aa r X i v : . [ m a t h . C O ] F e b On the stability of graph independence number
Zichao Dong ∗ Zhuo Wu † Abstract
Let G be a graph on n vertices of independence number α ( G ) such that every induced subgraphof G on n − k vertices has an independent set of size at least α ( G ) − ℓ . What is the largest possible α ( G ) in terms of n for fixed k and ℓ ? We show that α ( G ) ≤ n/ C k,ℓ , which is sharp for k − ℓ ≤ Background.
All graphs considered here are finite, undirected, and simple. For graph G andproperty P , the resilience of P measures how much one should change G in order to destroy P .Assume G has P , then the global resilience of P refers to the minimum number r such that byremoving r edges from G one can obtain a graph not having P , and the local resilience refers tothe minimum number r such that by removing at each vertex at most r edges one can obtain agraph not having P . For example, Tur´an’s theorem (see [13]) characterizes the global resilience ofhaving a k -clique in complete graphs, and Dirac’s theorem (see [2]) characterizes the local resilienceof having a Hamiltonian path in complete graphs. Moreover, the local resilience of various propertiesis extensively studied in [12].Note that both global resilience and local resilience focus on removing edges. What about removingvertices? As far as we are aware of, this vertex-removal version of resilience is never discussed. Todistinguish from removing edges, we shall always use the word stability when discussing removingvertices throughout this paper. ∗ Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213, USA. Supported in partby U.S. taxpayers through NSF CAREER grant DMS-1555149. † School of Mathematical Sciences, Peking University, Beijing 100871, China.
1o be more specific, we are going to study the resilience of graph independence number with respectto removing vertices. For vertices v , . . . , v k , denote the induced subgraph of G on V ( G ) \ { v , . . . , v k } by G \ { v , . . . , v k } . For non-negative integers k > ℓ ≥
0, a graph G ( V, E ) is called ( k, ℓ ) -stable , if forevery k vertices v , . . . , v k of G , α ( G \ { v , . . . , v k } ) ≥ α ( G ) − ℓ. That is, the independence number α ( G ) drops by at most ℓ after removing any k vertices from V ( G ).This is related to a classical problem of Erd˝os and Rogers. Extending the problem of studyingRamsey numbers, in [6] they defined the function f s,s + t ( n ) def = min (cid:8) max {| S | : S ⊆ V ( G ) and the induced subgraph G [ S ] contains no K s } (cid:9) , where the minimum is taken over all K s + t -free graphs G on n vertices. They established a lowerbound on Ramsey number R ( k, ℓ ) by arguing that f s,s +1 ( n ) ≤ n − ε ( s ) for some positive constant ε ( s ).Throughout the years, the upper bounds and lower bounds on f s,s + t ( n ) for different pairs ( s, t ) havebeen extensively studied (see [1, 8, 9, 10, 11, 4, 15, 3, 7]). The case t = s + 1 has received the mostattention. The best known bounds (see [1] and [3]) in this case are n / ≤ f s,s +1 ( n ) ≤ c ( s ) n / (log n ) c ( s ) . For a detailed survey on Erd˝os–Rogers function, see [5].The study of Erd˝os–Rogers function has focused on the case when s, t are fixed and n tends toinfinity. Our results imply the exact value of f s,s + t ( n ) when s > n/ Results.
Unlike the previous work on the Erd˝os–Rogers problem, in this paper we study the behav-ior of independence number on very large induced subgraphs of a graph. That is, given a ( k, ℓ )-stablegraph G ( V, E ), what can be said about α ( G ) in terms of | V ( G ) | as n → ∞ ? Our main result is thefollowing upper bound. Theorem 1.
Suppose G ( V, E ) is a ( k, ℓ ) -stable graph with | V ( G ) | = n and k > ℓ ≥ , then α ( G ) ≤ (cid:22) n − k + 12 (cid:23) + ℓ. (1)This upper bound on α ( G ) determines a new class of Erd˝os–Rogers function values. Theorem 2. f s,s + t ( n ) = n − t for every integers s, t, n such that s > ⌊ n − t +12 ⌋ and s + t ≤ n + 1 .
