A DISJOINT PATH PROBLEM IN THE ALTERNATING GROUP GRAPH EDDIE CHENG, LAZAROS KIKAS, SERGE KRUK Abstract. For the purpose of large scale computing, we are interested in linking computers into large interconnection networks. In order for these networks to be useful, the underlying graph must possess desirable properties such as a large number of vertices, high connectivity and small diameter. In this paper, we are interested in the alternating group graph, as an interconnection network, and the k-Disjoint Path Problem. We give a proof that the alternating group graph, An , has the (n − 2)-Disjoint Path Property. We close with a discussion on possible research stemming from this work.

Keywords: Interconnection networks, graphs, vertex disjoint paths 1. Introduction For the purpose of large scale computing, we are interested in linking computers/processors into large symmetric interconnection networks. The papers [1,2] list some desirable properties that we wish the underlying graph of our interconnection networks to have. These properties are vertex symmetry, large number of vertices, high connectivity and small diameter and degree. Informally, a graph is vertex symmetric if each vertex in the graph can be viewed as identical to the other vertices in the graph. Recall that the connectivity of a non-complete graph refers to the minimum number of vertices that must be deleted in order to disconnect the graph. In terms of interconnection networks, the deletion of a vertex can be viewed as equivalent to processor failure. Hence, it is desirable to have networks that allow for a large number of processor failure but still be operational. In graph theoretic terms, we want to be able to delete a large number of vertices and still have a connected graph. That is, we want the underlying graph to have high connectivity. Another measure of performance for interconnection networks deals with communication delay. When one sends a message from one processor to another in an interconnection network, the communication is never instantaneous. The message must be sent over a path in the graph. Hence, there will be some delay. The diameter of a graph is the maximum of the minimum distances between all pairs of vertices in the graph. We, therefore, want the graph to have a small diameter. 1

For a very long time the boolean n-cube has served as wonderful interconnection network model. However, in 1988 [1] introduced the star graph as a competitive alternative to the n-cube. The star graph is defined as follows. The vertex set is made up of the symmetric group Sn . The elements of this group are the permutations on the symbols 1, 2, 3, ..., n. Two permutations (vertices) are adjacent if and only if one can get from one permutation to the other by exchanging the first and kth symbol where k ∈ {2, 3, 4, ...n}. In the 1990’s both the split-star and the alternating group graph were introduced into the literature. Please see [4, 7]. The vertex set of the split-star is, again, the elements of the symmetric group Sn . Two permutations are adjacent if and only if one can get from one permutation to the other by either a 2-exchange or a 3-rotation. A 2-exchange is accomplished by exchanging the first and second symbol of the permutation. A 3-rotation is accomplished by rotating the first, second and kth symbol where k ∈ {3, 4, .., n} in either direction. The alternating group graph, An , is the subgraph of the split-star, Sn2 , induced by the even permutations. In this paper, we study the k-Disjoint Path Problem in the alternating group graph. 2. Motivation The alternating group graph, An , has as its vertex set of even permutations on the symbols 1, 2, 3, ..., n. It is an exercise in algebra to show that the alternating group is generated by the 3-cycles of the symmetric group Sn . Two vertices in An are adjacent if and only if one can get from one permutation to the other by a 3-rotation. A 3-rotation rotates the first, second and kth symbol where k ∈ {3, 4, ..., n} of the permutation in either direction. The rotation is accomplished by a 3-cycle. An is a graph with degree 2(n − 2), and has n!(n−2) edges. It is both vertex and edge 2 symmetric. We now show that An has a hierarchal structure. Consider the even permutations where, say 5, is fixed in the nth position. Now consider the subgraph of An induced by these permutations. It is clear that that the resulting subgraph is isomorphic to An−1 . Since there are n symbols it becomes clear that An has n copies of An−1 . We will refer to these subgraphs as substars throughout the rest of this paper. Note that this hierarchal structure is very useful for us as we prove our main result. For more details about the hierarchal structure please see [2, 7]. Consider the graph A4 in Figure 1. Let s1 = 1342, t1 = 4321, s2 = 3412 and t2 = 1423. Now suppose that we want to send a message from s1 to t1 and a message from s2 to t2 at the same time. Note that to send a message from s1 to t1 we can traverse the path 1342, 3241, 4321 and to send a message from s2 to t2 we can follow the path 3412, 4213, 1423. Further notice that these two paths are vertex disjoint. Hence, these two paths can 2

1342

2143 3241

4321

2431

1423

4132

3124

1234 2314

4213

3412

Figure 1. The Alternating Group Graph A4

be done at the same time without any interference or sharing of processor time and resources. We want to consider the following problem. Given k pairs of distinct nodes (s1 , t1 ), (s2 , t2 ),...,(sk , tk ) in a graph G, find k- vertex disjoint paths, one connecting each pair. This problem is called the k-Disjoint Path Problem. A graph G is said to have the k-Disjoint Path Property if one can find k-vertex disjoint paths for any selection of k pairs of distinct vertices. It has been shown that for k ≥ 3 the problem of finding the k-disjoint paths is N P -Hard. However, Watkins in [9] showed that a necessary condition for a graph have the the k-Disjoint Path Property is that the graph be (2k − 1)-connected. Work has been done on the k-Disjoint Path Problem and other interconnection networks. For example, Cheng and Lipman in [3] showed that the split-star graph has the (n − 1)-Disjoint Path Property. Another example is that Gu and Peng presented in [6], for the star graph an algorithm to compute the k-vertex disjoint paths where k = d n−1 2 e. 3

Since, An has degree 2n − 4 the best we can hope for is that the alternating group graph has the (n − 2)-Disjoint Path Property. Unfortunately, this is not true when n = 2. That is, A4 does not have the 2-Disjoint Path Property. For example, let s1 = 1342, t1 = 4213, s2 = 3412 and t2 = 2143 in Figure 1. Hence, the main result we wish to prove in this paper is that for n ≥ 5 the alternating group graph An has the (n − 2)-Disjoint Path Property. Due to the hierarchal structure of An , mathematical induction is the natural selection of method of proof to prove this result. In this paper, we concentrate on the nontrivial base case only. Since, An is an induced subgraph of the split-star it is not surprising that the induction step is a straightforward modification of the induction step given in [3]. The interested reader is directed to [8] for the modified induction step. Hence, what remains to be shown is that A5 has the 3-Disjoint Path Property. 3. The Proof for A5 We now wish to prove the base case of the following result. Theorem 3.1. The Alternating Group Graph, A5 , has the 3-Disjoint Path Property. Recall that Ckn counts the number of subsets of size k chosen from n elements. A brute force approach to the proof of Theorem 3.1 would involve C260 C258 C256 cases. Examination of all this cases may not be feasible, even with the help of a computer. Compare this to the base case for the splitstar where a brute force approach would involve C224 C222 C220 possible cases. Using the symmetry of the split-star the base case is much more manageable using a brute force approach and does not need the use of a computer. Please see [3]. Our proof is a combination of brute force and a careful examination of the symmetry of the graph A5 . Even so, we still need the help of a computer to prove one our propositions. However, this computer check involves minimal computation. Recall that the graph A5 is made up of 5 copies of A4 . We call these copies H1 , H2 , H3 , H4 and H5 substars of A5 . What we mean, specifically, by Hi is the substar of A5 that is made up of the even permutations where i is fixed in the 5th position. It is clear that between any two of these substars there are (5 − 2)! = 3! = 6 edges. Also remember that for any vertex v in say substar Hi , v has two neighbors that are not in the substar Hi . Not only that, but these two neighbors are themselves in different substars. We are now ready to proceed with the proof of Theorem 3.1. We have as sets the set S = {s1 , s2 , s3 } and the set T = {t1 , t2 , t3 }. We begin with the following lemmas. 4

