on the size of graphs labeled with a condition at distance two

On the Size of Graphs Labeled with a Condition at Distance Two

John P. Georges David W. Mahro

DEPARTMENT OF MATHEMATKS TRlNlTY COLLEGE

HARTFORD, CT 06106

ABSTRACT

A labeling of graph G with a condition at distance two is an integer labeling of V(G) such that adjacent vertices have labels that differ by at least two, and vertices distance two apart have labels that differ by a t least one. The lambda-number of G, A(G), is the minimum span over all labelings of G with a condition a t distance two. Let G(n, k) denote the set of all graphs with order n and lambda-number k. In this paper, we examine the sizes of graphs in G(n, k). We modify Chvatal's result on non-hamiltonian graphs to obtain a formula for the minimum size of a graph in G(n, k) , and w e use an algorithmic approach to obtain a formula for the maximum size. Finally, w e show that for any integer j between the maximum and minimum sizes there exists a graph with size j in G(n, k) . 0 1996 John Wiley & Sons, Inc.

1. INTRODUCTION

The problem of labeling the vertices of a graph with a condition at distance two is a recent variation of the well-known channel assignment problem (see Hale [6] and Roberts [9], [lo]), in which two transmitters interfere with one another if they are sufficiently close. In this variation, the vertices of a graph are labeled subject to constraints on the absolute difference between labels of adjacent vertices, and labels of vertices at distance two. Formally, we define a labeling of a graph G with a condition at distance two, hereafter called a labeling of G, to be an integer assignment L to the vertices of G such that

(i) lL(v) - L(w)l 2 2 if v and w are adjacent, and

(ii) ( L ( v ) - L(w)l 2 1 if the distance between v and w is two.

Journal of Graph Theory, Vol. 22, No. 1, 47-57 (1996) 0 1996 John Wiley & Sons, Inc. CCC 0364-9024/96/010047-11

48 JOURNAL OF GRAPH THEORY

Let .L (G) be the collection of all labelings of G. For any L E L (G), we define the span of L , denoted sL, to be the absolute difference between the maximum and minimum vertex assignments of L . We define the lambda-number of G , denoted A(G), to be minLEr ( G ) SL. If L E L (G) and SL = A(G), we say that L is a lambda-labeling of G. Without loss of generality, we assume that the smallest integer assigned by L is 0, and hence the largest is S L .

The problem of labeling graphs with a condition at distance two was first considered in a private communication between F. S. Roberts and J . R. Griggs in 1988, and subsequently studied in the seminal work of Griggs and Yeh [ 5 ] . Their investigation into the relationship between A(G) and other graph invariants led to general upper bounds of A2 + 2A and n + x(G) - 2 for A(G), as well as the conjecture that A(G) I A2 for all G with A 2 2. Griggs and Yeh also investigated the lambda-numbers of paths, cycles, trees, 3-connected graphs, and hypercubes.

Other researchers have since obtained additional results. Sakai [ 1 I ] considered the lambda- number of chordal graphs. Georges, Mauro, and Whittlesey [4] explored the relationship between the lambda-number of a graph G and the path-covering number of G'. They later used a coding theory approach in [I21 to improve the upper bound on the lambda number of the hypercube obtained by Griggs and Yeh. Chang and Kuo [2] investigated classes of interval graphs, established a polynomial-time algorithm that lambda-labels trees, and obtained an upper bound of A2 + A for A(G). Jonas [7] considered the lambda numbers of planar graphs and 3-regular graphs.

In this paper, we explore the relationship between the size of a graph and its lambda-number. In particular, we obtain explicit formulae for s,(n, k ) and s ~ ( n , k ) , the minimum and maximum sizes of graphs with order n and lambda-number k , respectively. We also show that for any integer j , s , ( n , k ) 5 j 5 sM(n ,k ) , there exists a graph G with order n, lambda-number k , and size j .

