Construction of Four Completely Independent Spanning Trees on

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES

arXiv:1705.01358v1 [math.CO] 3 May 2017

S. A. Mane, S. A. Kandekar, B. N. Waphare Center for Advanced Studies in Mathematics, Department of Mathematics, Savitribai Phule Pune University, Pune-411007, India.

[email protected]; [email protected]; [email protected] Abstract. Let T1 , T2 , ......., Tk be spanning trees in a graph G. If for any pair of vertices {u, v} of G, the paths between u and v in every Ti ( 1 ≤ i ≤ k) do not contain common edges and common vertices, except the vertices u and v, then T1 , T2 , ......., Tk are called completely independent spanning trees in G. The n−dimensional augmented cube, denoted as AQn , a variation of the hypercube possesses several embeddable properties that the hypercube and its variations do not possess. For AQn (n ≥ 6), construction of 4 completely independent spanning trees of which two trees with diameters 2n − 5 and two trees with diameters 2n − 3 are given.

Keywords: Completely independent spanning trees, Augmented cubes 1. Introduction Interconnection networks have been widely studied recently. The architecture of an interconnection networks is usually denoted as an undirected graph G. A graph G is a triple consisting of a vertex set V (G), an edge set E(G), and a relation that associates with each edge two vertices called its endpoints[18]. Many useful topologies have been proposed to balance performance and cost parameters. Among them, the hypercube Qn is one of the most popular topology and has been studied for parallel networks. Augmented cubes are derivatives of hypercubes (proposed by Choudam and Sunitha[4]) with good geometric features that retain some favorable properties of the hypercubes (since Qn ⊂ AQn ), such as vertex symmetry, maximum connectivity, routing and broadcasting procedures with linear time complexity. An n−dimensional augmented cube AQn can be formed as an extension of Qn by adding some links. For any positive integer n, AQn is (2n − 1)-regular and (2n − 1)-connected (except n = 3) graph with 2n vertices. Moreover, AQn possesses several embeddable properties that the hypercube and its variations do not possess. The main merit of augmented cubes is that their diameters are about half of those of the corresponding hypercubes. A tree T is called a spanning tree of a graph G if V (T ) = V (G). Two spanning trees T1 and T2 in G are edge-disjoint if E(T1 ) ∩ E(T2 ) = φ. For a given tree T and a given pair of vertices u and v of T , let PT (u, v) be the set of vertices in the unique path between u and v in T . Two spanning trees T1 and T2 are internally vertex disjoint if for any pair of vertices u and v of V (G), PT1 (u, v) ∩ PT2 (u, v) = {u, v}. Finally, the spanning trees T1 , T2 , ......., Tk of G are completely independent spanning trees (CISTs for short) if they are pairwise edge-disjoint and internally vertex disjoint. The study of CISTs was due to the early work of Hasunuma[7], he conjectured that there are k CISTs in any 2k−connected graph. P´ eterfalvi[16] gave counter example to disprove 2010 Mathematics Subject Classification. Primary 16W10; Secondary 06A06; 47L30. 1

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 2

Hasunuma’s conjecture. He showed that there exists a k−connected graph which does not contain two CISTs for each k ≥ 2. Pai et al[14] showed that the results are negative to Hasunuma’s conjecture in case of hypercube of dimension n ∈ {10, 12, 14, 20, 22, 24, 26, 28, 30}. Many authors provided a necessary condition of CISTs[1, 2, 5, 9, 12]. For more detail work on CISTs and their diameters see [3, 6, 7, 8, 9, 10, 11, 13, 17, 19]. Constructing CISTs has many applications on interconnection networks such as faulttolerant broadcasting and secure message distribution.In underlying graph of communication networks, we want vertices to be close together to avoid communication delays. Pai and Chang[15] provided a unified approach for constructing two CISTs in several hypercube-variant networks, in particular for an n−dimensional hypercube variant network, the diameters of the constructed CISTs were 2n − 1. They asked about the hypercube variant networks which they studied ”how to design algorithms to construct more than two CISTs in high dimensional hypercube-variant networks with smaller diameter?” Motivated by this question, we provide a construction of four CISTs in augmented cube AQn (n ≥ 6) of which two trees with diameters 2n − 3 and two trees with diameters 2n − 5 are constructed. Also, construction of n − 1 CISTs in augmented cube AQn for n = 3, 4, 5 is given which pointed out that the Hasunuma’s conjecture does hold in the case of AQn for n = 3, 4, 5. For undefined terminology and notation, see[18]

