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 inter- ested 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, A n , 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 sym- metry, 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 equiva- lent 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 interconnec- tion networks deals with communication delay. When one sends a message from one processor to another in an interconnection network, the commu- nication 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
Transcript
A DISJOINT PATH PROBLEM IN THE ALTERNATINGGROUP GRAPH
EDDIE CHENG, LAZAROS KIKAS, SERGE KRUK
Abstract. For the purpose of large scale computing, we are inter-ested in linking computers into large interconnection networks. Inorder for these networks to be useful, the underlying graph mustpossess desirable properties such as a large number of vertices, highconnectivity and small diameter. In this paper, we are interested inthe alternating group graph, as an interconnection network, and thek-Disjoint Path Problem. We give a proof that the alternating groupgraph, An, has the (n− 2)-Disjoint Path Property. We close with adiscussion on possible research stemming from this work.
For the purpose of large scale computing, we are interested in linkingcomputers/processors into large symmetric interconnection networks. Thepapers [1,2] list some desirable properties that we wish the underlying graphof our interconnection networks to have. These properties are vertex sym-metry, large number of vertices, high connectivity and small diameter anddegree. Informally, a graph is vertex symmetric if each vertex in the graphcan be viewed as identical to the other vertices in the graph. Recall thatthe connectivity of a non-complete graph refers to the minimum number ofvertices that must be deleted in order to disconnect the graph. In terms ofinterconnection networks, the deletion of a vertex can be viewed as equiva-lent to processor failure. Hence, it is desirable to have networks that allowfor a large number of processor failure but still be operational. In graphtheoretic terms, we want to be able to delete a large number of verticesand still have a connected graph. That is, we want the underlying graph tohave high connectivity. Another measure of performance for interconnec-tion networks deals with communication delay. When one sends a messagefrom one processor to another in an interconnection network, the commu-nication is never instantaneous. The message must be sent over a path inthe graph. Hence, there will be some delay. The diameter of a graph isthe maximum of the minimum distances between all pairs of vertices in thegraph. We, therefore, want the graph to have a small diameter.
1
For a very long time the boolean n-cube has served as wonderful in-terconnection network model. However, in 1988 [1] introduced the stargraph as a competitive alternative to the n-cube. The star graph is definedas follows. The vertex set is made up of the symmetric group Sn. Theelements 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 fromone permutation to the other by exchanging the first and kth symbol wherek ∈ {2, 3, 4, ...n}. In the 1990’s both the split-star and the alternating groupgraph were introduced into the literature. Please see [4, 7]. The vertex setof the split-star is, again, the elements of the symmetric group Sn. Twopermutations are adjacent if and only if one can get from one permutationto the other by either a 2-exchange or a 3-rotation. A 2-exchange is ac-complished by exchanging the first and second symbol of the permutation.A 3-rotation is accomplished by rotating the first, second and kth symbolwhere k ∈ {3, 4, .., n} in either direction. The alternating group graph, An,is the subgraph of the split-star, S2
n, induced by the even permutations. Inthis paper, we study the k-Disjoint Path Problem in the alternating groupgraph.
2. Motivation
The alternating group graph, An, has as its vertex set of even permuta-tions on the symbols 1, 2, 3, ..., n. It is an exercise in algebra to show thatthe alternating group is generated by the 3-cycles of the symmetric groupSn. Two vertices in An are adjacent if and only if one can get from onepermutation 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 ei-ther direction. The rotation is accomplished by a 3-cycle. An is a graphwith degree 2(n − 2), and has n!(n−2)
2 edges. It is both vertex and edgesymmetric.
We now show that An has a hierarchal structure. Consider the evenpermutations where, say 5, is fixed in the nth position. Now considerthe subgraph of An induced by these permutations. It is clear that thatthe resulting subgraph is isomorphic to An−1. Since there are n symbolsit becomes clear that An has n copies of An−1. We will refer to thesesubgraphs as substars throughout the rest of this paper. Note that thishierarchal structure is very useful for us as we prove our main result. Formore details about the hierarchal structure please see [2, 7].
Consider the graph A4 in Figure 1. Let s1 = 1342, t1 = 4321, s2 = 3412and 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 amessage from s1 to t1 we can traverse the path 1342, 3241, 4321 and to senda message from s2 to t2 we can follow the path 3412, 4213, 1423. Furthernotice that these two paths are vertex disjoint. Hence, these two paths can
2
1342 2143
3412 4213
2431 4321
1234 3124
2314
3241
4132 1423
Figure 1. The Alternating Group Graph A4
be done at the same time without any interference or sharing of processortime and resources.