2. PRELIMINARIES

Throughout this paper, graphs will be simple and loopless. We shall denote the vertex set and edge set of graph G by V(G) and E(G), respectively. The cardinality of any finite set S shall be denoted ISI; IV(G)l is called the order of G, and IE(G)I is called the size of G. The complement of G shall be denoted G'. We refer the reader to Bondy and Murty [l] for elaboration on terminology not covered in this section.

Let GI and G2 be graphs with disjoint vertex sets. The sum of GI and G2, denoted G I + G2, is the graph whose vertex set is given by V(Gl)UV(G2), and whose edge set is given by E(Gl)UE(Gz). The sum of r copies of G will be denoted rG. The join of GI and Gz, denoted GI V G2, is the graph obtained from GI + G2 by joining each vertex of GI to each vertex of Gz.

A path covering of C, denoted C(C) , is a collection of vertex-disjoint paths in G such that each vertex in V(G) is incident to a path in C(G). A minimum path covering of G is a path covering of G with minimum cardinality, and the path covering number of G, denoted c(G), is the cardinality of a minimum path covering of G. We note that G has a Hamilton path if and only if c ( G ) = 1.

The following result relates the lambda-number of a graph to the path covering number of its complement.

Theorem 2.1 (Georges-Mauro-Whittlesey [4]). (i) A(G) 5 n - 1 if and only if c(G') = 1.

ON THE SIZE OF GRAPHS 49

(ii) Let r be an integer, r 2 2. Then h(G) = n + r - 2 if and only if c(G') = r .

Let G be a graph with vertex set { v I , v 2 , . . . , v,} such that d i is the degree of vertex v , , and dl 5 d2 5 . . . 5 d,. The vector ( d l , d2 , . . . , d , ) is called the degree sequence of G. If ( d i , d:, . . . , d ; ) is the degree sequence of graph GI, then we say that G' is degree majnrized by G if and only if d: 5 d; for all i.

For 1 5 p < q / 2 let Cp,y denote the graph K p V ( K $ + Kq-2p) . Then the degree sequence of C p , y has

I

P9 i f l s i s p , d i = q - p - 1 , i f p + l S i s q - p ,

if q - p + 1 5 i 5 q . L - 1

Theorem 2.2 circuit, then H is degree majorized by some Cp,y.

(Chvhtal [3]). If H is a graph with order q 2 3 , and H does not have a Hamilton I

The next result, originally obtained by Ore, can also be derived from Theorem 2.2.

Theorem 2.3 Hamilton path is G is isomorphic to K,-, + K I .

Let K be a subgraph of G, and let L be a labeling of G. We define the labeling sequence l ( K , L ) to be the vector (mo, ml, m2, . . . , msL) , where mi is the number of vertices of K labeled i under L . For the case when K = G,m, is called the frequency of label i under L . Also for the case K = G, i is called a multiplicity in L if mi > 1, and i is called a hole in L if mi = 0. We denote the set of holes in a labeling L. of G by H ( G , L ) .

Let Ln(G) be the collection of lambda-labelings of G, and let G(n, k ) be the collection of all graphs with order n and lambda-number k . If G(n, k ) is nonempty, then the collection of all graphs in G(a , k ) of size sM(n, k ) is denoted M (n , k ) . Similarly, m(n, k ) is the collection of all graphs in G(n, k ) of size s,(n, k ) . Since it can be easily verified that G(n, k ) is nonempty if and only if 0 5 k 5 2n - 2, k # 1, we will restrict our attention to such n and k throughout the rest of this paper.

(Ore [7]). For n 2 2, the maximum size of a graph G with order n and no If n = 4, then G is isomorphic to either K1,3 or K3 + K1; otherwise,

I

3. MINIMUM SIZE OF A GRAPH IN G ( n , k )

In this section, we prove Theorem 3.1

0, if k = 0, ( k - 3 n + 2 ) (n-k-I) 6n-4

( k - 3 n + 2 ) (n-k-1) 2

2

, if k is even and n < k 5 5,

, if k is odd and n < k 5 y, if k is even, n < k , and k > @$,

if k is odd, n < k , and k > y, 1 k - 1 , if 2 5 k 5 n .

s m ( n , k ) = k(k+2) 8 '

( k - 1) (kf5) 8 '

Since K,' is the only graph in G(n, 0), then sm(n, 0) = 0. We henceforth restrict our attention to k 2 2. It will be convenient to consider the two cases 2 5 k 5 n and n < k 5 2n - 2.

