Y ⊆ V (G). Let S ⊆ V (G) be an (X,Y )-separator in G and let645
R denote the set of vertices reachable from X \ S in G− S (if X (resp. Y ) is a singleton set646
then S ∩X = ∅ (resp. S ∩ Y = ∅)). Then S is called an important (X,Y )-separator if it is647
inclusion-wise minimal and there is no (X,Y )-separator S′ such that |S′| ≤ |S| and R ⊂ R′,648
where R′ is the set of vertices reachable from X in G− S′.649
I Lemma 28 ( [18], Lemma 3.16). For any graph G (not necessarily chordal), W ⊂ V (G),650
and v ∈ V (G) \W , there is at most one important ({v},W )-separator of size exactly k651
inducing a clique.652
CVIT 2016
23:16 Quick Separation in Chordal and Split Graphs
While the lemma in [18] is stated as there are at most k important ({v},W )-separators of653
size at most k that induce a clique, the proof follows by proving the statement in Lemma 28.654
I Lemma 29. For a positive integer k, a chordal graph G, v ∈ V (G), and X ⊆ V (G) such655
that G[X] is a clique, there is at most one ({v}, X)-important separator of size k in G.656
The proof of Lemma 29 follows from Corollary 27 and Lemma 28.657
I Proposition 30 ( [3, 17]). Given a graph G and X,Y ⊆ V (G) and a positive integer k,658
the set Sk of all important (X,Y )-separators of G of size at most k can be computed in time659
O(|Sk| · k2 · (n+m)).660
We now step towards stating and proving our pushing lemma that, together with661
Lemma 29, leads to the design of a branching algorithm for MC on chordal graphs. Let662
(G,T, k) be an instance of MC where G is a chordal graph. Consider the clique forest, say663
(F, β), of G defined in Lemma 10. Without loss of generality, let G be a connected graph.664
Thus, it will be safe to assume that F is a tree. Root the tree F at an arbitrary node. Also,665
we will assume (G,T, k) to be reduced with respect to Reduction Rules 1 and 3 for the rest of666
this section. Note that Reduction Rule 1 can be applied in O(` · (n+m)) time and Reduction667
Rule 3 can be applied in O(k · (n+m)) time. Also both these rules are applied only once in668
the course of the algorithms. Thus, the application of these rules incur an additive running669
time of O((k + `) · (n + m)) to our algorithms. From Lemma 15, for each (si, ti) ∈ T , si670
and ti belong to exactly one bag of the clique-forest (F, β). We denote the unique bag of F671
containing si (resp. ti), as xsi(resp. xti). Let xlcai denote the bag which is the unique least672
common ancestor of xsi and xti in the rooted tree F .673
B Claim 31. Let (G,T, k) be an instance of MC where G is a chordal graph. Let S be any674
solution to the instance (G,T, k). Let (F, β) be a rooted clique-forest of G. For each pair675
(si, ti) ∈ T , every (si, ti)-separator contains an adhesion on the unique xsito xti path in F .676
The proof of the above claim follows from Lemma 11.677
I Lemma 32 (?, Pushing Lemma for MC on Chordal Graphs). Let (G,T, k) be an instance of678
MC where G is a chordal graph, and let (F, β) be a rooted clique forest of G as defined above.679
Let y denote the root bag of F . Let (sp, tp) ∈ T be such that xlcap is deepest in the rooted forest680
F , that is, xlcap is such that distF (y, xlcap ) = max{distF (y, xlcai ) : i ∈ [`]}, where distF (y, xlcai )681
denote the distance between y and xlcai in F . Then there is a solution to (G,T, k) that contains682
either an important ({sp}, β(xlcap ))-separator or an important ({tp}, β(xlcap ))-separator of size683
at most k.684
Proof. Let S be a solution to the instance (G,T, k). From Claim 31, S contains an adhesion685
on the unique xspto xlcap path or on the unique xtp to xlcap . Without loss of generality, let686
this adhesion, say A, be on the xspto xlcap path. Then A is an ({sp}, β(xlcap ))-separator. If687
A is an important ({sp}, β(xlcap ))-separator, we are done. Otherwise, let A∗ be the unique688
important separator of size |A| (from Lemma 29). Here we rely on the fact that β(xlcap )689
induces a clique in G. We claim that S∗ = (S \ A) ∪ A∗ is also a solution to (G,T, k).690
Observe that |S∗| = |S|. For the sake of contradiction, suppose that there exists (si, ti) ∈ T ,691
i 6= p, such that there exists an si to ti path in G − S∗. From Claim 31, since S was an692
(si, ti)-separator, S contains some adhesion, say A′, on the unique (xsi, xlcai )-path or on the693
unique (xti , xlcai )-path. Without loss of generality, let this path be the (xsi, xlcai )-path. If694
