IJCER (www.ijceronline.com) International Journal of computational Engineering research

Dr. A. Sudhakaraiah, V. Rama Latha, E. Gnana Deepika, T.Venkateswarulu/International Journal Of
Computational Engineering Research / ISSN: 2250–3005

To Find Strong Dominating Set and Split Strong Dominating Set of an
Interval Graph Using an Algorithm
Dr. A. Sudhakaraiah*, V. Rama Latha1, E. Gnana Deepika2, T.Venkateswarulu3
Department of Mathematics, S.V . University, Tirupati-517502, Andhra Pradesh, India.

Abstract:
Strong and weak domination arise naturally in certain practical situations. For example, consider a network of roads
connecting a number of locations. In such a network, the degree of a vertex v is the number of roads meeting at v . Suppose
deg u  deg v. Naturally, the traffic at u is heavier than that at v . If we consider the traffic between u and v , preference
should be given to the vehicles going from u to v . Thus, in some sense, u strongly dominates v and v weakly
dominates u . In this paper we present an algorithm to find a strong dominating set and split strong dominating set of an
interval graph which is connected.

Keywords: domination number, dominating set, Interval family, Interval graph, strong domination number , split
dominating set, spit strong dominating set, strong dominating set.

1. Introduction
We have defined a graph as a set and a certain relation on that set. It is often convenient to draw a “picture” of the graph. This
may be done in many ways usually one draws an interval graph corresponding to I for each vertex and connects vertex
u and vertex v with a directed arrow whenever uv is an edge. If both uv and vu are edges then some times a single line
joints u and v without arrows.
Let I  {I1 , I 2 ,....., I n } be the given interval family. Each interval i in I is represented by
[ai , bi ] , for i  1,2,....., n . Here ai is called the left endpoint and bi the right endpoint of the interval Ii . Without loss of generality
we may assume that all end points of the intervals in I which are distinct between 1and 2n. The intervals are labelled in the
increasing order of their right endpoints. Two intervals i and j are said to intersect each other, if they have non-empty
intersection. Interval graphs play important role in numerous applications, many of which are scheduling problems. They are a
subset of perfect graphs [1]. A graph G  (V , E ) is called an interval graph if there is a one-to-one correspondence between
V and I such that two vertices of G are joined by an edge in E if and only if their corresponding intervals in I intersect. That is,
if i  [ai , bi ] and j  [a j , b j ] , then i and j intersect means either a j  bi or ai  b j .
Let G be a graph, with vertex set V and edge set E . The open neighbourhood set of a vertex v V is
nbd (v)  {u V / uv  E}
The closed neighbourhood set of a vertex v V is
nbd[v]  nbd (v) {v}
A vertex in a graph G dominates itself and it‟s neighbours. A set D  V is called dominating set if every vertex in
 V  D  is adjacent to some vertex in D . The domination studied in [2-3]. The domination number  of G is the
minimum cardinality of a dominating set. The domination number is well-studied parameter. We can see this from the
bibliography [4] on domination. In [5] , Sampathkumar and Pushpa Latha have introduced the concept of strong domination
in graphs. Strong domination has been studied [6-67]. Kulli.V.R. et all [8] introduced the concept of split and non-split
domination in graphs. A dominating set D is called split dominating set if the induced subgraph  V  D  is disconnected.
The split domination number of  s of G is the minimum cardinality of a split dominating set. Let G  (V , E ) be a graph
and u, v V . Then u strongly dominates v if
(i) uv  E
(ii) deg v  deg u .
A set Dst  V is a strong dominating set of G if every vertex in V  Dst is strongly dominated by atleast one vertex in
Dst . The strong domination number  st (G) of G is the minimum cardinality of a strong dominating set. Define

IJCER | July-August 2012 | Vol. 2 | Issue No.4 |1026-1034 Page 1026


NI (i )  j , if bi  a j and there do not exist an interval k such that bi  ak  a j . If there is no such j , then define
NI (i)  null . Nsd (i) is the set of all neighbours whose degree is greater than degree of i and also greater than i. If there is no
such neighbor then define Nsd (i)  null. M (S) is the largest highest degree vertex in the set S. nbd+ (i) is the set of all
adjacent vertices which are greater than i. nbd- (i) is the set of all adjacent vertices which are less than i. d+ (i) is the number
of adjacent vertices which are greater than i. d- (i) is the number of adjacent vertices which are less than i.

