CHAPTER - 8
GRAPH EQUATIONS FOR LINE GRAPHS,
JUMP GRAPHS, MIDDLE GRAPHS,
SPLITTING GRAPHS AND
LINE SPLITTING GRAPHS
ABSTRACT
In this chapter, we solve graph equations L(G) = S(H), M(G) = S(H),
L(G) = LS(H), M(G) = LS(H), J(G) = S(H), M(G) = S(H), J(G) = LS(H)
and M(G) = LS(H). The equality symbol ‘=’ stands for an isomorphism
between two graphs.
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8.1. INTRODUCTION
For a graph G, 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
vertices adjacent to u. N(u) ={v/uve E(G)}.
For each vertex Ui of G, a new vertex u' is taken and the resulting
set of vertices is denoted by Vj(G).
The splitting graph S(G) of a graph G is defined as the graph
having vertex set V(G)uV|(G) and two vertices are adjacent if they
correspond to adjacent vertices of G or one corresponds to a vertex u{ of
Vj(G) and the other to a vertex wj of G and Wj is in N(uj). This concept
was introduced by Sampathkumar and Walikar in [12].
The open neighborhood N(ej) of an edge ej in E(G) is the set of
edges adjacent to ej. N(ej) = {ej / ej and ej are adjacent in G}.
For each edge ej of G, a new vertex ej' is taken and the resulting set
of vertices is denoted by Ei(G).
The line splitting graph LS(G) of a graph G is defined as the graph
having vertex set E(G)v;Ei(G) with two vertices are adjacent if they
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correspond to adjacent edges of G or one corresponds to element e/ of
E)(G) and the other to an element ej of E(G) where ej is in N(ej). This
concept was introduced by Kulli and Biradar in [10].
Akiyama, Hamada and Yoshimura [2] solved graph equations
L(G) = M(H), M(G) = T(H), M(G) = T(H) and L(G) =M(H). Also, they
got a solution for the graph equation L(G) = L(H) in [3]. In [7],
Cvetkovic and Simic obtained a bibliography of graph equations. We
solved graph equations L(G) = Q(H), L(G) = P(H), L(G) = Q(H) and
L(G) =P(H) in [4].
In this chapter, we solve the following graph equations :
(1) L(G) = S(H)
(5) J(G) = S(H)
(2) M(G) = S(H)
(6) M(G) = S(H)
(3) L(G) = LS(H)
(7) J(G)=LS(H)
(4) M(G) = LS(H)
(8) M(G) = LS(H)
Beineke has shown in [5] that a graph G is a line graph if and only
if G has none of the nine specified graphs Fj, i = 1,2,...,9 as an induced
subgraph as stated in Theorem 7.A and Fi is the complement of Fj as in
Theorem 7.B. We depict here three of the nine graphs which we use.
They are Fi = K
1,
, F3 and F5.
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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 u. of G. Hamada
and Yoshimura have proved in [8] that M(G) = L(G+).
8.2. THE SOLUTION OF L(G) = S(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 S(H).
ii.
H does not contain a cut-vertex, since otherwise, F is an
induced subgraph of S(H).
iii.
H does not contain a vertex which is adjacent to two
nonadjacent vertices, since otherwise, F is an induced subgraph
ofS(H).
iv.
H does not contain Cn, n > 4 as a subgraph since otherwise, F
is an induced subgraph of S(H).
It follows from above observations that H has no cut-vertices. We
consider the following cases:
Case 1. Suppose H is disconnected. Then components of H are K] or K2
or K3. Therefore, (nK|, 2nK2), n > 2; (nP5, nK2), n > 2; (nKj, nK3), n > 2;
(2mK2 u nP5, mK| u nK2), m, n > 1; (2mK2 u nKj, mKi u nK3), m,n > 1;
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(mP5 unKfy mK2unK3), m,n >1; and (2mK2 unP5 u/Kj, mKi unK2 u /K3),
m, n, / >
1
are the solutions.
Case 2. Suppose H is connected. Then H is K| or K2 or K3. The
corresponding G is 2K2 or P5 or K3 respectively.