Lemma 3.1. Let n = 5. Define Hij to be the subgraph of A5 induced by even permutations where i is fixed in the jth position. There exists j ∈ {3, 4, 5} such that S ∪ T 6⊆ Hij for i = 1, 2, 3, 4, 5. Proof. The proof is by contradiction. Assume that the conclusion is false. Then for every j ∈ {3, 4, 5} there is an i ∈ {1, 2, 3, 4, 5} such that S ∪ T ⊆ Hij . Hence, every element of S ∪ T will look like (−, −, x3 , x4 , x5 ). Thus, we get that |S ∪ T | = 1, a contradiction. ¤ Another lemma that we will prove of interest is the following. Recall that the graph A4 does not necessarily have the 2-disjoint path property. However, does there exist a condition where we are guaranteed that A4 does have the 2-Disjoint Path Property? The answer to this question is yes and the next lemma tells us the condition. Lemma 3.2. Let S = {s1 , s2 } and T = {t1 , t2 }. There are 2-disjoint paths from s1 to t1 and from s2 to t2 in A4 unless s1 , s2 , t1 and t2 are in a 4-cycle in this order. The proof of this lemma follows by a simple examination of cases an will be omitted. Another lemma that will be important to us is the following. In this lemma, we examine the structure of a vertex v of A4 and its neighbors Lemma 3.3. Let v be any vertex in A5 and suppose that v = (a, b, c, d, e). Then v is of distance one from Ha and Hb and of distance two from Hc and Hd . The number of 2-edge paths from v to the substars Hc and Hd are exactly two each. Proof. Simply examine the search tree of depth two rooted at v.

¤

Lemma 3.3 implies that if we take a vertex v from A5 then it has the following structure as seen in Figure 2. Before we give our next lemma let us consider the following. Let v be a vertex of A5 . Say v is in the the substar H5 . One of v’s adjacent vertices resides in another substar of A5 . We call this substar a substar neighbor of v. Lemma 3.4. Consider the substar Hj when j = 1, 2, 3, 4, 5 of the graph A5 . Let α, β ∈ {1, 2, 3, 4, 5} \ {j}. There are exactly two nodes v1 and v2 of Hj whose substar neighbors are Hα and Hβ . Proof. Let us consider, without loss of generality, the substar H5 . Let α and β be such that α 6= 5, β 6= 5 and α 6= β. We claim that there are exactly two vertices of H5 whose neighboring substars are Hα and Hβ . Let the vertex v1 have the form αβab5. The vertex v1 is certainly such a vertex in H5 since its two neighboring substars are Hα and Hβ . Also v2 of the form βαcd5 is another such vertex of H5 that has the desired property. The question is, how many possible v1 ’s and v2 ’s are there? Let us consider v1 . The number of possible permutations of the form v1 is two. But only 5

Ha

Hd

Hc +

+

+

+ v

+

+ +

+

+ +

Hb

Figure 2. The Structure Around v = abcde

half of them can be even. Hence, there can only be one v1 . By similar reasoning there can only be one v2 . Therefore, there can only be two such vertices. ¤ The proof of Theorem 3.1 proceeds as follows. We will be concerned with how the vertices of S ∪ T are distributed among the five substars of A5 . Define ψi = |V (Hi ) ∩ (S ∪ T )|. We look at the following 5P5 tuple:(ψ1 , ψ2 , ψ3 , ψ4 , ψ5 ). Notice that i=1 ψi = 6. So the cases that we need to consider are the following ten cases. These cases are (6, 0, 0, 0, 0), (5, 1, 0, 0, 0), (4, 2, 0, 0, 0), (4, 1, 1, 0, 0), (3, 3, 0, 0, 0), (3, 2, 1, 0, 0), (3, 1, 1, 1, 0), (2, 2, 2, 0, 0), (2, 2, 1, 1, 0), and (2, 1, 1, 1, 1). We go through each of these cases one at a time. 3.1. Case 1: (6, 0, 0, 0, 0). In this case we assume that S ∪ T ⊆ V (H1 ). This, however, violates Lemma 3.1. Therefore, this case cannot exist and so we are done. 3.2. Case 2: (5, 1, 0, 0, 0). In this case we shall show that (5, 1, 0, 0, 0), by re-coordination, can be reduced to one of the eight cases worked out below. 6

H3

H2 s3 t 3 + H1 s2

H4

t2 +

s1

t1

Figure 3. Nodes s2 and t2 Have Neighbors in Different Substars

Let us assume, without loss of generality, that s1 , s2 , s3 , t1 and t2 are in H5 and t3 is in H4 . Recall that H5 is isomorphic to A4 and thus, is made of four copies of A3 . The graph A3 is a triangle. Define Hi5 to be the subgraph of H5 where i is fixed in the fourth position. Clearly, Hi5 is isomorphic to A3 , H5 is made up of H15 , H25 , H35 , and H45 . Placing the five vertices in H5 , one cannot get more than three in any particular copy of A3 . By re-coordination, let Ji be a substar of A5 where i is fixed in the 4th position. So H25 is viewed now as part of J2 . Notice that t3 is assumed to be in H4 . But by the re-coordination t3 must be part of J1 , J2 , J3 , J4 or J5 . However, in H5 , Ji cannot have more than three vertices of S ∪ T . Considering H5 and H4 together Ji can have at most four vertices of S ∪ T . Thus, we are reduced to one of the eight cases below. 3.3. Case 3: (4, 2, 0, 0, 0). In this case we assume that H1 has four elements of S ∪ T and H2 has two elements of S ∪ T . Now note that if H2 contains a pair then H1 contains the other two pairs. So for this case, assume without loss of generality that s1 , t1 , s2 , and t2 are in H1 and s3 and t3 are in H2 . Clearly, we can route from s3 to t3 within H2 . Now consider 7

H2

H3

s3 t3

+ + s2 t2 s1 t1

H4

H1

Figure 4. Nodes s2 and t2 Have Neighbors in Same Substars

the neighboring substars of s2 and t2 . They may neighbor the same substar or different substars. See Figure 3 and Figure 4. In either case, it is clearly easy to route s2 to t2 . Now we can route from s1 to t1 . Since, we only need to avoid s2 and t2 , this can clearly be done as H1 is 4-connected. Now we consider the case where H2 does not contain a pair (si , ti ). Note that H1 must contain a pair. So let us assume, without loss of generality, that for this case we have that s1 , t1 , s2 and t3 are in H1 and s3 and t2 are in H2 . Consider the neighboring substars of s2 and t3 . We are only interested in the neighboring substars that are not H2 . So either s2 and t3 will have different substars or they will have the same substar neighbor. See Figure 5 and Figure 6. Let us look at the case where they are different as indicated in Figure 5. Let us say that t3 ’s neighbor is β in H4 . By Lemma 3.3, we can get from s3 to H4 in at most two steps, without touching t2 . We can therefore complete the path to t3 by going through β. Since, H2 is 4-connected we can route t2 to α in H3 , thus completing the path to s2 . Now we need to route from 8