We want to consider the following problem. Given k pairs of distinctnodes (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 Prob-lem. A graph G is said to have the k-Disjoint Path Property if one can findk-vertex disjoint paths for any selection of k pairs of distinct vertices. Ithas been shown that for k ≥ 3 the problem of finding the k-disjoint pathsis NP -Hard. However, Watkins in [9] showed that a necessary conditionfor 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 intercon-nection networks. For example, Cheng and Lipman in [3] showed that thesplit-star graph has the (n − 1)-Disjoint Path Property. Another exampleis that Gu and Peng presented in [6], for the star graph an algorithm tocompute the k-vertex disjoint paths where k = dn−1
2 e.3
Since, An has degree 2n − 4 the best we can hope for is that the alter-nating 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 PathProperty. For example, let s1 = 1342, t1 = 4213, s2 = 3412 and t2 = 2143in Figure 1. Hence, the main result we wish to prove in this paper is thatfor n ≥ 5 the alternating group graph An has the (n − 2)-Disjoint PathProperty.
Due to the hierarchal structure of An, mathematical induction is thenatural selection of method of proof to prove this result. In this paper,we concentrate on the nontrivial base case only. Since, An is an inducedsubgraph of the split-star it is not surprising that the induction step is astraightforward modification of the induction step given in [3]. The inter-ested reader is directed to [8] for the modified induction step. Hence, whatremains 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 PathProperty.
Recall that Cnk counts the number of subsets of size k chosen from n
elements. A brute force approach to the proof of Theorem 3.1 would involveC60
2 C582 C56
2 cases. Examination of all this cases may not be feasible, evenwith the help of a computer. Compare this to the base case for the split-star where a brute force approach would involve C24
2 C222 C20
2 possible cases.Using the symmetry of the split-star the base case is much more manageableusing 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 carefulexamination of the symmetry of the graph A5. Even so, we still need thehelp of a computer to prove one our propositions. However, this computercheck involves minimal computation.
Recall that the graph A5 is made up of 5 copies of A4. We call thesecopies 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 wherei is fixed in the 5th position. It is clear that between any two of thesesubstars there are (5 − 2)! = 3! = 6 edges. Also remember that for anyvertex v in say substar Hi, v has two neighbors that are not in the substarHi. Not only that, but these two neighbors are themselves in differentsubstars. We are now ready to proceed with the proof of Theorem 3.1. Wehave as sets the set S = {s1, s2, s3} and the set T = {t1, t2, t3}. We beginwith the following lemmas.
4
Lemma 3.1. Let n = 5. Define Hji 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⊆ Hj
i 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 ⊆Hj
i . 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. Recallthat 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 yesand the next lemma tells us the condition.
Lemma 3.2. Let S = {s1, s2} and T = {t1, t2}. There are 2-disjoint pathsfrom s1 to t1 and from s2 to t2 in A4 unless s1, s2, t1 and t2 are in a4-cycle in this order.
The proof of this lemma follows by a simple examination of cases an willbe omitted. Another lemma that will be important to us is the following. Inthis 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 areexactly 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 letus consider the following. Let v be a vertex of A5. Say v is in the thesubstar 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 graphA5. 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 αand β be such that α 6= 5, β 6= 5 and α 6= β. We claim that there areexactly two vertices of H5 whose neighboring substars are Hα and Hβ . Letthe vertex v1 have the form αβab5. The vertex v1 is certainly such a vertexin H5 since its two neighboring substars are Hα and Hβ . Also v2 of theform 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 considerv1. The number of possible permutations of the form v1 is two. But only
5
+
v
+
+
+
+
+
+
+
+
+
H
H
H
H
a
b
cd
Figure 2. The Structure Around v = abcde
half of them can be even. Hence, there can only be one v1. By similarreasoning there can only be one v2. Therefore, there can only be two suchvertices. ¤
The proof of Theorem 3.1 proceeds as follows. We will be concernedwith how the vertices of S ∪ T are distributed among the five substarsof A5. Define ψi = |V (Hi) ∩ (S ∪ T )|. We look at the following 5-tuple:(ψ1, ψ2, ψ3, ψ4, ψ5). Notice that
∑5i=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 thesecases 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 andso we are done.
3.2. Case 2: (5, 1, 0, 0, 0). In this case we shall show that (5, 1, 0, 0, 0), byre-coordination, can be reduced to one of the eight cases worked out below.
