Feedback Vertex Set in Split-Stars and Alternating Groups

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(1)Feedback Vertex Set in Split-Stars and Alternating Groups Cheng－Ju Hsu*. In－Jen Lin. Department of Computer Science, National Taiwan Ocean University, Keelung, Taiwan, Republic of China. Abstract The feedback vertex set F of a graph G is a subset of vertices such that the removal of F from G induces an acyclic subgraph. In this paper, we study the feedback vertex set problem on the directed and undirected spilt-stars and alternating group graphs separately. We give upper and lower bounds to the minimum feedback vertex set on the n-dimensional spilt-stars and n-dimensional alternating group graphs.. Keyword ： Feedback vertex set, Interconnection network, Split-stars, Alternating Groups.. *All correspondence should be addressed to Cheng － Ju Hsu, Department of Information Management, Chung Yu Junior College of Business Administration, 40, Yi 7th Road, Keelung , Taiwan, Republic of China. (Phone：886－02－24237785 ext 570, Fax：886－02－24237785 ext 570, Email：miranda@cc1.cyjcba.edu.tw). 1.

(2) 1. Introduction Let G = (V, E) be a graph with vertex set V(G) and edge set E(G), where E ⊆ V × V. We also let D = (V, A) be a directed graph with vertex set V(D) and arc set A(D), where A ⊆ V × V. An arc uv is said to be directed from u to v. In a graph G (V, E), a cycle is a graph with an equal number of vertices and edges whose vertices can be placed around a circle so that two vertices are adjacent if and only if they appear consecutively in the circle. Also, a graph with no cycle is called acyclic. The feedback vertex set F of a digraph D = (V, A) is a subset of vertices V ⊆ V whose removal from D, induces an acyclic subgraph D' = (V ' , A' ) where V ' = V \ V and A' = {< u, v >∈A | u, v∈ V ' }. A feedback vertex set with the minimum cardinality is called minimum feedback vertex set, and its cardinality denoted by µ(D). The feedback vertex set problem originated from applications in combinatorial circuit design, but have found their way into numerous other applications, such as deadlock prevention in operating system, constraint satisfaction, Bayesian inference in artificial intelligence, and graph theory. As an example, consider an interconnection network modeled by a graph, for which vertices represent processors and each edge < i, j> represents the request of processor i for a resource allocated to a processor j. If there is a cycle in such a graph, a deadlock occurs and every processor in the cycle will wait for the requested resource and will never release the resource already allocated to it. In order to solve the deadlock, one can remove some processors from the graph and put them in a waiting queue. Therefore, it is clear that we want to minimize the number of processors removed and make the graph acyclic. It is well known that the problem of finding a minimum feedback vertex set is NP-hard for general networks [4], but there exits polynomial solutions for particular graphs [ 6, 7, 8, 9,10]. In order to obtain polynomial solutions, one can restrict these problems to special classes of graphs, such as interval graphs, permutation graphs, etc. In this paper, we present results concerning feedback vertex set problem in both directed and undirected spilt-stars and alternating group graphs , which have recently been developed as a new model of the interconnection network for parallel and distributed computing systems. Jwo et al. [5] studied the alternating group graphs; Cheng et al. [2] studied a variant distributed processor architecture of the star graphs,. 2.

(3) which is known as the spilt-star. Cheng and Lipman [1] proposed an assignment of directions to the edges of the spilt-stars and the alternating groups. They also showed that resulting directed graphs are not only strongly connected, but, in fact, they have maximally arc-connected and have small diameters. The n-dimensional directed spilt-star S n2 is a directed graph, which has the set of n! permutations of an n-set as the vertex set. The vertices of the spilt-stars are in a one-to-one correspondence with n! permutations [p1,p2,…,pn] of the set N={1,2,…,n}, and two vertices u, v of S n are connected by an arc < u, v > if and only if the permutation of v can be obtained from u by either a 2-exchange or 3-rotation. Let u = [p1,p2,…,pn]. A 2-exchange interchanges the first symbol p1 with the second symbol p2 whenever p1＞p2 , i.e., v = [p2,p1,…,pn]. A 3-rotation rotates the symbols in positions 1, 2 and i for some i∈{3, 4,…, n}, i.e., v = [pi,p1,…pi-1, p2, pi+1,…, pn]. Figure 1 depicts an example of S n2 for n = 4. On the other hand we also denote the n-dimensional undirected spilt-star by Sn. The undirected spilt-star can be obtained form directed spilt-stars by letting each arc with bi-direction. For simplicity, we discard the arc’s directions of undirected spilt-stars, and Figure 2 depicts an example of Sn for n = 4.. 3.

(4) 4231. 2431. 4321. 3241. a. b. 3421. 2341. c. d. 1234. 4132. 1432. 2134. 1324. 4312. 3214. 3124. 1342. 3412. 2314. 4213. 2413. 4123. c. 1243. d 2143. 1423 a. b. Figure 1 : 4-dimensional directed spilt-star.. 4. 3142.

(5) 4231. 2431. 4321. 3241. a. b. 3421. 2341. c. d. 1234. 4132. 1432. 2134. 1324. 4312. 3214. 3124. 3142. 1342. 3412. 2314. 4213. 2413. 4123. 1243. c. d 2143. 1423 a. b. Figure 2 : 4-dimensional undirected spilt-star.. Let u = [p1, p2,…, pn] where pi ∈N for all 1≤ i ≤ n. Now, let Id denote the identity. 5.