H2

H 1

s3

s1 t1

t2 +

s2 t3

a

b +

H4

H3

Figure 5. The Nodes s2 and t3 Have Different Neighboring Substars

s1 to t1 . All we need to do in routing from s1 to t1 is to avoid s2 and t3 . This can be done as H1 is 4-connected. Let us look at the case where they are the same as indicated in Figure 6. Let us say that the neighbor of s2 is α and the neighbor of t3 is β, and both α and β reside in H3 . From α we can get to H4 in at most two steps. From t2 we can get to H4 in at most two steps. We, thus, can complete the path from s2 to t2 . Since, H2 is 4-connected, we can route from s3 to β thus completing the path to t3 . Now we need to route from s1 to t1 . All we need to do now is avoid s2 and t3 . This can be done as H1 is 4-connected. Hence, we have our three disjoint paths. 3.4. Case 4: (4, 1, 1, 0, 0). In this case we assume that H5 has four elements of S ∪ T , H4 has one element of S ∪ T and H3 has one element of S ∪ T . As in the previous case, if the element in H4 and the element in H3 form a pair then the four elements in H5 form the other two pairs. So without loss of generality let us assume that s1 , t1 , s2 and t2 are in H5 , s3 is in H4 , and t3 is in H3 . Now by Lemma 3.2, H5 has the 2-disjoint 9

H2

H1 s1

s3

t1 s2

t

2

t3 a

b H3 Figure 6. The Nodes s2 and t3 Have Same Neighboring Substars

path property unless s1 , s2 , t1 and t2 are on a 4-cycle in that order. So, if the pairs in H5 do not lie on a 4-cycle in the described order, then we are done in this case. But now let us assume that the two pairs in H5 do lie on a four cycle in the described order. There are six 4-cycles in H5 . One can check each one and see that the three disjoint paths exist. We do one here, the others being similar. One can see that there is a path for s3 to t3 completely contained in the substars H4 and H3 . Now let s1 = 13425, t1 = 42135, s2 = 34125 and t2 = 21435. Since s3 and t3 are in H4 and H3 respectively, we want to keep away from these substars. Observe that s1 has an neighbor in the substar H1 and t2 has a neighbor in H2 . Both H1 and H2 contain no elements of S ∪ T . So let α be the neighbor of s1 in H1 and let β be the neighbor of t1 in H2 . We complete the route from s1 to t1 by routing from α to β. The path from α to β is completely contained in the substars H2 and H1 , and is disjoint from the route from s3 to t3 . The route from s2 to t2 can be found, and this path does not touch s1 and t1 since H5 is 4-connected. 10

H4

H5

s1 s2

t1

H 3

s3 t2

t3

b

a H2

Figure 7. Nodes s2 and t3 Share a Common Substar

Now we assume that the elements of S ∪ T in H4 and H3 do not form a pair. Notice that H5 must contain a pair. So let us assume without loss of generality, that s1 , t1 , s2 , and t3 are in H5 , s3 is in H4 , and t2 is in H3 . We look at the neighboring substars of s2 and t3 . We have three cases to consider. The first case is that s2 and t3 both have the same neighboring substars and those substars are H4 and H3 . This makes things easy, since t2 is in H3 and s3 is in H4 . The paths from s2 to t2 and from s3 to t3 are easily found. The path from s1 to t1 has to avoid s2 and t3 . This can be done as H5 is 4-connected. The second case is that s2 and t3 both have the same neighboring substar, say H2 , but s2 ’s other substar neighbor is H4 , and t3 ’s other substar neighbor is H3 . Let α be the neighbor of s2 in H2 , and let β be the neighbor of t3 in H2 . It takes at most two steps to route from α to H3 . We can therefore complete the path from s2 to t2 . We can then route from β to H4 since H2 is 4-connected. Thus, we have a path from t3 to s3 . The path 11

H1

H5

H4

s1 t 1

s3 b

t3 s2

a

t 2

H 3

H 2

Figure 8. Nodes s2 and t3 Do Not Share a Common Substar

from s1 to t1 can be found since we need only to avoid s3 and t3 . This can be done as H5 is 4-connected. See Figure 7. The final case is that s2 and t3 have different substar neighbors. The solution becomes clear here. The path from s1 to t1 only needs to avoid s2 and t3 . This can be done as H5 is 4-connected. See Figure 8. 3.5. Case 5: (3, 3, 0, 0, 0). In this case we assume that the substars H1 and H2 have three elements of S ∪ T . We look at several cases and show that the three disjoint paths exist. First, let us consider the easy case. The easy case is that the substars H1 and H2 each contain a pair. So, without loss of generality, let us suppose that s1 , t1 , and s2 are in the substars H1 , and that s3 , t3 , and t2 are in H2 . There are three ”empty” substars that we can choose from. Notice that s2 can reach H3 , H4 , or H5 in at most two steps. Same thing for t2 . So pick which ever is available. Then complete the path from s2 to t2 . A path from s1 to t1 is easily found since in H1 we need only to avoid up to two vertices. The case from s3 to t3 is similar. See Figure 9. Hence, we are done with the easy case. 12

H1

H2

t1

t3

s1

s3 H5

H4

t2

s2

x

y

Figure 9. The Easy Case for (3, 3, 0, 0, 0)

Now we look at the case where H1 does not contain a pair. Then the other substar H2 cannot contain a pair. So assume, without loss of generality, that the vertices s1 , s2 and s3 are in H1 and t1 , t2 and t3 are in H2 . We consider several cases. There are two main cases that we shall look at. The first case is that there exists a pair (si , ti ) such that si and ti share a common substar neighbor. So for example, if the pair (s1 , t1 ) was such that s1 ’s neighboring substar was H1 and t1 ’s neighboring substar was also H1 we would then fall under this case. The other main case is that no pair (si , ti ) share a common substar. Let us consider the first main case. That is there exists a pair (si , ti ) such that si and ti share a common substar, say H4 . There are three cases here. The first case is that there is exactly one pair that shares H4 as a common substar neighbor. The second case is that there are exactly two pairs that share H4 as a common substar neighbor. And the third case is that there all three pairs share H4 as a common substar neighbor. We look at these three cases and solve each one. 13

H1

H 2 t3

s 3

t2

s 1 t1 s2 x

H4

H3

+

H 5 +

+ +

Figure 10. (s1 , t1 ) Share H4 as a Neighboring Susbstar

The first case that we look at is that there is exactly one pair that shares H4 as a common substar neighbor. Let us assume that this pair is (s1 , t1 ). Consider the neighboring substar of s2 and t2 . Both can’t have H4 as a neighboring substar, since that would violate the hypothesis of this case. So, let us assume that t2 ’s substar neighbor is H3 . We can route from s2 to either H3 or H5 in at most two steps. Choose which ever is available. Assume that we can route from s2 to H5 in two steps. See Figure 10. Then we can complete the path from s2 to t2 by routing in the substar H5 and H3 . The route from s1 to t1 is completed in the substar H4 . All we need to do is complete a path from s3 to t3 . Recall that there are six edges between the substars H1 and H2 . From Figure 10 one can see that the nodes s2 , α, s1 , t1 and t2 can touch at most five of these edges. So there is an edge available. We can route s1 to that edge without touching s1 , α or s2 . Same thing in H2 . This can be done because H1 and H2 are 4-connected. Hence, we are done with this case. The next case is that there are exactly two pairs that share the substar H4 as a common substar neighbor. Let us assume that these pairs are 14

