Parameterized and Exact Algorithms for Class
Domination Coloring
R. Krithika1 , Ashutosh Rai1 , Saket Saurabh1,2 ? , and Prafullkumar Tale1
1
The Institute of Mathematical Sciences, Chennai, India.
{ashutosh|rkrithika|saket|pptale}@imsc.res.in
2
University of Bergen, Norway.
Abstract. A class domination coloring (also called cd-coloring) of a
graph is a proper coloring such that for every color class, there is a
vertex that dominates it. The minimum number of colors required for
a cd-coloring of the graph G, denoted by χcd (G), is called the class
domination chromatic number (cd-chromatic number) of G. In this work,
we consider two problems associated with the cd-coloring of a graph.
(1) Given a graph G, find its cd-chromatic number. (2) Given a graph
G and integers k and q, can we delete at most k vertices such that
the cd-chromatic number of the resulting graph is at most q? For the
first problem, we give an exact algorithm with running time O∗ (2n )
using polynomial method, where the O∗ notation suppresses polynomial
factors. Also, we show that the problem is FPT when parameterized
by the number of colors q on graphs of girth at least 5 and admits a
kernel with O(q 3 ) vertices. For the second (deletion) problem, we show
NP-hardness for each q ≥ 2. Further, on split graphs (for which cdchromatic number can be obtained in polynomial time), we show that
it is NP-hard if q is a part of the input and FPT with respect to k and
q. For q ≥ 4, as recognizing graphs with cd-chromatic number at most
q is NP-hard on general graphs, the deletion problem is unlikely to be
FPT when parameterized by the size of deletion set on general graphs.
For the same parameterization, we show the fixed parameter tractability
on general graphs for the case when q ∈ {2, 3}. Our algorithms use the
known algorithms for finding a vertex cover and an odd cycle transversal
as a subroutine.
1
Introduction
Dominating Set and Graph Coloring are two of the very well studied problems in the field of algorithms. Given a graph G and an integer k, Dominating
Set problem asks whether ve can find a set of at most k vertices in G such
that every vertex in the graph either belongs to that set or has a neighbor
in it. A proper coloring of a graph G is where every vertex gets a color such
?
The research leading to these results has received funding from the European
Research Council under the European Union’s Seventh Framework Programme
(FP7/2007-2013) / ERC grant agreement no. 306992
that the graph induced on every color class is an independent set. The Graph
Coloring or Chromatic Number problem asks for the minimum number of
colors needed to properly color the graph. Both these problems have been studied extensively in the field of exact algorithms [3,15,16,22,23,31], approximation
algorithms [4,20,21,24], and parameterized algorithms [1,2,5,12] in various contexts.
Many variations of Chromatic Number problem have also been widely
studied, for example Edge-Chromatic Number, Achromatic Number, bChromatic Number etc. In the Dominator Coloring problem introduced
by Gera et al [18], we are asked to get a proper coloring of the graph with
minimum number of colors such that every vertex in the graph is neighbor to
all the vertices of at least one color class. Dominator Coloring is known to
be NP-complete when the number of colors is at least four, and is known to be
polynomial time solvable otherwise [17].
The problem we consider in this paper, namely Class Domination Coloring (cd-Coloring), is similar in flavour to the Dominator Coloring problem. In cd-Coloring, given a graph G and an integer k, we are asked whether
G can be properly colored by k colors such that for every color class, there exists
a vertex that dominates it. In other words, every color class is contained in the
neighbourhood of some vertex. The minimum number of colors needed to achieve
such a coloring is called the Class Domination Chromatic Number (cd-chromatic
number) of the graph and is represented by χcd (G). We say that a graph G is
q-cd-colorable if χcd (G) ≤ q. We define the problem formally as follows.
cd-Coloring
Input: A graph G.
Question: Is G q-cd-colorable?
Similar to Dominator Coloring, cd-Coloring is known to be NP-complete
when the number of colors is at least four and is polynomial time solvable otherwise [33]. In [33], the authors also give a characterization of graphs for which
the class domination chromatic number is three. The problem is also studied on
a number of restricted graph classes like Split Graphs, P4 -free graphs [33] and
middle and central graphs of K1,n , Cn and Pn [34].
In this work, we look at the cd-Coloring problem in the realm of exact
algorithms and parameterized complexity. The field of parameterized complexity
tries to provide efficient algorithms for NP-complete problems by going from the
classical view of single-variate measure of the running time to a multi-variate one.
It aims at getting algorithms with running time f (k)nO(1) , where k is an integer
measuring some aspect of the problem. The integer k is called the parameter,
and the problems which admit such an algorithm are called Fixed Parameter
Tractable (FPT). In most of the cases, the solution size is taken to be a parameter,
which means that this approach gives faster algorithms when the solution is
of small size. For more background, the reader is referred to the monographs
[10,11,13,28].
2
In the case of cd-Coloring, we observe that parameterizing by solution size
(the number of colors) is not of much use, as the problem is para-NP-hard (NPhard even for constant values of the parameter) and hence we can not expect the
problem to be FPT when parameterized by the solution size. Another direction
of questioning can be to find how close the given graph is to being q-cd-colorable.
That is, can we delete at most k vertices from the input graph G such that the
cd-chromatic number of the remaining graph is at most q?
cd-Partization
Input: Graph G, integers k and q
Question: Does there exist S ⊆ V (G), |S| ≤ k, such that χcd (G − S) ≤ q?
If q is fixed, then we refer to the problem as q-cd-Partization. From parameterized complexity point of view, this question is not interesting on general
graphs for values of q greater than three, as in those cases, an FPT algorithm
with deletion set size as parameter would give a polynomial time recognition
algorithm for graphs being q-cd-colorable. Hence, the deletion question makes
sense for deletion to graphs where the recognition problem is polynomial time
solvable. Keeping this in mind, given a graph G and integer k, we tackle the
questions of deleting at most k vertices such that χcd (G) ≤ q when value of q is
two or three.
Our contribution: We describe an O(2n n4 log n)-time algorithm for finding
the cd-chromatic number of a graph using polynomial method. We then show
that cd-Coloring is FPT when parameterized by the number of colors and
the treewidth of the input graph. Further, we show that the problem is FPT
when parameterized by the number of colors q on graphs of girth at least 5
and on chordal graphs. We also show a kernel with O(q 3 ) vertices. Moving on
to the deletion problems, we show that q-cd-Partization is NP-complete for
each q ≥ 2 and for q ∈ {2, 3}, the problem is FPT with respect to solution size
as the parameter. Our algorithms use the known parameterized algorithms for
finding a vertex cover and an odd cycle transversal as a subroutine. We also
show that cd-Partization remains NP-complete on split graphs and is FPT
when parameterized by the number of colors and solution size.
2
Preliminaries
Notations and definitions. All graphs considered here are finite, undirected
and simple. For a graph G, its vertex set is denoted by V (G) and its edge set is
denoted by E(G). For a vertex v ∈ V (G), its neighbourhood NG (v) is the set of
all vertices adjacent to it and its closed neighborhood is the set NG (v) ∪ {v}. The
degree of a vertex v is denoted by degG (v), which is the size of its neighborhood.
