2017 2nd International Conference on Communications, Information Management and Network Security (CIMNS 2017) ISBN: 978-1-60595-498-1
One-to-Many Node Disjoint Path Covers on WK-Recursive Networks
Lan-tao YOU
1, Yue-juan HAN
2,*, Xi WANG
3,2, Chen ZHOU
1and Rui GU
11Suzhou Industrial Park Institute Of Services Outsourcing, Suzhou, China
2School of Computer Science and Technology, Soochow University, Suzhou, China
3Suzhou Institute of Industrial Technology, Suzhou, China
*Corresponding author
Keywords: WK-recursive network, One-to-many, Disjoint path cover.
Abstract. The WK-recursive network denoted by K(d, t) has received much attention due to its many favorable properties. We use Oito denote the open vertex set of Ki(d, t−1) with 1 ≤ i ≤ d and let OI= {Oi|1 ≤ i ≤ d}. We prove that given any node μ and a set of distinct destination nodes T = {tj|1 ≤ j ≤ d−1}where tj∉ OI, and μ, tjare not in the same subgraph, there exist d−1 node-disjoint paths between μ and T whose union covers all the vertices of K(d, t) where d≥4 and t≥1.
Introduction
The WK-recursive network, which was proposed in [1], has two structural advantages: scalability and constant degree. We use K(d, t) to denote a WK-recursive network of level t, each of whose basic modules is a d-vertex complete graph, where d > 1 and t ≥ 1. Recently, much research has been devoted to the WK-recursive networks. For example, in [2], Fu proved that K(d, t) is Hamiltonian-connected where d ≥ 4. In [3], Fang et al. showed that the WK-recursive network is strictly 3 × 2t-1-edge-pancyclic for k ≥ 7 and t ≥ 1.
The one-to-many node-disjoint path problem is a fundamental problem in distributed computing systems. It’s described as follows: given a source node s and a set of distinct destination nodes T = {t1,
t2, …, tm}where s ∉ T and m ≥ 1, the m node disjoint path problem is to determine whether there exist m node disjoint paths between s and T, where each path connects the source node s with one of the destination node tj ∈ T(1 ≤ j ≤ m) [4]. Some examples of the one-to-many disjoint paths can be found in [5, 6, 7].
It is important for an interconnection network to efficiently route data among nodes. The node disjoint paths can be used to increase the transmission rate and enhance the transmission reliability [6, 7]. The one-to-many node disjoint path problem in WK-recursive networks has been considered by researchers. In [8], Bakhshi et al. developed an algorithm that presents d−1 node disjoint paths from an arbitrary node in K(d, t) to d−1 distinct target nodes. However, the union of all the disjoint paths does not cover all the vertices of K(d, t). In this paper, we prove the one-to-many node-disjoint path cover problem under certain restrictions. We prove that given any node μ and a set of distinct destination nodes T = {tj|1 ≤ j ≤ d−1}where tj∉ OI, and μ, tjare not in the same subgraph, there exist d−1 node-disjoint paths between μ and T whose union covers all the vertices of K(d, t) where d≥4 and
t≥1.
The rest of this paper is organized as follows. In Section 2, we formally define the WK-recursive network K(d, t) and present some basic properties of it. In Section 3, we give the proofs of our main results. In the end, we summarize the paper in Section 4.
Preliminaries
where d > 1 and t ≥ 1. A K(d, t) consists of d copies of K(d, t − 1) and each of them can be considered as a super- node. We define K(d, t) in terms of a graph as follows. Throughout this paper, we use network and graph, node and vertex, link and edge interchangeably.
Defnition 1. Each vertex of K(d, t) is labeled with a unique t-digit radix d number. Vertex
at-1at-2…a1a0 is adjacent to (1)at-1at-2…a1b, where b ≠ a0 and (2) at-1at-2…aj+1aj-1(aj)j-1 if aj≠ aj-1 and aj-1
= aj-2 = … = a0, where (aj)j-1 represents j − 1 consecutive aj. In addition, for any integer 0 ≤ j ≤ t − 1, (aj)tis called an open vertex, each open vertex is connected with an open edge which is reserved for further expansion.
O = {(aj)t|0 ≤ j ≤ t − 1}is the open vertex set of K(d, t). In this paper, we define o(i,0) or o(i, i) to be the
ith open vertex of K(d, t) and o(i, j) to be the open vertex of subgraph K(d, t − 1) where o(i, j)∈ Ki(d, t − 1) is connected to the vertex o(j, i) ∈ Kj(d, t − 1) with 1 ≤ i ≠ j ≤ d. We use Oito denote the open vertex set of Ki(d, t − 1) with 1 ≤ i ≤ d and use OI= {Oi|1 ≤ i ≤ d}to denote the open vertex set of all the subgraphs. The structures of K(5, 1) and K(5, 2) are shown in Figure 1.