H4

H1

H2

A

s1 s 2

t1

C D

B

t2 t3

s a 3

H 5

H 3

+

+

Figure 11. (s1 , t1 ) and (s2 , t2 ) Share H4 as a Common Substar Neighbor

(s1 , t1 ) and (s2 , t2 ). Now assume that neighbor of s1 in H4 is A, the neighbor of s2 in H4 is B, the neighbor of t1 in H4 is C and the neighbor of t2 in H4 is D. Now suppose that A, B, C and D do not lie on a 4-cycle. Then disjoint pairs from s1 to t1 and from s2 to t2 can be found within the substar H4 . The nodes s1 , s2 , t1 and t2 touch at most four of the six edges between the substars H1 and H2 . Thus, a path from s3 to t3 may be completed. Now let us assume that the nodes A, B , C and D do lie on a 4-cycle. Look at the neighboring substars of s3 and t3 . Both cannot have H4 as a neighboring substar since we would violate the hypothesis of this case. So assume that t3 has H3 as a neighboring substar. We can get from s3 to either H3 or H5 in at most two steps. Assume that we can get from s3 to H5 in two steps. See Figure 11. A route from s3 to t3 can be found by using the substars H5 and H3 . A route from s1 to t1 may be found by using the substar H4 . The nodes s3 , α , s1 , t1 and t3 touch at most five edges between H1 and H2 , and so a path from s2 to t2 may be completed. 15

H2

H1 s1

t1

s2

t 3 H3

s3

H

+

t2

4

+

H5

+

+

Figure 12. Situation 1

Consider Figure 11 and the case in the previous paragraph. Suppose that in the extreme case A = D and B = C. To solve this case one uses the exact same argument as discussed in the previous paragraph and so we are done with this subcase. Now we consider the case that all three pairs (s1 , t1 ), (s2 , t2 ) and (s3 , t3 ) share H4 as a common substar neighbor. Consider the other substar neighbor of the nodes in S ∪ T . Notice that s1 , s2 and s3 cannot all have H2 as a substar neighbor. If it did we would violate Lemma 3.4. So assume that t3 ’s substar neighbor is H5 . The node s3 can reach to either H3 or H5 in at most two steps. Let us say that s3 can be routed to H3 in two steps. We can find our disjoint paths in the usual way as described in the previous paragraphs. So we are done with this main case. Now we move on to the second main case where no pair share a common substar neighbor. Consider Figure 12, Figure 13, and Figure 14. We observe in our next theorem that only one of these situations can occur. 16

H2

H1

t1

s1

t2

s2

t 3

s3

H4

H3

+ + +

+

Figure 13. Situation 2

Proposition 3.1. Let s1 , s2 , s3 ∈ H1 and t1 , t2 , t3 ∈ H2 . Assume no pair (si , ti ) for i = 1, 2, 3 share a common substar neighbor. Then only one of situations depicted in Figure 12, Figure 13, or Figure 14 can occur. Proof. The proof of this theorem is through a simple computer check. We first wrote a computer program, using Matlab, that generates the pairs (s1 , t1 ), (s2 , t2 ) and (s3 , t3 ) with the property that no pair has a common substar neighbor. We then wrote a computer program that verifies that only one of the above situations can occur. Doing this for all possible pairs we verified the truth of our theorem. Matlab code and a description of our experiment are described in the Appendix of this paper. ¤ Having Proposition 3.1 in hand, we proceed in showing that in each situation one can find the desired disjoint paths. Let us consider situation 1 illustrated in Figure 12. To solve this case we can route s2 to the substar H5 in at most two steps and then complete the path to t2 . At most four edges between H3 and H4 will be eliminated. So complete the route for 17

H1

H2

si

tj Hk

y

Figure 14. Situation 3

s3 to t3 . At most four edges will be eliminated between H1 and H2 so complete the route from s1 to t1 Now consider situation 2 illustrated in Figure 13. To solve this case notice that we can finish the route from s2 to t2 . At most four edges will be eliminated between the substars H3 and H4 . So a path from s3 to t3 can be found. At most four edges will be eliminated between the substars H1 and H2 . So a path from s1 to t1 can be found. Now consider situation 3 as illustrated in Figure 14. Assume, without loss of generality, that s1 is adjacent to t2 and both share the same neighbor in H3 . The nodes s2 and t3 are adjacent and both same the share neighbor in H4 . And the nodes s3 and t1 are adjacent and both share the same neighbor in H5 . How do we resolve this situation and others like it? All are similar and are solved in the same way. For example suppose s1 = 24315 and is adjacent to t2 = 52314. Suppose s2 = 14235 and is adjacent to t3 = 51234. Finally, suppose s3 = 34125 and is adjacent to t1 = 53124. Here are the disjoint paths. From s1 to t1 we get t1 to 25134 to 54132. From 54132 we find a path to 45312 and then to s1 . From s2 to t2 we get 18

s2 to 42135 to 25134. From 25134 we can find a path to t2 since need only avoid t3 , t1 and 25134. Now for s3 to t3 . We go from t3 to 45231 to 43521 to 14523 to 45123 to s3 . We use this example now to prove our general case. Recall that each v in A5 has a a structure depicted in Figure 2. We first route s1 to t1 . The node s1 has a neighbor in H3 . The node t1 can get to H3 in two steps since t1 ’s neighbors are in substars H5 and H2 . Also t2 ’s neighboring substar is H3 , so there is path from t1 to H3 . Let α be the node in H2 that is along this path. Hence, we have path from s1 to t1 . Now we route s3 to t3 . The node s3 has a neighbor in the substar H5 and the node t3 has a neighbor in H4 . So the path from s3 to t3 can be found in the substars H4 and H5 . Now all that is left to do is construct a path from s2 to t2 . There are six edges between the substars H1 and H2 . The nodes s1 , s3 , t3 , α and t1 touch at most five of these edges. Hence, we may construct a path from s2 to t2 using the nodes of H1 and H2 since there is an edge available and H1 and H2 are 4-connected. Hence, we have our disjoint paths and we are done with this case. 3.6. Case 6: (3, 2, 1, 0, 0). In this case we have that the substar H1 has three elements of S ∪ T , H2 has two elements and H3 has one element. We start with the easy case. The easy case is that H1 contains a pair and the two elements in H2 form a pair. Let us assume, without loss of generality, that s1 , t1 and s2 are in H1 , s3 and t3 are in H2 and t2 is in H3 . To solve this case we look at the neighboring substars of s2 . There are two of these substars. Looking at all the possibilities it is clear that s2 will have neighbors in either H4 , H5 or H3 . If the neighboring substar is H3 then we can clearly route to t2 . The route from s3 to t3 is completely contained in the substar H2 . The route from s1 to t1 can be found as H1 is 4-connected. If the neighboring substars are H4 or H5 , the same type of argument holds again and we are done. The next case that we consider is that H1 has a pair but H2 does not. Let us assume without loss of generality that s1 , t1 and s2 are in H1 , and that s3 and t2 are in H2 and t3 is in H3 . We first want to route s2 to t2 . We look at the substar neighbors of s2 . There are two of them. If the immediate substar neighbors of s2 are H4 or H5 then we are done as follows. Let us assume that s2 ’s neighbor is in H4 . By Lemma 3.3 we can route from t2 to H4 in at most two steps. Thus, we can complete the from s2 to t2 . We can route from s3 to t3 since H2 is 4-connected. Finally, we can route from s1 to t1 again as H1 is 4-connected. See Figure 15. Now suppose that the immediate substar neighbors of s2 are H2 and H3 . Let α be the neighbor of s2 in H2 and β is the neighbor of s2 in H3 . Suppose that α 6= s3 . Then route s3 to one of the substars H3 , H4 , or H5 . This can be done in at most two steps. Then complete the path to t3 . To route from α to t2 can be done as H2 is 4-connected. Thus, we have a path from s2 to t2 . The from s1 to t1 can be found in H1 as H1 is 4-connected. See Figure 16. 19