2. Preliminaries The definition of the n−dimensional augmented cube is stated as the following. Let n ≥ 1 be an integer. The n-dimensional augmented cube, denoted by AQn , is a graph with 2n vertices, and each vertex u can be distinctly labeled by an n-bit binary string, u = u1 u2 ....un . AQ1 is the graph K2 with vertex set {0, 1}. For n ≥ 2, AQn can be recursively constructed by two copies of AQn−1 , denoted by AQ0n−1 and AQ1n−1 , and by adding 2n edges between AQ0n−1 and AQ1n−1 as follows: Let V (AQ0n−1 ) = {0u2 ....un : ui ∈ {0, 1}, 2 ≤ i ≤ n} and V (AQ1n−1 ) = {1v2 ....vn : vi ∈ {0, 1}, 2 ≤ i ≤ n}. A vertex u = 0u2 ....un of AQ0n−1 is joined to a vertex v = 1v2 ....vn of AQ1n−1 if and only if for every i, 2 ≤ i ≤ n either 1. ui = vi ; in this case an edge hu, vi is called a hypercube edge and we say v = uh , or 2. ui = vi ; in this case an edge hu, vi is called a complement edge and we say v = uc . Let Enh = {hu, uh i : u ∈ V (AQ0n−1 )} and Enc = {hu, uc i : u ∈ V (AQ0n−1 )}. See Fig.1.

0

s

00 s

AQ01 s

1 AQ1

10

000

s

s

AQ11

s

01 AQ2

001 s

101

100

s

s

AQ12

AQ02

s

s

s

11

010

011

s

111

AQ3 Fig. 1. The augmented cubes of dimension 1, 2 and 3

s

110

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 3

The following lemma is helpful to visualize edges of AQn .

Lemma 2.1 ([4]). Let G be the simple graph with vertex set V (G) = {a1 a2 ....an : ai = 0 or 1} where two vertices A = a1 a2 ....an and B = b1 b2 ...bn are joined iff there exists an integer k : 1 ≤ k ≤ n such that either (1) ak = bk and ai = bi , for every i, i 6= k, or (2) ai = bi , for 1 ≤ i ≤ k − 1 and ai = bi , for k ≤ i ≤ n.

3. Construction of CISTs in augmented cubes We need the following result given in [7] by Hasunuma.

Lemma 3.1 ([7] ). Let k ≥ 2 be an integer, T1 , T2 , ......., Tk are completely independent spanning trees in a graph G if and only if they are edge-disjoint spanning trees of G and for any v ∈ V (G), there is at most one Ti such that dTi (v) > 1. Pai and Chang[15] constructed two CISTs in several hypercube-variant networks, they proved the following result. Lemma 3.2 ([15] ). Let Gn be the n−dimensional variant hypercube for n ≥ 4 and suppose that T1 and T2 are two CISTs of Gn . For i ∈ {1, 2}, let T i be a spanning tree of Gn+1 constructed from Ti0 and Ti1 by adding an edge hui , vi i ∈ E(Gn+1 ) to connect two internal vertices ui ∈ V (Ti0 ) and vi ∈ V (Ti1 ). Then, T 1 and T 2 are two CISTs of Gn+1 . By using the same proof technique of above theorem one can state following corollary. Corollary 3.3 ( ). Let Gn be the n−dimensional variant hypercube for n ≥ 4 and suppose that Ti , for 1 ≤ i ≤ k (k < n) be k CISTs of Gn . Let T i be a spanning tree of Gn+1 constructed from Ti0 and Ti1 by adding an edge hui , vi i ∈ E(Gn+1 ) to connect two internal vertices ui ∈ V (Ti0 ) and vi ∈ V (Ti1 ). Then, T i are k CISTs of Gn+1 Firstly, we list out n − 1 CISTs in AQn for n = 3, 4. See Fig.2 and Fig.3. 000(1) ✈