Case 1. 2 I: k 5 n.


Lemma 3.2. For k I 2, s m ( k , k ) = k - 1 .

Pro05 By Theorem 2.1, G E G ( k , k ) if and only if c(GC) = 2. Consequently, K I , k - l E G ( k , k ) , which implies s m ( k , k ) 5 l E ( K l , k - ~ ) l = k - 1.

To show that s , ( k , k ) 2 k - 1, suppose to the contrary that IE(G)I < k - 1 for some G E G ( k , k ) . Then c(GC) = 2, and IE(G')I > ( i ) - ( k - 1) = ( k i l ) . But Theorem 2.3 then implies the contradiction c(G') = 1.

Proof of Theorem 3.1, Case 1. If n = k , then the result follows from Lemma 3.2. If n > k , then the graph G = K1,k-l + (n - k)K1 is in G ( n , k ) , and thus s,(n, k ) 5 IE(G)I = k - 1.

To show s,(n, k ) I k - 1, suppose to the contrary that sm(n, k ) < k - 1 for some integer n ,n > k . Let no be the smallest such integer, and suppose H E m ( n o , k ) . Since no - 3 I s,,(no, k ) = IE(H)I, then H is disconnected. Let Co, CI, . . . , C, be the components of H . Then h ( H ) = max, h(C,), which implies A(C,) = k for some j , O 5 j 5 x , and each of the other components of H is isomorphic to K I . Without loss of generality, we assume A(Co) = k . Let H' = Co + C1 + . . . C,-l. Then H' has order no - 1, size IE(H)I, and lambda-number k . Thus, s,(no - 1 , k ) 5 IE(H')l = lE(H)l = s , , (no,k) < k - 1. But if no - 1 = k , then Lemma 3.2 is contradicted. And if no - 1 > k , then the minimality of no is contradicted. Hence, s,(n, k ) must equal k - 1 for all n > k .

I

I Case 2. n < k 5 2n - 2.

Lemma 3.3. Hamilton path, but no Hamilton circuit.

Let G' be a graph with covering number c 2 2. Then G' V Kc-1 has a I

If G E m(n, k ) , n < k 5 2n - 2, then G' has a covering number c = k - n + 2 2 3. The problem of finding the size of G is thus equivalent to the problem of finding the maximum size of a graph ( G C ) with order n and covering number c 3. But this can be accomplished by an application of ChvBtal's result to the graph G' V K C P I , whose size will prove to be equal to the size of Cp,n+c-l for some p .

Lemma 3.4. concavity.

ProoJ: I Let Hf and Hg denote the respective graphs Kz-1 + Kn-c+l and ( K ; + K n f c - l - z j ) v

K , - ' + I , where j denotes the largest integer that is smaller than (n + c - 1)/2, and where c 2 3. We observe that Hf and Hi have order n and covering number c , which implies that H I and H2 are in G ( n , k ) .

Lemma 3.5. larger of IE(H;)I and IE(H2')I.

Since G E m(n, k ) , the size of G" V Kc- l is maximum among all graphs H" V K c - , , where H E G ( n , k ) . It thus suffices to show that IE(H" v Kc-l) l 5 max{lE(HT V K c - l ) l , IE(Hg V K,-,)I} for all H E G ( n , k ) , with equality for some H .

