Uniformly optimally reliable graphs

Uniformly Optimally Reliable Graphs

D. Gross, J. T. Saccoman

Department of Mathematics and Computer Science, College of Arts and Sciences,Seton Hall University, South Orange, New Jersey 07079-2696

Received 28 October 1997; accepted 17 December 1997

Abstract: A graph with n nodes and e edges, where the nodes are perfectly reliable and the edges failindependently with equal probability r, is said to be uniformly optimally reliable (UOR) if it has the greatestreliability among all graphs with the same number of nodes and edges for all values of r. UOR simplegraphs have been identified in the classes e Å n 0 1, e Å n , e Å n / 1, and e Å n / 2 (Boesch et al.,Networks 21 (199) 181–194). In this paper, we demonstrate that the UOR simple graphs in these classesare also UOR when the classes are extended to include multigraphs. q 1998 John Wiley & Sons, Inc.Networks 31: 217–225, 1998

1. INTRODUCTION AND PRELIMINARIES edges fail independently with equal probability r, 0 õ rõ 1. We define the so-called unreliability polynomial,P(G , r) Å ( e

iÅ1 miri(1 0 r) e0i , where mi is the numberLet G Å »V (G) , E(G) … denote a graph with node set

of edge-disconnecting sets of size i . The reliability of aV (G) Å {£1 , £2 , . . . , £n} and edge set E(G) Å {x1 , x2 ,graph, therefore, can be defined as R(G , r) Å 1 0 P(G ,. . . , xe}. We allow there to be more than one edge be-r) . A graph G is said to be uniformly optimally reliabletween a pair of nodes; if this occurs, we refer to each of(UOR) if for each H in its class P(G , r) ° P(H , r)them as a multiple edge. A double edge consists of two[respectively, R(G , r) ¢ R(H , r)] for all 0 õ r õ 1.multiple edges and a triple edge consists of three multipleBecause P(G , r) Å 1 whenever G is disconnected, itedges. If there is only one edge between a pair of nodes,suffices to consider only connected graphs.it is called a simple edge. The term edge will subsume

It is obvious that if a graph minimizes each term miboth a simple edge and a multiple edge. We use the termamong all graphs in its class then it will have the smallestmultigraph to refer to a graph that contains at least onevalue of P in its class for all r and, consequently, themultiple edge but no loops and the term simple graph tolargest value of R . Hence, it is UOR. While all knownrefer to a graph with neither multiple edges nor loops.examples of UOR graphs would indicate that the converseLet V(n , e) denote the class of all graphs (simple graphsis true, a proof of this fact has yet to be found.or multigraphs) with n nodes and e edges, and Vs(n , e) ,

In the 1991 paper ‘‘On the Existence of Uniformlythe class of all simple graphs with n nodes and e edges.Optimally Reliable Networks,’’ Boesch et al. demon-For other standard graph-theoretic notation and terminol-strated the existence of UOR simple graphs in the classesogy, we refer to Harary [3] .Vs(n , n 0 1), Vs(n , n) , Vs(n , n / 1), and Vs(n , n / 2)If the graph represents a connected network, we canby finding simple graphs that minimize mi for all i [1] .discuss a model in which nodes are perfectly reliable andWang covered the class Vs(n , n / 3) [5] .

Let G 0 x denote the deletion of an edge x from thegraph G , and G /x , the contraction of the edge x , that is,Correspondence to: D. Gross

q 1998 John Wiley & Sons, Inc. CCC 0028-3045/98/040217-09

217

8U1F 0815/ 8U1F$$0815 05-14-98 08:07:51 netwa W: Networks

218 GROSS AND SACCOMAN

the identification of the endnodes of x and the deletionof all resulting self-loops. The following theorem, whichis credited to Moskowitz [4] , applies to many graph in-variants. It is stated here in terms of the number of edge-disconnecting sets:

Theorem 1.1. (Factoring Theorem) Suppose that G is aconnected graph and x is a simple edge in G. For all i,1 ° i ° e, mi (G) Å mi01(G 0 x) / mi (G /x) .