From the above discussion, we conclude the following :
THEOREM 8.1. The following pairs (G,H) are all pairs of graphs
satisfying the graph equation L(G) = S(H):
(nP5, nK2), n > 1; (2nK2, nK,), n > 1; (nK*, nK3), n > 1; (2mK2 u nP5,
mK, u nK2), m, n > 1; (2mK2 u nKj, mKiUnK3), m, n > 1; (mPsUnK^,
mK2 u nK3), m, n > 1; and (2mK2 u nP5 u /K3, mK] u nK2 u /K3),
m, n, / > 1.
8.3. THE SOLUTION OF M(G) = S(H)
We have investigated the solutions (G,H) of the equation
L(G) = S(H) in Theorem 8.1. Among these solutions (2nK2, nKt), n > 1;
(nK^, nK3), n > 1; and (2mK2 u nK^, rnKj u nK3), m, n > 1 are of the
form (G+,H). Therefore, the solutions of the equation M(G) =S(H) are
(2nKj, nK,) n > 1, (nK3, nK3), n > 1 and (2mK,unK3, mK,unK3),
m, n > 1. Thus we have the following.
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THEOREM 8.2. The solutions (G,H) of the graph equation M(G) = S(H)
are (2nKl5 nKj) n > 1; (nK3, nK3), n > 1; and (2mKi u nK3, mKi u nK3),
m, n > 1.
8.4. THE SOLUTION OF L(G) = LS(H)
We observe that in this case H satisfies the following properties:
i.
H does not contain a component having more than one cutvertex since otherwise, Fj is an induced subgraph of LS(H).
ii.
H is not a complete graph Kn, n > 4, since otherwise, Fi is an
induced subgraph of LS(H).
iii.
H does not contain P4 as a subgraph since otherwise, Fj is an'
induced subgraph of LS(H).
iv.
H does not contain Ki>4 as an induced subgraph since otherwise,
F] is an induced subgraph of LS(H).
v.
H is not a cycle Cn, n > 4 since otherwise, Fj is an induced
subgraph of LS(H).
vi.
H does not contain a cut-vertex which lies on blocks other than K2.
From observation (i) 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 H is nK2, n > 1 or nK3, n > 1,
or mK2 u nK3, m, n > 1.
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ForH = nK2, n> 1,
G = 2nK2
For H = nK3, n > 1,
G= nK3+
For H = mK2 u nK3, m, n > 1, G = 2mK2 u nK3
Case 2. Suppose H has cut-vertices. We consider the following subcases :
Subcase 2.1. Assume H is connected. Then H is Ki,2 or K13. The
corresponding G is P5 or K3 respectively.
Subcase 2.2. Assume H is disconnected. Then each component of H has
atmost one cut-vertex. Then H is mKii2 u nK^ u /K2 u rK3, m > 1 and
n, /, r > 0 or mKi>2 u nK)>3 u IK2 u rK3, n > 1 and m, /, r > 0. In this case
(mP5 u (n+r) K3 u 2mK2, mK]>2 u nKIj3 u /K2 u rK3) is the solution.
From above discussions, we conclude the following:
THEOREM 8.3. The following pairs (G,H) are all pairs of graphs
satisfying the graph equation L(G) = Ls (H):
(2nK2, nK2), n > l;(nK;,nK3), n > 1; (2mK2unK3+, mK2unK3), m,n > 1;
(mP5 u (n+r)K3 u 2/K2, mK|2 u nK] 3u IK2 ^J rK3), m > 1, n, /, r > 0;
and (mP5u(n+r)K3 u 2/K2, mKi<2u nK1>3 u IK2u rK3), n > 1, m, /, r > 0.
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8.5. THE SOLUTION OF M(G)=LS(H)
Theorem 8.3 provides solutions of the equation L(G) = LS(H). Among
these solutions (2nK2, nK2), n > 1; (nK.3, nK3), n > 1; (2mK2 unKj,
mK2unK3), m,n >1; and (mP5u(n+r)K.3 u 2/K2, mK1>2 u nK!;3 u IK2 u rK3),
r, / > 0, n > 1 and m = 0 are of the form (G+,H). Therefore, solutions
of the equation M(G) = LS(H) are (2nKi, nK2), n > 1; (nK3, nK3),
n > 1; (2mKi u nK3, mK2 u nK3), m, n > 1; and ((n + r) K3 u 2/Kt,
nKi;3 vj /K2u rK3), n > 1 and r, / > 0. Now we state the following result.