Select H E G ( n , k ) . Since H" V K C P I has no Hamilton circuit by Lemma 3.3, then Theorem 2.2 shows that H" V Kc-l is degree majorized by some C p , n + c - ~ r 1 5 p < (n + c - 1)/2. In addition, since H" V Kc- l has at least c - 1 vertices of degree n + c - 2, then by (l), p 2 c - 1 . It thus follows that the size of H" V K c - , is bounded from above by the maximum size of C p , n + c - l , ~ - 1 5 p < (n + c - 1)/2. But by Lemma 3.4,

Let f ( p ) = lE(Cp,,)l, 1 5 p < q/2. Then f is quadratic in p with positive

By definition of C,,,, we have lE(C,,q)l = i ( 3 p 2 + p(1 - 2q) + q2 - q) .

Let G E rn(n ,k ) and let c(G") = c 2 3. Then the size of GC is equal to the

Pro05


this maximum occurs at either p = c - 1 or p = j . Since KC-1 V H ; = Cc-l,n+c-l and Kc-1 V H i = Cj,n+c-l, and since H1 and H2 are in G(n, k ) , the upper bound on the size of H' V Kc-l is achieved when H C is isomorphic to whichever of H f and HZ has the greater size. I

Proof of Theorem 3.1, Case 2. By Lemma 3.5, s,(n, k ) = (i) - max{lE(Hf) I , lE(Hi)I}. ) + ( ' - ! j+') + ( j - c + But the sizes of H t and H; are respectively ( n - g + l )

1) (n + c - j - l), where (;) = 0 if x < y. The result then follows from elementary algebra and the fact that c = k - n + 2.

We close this section by giving a characterization of those graphs that achieve minimum size.

and (n+c;1-2j

I

Theorem 3.6. Let H I and H2 be as defined above. Then

(i) For 2 5 k 5 n,

{Kl,k-I + (n - k)KI) , if k # 4, [ (K1,k-l + (n - k)KI ,Kf ,k - l + (n - k)K1}, if k = 4 . m(n,k) =

(ii) For n < k 5 2n - 2

{ H I } ,

{H2L otherwise.

if k is even and k < y, or k is odd and k < y, 6n -4 6n-I { H I , H ~ } , i f k = 7, o r k = 7,

Proof. (i) Ore [7] has shown that the maximum size of a graph with order n and path covering number 2 is and that this size is achieved only by Kn-l + I(, if n 3 2, n # 4. If n = 4, however, there are two graphs with order n, path covering number 2, and maximum size: K3 + K I and K I , ~ . Thus, m(n,n) = { K I , ~ , K F , J } for n = 4, and m(n,n) = { K I , ~ - I } otherwise. Now suppose G E m(n,k) , where n > k . Via arguments used in the proof in Theorem 3.1, Case I , we can show that G is the sum of components, all but one of which are copies of K I . The result then follows by Ore and induction on n.

(ii) Let C E m(n, k ) . Then the proof of Lemma 3.5 indicates that G" V K,-I has the same degree sequence as that of the larger of C c - ~ , n + c - ~ = K,-I V H f and Cj,n+c-l - K,-I V H i . But it can be verified that any two graphs with identical degree sequences as in (1) are isomorphic. Therefore, G is isomorphic to either H I or H2, and so Im(n, k)l 5 2. The result now follows from a comparison of the sizes of H I and H2. (We note that H I and H2 are isomorphic if k equals 2n - 3 or 2n - 2.)

-

I

4. M A X I M U M SIZE OF A GRAPH IN G(n ,k )

In this section, we prove


Theorem 4.1. Let n = a(k + 1) + r , a a non-negative integer, 0 5 r < k + 1. Then

if k = 0, if k = 2 and n 2 2,

u ( ( “1) - k) + (I), if 2 r - 2 5 k and 3 5 k 5 n - 1,

a ( ( ‘11) - k) + (I) - 2 r + k + 2, i f 2 r - 2 > k a n d 3 5 k 5 n - 1, ! s d n , k f =

I(;) - 2n + k + 2, if n - 1 5 k 5 2n - 2, k 2 3.

Since K: is the only graph in G(n,O), then sM(n,0) = 0. We also note that G(n,2) is the collection of matchings on n vertices, and hence s~ (11 ,2 ) equals Ln/2]. We may therefore restrict our attention to k 2 3. It will be convenient to consider the two cases n - 1 5 k 5 212 - 2 a n d 3 5 k 5 n - 1 .