Fig. 2. mi (G ) ¢ mi (G * ) ; m1 (G ) ú m1 (G * ) .The Factoring Theorem does not apply to factoring ona multiple edge, since contraction of a multiple edgewould result in a graph having more than one fewer edge into two edges. Then, mi (G) ¢ mi (G *) for all i, 2 ° i °than the original graph. If the class of graphs under con- e, and m1(G) ú m1(G *) . (See Fig. 2 for an illustration.)sideration was extended to include pseudographs, that is,

Proof. Let y be one of the edges formed in the subdivi-both multiple edges and loops are allowed, and the defini-sion. We factor on x in G and y in G *. Since G /x andtion of contraction was amended so that only the loopG * /y are isomorphic, mi (G /x) Å mi (G * /y) for all i .formed by the actual edge contracted was deleted, then theSince G 0 x is disconnected, mi (G 0 x) Å ( e01

i01 ) for allFactoring Theorem would hold. Henceforth, any factoringi . On the other hand, G * 0 y is connected, which impliesperformed is on simple edges in the graph. In the firstthat there may be some choices of the remaining edgesproposition, we show that the placement of a bridge in awhich do not disconnect. Thus, mi01(G * 0 y) ° ( e01

i01 )graph has no bearing on mi :for all i , and mi (G) Å mi01(G /x) / mi (G /x)¢ mi01(G *0 y) / mi (G * /y) Å mi (G *) . Since m1(G) is the number

Proposition 1. Let G be a graph with a bridge x, and of bridges, and G * has one fewer bridge than has G ,G * be the graph obtained by contracting the bridge x and m1(G) ú m1(G *) .adding a pendent edge y. Then, mi (G) Å mi (G *) for all Repeated application of Propositions 1 and 2 implyi, 1 ° i ° e. (See Fig. 1 for an illustration.) that, given any graph G having a bridge, there exists a

bridgeless graph G* in the same class such that P(G*, r)Proof. Factor on the bridge x and the pendent edge yõ P(G , r) , for all 0 õ r õ 1. The following is anin the respective graphs G and G *. Since G /x and G* /yimmediate consequence:are isomorphic, mi (G /x) Å mi (G * /y) for all i . The dele-

tion graphs G0 x and G *0 y are disconnected, so mi01(GProposition 3. If there exists a bridgeless graph G in0 x) Å mi01(G * 0 y) Å ( e01

i01 ) for all i . Thus, mi01(Gthe class V(n, e) such that P(G, r) ° P(H, r) for all0 x) / mi (G /x) Å mi01(G * 0 y) / mi (G * /y) , i.e.,bridgeless graphs H in the class V(n, e) and for all r,mi (G) Å mi (G *) for all i .0 õ r õ 1 , then G is UOR.Next, we establish that a graph containing a cycle and

a pendent edge will have mi at least as large as that of agraph in the same class having a longer cycle and one

2. THE CLASSES V(n , n 0 1)fewer pendent edge.AND V(n , n )

Proposition 2. Let G be a nonacyclic graph with a pen- There are no connected multigraphs in the class V(n , ndent edge x and let G * be the graph obtained by con- 0 1). Any connected simple graph is a tree, all of whichtracting the pendent edge x and subdividing a cycle edge have mi Å ( n01

i ) , i ¢ 1. Thus, all connected graphs inV(n , n 0 1) are UOR.

There are no connected bridgeless multigraphs in theclass V(n , n) . The only connected bridgeless simplegraph in this class is Cn , the cycle on n nodes. Thus, Cn

is the unique UOR graph in the class V(n , n) .

3. THE CLASS V(n , n / 1)

The removal of three or more edges from any graph inFig. 1. mi (G ) Å mi (G * ) . the class V(n , n / 1) always results in a disconnected


UNIFORMLY OPTIMALLY RELIABLE GRAPHS 219

graph; hence, mi Å ( n/1i ) for any graph in the class and

for all i ¢ 3. Thus, the only value that differentiates onebridgeless graph from another is m2 .

The unique UOR simple graph in the class Vs(n , n/ 1) is obtained in the following manner [1]:

STEP 1. Start with a triple-edge between two isolatednodes u and £.

