Double even star decomposition of complete bipartite graphs

Introduction

All basic terminologies from Graph Theory are used in this paper in the sense of Frank (Harary) [1] . Gnanadhas. N and Paulraj Joseph. J discussed on Continuous Monotonic Decomposition (CMD) of graphs in (Gnanadhas et al., 2000) [2] . Ebin Raja Merly. E and Gnanadhas. N introduced Arithmetic Odd ecomposition (AOD) in (Ebin Raja Merly et al., 2011) [3] 2. Double even decomposition Definition 2.1: An Even Decomposition is said to be Double Even Decomposition DED if the graph G can be decomposed into two copies of (G 2 , G 4 , …, G 2n ) for all n ∈ N of a graph G. Theorem 2.2: Any graph G admits Even Decomposition (G 2 , G 4 , G 6 , …, G 2n ), where G 2i = (V 2i , E 2i ) and |E(G 2i )| = 2i, (i = 1, 2,3, 4...., n), if and only if 𝑞 = 𝑛 ( 𝑛 + 1 ) for each n ∈ Z + . Remark 2.3: If any graph G admits DED (G 2 , G 4 , G 6 , …, G 2n ) where |E(G 2i )| = 2i, (i = 1, 2, 3, 4...., n), if and only if q = 2n(n + 1) for each n ∈ Z + .

Double even star decomposition (DESD) Definition 3.1 A Double Even Decomposition

A star is a complete bipartite graph of the form K 1,n and is denoted by S n . Clearly K m,n has mn edges. Theorem 3.4: A complete bipartite graph K 2 r + (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1] x -1) admits Double Even Star Decomposition (S 2 , S 4 , S 6 ,…, . S 2n ), if and only if n = 2 r + (Harary, 1969) [1] x -1, x, r ∈

N.

Proof: Assume K 2 r + (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1] x -1) admits DESD (S 2 , S 4 , S 6 ,…, . S 2n ) We know that q(K 2 r + (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1] x -1) ) = 2 r + (Harary, 1969) [1] . 2 x(2 r + (Harary, 1969) [1] x -1). Therefore 2 r + (Harary, 1969) [1] . 2 x (2 r + (Harary, 1969) [1] -1) = 2n(n+1), Implies n = 2 r + (Harary, 1969) [1] x -1 (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1]

x -1) admits DESD (S 2 , S 4 , S 6 ,…, . S 2n ) Applying induction on r. We proved the result is true when r = 1 Suppose the result is true when r = m. That is K 2 m + (Harary, 1969) [1] , 2x(2 m + (Harary, 1969) [1] x -1) admits DESD (𝑆2, 𝑆4, … , 𝑆2(𝑥2𝑚 + 1 -1)).

We prove that the result is true for r = m +1, that is to prove K 2 m + (Gnanadhas et al., 2000) [2] , 2x (2 m + (Gnanadhas et al., 2000) [2] x -1) admits DESD.

We have q(K2m + 2, 2x (2m + 2x -1)) = 2m + 2[2x(2m + 2x -1)] Is equal to 2x (Gnanadhas et al., 2000) [2] 2 2m+4 -2x2 m + (Gnanadhas et al., 2000) [2] Also, q(K2m + 1, 2 x (2m + 1x -1)) = 2m + 1[2x(2m + 1x -1)] = 2x2 22m + 2 -2x2m + 1 Therefore, q(K2m + 2, 2 x (2 m + 2x -1)) -q(K2m + 1, 2 x (2 m + 1x -1)) That is equal to 2x (Gnanadhas et al., 2000) [2] 2 2m + (Joseph Varghese et al., 2010) [4] -2x2 m + (Gnanadhas et al., 2000) [2] -2x (Gnanadhas et al., 2000) [2] 2 2m + (Gnanadhas et al., 2000) [2] + 2x2 m + (Harary, 1969) [1] = 3x (Gnanadhas et al., 2000) [2] . [2] 2 2m + (Ebin Raja Merly et al., 2011) [3] -x2 m + (Gnanadhas et al., 2000) [2] --------(Gnanadhas et al., 2000) [2] where n = 2 m + (Harary, 1969) [1] x -1 From (Harary, 1969) [1] and (Gnanadhas et al., 2000) [2] we have proved that [2] , 2 x (2 m + (Gnanadhas et al., 2000) [2] x -1) admits DESD (𝑆2, 𝑆4, …, 𝑆2(𝑥2𝑚 + 1 -1)).

