J. Comp. & Math. Sci. Vol. 1(5), 606-616 (2010)
Edge Decomposition of the Transformation graph Gxyz when xyz =+ + 1
B. BASAVANAGOUD and 2PRASHANT V. PATIL 1 Corresponding Author, Department of Mathematics, Karnatak University,Dharwad-580 003, e-mail*: b.basavanagoud@gmail.com bgouder1@yahoo.co.in 2 Department of Mathematics, S D M College of Engg. and Tech., Dharwad-580 002, e-mail: prashant66.sdm@gmail.com
ABSTRACTS The transformation graph G of G is the graph with vertex set
V (G ) E (G ) in which the vertices x and y are joined by an edge if one of the following conditions holds: (i) x, y V (G ) and x and y are adjacent in G, (ii) x, y E (G ) and x and y are adjacent in G, (iii) one of x and y is in V (G ) and the other is in E (G ) , and they are not incident in G. In this paper, for some standard class of graphs G, we establish the results on the decomposition of edges of the transformation graph G into some standard class of graphs. 2000 Mathematics Subject Classification: 05C38(05C99) Keywords: Transformation graph, Edge decomposition.
INTRODUCTION All graphs considered here are finite, undirected, simple and standard. We refer to4 for unexplained terminology
and notations. Let G = (V (G ), E (G )) be a graph. V (G ) and E (G ) are called the vertex set and edge sets of G respectively. For two vetrices u and v
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of G , if there is an edge e joining them, we say u and v are adjacent. In this case both u and v are end vertices of e , and u (or v ) and e are said to be incident. Two edges e and f are said to be adjacent if they have an end vertex in common. A graph is called hamiltonian if it has a spanning cycle. A spanning cycle is called hamiltonian cycle. A hamiltonian path in G is a path, which contains vertex of G . A decomposition of a graph G is a collection of subgraphs of G , whose edge set partition the edge set of G . The subgraphs of the decomposition are called the parts of the decomposition. A graph G is said to be F-decomposable or Fpackable if G has a decomposition in which all of its parts are isomorphic to graph F . The line graph L (G) of G is a graph whose vertex set is E (G ) in which two vertices are adjacent if and only if the edges are adjacent in G . The subdivision graph S1 (G ) of G is the graph with vertex set V (G ) E (G ) , two vertices of S1 (G ) are adjacent if and only if one of x, y is in V (G ) and other is in E (G ) and are incident in G . The total graph T (G ) of G is the graph whose vertex set is V (G ) E (G ) in which two vertices are adjacent if and only if they adjacent or incident in G . The complement of a graph G , denoted by G , is the graph with same vertex set as G and two vertices are adjacent if and only if they are not adjacent
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in G . Wu and Meng 6 generalized the concept of total graphs to a total transformation graph G xyz with x, y , z {,} , where G is precisely the total graph of G , and G is complement of G . Each of these eight kind of transformation graphs G xyz were studied in6. W e shall investigate the transformation graph G of a graph G . G is the graph with vertex set V (G ) E (G ) , in which two vertices u and v are joined by an edge in G if one of the following conditions holds: (i) u, v V (G ) and they are adjacent in G , (ii) u, v E (G ) , and they are adjacent in G , and (iii) one of u and v is in V (G ) and other is in E (G ) , and they are not incident in G . In this paper, for some standard class of graphs G , we decompose the edges of the transformation graph G into standard classes of graphs. Following results are used in the decomposition of the transformation graph G of G . THEOREM A.[3].(Bermond) (i) If n is even, K n can be decomposed into n/2 hamiltonian paths. (ii) If n is odd, K n can be decomposed into (n 1)/2 hamiltonian cycles. If n is even , K n can be decomposed into (( n 2)/2)C n cycles of length n and ( n/2) K 2 ‘s. Remark 1.[5] Let G be a graph of