STEP 2. Distribute n 0 2 nodes as equally as possibleamong the three edges. (See Fig. 3 for an illustration.)We refer to each of the three resulting paths joining u Fig. 3. UOR graph in Vs (7, 8) .

to £ as an edge-path.The only way to disconnect a UOR simple graph by

the removal of two edges is by deleting two of the edgesalong an edge-path. If there are j internal points on oneof these edge-paths, then the number of ways to discon- k Å 2: m2 Å

n 0 43

/ 1

2nect in this fashion is ( j/1

2 ) . These edge-paths are subdi-vided as evenly as possible, so if k Å (n 0 2) mod 3,then k of the edge-paths have (n 0 2)/3 internal nodes,while the other 3 0 k will have (n 0 2)/3 internalnodes. These observations are summarized in the follow- / 2

n 0 13

/ 1

2

Å n(n 0 1)6

.ing theorem:

By Proposition 3, we only need to compare the UORTheorem 3.1. The UOR simple graphs in the class Vs(n,simple graph to the bridgeless multigraphs in this class.n / 1) haveTo see that there are only two such multigraphs, namely,H1 , the cycle on n nodes with one of its edges doubled,and H2 , the cycle on n 0 1 nodes with a pendent double

m2 edge (see Fig. 4 for an illustration), we need only con-sider the graphs that result from the deletion of one ofthe multiple edges. Since m2(H1) Å ( n01

2 ) õ ( n012 ) / 1Å (3 0 k)

n 0 23

/ 1

2

/ k

n 0 23

/ 1

2

, Å m2(H2) , H1 is the best multigraph, so it is only neces-sary to compare it to the UOR simple graph. Indeed, for

k Å 0: m2(H1) 0 m2(UOR) Å (n 0 2)2

3ú 0,where k Å (n 0 2) mod 3 . Specifically,

when n ú 2;

k Å 1, 2: m2(H1) 0 m2(UOR) Å (n 0 3)(n 0 1)3

ú 0,k Å 0: m2 Å 3

n 0 23

/ 1

2

Å (n / 1)(n 0 2)6

.

when n ú 3.

Thus, the graph in Vs(n , n / 1) which is UOR amongall simple graphs remains UOR when we extend the class

k Å 1: m2 Å 2

n 0 33

/ 1

2

to include multigraphs.

4. THE CLASS V(n , n / 2)

The removal of four or more edges from any graph in/

n

3/ 1

2

Å n(n 0 1)6

. the class, V(n , n / 2) always results in a disconnectedgraph, so mi Å ( n/2

i ) for any graph in the class and for



m2Å (60 k)

n0 46

/ 1

2

/ k

n0 46

/ 1

2

,

specifically as follows:

k Å 0: m2 Å 6

n / 26

2

Å n 2 0 2n 0 812

.

Fig. 4. Multigraphs in V(7, 8) .

k Å 1: m2 Å 5

n / 16

2

/n / 7

6

2all i ¢ 4. Thus, the only values that differentiate onebridgeless graph from another are m2 and m3 .

The unique UOR simple graph in the class Vs(n , nÅ n 2 0 2n 0 3

12./ 2), n ¢ 4, is obtained in the following manner [1]:

STEP 1. Start with the graph K4 , the complete graph on4 nodes.

k Å 2: m2 Å 4

n

6

2

/ 2

n / 66

2

Å n 2 0 2n

12.STEP 2. Distribute n 0 4 nodes among the six edges as

follows:Let k Å (n 0 4) mod 6.If k Å 0, 1, 5 distribute the n 0 4 nodes among the

six edges of the K4 as evenly as possible; k Å 3: m2 Å 3

n 0 16

2

/ 3

n / 56

2If k Å 2, i.e., n 0 4 Å 6m / 2, insert m nodes in each

of the six edges of the K4 and the two remaining nodesin two independent edges (of the original K4) ;

Å n 2 0 2n / 112

.If k Å 4, i.e., n 0 4 Å 6m / 4, insert m / 1 nodes ineach of the six edges of the K4 and then remove two nodesfrom a pair of independent edges (of the original K4);

If k Å 3, i.e., n 0 4 Å 6m / 3, insert m nodes in eachk Å 4: m2 Å 2

n 0 26

2

/ 4

n / 46

2of the six edges of the K4 two of the remaining nodes intwo independent edges (of the original K4) , and the lastnode in any one of the remaining four edges (of theoriginal K4) . (See Fig. 5 for an illustration.) Å n 2 0 2n

12.