A′ ∩A = ∅, then we are done. We now show that A′ ∩A ⊆ A∗. Note that, if we prove this695
we are done.696
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:17
From Claim 31, A∗ contains an adhesion on the (xsp , xlcap )-path. Also, from Lemma 9,697
we conclude that A∗ is indeed an adhesion on the (xsp, xlcap )-path. Let A be the adhesion698
between bag xj and xj+1, that is, let A = β(xj)∩ β(xj+1). Let A′ = β(x`)∩ β(x`+1) and let699
A∗ = β(xr) ∩ β(xr+1). Let us define a < relation on the bags of F . We say xj < x` if xj is700
in the subtree rooted at x` in F . We say xj ≤ x` if either xj = x`, or xj < x`. Without loss701
of generality, let xj < xj+1, x` < x`+1 and xr < xr+1. Now we make the following claim.702
Claim: xj < xr.703
Proof. Suppose not. If xr = xj , then A∗ = A. Otherwise, if xr < xj , then set of vertices704
reachable from si in G−A∗ is a strict subset of the set of vertices reachable from si in G−A,705
thereby violating that A∗ is an important separator. J706
We divide the remaining proof of Lemma 32 in the following three cases.707
Case 1: xr ≤ x`: From the claim above, we have that xj ≤ xr ≤ xr+1 ≤ x`. Since708
A ∩ A′ ⊆ β(xj) and A ∩ A′ ⊆ β(x`), and F is a clique forest decomposition (which is709
a tree decomposition), therefore, A ∩ A′ ∈ xr and A ∩ A′ ∈ β(xr+1), we conclude that710
A ∩A′ ⊆ A∗ and hence A ∩A′ ⊆ S∗.711
Case 2: x` < xr: In this case, since xlcai 6< xlcap (because of the choice of the pair (sp, tp)),712
the unique xsi to xlcai path passes through xr and xr+1. Therefore, A∗ is also an adhesion713
on this path. Since A∗ ⊆ S∗, from Claim 31, we conclude that S is an (si, ti)-separator.714
Case 3: xr and x` are incomparable: In this case, since by the claim above, we have715
that xj < xr, xj and x` are also incomparable. Now, either A ∩A′ = ∅ or A ∩A′ ⊆ A∗716
by the connectivity property of forest-decompositions, and hence we are done.717
This finishes the proof of the lemma. J718
The algorithms of Theorem 3 and 4 are based on a branching algorithm that branches on719
important separators described in Lemma 32.720
Description of the algorithm for MC on chordal graphs (Algorithm 1): Let (G,T, k)721
be an instance of MC where G is a chordal graph. Let (F, β) be a rooted forest decomposition722
of G. From Lemma 32, there exists a pair (sp, tp) ∈ T such that there exists a solution723
S which contains either an important ({sp}, β(xlcap ))-separator of size at most k, or an724
important ({tp}, β(xlcap ))-separator of size at most k. Let Is (resp. It) be the collection of all725
important ({sp}, β(xlcap ))-separator (resp. ({tp}, β(xlcap ))-separator) of size at most k. The726
algorithm branches of the sets in Is ∪ It. That is, it reduces the instance (G,T, k) to the set727
of instances (G− I, T − I, k − |I|), where I ∈ Is ∪ It and T − I denotes the set of pairs of728
terminals with no vertex in I.729
Proof of Theorem 3. The correctness of Algorithm 1 follows from Lemma 32. To prove730
the theorem, we show that it runs in O(2k · (k3 + `) · (n+m)) time. Let T (k) denote the731
number of leaves in the branching tree rooted at an instance where the budget parameter732
is k. Since, from Lemma 29, there is a unique important ({sp}, β(xlcap ))-separator (and733
({tp}, β(xlcap ))-separator) of a fixed size, from the description of the algorithm we get the734
following recurrence: T (k) ≤ 2∑i∈[k] T (k − i), T (1) = 1. Using induction one can show735
that T (k) ≤ 2k+1. Thus, the number of nodes in the branching tree are at most 2 · 2k+1.736
Also the time spent at each node is equal to the time taken by Reduction Rules 1 and 3,737
which is O((k + `)(n + m)), plus time taken by the algorithm of Proposition 30, which is738
O(k3 · (n+m)) because Sk in the proposition has size at most k from Lemma 29. The desired739
running time thus follows. J740
Before we prove Theorem 4, we give the following reduction rule.741
CVIT 2016
23:18 Quick Separation in Chordal and Split Graphs
I Reduction Rule 8. Let (G,T, k) be an instance of MC. If there exists (si, ti) ∈ T such that742
there is no path from si to ti in G, then delete (si, ti) from T , that is the reduced instance is743
(G,T \ {(si, ti)}, k).744
The correctness of Reduction Rule 8 is easy to see and it can be applied exhaustively in745
O(`(n+m)) time.746
Proof of Theorem 4. For the proof of this theorem, observe that, after each branching step747