For a set S ⊆ V (G), the subgraph of G induced by S is denoted by G[S] and it
is defined as the subgraph of G with vertex set S and edge set {(u, v) ∈ E(G) :
u, v ∈ S} and the subgraph obtained after deleting S is denoted as G − S. The
girth of the graph G is the length of smallest induced cycle in G. A set D ⊆ V (G)
is said to be a dominating set of G if every vertex in V (G) \ D is adjacent to
3
some vertex in D. Dominating set is called total dominating set if every vertex in
V (G) is adjacent to some vertex in it. The set of integers {1, 2, . . . , k} is denoted
by [k]. A proper coloring of G with q colors is a function f : V (G) → [q] such
that for all (u, v) ∈ E(G), f (u) 6= f (v). For a proper coloring f of G with q
colors and i ∈ [q], f −1 (i) ⊆ V (G) is called a color class for the coloring f . The
chromatic number χ(G) of G is the minimum number of colors required in a
proper coloring of G. A clique is a graph which has an edge between every pair
of vertices. The clique number ω(G) of G is the size of a largest clique which is
a subgraph of G. A vertex cover is a set of vertices that contains at least one
endpoint of every edge in the graph. An odd cycle transversal is a set of vertices
whose deletion results in a bipartite graph.
Kernelization. A kernelization algorithm for a parameterized language L is a
polynomial time procedure which takes as input an instance (x, k), where k is
the parameter and returns an instance (x0 , k 0 ) such that (x, k) ∈ L if and only if
(x0 , k 0 ) ∈ L and |x0 | ≤ h(k) and k 0 ≤ g(k), for some computable functions h, g.
The returned instance is said to be a kernel for L and the function h is said
to be the size of the kernel. We say that L admits a polynomial kernel if h is
polynomial.
Treewidth. Let G be a graph. A tree-decomposition of a graph G is a pair
(T, X = {Xt }t∈V (T) ) such that
– ∪t∈V (T) Xt = V (G),
– for every edge xy ∈ E(G) there is a t ∈ V (T) such that {x, y} ⊆ Xt , and
– for every vertex v ∈ V (G) the subgraph of T induced by the set {t | v ∈ Xt }
is connected.
The width of a tree decomposition is maxt∈V (T) |Xt | − 1 and the treewidth of
G is the minimum width over all tree decompositions of G and is denoted by
tw(G).
Monadic Second Order Logic (MSO). The syntax of Monadic Second Order
Logic (MSO) of graphs includes the logical connectives ∨, ∧, ¬, ⇒, ⇔, variables
for vertices, edges, sets of vertices, sets of edges, the quantifiers ∀, ∃ that can be
applied to these variables, and the following five binary relations.
1. u ∈ U where u is a vertex variable and U is a vertex set variable;
2. d ∈ D where d is an edge variable and D is an edge set variable;
3. inc(d, u), where d is an edge variable, u is a vertex variable, and the interpretation is that the edge d is incident with the vertex u;
4. adj(u, v), where u and v are vertex variables and the interpretation is that
u and v are adjacent;
5. equality of variables representing vertices, edges, sets of vertices, and sets of
edges.
For a MSO formula φ, by ||φ|| we denote the length of the encoding of φ as
a string.
Theorem 1 (Courcelle’s theorem, [8,9]). Let φ be the property tht is expressable in Monadic Second Order Logic. Graph G is an n-vertex graph equipped
4
with evaluation of all the free variables of φ. Suppose the treewidth of G is tw.
Then there exists an algorithm that verifies whether φ is satisfied in G in time
f (||φ||, tw) · n, for some computable function f .
Polynomials and Fast Fourier Transform. A binary vector is a finite sequence of bits. For a binary vector B, val(B) represents an integer m of which
B is binary representation and in that case, we say that integer m and binary
string B are associated with each other. We can pad zeros to the left of binary
strings to make it of length n without changing the value of integer associated
with it. In this paper, all the binary vectors we consider are of length n and
we ensure that intergers under consideration are less than 2n . If B1 , B2 are binary vectors, then B1 + B2 represents the binary representation of the integer
val(B1 ) + val(B2 ).
Consider ordered set U = {u1 , u2 , . . . , un } of cardinality n. Like integers, we
define an unique bit string, called characteristic vector of length n associated
with each subset S of U . The characteristic vector ψ(S) of S with respect to U
is the binary vector with n bits whose j th bit, for 1 ≤ j ≤ n, is 1 if element uj
belongs to set S, and 0 otherwise. The characteristic vector can also be thought
of another way of storing or representing the set S. To obtain the cardinality
of set S from its characteristic vector B, we define Hamming weight of string
B. The hamming weight of a binary vector B is the number of 1s in B and it
is denoted by H(B). Observe that H(ψ(S)) = |S|. The hamming weight of an
integer is define as hamming weight of its binary representation.
To obtain fast running time for our exponential-time algorithm, we make use
of the fast algorithm for multiplying polynomials which is based on the Fast
Fourier Transform.
Lemma 1 ([32]). Two polynomials of degree at most d over any commutative
ring R can be multiplied using O(d · log d · log log d) additions and multiplications
in R.
We end the preliminaries section with following simple observations about
cd-coloring of a graph.
Observation 1 If graph G has two connected components C1 and C2 then
χcd (G) = χcd (G[C1 ]) + χcd (G[C2 ]).
Observation 2 If (G, k) is a YES instance of cd-coloring then graph G has
dominating set of size at most k.
3
Exact Algorithm for cd-Chromatic Number
Given a coloring on V (G), we can check in polynomial time whether it is a cdcoloring of a graph G or not. We can guess all possible colorings of V (G) with
n colors and for each coloring check whether it is a valid cd-coloring or not.
This brute force algorithm runs in 2O(n log n) time. In this section we present
5
an algorithm which runs in O∗ (2n ) time. The idea for this algorithm is inspired
from an exact algorithm for b-Chromatic Number presented in [29].
In this section we assume that there is a fixed ordering on the elements of
universe U . With every subset S of U we associate a unique binary vector ψ(S)
which is a characteristic vector of set S. Let p(z) be a polynomial in z. We define
a modified multiplication operator (?) in the following way.
(
z val(ψ(S1 ))+val(ψ(S2 )) if S1 ∩ S2 = ∅
val(ψ(S1 ))
val(ψ(S2 ))
z
?z
=
0
otherwise
We implement the ? operation using standard multiplication operation and
hamming weights of bit strings associated with exponents.
We will use a monomial in z to represent a set. As a natural extension, we
use univariate polynomials in z to represent a family of sets.
Definition 1 (Characteristic Polynomial of a Family). For a given set U
and a family F = {S
P1q , S2 , . . . , Sq } over U, the characteristic polynomial of F is
defined as p(z) := i=1 z val(ψ(Si )) .
We say that the polynomial p(z) contains the monomial z m , or that the
monomial z m is present in the polynomial p(z), if the coefficient of z m is not
zero in p(z). If p(z) is the characteristic polynomial of the family F, then p(z)
contains the monomial z val(ψ(S)) if and only if S is present in F. In the following
section, given certain monomial and a polynomial, we are interested in checking
whether that monomial exists in the P
polynomial or not.
q
i
For a given polynomial
p(z)
=
i=1 ai · z , we define the representative
Pq
i
polynomial p̃(z) := i=1 bi · z where bi = 1 if ai 6= 0 and bi = 0 if ai = 0.