(6) permutation of N. If pi＞pj where i＜j , the pair pi and pj constitute an inversion. A permutation is said to be even ( resp. odd) if its number of parity inversions is even ( resp. odd). Given a simple graph G and a simple graph H, an isomorphism from G to H is a bijection f :V(G)→V(H) such that uv ∈E(G) if and only if f(u) f(v)∈E(H). We say G is isomorphic to H, if there is an isomorphism from G to H. Let S n2 , E be the subgraph of S n2 induced by the set of even permutations. This is precisely the alternating group graph, An , introduced in [5]. Let S n2 , O be the subgraph of S n2 induced by the set of odd permutations. Then, S n2 , E is isomorphic to S n2 , O via a 2-exchange. Let An and An be n-dimensional undirected and directed alternating group graphs induced subgraph of Sn and S n2 with even permutations reactively. The remaining sections of this paper are organized as follows. In Section 2, we define some notations and study the feedback vertex set problem for the directed spilt-star. The upper and lower bounds to the minimum cardinality of the feedback vertex set for the n-dimensional directed spilt-star are given. In Section 3, we show the upper and lower bounds to the minimum cardinality of the feedback vertex set for the n-dimensional directed alternating group graphs. Section 4 and Section 5 are devoted to explore the existence of independent set vertices of spilt-stars and alternating group graphs to construct double rooted star as feedback vertex set of Sn and An. Finally, a concluding remark is given in the last section. 2. The Feedback Vertex Set of Directed Spilt-stars The n dimensional spilt-star is a regular graph with degree 2n－3, |V( S n2 )|＝n! and |E( S n2 )|＝(2n－3)n!/2. S n2 is recursively constructed by n copies of S n2−1 . Let Ni = N \ {i} for i＝1,2, and let N1,2＝N \ {1,2}, where N = {1,2,3…n}. We also let V( Sn )={[p1,p2,p3…pn] | pi ≠ pj and i, j ∈N }. We define X be a nonempty proper subset of the V( Sn ), and let Ex(X) to be the set of 2-exchange neighbors of X and R (X) to be the set of 3-rotation neighbors of X. Define δ(X) to be the set of arcs leaving X and ρ(X) to be the set of arcs entering X. 6.

(7) Let F = {[p1, p2, p3…pn] | p1＞p2 } and F ⊆ V( Sn ). Then F is a feedback. Lemma 1.. vertex set of Sn . Proof.. Let F be a subset of V( Sn ) with cardinality n! ⁄ 2. We want to show that F is a. feedback vertex set of Sn . Suppose, to the contrary, that F is not a feedback vertex set of Sn . Then a cycle C＝u1→u2→…→uk→u1 exists in Sn \F. ui,3,…, ui,n] ∈C. Since ui,1＜ui,2. ,. For each vertex ui＝[ui,1, ui,2,. ui has no 2-exchange neighbor. Therefore, ui+1 is a. 3-rotation neighbor of ui ,1≤ i ≤ k－1, and u1 is uk’s 3-rotation neighbor. Further, ui,1＝ ui+1,2 and uk,1 ＝u1,2. Since ui+1,1＜ui+1,2 , ui+1,1＜ui,1 . Thus, uk,1＜uk-1,1＜…＜u11＜uk,1 , which is a contradiction.. □. Let F ' ＝{[2,1,p3,…,pn] | pi ∈N1,2, for each 3 ≤ i ≤ n} and F ' ⊆ F. For. Lemma 2.. each vertex u∈ F ' , R ( u ) ⊆ F \ F ' and R(Ex( u )) ⊆ F \ F ' . Proof. F ' ⊆ V( Sn ). Let uv ∈ δ ( u ), then either v∈Ex( u ) or v∈R ( u ). If v ∈ R ( u ), it has the form [pi ,2,p3,.. pi-1,1, pi+1 ,…,pn] and pi ≥ 2, i∈N1,2. Since R ( u )∈F and R ( u ) ∉F ' , R ( u ) ⊆ F \ F ' . Otherwise, if v∈Ex ( u ) then v is the form of [1,2,p3,…,pn]. Since Ex ( u ) ⊄ F, Ex ( u ) ⊄ F \ F ' ,but for each vertex w∈R(Ex( u )), it has the form of [pi ,1,p3,.. pi-1,2, pi+1…,pn] and pi ≥ 1, pi≠2, i∈N1,2. . Thus, R(Ex( u )) ∉F ' , R(Ex(u)) ⊄ F \ F' .. □. Lemma 3. Let F ' ' = {[n,n－1,p3,…, pn] | pi ∈Nn,n-1, for each 3 ≤ i ≤ n} and F ' ' ⊆ F. For each vertex u∈ F ' ' , R ( u ) ⊆ F \ F ' ' . Proof. F ' ' ⊆ V( Sn ). Let vu ∈ ρ ( u ), then v∈R ( u ). If v∈R ( u ), it has the form [n－ 1,pi,p3,.. pi-1, n, pi+1 ,…,pn] and ( u ) ⊆ F \ F'' .. pi ≤ n－1, i∈N1,2. Since R( u )∈F and R ( u )∉ F ' ' , R. □. Based on the Lemmas 1, 2 and 3, we give the following algorithm to find the feedback vertex set for the directed split-stars. Algorithm FDS Input: A directed split-star Sn . Output: A feedback vertex set of Sn .. 7.