H2

H1

s3

s1 t1

H3

t3

t2 s2

y H 4

Figure 15. H1 Contains a Pair But H2 Does Not

Now suppose that α = s3 . Since the immediate substar neighbors of s2 are H2 and H3 , we can reach H4 and H5 in at most two steps. One path may be blocked by s1 and t1 , but then just pick the other. Let’s assume that it is H4 . The node t2 can be routed to H4 in at most two steps. Do this and complete the path to s2 . We can then clearly complete the paths for s1 to t1 and s3 to t3 as before. See Figure 17 The last case here is that the elements of contained in H1 and H2 do not form a pair. Let us assume, without loss of generality, that s1 , s2 and s3 are in H1 , and that t1 and t2 are in H2 and t3 in H3 . We look at the neighboring substars of s3 . In all possible cases the neighboring substars of s3 will be either H3 , H4 or H5 . If the neighbor is in H3 then we can clearly complete the path from s3 to t3 . Now route s2 to either H4 or H5 , and complete the path to t2 in H2 . Note that t2 can route to either H4 or H5 in at most two steps. Routing s1 to t1 can be accomplished as both H1 and H2 are 4-connected. Let us say now that both of s3 ’s neighboring substars are both H4 and H5 . Say we pick H4 . Let the neighbor of s3 be α in H4 . Now we can route from 20

H2

H1

s1

a

H3

t3

s3

t1

b s 2

t2

Figure 16. Node α 6= s3

s2 to either H4 or H5 in at most two steps. In the worst case, we are forced to route s2 to H4 . We get to H4 from s2 to γ to β. Notice that β is in H4 . Also notice that β 6= α. Now we can route β to H5 in at most two steps. Then route t2 to H5 in at most two steps. Complete the route from s2 to t2 . To complete the route from s3 to t3 , route α in H4 to t3 in H3 . This can be done as H4 is 4-connected. Now route s1 to t1 in H2 . The route has to miss at most three vertices in H1 and two vertices in H2 . This can be done as H1 and H2 are 4-connected. See Figure 18. We are done with this case. 3.7. Case 7: (3, 1, 1, 1, 0). In this case we must address several cases and their subcases. Let us assume first, that the substar H1 has three elements of S ∪ T . We first consider the easy case. Let us assume that of three elements in H1 , two of them form a pair. Hence, assume that s1 , t1 and s2 are in substar H1 , t2 is in the substar H2 and s3 is in H3 , and t3 is H4 . Recall that s2 has two neighbors outside the substar H1 . We are interested in these neighbors. If one of those neighbors are in H2 , or in H5 then we are done. We will consider the neighbor of s2 to be α and α is in H5 . The case 21

H1

H 3

H2

s1

s3

t

t1 s2

t +

3

2

H4

+

Figure 17. Node α = s3

of α in H2 is similar. Notice that we can route now from s1 to t1 without touching s2 as H1 is 4-connected. The path from s2 to t2 is completed by routing from α to t2 . The path from s2 to t2 is completely contained in the substars H1 , H5 and H2 and does not touch the path from s1 to t1 . The path from s3 to t3 can clearly be found and is obviously contained completely in the substars H3 and H4 . Please see Figure 19. Now it might be possible that a route from s2 to H5 will take two steps. If this happens we use a similar argument to find the disjoint paths. The next case that we consider is that none of the three elements in H1 form a pair. Hence, we shall assume that, without loss of generality, s1 , s2 and s3 are in the substar H1 , and that t1 is in H2 , t2 is in H3 and t3 is in H4 . We look at the substar neighbors of the element s3 . We consider the first case. Suppose that α is the neighbor of s3 in the substar H2 , and β is the neighbor of s3 in H3 . Suppose that at least one of α or β is not equal to t1 or t2 , respectively. Let β 6= t2 . Now from β we can route to t3 in H4 , without touching t2 . Thus, we complete the path from s3 to t3 . We can route from s1 to H5 in at most two steps. We are guaranteed that 22

H1

H2

s1

t1

s2 g

H3

t3

t 2

s3

b a

H 4

Figure 18. Special Case

s3 is not on one of these 2-paths because if it were we violate Lemma 3.3. From H5 we route to H2 and the complete the route to t1 . To route s2 to t2 we have at most three vertices to miss in H1 . This can be done as H1 is 4-connected. Thus, we have our three disjoint paths. The case in which α 6= t1 , but β = t2 is similar. See Figure 20. Now suppose that α = t1 in H2 and that β = t2 in H3 . Notice that at most one of s1 or s2 will have both substars H2 and H3 as neighbors. Suppose that s1 has γ as a neighbor in H2 . Note that γ cannot be α = t1 . It does not matter since we can complete the path to t1 easily. Now s3 can get to either H5 or H4 in two steps. Let us say that we are forced to H5 . Then we can complete the path to t3 by going from H5 to H4 . A path from s2 to t2 can be found because we need only to avoid three vertices in H1 . Hence, we are done. See Figure 21. Now suppose that the neighbor of s1 is in H3 . Again this neighbor will not be t2 . Let this neighbor in H3 be δ. Again to complete the path to t1 in H2 we can get from δ to H2 in at most two steps. The rest follows as before. Now suppose the neighbor of s1 are the substars H4 and H5 . Let γ be the neighbor of s1 in H5 . Now consider 23

s

H1

H2

1

t

t1

2

H3

H 4

s 3

t3

s2

a

H

5

Figure 19. Easy Subcase of Case 7

s2 . We can get from s2 to H2 or H3 in at most two steps. If we can reach H3 , then do so. If not, suppose that we are forced to H2 . Note that the vertex we reach in H2 will not be t1 . So we complete the route to H3 and thus to t2 . We can now complete the route from s1 to t1 . The path from s3 can be completed now since we need only to avoid up to three vertices in H1 . See Figure 22. Suppose that the neighboring substar of s3 is now H4 . Then a path from s3 to t3 can easily be found. We can get from s1 to H5 or H2 in at most two steps. If we can get to H2 then do so and complete the path. If not, and we are forced to H5 , then we complete the path to t1 via H5 . A path from s2 to t2 can be done since the path that needs to be found need only miss up to three vertices in H1 . Suppose s3 ’s neighboring substar is H5 . Then the same type of argument as before will give us our paths. Hence.we are done with this case. 3.8. Case 8: (2, 2, 2, 0, 0). In this case we have three subcases that we must consider. The first case is trivial. Let H1 , H2 and H3 each contain two elements of the set S ∪T . Assume that the two elements in each substar 24

H1 s + 1

H5

H4

+

t2

s2 b s3

a t1 H 2 Figure 20. Hard Case α 6= t1

form a pair. Further assume, without loss of generality, that the pair in Hi is (si , ti ). Then there is nothing to do. Clearly, the path from si to ti is completely contained in its substar Hi . So we are done with this case. In the next case we assume that H1 , H2 and H3 each contain two elements of the set S ∪ T and that only the two elements in H1 form a pair. Thus, let us assume without loss of generality that s1 , and t1 are in H1 , s2 and t3 are in H2 and s3 and t2 are in H3 . This case is not difficult to resolve. First, it is easy to see that we can find a path from s1 to t1 that is completely contained in the substar H1 . Now we need to find paths from s2 to t2 and from s3 to t3 . By the above Lemma 3.3 we know that we can reach from s2 to the substar H4 in at most two steps. Likewise, we can reach from t2 to H4 in most two steps. Note that we need not worry about t3 nor s3 touching these paths. Let the path from s2 to H4 be s2 αγ, where γ is in H4 . Let the path from t2 to H4 be t2 βδ, where δ is in H4 . Thus, the path from s2 to t2 is given by s2 αγ, ..., δβt2 . Note that this path is completely contained in the substars H2 , H3 and H4 , and so is disjoint to the path from s1 to t1 . Now we need a path from s3 to t3 . Recall that there 25