3. Algorithms
3.1. To find a Strong dominating set of an interval graph using an algorithm
Input: Interval family I  {I1 , I 2 ,....., I n } .
Output: Strong dominating set of an interval graph of a given interval family.
Step 1 : S1  nbd [1] .
Step 2 : S = The set of vertices in S1 which are adjacent to all other vertices in S1 .
Step 3 : Dst = The largest highest degree interval in S .
Step 4 : LI = The largest interval in Dst .
Step 5 : If N sd ( LI ) exists
Step 5.1 : a = M(Nsd ( LI )) .
Step 5.2 : b = The largest highest degree interval in nbd [a].
Step 5.3 : Dst  Dst {b} goto step 4.
end if
else
Step 6 : Find NI ( LI )
Step 6.1: If NI ( LI )  null goto step 7.
Step 6.2 : S2  nbd[ NI ( LI )] .
Step 6.3 : S 3 = The set of all neighbors of NI ( LI ) which are greater than or equal
to NI ( LI ) .
Step 6.4 : S 4 = The set of all vertices in S 3 which are adjacent to all vertices in S 3 .
Step 6.5 : c = The largest highest degree interval in S 4 .
Step 6.6 : Dst  Dst  {c} goto step 4.
Step 7 : End
3.2. To find a split strong dominating set(sdst) of an interval graph using an algorithm.
Input : Interval family I  {I1 , I 2 ,....., I n } .
Output : Split strong dominating set of an interval graph of an interval family I.
Step 1 : S1  nbd [1] .
Step 2 : S 2 =The set of all vertices in S1 which are adjacent to all other vertices in S1 .
Step 3 : SDst  {a} , where ��a‟ is the largest highest degree interval in S 2 .
Step 4 : Count = The number of pendent vertices or number of vertices with degree one in G.
Step 5 : If Count > 0 then goto step 7
Else
Step 6 : If there exists at least one edge (u, v) such that u  nbd  (a) & v  nbd  (a)
Step 6.1 : Count = Count + 1.
Step 6.2 : Take largest v .
Step 6.3 : SDst  SDst {v}  {a, v} .
Endif
Step 7 : LI = The largest interval in SD st .



Step 8 : If d  ( LI )  1 then (If already checked this condition for the same vertex skip)
Step 9 : If Count  1 then goto step 12
Else
Step 10 : If there exists atleast one edge (w, x) such that w  nbd  ( LI ) & x  nbd  ( LI )
Step 10.2 : Take largest x .
Step 10.3 : SDst  SDst {x} .
Endif
Step 11 : If d  ( x)  1
Step 12 : If N sd ( LI ) exists
Step 12.1 : a = M(Nsd (LI )) .
Step 12.2 : b = The largest highest degree interval in nbd [a].