Case 1. n - 1 5 k 5 2n - 2.

Let G E (n , k). Then by Theorem 2.1, the covering number of G c is c = k - n + 2. Furthermore, the size of G‘ is minimum over all graphs with order n and a covering number of c. It follows that G“ is a sum of c disjoint paths, and has size n - c. The number of edges in G is thus s M ( n , k ) = (;) - (n - c) = (;) - 2n + k + 2.

Corollary 4.2.

I If n - 1 5 k 5 2n - 3, then s ~ ( n , k ) < sM(n ,k + 1). 1

Case2. 3 5 k 5 n - 1 .

We begin with a number of lemmas

Lemma 4.3. Let G be a graph of order n and let L be a labeling of G such that SL 5 n - 1, I (G,L) is not of the form (m,O,m,O,.. .,in), lH(G,L)I = r 2 1, and no two holes in L are consecutive. Then there exists a graph G’ of order n and a labeling L’ of G’ such that sL = S L I , l(G’,L’) is not of the form (m,O,m,O,. . . , m ) , IH(G’,L’)I = r - 1 , no two holes in L’ are consecutive, and the sizes of G and G‘ are equal.

Proof. Throughout the proof, labels are presumed to be under L unless otherwise indicated. Since sL 5 n - 1 and IH(G, L)I 2 1, there exists at least one multiplicity in L . Let m, > 1

denote a maximum component of the labeling sequence I(G, L) . Suppose h is a hole, and suppose w is a vertex in V ( G ) with L(w) = i. If w is not adjacent

to any vertex with label h - 1 or h + 1, we obtain the desired graph G’ and labeling L’ by letting V(G’) = V ( G ) , E(G’) = E ( G ) , L’(w) = h, and L’(z) = L ( z ) for z # w. Thus, we may suppose that for every hole h and every vertex w in V ( G ) with L(w) = i , w is adjacent to some vertex labeled h - 1 or h + 1.

Suppose h is a hole such that either mh - 1 < m, or mh+ 1 < m,. With no loss of generality, we assume the former. Then there exist distinct vertices u and y such that L ( u ) = L ( y ) = i and u is not adjacent to any vertex with label h - 1. Thus, u must be adjacent to some vertex x with label h + 1. We obtain the desired graph G’ and labeling L’ by letting V(G’) =

V ( G ) , E(G’) = E(G) - { u x } ~ { u y } , L’(u) = h, and L’(z) = L ( z ) for z f u . Consequently, we may assume that for all holes h, mh-1 = mh+ 1 = m,. The constraint on l (G, L ) then implies that there exists a hole h* such that mh‘+2 > 0 or mh*-2 > 0.

With no loss of generality, we suppose that m, = mh*-l = mhw+l 2- mhs+2 > 0. Let x and y be vertices with L ( x ) = h* - 1 and L ( y ) = h* + 2, and suppose u is the vertex


with label h* + 1 to which x is adjacent. If every vertex labeled h* - 1 is adjacent to some vertex labeled h* + 2, then we may assume x and y are adjacent. Moreover, mh*-l = rnh*+2 = m,, so there exists vertex u # y with L(u) = h* + 2. Then the desired graph and labeling are given by V(G') = V ( G ) , E(G') = E(G) - { x y } u { u y } , L ' ( y ) = h", and L'(z) = L ( z ) for z # y . If there is some vertex f with label h* - 1 that is not adjacent to any vertex labeled h" + 2, then the desired graph and labeling are given by V(G') = V ( G ) , E(G') = E(G) - { x v } U { t y } , L'(x) = h*, and L'(z) = L ( z ) for z # x.

Lemma 4.4. Suppose H + X E G(n, k ) , where X may be empty. Suppose A E L A ( H + X ) , and Z(H,A) is an alternating vector of the form (rno,ml,rn2,m3 ,..., r n k - 1 , m k ) =

(rn, 0, rn, 0, rn, 0,. . .), where rn 3 2. Then H + X E 3 d ( n , k ) only if rn = 2 and k = 4.

