Capacitated domination faster than O(2n )
Marek Cygan1 , Marcin Pilipczuk1 , and Jakub Onufry Wojtaszczyk2
1
Dept. of Mathematics, Computer Science and Mechanics,
University of Warsaw, Poland
2
Institute of Mathematics, Polish Academy of Sciences
[cygan,malcin,onufry]@mimuw.edu.pl
Abstract. In this paper we consider the C APACITATED D OMINATING S ET problem — a generalisation of the D OMINATING S ET problem where each vertex v is
additionally equipped with a number c(v), which is the number of other vertices
this vertex can dominate. We provide an algorithm that solves C APACITATED
D OMINATING S ET exactly in O(1.89n ) time and polynomial space. Despite the
fact that the C APACITATED D OMINATING S ET problem is quite similar to the
D OMINATING S ET problem, we are not aware of any published algorithms solving this problem faster than the straightforward O∗ (2n )3 solution prior to this
paper which was stated as an open problem at Dagstuhl seminar 08431 in 2008
and IWPEC 2008.
We also provide an exponential approximation scheme for C APACITATED D OM INATING S ET which is a trade-off between the time complexity and the approximation ratio of the algorithm.
1
Introduction
The field of exact exponential-time algorithms for NP-hard problems has attracted a lot of attention in recent years (see Woeginger’s survey [21]). Many
difficult problems can be solved much faster than by the obvious brute-force
algorithm; examples include I NDEPENDENT S ET [13], D OMINATING S ET [12,
20] , C HROMATIC N UMBER [2] and BANDWIDTH [7, 10]. A few powerful techniques have been developed, including Measure & Conquer [12] and inclusion/exclusion principle applications [2, 3, 17]. However, there is still a bunch
of problems for which no faster solution than the obvious one is known. These
include S UBGRAPH I SOMORPHISM and C HROMATIC I NDEX which are mentioned as open problems in [1, 14].
In order to define the C APACITATED D OMINATING S ET problem let us
introduce some basic notions. Let G = (V, E) be an undirected graph. Given
F ⊂ E we write V (F ) to denote the set of all endpoints of the edges in F .
Given W ⊂ V by G[W ] we denote the subgraph induced by W . We say a
vertex v ∈ V dominates u ∈ V if u = v or uv ∈ E, i.e. a vertex dominates
3
By O∗ we denote standard big O notation but omitting polynomial factors.
itself and all its neighbours. By N [v] = {v} ∪ {u : uv ∈ E} we denote the
set of vertices dominated
S by v. We extend this notation to any subset W ⊂ V
by putting N [W ] = v∈W N [v], we say that a set W dominates a vertex u if
u ∈ N [W ]. The set N [W ] is called the closed neighbourhood of W .
The D OMINATING S ET problem asks for the smallest set that dominates
the whole V . In the C APACITATED D OMINATING S ET problem each vertex v is
additionally equipped with a number c(v), which is the number of other vertices
this vertex can dominate. Formally, we say that a set S ⊂ V is a capacitated
dominating set if there exists fS : V \ S → S such that fS (v) is a neighbour of
v for each v ∈ V \ S and |fS−1 (v)| ≤ c(v) for each v ∈ S. The function fS is
called a dominating function for the set S. The C APACITATED D OMINATING
S ET problem asks for the smallest possible size of a capacitated dominating set.
Note that for a given set S checking whether it is a capacitated dominating set is
a polynomial–time problem which can be solved using max–flow or maximum
matching techniques.
Finding an algorithm faster than O∗ (2n ) for D OMINATING S ET was an open
problem until 2004. Currently the fastest algorithm by van Rooij et al. runs in
O(1.5048n ) [20]. The problem of solving C APACITATED D OMINATING S ET
faster than the check-all-subsets O∗ (2n ) algorithm was posted by van Rooij in
2008 at Dagstuhl seminar [14] and on IWPEC 2008 open problem list. In this
paper we present an algorithm providing the positive answer to this question.