2o prove Theorem 1, the crucial step is to prove its special case when ( k, ℓ ) = (1 , Corollary 3.
Suppose G ( V, E ) with | V ( G ) | = n such that α ( G \ { v } ) = α ( G ) holds for an m -elementsubset V ′ of V , then α ( G ) ≤ ⌊ n − m/ ⌋ . Next, we explore the tightness of Theorem 1. We call a graph attaining equality in Theorem 1 tight ( k, ℓ ) -stable . For example, every balanced complete bipartite graph is a tight (1 , K k +1 is tight ( k, k . However, it is harder to findtight ( k, ℓ )-stable graphs on larger vertex sets. The next theorems are in that direction. Theorem 4.
Suppose n > k > ℓ ≥ .(i) For every ℓ , if k = ℓ + 1 , then there exists an n -vertex tight ( k, ℓ ) -stable graph for every n .(ii) For every ℓ , if k = ℓ + 2 , then there exists an n -vertex tight ( k, ℓ ) -stable graph for every n . Theorem 5.
For every ℓ ≥ , if k = ℓ + 3 , then there exists a sequence of ( ℓ + 3 , ℓ ) -stable graphs G m ( V m , E m ) with | V m | → ∞ such that α ( G m ) = | V m | / − O (cid:16)p | V m | (cid:17) . Paper organization.
This paper is split into two parts.In the first part, we prove the upper bounds (Theorem 1). We then apply them to prove Theorem 2and Corollary 3. These occupy Section 2 through Section 4.In the second part, we study the tightness of the upper bounds. We prove Theorems 4 and 5 inSections 5 and 6.Finally, we devote Section 7 to a partial characterization of tight (1 , , , Acknowledgments.
We are grateful to Boris Bukh for proposing this problem to us, and for manyhelpful suggestions on writing. We also want to thank Minghui Ouyang for help with computerprogramming, and for beneficial discussions. 3
Proof of Theorem 1 assuming ( k, ℓ ) = (1 , Let G ( V, E ) be a (1 , Y of V . Since | Y | = α ( G ), it suffices to show that Hall’s condition (see, e.g., [14]) holds from Y to V \ Y . To be specific,for a set A ⊆ V , define N ( A ) def = { u ∈ V \ A : u is a neighbor of some v ∈ A } . Our goal is to prove that | N ( Y ′ ) | ≥ | Y ′ | for every subset Y ′ ⊆ Y .Assume, for contradiction’s sake, that | N ( Z ) | < | Z | for some Z ⊆ Y . We may assume that Z isminimal. Choose an arbitrary z ∈ Z , which exists since Z = ∅ .Because α ( G ) = α ( G \ { z } ), we can find another maximum-sized independent set X with z / ∈ X .Define Z = X ∩ Z and Z = Z \ Z . Put U def = (cid:0) X \ N ( Z ) (cid:1) ∪ Z . We claim that U is independent and that | U | > α ( G ), which would be a contradiction.First, we show that U is independent. Note that both X and Z are independent, hence both X \ N ( Z ) and Z are independent. Since there is no edge between Z and X \ N ( Z ), we see that U is independent.Next, we show that U can also be written as (cid:0) X \ ( N ( Z ) \ N ( Z )) (cid:1) ∪ Z . It suffices to show that X \ ( N ( Z ) \ N ( Z )) = X \ N ( Z ). Suppose x ∈ X \ ( N ( Z ) \ N ( Z )) and z ∈ Z are adjacent. Then x ∈ N ( Z ) and hence x ∈ N ( Z ). Since Z ⊆ X and X is independent, there is no edge between Z and X , hence x / ∈ N ( Z ). Thus, x ∈ X while x ∈ N ( Z ) \ N ( Z ), which is a contradiction. Weconclude that X \ ( N ( Z ) \ N ( Z )) ⊆ X \ N ( Z ). Since N ( Z ) \ N ( Z ) ⊆ N ( Z ), the opposite inclusionholds as well. Hence U = (cid:0) X \ ( N ( Z ) \ N ( Z )) (cid:1) ∪ Z .Finally, we show that | U | > | X | = α ( G ). Note that Z ∩ X = ∅ by definition, we have | U | = | X \ ( N ( Z ) \ N ( Z )) | + | Z | ≥ | X | − | N ( Z ) \ N ( Z ) | + | Z | . It suffices to show that | N ( Z ) \ N ( Z ) | < | Z | . Since z ∈ Z \ X , so Z ( Z , hence | N ( Z ) | ≥ | Z | bythe minimality assumption of Z . Thus, | N ( Z ) \ N ( Z ) | = | N ( Z ) | − | N ( Z ) | ≤ | N ( Z ) | − | Z | < | Z | − | Z | = | Z | , hence | U | > α ( G ).This is the promised contradiction. Hence the proof is complete.4 Proofs of Theorem 1 and Corollary 3
In this section, we derive Theorem 1 and Corollary 3 from what we proved in Section 2.