Due to the form of Z(H, A), the subgraph H has order rnp, where p is the number of components of Z(H, A) that equal m. (If k is odd, p = ( k + 1)/2; otherwise, p = ( k / 2 ) + 1.) Thus, each of exactly p distinct, nonconsecutive labels is assigned, under A, to rn vertices in V ( H ) , which implies that the size of H is bounded from above by m(5). We will show that if k # 4 or rn # 2, then there exists a graph Y with order mp such that Y + X E G ( n , k )

Let H I be the complete graph on {uo, U I , u2,. . . , uk} minus the path uo, u1, u2,. . . , uk. Let H2 = (m - 2)Kp, and H3 be a graph of order 2 p - (k + 1). ( H I is isomorphic to Pi+I. The order of H3 is 0 or 1, depending on the parity of k.) The graph Y = H I + H2 + H3 has order (k + 1) + (m - 2)p + 2 p - (k + 1) = rnp, and size ((kll) - k ) + (m - 2 ) ( 5 ) . Additionally, since the lambda-number of H I is k , and the lambda-numbers of X , H2, and H3 are bounded from above by k , we have Y + X E G(n, k ) . But it is easily verified that if k # 4, then ((kll) - k ) + (m - 2 ) ( ; ) > m ( ; ) , implying IE(Y)l > IE(H)I.

If k = 4 and rn = 3, let Y = H I + Cq. Then Y is a graph with lambda-number 4 , order 9 = mp, and size 10 > m ( ; ) = 9. If k = 4 and m > 3, let Y = 2Hl + Kl + (rn - 4)K3. Then Y is a graph with lambda-number 4 , order 3m = rnp, and size 3m + 1 > m(;) = 3m. In either case, Y + X E G(n,4), and IE(Y)I > IE(H)I.

Lemma 4.5. For all k , 3 I k 5 n - 1, there exists G* with order n and L* E 1 (G*) such that SL* = k , H(G*, L * ) = 4, and IE(G*)I = s ~ ( n , k) .

1

Proof.

and IE(Y)l > IE(H)I.

I

Proof. If H ( G , A ) = 4, then we are done. Otherwise, let IH(G,A)l = r > 0. By Lemma 4.4,

1(G, A) = (2,0,2,0,2) or Z(G, A) is not of the form ( r n , O , m,O,. . . ,O,m). If I(G,A) = (2 ,0 ,2,0,2) , then n = 6, k = 4 , and s ~ ( n , k ) = 6 by inspection. But it can

be easily verified that the graph G* = P; + K I has order 6, size 6, and a labeling L" such that sL- = 4 and H(G*,L*) = 4 .

If 1(G, A) is not of the form (m,O, m, 0,. . . , m), then by Lemma 4.3 there exist a graph G ' and labeling L' of GI, such that sL1 = k , IH(G',L')I = r - 1, and IE(G')I = IE(G)I =

s ~ ( n , k ) . If r - 1 = 0, then we are done. Otherwise, an iterative application of Lemma 4.3 to GI and L1 will eventually lead to a graph G* with size s ~ ( n , k ) and a labeling L* of G* with no holes. 1

We next introduce an algorithm that may be applied to any graph G* with order ti and labeling L* such that H(G*,L*) = 4 and SL* 5 n - 1. The output of this algorithm will be a graph G such that V ( G ) = V(G*) , IE(G)I 2 IE(G")I,L* is a lambda-labeling of G, and G = A + X , where A is isomorphic to PSt.+1.

For each label i between 0 and sL*, let {v;, u;, . . . , uk , } denote the set of vertices labeled i under L*. Then the algorithm is as follows:

Let G E M ( n , k ) and A E L A ( G ) . Then A has no consecutive holes.


Line 1.

Line 2.

Line 3. is true of G’;

Set G := G*;

For each 0 5 i < j 5 SL such that j - i 2 2, do:

Determine which one of the following totally exhaustive, mutually exclusive cases

Case i . uiui E E(G*) .