(8) Method: Step 1: F＝{[p1, p2, p3…pn] | p1＞p2 } F ' ＝{[2,1,p3,…,pn] | pi ∈N1,2, for each 3 ≤ i ≤ n} F ' ' ＝{[n,n－1,p3,…, pn] | pi ∈Nn,n-1, for each 3 ≤ i ≤ n} Step 2:. S＝F / ( F ' ∪ F ' ' ).. Step 3:. output S.. From Lemma 1, Lemma 2 and Lemma 3, we can conclude that the upper bound to the minimum cardinality of the feedback vertex set for the n-dimensional directed spilt-star. Theorem 4. Proof.. µ( Sn ) ≤ n! ⁄ 2－2(n-2)!. For each vertex u∈ F ' , by Lemma 2, R( u ) ⊆ F \ F ' . Since R(Ex( u )) ⊆ F. \ F ' the existence of Ex( u), it just only makes paths not cycles. By Lemma 3, for each vertex v∈ F ' ' , R (v) ⊆ F \ F ' ' . Thus, the existence of F ' and F ' ' would not make any cycle in Sn . By Lemma 1, F is a feedback vertex set of V( Sn ) with cardinality n! ⁄ 2. So µ( Sn )≤ n! ⁄ 2－2(n-2)!.. □. In addition to give the upper bound to the minimum cardinality of the feedback vertex set for the n-dimensional directed spilt-star, we also give the lower bound to the minimum cardinality of the feedback vertex set for the n-dimensional directed spilt-star. Theorem 5. Proof.. n ≥ 4, µ( Sn ) ≥ n!/3. In S 4 , there exists 8 disjoint 3-cycles. In order to break cycles in S 4 , we have. to delete at least 8 vertices, a vertex for each disjoint cycle. The labels of deleted vertices are in the following: [3214], [3241], [3142], [3124], [4123], [4132], [4213] and [4231]. For |V( S 4 )|＝24 and µ( S 4 )＝8/24＝1/3. Again, since there are n! /4! copies of S 4 in Sn for n ≥ 4, and in each copy, we need to delete at least eight vertices. Then results µ( Sn ) ≥ (n!/4!)×8＝n!/3.. □. 3.The Feedback Vertex Set of Directed Alternating Group Graphs The n-dimensional directed alternating group graphs An2 is a directed graph, which is induced by the set of even ( resp. odd) permutations of Sn . It is a regular graph with. 8.

(9) degree 2(n－2). Since S n2 , E is isomorphic to S n2 , O via a 2-exchange, without loss of generality, we let An be the even permutation of Sn . The cardinality of vertex set and edge set of An is n! /2 and (n－2)n!/2. Alternating group graphs have a highly recursive structure. An is made up of n An −1 . Figure 3 and 4 depict examples of A3 and A4 , respectively.. 123. 231. 312 Figure 3: A3. 1342. 2143. 3241. 2431. 4321. 4132. 1423. 1234. 3124 2314. 3412. 4213. Figure 4: A4. 9.

(10) Lemma 6. Let F = {[p1,p2,p3,…,pn] | p1＞p2 } and F ⊆ V( An ). Then F is a feedback vertex set of An . Proof.. Let F be a subset of V( An ) with cardinality n! ⁄ 4. We want to show that F is a. feedback vertex set of An . Suppose, to the contrary, that F is not a feedback vertex set of An . Then a cycle C＝u1→u2→…→uk→u1 exists in An \ F.. For each vertex ui＝. [ui,1, ui,2, ui,3,…, ui,n] ∈C. Since ui,1＜ui,2 , ui has no 2-exchange neighbor. Therefore, ui+1 is a 3-rotation neighbor of ui ,1≤ i ≤ k－1, and u1 is uk’s 3-rotation neighbor. Further, ui,1＝ui+1,2 and uk,1 ＝u1,2. Since ui+1,1＜ui+1,2 , ui+1,1＜ui,1 . Thus, uk,1＜uk-1,1＜…＜u11＜ uk,1 , which is a contradiction. Lemma 7.. □. Let A' ⊆ V( An ) and A' = {[2,1, p3,…, pn] | pi ∈N1,2, for each 3 ≤ i ≤ n}, for. each vertex u∈ A' , R ( u ) ⊆ F \ A' . Proof. Let uw ∈ δ ( u ), w∈R ( u ) . If wi ∈R ( u ), it has the form [pi ,2,p3,.. pi-1,1, pi+1 ,…,pn] and pi ≥ 2, i∈N1,2 . Since R( u )∈F and R ( u )∉ A' , R ( u ) ⊆ F \ A' .. □. Lemma 8. Let A' ' ⊆ V( An ) and A' ' ＝{[n, n－1, p3,..., pn] | pi ∈Nn,n-1, for each 3 ≤ i ≤ n}, for each vertex v ∈A' ' , R ( v ) ⊆ F \ A' ' . Proof. Let. sv ∈ ρ ( v ), then s∈R ( v ) . If si ∈R ( v ), it has the form. pi-1, n, pi+1 ,…,pn] and \ A' ' .. [n－1,pi,p3,... pi ≤ n－1, i∈N1,2. Since R ( v )∈F and R ( v )∉ A' ' , R ( v ) ⊆ F. □. Based on the Lemmas 6, 7 and 8, we give the following algorithm for solving the feedback vertex set problem in the directed alternating group graphs. Algorithm FDA Input: A directed alternating group graphs An . Output: A feedback vertex set of An . Method: Step 1: F＝{[p1, p2, p3…pn] | p1＞p2 } F ' ＝{[2,1,p3,…,pn] | pi ∈N1,2, for each 3 ≤ i ≤ n} F ' ' ＝{[n,n－1,p3,…, pn] | pi ∈Nn,n-1, for each 3 ≤ i ≤ n}. 10.

