on mean distance in certain classes of graphs

7
On Mean Distance in Certain Classes of Graphs George R.T. Hendry* Aberdeen University, Scotland The mean distance, k(G), of a graph G is the arithmetic mean of the distances in G. Upper and lower bounds for the mean distance of a self-complementary graph of given order are obtained and the extremal graphs are determined. It is shown that if r 3 2 is rational, then there exist (1) graphs with arbitrarily low or high edge density, (2) bipartite graphs and (3) oriented graphs with mean distance equal to t, and also (4) trees and (5) tournaments with mean distance arbitrarily close to f. A number of open problems are mentioned. 1. INTRODUCTION AND NOTATION The mean distance [ 11, p(G), of a connected graph or strong digraph G of order n 2 2 is defined by p(G) = Xu.vEV(G&d~,~)/n (n - l), where dG(u,v)denotes the distance in G from u to v. Thus p(G) is the arithmetic mean of the distances in G. The eccentricity, ecG(v), of vertex v in G is defined by ecG(v) = maxuEV(G~dG(u,v). The maximum distance between two vertices of G is the diameter, diam(G), of G. We refer the reader to [4] for an excellent survey of results on mean distance in graphs and digraphs. The main results of this paper are presented in Section 2, where best possible upper and lower bounds for the mean distance of a self-complementary graph of given order are obtained and the extremal graphs are determined (Theorems 2.2 and 2.3). Theorem A [4, Theorem 91. For each pair of real numbers t 5 1 and E > 0, there exists a graph G such that Ip(G) - tl < E. Responding to a problem of Plesnik [4, p. 191, M. Truszczynski and the present author independently proved: Theorem B [3,5]. For each rational number t 3 1, there exist infinitely many graphs G such that p(G) = t. In [4], Plesnik proved the following result: *This work was supported by SERC Research Studentship No. 8331491 1 NETWORKS, Vol. 19 (1989) 451457 0 1989 John Wiley & Sons, Inc. CCC 0028-3045/89/030451-07$04.00

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Page 1: On mean distance in certain classes of graphs

On Mean Distance in Certain Classes of Graphs George R.T. Hendry* Aberdeen University, Scotland

The mean distance, k(G), of a graph G is the arithmetic mean of the distances in G. Upper and lower bounds for the mean distance of a self-complementary graph of given order are obtained and the extremal graphs are determined. It is shown that if r 3 2 is rational, then there exist (1) graphs with arbitrarily low or high edge density, (2) bipartite graphs and (3) oriented graphs with mean distance equal to t , and also (4) trees and (5) tournaments with mean distance arbitrarily close to f . A number of open problems are mentioned.

1. INTRODUCTION AND NOTATION

The mean distance [ 11, p(G), of a connected graph or strong digraph G of order n 2 2 is defined by p(G) = X u . v E V ( G & d ~ , ~ ) / n ( n - l ) , where dG(u,v) denotes the distance in G from u to v. Thus p(G) is the arithmetic mean of the distances in G. The eccentricity, ecG(v), of vertex v in G is defined by ecG(v) = maxuEV(G~dG(u,v). The maximum distance between two vertices of G is the diameter, diam(G), of G . We refer the reader to [4] for an excellent survey of results on mean distance in graphs and digraphs.

The main results of this paper are presented in Section 2, where best possible upper and lower bounds for the mean distance of a self-complementary graph of given order are obtained and the extremal graphs are determined (Theorems 2.2 and 2.3).

Theorem A [4, Theorem 91. For each pair of real numbers t 5 1 and E > 0, there exists a graph G such that Ip(G) - tl < E.

Responding to a problem of Plesnik [4, p. 191, M. Truszczynski and the present author independently proved:

Theorem B [3,5]. For each rational number t 3 1, there exist infinitely many graphs G such that p ( G ) = t .

In [4], Plesnik proved the following result:

*This work was supported by SERC Research Studentship No. 8331491 1

NETWORKS, Vol. 19 (1989) 451457 0 1989 John Wiley & Sons, Inc. CCC 0028-3045/89/030451-07$04.00

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452 HENDRY

In Section 111, several further results of types A and B are obtained for graphs with arbitrarily low or arbitrarily high edge density (Theorems 3 .1 and 3 . 2 ) , bipartite graphs (Theorem 3 . 3 , trees (Theorem 3.6), oriented graphs (Theorem 3.8) and tournaments (Theorem 3.9). Unfortunately, the proofs of these results involve routine and tedious calculations which, for the sake of brevity, are omitted. A number of open problems are mentioned.

