The Subgraph Summability Number of a Biclique

Transcription

1 The Subgraph Summability Number of a Biclique Sivaram Narayan Department of Mathematics Central Michigan University Mt. Pleasant, MI sivaram.narayan@cmich.edu Janae Eustice Russell* Central Michigan University Mt. Pleasant, MI Ken W. Smith Department of Mathematics Central Michigan University Mt. Pleasant, MI ken.w.smith@cmich.edu *This author thanks Central Michigan University for a Summer Research Scholarship to work on this project in summer 2001 when she was an undergraduate student. Abstract The subgraph summability number, σ(g), of a connected graph G is the largest integer defined by labeling the vertices of the graph so that the label sums of connected induced subgraphs cover the interval [1..σ(G)]. In this paper we examine the summability number of the biclique K r,s. 1 Introduction A vertex labeling α of a graph G assigns to each vertex x apositive integer α(x). Avertexlabelingnaturallyliftstoalabelingofsubsets 1

2 of vertices: if S is a set of vertices then define α(s) := s S α(s). Similarly, if H is a subgraph of G and V (H) isthesetofverticesof H, defineα(h) :=α(v (H)). We write x y to indicate that vertex x is adjacent to y and thus xy is an edge. In this paper we examine the family of connected induced subgraphs of a graph G and questions associated with the labelings of these subgraphs. The graphs in this paper will be simple graphs, without loops. For an introduction to the theory of graphs, and the basic terminology, see [1] or [2]. A nice summary of vertex labeling problems may be found in Gallian s papers ([3], [4].) If S is a set of vertices of G, wesays is a connected vertex set if the subgraph induced by S is connected. Set equal to the set of connected vertex sets. The number of connected induced subgraphs of a graph G will be denoted by c(g) :=. Let [1..N] representthesetofintegersfrom1ton inclusive. A labeling α of (the vertices of) a graph G is an N-labeling if [1..N] {α(h) :V (H) }. The largest integer N for which G has an N-labeling is called the subgraph summability number of G and is denoted by σ(g) (see [5], [6].) If, in addition, N = α(g), we say α is a restricted labeling and speak of the restricted summability number σ (G) =N. Clearly c(g) σ(g) σ (G). Given an N-labeling α of a graph G, theremaybesubgraphsh of G such that α(h) >N.The number of such graphs is the excess of the labeling α. We denote the excess by e(α) := {H : V (H) and α(h) >N}. If α is a restricted labeling then e(α) =0. There may also be subgraphs H 1 and H 2 such that α(h 1) = α(h 2) N. The redundancy, r(α), of the N-labeling α, isthe difference between the cardinality of {H : α(h) N} and N. If α is an N-labeling of a graph G then clearly c(g) =N + e(α)+r(α). Alabelingα is sharp if c(g) =σ (G). In a sharp labeling, e(α) = r(α) =0andsothereisaone-to-onecorrespondencebetweenthe integers [1..N] andtheconnectedsubgraphsofg. Anopenquestion in the study of summability numbers is the classification of graphs for which there is a sharp labeling. Examples 1. Let G be the cycle C 4,withverticesx 1 x 2 x 3 x 4 x 1. This graph has 13 connected induced subgraphs, so c(g) =13. The labeling α(x 1)=1, α(x 2)=2, α(x 3)=6, α(x 4)=4has redundancy zero and is a sharp labeling. So is the labeling 1, 3, 2, 7, in that order. 2. Consider the path of length three, with vertices labeled, in order, 3, 1, 5, and 2. The path P 3 has 10 connected subgraphs. This 9-labeling is optimal. It has excess equal to one (α(p 3)=11.) Soσ(P 3) 9. Another 9-labeling of P 3 is 1,