(11) Step 2:. S＝A / ( A' ∪ A' ' ).. Step 3:. output S.. Lemma 6, Lemma 7 and Lemma 8 can derive the upper bound to the minimum cardinality of the feedback vertex set for the n-dimensional directed alternating group graphs. Theorem 9. µ( An ) ≤ n! ⁄ 4－(n－2)! Proof. For each vertex u∈ A' , v∈ A' ' , by Lemma 7, R ( u ) ⊆ F \ A' , R ( v ) ⊆ F \ A' ' . Thus, the existence of A' and A' ' would not make any cycle in An . By Lemma 6, F is a feedback vertex set of V( An ) with cardinality n! ⁄ 4. So µ( An )≤ n! ⁄ 4－(n－2)!. □. We also give the lower bound to the minimum cardinality of the feedback vertex set for the n-dimensional directed alternating group graphs. Theorem 10. Proof.. n ≥ 4, µ( An ) ≥ n!/6. In A4 , there exists 4 disjoint 3-cycles. To break all cycles of A4 , we need to. prune at least 4 vertices, a vertex for each disjoint cycle. The labels of deleted vertices are in the following: [3241], [3124], [4132] and [4213]. For |V( A4 )|＝n!/2＝12 and µ( An )＝4/12＝1/3. Again, since there are n!/24 copies of A4 in An for n ≥ 4, and in each copy, we need to delete at least four vertices. Then results µ( An ) ≥ (n!/24)×4＝n!/6. □. 4. The Feedback Vertex Set of Undirected Spilt-stars An independent set in a graph G is a vertex set I ⊆ V(G) that contains no edge of G, that is to say G[ I ] has no edge. Let N ' , N ' ' ⊆ N, where N ' ＝{1,2,…, n / 2 } and N ' ' ＝{ n / 2 ＋1, n / 2 ＋2, …., n}. Lemma 11. Let I＝{[x,y, p3,…,pn] | x∈ N ' , y∈ N ' ' , pi ∈Nx,y and i＝3,4,…n} and vertex set I is an maximal independent vertex set of Sn. Proof.. Since I ⊆ V( Sn ), we immediately show that for any two vertices u, v ∈ I,. vertices u, v are not adjacent. Suppose, to the contrary, that I is not an independent. 11.

(12) vertex set of Sn. Then an edge uv exists in G[ I ]. Let u＝{[u1,u2,u3,…,un] | u1 ∈ N ' , u2 ∈ N ' ' and ui ∈Nu1,u2 and i＝3,4,…n } and v＝{[v1,v2,v3,…,vn] | v1 ∈ N ' , v2 ∈ N ' ' , vi ∈Nv1,v2 and i＝3,4,…n}. Hence, v is either the 2-exchange neighbor of u or a 3-rotation neighbor of u. If v is the 2-exchange neighbor of u, then v1＝u2, and v2＝u1. Since v1 ∈ N ' , u2 ∈ N ' ' and N ' ∩ N ' ' ＝ φ , it is a contradiction. Otherwise, v is a 3-rotation neighbor of u. Thus, v2＝u1or v1＝u2. Similarly, it contradicts that N ' ∩ N ' ' ＝ φ . Furthermore, we shall prove that I is maximal. Suppose, to the contrary, that I is not a maximal independent set of Sn. Then, there exists a vertex v∈V( Sn )\ I and I ∪ {v} is also a independent set of Sn . That is to say, u and v are nonadjacent, for each u∈I. Let u ＝{[u1,u2,u3,…,un] | u1 ∈ N ' , u2 ∈ N ' ' and ui ∈Nu1,u2 and i＝3,4,…n }. Since v∉I, v belongs to one of the following three vertex sets. (1) V ' ＝{[v1,v2,v3…,vn] | v1 ∈ N ' and v2 ∈ N ' , vi ∈Nv1,v2 and i＝3,4,…n}, (2) V ' ' ' ＝{[v1,v2,v3…,vn] | v1 ∈ N ' ' and v2 ∈ N ' ' , vi ∈Nv1,v2 and i＝3,4,…n}, (3) V ' ' ＝{[v1,v2,v3…,vn] | v1 ∈ N ' ' and v2 ∈ N ' , vi ∈Nv1,v2 and i＝3,4,…n}. Now, we discuss it according to the listed classes. Case 1:. v∈ V ' . Let w＝[v2,vi,…,vi-1,v1,vi+1,…,vn]∈N(v), where vi ∈ N ' ' . Then w∈I,. vw ∈E( Sn ). Which contradicts that v, w are nonadjacent, for each vertex w∈I. Case 2: v∈ V ' ' ' . The proof is similar to case 1. Case 3: v∈ V ' ' . Let w＝[v2,v1,…,vi-1,vi,vi+1,…,vn]∈N(v), where v2 ∈ N ' and v1 ∈ N ' ' . Then w∈I, vw ∈E( Sn ). It is a contradiction.. □. According to the size of N ' and N ' ' , we get the following result | I |＝(n2/4)(n－2)!, if n is even and | I |＝(n2－1/4)(n－2)!, if n is odd. Lemma 12.. Let L' = {[x, y, p3…,pn] | x,y∈ N ' , x＝1,3,5,…, n / 2 ,. y＝x+1, pi ∈Nx,y and i＝3,4,…n}, then L' is an independent set of Sn. Proof.. For each vertex u, v∈ L' , let u＝[x1, y1, p3…pn] and v＝[x2, y2, p3…, pn].. Suppose, to the contrary, that L' is not an independent vertex set of Sn. Then an edge uv exists in G[ L' ]. Hence, v is either the 2-exchange neighbor of u or a 3-rotation. 12.