We refer to the six paths joining the endnodes of theoriginal edges of the K4 as edge-paths. The only way todisconnect a UOR simple graph by the removal of twoedges is to delete two of the edges on an edge-path. On k Å 5: m2 Å

n 0 36

2

/ 5

n / 36

2the other hand, there are three different ways to disconnectby removing three edges: removing three edges from oneedge-path; removing two from one edge-path and one Å n 2 0 2n 0 3

12.

from a different edge-path; or removing exactly one edgefrom each of the three edge-paths emanating from an

(ii) The UOR simple graph in the class Vs(n, n / 2)original node of the K4 . The number of ways each canhas m3 Å a / b / c, where a is the number of waysbe done depends on how the n 0 4 nodes are inserted in

the six edges of K4 and is summarized in the following to remove three edges from one edge-path, b is thetheorem: number of ways to remove two edges from one edge-

path and one edge from a different edge-path, andTheorem 4.1. Let k Å (n 0 4) mod 6. c is the number of ways to remove exactly one edge

from each of the three edge-paths emanating from( i ) The UOR simple graph in the class Vs(n, n / 2) an original K4 node. The following chart summarizes

this information for m3:has



a b c

k Å 0 6

n / 26

3

6

n / 26

2F5n / 10

6 G 4Fn / 26 G3

k Å 1 5

n / 16

3

/n / 7

6

3

5

n / 16

2F5n / 11

6 G / n / 76

2F5n / 5

6 G 2Fn / 16 G2Fn / 7

6 G / 2Fn / 16 G3

k Å 2 4

n

6

3

/ 2

n / 66

3

4

n

6

2F5n / 12

6 G / 2

n / 66

2F5n / 6

6 G 4Fn

6G2Fn / 6

6 Gk Å 3 3

n 0 16

3

/ 3

n / 56

3

3

n 0 16

2F5n / 13

6 G / 3

n / 56

2F5n / 7

6 G 2Fn 0 16 G2Fn / 5

6 G / 2Fn 0 16 GFn / 5

6 G2

k Å 4 2

n 0 26

3

/ 4

n / 46

3

2

n 0 26

2F5n / 14

6 G / 4

n / 46

2F5n / 8

6 G 4Fn 0 26 GFn / 4

6 G2

k Å 5

n 0 36

3

/ 5

n / 36

3

n 0 36

2F5n / 15

6 G / 5

n / 36

2F5n / 9

6 G 2Fn 0 36 G2Fn / 3

6 G / 2Fn 0 36 GFn / 3

6 G2

The values of a / b / c are as follows: nonnegative integers. If s 0 r Å k ú 0 , then f (r , s)0 f (r / 1 , s 0 1) Å k 0 1 .

k Å 0:5n 3 / 3n 2 0 30n 0 32

54 Proposition 6. Let n be a fixed positive integer, and letr and s be nonnegative integers. Then,

k Å 1:5n 3 / 3n 2 0 18n 0 16

54 ( i) The expression ( r2) / ( n0r

2 ) is minimized when rÅ n /2 .

k Å 2:5n 3 / 3n 2 0 18n

54 ( ii) The expression ( r2) / ( s

2) / ( n0r0s2 ) is minimized

when r Å n /3 and s Å n /3 .

k Å 3:5n 3 / 3n 2 0 12n / 4

54

k Å 4:5n 3 / 3n 2 0 18n 0 16

54

k Å 5:5n 3 / 3n 2 0 36n

54.

Corollary 4. Let G be the UOR simple graph in Vs(n,n / 2) . Then, m2(G) ° (n 2 0 2n / 1)/12 and m3(G)° (5n 3 / 3n 2 0 12n / 4)/54.

There are four types of constructions that givebridgeless multigraphs in the class V(n , n / 2). Proposi-tion 6, which follows below, enables us to find the mostreliable multigraph within each type. We first state alemma which is used in the proof of the proposition:

Fig. 5. UOR graphs in Vs (n , n / 2), n Å 10, 11, . . . , 15.The large nodes are the k additional nodes.Lemma 5. Let f (r , s) Å ( r

2) / ( s2) , where r and s are



binatorial formulas for m2 and m3 , if in Step 2 of theconstruction an original edge was not subdivided, then ifthis edge was the one doubled in Step 3, it will be consid-ered an edge-path consisting of two multiple edges whichform a double edge; otherwise, the unsubdivided edge isconsidered an edge-path consisting of a single simpleedge. Let r be the number of simple edges on the edge-path containing the double-edge; s , the number of simpleedges on a second edge-path; and n 0 r 0 s , the numberof simple edges on the third edge-path.

