CHAPTER - VI
GRAPH EQUATIONS FOR LINE GRAPHS, MIDDLE
GRAPHS, TOTAL CLOSED NEIGHBORHOOD GRAPHS
AND TOTAL CLOSED EDGE NEIGHBORHOOD GRAPHS
ABSTRACT
In this paper, we solve the graph equations
L(G) = N1C(H),
M(G)
L(G) = Ntc(H),
= N1C(H), L(G)
= ENte(H),
M(G)
= Ntc(H),
Ufi) = ENte(H),
M(G) = EN,C(H) and M(G) = EN,C(H).
The symbol = stands for isomorphism between two graphs.
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6.1. INTRODUCTION
For a given graph operator (|), which graph is fixed under the
operator <J), i.e., 4>(G) = G ? It was known that for a connected graph G
L(G) = G if and only if G is a cycle (see [13]).
Cvetkovic and Simic [9] solved graph equations:
L(G) = T(H)
and
L(G) = T(H).
Akiyama, Hamada and Yoshimura [1] solved graph equations:
L(G) = M(H), M(G) = T(H)
M(G) = T(H), L(G) = M(H).
Also, they got a solution for the graph equation L(G) = L(H) in [2].
Further Akiyama, Koneko and Simic (see [3]) obtained solution for
graph equations:
L(G) = Gn, L(G) = Gn(orL(G) = G")andL(G) = (G).
Kulli and Patil [11] solved garaph equations:
L(G) = Tb(H), UG) s Tb(H)
L(G) - Tb(H), UG) = Tb(H).
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Basavanagoud and Mathad [4] solved graph equations:
L(G) = Q(H),
L(G) s Q(H)
M(G) = Q(H),
M(G)= Q(H)
L(G) = P(H),
L(G)= P(H)
M(G) = P(H),
M(G)= P(H).
In this chapter, we solve completely the following graph equations:
L(G) = N,C(H)
L(G) = Ntc(H)
M(G) = Ntc(H)
M(G) = Ntc(H)
L(G) = ENtc(H)
L(G) = ENtc(H)
M(G) = ENtc(H)
M(G) = ENtc(H).
That is, we obtain all pairs (G, H) of graphs satisfying the
equations (1) to (8).
The complement G of a graph G is the graph having the same set
of vertices as G and in which two vertices are adjacent if and only if they
are not adjacent in G.
The line graph L(G) of a graph G is the graph whose vertices can
be put in one-to one correspondence with the edges of G in such a way
that two vertices of L(G) are adjacent if and only if the corresponding
edges of G are adjacent (see [14]).
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The middle graph M(G) of a graph G is the graph whose edge set is
the union of vertices and edges of G in which two vertices are adjacent if
they are adjacent edges of G or one is a vertex and other is an edge
incident with it (see [10]).
A graph G+ is the endedge graph of a graph G if G+ is obtained
from G by adjoining an endedge U;u' at each vertex Uj of G (see [1]).
If G and H are graphs with the property that the identification of
any vertex of G with an arbitrary vertex of H results in a unique graph (up
to isomorphism) then we write G • H for this graph.
For a graph G, let V(G) and E(G) denote its vertex set and edge
set respectively.
The open-neighborhood N(u) of a vertex u in V(G) is the set of all
vertices adjacent to u.
N(u) = { v/uv € E(G) }.
The closed neighborhood N[u] of a vertex u in V(G) is given by
N[u] = {u}uN(u).
For each vertex uf of G, a new vertex u' is taken and the resulting
set of vertices is denoted by V|(G).
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The total closed neighborhood graph Ntc(G) of a graph G is defined
as the graph having vertex set V(G) u Vi(G) and two vertices are
adjacent if they correspond to adjacent vertices of G or one corresponds
to a vertex u' of Vi(G) and the other to a vertex Wj of G and Wj is in N[Uj]
(see [12]).
The open-neighborhood N^) of an edge e, in E(G) is the set of
edges adjacent to e(.
N(e;) = { ej / e, and ej are adjacent in G }.
The closed-neighborhood N[e; ] of an edge e( in E(G) is given by
N[ei] = {ei}uN(ei)
For each edge e, of G, a new vertex e' is taken and resulting set of
vertices is denoted by Ej(G).
For a graph G, we define the total closed edge neighborhood graph
ENtc(G) of a graph G as the graph having vertex set E(G) u Ej(G) with
two vertices are adjacent if they correspond to adjacent edges of G or one
corresponds to an element e' of Ei(G) and the other to an element
E(G), where e} is in N[eJ (see [5]).
In Figure 6.1, a graph G and its Ntc(G) and ENtc(G) are shown.
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of
HO
We need the following results.
THEOREM 6.A [6]. A graph G is a line graph if and only if none of the
graphs Fj, i = 1, 2,..9 of Figure 6.2 is an induced subgraph of G.