2. SELF-COMPLEMENTARY GRAPHS

Throughout this section G will denote a self-complementary graph of order n 2 4 with complementing permutation T . We refer to [ 2 ] for notation and basic results concerning self-complementary graphs. In particular, we note that: IV(G)I = 0 or 1 (mod 4); G is connected; A(G) =S n - 2; T may be chosen so that each cycle in T

has a length a power of 2 (with at most one cycle of length one and none of length two); for k 2 1, tk+ I is a complementing permutation and T*' is an isomorphism of G; ( x , y ) E E ( G ) if and only if ( - r U f l ( x ) , 72k"(y)) JZ E ( G ) and (T'~(X),T~& (y)) E E(G) , k 2 1; G has diameter 2 or 3.

Notation Let G* denote the graph with

V(G*) = {v E V ( G ) : ecG(v) = 3)

and

Theorem 2.1. G* is bipartite.

Proof. Let C be a component of G*. We need only show that C is bipartite. We may assume that there is a vertex u in C such that ( u , T ( u ) ) E E ( G ) , for otherwise we consider the complementing permutation T-' . We now prove two propositions from which the theorem follows easily.

If v E V ( C ) then (v,T(v)) E E(G) . (1)

Proof of ( I ) . Since C is connected, it is clearly sufficient to show that the result is true when (u,v) E E ( C ) . Therefore suppose (u,v) E E ( C ) . Since &(u,v) = 3 and ( u , T ( u ) ) E E ( G ) , we have (v,T(u)) E E ( G ) and so (T- ' (v ) ,u) E E(G) . Therefore (T- ' (v ) ,v ) E E ( G ) and so (v,T(v)) E E ( G ) , as required.

(2) If there is a path of length 2r 5 0 from w to v in C then (w,T(v)) E E(G) . Proof of ( 2 ) . We proceed by induction on r: the result following by (1) for r = 0.

Therefore suppose there is a w - v path P: w = vo,vl , . . . ,vZr = v of length 2r > 0 in C. By the inductive hypothesis, ( V ~ , T ( V ~ , - ~ ) ) E E(G) . Since &(v0,vI) = 3, it fol- lowsthat(v0,T(v2,1)) $Z E ( G ) a n d s o ( ~ - ~ ( v ~ ) , v ~ , - , ) E E ( G ) . SincedG(v,,,,v2,) = 3, it follows that ( T - ' ( V ~ ) , V ~ , ) JZ E ( G ) and so (vO,-r(vZr)) E E ( G ) . This completes the inductive step and hence the proof of ( 2 ) .

Now suppose that vI,v2, . . . . v ~ ~ + ~ Y ~ ( T 3 0) is an odd cycle in C. By ( 2 ) , G

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MEAN DISTANCE IN CERTAIN CLASSES OF GRAPHS 453

FIG. 1. A typical graph in %&.

contains the edges ( V ~ , T ( V ~ ~ - J ) and (vZr+ 1,~(v2r-1)). Therefore dC(vl,vzr+ ,) 6 2 which

Notation. For k 3 1 , let %44k denote the set of all graphs of the type shown in Figure 1 , where Gk is any graph of order k and the double lines connecting two circled graphs indicate that all edges between them are present.

Theorem 2.2. If G is a self-complementary graph of order n = 4k then 3/2 S p(G) S (13n - 12)/ (8n - 8). Furthermore, equality holds inthe lowerbound iff G has diameter two and in the upper bound iff G E %4t.

Since G has A(;) edges, we have

contradicts ( v ~ , v ~ ~ + ~ ) E E(G*). Therefore C, and hence G*, is bipartite.

Proof (Lower bound).