Step 12.3 : SDst  SDst {b} goto step 7.
end if
else
Step 13 : Find NI ( LI )
Step 13.1: If NI ( LI )  null goto step 14.
Step 13.2 : S2  nbd[ NI ( LI )] .
Step 13.3 : S 3 = The set of all neighbors of NI ( LI ) which are greater than or equal
to NI ( LI ) .
Step 13.4 : S 4 = The set of all vertices in S 3 which are adjacent to all vertices in S 3 .
Step 13.5 : c = The largest highest degree interval in S 4 .
Step 13.6 : Dst  Dst  {c} goto step 7.
Step 14 : End
4. Main Theorems.
Theorem 4.1 : Let G be an interval graph corresponding to an interval family I  {I1 , I 2 ,....., I n } . If i and j are any two
intervals in I such that i  Dst , where Dst is a minimum strong dominating set of the given interval graph G , j  1 and j is
contained in i and if there is at least one interval to the left of j that intersects j and at least one interval k  i to the right of j
that intersects j then  st (G)   sst (G).
Proof : Let G be an interval graph corresponding to an interval family I  {I1 , I 2 ,....., I n } . Let i and j be any two intervals in
I such that i  Dst , where Dst is a minimum strong dominating set of the given interval graph G , j  1 and j is contained in
i and Suppose there is at least one interval to the left of j that intersects j and at least one interval k  i to the right of j that
intersects j . Then it is obviously we know that j is adjacent to k in the induced sub graph  V  Dst  . Then there will be a
connection in  V  Dst  . Since there is at least one interval to the left of j that intersects j , there will be a connection in
 V  Dst  to it‟s left. In this connection we introduce another interval „h‟, which is to the right of j and i and also interest i
and j to Dst for disconnection in the induced subgraph  V  Dst  . We also formulated Split strong dominating set as
follows
SDst  Dst {h}  SDst  Dst  {h} .
Since Dst , h are disjoint  SDst  Dst  {h} Or Dst  {h} = SDst .
  st (G )  {h}   sst (G) .



  st (G)   sst (G) .
ILLUSTRATION
1 4 7 10
2 6 8
3 5 9
Figure 1. Interval Family
As follows an algorithm with illustration for neighbours as given interval family I.
We construct an interval graph G from an interval family I={1, 2,……,10}as follows
nbd 1  1, 2,3 , nbd  2  1, 2,3, 4 , nbd 3  1, 2,3, 4, 6 , nbd  4  2,3, 4,5, 6 ,
nbd 5  4,5, 6, 7 , nbd 6  3, 4,5, 6, 7,9 , nbd 7  5, 6, 7,8,9 , nbd 8  7,8,9,10 ,
nbd 9  6, 7,8,9,10 , nbd 10  8,9,10.
Nsd 1  2,3 , Nsd  2   3, 4 , Nsd  3  6 , Nsd  4   6 , Nsd  5  6 , Nsd  6   null, Nsd  7   null,
Nsd 8  9 , Nsd  9   null, Nsd 10   null.
NI 1  4, NI  2   5, NI  3  5, NI  4   7, NI  5  8, NI  6  8, NI  7   10,
NI 8  null, NI  9   null, NI 10   null.
Procedure for finding a strong dominating set( Dst) of an interval graph using an algorithm.
Input : Interval family I  {1, 2,......,10} .
Step 1 : S1  {1, 2,3} .
Step 2 : S  {1, 2,3} .
Step 3 : Dst  {3} .
Step 4 : LI  3 .
Step 5 : Nsd (3)  {6} .
Step 5.1 : a  M ( Nsd (3))  M ({6})  6 .
Step 5.2 : b  6 .
Step 5.3 : Dst  {3} {6}  {3,6} .
Step 6 : LI  6 .
Step 7 : NI (6)  8 .
Step 7.1 : S2  nbd [8] ={7,8,9,10} .
Step 7.2 : S3  {8,9,10} .
Step 7.3 : S4  {8,9,10} .
Step 7.4 : c  9 .
Step 7.5 : Dst  Dst {9}  {3,6} {9}  {3,6,9} .
Step 8 : LI  9 .
Step 9 : Nsd(9) = null and NI (9) = null.
Step 10 : End.
Output : {3,6,9} is the strong dominating set of an interval graph of given interval family.
Procedure for finding a split strong dominating set(SD st) of an interval graph using an algorithm.
Step 1 : S1  {1, 2,3} .
Step 2 : S2  {1, 2,3} .
Step 3 : SDst  {3} .
Step 4 : Count = 0.
Step 5 : There exists (2,4) such that 2  nbd  (3)  {1, 2} and 4  nbd  (3)  {4, 6} .