(13) neighbor of u. If v is the 2-exchange neighbor of u, then x1＝y2, and y1＝x2. Since x1, x2 are odd and y1, y2 are even, it is a contradiction. Otherwise, v is a 3-rotation neighbor of u. Thus, y2＝x1or x2＝y1. Similarly, it contradicts that x1, x2 are odd and y1, y2 are even. Therefore, L' is an independent set of Sn.. □. Lemma 13. Let L' ' ＝{u | v∈ L' , u∈Ex( v )}, then L' ' is an independent set of Sn. Proof.. Since L' ' is isomorphism to L' , thus L' ' is an independent set of Sn. An. independent edge set in a graph G is an edge set E ' ⊆ E(G) that each edge contains no common vertex of G, that is to say G[ E ' ] has no cycle. Lemma 14.. Let L' ' ＝{u | v∈ L' , u∈Ex( v )}, then G( L' ∪ L' ' ) is an independent. edge set in Sn. Proof.. For each vertex u, v∈ L' , s, w∈ L' ' . s is a 2-exchange neighbor of u and w is a. 2-exchange neighbor of v .We assume that us and vw have a common vertex . Let u＝ [x1, y1, p3…, pn], s＝[y1, x1, p3…, pn], v＝[x2, y2, p3…, pn] and w＝[y2, x2, p3…, pn]. Without loss generality let s＝v be the common vertex of us and vw . Then, y1＝x2 and x1 ＝ y2. Since u, v ∈ L'. and x1 ≠ y1 ≠ x2 ≠ y2. It is a contradiction.. □. A star is K1,n for some n ≥ 2. A doubled-rooted star (DRS) is the union of 2 K1,n, plus an edge between 2 vertices with maximum degrees. For example, Figure 5 (a) there are two K1,5 and Figure 5 (b) is double-rooted star constructed from 5 (a) with one more edge.. Figure 5 (a). 13.

(14) Figure 5 (b) Lemma 15. G( L' ∪ L' ' ∪ I ) is acyclic and G( L' ∪ L' ' ∪ I )is a union of disjoint double rooted stars. Proof.. Let u＝[p1, p2, p3…, pn]∈I. If p1 is odd (even), then there exists a unique. pi ∈ N ' and pi＝p1＋1 (pi＝p1－1) such that v＝[pi, p1, p3,..,pi-1, p2 pi+1,..,pn]∈ L' ' ( L' ), respectively. Then, v and its 2-exchange neighbor are the roots of a double-rooted star. Since each u∈I is uniquely connected to a double-rooted star, G( L' ∪ L' ' ∪ I ) is a union of disjoint double rooted stars. Thus, G( L' ∪ L' ' ∪ I ) is acyclic. □. Lemma 16. Let L' ' ' ＝ {[n, n－1, p3…, pn] | [n, n－1, p3…, pn] ∩ An, pi ∈Nn,n-1 , and i＝3,4,…n}, and I ' ⊆ I, I ' ＝ {[ p1, n, p3…,pn] | p1 ∈ N ' , pi ∈Np1,n and i＝3,4,…n}, then N( L' ' ' ) ∩ I ⊆ I '. Proof.. Let N( L' ' ' ) ∩ I＝Γ1 ∪ Γ2, where. Γ1＝{[n－1, pi, p3,…pi-1, n, pi+1…, pn] | pi ∈Nn-1,n, i＝3,4,…n} and Γ2＝{[pi, n, p3,…pi-1, n－1, pi+1…, pn] | pi ∈Nn-1,n, i＝3,4,…n} . Therefore, N( L' ' ' ) ∩ I＝(Γ1 ∩ I ) ∪ (Γ2 ∩ I ). Now, we want to computeΓi ∩ I, i＝1,2, to complete the proof. Since n－1 ∉N ' , (Γ1 ∩ I )＝ φ . To find Γ2 ∩ I, Γ2 ∩ I ⊆ I ' because pi ∈ N ' and n∈ N ' ' . Thus N( L' ' ' ) ∩ I ⊆ I '.. □. Lemma 17. For any two distinct vertex a, b∈ L' ' ' , let NI’ (a)＝N (a) ∩ I ', NI’ (b)＝N (b) ∩ I ' , then NI’ (a) ∩ NI’ (b)＝ φ . Proof.. Let a＝[n, n－1, a3,…,an] and b＝[n, n－1, b3,…,bn].Since. 14.