of Algorithm 1, there is no path between sp to tp. After each branching step of the algorithm,748
one can then apply Reduction Rule 8. Thus, in the resulting instance the size of the terminal749
set decreases by at least 1. Thus, if T (k, `) denotes the number of leaves in the branching tree750
rooted at an instance with ` terminal pairs and budget k, we get the following recurrence:751
T (k, `) ≤∑i∈[2k] T (k, `− 1), T (k, 0) = 1. This solves to T (k, `) ≤ (2k)`. Thus, the running752
time of 2O(` log k) · (n + m) + `(n + m) follows from Proposition 30 and the fact that the753
reduction rules can be applied in O((k + `)(n+m)) time. J754
6 FPT algorithms for Multicut on Split Graphs755
In this section, we design faster FPT algorithms for Multicut on split graphs. In particular,756
we prove Theorem 5, 6 and 7. Recall that a graph is a split graph if and only if its vertex set757
can be partitioned into two parts: C and I, such that the set C induces a clique and I is an758
independent set. It is known that given a split graph G, such a partition can be obtained in759
time O(n+m) [8]. Henceforth, we assume that the partition of the input split graph into a760
clique C and an independent set I is given to us. In what follows, we denote an instance of761
Multicut or Unrestricted Multicut on split graphs by (G = (C, I), T, k). Note that762
split graphs are also chordal graphs, hence the reduction rules designed for chordal graphs763
can also be applied on split graphs. We assume that the input instance (G = (C, I), T, k) is764
reduced with respect to Reduction Rule 1 and Reduction Rule 8. This exhaustive application765
of these reduction rules contribute O(` · (n+m)) to the running time of our algorithms.766
I Lemma 33. Let (G = (C, I), T, k) be an instance of MC on split graphs which is reduced767
with respect to Reduction Rules 1 and 8. Then, for each (si, ti) ∈ T , si, ti ∈ I.768
Proof. Since Reduction Rule 1 is not applicable, we remain to prove that it is not the case769
that there exists (si, ti) ∈ T , such that si ∈ C and ti ∈ I (or vice-versa; this case is symmetric).770
Suppose this happens. Then consider a shortest si to ti path, say P = siv1 . . . vrti, in G.771
Such a path exists because Reduction Rule 8 is not applicable. Also, since Reduction Rule 1772
is not applicable, we have that r > 1. Since ti ∈ I and vr ∈ N(ti), vr ∈ C. Since si, vr ∈ C773
and C is a clique, we have that (si, vr) ∈ E. Since P is a shortest path from si to ti, we774
conclude that r = 1, thereby contradicting the applicability of Reduction Rule 1. J775
I Lemma 34. If (G,T, k) is an instance of MC on split graphs that is reduced with respect776
to Reduction Rules 1 and 8, then for each (si, ti) ∈ T , the length of any shortest path from si777
to ti in G is 4. Also, the internal vertices of this shortest path belong to C.778
Proof. Let (si, ti) ∈ T . From Lemma 33, si, ti ∈ I. Let P be a shortest si to ti path779
in G. Since Reduction Rules 1 and 8 are not applicable, length of P is at least 4. Let780
P = siv1 . . . vrti. For the sake of contradiction, let r ≥ 3. From Lemma 33, since si, ti ∈ I,781
and v1 ∈ N(si), vr ∈ N(ti), we have that v1, vr ∈ C. Since C is a clique, we have that782
(v1, vr) ∈ E, thereby contradicting that P is a shortest si to ti path in G. J783
P. Misra, F. Panolan, A. Rai, S. Saurabh, R. Sharma 23:19
Let (G = (C, I), T, k) be an instance of MC where G is a split graph. For each (si, ti) ∈ T ,784
we associate a set of pairs of the vertices in C as follows. For each i ∈ [`], we now define785
Pi ⊆ C × C. A pair (u, v) ∈ C × C, u 6= v, belongs to Pi, if siuvti is a path in G.786
I Lemma 35. Let (G = (C, I), T, k) be an instance of MC on split graphs. For each787
(si, ti) ∈ T , let Pi be the collection of pairs of vertices of C, defined above. Then S is a788
multicut for the instance (G,T, k) if and only if S contains a vertex from each of the pairs in789
Pi, for each i ∈ [`].790
Proof. For the forward direction, let S ⊆ V (G) \ ∪i∈[`]{si, ti} be a multicut for the instance791
(G,T, k). For a Pi, consider any pair (u, v) ∈ Pi. If S ∩ {u, v} = ∅, then there exists the792
path siuvti in G− S, contradicting that S is a multicut for (G,T, k).793
For the backward direction, let S be a set that hits all the pairs in each Pi. For the794
sake of contradiction, suppose that there exists a path from si to ti in G− S. Then, from795