Hamming
projection of polynomial p(z) to integer h is defined as Hh (p(z)) :=
Pq
i
b
·
z
where bi = ai if H(i) = h and bi = 0 otherwise. For a positive
i=1 i
integer `, we inductively define polynomial p(z)` as p(z)` := p(z)`−1 ? p(z) and
as base case p(z)1 := p(z). For given p(z), r(z) of degree at most 2n , we compute
p(z) ? r(z) using Algorithm 3.1.
Algorithm 3.1: FastFourierMultiply
Input: Two polynomial p(z), r(z) of degree at most 2n
Output: p(z) ? r(z)
Initialize polynomials t(z) to 0
for (i, j) such that i + j ≤ n do
Compute qi (z) = Hi (p(z)) and qj (z) = Hj (r(z))
Compute qij (z) = qi (z) ∗ qj (z) using Lemma 1
q̃ij (z) = Hi+j (qij )
t0 (z) = t(z) + q̃ij (z)
t(z) ← Representative polynomial of t0 (z)
return t(z)
Lemma 2. Given two polynomials p(z) and r(z) of degree at most 2n , Algorithm 3.1 computes p(z) ? r(z) correctly in time O(2n n3 log n).
6
Proof. Correctness of the algorithm follows from the Lemma 2.1 presented in
[29]. By Lemma 1, one execution of fast fourier multiplication algorithm takes
O(2n n log n) time. The for loop runs n2 time resulting in total running time of
O(2n n3 log n).
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Corollary 1. Given a polynomial p(z) of degree at most 2n , one can compute
p(z)` in time O(2n n3 log n · l)
Now we prove the following generalized lemma which correlates existence of
partition of a set with presence of monomial associated with it.
Lemma 3. For a universe U and given a family F of its subsets, let p(z) be the
characteristic polynomial of family F. For any W ⊆ U , W is disjoint union of
` sets from F if and only if there exists a monomial z val(ψ(W )) in p(z)` .
Proof. (⇒) Let W be a disjoint union of S1 , S2 , . . . , S` such that Si ∈ F for all
i ∈ [`]. Consider the characteristic vectors of sets W, S1 , S2 , . . . , S` . For any j ∈
[n], if j th bit of vector φ(W ) is 1 if and only if there exists exactly one Si such that
j th bit of φ(Sj ) is one. This implies val(ψ(W )) = val(ψ(S1 ) + val(ψ(S2 )) + · · · +
val(ψ(S` )). For every Si there is a term z val(ψ(Si )) in p(z). Since Si ’s are mutually
disjoint, the monomial which is a result of z val(ψ(S1 )) ? z val(ψ(S2 )) ? · · · ? z val(ψ(S` ))
is present in p(z)` .
(⇐) We prove this using induction on `. Assume the statement holds for all
the integers which are smaller than ` i.e. if there exists a monomial z val(ψ(W ))
in p(z)`−1 then W can be partitioned into ` − 1 sets each belongs to F. The
statement is vacuously true for base case when ` = 1. If there exists a monomial
z val(ψ(W )) in p(z)` = p(z)`−1 ? p(z) then it must be product of two monomials,
say z val(ψ(W1 )) and z val(ψ(W2 )) , which were part of p(z)`−1 and p(z) respectively
and W1 ∩ W2 = ∅. By induction hypothesis, W1 can we written as disjoint union
of S1 , S2 , . . . , S`−1 such that Si ∈ F for all i ∈ [` − 1]. By construction, W2 is
in F and since W1 ∩ W2 = ∅, Si ∩ W2 is also an empty set for all i. Hence W
can be partitioned into S1 , S2 , . . . , S`−1 , W2 such that each of these sets belong
to F.
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We now are in position to prove the main theorem of this section.
Theorem 2. Given a graph G on n vertiecs as input, there is an algorithm
which finds the cd-chromatic number of G in O(2n n4 log n) time.
Proof. Given a graph G, we fix an arbitrary ordering on V (G). With V (G) as
universal set, we define family F of its subsets as follows.
F := {X ⊆ V (G)| X is an independent set and ∃ y s.t. X ⊆ N (y)}
Let p(z) be a characteristic polynomial family F. By Lemma 3, if there exists
a monomial z val(ψ(V (G))) in p(z)` then V (G) can be partitioned into ` sets each
belonging to F. Notice that every set in F is an independent set and there
exists a vertex which dominates it. F is collection of possible color classes in
7
cd-coloring of graph G. Hence the smallest integer ` for which there exists a
monomial z val(ψ(V (G))) in p(z)` is the cd-chromatic number of the graph G.
By Corollary 1, we can compute p(z)` in time O(2n n3 log n · l). Since cdchromatic number of graph is upper bounded by n, we get the desired running
time.
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4
FPT Algorithms for cd-Chromatic Number
Determining whether a given graph G has cd-chromatic number at most q is NPhard on general graphs for q ≥ 4. This implies that the cd-Coloring problem
parameterized by the number of colors is para-NP-hard on general graphs. Thus
this necessitates the search for special classes of graphs where cd-Coloring is
FPT. In particular, in this section we give FPT algorithms for cd-Coloring on
chordal graphs and graphs of girth at least 5.
4.1
Chordal Graphs
In this section we design FPT algorithm for the cd-Coloring problem parameterized by the number of colors. Towards this, we will use Courcelle’s powerful
theorem which interlinks fixed parameter tractability of certain graph property
with its expressibility in counting monadic second order logic (MSO) formula.
We can write many graph theoretical properties as MSO formula. Following are
three examples which we will use in writing MSO formula to check whether
graph has cd-chromatic number at most q. Let Q is set of integers from 1 to q.
– To check whether V1 , V2 , . . . , Vq is partition of V (G)
Part(V1 , V2 , . . . , Vq ) ≡ ∀u ∈ V (G)[∃i ∈ Q[(u ∈ Vi )∧(∀j ∈ Q[i 6= j ⇒ u 6∈ Vj )]]]
– To check whether a given set Vi is independent or not.
IndSet(Vi ) ≡ ∀u ∈ Vi [∀v ∈ Vi [¬adj(u, v)]]
– To check whether given vertex set Vi is dominated by some vertex or not.
Dom(Vi ) ≡ ∃u ∈ V (G)[∀v ∈ Vi [adj(u, v)]
We use φ(G, q) to denote the monadic second order logic formula which states
that G has cd-chromatic number at most q. We use the MSO formulas defined
above as macros in this formula.
φ(G, q) ≡ ∃V1 , V2 , . . . , Vq ⊆ V (G)[Part(V1 , V2 , . . . , Vq )∧
IndSet(V1 ) ∧ · · · ∧ IndSet(Vq ) ∧ Dom(V1 ) ∧ · · · ∧ Dom(Vq )]
It is easy to see that the length of the above formula is upper bounded by a
linear function of q. By applying Theorem 1 we obtain the following result.
Theorem 3. cd-Coloring parameterized by the number of colors and the treewidth
of the input graph is FPT.