H3

H1

t

s2

2

s 3x

H4

t

3

y s1

t1 g H2

H5

Figure 21. Case Where α = t1 and β = t2

are 6 edges between the substars H2 and H3 . The vertices s2 and α touch at most two of these edges. Likewise the vertices t2 and β touch at most two of these edges. Thus, at most four edges are touched leaving at least two edges available. Let x be a vertex in H2 and let y be a vertex in H3 such that x and y are connected by one of these available edges. Since H2 is 4-connected, there is path from t3 to x without touching the vertices s2 and α. Likewise, there is a path from s3 to y without touching the vertices t2 and β. Hence, the path from s3 to t3 can be completed. See Figure 23. In this final case we assume that the substars H1 , H2 and H3 contain two elements of the set S ∪ T and that none of the two elements in each substar form a pair. So let us assume, without loss of generality, that s1 and t2 are in H1 , s2 and t3 are in H2 and s3 and t1 are in H3 . We have two available substars H4 and H5 . We will make use of these substars. By Lemma 3.3, we know that we can reach from s1 to H4 in at most two steps. We need not worry about t2 touching one of these paths. If it does then just choose the other. Likewise, we can reach from t1 to H4 in at most two steps. Let the path from s1 to H4 be s1 αγ, where γ is in H4 . Let 26

H 1

H 5

s1

g

s2

s

3

+

H2 +

t 1

H 3

t2

t3

H

4

Figure 22. Case Where α = t1 and β = t2

the path from t1 to H4 be t1 βδ, where δ is in H4 . Thus, the path from s1 to t1 is given by s1 αγ, ..., δβ, t1 . This path is completely contained in the substars H1 , H3 and H4 . By Lemma 3.3, we can reach from t3 to H5 in at most two steps. We need not worry about s2 touching one of these paths since if it did we would just choose the other. Let the path from t3 to H5 be t3 ²ρ, where ρ is in H5 . Now we are in a position to compute the remaining paths. Between the substars H1 and H2 are six edges. The vertices s1 and α touch at most two of those edges. Likewise, the vertices t3 and ² touch at most two of these edges. Hence, at most four edges are touched leaving at least two untouched. Let x be a vertex in H1 , and y a vertex in H2 such that x and y are connected by one of these remaining edges. Since H1 is 4-connected there exists a path from t2 to x that does not touch the vertices s1 and α. Likewise, there exists a path from s2 to y that does not touch the vertices t3 and ². Hence, we can complete the path from t2 to s2 . This path is completely contained in H1 and H2 and does not touch the path from s1 to t1 . Now all we need is to show the existence of a disjoint path from s3 to t3 . Between the substars H3 and H5 are six 27

H3

H2

t2

t 3 y

x s2

s3

b

a

d g H 4

Figure 23. H1 Contains the Pair (s1 , t1 )

edges. The vertices t1 and β touch at most two of these edges. Thus, we have at least four edges available. Let z be a vertex in H3 touching one of these available edges. Since, H3 is 4-connected there is a path from s3 to z without touching the vertices t1 and β. Hence, we have a path from s3 to t3 that goes only through the substars H3 , H5 and H2 . All these paths are clearly disjoint. See Figure 24. 3.9. Case 9: (2, 2, 1, 1, 0). In this case we have three subcases that we must address. The first case is easy. Let the substars H1 and H2 each have two elements of the set S ∪ T . Assume, the two elements in H1 form a pair, and the two elements in H2 also form a pair. So for this case, we shall assume, without loss of generality, that the pair s1 and t1 are in the substar H1 and the pair s2 and t2 are in the substar H2 . Let us also assume that s3 in H3 , and that t3 is in H4 . It is very easy to see that we can find paths from s1 to t1 , s2 to t2 and s3 to t3 that are completely disjoint from one another. 28

H 2

H1 s1

H3

s 2

t2

s 3

y

x

t

t3 e

a

z

1

b

+ r g d H 5

H4

Figure 24. H1 , H2 and H3 Do Not Contain Pairs

The next case is easy as well. Assume that H1 and H2 have each two elements of S ∪ T . Assume that the two elements of H1 form a pair. Let us assume, without loss of generality, that this pair is (s1 , t1 ). Assume now, that s2 and t3 are in H2 , t2 is in H3 and s3 is in H4 . We can get a path from s1 to t1 completely contained in H1 . Now we need paths from s2 to t2 and s3 to t3 . By Lemma 3.3, we know that we can reach from s2 to H3 in at most two steps. We need not worry if t3 is on one of these paths. If it is just choose the other. Let the path from s2 to t2 be s2 α...t2 . Note that this path is completely contained in the substars H2 and H3 . Recall that there are six edges between the substars H2 and H4 . The vertices s2 and α touch at most two of these edges. Let y be a vertex touching one of the four remaining edges. Since, H2 is 4-connected then there exists a path from t3 to y without touching the vertices s2 and α. Hence, the path from s3 to t3 can be completed. Thus, we have our three disjoint paths. See Figure 25. This final case is the hardest, but still easy to resolve. Let us assume that the substars H1 and H2 each have two elements of S ∪ T . We assume 29

H3

H2 t3 s

2

t

2

a x

y

+ s 3 H

4

Figure 25. H1 Contains a Pair (s1 , t1 )

that the two elements in H1 and that the two elements in H2 do not form pairs. Hence, let us assume without loss of generality that s1 and t2 are in H1 , s2 and t3 are in H2 , t1 is in H3 , and s3 is in H4 . By Lemma 3.3, we can get from s1 to the substar H3 in at most two steps. Let us assume that the path from s1 to t1 is given by s1 α, ...t1 . Note that this path is completely contained in the substars H1 and H3 . Again, by Lemma 3.3, we can reach from t3 to H4 in at most two steps. Let us assume that the path from t3 to s3 is given by t3 β, ...s3 . Note that this path is completely contained in the substars H2 and H4 . Also, note that the paths between s1 and t1 and between s3 and t3 are disjoint. Finally, as before, we know that the vertex t2 does not touch the path from s1 to t1 and that the vertex s2 does not touch the path from t3 to s3 . Now all we need to do is find a path from s2 to t2 . We know that between the substars H1 and H2 are six edges. The vertices s1 and α touch at most two of these edges. Likewise, the vertices t3 and β also touch at most two of these edges. So at most four edges are eliminated, thus leaving at least two available. Let x be in H1 and let y be in H2 and x and y are connected by one of these available edges. Since 30

s s1

2

+ b

t2 a

H4

H 2

H1

y

x t3

s3

+

t H 3

Figure 26. H1 and H2 Do Not Have Pairs

H1 is 4-connected there exists a path from t2 to x without touching s1 and α. Likewise, there exists a path from s2 to y without touching t3 and β. Hence, we can complete that path from t2 to s2 . Therefore, we have our three disjoint paths and we are done. See Figure 26. 3.10. Case 10: (2, 1, 1, 1, 1). In this case we have two subcases that we must address. The first case is easy. It is the case where the two elements in a substar are an actual pair. Assume that this pair is (s1 , t1 ) and that it is in H1 . Let s2 be in H2 , t2 in H3 , s3 in H4 , and t3 in H5 . This done without loss of generality for this case. It is easy to see that we can find the three disjoint paths. The second case now assumes that the two elements in the substar is not a pair. So assume, without loss of generality, that s1 and t2 are in the substar H1 . Let s2 be in H2 , t1 in H3 , s3 in H4 and t3 in H5 . It is easy to see that we can find a path from s3 to t3 that is completely contained in the substars H4 and H5 . Now we need paths from s1 to t1 and from s2 to t2 . From Lemma 3.3 we know that we can get from s1 to the substar H3 in at 31