Note that at first glance breaking O∗ (2n ) barrier for the C APACITATED
D OMINATING S ET problem seems a hard task, since even the brute-force O∗ (2n )
algorithm involves matching or max–flow techniques. A standard approach to
improving exponential time algorithms is through fixed parameter tractability
(see eg. [18]). However from the parameterized point of view, Dom et al. [9]
showed that this problem is W [1]-hard when parameterized by both treewidth
and solution size, and Bodlaender et al. [4] showed that even the planar version
parameterized by the solution size is also W [1]-hard.
Superpolynomial approximation was recently considered as a way of coping
with hardness of approximation of different NP-hard problems. Results in this
field include a subexponential approximation algorithm for BANDWIDTH on
trees [11] and exponential approximation schemes for C HROMATIC N UMBER
[5, 6] or BANDWIDTH on arbitrary graphs [8, 15].
Our results. In Section 2 we provide an algorithm which solves the C A PACITATED D OMINATING S ET problem in O(1.89n ). The algorithm constructs
n
O∗ ( n/3
) = O(1.89n ) reductions of the input graph into a C ONSTRAINED
C APACITATED D OMINATING S ET problem instance (defined in Section 2.1),
each solvable in polynomial time. Section 2.3 tackles the exponential approximation of C APACITATED D OMINATING S ET. More precisely, we provide an
approximation algorithm that for a given c ∈ (0, 13 ), in time
n
∗
O (
) = O∗ ((1/(cc (1 − c)1−c ))n ),
cn
1
+c)-approximation in the case of c < 41 or a (2−3c)-approximation
computes a ( 4c
1
in the case of 4 ≤ c < 13 . This result should
be compared to the trivial approx
n
∗
imation scheme that works in O ( cn ) time too: iterate over all subsets of V
that have size at most cn or at least (1 − c)n and return the smallest feasible
solution found. However, this algorithm has an approximation factor of 1c − 1,
which is between 2× and 4× worse than our ratio. All algorithms in this paper
require polynomial space.
Our algorithm for C APACITATED D OMINATING S ET is somewhat similar to
one of the first algorithms to break O∗ (2n ) for the classical D OMINATING S ET
problem, namely the algorithm of Randerath and Schiermeyer [19]. Their algorithm also involves matching arguments and our algorithm, applied to D OMI NATING S ET , can be viewed as a simplification of their algorithm. However we
do not know whether their algorithm could be used to solve the C APACITATED
D OMINATING S ET problem.
2
C APACITATED D OMINATING S ET
2.1
C ONSTRAINED C APACITATED D OMINATING S ET
In this section we introduce a constrained version of the C APACITATED D OM INATING S ET problem, namely the C ONSTRAINED C APACITATED D OMINATING S ET problem, which can be solved in polynomial time.
The input of C ONSTRAINED C APACITATED D OMINATING S ET is an undirected graph G = (V, E), a set U ⊆ V and a capacity function c : V →
{0, . . . , n − 1}. We ask for a smallest capacitated dominating set S ⊆ V containing U such that each vertex outside U dominates at most one other vertex.
Formally we ask for a dominating function fS satisfying
|fS−1 (v)| ≤ 1
|fS−1 (v)|
for each
≤ c(v)
v ∈S\U
for each
v∈U
(1)
Let G = (V, E) with U ⊆ V and a capacity function c : V → {0, . . . , n−1}
be a C ONSTRAINED C APACITATED D OMINATING S ET instance. Consider the
following graph G0 = (V 0 , E 0 ):
– for any v ∈ V \ U we have v ∈ V 0 ;
– for any v ∈ U we have c(v) copies v1 , v2 , . . . , vc(v) of v in V 0 ;
– for any v ∈ V \ U and u ∈ U the edge ui v ∈ E 0 for all i iff uv ∈ E;
– for v, w ∈ V \ U we have vw ∈ E 0 iff vw ∈ E and c(v) + c(w) > 0;
– there are no edges of the form vi wj or vi vj for v, w ∈ U .
(E,0)
(D,2)
(F,0)
(M,0)
(L,1)
(K,0)
(H,0)
(A,2)
(B,3)
(C,2)
U={A,B,C}
H’
E’
D’
F’
K’
M’
L’
A01
A02
B10
B20
B30
C10
C20
Fig. 1. From the constrained capacitated dominating set {A, B, C, D, L} (where U =
{A, B, C}) to a matching. By a pair (X, i) we denote a vertex X with its capacity c(X) = i.