8
As the graph gets more structured, we expect many NP-hard problem to get
easier in some sense on the restricted class of graphs having that structure. For
example, Chromatic-Coloring is NP-hard on general graphs but it is polynomial time solvable on chordal graphs. As we have mentioned, cd-Coloring
problem is NP-hard even on the chordal graphs. However, we show that the
problem is FPT on chordal graphs parameterized by the number of colors.
Theorem 4. cd-Coloring parameterized by the number of colors is FPT on
chordal graphs.
Proof. For a chordal graph G, tw(G) = ω(G) − 1 where tw(G) denotes the
treewidth of G and ω(G) is the size of maximum clique in the graph G [19].
Since, cd-coloring of a graph is also a proper coloring of the graph we have
that no two vertices in a maximum clique of G can share a color class. Thus, if
ω(G) ≥ k then we can conclude that (G, k) is NO instance of cd-Coloring.
Hence, we can assume that ω(G) ≤ k which implies tw(G) ≤ k. This together
with Theorem 3 imply that the cd-Coloring problem parameterized by the
number of colors is FPT on chordal graph.
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4.2
Graphs with girth at least 5
Kaminski and Lozin [27] showed that k-Coloring is NP-complete for graphs of
girth at least g for any fixed k ≥ 3 and g ≥ 3. In particular, Chromatic
Number problem is para-NP-hard for graphs of girth at least 5. In this section,
we demonstrate that cd-coloring on the same class of graphs is FPT.
By Observation 1, we can assume that the input graph G is connected. Once
we know that graph is connected, we can define cd-coloring of graph as - minimum number of colors k such that every color class is contained in open neighbourhood of a some vertex. In other words, we do not allow a vertex to dominate itself. One can verify that two definition of cd-coloring are identical on
connected graphs. We now define a notion of total-dominating set of a graph G.
A set S ⊆ V (G) is called total-dominating set if V (G) = ∪v∈S N (v). That is, for
every vertex v ∈ V (G), there exists a vertex u ∈ S, u 6= v, such that v ∈ N (u).
Our interest in total-dominating set is because of its relation to cd-coloring in
graphs that do not contain triangles, that is, g(G) ≥ 4. In particular, we show
the following lemma.
Lemma 4. For a given graph G with g(G) ≥ 4, cardinality of minimum total
dominating set is equal to cd-chromatic number of the graph.
Proof. Let φ be a cd-coloring of G that uses the minimum number of colors
and let V1 , . . . , Vq be the color classes. Since φ is a cd-coloring, for every color
class Vi there exists a vertex vi such that Vi ⊆ N (vi ). Let X denotes the set
of these vertices. The set X has cardinality at most k and by definition it is a
total dominating set of the graph G. Hence, the cardinality of a minimum total
dominating set of a graph is at most the cd-chromatic number of the graph.
9
Suppose X = {v1 , v2 , . . . , vq } is a minimum total dominating set of graph.
We construct a cd-coloring of graph using at most q colors. We define the color
classes in the following way – Let V1 = N (v1 ) and for i = 2, . . . , q, define
Vi = N (vi ) \ (V1 ∪ V2 ∪ · · · ∪ Vi−1 ). Notice that V1 , . . . , Vq forms a partition of
V (G). Since, g(G) ≥ 4, we have that each Vi is an independent set. Furthermore,
since X is a total dominating set we have a vertex vi ∈ X such that Vi ⊆ N (vi ).
Hence, this gives a cd-coloring of the graph G. This implies that the cd-chromatic
number of graph is at most the cardinality of a minimum total dominating set.
Now the lemma follows by combining the above two inequalities.
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Lemma 4 shows that to show that cd-Coloring is FPT on graphs of g(G) ≥ 4
it is enough to find a total dominating set of size at most k. This leads to the
Total Dominating Set problem. For a given graph G and an integer k Total
Dominating Set problem asks whether there exists a total dominating set of
size at most k. Observe that we can test whether a given graph G has a total
dominating set of size at most k by enumerating all subsets S of V (G) of size at
most k and checking whether it forms a total-dominating set in polynomial time.
This immediately gives an algorithm with running time nO(k) for cd-Coloring
on graphs with girth at least 5. It is not hard to modify the reduction given in [30]
to show that Total Dominating Set is W [2] hard on bipartite graphs. Thus,
Lemma 4 implies that even cd-Coloring is W [2] hard on bipartite graphs.
Hence, if we need to show that cd-Coloring is FPT, we must assume that
the girth of the input graph is 5. In the rest of the section we show that cdColoring is FPT on graphs with girth 5 by showing that Total Dominating
Set is FPT on those graphs. Before proceeding, we note some simple properties
of graphs with girth at least 5.
Observation 3 For a given graph G, if g(G) ≥ 5 then for any v in V (G), N (v)
is an independent set and for any u, v in V (G), |N (v) ∩ N (u)| ≤ 1.
Raman and Saurabh [30] defined a variation of Set Cover problem, namely,
Bounded Intersection Set Cover. An input to the problem consists of a
universe U, a collection F of subsets of U and a positive integer k with the
property that for any two Si , Sj in F, |Si ∩ Sj | ≤ c for some constant c and
the objective is to check whether there exists a sub-collection F0 of F of size
at most k such that ∪S∈F0 = U. In the same paper the authors proved that
the Bounded Intersection Set Cover is FPT when parameterized by the
solution size. Total Dominating Set(G, k) on graph with girth at least 5
can be reduced to Bounded Intersection Set Cover with U = V (G) and
F = {N (v)| ∀v ∈ G}. By Observation 3, we can fix the constant c to be 1. Hence
we have the following lemma.
Lemma 5. Total Dominating Set problem is FPT parameterized by the solution size on the graphs with girth at least 5.
We now prove that the problem has a polynomial kernel and use it to design
a faster FPT algorithm.
10
Lemma 6. Total Dominating Set problem admits a kernel of O(k 3 ) vertices
on the class of graphs with girth at least 5.
Proof. We start the proof with following claim which says that every high degree
vertex should be included in every total dominating set of size at most k.
Claim. For a given graph G such that g(G) ≤ 5 and vertex u, if deg(u) ≥ k + 1
then any total dominating set of size at most k contains u.
Proof. We prove this statement via contradiction. Suppose there exists a total
dominating set X of G of size at most k which does not contain u. Since N (u) is
dominated by X, has cardinality at least k+1, and no vertex can dominate itself,
by Pigeon Hole Principle, there exists a vertex, say w, in X which is neighbor
to at least two vertices, say, v1 , v2 in N (u). This implies that w, v1 , v2 , u forms
a cycle of length 4, contradicting the fact that girth of G is at least 5.
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Construct a tri-partition of V (G) as follows:
H = {u | deg(u) ≥ k + 1};
J = {v | v ∈
/ H, ∃u ∈ H such that uv ∈ E(G)};
R = V (G) \ (H ∪ J)
By the above claim, H is subset of every total dominating set of size at most k.
Hence, the cardinality of H is upper bounded by k. Notice that there is no edge
across set H and R. Thus, the vertices in R needs to be dominated by at most
k vertices in J ∪ R. But the degree of vertices in J ∪ R is at most k and hence
|R| ≤ O(k 2 ). Since the degree of R is bounded above by k, |J ∩ N (R)| is upper
bounded by O(k 3 ). We will now bound size of J ? = J \ N (R). For that, we first
apply the following reduction rule on the vertices in J ? .