Therefore the result is true for r = m+1. Hence, K 2 r + (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1] x -1) admits Double Even Star Decomposition (S 2 , S 4 , S 6 ,…, . S 2n ). Theorem 3.5: A complete bipartite graph K 2 r + (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1] x + 1) admits Double Even Star Decomposition (S 2 , S 4 , S 6 ,…, . S 2n ), if and only if n = 2 r + (Harary, 1969) [1] x, x, r ∈ N.

Proof: Assume K 2 r + (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1] x + 1) admits DESD (S 2 , S 4 , S 6 ,…, . S 2n ) We know that q(K 2 r + (Harary, 1969) [1] , 2 x(2 r + (Harary, 1969) [1] x + 1) ) = 2 r + (Harary, 1969) [1] . 2 x(2 r + (Harary, 1969) [1] x + 1). Therefore 2 r + (Harary, 1969) [1] . 2 x (2 r + (Harary, 1969) [1] x + 1) = 2n(n + 1), Implies n = 2 r + (Harary, 1969) [1] x Conversely, To prove K 2 r + (Harary, 1969) [1] , 2 x (2 r + (Harary, 1969) [1]

x + 1) admits DESD (S 2 , S 4 , S 6 ,…, . S 2n ) Applying induction on 'r'. We proved the result is true when r = 1 Suppose the result is true when r = m. That is K 2 m + (Harary, 1969) [1] , 2 x (2 m + (Harary, 1969) [1] x + 1) admits DESD (𝑆2, 𝑆4, … , 𝑆2(𝑥2𝑚 + 1)).

We prove that the result is true for r = m +1, that is to prove K 2 m + (Gnanadhas et al., 2000) [2] , 2x(2 m + (Gnanadhas et al., 2000) [2] x + 1) admits DESD.

We have q(K2m + 2, 2 x (2 m + 2x + 1)) = 2m + 2[2x(2m + 2x + 1)] Is equal to 2x (Gnanadhas et al., 2000) [2] 2 2m + (Joseph Varghese et al., 2010) [4] + 2x2 m + (Gnanadhas et al., 2000) [2] Also, q(K2m + 1, 2x(2m + 1x + 1)) = 2m + 1[2x(2m + 1x + 1)]. Is equal to 2x (Gnanadhas et al., 2000) [2] 2 2m + (Gnanadhas et al., 2000) [2] + 2x2 m + (Harary, 1969) [1] Therefore, q(K2m + 2, 2 x (2 m + 2x + 1)) -q(K2m + 1, 2 x (2 m + 1x + 1)) That is equal to 2x (Gnanadhas et al., 2000) [2] 2 2m + (Joseph Varghese et al., 2010) [4] + 2x2 m + (Gnanadhas et al., 2000) [2] -2x (Gnanadhas et al., 2000) [2] 2 2m + (Gnanadhas et al., 2000) [2] -2x2 m + (Harary, 1969) [1] = 4.2x (Gnanadhas et al., 2000) [2] 2 2m + (Gnanadhas et al., 2000) [2] -2x (Gnanadhas et al., 2000) [2] 2 2m + (Gnanadhas et al., 2000) [2] -2.2x.2 m + (Harary, 1969) [1] + 2x. 2 m + (Harary, 1969) [1] Implies 3x (Gnanadhas et al., 2000) [2] .2 2m + (Ebin Raja Merly et al., 2011) [3] + x.2 m + (Gnanadhas et al., 2000; Harary, 1969) [2,1] Now, 2[q(S 2n + 2 ) + q(S 2n + 4 ) + … +q(S 4n )] Equal to 2[2n+2 + 2n+4 + …+4n = 2(3n (Gnanadhas et al., 2000) [2] + n) = 3x (Gnanadhas et al., 2000) [2] 2 2m + (Ebin Raja Merly et al., 2011) [3] + x2 m + (Gnanadhas et al., 2000) [2] ------------(Gnanadhas et al., 2000) [2] where n = 2 m + (Harary, 1969) [1] x From (Harary, 1969) [1] and (Gnanadhas et al., 2000) [2] we have proved that q(K2m + 2, 2 x (2m + 2x + 1)) -q(K2m + 1, 2x(2m + 1x + 1)) = 2[q(S2n + 2) + q(S2n + 4) + … +q(S4n)]. Therefore, q(K2m + 2 ,2 x (2m + 2x + 1)) = q(K2m + 1, 2 x (2m + 1x + 1)) + 2[q(S2n + 2) + q(S2n+4) + … +q(S 4n )]. Therefore, K 2 m + (Gnanadhas et al., 2000) [2] , 2 x (2 m + (Gnanadhas et al., 2000) [2] x + 1) admits DESD (𝑆2, 𝑆4, … , 𝑆2(𝑥2𝑚 + 1)).