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V (G ) = n and E (G ) = m . Then V (G ) = n m , d G ( x) = m , for any x V (G ) and d (e) = n 4 d (u ) d (v ) , for any G e = uv E (G ) . Remark 2.[4] A connected graph is isomarphic to its line graph if and only if it is a cycle. Remark 3.[4] A graph G is the line graph of Kn if and only if 1) G has n(n - 1)/2 vertices, 2) G is regular of degree 2(n - 2) , 3) Every two nonadjacent points are mutually adjacent to exactly four points, 4) Every two adjacent points are mutually adjacent to exactly p - 2 points. Remark 4. If G = K1, n , then the line graph L(G) of G is the complete graph Kn. Remark 5.[1] For any graph G, G and
L (G ) are the edgedisjoint subgraphs of
RESULTS First, partition of edges of G into edges of G, L(G) and stars for a given graph G. Proposition 2.1. Let G be a (p, q) graph. Then the edges of G can be partitioned
E (G ) ,
edges of the line graph of G and the remaining edges corresponding to the point vertices are the edges of G. Hence the proposition. Next Theorm gives the partition of edges of G into edges of G, a regular graph and stars, when G is a m-regular hamiltonian graph. Theorem 2.1 Let G be a m-regular hamiltonian graph with n vertices. Then the edges of G ++ – can be partitioned into G, (2(m -1)) regular graph on mn/2 vertices, (n - 2) regular graph on 2n vertices and (( mn/2) n) times K1, n 2 . Proof. Let G be a m-regular hamiltonian graph with n vertices. Then the number of edges in G is mn /2 . Therefore in G , n point vertices are of degree mn /2 and
mn/2 line vertices of degree ( n 2m 4) . Since, line graph L(G) of G is the
G .
into
vertices there are q times K1, p 2 and the
E ( L(G )) and q times
E ( K1, p 2 ) . Proof. In G , each line vertex is adjacent to (p - 2) point vertices and also with the line vertices which are adjacent edges in G. Therefore corresponding to q line
subgraph of G and each line vertex is adjacent to 2(m -1) line vertices, hence edges of G can be partitioned into 2(m -1) regular graph on mn/2 line vertices and the remaining edges can be partitioned as below. Consider n point vertices and n line vertices, which forms a cycle in G. Let H be a subgraph of G formed by 2n vertices. In H, degree of a point vertex is m + (n - 2) and degree of line vertex is (n - 2). Therefore edges of this subgraph can be partitioned into edges of G and (n - 2)
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-regular graph on 2n vertices. Now the remaining line vertices are ((mn/2) - n), which are adjacent to exactly (n - 2) point vertices. Therefore edges of G can be partitioned into 2(m - 1) -regular graph on mn/2 vetices, edges of G, (n - 2) -regular graph on 2n vertices and (mn/2 - n) times
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Proof. In H the central vertex of G is not adjacent to any line vertices and each line vertex is adjacent to exactly n 1 point vertices and n 1 line vertices. Hence H can be partitioned into n times K1,n , one
K1,n and one K n .
K1, n 2 .
Proposition 2.3. Let G = K1, n and
Corollary 2.1.1 Let G be a cycle Cn. Then
H = G . Consider the point vertices of
the edges of G can be partitioned into 2Cn and (N - 2) -regular graph on 2n vertices. Proof. In Theorem 2.1, take m = 2 . The number of edges of G is n. The number of
G, which are not central vertices, then edges of H can be partitioned into n times
vertices of G is 2n and the degree of point vertices and line vertices is n in G . Edges of
G can be partitioned into
G = C n , L(G) = C n and (n - 2) -regular graph on 2n vertices. Next, we give the edge decomposition of
G , when G = C n , Pn , K n , K n ,n , K 1, n . Proposition 2.2. Let
G = K1, n and
K1,n and on line vertices one K n . Proof. In H, each point vertex, which is not a central vertex of G is adjacent to exactly n 1 line vertices and one point vertex which is the central vertex of G, and each line vertex is adjacent to exactly n 1 line vertices. Hence H can be partitioned into n times K1,n and one K n . In the following theorem we partition the edges of G into cycles of different lengths, when G is cycle.
partitioned into n times K1, n 1 , such that
Theorem 2.2 Let G = C n and H = G . Then the edges of H can be partitioned into
any two K1, n 1 have
n 2 common
(a) 2C n , (( n 3)/2)C 2 n and nK 2 , if n is odd.
vertices, one K1,n , such that K1, n 1 and
(b) 2C n and (( n 2)/2)C 2 n , if n is even.