Reduction Rule 1 For u, v ∈ J ? , if N (u) ∩ H ⊆ N (v) ∩ H then delete u.
The correctness of this reduction follows from the observation that all the vertices
in J have been dominated by the vertices in H. The only reason any vertex in
J ? is part of a total dominating set is because that vertex is used to dominate
some vertex in H. If this is the case then the vertex u in the solution can be
replaced by the vertex v. In the reverse direction, if X is a total dominating
set of G \ u, then there exists x ∈ X such that v and x are adjacent in graph
G − {u}. Since N (u) ∩ H is contained in N (v) ∩ H, u and x are adjacent in G.
Hence u is dominated by x and X is total dominating set of G \ u.
All that remains is to bound the size of J ? . We partition J ? into two sets
namely J1 and J2 . The set J1 is the set of vertices which are adjacent to exactly
one vertex in X whereas each vertex in J2 is adjacent to at least two vertices in
H. After exhaustive application of Reduction Rule 1, no two vertices in J1 can
be adjacent to one vertex in H and hence |J1 | ≤ |H| ≤ k. Any vertex in J2 is
adjacent to at least two vertices in H. For every vertex u in J2 , we assign a pair
of vertices in H to which u is adjacent. By Observation 3, no two vertices in J2
11
can be
assigned to the same pair and hence cardinality of J2 is upper bounded
by k2 ≤ k 2 . Combining all the bounds, we get a kernel with O(k 3 ) number of
vertices.
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Combining Lemmata 4 and 6 we obtain the following theorem.
Theorem 5. There exists an algorithm for cd-Coloring running in 2O(q log q) nO(1)
time and admits a O(q 3 ) sized vertex kernel on the graphs with girth at least 5.
5
Complexity of CD-Partization
In this section, we study the complexity of cd-Partization. As recognizing
graphs with cd-chromatic number at most q is NP-hard on general graphs for
q ≥ 4, the deletion problem is also NP-hard on general graphs for such values of
q. For q = 1, the problem is trivial as χcd (G) = 1 if and only if G is the graph
on one vertex. In this section, we show NP-hardness for q ∈ {2, 3}. We remark
that G = {G | χcd ≤ q} is not a hereditary graph class and so the generic result
of Lewis and Yannakakis [25] does not imply the claimed NP-hardness.
5.1
Para-NP-hardness in General Graphs
Consider the following problem.
Partization
Input: Graph G, integers k and q
Question: Does there exist S ⊆ V (G), |S| ≤ k, such that χ(G − S) ≤ q?
Once again if q is fixed, we refer to the problem as q-Partization. Observe
that the classical NP-complete problems Vertex Cover [14] and Odd Cycle
Transversal [14] are 1-Partization and 2-Partization, respectively. Now,
we proceed to show the claimed hardness.
Theorem 6. q-cd-Partization is NP-complete for q ∈ {2, 3}.
Proof. The problem is in NP as determining if the cd-chromatic number of a
graph is at most q ∈ {1, 2, 3} is polynomial-time solvable. Given an instance
(G, k) of q-Partization where q ∈ {1, 2}, we construct the instance (G0 , k) of
(q + 1)-cd-Partization as follows: G0 is obtained from G by adding a new
vertex v adjacent to every vertex in V (G) and adding k + q + 2 new vertices
v1 , · · · , vk+q+2 adjacent to v. We claim that G has a set of k vertices whose
deletion results in a q-colorable graph if and only if G0 has a set of k vertices
whose deletion results in a (q + 1)-cd-colorable graph.
Consider a set S of k vertices such that χ(G − S) ≤ q. Then, G0 − S is
(q + 1)-cd-colorable as a new color can be assigned to v and any of the q colors
of G − S can be assigned to v1 , · · · , vk+q+2 . The color class containing v is a
singleton set. This class is dominated by all vertices in G0 − (S \ {v}). Further,
v dominates each of the other q color classes as v is a universal vertex in G0 .
12
Conversely, let S 0 ⊆ V (G0 ) be a minimal set of at most k vertices such
that χcd (G0 − S 0 ) ≤ q + 1. Now, if v ∈ S 0 , then vertices v1 , · · · , vk+q+2 are
isolated in G − {v} implying that either |{v1 , · · · , vk+q+2 } ∩ S 0 | ≥ k + 1 or
χcd (G0 − S 0 ) > q + 1. So, we can assume that v ∈
/ S 0 . Further, as S 0 is minimal,
0
it follows that {v1 , · · · , vk+q+2 } ∩ S = ∅. Also, as v is a universal vertex in G0 ,
we have that χ(G − (S 0 \ {v})) ≤ q. So, S 0 is a subset of V (G) of size at most k
such that G − S 0 is q-colorable.
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5.2
NP-hardness and Fixed-Parameter Tractability in Split Graphs
A graph is a split graph if its vertex set can be partitioned into a clique and an
independent set. As split graphs are perfect (clique number is equal to the chromatic number for every induced subgraph), we have the following observation.
Observation 7 A split graph G is r-colorable if and only if ω(G) ≤ r.
The following result is known for the corresponding deletion problem.
Theorem 8 ([7,35]). Partization on Split Graphs is NP-complete.
This hardness was shown by a reduction from Set Cover [14]. We modify this
reduction to show that cd-Partization is NP-complete on split graphs. The
problem is in NP as the cd-chromatic number of a split graph can be obtained
in polynomial time due to the following result.
Theorem 9 ([33]). If G is a connected split graph G, then ω(G) = χcd (G).
2
Furthermore, there is an O(|V (G)| ) time algorithm that returns a minimum
cd-coloring of G.
Theorem 10. cd-Partization on split graphs is NP-hard.
Proof. Consider a Set Cover instance (U, F, k) where U = {x1 , · · · , xn } is a
finite set and F is a family {S1 , · · · , Sm } of subsets of U . The problem is to
determine if there is a collection of at most k sets in F such that each element
of U is in at least one set of the collection. The corresponding instance of cdPartization is (G, k 0 = m − k, q = k + 1) where G is a split graph on the
vertex set C ∪ I ∪ {w0 , w1 , · · · , wk+k0 +2 } where C = {ui | Si ∈ F} and I =
{vi | xi ∈ U }. Also, (vi , uj ) ∈ E(G) if and only if xi ∈
/ Sj and w0 is adjacent to
every vertex in C ∪ I ∪ {w1 , · · · , wk+k0 +2 }. Further, I ∪ {w1 , · · · , wk+k0 +2 } and
C induce an independent set and a clique, respectively, in G. We claim that a
set F 0 ⊆ F of size k is a set cover if and only if G − S 0 is q-cd-colorable where
S 0 = {ui ∈ C | Si ∈ F \ F 0 } and |S 0 | = k 0 .
Consider a set cover F 0 ⊂ F of size k. If there is a clique Q (without loss
of generality assume w0 ∈ Q) of size k + 2 in G − S 0 , then Q must contain an
element vi ∈ I that is adjacent to k vertices in C \ S 0 . However, since F 0 is a set
cover, vi is non-adjacent to at least one uj in C \ S 0 leading to a contradiction.