Proof of Theorem 1.
The proof is by induction on ℓ and k .Suppose ℓ = 0, and we are going to show that α ( G ) ≤ (cid:22) n − k + 12 (cid:23) . (2)(2) holds when k = 1 since we proved it in Section 2. Assume (2) is established for k −
1, and suppose G ( V, E ) is a ( k, | V | = n . For any vertex v ∈ V , by definition we have that G \ { v } is ( k − , α ( G \ { v } ) = α ( G ), so the induction hypothesis tells us that α ( G ) = α ( G \ { v } ) ≤ (cid:22) ( n − − ( k −
1) + 12 (cid:23) = (cid:22) n − k + 12 (cid:23) , which concludes the inductive proof of (2).Assume (1) is established for ℓ −
1, and suppose G ( V, E ) is a ( k, ℓ )-stable graph with | V | = n .If there exists v , . . . , v k − ∈ V such that α ( G ′ ) = α ( G ) − ℓ , here G ′ def = G \ { v , . . . , v k − } . Then bydefinition G ′ is a (1 , α ( G ) ≤ α ( G ′ ) + ℓ = (cid:22) n − ( k − (cid:23) + ℓ = (cid:22) n − k + 12 (cid:23) + ℓ. Otherwise, α ( G ) drops by at most ℓ − k − V , and bydefinition G is ( k − , ℓ − α ( G ) ≤ (cid:22) n − ( k −
1) + 12 (cid:23) + ( ℓ −
1) = (cid:22) n − k (cid:23) + ℓ. Putting these together gives us that (1) holds for k + 1, which concludes the inductive proof. Proof of Corollary 3.
Let S be the set of vertices v satisfying α ( G \ { v } ) = α ( G ). By the assumption | S | ≥ m . Let S = V ( G ) \ S be the complement of S . A key observation is that every maximum-sizedindependence set of G contains S .Define S def = { v ∈ S : N ( v ) ∩ S = ∅ } , S def = { v ∈ S : N ( v ) ∩ S = ∅ } . Let G be the induced subgraph of G on S , and we claim that α ( G \ { v } ) = α ( G ) for every vertex v of S . In fact, for v ∈ S , suppose X is an independent set of size α ( G ) in G \ { v } . Then X def = X ∩ S G of size α ( G ), for otherwise there exists some independent set X ′ of G such that | X ′ | > | X | , hence X ′ def = X ′ ∪ S is an independent set in G with | X ′ | > α ( G ), whichis impossible. Thus, the existence of X verifies our claim.Since every maximum-sized independence set of G contains S , we have α ( G ) = | S | + α ( G ), henceit follows from what we proved in Section 2 that α ( G ) = | S | + α ( G ) ≤ n − | S | + | S | / ≤ n − | S | + | S | / ≤ n − m/ . We remark that the bound in Corollary 3 is tight, as witnessed by a disjoint union of a balancedcomplete bipartite graph K ⌊ m/ ⌋ , ⌊ m/ ⌋ and an independent set of size n − ⌊ m/ ⌋ (possibly empty). By considering the graph complements, there is no difference between cliques and independent setsin the definition of f s,s + t . For convenience we replace all cliques by independent sets.Define G mn def = { G : G is a graph on n vertices with α ( G ) ≤ m − } . Our goal is to verify that • on the one hand, there is H ∈ G s + tn such that α ( H \ { u , . . . , u t − } ) ≥ s for every t − u , . . . , u t − of H ; • on the other hand, for every G ∈ G s + tn , there are t vertices v , . . . , v t of G such that α ( G \ { v , . . . , v t } ) ≤ s − . The first part is seen by considering any n -vertex graph H with α ( H ) = s + t −
1. Such an H exists because s + t ≤ n + 1. Note that the second part is trivial when α ( G ) ≤ s −
1, so we assume that α ( G ) = s − m for some positive integer m ≤ t . The key is to observe that G is not ( t, m − G were ( t, m − s − m = α ( G ) ≤ (cid:22) n − t + 12 (cid:23) + ( m − , hence s ≤ ⌊ n − t +12 ⌋ , which is a contradiction. Now that G is not ( t, m − t vertices v , . . . , v t of G such that α ( G \ { v , . . . , v t } ) ≤ α ( G ) − m = s − , which concludes the proof of the second part. The proof of Theorem 2 is complete.6 Proof of Theorem 4
With the help of the lemma below, the general ( k, ℓ )-stability in Theorem 4 can be reduced to thesimpler ( k − ℓ, Lemma 6.