(15) N I’(a), N I’(b) ⊆ I ' , therefore NI’ (a)＝[ai, n, a3,.. ai-1, n－1, ai+1…, an], ai ∈N ' and NI’ (b) ＝[bj, n, b3,…bj-1, n－1, bj+1…, bn], bj ∈ N ' .Let for any s be the common neighbor of a and b, then s＝[pk, n, p3,…pk-1, n－1, pk+1…, pn], pi ∈ N ' .For such the position of n－1 that i＝j＝k and ai＝pk＝bj, ai-1＝pk-1＝bj-1, ai+1＝pk+1＝bj+1 . It implies that a＝b, but it contradicts to a≠b, thus NI’ (a) ∩ NI’ (b)＝ φ .. □. Lemma 18. G( L' ∪ L' ' ∪ L' ' ' ) contains no cycle in Sn. Proof.. Let v＝[n, n－1, p3,…, pn] ∈ L' ' ' . For each u∈N(v), there are three possible. forms of u in the following. Case(1): u＝[n－1, n, p3,…, pn]. Since n－1 and n ∉N ' , u ∉L' ∪ L' ' . Case(2): u＝[n－1, pi , p3,…pi-1, n, pi+1…, pn]. Since n－1 ∉N ' , u ∉L' ∪ L' ' . Case(3): u＝[pi, n, p3,…pi-1, n－1, pi+1…, pn]. Since n ∉N ' , u ∉L' ∪ L' ' .. □. Lemma 19. If u∈ L' ' ' then u can connect to at most one vertex v in each double rooted star of G( L' ∪ L' ' ∪ I '). Proof. We defineΠi , i＝1,3,5,… n / 2 , denote the set of double rooted star which. roots are labeled with [i, i+1, p3,…, pn] and [i+1, i, p3,…, pn]. Let u＝[n, n－1, p3,…, pn] ∈ L' ' ' . For each v∈N(u), there are three possible forms of v in the following. Case(1): v＝[n－1, n, p3,…, pn]. Since n－1 ∉N ' , v ∉L' ∪ L' ' ∪ I ' . That is to say u does not adjacent with any vertex of G( L' ∪ L' ' ∪ I '). Case(2): v＝[n－1, pi , p3,…pi-1, n, pi+1…, pn]. This proof is similar to case (1). Case(3): v＝[pi, n, p3,…pi-1, n－1, pi+1…, pn]. If v is in no DRS, then u does not adjacent with any DRS. v∈ Πpi if pi is odd and v∈ Πpi－1, otherwise. For eachΠi, since the second symbol of each roots is less than n, v is not a root. By the construction ofΠi, there are exactly two leaves. v1＝[i, n,. x3, x4,…xn] and v2＝[i＋1, n, x3, x4,…xn] with the permutation that the second symbol is n.. since. v1 ≠ v2. ,. either. v1. or. v2. □. Theorem 20. G( L' ∪ L' ' ∪ I ∪ L' ' ' ) is acyclic.. 15. is. the. only. neighbor. of. u..

(16) Proof. By lemma 15 G( L' ∪ L' ' ∪ I ) is acyclic and by Lemma 19, u can connect to at most one vertex in each DRS. Thus, there is no cycles in G( L' ∪ L' ' ∪ I ∪ L' ' ' ). □. Based on the Theorem 20, we give the following algorithm for solving the feedback vertex set problem in the undirected split-stars. Algorithm FUS Input: An undirected split-star Sn. Output: A feedback vertex set of Sn. Method: Step 1: I＝{[x,y, p3,…,pn] | x∈ N ' , y∈ N ' ' , pi ∈Nx,y and i＝3,4,…n}. L' ＝{[x, y, p3…,pn] | x, y∈ N ' , x＝1,3,5,…, n / 2 , y＝x+1, pi ∈Nx,y and i ＝3,4,…n}. L' ' ＝{u | v∈ L' , u∈Ex( v )}. L' ' ' ＝{[n, n－1, p3…, pn] | [n, n－1, p3…, pn] ∩ An, pi ∈Nn,n-1 , and i＝ 3,4,…n}. Step 2:. S＝I ∪ L' ∪ L' ' ∪ L' ' ' .. Step 3:. output S.. Since G( L' ∪ L' ' ∪ I ∪ L' ' ' ) is acyclic, G( L' ∪ L' ' ∪ I ∪ L' ' ' ) is a feedback vertex set, we immediately have the following result. Theorem 21. µ( Sn ) ≤ n!－[(n2+2n/4)(n－2)! + (n－2)!/2], if n is even. µ( Sn ) ≤ n!－[(n2－1+2n /4)(n－2)! + (n－2)!/2], if n is odd. Any connected acyclic graph must be a tree. To determine a given simple graph G is acyclic or not, we make use of the relationship between number of vertices and edges in each component of G. Furthermore, the following lemma applied to find the lower bound of the undirected spilt-stars. Lemma 22. Let G be a simple graph. G is cyclic, if |V(G)|≤ |E(G)|. Proof.. Without loss of generality, we may assume G is connected. Otherwise there is a. cycle in a small component (by induction). If G does not contains a cycle, tree then |E(G)|=|V(G)|－1.. □. 16. the G is a.

(17) An edge is called outer-edge if the endvertices of this edge belong to two different substars and the cardinality of outer-edges of some vertex v is the outer-degree of v. Otherwise, an edge is called inner-edge if the endvertices of this edge belong to the same substars and the cardinality of inner-edges of some vertex v is the inner-degree of v. For example, Figure 2 shows the 24 outer-edges in S 4. Let we denote the degree of v in graph G by degG(v). Lemma 23. µ( S4 ) ≥ 11. Proof.. The 4-dimensional split-star graph S4 can be recursively constructed by four. 3-dimensional split-star as its subgraph, named S3, and each S3 contains two vertex disjoint 3-cycles. To count the cardinality of the 3-cycles in S4, it can be seen that since each vertex in S4 incidents with two 3-cycles and each 3-cycle is repeatedly counted three times, there are totally sixteen 3-cycles in S4. For each vertex disjoint 3-cycle, we must delete at least one vertex to ensure the acyclic, and so this vertex results in two 3-cycles be broken. Then each vertex in S3 is forced to loss its degree by 2, for the removal of the two cycles, which the vertex belongs. Therefore, each vertex has one or two inner-degree less than the original vertex, because they have to adjacent to at least one of the two vertices deleted. We find the out-degree of each vertex in S4 is two. Since each vertex incidents with two 3-cycles and if we delete eight vertices for each vertex disjoint 3-cycle then we discredit sixteen 3-cycles. So the out-degree less than or equals to one for each vertex u in the remaining graph. It is clear deg(u) ≤ 2 ＋ 1＝ 3 and µ ( S 4 ) ≥ 8 . After we prune eight vertices in S4 , there are at least 20 edges and 16 vertices left. By Lemma 22, there still exists cycles in the remaining graph. Then, we further delete vertices to let the remaining graph to be acyclic. For each vertex u with degree 3, since u is adjacent with at most one vertex in S3 with degree 3, there are at most two neighbors of u with degree 3. So, we first remove one of the eight vertices, v1, with degree 3, and then there are at least five vertices with degree three exist in the remaining graph. Furthermore, we delete another vertex v2 with degree 3. There remain two vertices with degree three. Again, we can cancel one of these two vertices to break all the cycles of the remaining graph. Thus, the remaining graph is acyclic and µ( S4 ) ≥ 11. Theorem 24. Proof. □. n ≥ 4, µ( Sn ) ≥ (11/24)n!. In order to break cycles in S4, we have to delete at least 11 vertices. The labels. 17.