H = G . Then the edges of H can be
K1,n have n 1 common vertices and one complete graph K n , such that the vertices of K n and centers of K1, n 1 are common vertices.
Proof. Let G = C n . Then H has n point vertices with degree n and n line vertices with degree n. Case 1. When n is even. Let
v1 , v 2 ,...., v n be the point vertices
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e12 , e23 ,...., e( n 1) n , be the line vertices in H.
v j , j = 1,2,3,...., n . Consider the point vertex v1 . v1 is
In H each v j is adjacent to n 2 line vertices
adajacent to
e12 , e23 ,...., e( j 2)( j 1) , e( j 1)( j 2) ,...., e( n 1) n , en1.
combine these into (e23 , e34 ) ; (e45 , e56 ) ;....;
Combining these line vertices in two’s as
(e( n 2)( n 1) , e( n 1) n ) . We get the ( n 2)/2
n is even. There are ( n 2)/2 such
cycles of length 2 n in G as follows.
collections, which are adjacent to
v
v
1
e
v 2
e
12
23
e
3
34
v
e23 , e34 ,...., e( n 1) n . Now
v
4
e
45
e
n- 2
(n -2 )( n- 1)
v
e
n -1
v
n
e
n1
(n -1 )n
1. v1e( n 1) n v2 en1v3e12 v4 .....vn e( n 2)( n 1) v1 . 2. v1e( n 3)( n 2) v2 e( n 2)( n 1) v3 e( n 1) n v4 en1v5 e12 v6 e23 .....vn e( n 4)( n 3) v1 . 3. v1e( n 5)( n 4) v2 e( n 4)( n 3) v3e( n 3)( n 2) v4 e( n 2)( n 1) v5e( n 1) n v6 en1v7 e12 .....vn e( n 6)( n 5) v1 . .
.
.
.
.
.
(n-2)/2. v1e34 v2 e45 v3e56 v4 e67 .....vn e23v1 . From this construction and by Remark 2, it follows that edges of G is the union of edges of G = C n , edges of L(G ) = C n and the edges of cycles (( n 2)/2)C 2 n . Journal of Computer and Mathematical Sciences Vol. 1, Issue 5, 31 August, 2010 Pages (528-635)
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Case 2. When n is odd. There are n point vertices
v1 , v 2 ,...., v n
and
n
line
e12 , e23 ,...., en1 . Combine
vertices
( n 1) line
vertices into group of two's. v1 is adjacent to
e23 , e34 ,....., e( n 1) n . Leaving e( n 1) n
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if n is odd. (b)
K 1, n , (( n 2)/2)C n , (( n 2)/2 )C 2 n
and (3n/2) K 2 , if n is even. Proof. Let G = K1, n . Then G has n 1 point vertices each of degree n and n line vertices with degree
2(n 1) . Let
combine these into
v1 , v2 ,....vn , v be the point vertices and
(e23 , e34 ); (e45 , e56 )....; (e( n 3)( n 2) , e( n 2)( n 1) ) .
e1 = vv1 , e2 = vv2 ,....en = vvn E (G ) be the
Similarly v2 is adjacent to
n line vertices, where v is the central node of G. Case 1. When n is odd.
(e34 , e45 ); (e56 , e67 )....; ( e( n 2)( n 1) , e( n 1) n ) and
en1. As in Case 1, there are (( n 1) 2 ) / 2 cycles of length 2 n and nK 2 's,
In G , each v j is adjacent to n 1 line vertices e1 , e 2 ,...., e j 1 , e j 1 ,....., en and is
v1e( n 1) n , v2 en1 , v3 e12 ,.....vn e( n 2)( n 1) . be
adjacent to one point vertex v . Combining these line vertives into two by two. Thus
partitioned into G = C n , L(G ) = C n and
there are ( n 1)/2 such collections , which
((n 3)/2)C 2 n and nK 2 .
are adjacent to v j , for j = 1,2,...., n .