(1/2) (;) ’ 1 + (1/2) (‘i) * 2 - _ 3 (‘i) 2

- P(G) 3

with equality iff G has diameter two. (Upper bound) Since diam(G) S 3, all distances in G equal 1, 2 or 3. Furthermore,

since the number of distances in G which equal three is IE(G*)I, we need only find an upper bound on (E(G*)I. Since A(G) S n - 2 and diam(G) S 3, every vertex of G has eccentricity two or three. Let Az and A, denote the sets of vertices in G having eccentricity two and three, respectively, and let A, and A, be the corresponding sets in G. Since it is easily seen that v E A3 implies v E AZ, we have [A3[ S lAzl = IAzl = 4k - IA31. Therefore (V(G*)I = (A3[ S 2k. By Theorem 2.1, G* is triangle-free and so, using Turdn’s Theorem, we have 1E(G*)I 6 a IV (C*)l 6 k2. Consequently

(1/2)(1) * 1 + ((1/2)(2) - k2) * 2 + k2 * 3 13n - 12 . (1)

Conversely, suppose that G is a self-complementary graph of order n = 4k such that F ( G ) = (13n - 12)/ (8n - 8). We will show that G E %4k. Since equality must hold in each inequality used in establishing ( l ) , it follows that G* has exactly 2k vertices and k2 edges. Therefore, again by TurBn’s Theorem, G*

Let us assume that the complementing permutation T of G is the product of t 3 1 disjoint cycles Cl,Cz, . . . ,C,, where the cycle Ci = (44 + 1 , 4ai + 2, . . . ,4bi) has length 4ki with bi = Zjj=Ikj and ai = bi - k i . We call this the standard form. (2) If u,v E A, lie on the same cycle C of T, then the distance from u to v around C is even.

- - (9 8n - 8 P(G) 6

Kk,&.

Proof of (2) . Suppose that the distance from u to v around C is 2t + 1, for some t 3 0. Then cr = -?‘+I is also a complementing permutation of G and a(u) = v.

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454 HENDRY

Since A3 C A2, we have 3 = ecG(u) = ecc(v) = 2. This contradiction completes the proof of (2).

Therefore, for 1 S i c t , C, contains exactly 2k, vertices of A3 and these vertices all have the same parity.

(3) If d,(x,y) = 3, then for i 3 0, dG(x,-r4'+'(y)) S 2. Proof of (3). Suppose that dG(x,y) = dG ( x , ~ ~ " ~ ( y ) ) = 3. Since G contains ex-

actly one of the edges (y,?"(y)) and (g '+I (y) , ~ ~ ' + ' ( y ) ) , it follows that ( X , T ~ " ' ( Y ) )

$Z E ( G ) . Therefore G contains the edges (T-~ ' - ' ( x ) , y ) and ( ~ ~ ' ~ ' ( x ) , ~ ~ ' ' ~ ( y ) ) and consequently contains neither of the edges ( T - ~ ' - I ( x ) ~ ) and ( T ~ ' + ' (x),x). This con- tradiction establishes (3).

Therefore each vertex of A3 is at distance three from at most k, vertices of A3 on C,, 1 c i c t , and so

r

kz = IE (G*)l S kf + 2 2k,k, = k, = k2 (4) I = I I S K j S : ) .

Since equality must hold throughout (4), we conclude that each vertex of A3 is at distance three from exactly k, vertices of A3 on C,, 1 S i S t.

In view of these remarks and statements (3) and (4), it is clear that, by suitably rotating and relabeling the vertices in each C, , we may ensure that

(i) T is in standard form, (ii) all the vertices in A3 are odd-labeled and (iii) for 2 c i c t , dG(1,4a, + 1) S 2.

( 5 ) d,(x,y) = 3 iff x = y We are now prepared to show that G E %4L. Partition V ( G ) into four sets A, B, C ,

The point of all this is that we now have 1 (mod 2) and x - y = 2 (mod 4).

and D as follows:

A = (4i + 1 : 0 c i S k - 1) = {u: u = 1 (mod 2), d(1 ,u) S 2},

B = {u: u = O (mod 2), d(1,u) = I},

C = {u: u = O (mod 2), d(1,u) = 2),

and

D = {4i + 3: 0 < i S k - 1) = { u: d(1,u) = 3).

Now there are two cases to consider, depending on whether or not (1,2) E E ( G ) . However, since the two proofs run parallel, we will only treat the case in which ( 1 2 ) E E ( G ) .

( 6 ) For 1 S i S t , (4a, + 1,4u, + 2) E E ( G ) . Proof of (6). By assumption, this is true for i = 1. So consider i , 2 S i S f .