(18) of deleted vertices are in the following: [3124], [3142], [4123], [4132], [3214], [3241], [4213], [4231], [3412], [3421] and [4312]. For |V(S4)|＝24 and µ(S4)＝11. Again, since there are n! /4! copies of S4 in Sn for n ≥ 4, and in each copy, we need to delete at least eleven. vertices.. Then. results. µ(Sn). ≥. (n!/4!)×11. ＝. (11/24)n!.. □. 5. The Feedback Vertex Set of undirected Alternating Group Graphs The constructions of undirected alternating group graphs are the same as directed alternating group graphs except that the direction of every edge is bi-directional. For simplicity, we discard the arc’s directions of undirected alternating group graphs. Figure 6 depicts example of A4 . 1342. 2143. 3241. 2431. 4321. 4132. 1423. 1234. 3124 2314. 3412. 4213. Figure 6: A4 Here, we also use the result of independent set of spilt-star to implement the alternating group graphs. Lemma 25. Let I ＝{[a,b, p3…pn] | a∈ N ' ,by∈ N ' ' , pi ∈Na,b and. 18.

(19) i＝3,4,…n} and vertex set I is an maximal independent vertex set of An. Proof.. Since I ⊆ V( An ), we immediately show that for any two vertices u, v ∈ I,. vertices u, v are not adjacent. Suppose, to the contrary, that I is not an independent vertex set of Sn. Then an edge uv exists in G[ I ]. Let u＝{[u1,u2,u3…un] | u1 ∈ N ' , u2 ∈ N ' ' and ui ∈Nu1u2 and i＝3,4,…n } and v＝{[v1,v2,v3…vn] | v1 ∈ N ' , v2 ∈ N ' ' , vi ∈ Nv1v2 and i＝3,4,…n}. Hence, v is either the 2-exchange neighbor of u or a 3-rotation neighbor of u. If v is the 2-exchange neighbor of u, then v1＝u2, and v2＝u1. Since v1 ∈ N ' , u2 ∈ N ' ' and N ' ∩ N ' ' ＝ φ , it is a contradiction. Otherwise, v is a 3-rotation neighbor of u. Thus, v2＝u1or v1＝u2. Similarly, it contradicts that N ' ∩ N ' ' ＝ φ . Furthermore, we shall prove that I is maximal. Suppose, to the contrary, that I is not a maximal independent set of An. Then, there exists a vertex v∈V( An )\ I and I ∪ {v} is also a independent set of An . That is to say, u and v are nonadjacent, for each u∈I. Let u. ＝ {[u1,u2,u3…un] | u1 ∈ N ' , u2 ∈ N ' ' and ui ∈ Nu1,u2 and i＝3,4,…n }. Since v∉ I, v belongs to one of the following three vertex sets. (1) V ' ＝{[v1,v2,v3…vn] | v1 ∈ N ' and v2 ∈ N ' , vi ∈Nv1,v2 and i＝3,4,…n}, (2) V ' ' ' ＝{[v1,v2,v3…vn] | v1 ∈ N ' ' and v2 ∈ N ' ' , vi ∈Nv1,v2 and i＝3,4,…n}, (3) V ' ' ＝{[v1,v2,v3…vn] | v1 ∈ N ' ' and v2 ∈ N ' , vi ∈Nv1,v2 and i＝3,4,…n}. Now, we discuss it according to the listed classes. Case 1:. v∈ V ' . Let w＝[v2,vi,…,vi-1,v1,vi+1,…,vn]∈N(v), where vi ∈ N ' ' . Then w∈I,. vw ∈E( Sn ). Which contradicts that v, w are nonadjacent, for each vertex w∈I. Case 2: v∈ V ' ' ' . The proof is similar to case 1. Case 3: v∈ V ' ' . Let w＝[v2,v1,…,vi-1,vi,vi+1,…,vn]∈N(v), where v2 ∈ N ' and v1 ∈ N ' ' . Then w∈I, vw ∈E( An ). It is a contradiction.. □. Based on the theorem 20 and Lemma 25, we give the following algorithm for solving the feedback vertex set problem in the undirected alternating group graphs. Algorithm FUA Input: An undirected alternating group graphs An. Output: A feedback vertex set of An.. 19.