Fig. 6. Type1 multigraphs in V(7, 9) . The only way to disconnect a Type1 multigraph by theremoval of two edges is to remove two simple edges fromthe same edge-path. Thus, m2 Å ( r

2) / ( s2) / ( n0r0s

2 ) . By( iii) The expression ( r2) / ( n0r

2 ) / r /2 is minimized whenpart ( ii ) of Proposition 6, we know that m2 is minimizedr Å n /2 .when rÅ n /3 and s Å n /3 , i.e., the number of simple( i£) The expression ( r

2) / ( s2) / ( n0r0s

2 ) / r /2 is mini-edges on each edge-path is as equal as possible.mized when r Å n /3 and s Å n /3 .

There are three ways to disconnect by removing threeedges: remove any three simple edges; remove two simpleProof. Parts ( i) and (ii) are well known and the proofedges from the same edge-path and one of the multipleof ( iv) is similar to that of ( iii ) , so we only give theedges from the double-edge; or remove the double-edgeproof of (iii ) .and a simple edge from the edge-path that contained the(iii ) Let s Å n 0 r , and define the function F(r , s)double-edge. Hence, m3 Å (n

3) / 2(( r2) / (s

2) / (n0r0s2 ))Å f (r , s) / r /2, where f (r , s) is as defined in Lemma

/ r . Since the first summand is constant for all graphs5. Trivially F(r , s) ° F(s , r) , whenever r ° s , soof Type1, applying part ( iv) of Proposition 6 yields thatwithout loss of generality, we assume that r ° s . Assumem3 is also minimized when r Å n /3 and s Å n /3 .that F(r , s) is minimized and r õ n /2 . Then, since r

Thus, the best graph of this type, which we call M1 ,/ s Å n , s ú n /2 . Thus, s 0 r ¢ 2. Lemma 5 impliesis constructed as follows: Begin with three edges betweenthat f (r , s) 0 f (r / 1, s 0 1) ¢ 1. Hence, F(r , s)a pair of nodes and distribute n 0 2 nodes as evenly as0 F(r / 1, s 0 1) Å ( f (r , s) / (r /2)) 0 ( f (r / 1, spossible among the three edges. Next double an edge on0 1) / (r / 1)/2)¢ 1

2, which contradicts the assumptionan edge-path that has the most nodes.that F(r , s) is minimized. Therefore, r Å n /2 .

The values of m2 and m3 depend on the value of kÅ n mod 3. Specifically,

Next, we describe the four constructions leading tobridgeless multigraphs in V(n , n/ 2). For each construc-tion type, we determine the best multigraph, then find the

k Å 0: m2 Å 3

n

3

2

Å n 2 0 3n

6,best one from among the four, and finally compare the

best multigraph to the UOR simple graph in V(n , n / 2).A multigraph is of Type1 if it is constructed in the

following manner:m3 Å Sn

3D / 6

n

3

2

/ n

3Å n 3 0 n 2 0 2n

6;

STEP 1. Start with a three edges between a pair of nodes.

STEP 2. Insert n 0 2 nodes randomly in the three edges.

STEP 3. Double one of the edges.k Å 1: m2 Å 2

n 0 13

2

/n / 2

3

2

Å n 2 0 3n / 26

,Type1 multigraphs include M2 : the n-cycle with oneedge tripled (i.e., insert the nodes into only one edge anddouble one of the original edges); M3 : the n-cycle withtwo edges doubled (i.e., insert the nodes into only one m3 Å Sn

3D / 2m2 /n 0 1

3Å n 3 0 n 2 0 2n / 2

6;

edge and double one of the newly-formed edges); M4 :the chordal n-cycle with a double-edge on the cycle (i.e.,insert the nodes into two of the original edges and doubleone of the newly formed edges); and M5 : the n-cycle k Å 2: m2 Å

n 0 23

2

/ 2

n / 13

2

Å n 2 0 3n / 26

,with a double chord (i.e., insert the nodes into two of theoriginal edges and double the other original edge). (SeeFig. 6 for an illustration.)

m3 Å Sn

3D / 2m2 /n 0 2

3Å n 3 0 n 2 0 2n

6.We refer to the resulting paths between the original

two nodes as edge-paths. For ease of describing the com-



the three edges. The pendent double edge may be addedanywhere.