Therefore the result is true for r = m + 1. Hence, K 2 r + (Harary, 1969) [1] , 2 x(2 r + (Harary, 1969) [1] x -1) admits Double Even Star Decomposition (S 2 , S 4 , S 6 ,…, . S 2n ) Theorem 3.6: A complete bipartite graph 𝐾2,𝑥(𝑥+1)admits Double Even Star Decomposition (S 2 , S 4 ,S 6 , …, S 2n ), x ∈ N if and only if n = x.

Proof: Assume 𝐾2,𝑥(𝑥+1) admits DESD (S 2 , S 4 ,S 6 , …, S 2n )

We know that q(𝐾2,𝑥(𝑥+1)) = 2x(x + 1). Therefore 2x(x + 1) = 2n(n+1), Implies n = x. Conversely, To prove K 2, x(x + 1) admits DESD (S 2 , S 4 ,S 6 , …, S 2n ), x

Let T 1 denotes the star S x(x + 1) of K 2, x(x + 1) rooted at u 1 . Clearly q(T 1 ) = x(x + 1).

Then T 1 = ∪ 𝑆2𝑖1 , where 2i 1 = 2, 4, 6, . . ., 2n and Σ 2i 1 = x(x + 1).

Thus the decomposition of T 1 is (S 2 , S 4 , S 6 , …, S 2n ). Let T 2 denotes the star S x(x + 1) of K 2, x(x + 1) rooted at u 2, Clearly q(T 2 ) = x(x + 1), then T 2 = ∪ 𝑆2𝑖2 , where 2i 2 = 2, 4, 6, . . ., 2n and Σ 2i 2 = x(x + 1).

Thus the decomposition of T 2 is (S 2 , S 4 , S 6 , …, S 2n ). The decomposition of T 1∪ T 2 is DESD (S 2 , S 4 ,S 6 , …, S 2n ) , Which is the required Double Even Star Decomposition of K 2, x(x + 1) . Illustration 3.7. K 2, 12 admits DESD.

Conversely, to prove K 4, 8 x (Gnanadhas et al., 2000) [2] -2x admits DESD (S 2 , S 4 , ..., (Gnanadhas et al., 2000) [2] -2x } be the partition of V(K (Gnanadhas et al., 2000) [2] ).

4, 8 x -2x

Let T 1 denotes the star S (Gnanadhas et al., 2000) [2] of K (Gnanadhas et al., 2000) [2] rooted at u Clearly q(T)= 8x (Gnanadhas et al., 2000) [2] -2x, 8 x -2x 4, 8 x -2x 1, 1 then T 1 = ∪𝑆𝑅 2𝑖 1 , where, 2i 1 = 2,4,6, …, , ,..., , ,..., 2n and ∑ 2i 1 = 8x -2x 𝑛-7 𝑛-3 3𝑛+3 3𝑛+7 2 2 2 2 2 T 1 can be decomposed into (S 2 , S 4 , ..., 𝑆𝑛 -7 , 𝑆𝑛 -3 ,..., 𝑆3𝑛 +3 , 𝑆3𝑛 +7 , ..., S 2n ).