6
s
HH
HH
23
14
s
s
t
t
t
+
+
3
3
22
11
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 arein H5 and t3 is in H4. Recall that H5 is isomorphic to A4 and thus, ismade of four copies of A3. The graph A3 is a triangle. Define Hi5 to bethe subgraph of H5 where i is fixed in the fourth position. Clearly, Hi5 isisomorphic to A3, H5 is made up of H15, H25, H35, and H45. Placing thefive vertices in H5, one cannot get more than three in any particular copyof A3. By re-coordination, let Ji be a substar of A5 where i is fixed in the4th position. So H25 is viewed now as part of J2. Notice that t3 is assumedto 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 ele-ments 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, as-sume without loss of generality that s1, t1, s2, and t2 are in H1 and s3 andt3 are in H2. Clearly, we can route from s3 to t3 within H2. Now consider
7
t
H H
H H
s
s
s
t
t
3
3
2
2
1
1
2 3
1 4
+
+
Figure 4. Nodes s2 and t2 Have Neighbors in Same Substars
the neighboring substars of s2 and t2. They may neighbor the same substaror different substars. See Figure 3 and Figure 4. In either case, it is clearlyeasy to route s2 to t2. Now we can route from s1 to t1. Since, we only needto 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). Notethat 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 t2are in H2. Consider the neighboring substars of s2 and t3. We are onlyinterested in the neighboring substars that are not H2. So either s2 andt3 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 completethe path to t3 by going through β. Since, H2 is 4-connected we can routet2 to α in H3, thus completing the path to s2. Now we need to route from
8
t
H H
H H
s
s
s
tt
+
a b
1 1
2
3
3
2
1 2
3 4
+
Figure 5. The Nodes s2 and t3 Have Different Neighbor-ing 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, andboth 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 completethe path from s2 to t2. Since, H2 is 4-connected, we can route from s3 to βthus completing the path to t3. Now we need to route from s1 to t1. All weneed 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 ele-ments of S ∪ T , H4 has one element of S ∪ T and H3 has one element ofS ∪ T . As in the previous case, if the element in H4 and the element inH3 form a pair then the four elements in H5 form the other two pairs. Sowithout 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
s
H H
H
s
t
s
t
t
a
b
1
1
22
3
3
12
3
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, ifthe pairs in H5 do not lie on a 4-cycle in the described order, then we aredone in this case. But now let us assume that the two pairs in H5 do lieon a four cycle in the described order. There are six 4-cycles in H5. Onecan check each one and see that the three disjoint paths exist. We do onehere, the others being similar. One can see that there is a path for s3 tot3 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 t1by routing from α to β. The path from α to β is completely contained inthe substars H2 and H1, and is disjoint from the route from s3 to t3. Theroute from s2 to t2 can be found, and this path does not touch s1 and t1since H5 is 4-connected.
10
a
H H H
H
s t
s t
b
s
t
1 1
2 3
3
2
5 4 3
2
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 forma pair. Notice that H5 must contain a pair. So let us assume without lossof 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 toconsider.
The first case is that s2 and t3 both have the same neighboring substarsand those substars are H4 and H3. This makes things easy, since t2 is inH3 and s3 is in H4. The paths from s2 to t2 and from s3 to t3 are easilyfound. The path from s1 to t1 has to avoid s2 and t3. This can be done asH5 is 4-connected.
The second case is that s2 and t3 both have the same neighboring sub-star, say H2, but s2’s other substar neighbor is H4, and t3’s other substarneighbor is H3. Let α be the neighbor of s2 in H2, and let β be the neighborof t3 in H2. It takes at most two steps to route from α to H3. We cantherefore complete the path from s2 to t2. We can then route from β toH4 since H2 is 4-connected. Thus, we have a path from t3 to s3. The path
11
t
H H H
H H
s
s
s
t
ta
b
1 1
3
2
3
2
5 1 4
2 3
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 canbe done as H5 is 4-connected. See Figure 7.
The final case is that s2 and t3 have different substar neighbors. Thesolution 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 showthat 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 supposethat s1, t1, and s2 are in the substars H1, and that s3, t3, and t2 are inH2. There are three ”empty” substars that we can choose from. Noticethat 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. Apath from s1 to t1 is easily found since in H1 we need only to avoid up totwo vertices. The case from s3 to t3 is similar. See Figure 9. Hence, we aredone with the easy case.
12
s
H H
H H
s
t
s
t
t
x y
1
1
2
3
3
2
1 2
4 5
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 theother substar H2 cannot contain a pair. So assume, without loss of gener-ality, 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 thatthere exists a pair (si, ti) such that si and ti share a common substarneighbor. So for example, if the pair (s1, t1) was such that s1’s neighboringsubstar was H1 and t1’s neighboring substar was also H1 we would thenfall under this case. The other main case is that no pair (si, ti) share acommon 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 caseshere. The first case is that there is exactly one pair that shares H4 as acommon substar neighbor. The second case is that there are exactly twopairs that share H4 as a common substar neighbor. And the third case isthat there all three pairs share H4 as a common substar neighbor. We lookat these three cases and solve each one.