Theorem 2.3 Let G = Pn and H = G , then the edges of H can be partitioned into
Consider v1 , combine the edges as
(a) Pn , Pn 1 , (( n 3)/2) P2 n 1 and (n 1) K 2
be obtained as follows:
Therefore the edges of
G
can
if n is odd. (b)
(e2 , e3 ); (e4 , e5 );.....(en 1 , en ) . The ( n 1)/2 cycles of length 2 n in G can 1. v1e3v2 e4 v3e5 .....en vn 1e1vn e2 v1 .
Pn , Pn1 , ((n 2)/2) P2 n1 if n is even.
Proof. Similar to Theorem 2.2.
2. v1e5 v2 e6 v3e7 .....en vn 3e1vn 2 e2 vn 1e3vn e4 v1 . .
.
cycles and stars.
.
.
Theorem 2.4 If G = K1, n , then the edges
((n-1)/2) v1en v2 e1v3 e2 .....vn en 1v1 .
Next we partition the edges of K1,n into
of G can be partitioned into (a) K 1, n , ( ( n 1)/2 )C n and (( n 1)/2)C 2 n ,
From this construction and from the definition of G , it follows that edges of
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G can be partitioned into G = K1, n ,
length n and (n/2) K 2 's. Thus by Theorem
( n 1)/2 cycles of length 2 n and by
A,
Remark 4, a complete graph K n on line
K 1, n , (( n 2)/2)C n , (( n 2)/2 )C 2 n and
vertices, which can be further partitioned
(3n/2) K 2 's.
into
( n 2)/2 cycles of length n, by
Theorem A. Case 2. When n is even. Since each point vertex v j in G is adjacent to n 1 line vertices,
e1 , e 2 ,...., e j 1 , e j 1 ,....., en therefore among these, n 2 line vertices can be grouped into pairs. Consider v1 . v1 is adjacent to n 1 line vertices e2 , e3 ,...., en 1 , en . Leaving en , we get ( n 2)/2 pairs
(e2 , e3 ); (e4 , e5 );.....; (e n2 , e n1 ) . Similarly
v2 is adjacent to e1 , e3 ,...., en1 , en leaving e1 , we get ( n 2)/2 pairs (e3 , e 4 ); (e5 , e6 );.....; (e n1 , e n ) . Therefore as in Case 1, there are
can be partitioned into
G
In next theorems we partition
( K n ) into cycles, stars and regular graphs. Lemma 1. If G = K n , then the edges of the line graph L (G ) of G can be partitioned into (a) (( n 1)/2)C n and nK 1,2( n 3) , when n is odd. (b)
(( n 2)/2 )C n , nK 1,2( n 4) and (n/2) K1,2( n 2) , when n is even.
G = K n . Then L (G ) is a
Proof. Let
2(n 2) -regular graph on (n( n 1)) /2 vertices. Case 1. When n is odd. By
A, K n can
Theorem
be
decomposed into (( n 1)/2)C n , hence by
(n 2)/2 cycles of length 2 n and nK 2 's
Remark
given by v1en , v2 e1 , v3e2 .....vn en 1 . Hence by
into (( n 1)/2)C n . Consider n vertices
definition of
2,
L (G )
is
partitioned
G and from above
which form a cycle C n is L (G ) , each of
construction G can be partitioned into
these n vertices are adjacent with
G = K1, n , ( n 2)/2 cycles of length
2(n 2) 2 = 2( n 3) vertices of other
2n, nK 2 and one complete graph K n on
cycles in L( K n ) , hence nK 1,2( n 3) .
line vertices. By Remark 4, which can be
Case 2. When n is even.