We show a correspondence between feasible solutions of C ONSTRAINED
C APACITATED D OMINATING S ET in G and matchings in G0 .
Lemma 1. Let S be a capacitated dominating set in G with a dominating function fS satisfying condition 1. Then one may construct in polynomial time a
matching φ(S, fS ) in G0 satisfying |φ(S, fS )| = |V | − |S|.
Proof. Let us define the matching φ(S, fS ) as follows:
– for each v ∈
/ S such that fS (v) ∈
/ U add vfS (v) to φ(S, fS );
– for each v ∈
/ S such that u = fS (v) ∈ U add vui to φ(S, fS ), where ui
is a copy of u in G0 and different copies ui are chosen for different vertices
v with fS (v) = u (note that |fS−1 (u)| ≤ c(u), so there are enough vertices
ui ).
Note that every vertex v ∈ V \ S is an endvertex of an edge in the matching
φ(S, fS ). The second endvertex is fS (v) (in the case fS (v) ∈
/ U ) or a copy of
fS (v) in G0 (in the case fS (v) ∈ U ). Note that by condition 1, every vertex
w ∈ S \ U is an endvertex of at most one chosen edge, thus φ(S, fS ) is indeed a
matching. Moreover, every edge in φ(S, fS ) has exactly one endpoint in V \ S.
Therefore |φ(S, fS )| = |V | − |S|.
t
u
Lemma 2. Let M be a matching in G0 . Then one may construct in polynomial
time a feasible solution ψ(M ) to the C ONSTRAINED C APACITATED D OM INATING S ET problem with dominating function fψ(M ) satisfying |ψ(M )| =
|V | − |M |.
Proof. Consider the following capacitated dominating set ψ(M ) with dominating function fψ(M ) :
– U ⊆ ψ(M );
– for u ∈ U and for each i such that ui v ∈ M , we take fψ(M ) (v) = u;
– for any edge vw ∈ M , where v, w 6∈ U one of the endpoints (say v) has to
satisfy c(v) > 0, we add v to ψ(M ) and set fψ(M ) (w) = v;
– for any v 6∈ U which is not an endpoint of any edge in M we add v to ψ(M ).
It is easy to verify that the above procedure does indeed give a feasible solution to C ONSTRAINED C APACITATED D OMINATING S ET. We have |ψ(M )| =
|V |−|M | since for each edge in M , exactly one of its endpoints does not belong
to ψ(M ).
t
u
We conclude this section with the following theorem.
Theorem 3. The C ONSTRAINED C APACITATED D OMINATING S ET problem
can be solved in polynomial time.
Proof. By Lemmas 1 and 2, to find the solution of the C ONSTRAINED C APAC ITATED D OMINATING S ET problem it is enough to find any maximum matching
in G0 , which can be done in polynomial time (see e.g. [16]).
2.2
From C ONSTRAINED C APACITATED D OMINATING S ET to
C APACITATED D OMINATING S ET
Let us start with the following simple observation. Let S be any capacitated
dominating set and let fS be a dominating function for S. Let
US = {v ∈ S : |fS−1 (v)| ≥ 2}.
We have
X
v∈S
1 + |fS−1 (v)| = |S| +
X
v∈S
|fS−1 (v)| = n,
thus in particular |US | ≤ n/3. Moreover, S with the function fS is a feasible solution for the C ONSTRAINED C APACITATED D OMINATING S ET instance with
the graph G and the set US . Therefore the following algorithm solves C APAC ITATED D OMINATING S ET:
1. For each U ⊆ V satisfying |U | ≤ n/3 solve the C ONSTRAINED C APACI TATED D OMINATING S ET instance with graph G and subset U .
2. Return the smallest capacitated dominating set from the constructed C ON STRAINED C APACITATED D OMINATING S ET instances.
The C ONSTRAINED C APACITATED D OMINATING S ET problem can be
solved in polynomial time and there are
dn/3e X
k=0
n
k
=O
∗
n
= O(1.89n )
dn/3e
possible sets U (i.e. sets of cardinality at most n/3), thus the whole algorithm
works in O(1.89n ) time.