3 3, 3, and 2, in that order. This labeling has excess zero and redundancy one. Thus σ (P 3) 9. One can show that there is no 10-labeling of P 3 and so σ(p 3)=σ (P 3)=9. 3. The path P 6 has 28 connected induced subgraphs. Label the seven vertices of the path of length six with the integers 8, 7, 2, 3, 1, 10, 8, in that order. This is a 24-labeling of P 6 with excess 3 and redundancy 1. This is the only 24-labeling; there is no 25-labeling for P 6.Thereforeσ(P 6)=24. Note that α(p 6)=39, so the restricted summability number of P 6 is lower than 24. Indeed σ (P 6)is23;thelabelings[2, 3, 3, 4, 9, 1, 1] and [2, 3, 2, 6, 6, 3, 1] are 23-labelings for P Consider the bipartite graph K 3,3. Label the vertices of one coclique 1, 2, 23 and the vertices of the other coclique 3, 6, 13. This is a 48-labeling with redundancy 7. We will show this labeling is optimal and that σ (K 3,3) =48. 2 The restricted summability number of a biclique Abiclique(orcompletebipartitegraph)K r,s has two disjoint sets of vertices, A = {a 1,...,a r} of size r and B = {b 1...,b s} of size s such that each set consists of pairwise nonadjacent vertices and two vertices belonging to different sets are adjacent. Penrice (in [5]) showed that σ(k 1,s) =2 s +2. This labeling may be achieved by labeling the vertices of the s-coclique B by powers of 2: 1, 2, 4,...,2 s 1 and then labeling the single vertex a 1 of degree s with the number 3. Let T be any of the 2 s subsets of B (including the empty set.) T {a 1} is a connected vertex set. In addition, the singletons {b j} are also in. Thus the number of connected induced subgraphs of K 1,s is 2 s + s and so Penrice s solution has redundancy s 2. (The repeated values are powers of 2: 2 i, 2 i s 1.) Note that c(k 1,s) growsexponentiallyins. Penricedemonstrated that no N-labeling has smaller redundancy using an S + T argument. Suppose S and T are disjoint subsets of B such that α(s) = α(t). For any U B, U disjoint from S T, the sets (S {a 1}) U and (T {a 1}) U have the same label sum. Thus there are at least 2 s ( S + T ) pairs of subgraphs with the same value and so r(α) 2 s ( S + T ). If S + T is small, we would expect 2 s ( S + T ) to be greater than s 2. We will use this argument repeatedly in this paper. We begin with two sets S and T such that α(s) =α(t). Call them an initial redundant pair. Beginning with such a pair, we create additional redundant pairs (S U) and(t U) toforcetheredundancytobe large.

4 Our goal is to prove the following theorem. Theorem 1. Let G be the complete bipartite graph K r,s, withs r 2. Then σ (K r,s) =3(2 r+s 2 ) 2 r Note that c(k r,s) =(2 r 1 )(2 s 1 )+r + s =4(2 r+s 2 ) 2 r 2 s + r + s +1. (Choose any nonempty subset of A and nonempty subset of B; thischoicewillgiveaconnectedvertexset. Indeedallconnected vertex sets of size 2 or greater are formed this way. To this list we add the singletons, the r + s vertices.) If G n = K n,n then σ (G n) c(g n) approaches 3 as n goes to infinity. This 4 is the first example known to the authors of a sequence of graphs where σ (G n) c(g n) is bounded away from 1. We conjecture that σ (K r,s) =σ(k r,s). Given a labeling α, ordertheverticesoftheseta so that α(a i) α(a i+1), 1 i r 1andsimilarlyordertheverticesofB so that α(b j) α(b j+1), 1 j s 1. We will abbreviate the labeling by writing α := (α(a 1),α(a 2),...,α(a r); α(b 1),α(b 2),...,α(b s)). For example, an optimal labeling of the 4-cycle K 2,2 is α = (1, 2; 3, 7). In an earlier example, we found σ(k 3,3) =48usingthe labeling (1, 2, 23; 3, 6, 13). Similarly, the labeling (1, 2; 3, 6, 12, 25) gives σ(k 2,4) =49. We first let r =2andconstructtheoptimumlabelingforK 2,s. We assume one coclique has vertices A = {a 1,a 2} and the other has vertices B := {b 1,b 2,...,b r}. Set J s := 3(2 s ) 1. Lemma 2. Let G be the complete bipartite graph K 2,s, with s 2. Then σ(k 2,s) =3(2 s )+1=J s +2. In addition, if s 3, any 3(2 s )-labeling of K 2,s can be transformed into a (3(2 s )+1)-labeling byraisingbyone the value ofthe largest label. Proof of Lemma 2. We label a 1,a 2 with the integers 1 and 2; we label b 1,b 2,...b s 1 with the integers 3, 6, 12,...,3 2 s 2,andfinallylabelb s with 3 2 s We first show that the labels of the connected induced subgraphs cover the interval [1..3(2 s )+1]. Note that each integer of the form 3 2 i 1, 1 i s 1isthevalueofasinglevertexb i and also the value of a subgraph on the vertices a 1,a 2,b 1,...b i 1.