001(2) ✈

111(8)

✈

✘✈ ✘✘ ✘✘✘ ③

010(3)

010(3) ③

011(4)

000(1) 101(6)

✈ 011(4)

001(2) ✈

✈

101(6) ✈

✈100(5)

✈

100(5)

③

✈ 110(7)

111(8)

Fig.2. Two CISTs in AQ3

③

✈ 110(7)

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 4

0011(4) 0000(1) s

1111(16) s

1001(10) s

1101(14) s

s0110(7)

s

0010(3)

①

0101(6) 1100(13) s

0111(8) s ①

0100(5)

①1101(14)

1100(13) s

0100(5) s

①

0011(4)

1000(9) s

PP PP①

0101(6) s

1001(10) s

1101(14) s

1100(13) 1111(16) s ①

0111(8)

1011(12)s

1011(12) s

1000(9) s

①

1010(11)

0001(2)

1110(15) s

0010(3)

0001(2)

① s

s

0101(6) s

①

1011(12)

1010(11)

①

1001(10)

0010(3) s

s

s

①

①

1110(15) s

0000(1)

s

s

①1000(9)

0011(4) s

1010(11)

0110(7)

1110(15) 0100(5) s

①

0000(1) s

1111(16) Fig.3. Three CISTs in AQ4

0001(2)

s

0111(8) s ①

0110(7)

Note : To make things readable, we denote vertices of AQ5 by using numbers 1, 2, ....., 32. For example the vertex 00000 will be denoted by 1 and sometime will be written as 00000(1). Also, we will use short forms to denote edges of AQ5 , for example the edge h00000(1), 00001(2)i will be denoted by h1, 2i. For every i, InV (Ti ) denotes set of internal vertices of the tree Ti , for 1 ≤ i ≤ 4. Vertex v ∈ V (Ti ) is called internal if dTi (v) ≥ 2. Let V (AQ5 ) = {00000(1), 00001(2), 00010(3), 00011(4), 00100(5), 00101(6), 00110(7), 00111(8), 01000(9), 01001(10), 01010(11), 01011(12), 01100(13), 01101(14), 01110(15), 01111(16), 10000(17), 10001(18), 10010(19), 10011(20), 10100(21), 10101(22), 10110(23), 10111(24), 11000(25), 11001(26), 11010(27), 11011(28), 11100(29), 11101(30), 11110(31), 11111(32)}. Now for n = 5, we construct four trees T1 , T2 , T3 and T4 as shown in Fig.4(a), Fig.4(b), Fig.4(c) and Fig.4(d) respectively.

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 5

s01001(10)

00110(7)✉

s

00101(6)

s00111(8)

01011(12) s

s 11011(28) ✉

00011(4) s

00100(5)

00001(2) s 00010(3) s

✉

00000(1)

10001(18) s

s10100(21) s 01110(15)

s01101(14) ✉ ❍ ❍❍ 01111(16) ❍ s 01100(13)

✉

s10111(24)

10000(17)

10011(20)s

11000(25) ✉ 01010(11) s

s10010(19) ✉

11001(26) s 11010(27) 10110(23) s 11100(29) s

s 01000(9)

s10101(22)

✏ s11101(30) ✏✏ ✏ ✏ ✉ s11111(32)

11110(31)

Fig. 4(a). Spanning Tree T1 in AQ5

s 00100(5)

10111(24) s 11000(25) s

✈

00111(8)

00000(1) s

✈

11100(29) s

✈

00001(2)

01010(11) s ✈

01000(9) s 11011(28) s 10011(20) s 10000(17) s

s 00101(6) s01110(15)

01001(10) s 00110(7)

10100(21) ✈ ✈ 10110(23) 11001(26)

✈

01101(14) s s

s 01011(12) s 01100(13)

00011(4)

00010(3) s

s01111(16)

s10001(18) s10101(22) s11110(31) s 10010(19)