further partitioned into ( n 2)/2 cycles of
By
Theorem
A, K n can
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be
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B. Basavanagoud et al., J. Comp. & Math. Sci. Vol. 1(5), 606-616 (2010)
((n 2)/2)C n and
decomposed into
((n 2)/2)C n , ((n 2)/2)((n 2)/2)C 2n
(n/2) K 2 . Hence by Remark 2. L (G) can
and L( K n ) . But by Lemma 1, L( K n ) is
be partitioned into (( n 2)/2)C n . Consider
partitioned into (( n 2)/2 )C n , nK 1,2( n 4)
n vertices which forms a cycle C n in L (G ) ,
and (n/2) K1,2( n 2) .
each of these n vertices are adjacent with
2( n 2) 4 = 2( n 4) vertices of other
Now consider (n/2) K 2 's on G. In
cycles in L (G ) , thus nK1,2( n 4) and the
G , (n/2) line vertices are adjacent to exactly n 2 point vertices. Therefore, in
remaining (n/2) K 2 's in G, will have (n/2)
G there are ( n/2) K1,n 2 and (n/2) K 2 .
vertices in L (G ) which are adjacent to
Hence from the above,
2( n 2) vertices of
partitioned into
L (G ) , hence
G can be
( n/2) K1,2( n 2) .
(n 2)Cn , ((n 2) 2 /4)C2 n , nK1,2( n 4) ,
Theorem 2.5 If G = K n , then the edges
(n/2) K1,2( n 2) , (n/2) K1, n 2 and (n/2) K 2 .
of G can be partitioned into
Case 2. When n is odd.
2
(a) (n 2)C n , ((n 2) /4)C2 n , nK1,2( n 4) ,
(n/2) K1,2( n 2) , (n/2) K1, n 2 and
(n/2) K 2 ,
when n is even. (b) (n 1)C n , (( n 1)( n 3)/4)C2 n , nK 1,2( n 3)
The edges of
K n can be
partitioned into (( n 1)/2)C n . Therefore by Proposition 1. and Theorem 2.2, edges of
G can be partitioned into ((n 1)/2)C n ,
((n 1)/2)((n 3)/2)C 2 n , (n(n 1)/2) K 2
and (n(n 1)) K 2 , when n is odd.
and L( K n ) . But by Lemma 1, L( K n ) can
Proof. Let G = K n . Then
be partitioned into (( n 1)/2 )C n , nK 1,2( n 3) .
G has n point vertices, each of degree (n( n 1)/2) and (n( n 1)/2) line vertices each of degree 3( n 2) . Case 1. When n is even.
Thus, G
can be partitioned into
(n 1)C n , (( n 1)( n 3)/4 )C 2 n , nK 1,2( n 3) and (n(n 1)) K 2 .
The edges of K n can be partitioned
Theorem 2.6 If G = K n with n > 3 , then
into (( n 2)/2)C n and (n/2) K 2 's. Therefore
the edges of G can be partitioned into
by Proposition 2.1, and by Theorem 2.2,
(a) ( n 1)/2 times (n 2) -regular graph
edges of G can be partitioned into
on 2 n vertices, with n point vertices are
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B. Basavanagoud et al., J. Comp. & Math. Sci. Vol. 1(5), 606-616 (2010)
common to all of these regular graphs,
2(n 2) -regular graph on
n( n 1)/2
of G , K n is also a subgraph of G . Case 2. When n is even.
vertices and K n , when is odd.
Edges of K n can be partitioned into
(b) ( n 2)/2 times (n 2) -regular graph
( n 2)/2 cycles of length n and (n/2)K 2 .
on 2 n vertices with n point vertices are common to all of these regular
In G each line vertex is adjacent to
graphs, 2(n 2) -regular
graph
on
n( n 1)/2 vertices, K n and ( n/2) K1,n 2 , when n is even. Proof. Let G = K n . Then G
( n 2) point vertices and also with 2( n 2) line vertices. Now, consider n line vertices which forms n-cycles in G. These n line vertices and n point vertices forms
has n
a ( n 2) -regular graph on 2n vertices.