Lemma 34, there is an si to ti path siuvti such that u, v ∈ C. From the definition of Pi,796
(u, v) ∈ Pi. This contradicts that S hits all pairs of vertices in Pi. J797
Proof of Theorem 5. Let (G = (C, I), T, k) be an instance of MC where G is a split graph.798
Construct a graph G′ with vertex set V (G′) = C and, (u, v) ∈ E(G′) if (u, v) ∈ Pi, for some799
i ∈ [`]. Observe that (u, v) ∈ Pi if and only if u, v ∈ N(si) ∩N(ti) which can be checked in800
linear time for a pair (si, ti). So the total time for constructing the graph G′ is O(` · (n+m))801
Then from Lemma 35, S is a multicut for (G,T, k) if and only if S is a vertex cover for802
(G′, k). Since Vertex Cover can be solved in time O(1.2739k + kn) [2] and the reduction803
rules can be applied in O(` · (m+ n)) time, the theorem follows. J804
I Lemma 36. Let (G = (C, I), T, k) be an instance of MC that is reduced with respect to805
Reduction Rules 1 and 8. Let S be a multicut for (G = (C, I), T, k). For any (si, ti) ∈ T ,806
either N(si) ⊆ S or N(ti) ⊆ S.807
Proof. For the sake of contradiction, suppose that there exists u ∈ N(si) and v ∈ N(ti), such808
that u, v 6∈ S. From Lemma 33, since si, ti ∈ I, u, v ∈ C. Since C is a clique, (u, v) ∈ E(G)809
and si, ti 6∈ S. Therefore, siuvti is a path in G− S, contradicting that S is a multicut for810
(G = (C, I), T, k). J811
Proof of Theorem 6. We design a branching algorithm for MC on split graphs. Let (G =812
(C, I), T, k) be an instance of MC that is reduced with respect to Reduction Rules 1 and 8. If813
T = ∅, then return that (G,T, k) is a Yes instance. Otherwise, pick a pair (si, ti) ∈ T . From814
Lemma 36, we know at least one of the following definitely hold: either N(si) belongs to the815
solution or N(ti) belongs to the solution. Thus, we branch on the following two instances:816
(G−N(si), T1 = T − si, k − |N(si)|) and (G−N(ti), T2 = T − ti, k − |N(ti)|), where T − si817
(similarly T − ti) denote the set of terminal pairs in T that do not contain si (or ti). Since818
the deletion of the neighbours of si (resp. ti) isolates si (resp. ti) in the resulting graph, the819
correctness of the algorithm follows from Lemma 36. Also, |T1|, |T2| < `, as (si, ti) ∈ T but,820
(si, ti) 6∈ T1 and (si, ti) 6∈ T2.821
Since we stop when T = ∅, the depth of the branching tree of this branching algorithm822
is at most `. Since at each time, we branch in two branches, the number of leaves in this823
branching tree is at most 2`. Also, the time taken at each node (which is equal to checking824
if T is empty and if it is not empty, then computing the two instances to recurse on) is825
O(n+m), and the reduction rules can be applied in O(` · (n+m)) time, we get an algorithm826
with running time O(2` · ` · (n+m)). J827
CVIT 2016
23:20 Quick Separation in Chordal and Split Graphs
The algorithm of Theorem 7 is similar to the algorithm of Theorem 6, except in this case828
we also branch on the possibilities of including si or ti in the solution.829
Proof of Theorem 7. Let (G,T, k) be an instance of UMC. If T = ∅, then return that830
(G,T, k) is a Yes instance. Otherwise, pick a pair (si, ti) ∈ T . We branch into the following831
four cases: either si belongs to the solution, or ti belongs to the solution, or when neither832
si nor ti belongs to the solution, Reduction Rules 1 and 8 can be applied for the pair833
(si, ti). Then in this case, from Lemma 36, either N(si) belongs to the solution or N(ti)834
belongs to the solution. Thus, corresponding to the above four cases we recurse on the835
following four instances: (G − {si}, T1 = T − si, k − 1), (G − {ti}, T2 = T − ti, k − 1),836
(G−N(si), T1 = T −si, k−|N(si)|) and (G−N(ti), T2 = T − ti, k−|N(ti)|). The correctness837
of this algorithm again follows from the exhaustiveness of the branching cases and Lemma 36.838
Again, |T1|, |T2| < `.839
Since we stop when T = ∅, the depth of the branching tree of this branching algorithm840
is at most `. Since at each time, we branch in four branches, the number of leaves in this841
branching tree is at most 4`. Also, the time taken at each node is O(n+m) and reduction842
rules are applicable in time O(` · (n+m)). Thus, we get an algorithm with running time843
O(4` · ` · (n+m)). J844
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CVIT 2016
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