H 2

H1 +

t2 y

s2 s1

a

b t

1

H 3

Figure 27. Second Case for (2, 1, 1, 1, 1)

most two steps. There are two such paths so we need not worry if t2 is on one of these paths. If it is just choose the other. Let the path from s1 to t1 be s1 αβγ...t1 . Note that this path is completely contained in the substars H1 and H3 . Now note that vertices s1 and α are in the substar H1 . Recall that there are six edges between the substars H1 and H2 . The vertices s1 and α touch at most two of these edges. Let y be a vertex touching one of the four remaining. Since, H1 is 4-connected there is a path from s2 to y without touching s1 and α. Hence, we can complete the path from s2 to t2 . We therefore have our three disjoint paths. See Figure 27. Hence, we are done with this case. Putting these cases together establishes Theorem 3.1 and so we are done. 4. Conclusion The object of this paper was to show that the alternating group graph An has the (n−2)-Disjoint Path Property. This result is not just interesting theoretically, but as described in the motivation section, this result also has a practical aspect as well. 32

A question to ask is what is the next step in our research. One thing to consider is that the proof that we presented is an existence proof. What we would like to do now is to develop an algorithm that will actually construct the (n − 2)-disjoint paths. A first place to start would be at the base case level and then try to generalize. Another question to consider are directed versions of the alternating group graph. Work has been done in the literature to orient the alternating group graph. Please see [5]. It would be interesting to develop a directed version of the k-Disjoint Path Problem. Such a problem also has a firm footing in the real world of interconnection networks. Consider an interconnection network in which something happens to cause the communication links to only work one way. Can the interconnection network still be operational and can we write down conditions in which the k-disjoint paths still be found? There are many other things to consider in the study of these type of networks. This paper only represents a beginning. 5. Appendix In this appendix we outline the experiment done for the case (3, 3, 0, 0, 0) discussed in the previous section. In the second part of our main case we assumed that no pair (si , ti ) had a common substar. We conjectured that our case gave rise to only one of three possible situations. The experiment we ran verified our conjecture. The outline of this Appendix is as follows. We first give the Matlab code of our programs. A brief description of each program is given. From there we describe, in general, our experiment. We end with a particular instance of our experiment. 5.1. Matlab Code. Given a node t1 we want a listing of all possible s1 ’s that fit our hypothesis. That is a s1 is a candidate if and only if s1 does not share a common substar neighbor with t1 . The following Matlab code does this. If the reader is interested in the obtaining the Matlab source code please contact the author at [email protected]. function [v]=candid(t1,s) v=[] i=1 a=t1(1); b=t1(2) while(i~=13) if (s(i,1)==a or s(i,2)==a or s(i,1)==b or s(i,2)==b) v=v else v=[v;s(i,:)]; end i=i+1 33

end v The following function compares two vectors and see if they are equal. function t=compare(x,y) if (x==y) t=1; else t=0; end The following function does a left rotation as described in Chapter 2. function v=leftr(x); v=x; temp=v(2); v(2)=v(5) v(5)=v(1) v(1)=temp In a similar way we have the right rotation. function v=rightr(x) v=x temp=v(1) v(1)=v(5) v(5)=v(2) v(2)=temp The following function takes on the pairs and outputs a 1 if it fits any one of the three situations described in Chapter 6. function y=allca(s1,t1,s2,t2,s3,t3) y=0 y=case1(s1,t1,s2,t2,s3,t3) if y==0 y=case2(s1,t1,s2,t2,s3,t3) end if y==0 y=case3(s1,t1,s2,t2,s3,t3) end The following three functions determine if the pairs (si , ti ) for i = 1, 2, 3 fit into one of three situations hypothesized in our theorem. The function y = case1 corresponds to situation number 2. The function y = case2 corresponds to situation number1. And the funtion y = case3 corresponds to situation 3. function y=case1(s1,t1,s2,t2,s3,t3) a1=s1(1) ; a2=s1(2) b1=t1(1); b2=t1(2) 34

c1=s2(1); c2=s2(2); d1=t2(1); d2=t2(2); e1=s3(1); e2=s3(2); f1=t3(1); f2=t3(2); y=0; if ((c1==e1) || (c1==e2) || (c2==e1) || (c2==e2)) & ((d1==f1) || (d1==f2) || (d2==f1) || (d2==f2)) y=1; if (y==0) if ((a1==e1) || (a1==e2) || (a2==e1) || (a2==e2)) & ((b1==f1) || (b1==f2) || (b2==f1)|| (b2==f2)) y=1; end; end; if (y==0) if ((a1==c1) || (a1==c2) || (a2==c1) || (a2==c2)) & ((b1==d1) || (b1==d2) || (b2==d1)|| (b2==d2)) y=1; end end function y=case2(s1,t1,s2,t2,s3,t3) n1s1=leftr(s1); n2s1=rightr(s1); n1s2=leftr(s2); n2s2=rightr(s2); n1s3=leftr(s3) n2s3=rightr(s3) n1t1=leftr(t1) n2t1=rightr(t1) n1t2=leftr(t2) n2t2=rightr(t2) n1t3=leftr(t3) n2t3=rightr(t3) y=0; T=[n1s2;n2s2;n1s3;n2s3;n1t2;n2t2;n1t3;n2t3] T1=[n1s2(5),n2s2(5),n1s3(5),n2s3(5),n1t2(5), n2t2(5),n1t3(5),n2t3(5)] J=find(T1==2) I=size(J) if (I(2)==2) tr=compare(T(J(1),:),T(J(2),:)); 35

if (tr==0) k=sum(find(T1==1)); l=sum(find(T1==3)); if ( (k>=1) & (l>=1)) y=1; end end end if (y==0) J=find(T1==1) I=size(J); if(I(2)==2); tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==2)); l=sum(find(T1==3)); if ( (k>=1) & (l>=1)) y=1; end end end end if (y==0) J=find(T1==3); I=size(J); if(I(2)==2); tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==2)); l=sum(find(T1==1)); if ( (k>=1) & (l>=1)) y=1; end end end end if (y==0) T=[n1s2;n2s2;n1s3;n2s3;n1t2;n2t2;n1t3;n2t3] T1=[n1s1(5),n2s1(5),n1s3(5),n2s3(5),n1t1(5), n2t1(5),n1t3(5),n2t3(5)] J=find(T1==2) I=size(J) if (I(2)==2) 36

tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==1)); l=sum(find(T1==3)); if ( (k>=1) & (l>=1)) y=1; end end end if (y==0) J=find(T1==1) I=size(J); if(I(2)==2); tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==2)); l=sum(find(T1==3)); if ( (k>=1) & (l>=1)) y=1; end end end end if (y==0) J=find(T1==3); I=size(J); if(I(2)==2); tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==2)); l=sum(find(T1==1)); if ( (k>=1) & (l>=1)) y=1; end end end end if(y==0) T=[n1s2;n2s2;n1s3;n2s3;n1t2;n2t2;n1t3;n2t3] T1=[n1s1(5),n2s1(5),n1s2(5),n2s2(5),n1t1(5), n2t1(5),n1t2(5),n2t2(5)] J=find(T1==2) I=size(J) 37