Thus, S 0 has a non-empty intersection with every (k + 2)-clique in G. As G is
13
a split graph, it is (k + 1)-colorable due to Observation 7. Further, G − S 0 is
(k + 1)-cd-colorable as the color class containing {w0 } is a singleton set (since it
is an universal vertex) which is dominated by itself and the other color classes
are dominated by w0 .
Conversely, consider a minimal subset S 0 of k 0 vertices such that G − S 0 is
(k+1)-cd-colorable. Now, if w0 ∈ S 0 , then vertices w1 , · · · , uk+k0 +2 are isolated in
G − {w0 } implying that either |{w1 , · · · , wk+k0 +2 } ∩ S 0 | ≥ k 0 + 1 or χcd (G − S 0 ) >
k + 1. So, we can assume that w0 ∈
/ S 0 . Further, as S 0 is minimal, it follows that
0
{w1 , · · · , wk+k0 +2 } ∩ S = ∅. Now, all vertices in S 0 must belong to C. If there
exists vi ∈ S 0 ∩ I, there is a clique of size k + 2 in G − S 0 as C is a clique. Also,
no vertex in I is adjacent to all nodes in C \ S 0 as if there is such a vertex vi
then there is a (k + 2)-clique in G − S 0 . Thus, every vertex in I is nonadjacent
to at least one element in C \ S 0 implying that {si ∈ F | ui ∈ C \ S 0 } is a set
cover of (U, F) of size at most k.
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As Set Cover parameterized by solution size is W[2]-hard [10], we have the
following result.
Corollary 2. cd-Partization on split graphs parameterized by q is W[2]-hard.
Now, we show that the problem is FPT with respect to q and k.
Theorem 11. cd-Partization on split graphs is FPT with respect to parameters q and k. Furthermore, the problem does not admit a polynomial kernel
unless NP ⊆ coNP/poly.
Proof. Compute a maximum clique Q of G in polynomial time. If |Q| ≤ q, then
the input instance is an YES instance as χcd (G) ≤ q from Theorem 9. Otherwise,
choose an arbitrary subset of size q + 1 from Q. Since any solution contains at
least one of the q + 1 vertices, a straightforward branching algorithm runs in
O∗ ((q +1)k ) time. Now, we move on to the kernelization hardness. Set Cover is
known not to admit a polynomial kernel when parameterized by the solution size
k 0 and family size m as combined parameters unless NP ⊆ coNP/poly [10]. The
reduction in Theorem 10 produces instances of cd-Partization where solution
size k is m − k 0 and q is k 0 + 1 implying that q + k is m + 1. Therefore, an
(q + k)O(1) kernel for cd-Partization implies an mO(1) kernel for Set Cover
which is unlikely.
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6
Deletion to 3-cd-Colorable Graphs
In a graph G, an edge e = (u, v) is said to be a dominating edge if N (u)∪N (v) =
V (G). Let N [v] denote the set V (G) \ N [v]. The following characterization of
3-cd-colorable graphs is known.
Theorem 12 ([33]). A connected graph G satisfies χcd (G) ≤ 3 if and only if
G is one of the following types.
(Type 0) G is a graph on at most 3 vertices.
14
(Type 1) G is a bipartite graph with a dominating edge.
(Type 2) G has a vertex v such that G − v is a bipartite graph with a dominating
edge.
(Type 3) G has an ordered pair (x, y) of adjacent vertices such that,
– V (G) = {x, y} ] X ] Y ,
– G[X ∪ {y}] is a bipartite graph with at least one edge,
– Y ∪ {x} is an independent set, Y ∪ {x} ⊆ N (y) and X ∪ {y} ⊆ N (x).
(Type 4) G has an ordered set (x, y, z) of vertices inducing a triangle such that,
– V (G) = {x, y, z} ] X ] Y ] Z,
– X ⊆ N (x), Y ⊆ N (y) and Z ⊆ N (z),
– X ∪ {y}, Y ∪ {z} and Z ∪ {x} are independent sets.
(Type 5) G has an ordered triple (x, y, z) of vertices such that,
–
–
–
–
V (G) = {x, y} ] X ] Y ] Z,
z ∈ X ∪ Y , (x, y) ∈
/ E(G) and (x, z), (y, z) ∈ E(G),
X ⊆ N (x), Y ⊆ N (y) and Z ⊆ N (z),
X, Y , Z ∪ {x} and Z ∪ {y} are independent sets.
We refer to the ordered sets in Types 3, 4 and 5 as dominators. In [33], they
are called as d-pair, cd-triangle and NB-triplet respectively. Now, we proceed
to solve 3-cd-Partization. Let G be the input graph on n vertices, m edges
and k be a positive integer. Consider a set S ⊆ V (G) such that H = G − S
is 3-cd-colorable. Then, H is of one of the types listed in Theorem 12. Before
we proceed to describe algorithms for each of these types, we list the following
well-known results on Vertex Cover and Odd Cycle Transversal that we
use in our algorithms.
Theorem 13 ([6]). Given a graph G and a positive integer k, there is an algorithm running in O∗ (1.2738k ) time that determines if G has a vertex cover of
size at most k or not.
Theorem 14 ([26]). Given a graph G and a positive integer k, there is an
algorithm running in O∗ (2.3146k ) time that determines if G has an odd cycle
transversal of size at most k or not.
6.1
Deletion to Types 0, 1 and 2
It is trivial to check if G has a solution whose deletion results in a graph H with
at most 3 vertices. So, deletion to Type 0 is easy. Now, suppose H is of Type
15
1. Then, we need to identify an edge of G that is a dominating edge for H. We
describe an algorithm based on this observation.
Algorithm 6.1: Deletion-to-Type1(G, k)
Input : A graph G and a positive integer k.
Output: S ⊆ V (G), |S| ≤ k such that G − S is of Type 1 (if one exists).
1
2
3
4
for each edge (x, y) in G do
Let X 0 = N (x) ∩ N [y] and Y 0 = N (y) ∩ N [x].
Let S 0 be V (G) \ (X 0 ∪ Y 0 ) and decrease k by |S 0 |.
for each k1 and k2 such that k1 + k2 ≤ k do
Compute a vertex cover S1 of G[X 0 ] with |S1 | ≤ k1 (if one exists).
/* (X 0 \ S1 ) ∪ {y} is an independent set */
Compute a vertex cover S2 of G[Y 0 ] with |S2 | ≤ k2 (if one exists).
/* (Y 0 \ S2 ) ∪ {x} is an independent set */
if S1 and S2 are non-empty sets then
return S 0 ∪ S1 ∪ S2
Lemma 7. Algorithm 6.1 runs in O∗ (1.2738k ) time.
Proof. The outer loop (step 1) is executed at most m times (once for each edge)
and the inner loop (step 2) is executed at most k 2 times. Let (x, y) be an edge in
G. We need to extend {x} and {y} into independent sets Y and X respectively,
such that X is dominated by x and Y is dominated by y. Clearly, neighbors of
x and non-neighbors of y cannot be in Y . Similarly, neighbors of y and nonneighbors of x cannot be in X. No common neighbor of x and y can be in
either X or Y . Thus, the candidates for X and Y are X 0 = N (x) ∩ N [y] and
Y 0 = N (y)∩N [x] respectively. All vertices in V (G)\(X 0 ∪Y 0 ) are in any solution.