2 2 2 2 Let T 2 denotes the star S (Gnanadhas et al., 2000) [2] of K (Gnanadhas et al., 2000) [2] rooted at u Clearly q(T) = 8x (Gnanadhas et al., 2000) [2] -2x, 8 x -2x 4, 8 x -2x 2, 2 then T 2 = ∪𝑆𝑅 2𝑖 2 , Where 2i 2 = 2,4,6, …, 𝑛-7 , 𝑛-3,..., 3𝑛+3 , 3𝑛+7 ,..., 2n and ∑2i 2 = 8x (Gnanadhas et al., 2000) [2] -2x 2 2 2 2 T 2 can be decomposed into (S 2 , S 4 , ..., 𝑆𝑛 -7 , 𝑆𝑛 -3 ,..., ,𝑆3𝑛 +3 , 𝑆3𝑛 +7 , ..., S 2n ).

2

Let T 3 denotes the star S (Gnanadhas et al., 2000) [2] of K (Gnanadhas et al., 2000) [2] rooted at u , Clearly q(T) = 8x (Gnanadhas et al., 2000) [2] -2x, 8 x -2x 4, 8 x -2x 3 3 then T 3 = ∪𝑆𝑅 2𝑖 3 , where 2i 3 = 𝑛+1 , 𝑛+5

, …, 3𝑛-1 , and ∑2i 3 = 8x (Gnanadhas et al., 2000) [2] -2x. 2 2 2

Thus the decomposition of T 3 is , , ..., Let T 4 denotes the star S 8 x (Gnanadhas et al., 2000) [2] -2x of K 4, 8 x (Gnanadhas et al., 2000) [2] -2x rooted at u 4 , clearly q(T 4 ) = 8x (Gnanadhas et al., 2000) [2] -2x,

That is 4(8x (Gnanadhas et al., 2000) [2] + 2x) = 2n(n + 1) . Implies 2(4x + 1)(4x) = 2n(n + 1) Implies n = 4x Conversely, to prove 𝐾 4, 8𝑥 2 + 2𝑥 admits DESD (S 2 , S 4 , ..., S 2n ) Let V 1 = { u 1 , u 2 , u 3 , u 4 } V 2 = {v 1 , v 2 , v 3 , ..., v (Gnanadhas et al., 2000) [2] } be the partition of V(K (Gnanadhas et al., 2000) [2] ). 8 x +2x 4, 8 x +2x Let T 1 denotes the star S 2 8 x +2x of K 2 4, 8 x +2x rooted at u 1, Clearly q(T 1 ) = 8x (Gnanadhas et al., 2000) [2] + 2x, then T 1 = ∪𝑆2𝑖1

where, 2i 1 = 2,4,6, …, 𝑛 , 𝑛+4,..., 3𝑛+4 , 3𝑛+8 ,..., 2n and ∑ 2i 1 = 8x (Gnanadhas et al., 2000) [2] + 2x 2 2 2 2 T 1 can be decomposed into (S 2 , S 4 , ..., 𝑆𝑛 , 𝑆𝑛 +4 ,..., 𝑆3𝑛 +4 , 𝑆3𝑛 Clearly q(T 2 ) = 8x (Gnanadhas et al., 2000) [2] +2x, then T 2 = ∪𝑆2𝑖2

Where, 2i 2 = 2,4,6, …, 𝑛 , 𝑛+4,..., 3𝑛+4 , 3𝑛+8 ,..., 2n and ∑2i 2 = 8x (Gnanadhas et al., 2000) [2] + 2x . 2 2 2 2 T 2 can be decomposed into (S 2 , S 4 , ..., 𝑆𝑛 , 𝑆𝑛 +4 ,..., 𝑆3𝑛 +4 , 𝑆3𝑛 +8 , ..., S 2n ). Let T 3 denotes the star S

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7. How to cite this article: Suthiesh Goldy J. Double even star decomposition of complete bipartite graphs. International Journal of Statistics and Applied Mathematics. 2026;11(10):15-19. doi: h ttps://www.doi.org/10.22271/maths.2026.v11.i10a.2711