13
t
HH
H
H H
s
s
s
t
t
3
1
2 x
3
2
1
12
5
4 3
+
+
+
+
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 sharesH4 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 aneighboring 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. Thenwe can complete the path from s2 to t2 by routing in the substar H5 andH3. The route from s1 to t1 is completed in the substar H4. All we need todo is complete a path from s3 to t3. Recall that there are six edges betweenthe 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 edgeavailable. We can route s1 to that edge without touching s1, α or s2. Samething 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 substarH4 as a common substar neighbor. Let us assume that these pairs are
14
B
H H H
H H
s
s
s
ACD
t
t
t
+ +
1
2
3
1
2
3
1 4 2
5 3
a
Figure 11. (s1, t1) and (s2, t2) Share H4 as a CommonSubstar Neighbor
(s1, t1) and (s2, t2). Now assume that neighbor of s1 in H4 is A, the neighborof s2 in H4 is B, the neighbor of t1 in H4 is C and the neighbor of t2 in H4 isD. Now suppose that A, B, C and D do not lie on a 4-cycle. Then disjointpairs 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 betweenthe 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 aneighboring substar since we would violate the hypothesis of this case. Soassume that t3 has H3 as a neighboring substar. We can get from s3 toeither 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 foundby using the substars H5 and H3. A route from s1 to t1 may be found byusing the substar H4. The nodes s3, α , s1, t1 and t3 touch at most fiveedges between H1 and H2, and so a path from s2 to t2 may be completed.
15
t
HH
H HH
s
s
s
t
t
+
+
++
1
2
3
1
23
12
3 54
Figure 12. Situation 1
Consider Figure 11 and the case in the previous paragraph. Supposethat in the extreme case A = D and B = C. To solve this case one usesthe exact same argument as discussed in the previous paragraph and so weare 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 neigh-bor of the nodes in S ∪ T . Notice that s1, s2 and s3 cannot all have H2 asa substar neighbor. If it did we would violate Lemma 3.4. So assume thatt3’s substar neighbor is H5. The node s3 can reach to either H3 or H5 in atmost two steps. Let us say that s3 can be routed to H3 in two steps. Wecan find our disjoint paths in the usual way as described in the previousparagraphs. So we are done with this main case.
Now we move on to the second main case where no pair share a commonsubstar neighbor. Consider Figure 12, Figure 13, and Figure 14. We observein our next theorem that only one of these situations can occur.
16
t
HH
HH
s
s
s
t
t
+
++
+
12
34
1
2
3
1
2
3
Figure 13. Situation 2
Proposition 3.1. Let s1, s2, s3 ∈ H1 and t1, t2, t3 ∈ H2. Assume nopair (si, ti) for i = 1, 2, 3 share a common substar neighbor. Then only oneof situations depicted in Figure 12, Figure 13, or Figure 14 can occur.
Proof. The proof of this theorem is through a simple computer check. Wefirst 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 commonsubstar neighbor. We then wrote a computer program that verifies thatonly one of the above situations can occur. Doing this for all possible pairswe verified the truth of our theorem. Matlab code and a description of ourexperiment are described in the Appendix of this paper. ¤
Having Proposition 3.1 in hand, we proceed in showing that in eachsituation one can find the desired disjoint paths. Let us consider situation1 illustrated in Figure 12. To solve this case we can route s2 to the substarH5 in at most two steps and then complete the path to t2. At most fouredges between H3 and H4 will be eliminated. So complete the route for
17
H
H H
s t
y
1 2
k
i j
Figure 14. Situation 3
s3 to t3. At most four edges will be eliminated between H1 and H2 socomplete the route from s1 to t1
Now consider situation 2 illustrated in Figure 13. To solve this casenotice that we can finish the route from s2 to t2. At most four edges willbe eliminated between the substars H3 and H4. So a path from s3 to t3can be found. At most four edges will be eliminated between the substarsH1 and H2. So a path from s1 to t1 can be found.