if (I(2)==2) tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==1)); l=sum(find(T1==3)); if ( (k>=1) & (l>=1)) y=1; end end end if (y==0) J=find(T1==1) I=size(J); if(I(2)==2); tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==2)); l=sum(find(T1==3)); if ( (k>=1) & (l>=1)) y=1; end end end end if (y==0) J=find(T1==3); I=size(J); \if(I(2)==2); tr=compare(T(J(1),:),T(J(2),:)); if (tr==0) k=sum(find(T1==2)); l=sum(find(T1==1)); if ( (k>=1) & (l>=1)) y=1; end end end end end function y=case3(s1,t1,s2,s3,t3) n1s1=leftr(s1) 38

n2s1=rightr(s1) n1s2=leftr(s2) n2s2=rightr(s2) n1s3=leftr(s3) n3s3=rightr(s3) n1t1=leftr(t1) n2t1=rightr(t1) n1t2=leftr(t2) n2t2=rightr(t2) n1t3=leftr(t3) n2t3=rightr(t3) y=0; if (n1s1(5)==4) x1=compare(n2s1,n1t2); x2=compare(n2s1,n2t2); x3=compare(n2s1,n1t3); x4=compare(n2s1,n2t3); else x1=compare(n1s1,n1t2); x2=compare(n1s1,n2t2); x3=compare(n1s1,n1t3); x4=compare(n1s1,n2t3); end if (n1s2(5)==4) x5=compare(n2s2,n1t1); x6=compare(n2s2,n2t1); x7=compare(n2s2,n1t3); x8=compare(n2s2,n2t3); else x5=compare(n1s2,n1t1); x6=compare(n1s2,n2t1); x7=compare(n1s2,n1t3); x8=compare(n1s2,n2t3); end if (n1s3(5)==4) x9=compare(n2s3,n1t2); x10=compare(n2s3,n2t2); x11=compare(n2s3,n1t1); x12=compare(n2s3,n2t1); else x9=compare(n1s3,n1t2); 39

x10=compare(n1s3,n2t2); x11=compare(n1s3,n1t1); x12=compare(n1s3,n2t1); end y1= x1 || x2 || x3 || x4 y2= x5 || x6 || x7 || x8 y3= x9 || x10 || x11 || x12 z=y1 & y2 & y3; if z==1 y=1; end

5.2. Experiment Description. In this section we describe,in general, our experiment. The purpose of our experiment is to provide a computer check of our main theorem in Case(3, 3, 0, 0, 0). Our theorem assumes that given (si , ti ) for i = 1, 2, 3, no pair share a common substar as a neighbor. The first part of the experiment involves generating the pairs (si , t) for i = 1, 2, 3. Due to vertex symmetry of A5 , we let t1 = 31524 throughout the whole experiment. The nodes s1 , s2 , and s3 come from H5 . The nodes of H5 are 13425, 21435, 32415, 24315, 43215, 41325, 14235, 12345, 31245, 23145, 34125 and 42135. We put these nodes in the matrix s of our code. The nodes t1 , t2 , and t3 come from the substar H4 . These nodes are 12534, 23514, 52314, 35214, 15324, 51234, 21354, 13254, 32154, 53124, and 25134. We put these nodes in matrixt. By using the function candid(t1 , s) we get two possible nodes for s1 . The node s1 = 24315 or s1 = 42135. Hence, we must run two sets of experiments. The first set of experiments involve s1 = 24315 and t1 = 31524 and the second set of experiments involve s1 = 42315 and t1 = 31524. The node t2 is selected one at a time from the matrix t excluding t1. All potential s2 ’s are found using candid(t2 , s), where the matrix s contains the nodes of H5 excluding s1 . This gives us a set of (s1 , t1 ), (s2 , t2 ). For each one of these elements we run the following code. function y=experiment(alpha) The parameter, alpha, is a potential s2 for t2=[5,3,1,2,4] s1=[4,2,3,1,5] t1=[3,1,5,2,4]

s are all nodes of H5 excluding s1 40

and s2.

t are all the node of H4 excluding t2 and t1.

i=1; while (i < 2) y=0; t3=t(i,:) v=candid(t3,s); This line gets all candidates of s_3. nrow=K(1); for h=1:nrow s3=v(h,:); y=allca(s1,t1,s2,t2,s3,t3); Here we check to see if we are in one of the cases. We output one if we are in one of the cases. if (y==0) s1; t1; s2; t2; s3; t3 ; Output the pairs that fail to meet case 1, case 2, or case 3. end end =i+1; end Running for all possible pairs, we verified our theorem and so we are done.

5.3. A Particular Case. The following code illustrates the experiment that we did for our verification. We fixed t1 = 31524, due the vertex symmetry of An . From there we determined possible candidates for s1 using the code candid(t1, s). Here s are the nodes of H5 . A possible s1 = 42135. We vary t2 from the nodes of H4 . In the code below we let t2 = 23514. We again used candid(t2 , s) to determine possible s2 ’s. Here s are the nodes of H5 . A possible s2 is s2 = 14235. In the code below s are the nodes of H5 with s1 and s2 excluded. The vector t are the nodes of H4 with t1 and t2 excluded. From s and t possible t3 ’s and s3 ’s are determined. Once we do this we use the function allca to see if we are in one of the situations. If the output is 1 we know we are in one of the situations. Otherwise no and then we list the nodes s1 ,t1 ,s2 ,t2 ,s3 ,t3 . In our experiments we found 41

we were always in one of our situations. Here is the Matlab code for our calculation. s1=[4,2,1,3,] t1=[3,1,5,2,4] s2=[1,4,2,3,5] t2=[2,3,5,1,4] s; as described above t; as described above i=1; while (i<10) t3=t(i,:) v=candid(t3,s) K=size(v) nrow=K(1) for h=1:nrow s3=v(h,:) y=allca(s1,t1,s2,t2,s3,t3) if (y==0) s1 t1 s2 t2 s3 t3 end end i=i+1 end y References [1] S. B. Akers, D. Harel, and B. Kirshnamurthy. The star graph: An attractive alternative to the n-cube. Proceedings of the International Conference on Parallel Processing, pages 393–400, 1987. [2] S. B. Akers and B. Kirshnamurthy. A group theoretic model for symmetric interconnection networks. IEEE Transactions on Computers, 38(4):555–566, 1989. [3] E. Cheng and M. Lipman. Disjoint paths in split-stars. Congressus Numerantium, 137:47–63, 1999. [4] E. Cheng, M. Lipman, and H. A. Park. Super connectivity of star graphs, alternating group graphs and split-stars. Ars Combinatoria, 59:107–116, 2001. [5] S. C. Chern, J. S. Jwo, and T. C. Tuan. Uni-directional alternating group graphs. In Computing and Combinatorics, volume 959 of Lecture Notes in Computer Science, pages 490–495. Springer, 1995. [6] Q. Gu and S. Peng. An efficient algorithm for k-disjoint paths in star graphs. Information Processing Letters, 67:283–287, 1998. 42

[7] J. S. Jwo, S. Lakshimivarahan, and S. K. Dhall. A new class of interconnection network based on the alternating group. Networks, 23:315–325, 1993. [8] Lazaros D. Kikas. Interconnection Networks and the k-Disjoint Path Problem. PhD thesis, Oakland University, 2004. [9] M. E. Watkins. On the existence of certain disjoint arcs in graphs. Duke Mathematics Journal, 35:231–246, 1968. Eddie Cheng, Department of Mathematics and Statistics, Oakland University, Rochester, MI, 48309, USA E-mail address: [email protected] Lazaros Kikas, Corresponding Author, Department of Mathematics and Computer Science, University of Detroit Mercy, Detroit, MI, 48221, USA E-mail address: [email protected] Serge Kruk, Department of Mathematics and Statistics, Oakland University, Rochester, MI, 48309, USA E-mail address: [email protected]

43