✈

s 11111(32) 11101(30) Fig. 4(b). Spanning Tree T2 in AQ5

11010(27)

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 6

01010(11)r

r 00000(1)

01011(12)r ✉

01111(16) r

01000(9)

00001(2) r

r 00111(8)

r 00100(5)

00110(7) r

01110(15)

10001(18) r

✉

11110(31)r 11001(26) r

✉

✉

01101(14)

11100(29)

r 11000(25)

✉

r 11101(30)

11111(32) r

00011(4)

r

r 01001(10)

r10100(21)

11011(28)

11010(27) r

r 00101(6)

✉

01100(13)

r 00010(3)

✉

r

10111(24)

10011(20)

10110(23) r

✉ r10101(22) 10010(19) 10000(17) r Fig. 4(c). Spanning Tree T3 in AQ5

t 01111(16)

01001(10) t 01100(13) t 00011(4) t 00110(7) t 00001(2)

01011(12)

t

01110(15)

t00100(5) ✈

00101(6)

01010(11)

t

✈

10011(20) t

✈

t11101(30) ✈

10001(18) 10101(22)

t

t 11011(28)

01101(14) t 00010(3) ✈

10010(19) t

11001(26)

✈

t10100(21) t10110(23)

11110(31) t 01000(9)

t

✈

10111(24)

t11000(25) t 10000(17)

00000(1) t

✈

t

11100(29) 11111(32) Fig. 4(d). Spanning Tree T4 in AQ5

t 00111(8) t11010(27)

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 7

Here, we observe that InV (T1 ) = {1, 5, 7, 16, 17, 25, 27, 31}, InV (T2 ) = {2, 4, 8, 10, 21, 23, 26, 30} InV (T3 ) = {9, 13, 14, 15, 19, 20, 28, 29} InV (T4 ) = {3, 6, 11, 12, 18, 22, 24, 32} are such that InV (Ti ) ∩ InV (Tj ) = φ, for i 6= j and 1 ≤ i, j ≤ 4 Also, observe E(T1 ) = {h1, 2i, h1, 3i, h1, 5i, h1, 16i, h1, 17i, h4, 5i, h5, 7i, h5, 12i, h5, 21i, h5, 28i, h6, 7i, h7, 8i, h7, 10i, h9, 25i, h11, 27i, h13, 16i, h14, 16i, h15, 16i, h17.18i, h17, 20i, h17, 24i, h17, 25i, h19, 27i, h22, 27i, h23, 31i, h25, 27i, h26, 27i, h27, 31i, h29, 31i, h30, 31i, h31, 32i} E(T2 ) = {h1, 4i, h2, 3i, h2, 4i, h2, 6i, h2, 10i, h4, 8i, h4, 12i, h4, 13i, h4, 29i, h5, 8i, h7, 23i, h8, 16i, h8, 24i, h8, 25i, h9, 10i, h10, 11i, h10, 15i, h10, 23i, h14, 30i, h17, 21i, h18, 23i, h19, 30i, h20, 21i, h21, 23i, h21, 28i, h22, 23i, h23, 26i, h26, 30i, h26, 31i, h27, 30i, h30, 32i} E(T3 ) = {h1, 9i, h2, 15i, h3, 14i, h4, 20i, h5, 13i, h6, 14i, h7, 15i, h8, 9i, h9, 11i, h9, 12i, h9, 13i, h9, 16i, h10, 14i, h13, 14i, h13, 15i, h13, 29i, h15, 18i, h15, 31i, h17, 19i, h19, 20i, h19, 22i, h19, 23i, h20, 24i, h20, 29i, h21, 29i, h25, 29i, h26, 