Now consider situation 3 as illustrated in Figure 14. Assume, withoutloss of generality, that s1 is adjacent to t2 and both share the same neighborin H3. The nodes s2 and t3 are adjacent and both same the share neighborin H4. And the nodes s3 and t1 are adjacent and both share the sameneighbor in H5. How do we resolve this situation and others like it? All aresimilar and are solved in the same way. For example suppose s1 = 24315and is adjacent to t2 = 52314. Suppose s2 = 14235 and is adjacent tot3 = 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 onlyavoid t3, t1 and 25134. Now for s3 to t3. We go from t3 to 45231 to 43521to 14523 to 45123 to s3. We use this example now to prove our generalcase. Recall that each v in A5 has a a structure depicted in Figure 2. Wefirst route s1 to t1. The node s1 has a neighbor in H3. The node t1 canget to H3 in two steps since t1’s neighbors are in substars H5 and H2. Alsot2’s neighboring substar is H3, so there is path from t1 to H3. Let α bethe 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 andthe node t3 has a neighbor in H4. So the path from s3 to t3 can be foundin the substars H4 and H5. Now all that is left to do is construct a pathfrom s2 to t2. There are six edges between the substars H1 and H2. Thenodes s1, s3, t3, α and t1 touch at most five of these edges. Hence, we mayconstruct a path from s2 to t2 using the nodes of H1 and H2 since thereis an edge available and H1 and H2 are 4-connected. Hence, we have ourdisjoint 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 hasthree elements of S ∪T , H2 has two elements and H3 has one element. Westart with the easy case. The easy case is that H1 contains a pair and thetwo 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. Tosolve this case we look at the neighboring substars of s2. There are two ofthese substars. Looking at all the possibilities it is clear that s2 will haveneighbors in either H4, H5 or H3. If the neighboring substar is H3 then wecan clearly route to t2. The route from s3 to t3 is completely contained inthe 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 holdsagain 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, andthat s3 and t2 are in H2 and t3 is in H3. We first want to route s2 tot2. We look at the substar neighbors of s2. There are two of them. Ifthe immediate substar neighbors of s2 are H4 or H5 then we are done asfollows. Let us assume that s2’s neighbor is in H4. By Lemma 3.3 we canroute from t2 to H4 in at most two steps. Thus, we can complete the froms2 to t2. We can route from s3 to t3 since H2 is 4-connected. Finally, wecan route from s1 to t1 again as H1 is 4-connected. See Figure 15. Nowsuppose that the immediate substar neighbors of s2 are H2 and H3. Let αbe the neighbor of s2 in H2 and β is the neighbor of s2 in H3. Suppose thatα 6= s3. Then route s3 to one of the substars H3, H4, or H5. This can bedone in at most two steps. Then complete the path to t3. To route from αto 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
t
H H H
H
s
s
s
tt
y
1 2 3
4
1
1
2
3
2
3
Figure 15. H1 Contains a Pair But H2 Does Not
Now suppose that α = s3. Since the immediate substar neighbors of s2 areH2 and H3, we can reach H4 and H5 in at most two steps. One path maybe blocked by s1 and t1, but then just pick the other. Let’s assume that itis H4. The node t2 can be routed to H4 in at most two steps. Do this andcomplete the path to s2. We can then clearly complete the paths for s1 tot1 and s3 to t3 as before. See Figure 17
The last case here is that the elements of contained in H1 and H2 donot form a pair. Let us assume, without loss of generality, that s1, s2 ands3 are in H1, and that t1 and t2 are in H2 and t3 in H3. We look at theneighboring 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 froms3 to t3. Now route s2 to either H4 or H5, and complete the path to t2in 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
s
H H H
s
s
ta
t
t
b
1 2 3
1
1
2
3
2
3
Figure 16. Node α 6= s3
s2 to either H4 or H5 in at most two steps. In the worst case, we are forcedto 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 tot2. To complete the route from s3 to t3, route α in H4 to t3 in H3. Thiscan be done as H4 is 4-connected. Now route s1 to t1 in H2. The routehas to miss at most three vertices in H1 and two vertices in H2. This canbe done as H1 and H2 are 4-connected. See Figure 18. We are done withthis case.
3.7. Case 7: (3, 1, 1, 1, 0). In this case we must address several cases andtheir subcases. Let us assume first, that the substar H1 has three elementsof S ∪ T . We first consider the easy case. Let us assume that of threeelements in H1, two of them form a pair. Hence, assume that s1, t1 ands2 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 interestedin these neighbors. If one of those neighbors are in H2, or in H5 then we aredone. We will consider the neighbor of s2 to be α and α is in H5. The case
21
t
H
H H
H
s
s
s
t
t
+
+
1 2 3
4
1
1
2
3
2
3
Figure 17. Node α = s3
of α in H2 is similar. Notice that we can route now from s1 to t1 withouttouching s2 as H1 is 4-connected. The path from s2 to t2 is completed byrouting from α to t2. The path from s2 to t2 is completely contained inthe 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 containedcompletely in the substars H3 and H4. Please see Figure 19. Now it mightbe possible that a route from s2 to H5 will take two steps. If this happenswe 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 isin H4. We look at the substar neighbors of the element s3. We considerthe first case. Suppose that α is the neighbor of s3 in the substar H2, andβ is the neighbor of s3 in H3. Suppose that at least one of α or β is notequal to t1 or t2, respectively. Let β 6= t2. Now from β we can route tot3 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
t
H H H
H
s
s
s
gt
t
a
b
1 2 3
4
1
2
3
3
2
1
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 tot2 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 whichα 6= t1, but β = t2 is similar. See Figure 20.