[image:2.595.137.462.262.442.2](a) (b)
Figure 1. The graphs (a) K(5, 1) and (b) K(5, 2).
A path ⟨v1, v2, …, vk⟩is an ordered list of distinct vertices such that viand vi+1 are adjacent for 1 ≤ i
≤ k − 1. A path from node x to node y is abbreviated as an x − y path. An x − x path, which has length 0, degenerates to a node x. We define a path ⟨x, x⟩to be a virtual path in this paper. Let P be a node set, the Algorithm First(P) is to retrieve the first element of P and it will also remove the first element from P. Let S and T be two path sets, the operation S + T is a set can be obtained as followings:union
S and T to get a new set, and in this new set, if path s and t (s∈S and t∈T) have one same end node, then s and t are delete from this set, and s+t (s will be connected to t as a new path s+t) are added into this set.
One-to-Many Node Disjoint Path Covers of K(d, t)
In this section, we study our main results Theorem 9. In order to prove the results, we first present Lemmas 1–7.
Lemma 1. (See [2].) The WK-recursive network is Hamiltonian-connected where d ≥ 4 and t ≥ 1,
and the algorithm for the lemma is HamPath(K(d, t), μ, ν).
Lemma 2. Given K(d, t) and its d subgraphs Ki(d, t − 1), where t ≥ 2, d ≥ 4 and 1 ≤ i ≤ d, let G = Kw1(d, t−1)∪Kw2(d, t−1)∪…∪Kwm(d, t−1) with 2 ≤ m ≤ d−1,1 ≤ j ≤ m and 1 ≤ wj ≤ d. For any two distinct vertices μ and ν in different subgraphs of G, there is a Hamiltonian path between μ and ν in G.
Lemma 3. (See [2].) Let A = at-1at-2…a0and B = bt-1bt-2…b0be two distinct nodes in K(d, t) where
d ≥ 4. Given two distinct integers c, e ∈{1,2,…, d} such that {c, e} ≠ {at-1, bt-1}, then there exist two
disjoint paths PA,X and PB,Y such that V (PA,X) ∪ V (PB,Y) = V (K(d, t)) where {X, Y }= {(c)t,(e)t}. Lemma 4. Let O = {o(i,0)|1 ≤ i ≤ d} be the open vertex set of K(d,1) where d ≥ 4, Sr= {si|1 ≤ i ≤ r}
⊂ V (K(d,1)) with 2 ≤ r ≤ d − 1 be the set of source nodes and Tr= {ti|1 ≤ i ≤ r}⊂ O be the set of destination nodes where Sr≠ Tr. Then there exist r node disjoint paths between S and T whose union covers all the vertices of K(d,1).
Proof. Since Sr ≠ Tr, we can find two nodes x ∈ Sr and y ∈ Trwhere x ≠ y. Since K(d,1) is a complete graph, we can construct r − 1 node disjoint paths between Sr − {x}and Tr − {y}. Again, we can construct one path between x and y which covers all the vertices of V (K(d,1))−Sr−Tr+{x, y}. Hence there exist r node disjoint paths between Srand Trin K(d,1) where 2 ≤ r ≤ d − 1. □
Lemma 5. Let O = {o(i,0)|1 ≤ i ≤ d} be the open vertex set of K(d, t) where d ≥ 4 and t ≥ 2, I be the
internal vertex set of K(d, t), Sr= {si|1 ≤ i ≤ r}⊂I with 2 ≤ r ≤ d−1 be the set of source nodes and Tr = {ti|1 ≤ i ≤ r} ⊂O be the set of destination nodes. Then there exist r node disjoint paths between Sr and Tr whose union covers all the vertices of K(d, t).
Proof. The lemma can be proved by Lemma 1, Lemma 2, Lemma 3 and Lemma 4, however, due to the page limitation, we omit the proof. The algorithm ManyToMany(K(d, t), S, T) for this lemma is
also omitted here. □
Lemma 6. Let μ∈ V (K(d,1)) and T = {o(wi,0)|1 ≤ i ≤ d − 1 and 1 ≤ wi ≤ d}⊂ V (K(d,1)) where d ≥ 4. Then there exist d−1 node disjoint paths between μ and T whose union covers all the vertices of K(d,1).
Proof. Let x ∈ T where x ≠ μ. Since K(d,1) is a complete graph, there exist r − 1 node disjoint paths between μ and T −{x}. Again we can construct one path between μ and x which covers all the vertices of V (K(d,1))−T +{x}. Hence there exist r node disjoint paths between μ and T whose union covers all the vertices of K(d,1). □ Lemma 7. Let μ ∈ V (K(d, t)) and T = {o(wi,0)|1 ≤ i ≤ d − 1 and 1 ≤ wi ≤ d}⊂ V (K(d, t)) where d ≥ 4. Then there exist d−1 node disjoint paths between μ and T whose union covers all the vertices of K(d, t).