Since dG(1,4u, + 3) = 3 and (1,2) E E ( G ) , we have (2,4u, + 3) E E ( G ) and so (1,4a, + 2) E E ( G ) . Therefore (4u, + 2,&, + 3) E E ( G ) and (6) follows.

(7) If 2 < 2r c 4k then (1,2r) E E ( G ) iff 2r = 2 (mod 4). Proofof(7). W.1.o.g. let us assume that 2r is on cycle C,: S O ~ U , + 2 e 2r S 4b,. First suppose that (1,2r) E E ( G ) . Since (4u, + 1,4a, + 2) E E ( G ) , it follows that

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MEAN DISTANCE IN CERTAIN CLASSES OF GRAPHS 455

(2r - 1,2r) E E(G) and hence that dG(1,2r - 1) S 2. Therefore, by (3, 2r = 2 (mod 4).

Conversely, suppose that 2r = 2 (mod 4) and (1,2r) E E(G) . Therefore (2,2r + 1) E E ( G ) and, since (1,2) E E(G) , we have dG(l,2r + 1) S 2, which contradicts (5). This completes the proof of (7).

By(7), we have B = (4i + 2: 0 S i S k - 1)andC = (42' + 4: 0 S i S k - 1). Now the mappings -? I,.,: A + D, * I B : B + C and - T I A : A + B , respectively, show that <A> = <D>, <B> <C> and <A> = <B>. By definition, there are no edges from B to D. Applying -?, T , and 9, it follows that there are no edges from A to C and that all edges from D to C and from A to B are present. If 4i + 2 E B and 4j E C then, by (3, dc(4i + 1,4j - 1 ) = 3 and so (4i + 1,4j - 1) JZ E ( G ) and (4i + 2,4j) E E ( G ) . Therefore all edges from B to C are present and, applying T ,

there are no edges from D to A. Therefore G E %4k. Since it is easily verified that each graph in %4k is self-complementary and has mean distance ( 13n - 12)/(8n - 8),

w

Notation. For k 5 1 , let %4k+ be the set of graphs which can be obtained from a graph in %4k by adding a new vertex adjacent to all vertices in the two copies of Gk as shown in Figure 1 .

Theorem 2.3. If G is a self-complementary graph of order n = 4k + 1 then 3/2 s p(G) S (13n - 1)/8n. Furthermore, equality holds in the lqwer bound iff G has diameter two and in the upper bound iff G E %4k+ I .

The proof for the lower bound is the same as in the case n = 4k. We now establish the upper bound. Let 0 be the vertex of G such that T( 0) = 0. So deg( 0) = 2k and ecc(0) = 2. Since G - 0 is a self-complementary graph of order 4k, we have

the proof of Theorem 2.2 is complete.

Proof.

~ { u . " ) c v ( c - o ) &(u,v) + C u E V ( G - O ) d(O&) I )

( 1 ) p(G) =

(f)(52k - 12)/(32k - 8) + 6k s (4k: 9

1311 - 1 8n '

- -

Conversely, suppose that p(G) = (13n - 1)/8n. Since equality must hold through-

(i) p(G - 0) = (52k - 12)/(32k - 8) and (ii) for u,v E V(G - 0 ), dG(u,v) = d c a ( u , v ) .

out ( l ) , we have

By (i) and Theorem 2.2, it follows that G - 0 E %4k. We may assume w.1.o.g. that T I V ( G - o ) , which is a complementing permutation of G - 0, is in standard form and that a vertex has eccentricity three in G - 0 if and only if it is odd. Since 0 is

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456 HENDRY

adjacent to alternate vertices in each cycle of T , it follows from (ii) that 0 is only

Problem 2.4.

Problem 2.5. graph G of order 4k.

Problem 2.6. there exist a self-complementary graph of order 4k such that p(G) = r/($k)?

adjacent to even vertices. Therefore G E %4L+ as desired.

Is G* an (equipartite) subgraph of K k . k if n = 4k?

Characterize graphs H such that H = G* for some self-complementary

For k 2 2 and each integer r , 12k2 - 3k S r S 13k2 - 3k, does

3. FURTHER RESULTS AND PROBLEMS

The edge densify, p(G), of a graph G is defined by p(G) = ~ E ( G ) ~ / ( l " ~ G ) ~ ) . Here we consider the question of whether there exist graphs with prescribed mean distance having arbitrarily low or high edge density. The proofs of the next two results are similar to that used in [3].