Now suppose that α = t1 in H2 and that β = t2 in H3. Notice thatat 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 canget 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 froms2 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 mosttwo steps. The rest follows as before. Now suppose the neighbor of s1 arethe substars H4 and H5. Let γ be the neighbor of s1 in H5. Now consider
23
t
H H H
H
H
s
s
st
t
a
1 2 3 4
5
1 1
2
23 3
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 reachH3, then do so. If not, suppose that we are forced to H2. Note that thevertex we reach in H2 will not be t1. So we complete the route to H3 andthus to t2. We can now complete the route from s1 to t1. The path froms3 can be completed now since we need only to avoid up to three verticesin H1. See Figure 22.
Suppose that the neighboring substar of s3 is now H4. Then a path froms3 to t3 can easily be found. We can get from s1 to H5 or H2 in at mosttwo 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 pathfrom s2 to t2 can be done since the path that needs to be found need onlymiss up to three vertices in H1.
Suppose s3’s neighboring substar is H5. Then the same type of argumentas 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 wemust consider. The first case is trivial. Let H1, H2 and H3 each containtwo elements of the set S∪T . Assume that the two elements in each substar
24
t
HH H
H
s
s
sb
a
t
++
15 4
2
1
2
3
1
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 iscompletely 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 ele-ments 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 toresolve. First, it is easy to see that we can find a path from s1 to t1 that iscompletely contained in the substar H1. Now we need to find paths froms2 to t2 and from s3 to t3. By the above Lemma 3.3 we know that we canreach from s2 to the substar H4 in at most two steps. Likewise, we canreach from t2 to H4 in most two steps. Note that we need not worry aboutt3 nor s3 touching these paths. Let the path from s2 to H4 be s2αγ, whereγ is in H4. Let the path from t2 to H4 be t2βδ, where δ is in H4. Thus,the path from s2 to t2 is given by s2αγ, ..., δβt2. Note that this path iscompletely contained in the substars H2, H3 and H4, and so is disjoint tothe path from s1 to t1. Now we need a path from s3 to t3. Recall that there
25
s
H H H
HH
s
s
t
yx
g
t
t
1 3 4
2 5
2
1
3
2 3
1
Figure 21. Case Where α = t1 and β = t2
are 6 edges between the substars H2 and H3. The vertices s2 and α touchat most two of these edges. Likewise the vertices t2 and β touch at mosttwo of these edges. Thus, at most four edges are touched leaving at leasttwo 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 verticest2 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 containtwo elements of the set S ∪ T and that none of the two elements in eachsubstar 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 havetwo available substars H4 and H5. We will make use of these substars. ByLemma 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 thenjust choose the other. Likewise, we can reach from t1 to H4 in at mosttwo steps. Let the path from s1 to H4 be s1αγ, where γ is in H4. Let
26
s
H H
HHH
s
s
t t t
g
+
+
1 5
2 3 4
1
2 3
1 2 3
Figure 22. Case Where α = t1 and β = t2
the path from t1 to H4 be t1βδ, where δ is in H4. Thus, the path froms1 to t1 is given by s1αγ, ..., δβ, t1. This path is completely contained inthe 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 thesepaths since if it did we would just choose the other. Let the path fromt3 to H5 be t3ερ, where ρ is in H5. Now we are in a position to computethe remaining paths. Between the substars H1 and H2 are six edges. Thevertices s1 and α touch at most two of those edges. Likewise, the verticest3 and ε touch at most two of these edges. Hence, at most four edges aretouched leaving at least two untouched. Let x be a vertex in H1, and y avertex in H2 such that x and y are connected by one of these remainingedges. Since H1 is 4-connected there exists a path from t2 to x that doesnot touch the vertices s1 and α. Likewise, there exists a path from s2 to ythat does not touch the vertices t3 and ε. Hence, we can complete the pathfrom t2 to s2. This path is completely contained in H1 and H2 and doesnot touch the path from s1 to t1. Now all we need is to show the existenceof a disjoint path from s3 to t3. Between the substars H3 and H5 are six
27
a
H H
H
t
sx
g
d
y
b
t
s
2 3
4
3
2
2
3
Figure 23. H1 Contains the Pair (s1, t1)
edges. The vertices t1 and β touch at most two of these edges. Thus, wehave at least four edges available. Let z be a vertex in H3 touching one ofthese available edges. Since, H3 is 4-connected there is a path from s3 to zwithout touching the vertices t1 and β. Hence, we have a path from s3 tot3 that goes only through the substars H3, H5 and H2. All these paths areclearly disjoint. See Figure 24.