It follows immediately from the discussion above thatm2(M6) 0 m2(M1) Å 1, and m3(M6) 0 m3(M1) Å n 0 rú 0. Therefore, M1 is the best multigraph among allmultigraphs of Type1 or Type2.

A multigraph is of Type3 if it is constructed in thefollowing manner:

STEP 1. Start with a path of length 2.

STEP 2. Double each edge.Fig. 7. Type2 multigraphs in V(7, 9) .

STEP 3. Insert n 0 3 nodes randomly in the double edges.

STEP 4. Double an edge.Thus, for any value of n , we obtain (n 2 0 3n) /6If the nodes are inserted in both of the double edges° m2(M1) ° (n 2 0 3n / 2)/6 and (n 3 0 n 2 0 2n) /6

formed in Step 2, the resulting multigraph, M8 , is a bicycle° m3(M1) ° (n 3 0 n 2 0 2n / 2)/6.( i.e., two cycles with a common node) with one doubleA multigraph is of Type2 if it is constructed in theedge. If the nodes were inserted in only one of the doublefollowing manner:edges formed in Step 2, then if one of the multiple edges

STEP 1. Start with three edges between a pair of nodes.from the other double edge was the one doubled in Step4, we obtain M9 , an (n 0 1)-cycle with a pendent tripleSTEP 2. Insert n 0 3 nodes randomly in the three edgesedge; otherwise, the graph obtained, M10 , is an (n 0 1)-so that at least two of the edges receive nodes.cycle with one edge doubled and a pendent double edge.

STEP 3. Connect a new node to any other node using two(See Fig. 8 for an illustration.)

edges, forming a pendent double edge.For ease of describing the combinatorial formulas for

If only two of the original three edges are subdivided,m2 and m3 , if the multigraph has a pendent double edge,

the resultant multigraph, M7 , is a chordal (n 0 1)-cyclewe treat it as a cycle consisting of two simple edges, and

with a pendent double edge. (See Fig. 7 for an illustra-if the multigraph has a pendent triple edge, we consider

tion.)it to be a cycle of length 2 with one double edge and one

We refer to the three paths between the original twosimple edge. Let r be the number of simple edges on the

nodes as edge-paths. For ease of describing the combina-cycle containing the double edge; then, n 0 r is the num-

torial formulas for m2 and m3 , if in the construction oneber of simple edges on the other cycle.

of the original three edges is not subdivided, it will beThe only way to disconnect a Type3 multigraph by the

considered an edge-path consisting of a single simpleremoval of two edges is to remove two simple edges on

edge. Let r be the number of simple edges on the edge-one of the cycles. Thus, m2 Å ( r

2) / ( n0r2 ) . By part ( i) of

path containing the double edge; s , the number of simpleProposition 6, m2 is minimized when r Å n /2 .

edges on a second edge-path; and, therefore, n 0 r 0 s ,There are three ways to disconnect a Type3 multigraph

the number of simple edges on the third edge-path.by removing three edges: remove any three simple edges,

There are two ways to disconnect a Type2 multigraphremove two simple edges from the same cycle and a

by the removal of two edges: remove two simple edgesfrom the same edge-path or remove the pendent doubleedge. Thus, m2 Å ( r

2) / ( s2) / ( n0r0s

2 ) / 1. Again, by part( ii ) of Proposition 6, m2 is minimized when r Å n /3and s Å n /3 , that is, the n 0 3 nodes should be insertedas evenly as possible along the edge paths.

There are three ways to disconnect by a Type2multigraph by the removal of three edges: remove anythree simple edges, remove two simple edges from thesame edge-path and one of the multiple edges, or removethe pendent double edge and any other edge. Hence, m3

Å ( n3) / 2(( r

2) / ( s2) / ( n0r0s

2 )) / n . Once again, m3 isminimized when r Å n /3 and s Å n /3 .