Step 5.1 : Count = Count +1 .
Step 5.2 : SDst  {3} {4}  {3, 4} .
Step 6 : LI  4 .
Step 7 : Nsd (4)  {6} .
Step 5.1 : a  M ( Nsd (4))  M ({6})  6 .
Step 5.2 : b  6 .
Step 5.3 : SDst  {3, 4} {6}  {3, 4,6} .
Step 8 : LI  6 .
Step 9 : NI (6)  8 .
Step 9.1 : S2  nbd [8] ={7,8,9,10} .
Step 9.2 : S3  {8,9,10} .
Step 9.3 : S4  {8,9,10} .
Step 9.4 : c  9 .
Step 9.5 : Dst  Dst {9}  {3,6} {9}  {3,6,9} .
Step 10 : LI  9 .
Step 12 : End.
Output : {3,4,6,9} is the split strong dominating set of an interval graph of given interval family.
Dst  {3,6,9} .
SDst  {3, 4,6,9} .
Dst  SDst .
 st (G)   sst (G) .
Theorem 4.2 : Let Dst be a strong dominating set of the given interval graph G corresponding to an interval family
I  {I1 , I 2 ,....., I n } . If i and j are any two intervals in I such that j is contained in i and if there is no other interval k  i
that intersects j then the strong dominating set Dst is also a split strong dominating set of an interval graph G.
Proof : Let I  {I1 , I 2 ,....., I n } be an interval family and G is an interval graph corresponding to I . Let i and j be any two
intervals in I such that j is contained in i . If there is no interval k  i that intersect j . Then clearly i lies in the strong
dominating set Dst . Further in induced subgraph  V  Dst  the vertex j is not adjacent to any other vertex and then j
becomes as an solated vertex in induced sub graph  V  Dst  . There is a disconnection in  V  Dst  .Hence the strong
dominating set, which we considered is split strong dominating set.Hence we follows an algorithm to find strong dominating set
and split strong dominating set of an interval graph with an illustration.

ILLUSTTRATION
2 4 7
1 3 6 9
5 8

Figure 2.Interval Family

As follows an algorithm with illustration for neighbours as given interval family I.
We construct an interval graph G from an interval family I={1, 2,……,9}as follows
nbd 1  1, 2 , nbd  2  1, 2, 3 , nbd 3  2,3, 4,5 , nbd  4  3, 4,5, 6 , nbd 5  3, 4,5, 6 ,
nbd 6  4,5, 6, 7,8 , nbd 7   6, 7,8,9 , nbd 8  6, 7,8, 9 , nbd 9  7,8,9
Nsd 1  2 , Nsd  2   3 , Nsd  3  null, Nsd  4   6 , Nsd  5  6 , Nsd  6   null,
Nsd  7   null, Nsd 8  null, Nsd  9   null.
NI 1  3, NI  2   4, NI  3  6, NI  4   7, NI  5  7, NI  6   9, NI  7   null,



NI 8  null, NI  9   null.