(20) Method: Step 1: I＝{[x,y, p3,…,pn] | x∈ N ' , y∈ N ' ' , pi ∈Nx,y and i＝3,4,…n}. L' ＝{[x, y, p3…,pn] | x, y∈ N ' , x＝1,3,5,…, n / 2 , y＝x+1, pi ∈Nx,y and i ＝3,4,…n}. L' ' ＝{u | v∈ L' , u∈Ex( v )}. L' ' ' ＝{[n, n－1, p3…, pn] | [n, n－1, p3…, pn] ∩ An, pi ∈Nn,n-1 , and i＝ 3,4,…n}. Step 2:. S＝I ∪ L' ∪ L' ' ∪ L' ' ' .. Step 3:. output S.. | I |＝(n2/4)(n－2)!/2, if n is even and | I |＝(n2－1/4)(n－2)!/2, if n is odd. By Theorem 20, we have shown that G( L' ∪ L' ' ∪ I ∪ L' ' ' ) is acyclic. Since G( L' ∪ L' ' ∪ I ∪ L' ' ' ) is acyclic, G( L' ∪ L' ' ∪ I ∪ L' ' ' ) is a feedback vertex set, we immediately have the following result. Theorem 26. µ( An ) ≤ n!/2－[(n2/4)(n－2)!/2+ (n－2)!/4], if n is even. µ( An ) ≤ n!/2－[(n2－1/4)(n－2)!/2+(n－2)!/4], if n is odd. We also make use of the relationship between number of vertices and edges in each component of A4 to find the lower bounds of the undirected alternating group graphs. To decide the lower bound of the feedback vertex number, By Lemma 22 and Lemma23, an important observation is established as follows. In A4, the graph is exactly covered by 4 disjoint vertex 3-cycles. To break all cycles of A 4, we discard four vertices, a vertex for each disjoint cycle, at first. And cycle does not survive in any A 3. It is clear that µ ( A4 ) ≥ 4 , and since there are 24 edges in A 4 , it is easy to see that there are at least 8 edges left. To cut the remaining cycles, one edge should be pruned at least, because there are 8 vertices survived in the remaining graph. Therefore, it is necessary to remove one vertex in the remaining graph to break all cycles of A4. Thus, µ ( A4 ) ≥ 5 and the lower bound of An is built as follows. Theorem 27. n ≥ 4, µ( An ) ≥ (5/24)n! Proof. To break cycles in A4, we have to omit at least 5 vertices. The labels of deleted. vertices are in the following: [3124], [4132], [3241], [4213] and [3412]. For |V(A4)|＝12. 20.

(21) and µ(A4)＝5. Again, since there are n! /4! copies of A4 in An for n ≥ 4, and in each copy, we need to delete at least five vertices. Then results µ(An) ≥ (n!/4!)×5＝(5/24)n!. □. 6. Concluding Remarks A recent line of research on polynomially solvable cases focuses on special undirected graphs having bounded degree and that are widely used as connection networks, namely meshes and toroidal meshes, Butterflies, toroidal butterflies, and hypercubes. In meshes and toroidal meshes, Luccio [10] obtained the upper bounds on the size of the minimum feedback vertex set. These bounds either match the lower bounds or are very close to them. For butterfly graphs, Luccio [10] found both bounds to the size of a minimum feedback vertex set. Similar results to those obtained for butterflies can also be obtained for toroidal butterflies. Spilt-stars, an alternative to the star graphs, are companion graphs to alternating group graphs. These graphs have many advantages over the n-cubes. Recently, Cheng et al. [1] proposed an orientation to the spilt-stars and the alternating group graphs. They showed that the oriented graphs are maximally arc-connected and have small diameters. In this thesis, we study the feedback vertex set problem on directed and undirected spilt-stars and alternating group graphs separately. At the first part, the upper and lower bounds to the feedback vertex set for the directed spilt-stars and alternating group graphs, respectively, are determined. At the second part, we give the both bounds to the undirected spilt-stars and alternating group graphs by expanding maximal independent sets, respectively, to decide the feedback vertex sets. In the construction of the remaining graph, discard the feedback vertex set from the given undirected graph, we add a specified maximal independent set with undirected other vertices. However, the independent set we used is not maximum. Further, a natural question to ask a maximum independent set to increase the size of feedback vertex set is our next research. And we can also study the feedback vertex set for the other topologies such as multi-mesh or star graph.. 21.

(22) 7. Reference [1]Eddie Cheng and Marc J. Lipman, Orienting spilt-stars and alternating group graphs, Networks, Vol. 35, pp. 139-144, 200. [2]E. Cheng, M. J Lipman, and H.A. Park, An attractive variation of the star graphs: spilt-stars, Technical Report No.98－3, Oakland University, 1998. [3]Riccardo Focardi and Flaminia L. Luccio, Minimum Feedback Vertex Set in K-Dimensional Hyercubes. Technical Report 26 October 1999. [4]M. R. Garey and D.S. Johnson. Computers and Intractability. Freeman, San Francisco 1979. [5]J. S. Jwo, S. Lakshmivarahan, and S.K. Dhall, A new class of interconnection networks based on the alternating group, Networks 23(1993), 315－326. [6]Y. D. Liang. On the feedback vertex set in permutation graphs. Information Processing Letters, 52, (3), 123－129, 1994. [7]Y. D. Liang and M. S. Cheng. Minimum Feedback Vertex Set in Cocomparability Graphs and Convex Bipartite Graphs. Acta Informatica, 34, 337－346, 1997. [8]E. L. Llyod and M. L. Soffa. On Locating Minimum Feedback Vertex Sets. Journal of Computer and System Sciences, 37, 292－311, 1988. [9]C. Lu and C. Tang. A linear-time algorithm for the weighted feedback vertex problem in interval graphs. Information Processing Letters, 61, (2), 107－112, 1997. [10]F. L. Luccio . Exact Minimum Feedback Vertex Set in Meshes and Butterflies. Information Processing Letters, 66, (2), 59－64, 1998.. 22.

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