Thus, the best graph of this type, M6 , is constructedas follows: Start with three edges between a pair of nodes

Fig. 8. Type3 multigraphs in V(7, 9) .and distribute n 0 3 nodes as evenly as possible among



multiple edge, or remove the double edge and a simpleedge from the cycle containing the double edge. So, m3

Å ( n3)/ 2(( r

2)/ ( n0r2 ))/ r , which by part ( iii ) of Proposi-

tion 6 is also minimized by r Å n /2 . Thus, the bestgraph of this type, M8 , is obtained as follows: Start witha path of length two, double each edge, insert the n 0 3nodes alternately into each double-edge, and double anedge on the resulting cycle having the most nodes.

The values of m2 and m3 depend on the value of kÅ n mod 2. Specifically,

Fig. 9. Type4 multigraphs in V(7, 9) .k Å 0: m2 Å 2

n

2

2

Å n 2 0 2n

4,

from the same cycle or remove the double edge. Thus,m3 Å Sn

3D / 4

n

2

2

/ n

2Å n 3 0 n

6; m2 Å ( r

2) / ( n0r2 ) / 1. By part ( i) of Proposition 6, m2

is minimized when r Å n /2 .There are three ways to disconnect a Type4 multigraph

by removing three edges: remove any three simple edges,remove two simple edges from the same cycle and a multi-k Å 1: m2 Å

n 0 12

2

/n / 1

2

2

Å n 2 0 2n / 14

,ple edge, or remove the double edge and any other edge.Therefore, m3Å (n

3)/ 2(( r2)/ (n0r

2 ))/ n . Again, applyingProposition 6(i), m3 is minimized when r Å n /2 .

m3 Å Sn

3D / 2m2 /n 0 1

2Å n 3 0 n

6. Thus, the best multigraph of Type4, M11 , is obtained

when the nodes are inserted as evenly as possible intotwo of the double edges. It follows immediately from thediscussion above that m2(M8) õ m2(M11) , and m3(M8)Thus, for any value of n , we obtain m2(M8) ¢ (n 2

õ m3(M11) , so M8 is the best multigraph among all0 2n) /4 and m3(M8) Å (n 3 0 n) /6.multigraphs of Type3 or Type4.Finally, a multigraph is of Type4 if it is constructed in

We now compare the values of m2 and m3 for thethe following manner:multigraphs M1 and M8 . Since m2(M1) ° (n2 0 3n / 2)/

STEP 1. Start with a tree on four nodes. 6 and m2(M8) ¢ (n2 0 2n)/4, m2(M8) 0 m2(M1) ¢ [(n0 2)(n / 2)]/12 ú 0 when n ú 2. Since m3(M1) ° (n3

STEP 2. Double each edge. 0 n2 0 2n / 2)/6 and m3(M8) Å (n3 0 n)/6, m3(M8)0 m3(M1) ¢ [(n / 2)(n 0 1)]/6 ú 0 when n ú 1.STEP 3. Insert n 0 4 nodes randomly in the double edges,Hence, M1 is the best multigraph from among all fourleaving at least one double edge unchanged.types.If the nodes were inserted in two of the double edges,

Before we compare M1 to the UOR simple graph inthe resulting multigraph is either a bicycle with a pendentV(n , n / 2), we must demonstrate that the constructionsdouble edge (M11 , M12 , or M13) or two cycles joinedhave included all possible bridgeless multigraphs in theby a double bridge; else, the multigraph consists of an (nclass. To this end, we consider the graphs formed by0 2)-cycle with either two pendent double edges (M14

deleting one multiple edge from a bridgeless multigraphor M15) or a pendent doubled path of length two (M16) .in V(n , n / 2), that is, the graphs in V(n , n / 1) which(See Fig. 9 for an illustration.)have at most one bridge. The bridgeless multigraphs inFor ease of describing the combinatorial formulas forV(n , n / 2) are obtained by adding a multiple edgem2 and m3 , if the nodes are inserted in only one of theanywhere if the graph in V(n , n / 1) is bridgeless; other-double edges formed in Step 2, then one of the otherwise, the bridge must be doubled. We let H denote thedouble edges is considered to be a cycle consisting ofappropriate graph in V(n , n / 1).two simple edges, while the other is treated as a double

If H is a cycle with a double edge, then the additionedge (see M14 , M15 , M16 in Fig. 9) . Let r be the numberof another multiple edge results in either a cycle with twoof simple edges on one cycle, and n 0 r , the number ofdouble edges or a cycle with a triple edge, both of Type1.simple edges on the other cycle.