Suppose n > k > ℓ ≥ . If G is an n -vertex tight ( k, ℓ ) -stable graph, then G ⊔ K , thedisjoint union of G and an isolated vertex, is an ( n + 1) -vertex tight ( k + 1 , ℓ + 1) -stable graph.Proof. Note that G is tight ( k, ℓ )-stable. Hence α ( G ⊔ K ) = α ( G ) + 1 = (cid:18)(cid:22) n − k + 12 (cid:23) + ℓ (cid:19) + 1 = (cid:22) ( n + 1) − ( k + 1) + 12 (cid:23) + ( ℓ + 1) , which attains equality in Theorem 1. So it suffices to prove that G ⊔ K is ( k + 1 , ℓ + 1)-stable.Let v ′ be the isolated vertex in G ⊔ K , and suppose a subset of k + 1 vertices { v , . . . , v k +1 } areremoved from V ( G ⊔ K ). If v ′ is removed, we may assume v k +1 = v ′ , hence by the stability of G , α ( G ⊔ K \ { v , . . . , v k +1 } ) = α ( G \ { v , . . . , v k } ) ≥ α ( G ) − ℓ = α ( G ⊔ K ) − ( ℓ + 1) . Otherwise, v , . . . , v k +1 ∈ V ( G ), then by the stability of G , α ( G ⊔ K \ { v , . . . , v k +1 } ) = α (cid:0) ( G \ { v , . . . , v k } ) \ { v k +1 } (cid:1) + 1 ≥ α ( G \ { v , . . . , v k } ) ≥ α ( G ) − ℓ = α ( G ⊔ K ) − ( ℓ + 1) . Thus, G ⊔ K is ( k + 1 , ℓ + 1)-stable, and the proof is complete.Evidently, Lemma 6 helps reduce the proof of Theorem 4(i), Theorem 4(ii), and Theorem 5 ingeneral to the proof of their special cases when ( k, ℓ ) = (1 , , , Proof of Theorem 4.