5 Given an integer X in this interval, we create a connected vertex set W as follows. If X 3 2 s 1 +1, then choose the vertex b s to be in W and replace X by X 1 := X (3 2 s 1 +1). Otherwise X 1 = X. If X 1 3(2 s 2 ), place b s 1 in the vertex set W and set X 2 = X 1 3(2 s 2 ); otherwise X 2 = X 1. We will continue in this manner until we achieve an integer X s 2 which is 0, 1, or 2. If X s 2 is 1 or 2, place a 1 or a 2 into W. If X s 2 is zero, choose the smallest value of i such that b i is in W and replace b i by a 1,a 2,b 1,...,b i 1. The vertex set W so created will have value X and will be connected. This labeling demonstrates that σ (K 2,s) 3(2 s )+1. Is this labeling maximal? The redundancy is s 2. We will show that any other restricted labeling must have redundancy at least this large. Any labeling will require a vertex with label 1. We first note that there cannot be two vertices labeled 1 in an optimum labeling. (This is an example of Penrice s S + T argument.) If there are two adjacent vertices labeled with 1 (say a 1 and b 1)then{a 1,a 2,b 2} and {b 1,a 2,b 2} form an initial redundant pair of vertices for the labeling α. This generates 2 s 1 redundant pairs {{a 1,a 2} T, {b 1,a 2} T} where T is any subset {b i :3 i s}. Thus the redundancy is at least 2 s 2 > s 2 if s > 2. Similarly, if α(a 1) = α(a 2) = 1 then {a 1}, {a 2} form an initial redundant pair and r(α) 2 s ; if α(b 1)=α(b 2)=1then{a 1,b 1}, {a 1,b 2} form an initial redundant pair and r(α) 2 s 2. Thus in an optimum labeling there is only one vertex labeled by 1 and so there must be a vertex labeled 2. Now we examine a certain 4-cycle x 1 x 2 x 3 x 4 x 1 containing the vertices labeled 1 and 2 with α(x 1)=1. We consider two cases, depending on whether there is a vertex of the graph labeled 3. (Case 1.) Suppose there is a vertex labeled 3. We cannot have the vertices with labels 1, 2, and 3 all in the set B for if α(b 1)= 1,α(b 2)=2, and α(b 3)=3then{a 1,b 1,b 2} and {a 1,b 3} are an initial redundant pair and r(α) 3 2 s 3. If there is a vertex labeled 3, it is on a 4-cycle containing the vertices labeled 1 and 2. If α(x 2) = 2 and α(x 3) = 3 then {x 1,x 2} and {x 3} form an initial redundant pair and r(α) 2 s Similarly, if α(x 2)=2 and α(x 4)=3then{x l,x 2} and {x 4} form an initial redundant pair and force r(α) tobelarge. Thereforeα(x 3)=2andα(x 2)=3,that is, the graph has a path with vertex labels 1, 3, 2inthatorder. Suppose a 4-cycle has label (1, 3, 2,x 4). If α(x 4) 5thentherearetwosubgraphsofthe4-cyclewiththe same label sum and each of these subgraphs includes vertices from both cocliques. Thus, given any T B, T disjoint from the 4-cycle,