Procedure for finding a strong dominating set( D st) of an interval graph using an algorithm.
Step 1 : S1  nbd[1]  {1, 2} .
Step 2 : S  {1, 2} .
Step 3 : Dst  {2} .
Step 4 : LI  2 .
Step 5 : Nsd (2)  {3} .
Step 5.1 : a  M ( Nsd (2))  M ({3})  3 .
Step 5.2 : b  5 .
Step 5.3 : Dst  {2} {5}  {2,5} .
Step 6 : LI  5 .
Step 7 : Nsd (5)  {6} .
Step 7.1 : a  M ( Nsd (5))  M ({6})  6 .
Step 7.2 : b  6 .
Step 7.3 : Dst  {2,5} {6}  {2,5,6} .
Step 8 : LI  6 .
Step 9 : NI (6)  9 .
Step 9.1 : S2  nbd [9] ={7,8,9} .
Step 9.2 : S3  {9} .
Step 9.3 : S4  {9} .
Step 9.4 : c  9 .
Step 9.5 : Dst  Dst {9}  {2,5,6} {9}  {2,5,6,9} .
Step 10 : LI  9 .
Step 11: Nsd(9) = null and NI (9) = null.
Step 12 : End.
Output : {2,5,6,9} is the strong dominating set of an interval graph of given interval family.
Step 1: S1  nbd[1]  {1, 2} .
Step 2 : S  {1, 2} .
Step 3 : SDst  {2} .
Step 4 : Count = 1.
Step 5 : LI  2 .
Step 6 : d  (2)  1 .
Step 6.1 : Count = 1 + 1 = 2.
Step 7 : Nsd (2)  {3} .
Step 7.1 : a  M ( Nsd (2))  M ({3})  3 .
Step 7.2 : b  5 .
Step 7.3 : Dst  {2} {5}  {2,5} .
Step 8 : LI  5 .
Step 9 : Nsd (5)  {6} .
Step 9.1 : a  M ( Nsd (4))  M ({6})  6 .
Step 9.2 : b  6 .



Step 9.3 : SDst  {2,5} {6}  {2,5,6} .
Step 10 : LI  6 .
Step 11 : NI (6)  9 .
Step 11.1 : S2  nbd [9] ={7,8,9} .
Step 11.2 : S3  {9} .
Step 11.3 : S4  {9} .
Step 11.4 : c  9 .
Step 11.5 : Dst  Dst {9}  {2,5,6} {9}  {2,5,6,9} .
Step 12 : LI  9 .
Step 14 : End.
Dst  {2,5,6,9} , SDst  {2,5,6,9} .
Dst  SDst .
Theorem 4.3 : Let I  {I1 , I 2 ,....., I n } be an interval family and Dst is a strong dominating set of the given interval graph G. If
i , j , k are any three consecutive intervals such that i  j  k and if j  Dst , and i intersects j , j intersect k and i does
not intersect k then Dst  SDst .
Proof : Suppose I  {I1 , I 2 ,....., I n } be an interval family . If i , j , k be three consecutive intervals such that i  j  k and i
intersect j , j intersect k , but i does not intersect k . Suppose j  Dst , where Dst is a strong dominating set. Then i and k
are not adjacent in the induced subgraph  V  Dst  . There exists a a disconnection between i and k . That is, there is no m  I ,
m  k such that m intersects k . If possible suppose that such an m exists, then since m  k we must have
m  i  j  k (m  k ). Now m intersects k implies i and j also intersect. Then there is a path between i and k and are
adjacent. This is a contradiction to hypothesis. So such a m does not exists. Hence we get disconnection. Hence Dst is also a split
strong dominating set of the given interval graph G . As usual as follows an algorithm to find a strong dominating set and split
strong dominating set of an interval graph G .
ILLUSTRATION
4 8
6
2 5 9
1 3 7

Figure 3.Interval Family I.
We construct an interval graph from an interval family I={1,2,3,4,5,6,7,8,9} as follows.
nbd 1  1, 2 , nbd  2  1, 2,3, 4 , nbd 3  2,3, 4,5 , nbd  4  2,3, 4,5 , nbd 5  3, 4,5, 6 ,
nbd 6  5, 6, 7,8 , nbd 7  6, 7,8,9 , nbd 8  6, 7,8,9 , nbd 9  7,8,9.
Nsd 1  2 , Nsd  2   null, Nsd  3  null, Nsd  4   null, Nsd  5  null, Nsd  6   null,
Nsd  7   null, Nsd 8   null, Nsd  9   null.
NI 1  3, NI  2   5, NI  3  6 , NI  4   6, NI  5  7, NI  6   9, NI  7   null,
NI 8  null, NI  9   null.
Procedure for finding a strong dominating set( D st) of an interval graph using an algorithm.
Step 1 : S1  nbd[1]  {1, 2} .
Step 2 : S  {1, 2} .
Step 3 : Dst  {2} .
Step 4 : LI  2 .
Step 5 : NI (2)  5 .