3.9. Case 9: (2, 2, 1, 1, 0). In this case we have three subcases that wemust address. The first case is easy. Let the substars H1 and H2 each havetwo elements of the set S ∪ T . Assume, the two elements in H1 form apair, and the two elements in H2 also form a pair. So for this case, we shallassume, without loss of generality, that the pair s1 and t1 are in the substarH1 and the pair s2 and t2 are in the substar H2. Let us also assume thats3 in H3, and that t3 is in H4. It is very easy to see that we can find pathsfrom s1 to t1, s2 to t2 and s3 to t3 that are completely disjoint from oneanother.
28
g
HH H
H H
st
a
x y
s
t
e
s
t
z b
+
r
d
12 3
4 5
12 2
3
3
1
Figure 24. H1, H2 and H3 Do Not Contain Pairs
The next case is easy as well. Assume that H1 and H2 have each twoelements of S ∪T . Assume that the two elements of H1 form a pair. Let usassume, 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 pathfrom s1 to t1 completely contained in H1. Now we need paths from s2 tot2 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. Ifit is just choose the other. Let the path from s2 to t2 be s2α...t2. Notethat this path is completely contained in the substars H2 and H3. Recallthat 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 oneof the four remaining edges. Since, H2 is 4-connected then there exists apath from t3 to y without touching the vertices s2 and α. Hence, the pathfrom 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 assumethat the substars H1 and H2 each have two elements of S ∪ T . We assume
29
y
H H
H
t
s
ax
+
s
t
2 3
4
3
22
3
Figure 25. H1 Contains a Pair (s1, t1)
that the two elements in H1 and that the two elements in H2 do not formpairs. Hence, let us assume without loss of generality that s1 and t2 are inH1, s2 and t3 are in H2, t1 is in H3, and s3 is in H4. By Lemma 3.3, we canget from s1 to the substar H3 in at most two steps. Let us assume that thepath from s1 to t1 is given by s1α, ...t1. Note that this path is completelycontained in the substars H1 and H3. Again, by Lemma 3.3, we can reachfrom t3 to H4 in at most two steps. Let us assume that the path from t3to s3 is given by t3β, ...s3. Note that this path is completely contained inthe substars H2 and H4. Also, note that the paths between s1 and t1 andbetween s3 and t3 are disjoint. Finally, as before, we know that the vertext2 does not touch the path from s1 to t1 and that the vertex s2 does nottouch 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. Thevertices s1 and α touch at most two of these edges. Likewise, the verticest3 and β also touch at most two of these edges. So at most four edges areeliminated, thus leaving at least two available. Let x be in H1 and let ybe in H2 and x and y are connected by one of these available edges. Since
30
y
HH H
H
s
t
xa
s
b
t
+
s
+
t
1 2 4
3
1
2
2
33
Figure 26. H1 and H2 Do Not Have Pairs
H1 is 4-connected there exists a path from t2 to x without touching s1 andα. 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 ourthree 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 wemust address. The first case is easy. It is the case where the two elementsin a substar are an actual pair. Assume that this pair is (s1, t1) and thatit is in H1. Let s2 be in H2, t2 in H3, s3 in H4, and t3 in H5. This donewithout loss of generality for this case. It is easy to see that we can findthe three disjoint paths.
The second case now assumes that the two elements in the substar isnot a pair. So assume, without loss of generality, that s1 and t2 are in thesubstar H1. Let s2 be in H2, t1 in H3, s3 in H4 and t3 in H5. It is easy tosee that we can find a path from s3 to t3 that is completely contained in thesubstars 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
a
HH
H
t
y
s
b
t
s
+
12
3
2
12
1
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 onone of these paths. If it is just choose the other. Let the path from s1 to t1be s1αβγ...t1. Note that this path is completely contained in the substarsH1 and H3. Now note that vertices s1 and α are in the substar H1. Recallthat 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 ofthe four remaining. Since, H1 is 4-connected there is a path from s2 to ywithout touching s1 and α. Hence, we can complete the path from s2 tot2. We therefore have our three disjoint paths. See Figure 27. Hence, weare 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 graphAn has the (n−2)-Disjoint Path Property. This result is not just interestingtheoretically, but as described in the motivation section, this result also hasa practical aspect as well.
32
A question to ask is what is the next step in our research. One thing toconsider is that the proof that we presented is an existence proof. What wewould like to do now is to develop an algorithm that will actually constructthe (n− 2)-disjoint paths. A first place to start would be at the base caselevel and then try to generalize.
Another question to consider are directed versions of the alternatinggroup graph. Work has been done in the literature to orient the alter-nating group graph. Please see [5]. It would be interesting to develop adirected version of the k-Disjoint Path Problem. Such a problem also hasa firm footing in the real world of interconnection networks. Consider aninterconnection network in which something happens to cause the commu-nication links to only work one way. Can the interconnection network stillbe operational and can we write down conditions in which the k-disjointpaths still be found?