For n ≥
2, denote by K n the complete bipartite graph K n/ ,n/ when n is even,or the complete tripartite graph K ⌊ n/ ⌋ , ⌊ n/ ⌋ , when n is odd. Then α ( K n ) = ⌊ n/ ⌋ . After removingany one vertex v , at least one ⌊ n/ ⌋ -vertex part is left untouched, so α ( K n \ { v } ) = ⌊ n/ ⌋ = α ( K n ).Hence K n is always (1 , n -vertex (2 , n ≥ cycles C n and wheels W n . Here an n -wheel refers to an ( n − C n − with anothervertex connected to every vertex of the cycle. For example, C and W are shown in the figure below: Figure 1:
Graphs C and W .7e claim that C n is tight (2 , n is odd, and W n is tight (2 , n is even.Denote the n -vertex path graph (i.e. V = { v , . . . , v n } and E = { ( v , v ) , . . . , ( v n − , v n ) } ) as P n ,evidently α ( P n ) = ⌈ n/ ⌉ = ⌊ ( n + 1) / ⌋ . When n is odd, obviously α ( C n ) = ( n − /
2. Supposetwo disjoint paths P a and P b are left after vertices u, v being removed from C n (note that a, b can bezero), then a + b = n −
2, and α ( C n \ { u, v } ) = (cid:22) a + 12 (cid:23) + (cid:22) b + 12 (cid:23) = a + b + 12 = n −
12 = α ( C n ) , since a and b are of different parity. So C n is tight (2 , n is even, obviously α ( W n ) = α ( C n − ). Suppose two vertices u, v are removed from W n , then at most two vertices are removedfrom the induced subgraph C n − . Since C n − is (2 , α ( C n − ) ≤ α ( W n \ { u, v } ) ≤ α ( W n ) = α ( C n − ) = n − , hence W n is tight (2 , n -vertex tight (2 , , G is a 6-vertex tight (3 , α ( G ) = ⌊ (6 − / ⌋ = 2, so there is no trianglesubgraph K in the complement graph G . Also, if three vertices of G form a triangle, by removingthe other 3 vertices the independence number drops to 1, which contradicts with the (3 , K in G . Now we reach a contradiction with the fact that Ramseynumber R (3 ,
3) = 6, so no 6-vertex tight (3 , We are going to construct a sequence of (3 , G m ( V m , E m ) with | V m | → ∞ such that α ( G m ) = | V m | / − O (cid:16)p | V m | (cid:17) . Note that the existence of such a sequence directly implies Theorem 5 by Lemma 6.Suppose m ∈ N ( m ≥ V m def = Z / (2 m + 2 m ) Z , E m def = { ( i, j ) : | i − j | ≡ m or m + 1 (mod 2 m + 2 m ) } , we claim that G m def = ( V m , E m ) is (3 , α ( G m ) = m . Since m = (2 m + 2 m ) / − O (cid:16)p m + 2 m (cid:17) , First, we prove that α ( G m ) = m . It is easily seen that the vertices0 , m + 1 , m + 2 , · · · , ( m − m + 1) , , m + 2 , m + 3 , · · · , ( m − m + 1) + 1 , , m + 3 , m + 4 , · · · , ( m − m + 1) + 2 , ... ... ... . . . ... m − , m, m + 1 , · · · , ( m − m + 1) + m − m in G m . Hence, it suffices to show that α ( G m ) ≤ m .We prove this by contradiction. Suppose X ⊂ V m is an independent set of size m + 1. Partitionthe vertices into m groups of size 2 m + 2 each as follows: V def = { , m + 0 , m + 0 , · · · , (2 m + 1) m + 0 } ,V def = { , m + 1 , m + 1 , · · · , (2 m + 1) m + 1 } ,V def = { , m + 2 , m + 2 , · · · , (2 m + 1) m + 2 } , ... V m − def = { m − , m − , m − , · · · , (2 m + 2) m − } . Note that | X ∩ V j | ≤ m + 1 since every two consecutive points (including the first and the last) areadjacent in G m . Thus, by pigeonhole principle | X ∩ V j | = m + 1 for some j . Due to the obvioustranslation symmetry in G m , we may assume without loss of generality that X ∩ V = { , m, m, . . . , m } . Since X is an independent set, from the construction of E m we see that m, m, m, . . . , (2 m + 1) m / ∈ X,m + 1 , m + 1 , m + 1 , . . . , (2 m + 1) m + 1 / ∈ X,m − , m − , m − , . . . , (2 m + 1) m − / ∈ X. The key is to pair up every other vertex of G m as follows: { , m + 2 } , { m + 1 , m + 2 } , · · · , { (2 m · m + 1 , (2 m + 1) m + 2) } , { , m + 3 } , { m + 2 , m + 3 } , · · · , { (2 m · m + 2 , (2 m + 1) m + 3) } , ... ... . . . ... { m − , m − } , { m − , m − } , · · · , { (2 m + 1) m − , (2 m + 2) m − } . G m . Hence they are not in X simultaneously. So | X | ≤ | X ∩ V | + m + 1) + ( m − m + 1) = m − , which contradicts with | X | = m + 1. Thus, α ( G ) = m , as desired. Then, we show that G m is (3 , Think about V ( G m ) as 2 m + 2 m evenly distributedpoints on a circle. Here two points are adjacent in G m if and only if they form an interval of length m or m + 1 on the circle. We are going to prove that no matter which 3 points are deleted, we canstill pick m points such that no two of them are adjacent to each other in G m .Suppose 3 points u, v, w (clockwise in this order) are deleted, and the circle is cut into threepieces with a, b, c consecutive points in u ⌢ v, v ⌢ w, w ⌢ u , respectively ( a, b, c can be zero), then a + b + c = 2 m + 2 m −
3. Here u ⌢ v refers to the clockwise arc from u to v on the circle. The idea isto chop these pieces into many (2 m + 1)-consecutive-point groups and three small consecutive-pointgroups, then to pick some m consecutive points out of each (2 m + 1)-consecutive-point group, andto deal with the remaining small groups carefully. Let a ′ def = a mod 2 m + 1, b ′ def = b mod 2 m + 1, and c ′ def = c mod 2 m + 1.Case 1. a ′ + b ′ + c ′ = m −
3. Pick the first m points clockwise in each normal group. We havepicked m points in total, and every two of them are either among some m consecutive points or areat least m + 2 apart from each other, so these points form an independent set of size m .Case 2. a ′ + b ′ + c ′ = 3 m −
2. By symmetry we may assume without loss of generality that a ′ ≥ m and b ′ ≤ m −
1. Chop u ⌢ v clockwise into a ′ , m + 1 , . . . , m + 1-point groups, chop v ⌢ w into2 m + 1 , . . . , m + 1 , b ′ -point groups, chop w ⌢ u into 2 m + 1 , . . . , m + 1 , c ′ -point groups. Pick the first m points clockwise in the a ′ -point group. Starting from here and moving clockwise on the circle, wepick the first available m points from each normal group respectively (here available means forming agap of length at least m +2 from the last point that has been picked). So far we have picked m points.Suppose there are x points (including u ) between the first and the last point we picked. It suffices toshow that x ≥ m + 1. Suppose the gap (consecutive unpicked points on the circle) containing v has y points inside, and the gap containing the b ′ -point group and w has z points inside, then x + y + z = (2 m + 2 m ) − m − ( m − m + 1) = 4 m + 3 . Hence it suffices to show that y + z ≤ m + 2. Note that a < m , the first point clockwise ineach regular group in u ⌢ v is not picked. Hence, at most m points counterclockwise before v are10npicked, so y ≤ m + 1. As for z , the points in the gap consists of w and every point from the b ′ -point group and some points from a normal group, so z ≤ b ′ + ( m + 1) = b ′ + m + 2. Thus, y + z ≤ b ′ + 2 m + 3 ≤ m + 2 since b ′ ≤ m −
1, so the m points we picked form an independent set.Case 3. a ′ + b ′ + c ′ = 5 m −
1. The fact a ′ , b ′ , c ′ ≤ m imply that m − ≤ a ′ , b ′ , c ′ ≤ m . Bysymmetry we may assume a ′ = min { a ′ , b ′ , c ′ } . Chop u ⌢ v clockwise into 2 m + 1 , . . . , m + 1 , a ′ -pointgroups, v ⌢ w into b ′ , m + 1 , . . . , m + 1-point groups, w ⌢ u into 2 m + 1 , . . . , m + 1 , c ′ -pointgroups. Then we pick points as below: • Starting from u and moving clockwise, we pick the ( m + 1)’th through the 2 m ’th point fromeach normal group on u ⌢ v . • Starting from v and moving clockwise, we pick the first m points from the b ′ -point group (recallthat b ′ ≥ m ). Moving clockwise on v ⌢ w , we pick the first available m points from each normalgroup. • Starting from u and moving counterclockwise, we pick the first m points from the c ′ -point group(recall that c ′ ≥ m ). Moving counterclockwise on w ⌢ u , we pick the first available m pointsfrom each normal group.So far we have picked m points. Note that a ′ ≥ m −