Proof. We can use the mathematical induction on t to prove this lemma. Due to the page limitation, we omit the proof. We propose an algorithm OneToMany(K(d, t), μ, T) to find d − 1 node disjoint paths between μ and T. □ Algorithm 1 OneToMany(K(d, t), μ, T)
Input:
A node μ and a set of d − 1 nodes T = {o(wi,0)|1 ≤ i ≤ d − 1 and 1 ≤ wi ≤ d}. Output:
d − 1 node disjoint paths between μ and T. 1: if (t = 1) then
2: Generate P1, P2, …,and Pd-1 by Lemma 6;
3: return P; 4: end if
5: Get x where μ ∈ Kx(d, t − 1); 6: if (x ∈ {wi|1 ≤ i ≤ d − 1}) then
7: Find y where y ∈ {1,2, …, d}and y ∉{wi|1 ≤ i ≤ d − 1}; 8: Find z where z ≠ x and z ∈{wi|1 ≤ i ≤ d − 1};
9: R[z] ← HP(Ky(d, t − 1) ∪ Kz(d, t − 1), o(y, x), o(z,0));
10: Put o(x, y) into Tx; 11: T ← T − {o(z,0)};
12: end if
13: for each t in T do
15: Put o(x, y) into Tx;
16: R[y] ← HamPath(Ky(d, t − 1), o(y, x), t);
17: end for
18: P ← OneToMany(Kx(d, t − 1), μ, Tx); 19: return P + R;
Theorem 8. Let μ be any vertex in K(d,1) and T = {t1, t2, …, tm}⊂ V (K(d,1)) where μ ∉ T and m = d−1. Then there exist d − 1 node disjoint paths between μ and T whose union covers all the vertices of K(d, t).
Proof. Since K(d, 1) is a complete graph, obviously, there exist m node disjoint paths {Pi|1 ≤ i ≤ m} between μ and T, where Pi=⟨μ, ti⟩. □
Theorem 9. Let μ ∈ V (K(d, t)) with d ≥ 4 and T = {ti|1 ≤ i ≤ d − 1}⊂ V (K(d, t)) where ti ∉ OI , and μ, ti are not in the same subgraph. Then there exist d − 1 node disjoint paths between μ and T whose union covers all the vertices of K(d, t).
Proof. We can use the mathematical induction on t to prove this lemma. By Theorem 8, this theorem holds for K(d,1). We propose an algorithm OneToManyDPC(K(d, t), μ, T) to find d − 1 node disjoint paths between s and T. □ Algorithm 2 OneToManyDPC(K(d, t), μ, T)
Input:
A node μ and a set of d − 1 nodes T = {t1, t2, …, td-1}.
Output:
d − 1 node disjoint paths between μ and T whose union covers all the vertices of K(d, t). 1: if (t = 1) then
2: Generate P1, P2, …,and Pd-1 by Theorem 8;
3: return P; 4: end if
5: Get x where μ ∈Kx(d, t − 1); 6: for each t in T do
7: Get y where t ∈ Ky(d, t − 1); 8: Put t into Uy;
9: end for
10: for i = 1; i ≤ d; i + + do 11: if (|Ui| = 1) then 12: t ← First(Ui);
13: R[i] ← HamPath(Ki(d, t − 1), o(i, x), t);
14: visited[i] ← true; 15: end if
16: if (i ≠ x) then 17: Put o(x, i) into Tx; 18: end if
19: end for
20: for i = 1; i ≤ d; i + + do
21: if (|Ui| > 1 and visited[i] = false) then 22: t ← First(Ui);
23: Put o(i, x) into Ti; 24: visited[x] ← true; 25: for each t in T do
26: Find y where visited[y] = false;
27: R[y] ← HamPath(Ky(d, t − 1), o(y, x), o(y, i));
28: Put o(i,y) into Ti; 29: end for
32: end for
33: P ← OneToMany(Kx(d, t − 1), μ; Tx); 34: return P + R + Q;
Conclusions
In [8], Bakhshi et al. developed an algorithm that presents d−1 node-disjoint paths from an arbitrary node in K(d, t) to d−1 distinct target nodes. However, the union of all the disjoint paths does not cover all the vertices of K(d, t). In this paper, we prove the one-to-many node-disjoint path cover problem under certain restrictions. We prove that given any node μ and a set of distinct destination nodes T = {tj|1 ≤ j ≤ d−1}where tj∉ OI, and μ, tjare not in the same subgraph, there exist d − 1 node-disjoint paths between μ and T whose union covers all the vertices of K(d, t) where d ≥ 4 and t ≥ 1.
Acknowledgment
This paper is supported by National Natural Science Foundation of China (61602333, 61702351), Natural Science Foundation of the Jiangsu Higher Education Institutions of China (16KJB520050, 17KJB520036) and China Postdoctoral Science Foundation (172985).
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