28i, h27, 28i, h28, 29i, h28, 32i, h29, 30i} E(T4 ) = {h1, 32i, h2, 18i, h3, 4i, h3, 7i, h3, 11i, h5, 6i, h6, 8i, h6, 11i, h6, 27i, h9, 24i, h10, 12i, h11, 12i, h11, 14i, h11, 15i, h11, 22i, h12, 13i, h12, 16i, h12, 28i, h17, 32i, h18, 19i, h18, 20i, h18, 22i, h18, 26, i, h18, 31i, h21, 22i, h22, 24i, h22, 30i, h23, 24i, h24, 25i, h24, 32i, h29, 32i} are such that E(Ti ) ∩ E(Tj ) = φ, for i 6= j and 1 ≤ i, j ≤ 4. According to Lemma 3.1, above constructed trees T1 , T2 , T3 and T4 are CISTs on AQ5 . Theorem 3.4. Let n ≥ 6 be an integer. There exist four completely independent spanning trees of which two are with diameter 2n − 3 and two are with diameter 2n − 5, in augmented cube AQn . Proof. By using above constructed four CISTs in AQ5 and Corollary 3.3, we get four CISTs in AQn for n ≥ 6. As we want vertices to be close together to avoid communication delays. So, we will concentrate on diameters of above trees. Above constructed trees T1 , T2 have diameters 8 and T3 , T4 have diameter 6 in AQ5 . We first consider the construction of four CISTs in AQ6 . It is sufficient to show the construction of only one tree. Consider tree T1 of AQ5 , having longest path of length 8 and central vertex 10000(17). Now, by prefixing 0 and 1 to this vertex we get central vertex of T10 and T11 respectively. Means, we select u1 = 010000 ∈ V (T10 ) and v1 = 110000 ∈ V (T11 ). Then, T 1 is CIST with diameter 9 on AQ6 . Constructing in the similar manner we get T 2 , T 3 , T 4 CISTs with diameters 9, 7, 7 respectively on AQ6 . Let AQn+1 (n ≥ 6) be decomposed into two augmented cubes say AQ0n and AQ1n with vertex set say {x0i : 1 ≤ i ≤ 2n } and {x1i : 1 ≤ i ≤ 2n } respectively. Denote by T10 , T20 , T30 , T40 the CISTs with diameter 2n − 3, 2n − 3, 2n − 5, 2n − 5 respectively in AQ0n . Let the identical corresponding CISTs in AQ1n be denoted by T11 , T21 , T31 , T41 . Now, we will prove by induction that the diameters of T 1 , T 2 , T 3 , T 4 ) CISTs in AQn+1 (n ≥ 6) are 2n − 1, 2n − 1, 2n − 3, 2n − 3 respectively. It is sufficient to prove result for a single tree say T10 . Let P10 = x01 − x02 − .......x0n−1 − x0n − .....x02n−2 be the longest path naturally of length 2n − 3