2.3
Approximating C APACITATED D OMINATING S ET
It is known that D OMINATING S ET is as hard to approximate as S ET C OVER
and since C APACITATED D OMINATING S ET generalizes D OMINATING S ET it
is hard to approximate it as well. Thus there probably does not exists a polynomial time algorithm solving C APACITATED D OMINATING S ET with a constant
approximation ratio. If we do not have enough time to obtain an exact solution
for the C APACITATED D OMINATING S ET problem we can use the following
constant approximation scheme. Instead of investigating all subsets U ⊆ V satisfying |U | ≤ n/3 we can check only smaller sets, namely |U | ≤ cn for some
constant c ∈ (0, 13 ). Thus the approximation algorithm has the following form:
1. For each U ⊆ V satisfying |U | ≤ cn solve the C ONSTRAINED C APACI TATED D OMINATING S ET instance with graph G and subset U .
2. Return the smallest capacitated dominating set from the constructed C ON STRAINED C APACITATED D OMINATING S ET instances.
Theorem
any fixed constant c ∈ (0, 13 ) the described algorithm runs in
4. For
n
∗
∗
O ( cn ) = O ((1/(cc (1 − c)1−c ))n ) time and polynomial space. For c ≤ 1/4
1
the approximation ratio is at most ( 4c
+ c) and for c ≥ 1/4 the approximation
ratio is at most 2 − 3c.
Proof. For each subset U ⊆ V the algorithm uses polynomial time only, thus
the time bound follows directly from
the Stirling formula which can be used to
n
bound the number of subsets cn
.
To calculate the approximation ratio let us consider some optimal solution
OP T ⊆ V together with a function fOP T : V \ OP T → OP T . By OP T0 ,
OP T1 and OP T2 let us denote subsets of OP T containing vertices which dominate exactly zero, exactly one and at least two vertices from V \ OP T , according to fOP T , respectively. By m, m0 , m1 , m2 we denote the cardinalities of sets
OP T , OP T0 , OP T1 and OP T2 respectively.
We may assume that m2 > cn since otherwise our algorithm finds the optimal solution. By s let us denote the average number of vertices from V \ OP T
which a vertex from OP T2 dominates, i.e. s = (n − m − m1 )/m2 . Since our
algorithm checks all subsets U ⊆ V satisfying |U | ≤ cn it obviously considers the subset U0 ⊆ OP T2 , |U0 | ≤ cn containing vertices which dominate the
largest number of vertices from V \ OP T . For this particular subset U0 note
that there exists a feasible solution to C ONSTRAINED C APACITATED D OMI NATING S ET of size m + (m2 − cn)(s − 1): we take OP T and for each vertex
−1
v ∈ OP T2 \ U0 we take all but one vertices from fOP
T (v) since then each
vertex outside U0 dominates at most one other vertex.
Thus the approximation ratio can be bounded by α = 1+(m2 −cn)(s−1)/
m. Note that if we keep m2 fixed and increase m0 and m1 , the approximation
ratio decreases — we increase m and decrease s — therefore w.l.o.g. we may
assume m0 = m1 = 0. Denoting x = n/m2 , we obtain α ≤ 1+(1−cx)(x−2)
for 3 ≤ x ≤ 1c . The bound for α is a concave function of x with maximum at
1
1
+ 1. However when c ≥ 14 we have x0 ≤ 3. This gives α ≤ 4c
+ c for
x0 = 2c
1
1
1
c ≤ 4 and α ≤ 2 − 3c for 4 ≤ c ≤ 3 .
t
u
In Table 1 we gather a few examples of obtained approximation ratios and
running times. We compare the running times of our approximation scheme
with the running times of the trivial approximation algorithm needed to obtain
the same approximation ratio. This algorithm iterates over all subsets of V that
have size at most cn or at least (1−c)n and returns the smallest feasible solution
found.
c
ratio
time
trivial time
1/3
1 (exact)
O(1.89n )
O? (2n )
1/4
5/4
O(1.76n )
O(1.99n )
1/6
5/3
O(1.57n )
O(1.94n )
1/10
13/5
O(1.39n )
O(1.81n )
0.02506 . . .
10
O(1.13n )
O(1.36n )
Table 1. Sample c values, approximation ratios and running times of the algorithm described in
Theorem 4 and running times of the trivial approximation algorithm with the same approximation
ratio.
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