There are many other things to consider in the study of these type ofnetworks. 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 weassumed that no pair (si, ti) had a common substar. We conjectured thatour case gave rise to only one of three possible situations. The experimentwe 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 eachprogram is given. From there we describe, in general, our experiment. Weend with a particular instance of our experiment.
5.1. Matlab Code. Given a node t1 we want a listing of all possible s1’sthat fit our hypothesis. That is a s1 is a candidate if and only if s1 doesnot share a common substar neighbor with t1. The following Matlab codedoes this. If the reader is interested in the obtaining the Matlab sourcecode please contact the author at [email protected] [v]=candid(t1,s)v=[]i=1a=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=velsev=[v;s(i,:)];
endi=i+1
33
endv
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=xtemp=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 anyone of the three situations described in Chapter 6.function y=allca(s1,t1,s2,t2,s3,t3)y=0y=case1(s1,t1,s2,t2,s3,t3)if y==0y=case2(s1,t1,s2,t2,s3,t3)
endif y==0y=case3(s1,t1,s2,t2,s3,t3)
end
The following three functions determine if the pairs (si, ti) for i = 1, 2, 3fit into one of three situations hypothesized in our theorem. The functiony = case1 corresponds to situation number 2. The function y = case2corresponds to situation number1. And the funtion y = case3 correspondsto situation 3.function y=case1(s1,t1,s2,t2,s3,t3)a1=s1(1) ; a2=s1(2)b1=t1(1); b2=t1(2)
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))
if (n1s2(5)==4)x5=compare(n2s2,n1t1);x6=compare(n2s2,n2t1);x7=compare(n2s2,n1t3);x8=compare(n2s2,n2t3);elsex5=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);elsex9=compare(n1s3,n1t2);
5.2. Experiment Description. In this section we describe,in general, ourexperiment. The purpose of our experiment is to provide a computer checkof 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) fori = 1, 2, 3. Due to vertex symmetry of A5, we let t1 = 31524 throughoutthe whole experiment. The nodes s1, s2, and s3 come from H5. The nodesof 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 ofexperiments. The first set of experiments involve s1 = 24315 and t1 = 31524and 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. Allpotential s2’s are found using candid(t2, s), where the matrix s containsthe nodes of H5 excluding s1. This gives us a set of (s1, t1), (s2, t2). Foreach one of these elements we run the following code.
function y=experiment(alpha)The parameter, alpha, is a potential s2for 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 s140
and s2.
t are all the node of H4 excludingt2 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:nrows3=v(h,:);y=allca(s1,t1,s2,t2,s3,t3); Here we check to seeif 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 thatfail 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 aredone.
5.3. A Particular Case. The following code illustrates the experimentthat we did for our verification. We fixed t1 = 31524, due the vertexsymmetry of An. From there we determined possible candidates for s1 usingthe 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. Weagain used candid(t2, s) to determine possible s2’s. Here s are the nodesof H5. A possible s2 is s2 = 14235. In the code below s are the nodes ofH5 with s1 and s2 excluded. The vector t are the nodes of H4 with t1 andt2 excluded. From s and t possible t3’s and s3’s are determined. Once wedo 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 noand 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 ourcalculation.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 abovet; as described abovei=1;while (i<10)t3=t(i,:)v=candid(t3,s)K=size(v)nrow=K(1)for h=1:nrows3=v(h,:)y=allca(s1,t1,s2,t2,s3,t3)if (y==0)s1t1s2t2s3t3
endendi=i+1endy
References
[1] S. B. Akers, D. Harel, and B. Kirshnamurthy. The star graph: An attractive al-ternative to the n-cube. Proceedings of the International Conference on ParallelProcessing, pages 393–400, 1987.
[2] S. B. Akers and B. Kirshnamurthy. A group theoretic model for symmetric intercon-nection 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, alternatinggroup 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. InComputing 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. Infor-mation Processing Letters, 67:283–287, 1998.
42
[7] J. S. Jwo, S. Lakshimivarahan, and S. K. Dhall. A new class of interconnectionnetwork based on the alternating group. Networks, 23:315–325, 1993.
[8] Lazaros D. Kikas. Interconnection Networks and the k-Disjoint Path Problem. PhDthesis, Oakland University, 2004.
[9] M. E. Watkins. On the existence of certain disjoint arcs in graphs. Duke MathematicsJournal, 35:231–246, 1968.
Eddie Cheng, Department of Mathematics and Statistics, Oakland Univer-sity, Rochester, MI, 48309, USA