Theorem 3.1. G such that p(G) = rand p(Gj < E.

Theorem 3.2. Let r > 1 be rational and E > 0 be real. Then there exists a graph G such that p(G) = rand p(G) > 1 - E.

For rational r 2 1, define A(r) = inf{p(G): p(G) = r } and B ( r ) = sup(p(G): p(G) = t } . Obviously, A(1j = 1 and, by Theorems 3.1 and 3.2, A(t) = 0, r 2 2, andB(r) = 1, t 1.

Problem 3.3.

Problem 3.4. G with p(G) = t and p(G) = s?

Theorem 3.5. graphs B such that p ( B ) = r .

Let t 2 2 be rational and E > 0 be real. Then there exists a graph

Determine A ( r ) , 1 < r < 2.

For rationals t 2 1 and s, A(t) < s < B ( t ) , does there exist a graph

For each rational number t 2 3/2, there exist infinitely many bipartite

To prove Theorem 3.5, we consider bipartite graphs which are subgraphs of the graph obtained by identifying the final edge of a path of order 4p with an edge of K9+ I ,9 + I and which are supergraphs of the tree obtained from this graph by deleting the $ edges not incident with vertices of the path. To prove the next result, we consider trees obtained by identifying the end vertex of a path with the central vertex of a star.

Theorem 3.6. Let t 2 2 and E > 0 be real numbers. Then there exists a tree T such that Ip(T) - rI < E.

Problem 3.7. Let t 2 2 be rational. Does there exist a tree T with p ( T ) = r? P. Winkler [6] has recently informed me that he has solved this problem. If G* is the symmetric digraph associated with the undirected graph G then

p( G*) = p( G). Therefore there is no interest in considering mean distance in digraphs in general. So we only deal with digraphs which have no cycle of length 2, i.e., oriented graphs.

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MEAN DISTANCE IN CERTAIN CLASSES OF GRAPHS 457

Theorem 3.8. Let t 3 312 be rational. Then there exist infinitely many oriented graphs G such that p(G) = t .

Gp,q, where Hp,q is the oriented graph with vertex set Ui’.= l { ~ i r ~ i } U U:= {wi,xi} and arc set consisting of the 2p - 1 + q directed 4-cycles uiui+ Ivi vi+ ,ui, 1 s i < 2p - 1, and v2pwiu2pxiv2p, 1 S i s q, and the q arcs w j i , 1 S i S q, and where Gp.q is the oriented graph obtained from Hp,q by adding the arcs w,xj, 1 < i , j s q, i # j , and also the arcs of a diameter two tournament of order q on each of the sets {w,: 1 6 i s q} and {x i : 1 G i S q}.

Let Tq denote any diameter two tournament of order q and let Tp denote the tour- nament of diameter p - 1 with vertex set {vi: 1 s i s p } and arc set {vivi+ I : 1 < i < p - 1) U {vivj: 1, S j , j + 2 < i S p } . The tournaments we need in the proof of Theorem 3.9 are those obtained from Tq U Tp by adding all arcs from V(TJ to V(Tp) - {vp} and all arcs from vp to V(T,).

Theorem 3.9. Let t 5 3/2 and E > 0 be real. Then there exists a tournament T such that Ip(T) - tl < E.

Problem 3.10.

To prove this result, we consider oriented graphs G such that Hp,q C G

Let t 5 3/2 be rational. Does there exist a tournament T such that K(T) = t?

REFERENCES [l] J . K. Doyle and J. E. Graver, Mean distance in a graph. Discrete Math. 17 (1977) 147-

154. ._ .

121 R. A. Gibbs, Self-complementary graphs. J. Cornbinatorid Theory Ser. B 16 (1974) 106 123.

[3] G. R. T. Hendry, Existence of graphs with prescribed mean distance. J. Graph Theory 10

[4] J. Plesnk, On the sum of all distances in a graph or digraph. J . Graph Theory 8 (1984)

[5] M. Truszczynski, A graph with mean distance being a given rational. Demonstratio Math.

[6] P. Winkler (personal communication).

(1986) 173-175.

1-21.

18 (1985) 619-620.

Received August, 1986 Accepted June, 1988