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 8

in tree T10 . And P11 = x11 − x12 − .......x1n−1 − x1n − .....x12n−2 be its corresponding path in corresponding tree T11 . The vertex x0n is in the center of the path P10 hence any vertex on tree T10 will be within a distance n − 1 from the vertex x0n , means any vertex x0i ∈ V (T10 ), dT10 (x0i , x0n ) ≤ n − 1. Similarly, any vertex x1i ∈ V (T11 ), dT11 (x1i , x1n ) ≤ n − 1. Now, consider tree say T 1 be a spanning tree of AQn+1 constructed from T10 and T11 by adding an edge hx0n , x1n i ∈ E(AQn+1 ) to connect two internal vertices x0n ∈ V (T10 ) and x1n ∈ V (T11 ). Then, T 1 is with diameter 2n − 1. As V (T 1 ) = V (T10 ) ∪ V (T11 ), dT 1 (x0i , x1j ) = dT10 (x0i , x0n ) + 1 + dT11 (x1n , x1j ) ≤ (n − 1) + 1 + (n − 1) = 2n − 1. Constructing in the similar manner we get T 2 , T 3 , T 4 CISTs with diameters 2n − 1, 2n − 3, 2n − 3 respectively on AQn+1 .  Concluding remarks. In this paper, we have proposed a construction of four CISTs in augmented cube AQn (n ≥ 6) of which two trees with diameters 2n − 3 and two trees with diameters 2n − 5. Our results provide n − 1 CISTs in augmented cube AQn (n = 3, 4, 5) and thus we pointed out that Hasunuma’s conjecture does hold for AQn when n = 3, 4, 5. As connectivity of augmented cubes is comparatively higher than other variants of hypercubes, an interesting problem is whether Hasunuma’s conjecture is true for AQn (n ≥ 6) if so then how to derive an algorithm that construct n − 1 CISTs in AQn (n ≥ 6)? Acknowledgment: The first author gratefully acknowledges the Department of Science and Technology, New Delhi, India for the award of Women Scientist Scheme for research in Basic/Applied Sciences. References [1] T. Araki, Diracs condition for completely independent spanning trees, J. Graph Theory 77(2014)171 − 179. [2] H.-Y. Chang, H.-L. Wang, J.-S. Yang, J.-M. Chang, Anote on the degree condition of completely independent spanning trees, IEICE Trans. Fundam. Electron. Commun. Comput. Sci. 98 − A(2015)2191 − 2193. [3] B. Cheng, J. Fan, D. Wang, J. Yang, Areliable broadcasting algorithm in locally twisted cubes, in: Proc. 2nd Int. Conf. on Cyber Security and Cloud Computing, 2015, pp.323 − 328. [4] S.A. Choudum, V. Sunitha, Augmented cubes, Networks 40(2)(2002)71 − 84. [5] G. Fan, Y. Hong, Q. Liu, Ores condition for completely independent spanning trees, Discrete Appl. Math. 177(2014)95 − 100. [6] Z. Ge, S.L. Hakimi, Disjoint rooted spanning trees with small depths in de Bruijn and Kautz graphs, SIAM J. Comput. 26(1997)79 − 92. [7] T. Hasunuma, Completely independent spanning trees in the underlying graph of a line digraph, Discrete Math. 234(2001)149 − 157. [8] T. Hasunuma, Completely independent spanning trees in maximal planar graphs, in: Proc. 28th Int. Workshop on Graph-Theoretic Concepts in Computer Science, WG 2002, in: Lecture Notes in Comput. Sci., vol.2573, 2002, pp.235 − 245. [9] T. Hasunuma, Minimum degree conditions and optimal graphs for completely independent spanning trees, in: Proc. 26th Int. Workshop on Combina-torial Algorithms, IWOCA 2016, in: Lecture Notes in Comput. Sci., vol.9538, 2016, pp.260 − 273. [10] T. Hasunuma, C. Morisaka, Completely independent spanning trees in torus networks, Networks 60(2012)59 − 69. [11] T. Hasunuma, H. Nagamochi, Independent spanning trees with small depths in iterated line digraphs, Discrete Appl. Math. 110(2001)189 − 211.

CONSTRUCTION OF FOUR COMPLETELY INDEPENDENT SPANNING TREES ON AUGMENTED CUBES 9

[12] X. Hong, Q. Liu, Degree condition for completely independent spanning trees, Inform. Process. Lett. (2016)644 − 648. [13] M. Matsushita, Y. Otachi, T. Araki, Completely independent spanning trees in (partial) k-trees, Discuss. Math. Graph Theory 35(2015)427 − 437. [14] K.-J. Pai, J.-S. Yang, S.-C. Yao, S.-M. Tang, J.-M. Chang, Completely independent spanning trees on some interconnection networks, IEICE Trans. Inf. Syst. 97−D (2014)2514 − 2517. [15] K.-J. Pai, J.-M. Chang, Constructing two completely independent spanning trees in hypercube-variant networks, Theoret. Comput. Sci. 652(2016)28 − 37. [16] F. P´ eterfalvi, Two counterexamples on completely independent spanning trees, Discrete Math. 312(2012)808 − 810. [17] Y. Wang, J. Fan, X. Jia, H. Huang, An algorithm to construct independent spanning trees on parity cubes, Theoret. Comput. Sci. 465(2012)61 − 72. [18] Douglas. B. West, Introduction to Graph theory, Second Edition, Prentice-Hall of India, New Delhi, 2002. [19] J.-S. Yang, J.-M. Chang, S.-M. Tang, Y.-L. Wang, Reducing the height of independent spanning trees in chordal rings, IEEE Trans. Parallel Distrib. Syst. 18(2007), 644 − 657.

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