Let k 0 = k − |V (G) \ (X 0 ∪ Y 0 )|. Observe that the following are equivalent.
– G has a 3-cd-partization solution S of size at most k such that G − S is of
Type 1 with (u, v) as a dominating edge.
– There exists integers k1 and k2 with k1 + k2 ≤ k 0 such that G[X 0 ] has a
vertex cover of size at most k1 and G[Y 0 ] has a vertex cover of size at most
k2 .
Steps 3 and 4 take O∗ (1.2738k ) time from Theorem 13. Thus, the overall running
time is O∗ (1.2738k ).
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Suppose H is of Type 2. Then, for each vertex v of G, we simply run Algorithm
6.1 on G − {v} with parameter k.
Algorithm 6.2: Deletion-to-Type2(G, k)
Input : A graph G and a positive integer k.
Output: S ⊆ V (G), |S| ≤ k such that G − S is of Type 2 (if one exists).
1
for each vertex x in G do
Deletion-to-Type1(G − {x}, k).
16
Lemma 8. Algorithm 6.2 runs in O∗ (1.2738k ) time.
Proof. As Algorithm 6.2 calls Algorithm 6.1 at most n times, its running time
is O∗ (1.2738k ).
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6.2
Deletion to Type 3
Suppose H is of Type 3 with dominator (x, y). Then, the following holds.
Observation 15 ([33]) NH [x] is an independent set and NH [x] ⊆ NH (y). Further, NH (x) induces a bipartite graph with at least one edge.
This observation leads to the following algorithm.
Algorithm 6.3: Deletion-to-Type3(G, k)
Input : A graph G and a positive integer k.
Output: S ⊆ V (G), |S| ≤ k such that G − S is of Type 3 (if one exists).
1
2
3
4
for each ordered pair (x, y) of adjacent vertices in G do
Let Y 0 = N (y) ∩ N [x] and X 0 = N (x).
Let S 0 be V (G) \ (X 0 ∪ Y 0 ) and decrease k by |S 0 |.
for each k1 and k2 such that k1 + k2 ≤ k do
Compute a vertex cover S1 of G[Y 0 ] with |S1 | ≤ k1 (if one exists).
/* (Y 0 \ S1 ) ∪ {x} is an independent set */
Compute a minimal odd cycle transversal S2 of at most k2 vertices
(if one exists) in G[X 0 ] such that y ∈
/ S2 .
if S1 and S2 are non-empty sets then
return S 0 ∪ S1 ∪ S2
Lemma 9. Algorithm 6.3 runs in O∗ (2.3146k ) time.
Proof. The outer loop (step 1) is executed at most 2m times (as there are two
ordered pairs for each edge) and the inner loop (step 2) is executed at most k 2
times. Consider an edge (x, y) in G. If (x, y) is a dominator in H, then we need
to extend {x} into an independent set Y that is dominated by y and extend {y}
into an induced bipartite graph B (with at least one edge) such that V (B) is
dominated by x. Observe that Y contains only neighbors of y and V (B) contains
only neighbors of x. Further, a neighbor of y that is not adjacent to x cannot
be in V (B) and a neighbor of y that is adjacent to x cannot be in Y . Thus, the
candidates for V (B) and Y are X 0 = N (x) and Y 0 = N (y) ∩ N [x] respectively.
All vertices in V (G)\(X 0 ∪Y 0 ) are in any solution. Let k 0 = k−|V (G)\(X 0 ∪Y 0 )|.
Now, it is not hard to see that the following are equivalent.
– G has a 3-cd-partization solution S of size at most k such that G − S is of
Type 3 with (x, y) as a dominator.
– There exists integers k1 and k2 with k1 + k2 ≤ k 0 such that G[Y 0 ] has a
vertex cover of size at most k1 and G[X 0 ] has an odd cycle transversal of size
k2 not containing y such that the resultant bipartite graph is non-edgeless.
17
Clearly step 3 takes O∗ (1.2738k ) time. For step 4, we need to find a minimal odd
cycle transversal that excludes vertex y. We construct a graph G0 obtained from
G[X 0 ] by deleting y and adding a new vertex yij for each pair yi , yj of neighbors
of y; further yij is adjacent to yi and yj . Now, any odd cycle transversal of G0
can be modified to one that excludes each vertex in {yij ∈ V (G0 ) | yi , yj ∈ N (y)}
without increasing the size. This is due to the fact that any induced odd cycle
through yij is an induced odd cycle through yi and yj . Therefore, the following
are equivalent.
– G[X 0 ] has a minimal odd cycle transversal of size at most k2 not containing
y.
– G0 has a minimal odd cycle transversal of size at most k2 .
Now, by using Theorem 14, it follows that step 4 takes O∗ (2.3146k ) time and
this gives us the claimed running time of the algorithm.
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6.3
Deletion to Type 4
Suppose H is of Type 4 and has (x, y, z) as a dominator. Then, we have the
following observation.
Observation 16 ([33]) NH (x) ∩ NH (y) ∩ NH (z) = ∅ and NH [x] ∩ NH [y] ∩
NH [z] = ∅. Further, X = NH (x) ∩ NH [y], Y = NH (y) ∩ NH [z] and Z = NH (z) ∩
NH [x].
Now, we have the following algorithm.
Algorithm 6.4: Deletion-to-Type4(G, k)
Input : A graph G and a positive integer k
Output: S ⊆ V (G), |S| ≤ k such that G − S is of Type 4 (if one exists)
1
2
3
4
5
for each ordered triple (x, y, z) of pairwise adjacent vertices of G do
Let X 0 = N (x) ∩ N [y], Y 0 = N (y) ∩ N [z] and Z 0 = N (z) ∩ N [x].
Let S 0 be V (G) \ (X 0 ∪ Y 0 ∪ Z 0 ) and decrease k by |S 0 |.
for each k1 , k2 and k3 such that k1 + k2 + k3 ≤ k do
Compute a vertex cover S1 of G[X 0 ] with |S1 | ≤ k1 (if one exists).
/* (X 0 \ S1 ) ∪ {y} is an independent set */
Compute a vertex cover S2 of G[Y 0 ] with |S2 | ≤ k2 (if one exists).
/* (Y 0 \ S2 ) ∪ {z} is an independent set */
Compute a vertex cover S3 of G[Z 0 ] with |S3 | ≤ k3 (if one exists).
/* (Z 0 \ S3 ) ∪ {x} is an independent set */
if S1 , S2 and S3 are non-empty sets then
return S 0 ∪ S1 ∪ S2 ∪ S3
Lemma 10. Algorithm 6.4 runs in O∗ (1.2738k ) time.
18
Proof. The outer loop (step 1) is executed at most n3 times and the inner loop
(step 2) is executed at most k 3 times. Consider a triangle {x, y, z} in G. If (x, y, z)
is a dominator in H, then we need to extend {x}, {y}, {z} into independent sets
Y , Z, X dominated by y, z and x respectively. Thus, the candidates for X, Y
and Z are sets X 0 = N (x) ∩ N [y], Y 0 = N (y) ∩ N [z] and Z 0 = N (z) ∩ N [x]. All
vertices in S 0 = V (G)\(X 0 ∪Y 0 ∪Z 0 ) are in any solution. Let k 0 = k −|V (G)\S 0 |.