6 we may create two subgraphs with the same label sum and force r(α) 2 s If x 4 is labeled 6 then x 4 must be in B or we again have 2 s 2 redundant subgraph labels. But the labeling α =(1, 2; 3, 6,...)is fruitful and we will look at this further (below.) If x 4 =7thenwehaveasharplabeling(1, 3, 2, 7) of the 4-cycle on the vertices {x i}. Weattempttoextendthistoagoodlabelingof K 2,3. If the vertices of A = {a 1,a 2} are labeled 1 and 2 while b 1,b 2 are labeled 3 and 7 then there will not be a connected subgraphs with value α(b 3)+7 and so we must label b 3 with an integer less than 7. This contradicts our assumption about the ordering of the labels of the set B. If a 1,a 2 are labeled 3 and 7 while b 1,b 2 are labeled 1 and 2 then α(b 3) 11. If α(b 3) 11, then the redundancy is at least three and the three redundant pairs force r(α) > 2 s 2. If α(b 3)=11thenwehavea24-labelingforK 2,3. This labeling has redundancy two; two connected subgraphs bear the label 11 and another pair bear the label 21. This is not an optimal labeling for K 2,3 and if s>3, this labeling forces redundancy r(α) 2 s 3 +1> s 2. We digress for a moment to examine this labeling more carefully. Occasionally an optimal labeling of a graph G induces a less than optimal labeling on a subgraph. Is it possible to get good labelings for K r,s using this 24-labeling of K 2,3? In particular, is it possible that this labeling, where N = σ(k 2,3) 1, lifts to other labelings with value σ(k r,s) 1? For example, could this labeling (3, 7; 1, 2, 11) lead to a 48-labeling of K 2,4? No, for if α(b 4)=24thenwewillnothavesubgraphslabeled 25 or 26. Or could this lead to a 47-labeling of K 3,3? No, for there will not be a subgraph with label 30. So, although the labeling (3, 7; 1, 2, 11) is a 24-labeling of K 2,3, missingbyonetheoptimal value of N, anyn-labeling of a larger graph K r,s which has this subgraph label, will be far from optimal. We return to examining the case where (α(x 1) = 1,α(x 2) = 3,α(x 3)=2,α(x 4)) is the labeling of a 4-cycle. We have noted that α(x 4)=7completesasharplabelingforthe4-cyclebutthatthis labeling does not extend to a labeling of K 2,s for s 3. If we label x 4 with 8 or higher this will force the label of the vertex b 3 to be seven. If x 4 B and α(b 3)=7thenwehaveviolatedourassumptionabout the ordering of the labels of B. If x 4 B and α(b 3)=7thenthe redundancy will be too high. We conclude then that the only possible labeling for the 4-cycle x 1,x 2,x 3,x 4 is 1, 3, 2, 6, in that order, and the vertices x 1,x 3 are in the set A = {a 1,a 2} while x 2,x 4 are in the set B. The vertices of

7 A are labeled 1 and 2 while the first two vertices of B are labeled 3 and 6. We also conclude that (in this case, where there is a vertex labeled 3) any other labeling of a 1,a 2,b 1,b 2 leads to a value of N which in general differs from σ(k 2,s) bymorethan1. Now we assume the label is (1; 2; 3, 6,...)andconsidertheother vertices in B. Supposes>3andα(b 3)=X. If X<12 then there are redundant labelings in the subgraph on a 1,a 2,b 1,b 2,b 3, redundancies that lift to a family of redundancies in the larger graph G. IfX>12 then the integer X +α(b 4)willnotbealabelofasubgraph. (Ifweare too greedy in our labeling of b 3,wewillnotbeabletoappropriately label b 4.) Thus the vertex b 3 must be labeled 12. An argument by induction then shows that α(b j) 3 2 j 1 as long as j<s.however, the vertex b s may be labeled 3 2 s 1 +1 since there is not vertex b s+1 to worry about. Note that labeling b s+1 by 3 2 s 1 will give a 3(2 s )-labeling of K 2,s, oneshortofoptimal. (Case 2, there is no vertex labeled 3.) On the other hand, if there is no vertex labeled 3 then the vertices labeled by 1 and 2 must be adjacent and some vertex in the graph must be labeled 4. Now we have a 4-cycle x 1 x 2 x 3 x 4 x 1 such that α(x 1)=1,α(x 2)=2andeitherα(x 3)=4orα(x 4)=4. If α(x 3) = 4 then the connected vertex set {x 2,x 3} has value 6andavertexwithlabel5isnecessary. Iftheverticeslabeledby 1, 5 and 6 are all in B then we may join them to a 1 to create a redundant pair and force redundancy at least 3 2 s 3. If instead we have α(x 4)=5thentheredundantpair{x 1,x 4}, {x 2,x 3} force redundancy at least 2 s 2.(Asinanearliercase,onecanshowthat this 12-labeling of K 2,2 does not lift to a 24-labeling of K 2,3 and so in general the redundancy will be considerably higher.) If α(x 4)=4thentheconnectedvertexset{x 1,x 4} has value 5 and a vertex with label 6 is necessary. Thus we must give the 4-cycle on the x i asharplabeling1, 2, 6, 4. Can we extend the label (α(a 1),α(a 2); α(b 1),α(b 2)) = (1, 6; 2, 4) to a good labeling of K 2,s? The value α(b 3)+4 is not possible for any subgraph unless α(b 3) 9, so again we will get high redundancy unless the vertices labeled 1 and 6 are in B. Can we extend the label (α(a 1),α(a 2); α(b 1),α(b 2)) = (2, 4; 1, 6) to a good labeling of K 2,s? Yes. Once again we consider the vertex b 3 and argue that it must be labeled with 12. The inductive argument here is simpler; each vertex b j must be labeled 3 2 j 1,includingthelastvertexb s which is labeled 3 2 s 1. We are led to the labeling (2, 4; 1, 6, 12, 24,...,3 2 s 1 ) This offers a competing labeling for K 2,s; thisisalsoa3(2 s )+1 labeling of the graph. We may, if we wish, obtain a 3(2 s )-labeling of K 2,s by instead