Case ii. There exist q, t 2 2 such that the edges uiui and u6.r are in E(G*).

Case iii. There is no edge from u; to any vertex labeled j , and there exists q L 2 such

There is no edge from u:’ to any vertex labeled i , and there exists r 2 2 such

There is no edge from u:’ to any vertex labeled i, and there is no edge from u;

Manipulate edges of G as follows:

that edge u;ui is in E(G*).

that uiu: is in E(G*). Case iv.

Case v. to any vertex labeled j .

Line 4.

if Case i holds, take no action; if Case ii holds, delete edges uiui and uiu; , and insert edges uiu{ and U ~ U : .

if Case iii holds, delete edge udui, and insert edge uiui .

if Case iv holds, delete edge uiu;, and insert edge uiu:’. if Case v holds, insert edge uiuf

Line 5.

Line 6.

Set G equal to the graph that results from executing Line 4;

If Line 2 is exhausted, set A equal to the subgraph of G on {up, u1, u t , . . . , us,}, then quit. Otherwise, update integers i and j , then go to Line 3. I

We make the following three observations:

1. A is isomorphic to P:L- and has order and size SL* + 1 and ( sLmll) - sL*, respec-

2. L* is a labeling of G , and the lambda-number of A is S L ~ . Thus, L* is a lambda-labeling

3. IE(G)I 2 IE(G*)I, with equality if and only if Case v is never realized.

tively. The order of X is n - (SL* + 1).

of G.

Lemma 4.6. For all k , 3 9 k 9 n - 1, there exist G E %(n , k ) and A E Lh(G) such that G = A + X, where A has order k + 1 , size ( ‘ ; I ) - k , and H ( G , A ) = 4 .

By Lemma 4.5, there exist a graph G* with order n and a labeling L* of G* such that sLm = k , s M ( n , k ) = IE(G*)I, and H(G*, L * ) = 4 . Applying the algorithm described above to G* and L* then yields a graph G = A + X such that the order of G is n and the lambda-number of G is SL* = k (observation 2). Hence, G E G ( n , k ) . Since IE(G)I 2 IE(G*)( (observation 4) and (E(C*)( = sM(n, k ) , then C E 34 ( n , k ) . Finally, we note that the order and size of A are respectively k + 1 and ( ‘ ; I ) - k (observation l ) , and h = L* is a lambda-labeling of G (observation 2 ) with no holes.

Lemma 4.7.

Proof.

For all k , 2 5 k 5 2n - 3, s M ( n , k ) < s M ( n , k + I ) .

Corollary 4.2 addresses the situation in which n - 1 5 k 5 2n - 3. Proof. We consider the cases k = 2 and 2 < k < n - 1 separately, noting that


Case i. k = 2. Since s ~ ( n , 2 ) = 1n/2J, it thus suffices to show that there exist a graph G with order n and a lambda-labeling L of G with S L = 3 such that the size of G is greater than n/2.

Let n = 4a + r , where a is a non-negative integer, and 0 5 r 5 3. (Note that a = 0 implies r = 3.) We consider the graph G = H + M , where H is isomorphic to a ( P i ) and M is isomorphic to P,. Then A(G) = 3, and the size of G is 3a + max(0,r - 1 } , which is strictly greater than n/2. Hence, sM(n,2) < sM(n, 3) .

2 < k < n - 1. From Lemma 4.6, there exist G E 3f (n, k ) and A E L*(G) such that H ( G , A) = 4 . For fixed i, 0 I i < k , let u and w be vertices of G that are assigned labels i and i + 1 under A, respectively. We produce a new graph G' and a labeling L' of GI as follows: V ( G ' ) = V ( G ) , E ( G ' ) = E ( G ) U { u w } , and

Case ii.

if A(x) 5 i, A ( x ) + 1, otherwise.