Then, the following are equivalent.
– G has a 3-cd-partization solution S of size at most k such that G − S is of
Type 4 with (x, y, z) as a dominator.
– There exists integers k1 , k2 and k3 with k1 + k2 + k3 ≤ k 0 such that G[Y 0 ]
has a vertex cover of size at most k1 , G[Z 0 ] has a vertex cover of size at most
k2 and G[Z 0 ] has a vertex cover of size at most k3 .
Steps 3, 4 and 5 take O∗ (1.2738k ) time from Theorem 13 and the overall running
time is O∗ (1.2738k ).
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6.4
Deletion to Type 5
Suppose H is of Type 5 and has (x, y, z) as a dominator. Then, we have the
following observation.
Observation 17 ([33]) NH [x] ∩ NH [y] is an independent set. Further, z ∈
NH (x)∪NH (y) and NH [x]∩NH [y] ⊆ NH (z). Moreover, in G−Z, N [x]∪N [y] =
V (G − Z), N (x) \ N (y) ⊆ X and N (y) \ N (x) ⊆ Y .
Now, we have the following algorithm.
Algorithm 6.5: Deletion-to-Type5(G, k)
Input : A graph G and a positive integer k
Output: S ⊆ V (G), |S| ≤ k such that G − S is of Type 5 (if one exists)
1
2
3
4
for each ordered triple (x, y, z) of vertices of G such that (x, y) ∈
/ E(G)
and (x, z), (y, z) ∈ E(G) do
Let Z 0 be the set N (z) ∩ (N [y] ∩ N [x]).
Let S 0 be (N (y) ∩ N (z)) \ N (x).
Let B 0 be the set {z} ∪ (((N (x) ∪ N (y)) \ S 0 ).
Let S 00 be V (G) \ (Z 0 ∪ B 0 ) and decrease k by |S 00 |.
for each k1 and k2 such that k1 + k2 ≤ k do
Compute a vertex cover S1 of G[Z 0 ] with |S1 | ≤ k1 (if one exists).
/* (Z 0 \ S1 ) ∪ {x, y} is an independent set */
Compute a minimal odd cycle transversal S2 of G[B 0 ] with
|S2 | ≤ k2 not containing z (if one exists) such that the resultant
bipartite graph has a bipartition (X, Y ) such that X ⊆ N (x),
Y ⊆ N (y) and z ∈ Y .
if S1 and S2 are non-empty sets then
return S 00 ∪ S1 ∪ S2
19
Lemma 11. Algorithm 6.5 runs in O∗ (2.3146k ) time.
Proof. Consider an ordered triple (x, y, z) of vertices in G. If (x, y, z) is a dominator in H, then we need to extend {x, y} into an independent set Z that is
dominated by z and extend {z} into a bipartite graph B with bipartition (X, Y )
such that X is dominated by x and Y is dominated by y. Thus, the candidates
for Z and V (B) are Z 0 = N (z) ∩ (N [y] ∩ N [x]) and B 0 = {z} ∪ ((N (x) ∪ N (y)) \
(N (y) ∩ N (z)) \ N (x)) respectively. All vertices in V (G) \ (Z 0 ∪ B 0 ) are in any
solution. Let k 0 = k − |V (G) \ (Z 0 ∪ B 0 )|. The following are equivalent.
– G has a 3-cd-partization solution S of size at most k such that H = G − S
is of Type 5 with (x, y, z) as a dominator.
– There exists integers k1 and k2 with k1 + k2 ≤ k 0 such that G[Z 0 ] has a
vertex cover of size at most k1 and G[B 0 ] has an odd cycle transversal of size
k2 not containing z such that the resultant bipartite graph has a bipartition
(X, Y ) such that X ⊆ N (x), Y ⊆ N (y) and z ∈ Y .
Step 3 takes O∗ (1.2738k ) time. For step 4, we need to find a minimal odd cycle
transversal that excludes vertex z. For this we can use the construction described
in Theorem 10. Let G0 be the graph obtained from G[B 0 ] by deleting z and
adding a new vertex zij for each pair zi , zj of neighbors of z, adjacent to zi and
zj . Now, a minimal odd cycle transversal of G0 corresponds to a minimal odd
cycle transversal of G[B 0 ] not containing z. However, we also need the additional
constraint that such an odd cycle transversal results in a bipartite graph B which
has a bipartition (X, Y ) such that X ⊆ N (x) and Y ⊆ N (y). The possible
vertices in B are from the set {z} ∪ (((N (x) ∪ N (y)) \ (N (y) ∩ N (z)) \ N (x)).
The following observations on vertices from this set are easy to verify.
– Nx = N (x) \ (N (y) ∪ N (z)) cannot be dominated by y and Ny = N (y) \
(N (x) ∪ N (z)) cannot be dominated by x.
– Nzx = (N (x) ∩ N (z)) \ N (y) and Nxyz = N (x) ∩ N (y) ∩ N (z) cannot be in
a part of the bipartition that contains z.
It follows that we need an odd cycle transversal (of size at most k2 ) of G[B 0 ] after
deleting which the resultant bipartite graph has a 2-coloring in which any vertex
from P = {z} ∪ Ny receives color 1 and any vertex from Q = Nx ∪ Nzx ∪ Nxyz
receives color 2. This is achieved by constructing graph G00 from G0 by adding an
independent set IP of k2 +1 new vertices each of which is adjacent to every vertex
in P and an independent set IQ of k2 + 1 new vertices each of which is adjacent
to every vertex in Q. Further, every vertex in IP is adjacent to every vertex
in IQ . Consider a minimal odd cycle transversal O of size k2 of G00 . Clearly, O
excludes at least one vertex from both IP and IQ . Also, we can also assume that
O ∩ (IP ∪ IQ ) = ∅ by the minimality of O. As any two vertices p ∈ IP and q ∈ IQ
are adjacent, vertices from IP and vertices from IQ are in different parts in any
bipartition, thus forcing the same for the sets P and Q. Hence, the following are
equivalent.
– G[B 0 ] has a minimal odd cycle transversal of size at most k2 not containing
z such that the resultant bipartite graph has a bipartition (X, Y ) such that
X ⊆ N (x), Y ⊆ N (y) and z ∈ Y .
20
– G00 has a minimal odd cycle transversal of size at most k2 .
Now, using Theorem 14, it follows that step 4 takes O∗ (2.3146k ) time and the
overall running time is dominated (upto polynomial factors) by this computation.
t
u
From Lemmata 7,8,9,10 and 11, we have the following result.
Theorem 18. Given a graph G and an integer k, there is an algorithm that
determines if there is a set S of size k whose deletion results in a graph H with
χcd (H) ≤ 3 in O∗ (2.3146k ) time.
7
Concluding Remarks
In this work, we explored the parameterized complexity of the cd-Coloring
problem. We gave exact and FPT algorithms for the problem on general graphs
as well as on some restricted classes of graphs. We also gave a polynomial kernel
for the problem on graphs of girth at least 5, but the kernelization complexity
of cd-Coloring still remains vastly unexplored on restricted classes of graphs
and can be a fruitful direction in future research.
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