THEOREM 6.B [8]. A graph G is the complement of a line graph if and
only if none of the nine graphs Ff, i = 1, 2, ..., 9 of Figure 6.3 is an
induced subgraph of G.
THEOREM 6.C [10]. Let G be any graph. Then L(G+) = M(G).
6.2. THE SOLUTION OF L(G) = Ntc(H)
Any graph H which is a solution of the above equation, satisfies the
following properties:
i)
H must be a line graph, since H is an induced subgraph of N,C(H).
ii)
H does not contain a cut-vertex, since otherwise, Fi would be an
induced subgraph of Nte(H).
iii)
H does not contain a component having more than two vertices,
since otherwise, Fi would be an induced subgraph of Ntc(H).
It is not difficult to see from observation (ii) that H has no
cut-vertices. We consider the following cases:
Case 1. Suppose H is connected. Then H is Ki or K2. The corresponding
G is Kii2 or K3 • K2 respectively.
Case 2. Suppose H is disconnected. Then H is nKi or nK2. The
corresponding G is nK!i2 or n(K3 • K2) respectively.
ill
Figure 6.2
112
Figure 6.3.
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From the above discussion, we conclude the following:
THEOREM 6.1. The following pairs (G, H) are all pairs of graphs
satisfying the graph equation L(G) = Ntc(H):
(nK)i2, nKO, n > 1 ; and (n(K3 • K2), nK2), n > 1.
6.3. THE SOLUTION OF L(G)=Ntc(H)
First, we observe that in this case H satisfies the following
properties:
i)
If H has atleast one edge, then it is connected, since otherwise,
F, and F2 are induced subgraphs of Ntc(H).
ii)
H does not contain a path P4 as an induced subgraph, since
otherwise, Fj is an induced subgraph of Ntc(H),
iii)
H does not contain C„ , n > 5 as an induced subgraph, since
otherwise, F, would be an induced subgraph of N,C(H).
iv)
H does not contain more than one cut-vertex, since otherwise,
F* would be an induced subgraph of Ntc(H).
v)
H does not contain Ki>4 as an induced subgraph, since
otherwise, F3 would be an induced subgraph of Ntc(H).
vi)
H does not contain a cut-vertex which lies on blocks other than
K2, since otherwise, F2 is an induced subgraph of Ntc(H).
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Thus H has atmost one cut-vertex. We consider the following cases:
Case 1. Suppose H has exactly one cut-vertex. Then H is Ki,2 or K)i3.
Corresponding G is (C4 • K2) u K2 or (K4 • K2) u K2 respectively.
Case 2. Suppose H has no cut-vertices. We consider the following
subcases:
Subcase 2.1. H = K„. In this case (K,,„ u nK2 , K„), n > 1 and
(K3
u3K2,
K3) are the solutions.
Subcase 2.2. H = Km,n. Then from observation (v), (C4 • K4 , K2,3) and
(K4 • K4, K3,3) are the solutions.
Subcase 2.3. H is neither a complete graph nor a complete bipartite
graph. From observation (iii), H is Cn , n < 4 or K4 - x, where x is any
edge of K4. In this case the solutions are (Kj,3 u 3K2 , C3), (K3u 3K2 ,C3),
(C4 • C4, C4) and (G\ K4 - x) where G' is the graph shown in Fig. 6.4
are the solutions.
Figure 6.4.
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Thus we have the following
THEOREM 6.2. The following pairs (G, H) are all pairs of graphs
satisfying the graph equation L(G) = Ntc(H):
«C4 • K2) u K2, K,i2); ((IQ • K2) uK2 , K,t3); (K,,n u nK2, Kn), n > 1;
(K3 u 3K2 , K3); (C4 • IQ ,K2j); (IQ • IQ , K3i3); (C4 • C4 , C4); and
(GMQ-x), where x is any edge of IQ and G' is the graph shown in Fig. 6.4.
6.4. THE SOLUTION OF M(G) = Nte(H)
Theorem 6.1 gives solutions of the graph equation L(G) = Ntc(H).
But none of these is of the form (G+, H). Hence, there is no solution of the
equation M(G) = N,C(H).
Now, we state the following result.
THEOREM 63. There is no solution of the graph equation M(G) = Ntc(H).
6.5. THE SOLUTION OF M(G)=Ntc(H)
Theorem 6.2 gives solution of the equation L(G) = Ntc(H). But
none of these is of the form (G+, H). Therefore there is no solution of the
graph equation M(G) = N,C(H).
Now, we state the following result.
THEOREM 6.4. There is no solution of the graph equation M(G) = Ntc(H).