THEOREM 8.4. The solutions (G,H) of the graph equation
M(G) = LS(H) are (2nKb nK2), n > 1; (nK3, nK3), n > 1; (2mK, u nK3,
mK2 u nK3), m, n > 1; and ((n+r)K3 u 2/Ki, nKjj3 u IK2 u rK3), n > 1 and
r, / > 0.
8.6. THE SOLUTION OF J(G) = S(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,
F3 is an induced subgraph of S(H).
ii.
H does not contain a cut-vertex, since otherwise, F3 is an
induced subgraph of S(H).
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iii.
If H is a block, then it is a complete graph since otherwise, Fs is
an induced subgraph of S(H).
It follows from observations (i), (ii) and (iii), that H is nKl5 n > 1 or
Kn, n > 2. The corresponding G is Kii2n or Kfy - v, where v is a pendant
vertex of Kfy which is adjacent to the vertex of maximum degree
respectively.
Hence equation 8.6 is solved and solutions are given in the
following theorem.
THEOREM 8.5. The following pairs (G,H) are all pairs of graphs
satisfying the graph equation J(G) = S(H):
(K1>2n, nK]), n > 1; and (Kfy - v, Kn), n > 2 where v is a pendant
vertex of Kfy which is adjacent to the vertex of maximum degree.
8.7. THE SOLUTION OF M(G) = S(H)
Theorem 8.5 gives solutions for the equation J(G) = S(H). But
none of these is of the form (G+,H). Hence, there is no solution of the
equation M(G) = S(H).
Thus we have the following result.
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THEOREM 8.6. There is no solution of the graph equation
M(G) = S(H).
8.8. THE SOLUTION OF J(G) = LS(H)
In this case H satisfies the following properties:
i.
H does not contain more than one cut-vertex, since otherwise,
Fs is an induced subgraph of LS(H).
ii.
H does not contain a cut-vertex, which lies on blocks otherthan
K2, since otherwise, Fs is an induced subgraph of LS(H).
iii.
H does not contain a cycle Cn, n > 4, since otherwise, Fs is an
induced subgraph of LS(H).
iv.
If H is disconnected graph then every component of H is K2,
since otherwise, F3 is an induced subgraph of LS(H).
We consider the following cases.
Case 1. Suppose H is disconnected. Then from observation (vi), H is nK2,
n > 2. The corresponding G is 2nK2.
Case 2. Suppose H is connected. Then from observation (i), H has atmost
one cut-vertex. We consider the following subcases.
Subcase 2.1. Assume H has the cut-vertex. Then from observation (ii), H
is K) n, n > 2. The corresponding G is K^n - v, where v is a pendant
vertex adjacent to the vertex of maximum degree.
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Subcase 2.2. Assume H is a block. Then H is K2 or K3. The
corresponding G is Ku or
- v, where v is a pendant vertex of
adjacent to a vertex of maximum degree.
Thus equation 8.8 is solved and we have the following.
THEOREM 8.7. The following pairs (G,H) are all pairs of graphs
satisfying the graph equation J(G) = LS(H): (2nK2, nK2), n > 2;(Ki>2, K2);
(K^-VjKj), where v is a pendant vertex adjacent to the vertex of
maximum degree; and (K^n - v,K, n), n > 2, where v is a pendant vertex
adjacent to the vertex of maximum degree.
8.9. THE SOLUTION OF M(G) = LS(H)
Theorem 8.7 gives solutions for the equation J(G) = LS(H).
Among these (2nK2, nK2), n > 2 is of the form (G+,H). Therefore, the
solution of the equation M(G) = LS(H) is (2nKh nK2), n > 2.
Now, we state the following result.
THEOREM 8.8. The solutions of the graph equation M(G) = LS(H) are
(2nK,,nK2), n > 2.
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