Then G ' has order n and size s ~ ( n , k ) + 1. Furthermore, the lambda-number of GI is greater than k by our choice of G. Since S L I = k + 1, then the lambda-number of GI is k + 1. Hence, s ~ ( n , k ) < IE(G')l 5 s u ( n , k + 1). I

Proof of Theorem 4.1, Case 2. Let n = a ( k + 1 ) + r , where a is a positive integer, and 0 5 r < k + 1. We proceed by induction on a.

If a = 1, then Lemma 4.6 implies that there exists G = A + X E N(n, k ) such that the order of X is r , and the order and size of A are k + 1 and ( ';I) - k , respectively. We can thus find s ~ ( n , k ) by summing ( ':I) - k with the largest possible size of a graph X whose order is r and whose lambda-number is less than or equal to k . If 2r - 2 5 k , then the maximum size of X is (;), since the complete graph on r vertices has lambda-number 2r - 2 5 k . If 2r - 2 > k , then Lemma 4.7 and Case 1 of Theorem 4.1 imply that the maximum size of X is ( i ) - 2r + k + 2. Thus, the theorem is true if a = 1.

Now suppose the theorem is true for a = ao,ao 2 1 . Let n = (a0 + l ) ( k + 1) + r . By Lemma 4.6, there exists G = A + X E 3f ( n , k ) such that the order and size of A are k + 1 and ( ' : I ) - k , respectively, and the order of X is n - ( k + 1) = ao(k + 1) + r . We can thus find s ~ ( n , k ) by summing ( ';I) - k with the largest possible size of a graph X whose order is ao(k + 1) + r and whose lambda-number is less than or equal to k . But by Lemma 4.7 and the inductive hypothesis, the largest possible size of such a graph is

ao(( "1) - k ) + ( $ if 2r - 2 5 k ,

U O ( ( ',I) - k ) + (I> - 2r + k + 2, if 2r - 2 > k . sM(ao(k + 1 ) + r , k ) =

Thus, sM(n , k ) = ( ":') - k + sM(uo(k + 1) + r , k ) , giving the desired bound, and proving Theorem 4.1, Case 2. I

If n - 1 5 k , then the proof of Theorem 4.1, Case 1 shows that M ( n , k ) is the collection of graphs whose complements are sums of k - n + 2 disjoint paths. If k < n - 1, then the characterization of M ( n , k ) is an open question.


5. SIZES OF GRAPHS IN G(n,k)

We conclude our discussion by showing the existence of a graph with any size between s,(n, k ) and s H ( n , k ) .

Theorem 5.1. For any integer j , s,(n, k ) 5 j 5 s ~ ( n , k ) , there exists a graph G E G(n, k ) with size j .

Proof.

Case I .

We consider the two cases n > k and n 9 k .

n > k . By the proof of Lemma 4.6, there exists a graph H E 312 ( n , k ) such that H has a component A that is isomorphic to Pi+ 1 . By Theorem 3.6, there exists a graph H ' E m(n, k ) such that H ' is isomorphic to K1,k-l + ( n - k ) K I . Let A' be a subgraph of HI such that A' is isomorphic to Kl,kPl + K1. We observe that A' is a subgraph of A, and that HI is a subgraph of H . We construct a graph G with order n and size j by adding j - s,(n, k ) edges from E ( H ) - E ( H ' ) to E ( H ' ) . Since HI is a subgraph of G, and G is a subgraph of H , then k = h ( H 1 ) 9 h ( G ) 5 h ( H ) = k . Hence G E G(n, k ) .

n 4 k . Let H E m ( n , k ) . Then the path covering number of H' is c ( H ' ) =

k - n + 2. Let X be a minimum path covering of H' . We now produce a graph G with order n and size j by adding j - s,(n, k ) edges from E ( X ' ) - E ( H ) to E ( H ) . Since X is a subgraph of G", which in turn is a subgraph of H ' , then c = c ( X ) 2 c(G') 2 c(H') = c ; hence, h(G) = k , and G E G(n, k ) .

Case 2.

I

ACKNOWLEDGMENT

The authors wish to thank the referees for their many suggestions, which resulted in an improved paper.

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