Step 5.1 : S2  nbd [5] ={3,4,5,6} .
Step 5.2 : S3  {5, 6} .
Step 5.3 : S4  {5,6} .
Step 5.4 : c  6 .
Step 5.5 : Dst  Dst {6}  {2} {6}  {2,6} .
Step 6 : LI  6 .
Step 7 : NI (6)  9 .
Step 7.1 : S2  nbd [9] ={7,8,9} .
Step 7.2 : S3  {9} .
Step 7.3 : S4  {9} .
Step 7.4 : c  9 .
Step 7.5 : Dst  Dst {6}  {2,6} {9}  {2,6,9} .
Step 8 : LI  9 .
Step 9: Nsd(9) = null and NI (9) = null.
Step 12 : End.
Output : {2,6,9} is the strong dominating set of an interval graph of given interval family.

Step 1 : S1  nbd[1]  {1, 2} .
Step 2 : S  {1, 2} .
Step 3 : SDst  {2} .
Step 4 : Count = 1.
Step 5 : LI  2 .
Step 6 : d+(2) = 1.
Step 6.1 : Count = 1 + 1 = 2.
Step 7 : Nsd (2)  {3} .
Step 7.1 : a  M ( Nsd (2))  M ({3})  3 .
Step 7.2 : b  5 .
Step 7.3 : Dst  {2} {5}  {2,5} .
Step 8 : LI  5 .
Step 9 : Nsd (5)  {6} .
Step 9.1 : a  M ( Nsd (4))  M ({6})  6 .
Step 9.2 : b  6 .
Step 9.3 : SDst  {2,5} {6}  {2,5,6} .
Step 10 : LI  6 .
Step 11 : NI (6)  9 .
Step 11.1 : S2  nbd [9] ={7,8,9} .
Step 11.2 : S3  {9} .
Step 11.3 : S4  {9} .
Step 11.4 : c  9 .
Step 11.5 : Dst  Dst {9}  {2,5,6} {9}  {2,5,6,9} .
Step 12 : LI  9 .
Step 13 : Nsd(9) = null and , NI (9) = null.
Step 14 : End.



Dst  {2,5,6,9}
SDst  {2,5,6,9} .
 Dst  SDst .
5. Conclusions
In this paper we introduced an algorithm for finding strong dominating set and split strong dominating set of an interval graph
which is connected.

6. Acknowledgements
The authors are very grateful to the referees for many valuable suggestions and corrections which have helped to significantly
improve the quality of the presentation of the paper.. This work was supported by S.V. University, Tirupati-517502, Andhra
Pradesh, India.

References
[1] M.C. Golumbic, “Algorithmic graph theory and perfect graphs” ,Academic press ,1980.
[2] T. W. Haynes, S.T. Hedetniemi and P.J.Slater, Fundamentals of domination in graphs, Marcel Dekker, Inc., New
York (1998).
[3] T. W. Haynes, S.T. Hedetniemi and P.J.Slater, Domination in Graphs: advanced topics , , Marcel Dekker, Inc., New
York (1998).
[4] S.T. Hedetniemi and R.C. Laskar, 1990, 86, 257-277
[5] E. Sampathkumar, L.Pushpa Latha , 1996, 161, 235-242
[6] J.H. Hahingh, M.A.Henning, 1998, 26 ,73-92.
[7] D.Rautenbach, 2000, 215 , 201-212.
[8] Kulli.V. R. and Janakiram . B, 2000,Vol.19. No.2, pp. 145-156.


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