8 setting α(b s)=3(2 s 1 ) 1; this is the only way to obtain a labeling which is one less than optimal. This proves the lemma. Proof of Theorem 1. We are now ready to prove the main theorem. As before, assume the vertices a 1,...,a r and b 1,...,b s are ordered so that α(a j+1) α(a j)andsimilarlyα(b j+1) α(b j). The optimum labeling for K r,s is obtained by first labeling the subgraph K 2,s as in case 1: α =(1, 2; 3, 6,...,3(2 s 2 ), 3(2 s 1 )+1). This is a J s +2-labeling. Wethenlabelthe remainingverticesina by setting α(a j)=2 j 3 J s, 3 j r. We show that this labeling is maximal by proving that any other labeling has a larger redundancy. The proof is by induction on r, the size of the coclique A. We know the result is true for r = 2. For r 3, set M := 2 r 3 J s +2 = 3(2 r+s 3 ) 2 r 3 +2 and assume that σ (K r 1,s) =M. In addition, justified by the proof of Lemma 2, we assume that any M 1labelingofK r 1,s cannot have two vertices of B labeled 1 and 2. We wish to show that σ(k r,s)=2m 2andthattheoptimal labeling occurs when a r is labeled M 2andthesubgraphK r 1,s receives the optimal labeling of M. Consider the graph K r,s and suppose it has an N-labeling α where N>2M 2. Set X := α(a r)andy := N X. We separate the proof into three cases. Case 1. α(k r 1,s) =Y > M. By our inductive assumption, the labeling on the subgraph K r 1,s does not cover the interval [1..M]. Thus X must be the smallest integer in [1..M] notin{α(h)} where H ranges over the subgraphs of K r 1,s. If 2X is in {α(h)} then 2X Y. Now N = X + Y 3 Y. 2 However, since we are examining the restricted summability number, X must be the only integer in the interval [1..Y ]notrepresentedby asubgraphofk r 1,s and so the interval [1..Y ]isboundedaboveby c(k r 1,s)+1=2 r+s 1 2 r 1 2 s + r + s +1. But we are assuming that N 2M 1 =3(2 r+s 2 ) 2 s Thus 3(2 r+s 2 ) 2 s 2 +3 X + Y 3 Y (2r+s 1 2 r 1 2 s + r + s +1), which is impossible if s r>2. On the other hand, if 2X is not in {α(h)} then since no subgraph has label 2X, wemusthave2x>2m 1andthereforeX M. Since X was the smallest integer not represented by a connected subgraph of K r 1,s, α must be an M 1labelingonK r 1,s and X = M. However, no such labeling allows α(b 1)=1andα(b 2)=2. Thus X +1 or X +2isnot represented byaconnected subgraph of K r,s.

9 Case 2. Suppose α(k r 1,s) =M. Then X M. In order to obtain graphs with labels in the interval M +1,...,2M 1, the labeling on K r 1,s must then be an M-labeling and yet, to represent the integers X +1and X +2, we must have α(b 1)=1andα(b 2)=2, contradicting our inductive assumption. Case 3. Suppose α(k r 1,s) <M.Then X M. But if X M +1, there is no subgraph with label M and so we must have X = M while α(k r 1,s) =M 1. To obtain the values X +1 and X +2we must have α(b 1)=1andα(b 2)=2, while still having an M 1labelingonK r 1,s. Againthiscontradictsourassumption. We conclude, therefore, that the optimum labeling is the 2M 2 labeling given above.. Comment. We have shown that σ (K r,s)=3(2 r+s 2 ) 2 r Is it possible to label K r,s so that σ(k r,s) > 3(2 r+s 2 ) 2 r 2 +2? References [1] Bollobás Béla, Modern Graph Theory, Springer-Verlag, [2] Diestel, Reinhard, Graph Theory, Springer-Verlag, [3] Gallian, Joseph, A survey: recent results, conjectures, and open problems in labeling graphs. J. Graph Th. 13, , [4] Gallian,Joseph, A dynamic survey of graph labeling. Electron. J. Combin., #DS6, [5] Penrice, Stephen, Some new graph labeling problems: a preliminary report, DIMACS, preprint, [6] West, Douglas, SIAM Activity Group newsletter in Discrete Mathematics, Open Problem # 22, 1996.