point vertices each of degree n( n 1)/2
Hence, corresponding to ( n 2)/2 cycles
and n( n 1)/2 line vertices each of degree
of length n in G, there exists ( n 2)/2
3( n 2) .
times ( n 2) -regular graph on 2n vertices
Case 1. When n is odd.
such that n point vertices are in common to these regular graphs. Also each line
Edges of K n can be partitioned into
( n 1)/2 cycles of length n. In G , each line vertex is adjacent to ( n 2) point vertices and 2( n 2) line vertices. Now,,
vertex is adjacent with 2( n 2) line vertices. Corresponding to n( n 1)/2 line vertices, there exists
2( n 2) -regular
consider n line vertices which form a n-
graph. Now, consider (n/2)K 2 's in G. In
cycle in G and n point vertices in G . These 2n vertices form a ( n 2) -regular
G these n/2 line vertices are adjacent to ( n 2) point vertices. Therefore in G ,
graph on 2n vertices. Corresponding to
there are
( n 1)/2 exists
cycles of length n in G, there
( n 1)/2 times ( n 2) -regular
graph on 2n vertices such that n point vertices are common in all of these regular graphs. Also each line vertex is adjacent with 2( n 2) line vertices. Corresponding to these n( n 2)/2 line vertices there exists s
( 2(n 2)) -regular graph. And by definition
( n/2) K1,( n 2) and also by
definition of G , K n
is also induced
subgraph of G . Theorem 2.7 If G = K n , n , edges of G can be partitioned into (a) nC2 n , (n( n 1)/2)C4 n and
(n 2 (n 2) 4) K1, 4 if n is even.
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B. Basavanagoud et al., J. Comp. & Math. Sci. Vol. 1(5), 606-616 (2010)
(b)
2
2
(n 1)C2n , ((n 1) /2)C4n , n(n 1) 4 K1,4, nK1, 2 ( n 1) and nK2 if n is odd. Proof. Let G = K n , n . Then G has 2n point vertices, each of degree n2 and n2 line vertices each of degree (4n 4) .
615
(2(n 1)/2)C 2n , ((n 1)/2(2n 2)/2)C4 n and by considering two cycles of lenth 2n from G, each line vertex corresponding to one C2 n is adjacent to 4 line vertices of other C2 n in G . Hence 2nK1, 4 . But in
(n 1) 2 combinations of 2
Case 1. When n is even.
G, there are
The edges of K n , n can be partitioned into
two cycles of length
(n/2)C2n . Therefore by Theorem 2.2,
n(n 1)(n 3) 4 K1, 4 .
2n. Therefore
egdes of (C 2n ) can be partitioned into
Now, consider the edges of nK2
2C2 n , ((2n 2)/2)C4 n . Hence G can be
in G, in G n line vertices are adjacent to
partitioned into
4 line vertices of each C2 n therefore
2(n/2)C2 n , (n/2)((2n 2)/2)C 4n
n(n 1) / 2K1,4
and by considering two cycles of lenth 2n from G, each line vertex corresponding to
are adjacent with (2n 2) point vertices.
one C2 n is adjacent to 4 line vertices of other C2 n in G . Hence 2nK1, 4 . But in
n 2 combinations of two G, there are 2 cycles
of
length
2n.
Therefore
(n 2 (n 2) 4) K1, 4 . Hence G can be
and these n line vertices
Hence G can be partitioned into
(n 1)C 2n ,
((n 1) 2 /2)C4 n , n(n 1) 2 4 K1,4 , nK1, 2 ( n 1) and nK2 . REFERENCES 1. B. Basavanagoud, Keerti M. and Malghan S.H., On trnasformation
((n 1)/2)C2 n , nK 2 .Therefore by Theorem
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2.2, (C 2n ) can be partitioned into
G xyz when xyz = , (Proc) Graph
partitioned into nC2 n , ( n( n 1)/2)C 4 n and
(n 2 (n 2) 4) K1, 4 . Case 2. when n is odd. The edges of K n.n can be partitioned into
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