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6.6. THE SOLUTION OF L(G) = ENtc(H)
In this case, H satisfies the following properties:
i)
H does not contain a cycle Cn , n > 3 as a subgraph, since
otherwise, Fj is an induced subgraph of ENtc(H).
ii)
H does not contain a component having more than one
cut-vertex, since otherwise, Fi is an induced subgraph of ENtc(H).
iii)
The maximum degree of H does not exceed two, since
otherwise, Fj is an induced subgraph of ENtc(H).
iv)
H does not contain a cut-vertex which lies on more than two
blocks, since otherwise, Fi is an induced subgraph of ENtc(H).
v)
H does not contain a cut-vertex which lies on a block other than
K2, since otherwise, Fi is an induced subgraph of ENtc(H).
From observation (ii), it follows that every component of H has
atmost one cut-vertex. We consider the following cases:
Case 1. Suppose H has no cut-vertices. Then from observation (i), H is
nK2, n > 1. The corresponding G is nKi,2, n > 1.
Case 2. Suppose H has cut-vertex. We consider the following subcases:
Subcase 2.1. Assume H is connected. Then H is Kj 2. The corresponding
G is K.3 • K2.
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Subcase 2.2. Assume H is disconnected. Then H is nKi,2 u mK2, m > 0,
n > 1. The corresponding G is n(K3 • K2) u mK)i2.
From above discussions, we conclude the following:
THEOREM 6.5. The following pairs (G, H) are all pairs of graphs
satisfying the graph equation L(G) = ENte(H):
(nK1>2, nK2), n > 1; (K3 • K2, K,.2); and
(n(K3 • K2)
u
mK|i2, nKj^ u mK2), m > 0, n > 1.
6.7. THE SOLUTION OF Z(G) = ENtc(H)
In this case, H satisfies the following properties:
i)
If H is disconnected, then it has atmost three components, each of
which is K2, since otherwise, F, is an induced subgraph of
ENte(H).
ii)
H is not a path Pn , n > 5, since otherwise, Fj is an induced
subgraph of ENtc(H).
iii)
H does not contain Cn, n > 5, since otherwise, F2 is an induced
subgraph of ENtc(H).
iv)
H is not a complete bipartite graph Km,n , for m > 3 or n > 3,
since otherwise, F2 is an induced subgraph of ENtc(H).
v)
H does not contain more than two cut-vertices, since otherwise,
1^ is an induced subgraph of ENtc(H).
Thus H has atmost two cut-vertices. We consider the following cases:
Case 1. If H has exactly one cut-vertex, then H is K),n, n > 1 or K3 • K2.
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For H = Ki,n, n > 1, G = Ki,„ u nK2
For H = K3 • K2, G is a graph as shown in Figure 6.4.
Case 2. If H has exactly two cut-vertices. Then H is a path P4.
Corresponding G is (C4 • K2) u K2.
Case 3. If H has no cut-vertices. We consider the following subcases:
Subcase 3.1. If H is disconnected. Then from observation (i), H is nK2, n < 3.
For n = 1, H = K2 and G = 2K2
For n = 2, H = 2K2 and G = C4
For n = 3, H = 3K2 and G = K4.
Subcase 3.2. If H is connected. We consider the following subcases.
Subcase 3.2.1. H = K„. In this case, it follows from observation (iii), that
(2K2, K2), (K3 u 3K2, K3), (K,i3 u 3K2, K3) and (G\ K4) where G' is the
graph shown in Fig. 6.5 are the solutions.
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Subcase 3.22. H = Km,n. Then from observation (iv), (2K2 , Kjj),
(Ki,2 u 2K2, K],2) and (C4 • C4 , K2i2) are the solutions.
Subcase 3.2.3. H is neither a complete graph nor a complete bipartite graph.
From observation (iii), H is C„, n < 4 or IQ - x, where x is any edge of K4.
In this case (K3 u 3K2, C3), (K)i3 u 3K2, C3), (C4 • C4, C4) and (G', IQ - x),
where G' is the graph as shown in Fig. 6.6, are the solutions.
Thus the graph equation is solved and we have the following:
THEOREM 6.6. The following pairs (G, H) are all pairs of graphs
satisfying the graph equation L(G) = ENtc(H):
G' is the graph as shown in Fig. 6.4; (G\ IQ), where G' is the graph
as shown in Fig. 6.5; and (G', IQ - x), where G' is the graph as shown
in Fig. 6.6.
Figure 6.6.
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6.8.THE SOLUTION OF M(G) = ENtc(H)
Theorem 6.5 gives solutions of the equation L(G) = ENtc(H). But
none of these is of the form (G+, H). Hence there is no solution of the
equation M(G) = ENtc(H).
Thus we obtain the following result.
THEOREM 6.7. There is no solution of the graph equation M(G) = ENtc(H).
6.9. THE SOLUTION OF M(G)= ENtc(H)
Theorem 6.6 gives the solution of the graph equation L(G) = EN,C(H).
Among these only one solution (2K2 , K2) is of the form (G+, H). Therefore,
the solution of the equation M(g) = ENtc(H) is (2Kj, K2).
Thus we have the following result.
THEOREM 6.8. There is only one solution (2Kj , K2) of the graph
equation M(g) = ENtc(H).
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■ W
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