If H is a cycle with a pendent double edge, then addingThere are two ways to disconnect a Type4 multigraphby removing two edges: either remove two simple edges another multiple edge results in either a cycle with a



pendent triple edge or a cycle with a double edge and apendent double edge, both of Type3.

If H is a simple bridgeless graph, then it is either achordal n-cycle, a bicycle, or a graph which is three paths,each of length at least two, between two nodes. If H is achordal cycle, doubling any edge yields a graph of Type1.If H is a bicycle, doubling any edge yields a graph ofType3. If H consists of three paths between a pair of

Fig. 10. The multigraph M1 for n Å 4.nodes, doubling any edge yields a graph of Type1.If H is a graph formed by adding a pendent edge to a

simple bridgeless, it must be either a chordal n-cycle with (see Fig. 10). For this multigraph m2 Å 1 and m3 Å 7,a pendent edge, a bicycle with a pendent edge, or a graph so the UOR simple graph is better when n Å 4, and,which is three paths, each of length at least two, between therefore, it is best over all graphs in V(n , n / 2).two nodes, with a pendent edge. Since the graph in V(n ,n / 2) which results from adding the edge to H must bebridgeless, the pendent edge must be doubled, thereby 5. CONCLUSIONresulting in a graph of Type2, Type4, or Type2, respec-

The work of Boesch et al. [1] gave the constructions fortively.the UOR simple graphs in sparse classes which have nIf H has a bridge between two cycles, then the only0 1, n , n / 1, and n / 2 edges. We have shown thatchoice is to double the bridge, which gives a graph ofwhen the class is enlarged to include multigraphs theType4.graphs they found are still UOR. The class V(n , n / 3)If H is a multigraph with a pendent edge, then it iswas covered by Wang [5]. Consideration of bridgelessone of the following: a cycle with a double edge and amultigraphs in this class involves many more instances.pendent edge, a cycle with a double pendent edge and aWe conjecture that the UOR graphs Wang found are UORsingle pendent edge (emanating from either the same oreven when the class is expanded to include multigraphs.different nodes) , or a cycle with a pendent path of length

two having one of the paths edges doubled. In all thesecases, adding another multiple edge requires doubling the

REFERENCESbridge. This yields, respectively, graphs of Type3, Type4,Type4, or Type4.

[1] Boesch, Li, and Suffel, On the existence of uniformlyFinally, it remains to compare M1 to the UOR simple most reliable networks. Networks 21 (1991) 181–194.

graph. Since m2(UOR)° (n 2 0 2n / 1)/12 and m2(M1) [ 2] C. Cheng, Maximizing the total number of spanning treesú (n2 0 3n)/6, m2(M1) 0 m2(UOR) ú (n2 0 4n 0 1)/12 in a graph: Two related problems in graph theory andú 0 for n ú 4. Since m3(UOR) ° (5n 3 / 3n 2 0 12n optimization design theory. J. Combin. Theory 31 (1981)/ 4)/54 and m3(M1) ¢ (n 3 0 n 2 0 2n) /6, m3(M1) 240–248.0 m3(UOR) ¢ (2n 3 0 6n 2 0 3n 0 2)/27 ú 0 for n ú [ 3] F. Harary, Graph Theory. Addison-Wesley, Reading, MA3. Thus, the UOR simple graph is better for n ú 4. When (1971).n Å 4, the UOR simple graph is just K4 , which has m2 [ 4] F. Moskowitz, The analysis of redundancy networks.Å 0 and m3 Å 4, while M1 is the multigraph obtained by AIEE Trans. Comm. Electron. 39 (1958) 627–632.starting with a triple edge, inserting a node in two of the [ 5] G. Wang, A proof of Boesch’s conjecture. Networks 24

(1994) 277–284.edges, and then doubling one of the newly formed edges


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Uniformly optimally reliable graphs