1, and the point counterclockwise before the a ′ -point group is not picked, we see that the gap containing v has at least m + 1 points inside. Thenit suffices to show that the gap containing w has at least m + 1 points inside. Still, suppose the gapcontaining u, v, w has x, y, z points inside, respectively, then x + y + z = 4 m + 3 as before. Note that x = m + 1 according to how we picked the points, then z ≥ m + 1 if and only if y ≤ m + 1, which isequivalent to a ′ ≤ m −
1, which is true since a ′ = min { a ′ , b ′ , c ′ } . Thus the m points we picked forman independent set.By combining the three cases, we conclude that G m is (3 , k = ℓ + 4. In fact, every graph in G m ( V m , E m ) with V m def = Z / (2 m + 2 m + 1) Z , E m def = { ( i, j ) : | i − j | ≡ m or m + 1 (mod 2 m + 2 m + 1) } is (4 , More on tight (1 , , As we have seen in the proof of Theorem 4, there is an n -vertex (1 , n ≥ n -vertex (2 , n ≥
3. In this section, we care about the uniqueness ofsuch tight stable graphs. That is, for some given n , we want to know whether every n -vertex tight(1 , K n and whether every n -vertex (2 , C n or W n .For tight (1 , n ≥ M n the disjoint union of n/ n is even, and the disjoint union of ⌊ n/ ⌋ − n is odd. Evidently, the graph M n is tight (1 , n . Moreover,every n -vertex graph that is sandwiched between M n and K n is also tight (1 , n -vertex ( n ≥
3) path P n is another class of tight (2 , , , n is even (assume n ≥ V k def = Z / k Z , E k def = { ( i, j ) : | i − j | ≡ k (mod 2 k ) } ,G k def = ( V k , E k ) is tight (2 , k ≥
3. However, we are unaware of any n -vertex tight(2 , C n when n is odd. We doubt that C n is the only n -vertextight (2 , C n , W n , and the tight (2 , , Characterization of tight (1 , One might wonder whether a tight (1 , G (especially when | V ( G ) | is even) is always sandwiched between a perfect matching and acomplete bipartite graph (i.e. M n ⊆ G ⊆ K n ). In fact, M n ⊆ G is already shown by Hall’s theoremin the proof in Section 2, while G ⊆ K n does not necessarily happen. An example is the graph below. Figure 2:
A tight (1 , G with G
6⊆ K n .12evertheless, we can still prove tight upper and lower bounds on number of edges: Theorem 7.
Suppose G ( V, E ) is a tight (1 , -stable graph with | V ( G ) | = n , then | E ( M n ) | ≤ | E ( G ) | ≤ | E ( K n ) | . Proof.
We consider the cases when n is even and when n is odd separately.Case 1. n is even. Suppose n = 2 k , then α ( G ) = k implies that the complement graph G containsno ( k + 1)-clique. Hence Tur´an’s theorem implies that | E ( G ) | ≤ | E ( K ,..., ) | = 2 k − k. Here K ,..., refers to the complete k -partite graph in which each part contains 2 vertices. So | E ( G ) | ≥ (cid:18) k (cid:19) − (2 k − k ) = k = | E ( M n ) | , which gives the desired lower bound.For the upper bound, if there is a vertex v with deg( v ) ≥ k + 1, then v is not contained in anymaximum-sized independent set since α ( G ) = k . Hence G \ { v } is (1 , α ( G \ { v } ) = k , which contradicts Theorem 1. Thus, every vertex of G is of degree at most k , hence | E ( G ) | ≤ · k · k = k = | E ( K n ) | , which gives the desired upper bound.Case 2. n is odd. Suppose n = 2 k + 1, then α ( G ) = k implies that the complement graph G contains no ( k + 1)-clique. Hence Tur´an’s theorem implies that | E ( G ) | ≤ | E ( K ,..., , ) | = 2 k − . Here K ,..., , refers to the complete k -partite graph in which k − | E ( G ) | ≥ (cid:18) k + 12 (cid:19) − (2 k −
2) = k + 2 = | E ( M n ) | , which gives the desired lower bound.For the upper bound, if there is a vertex v with deg( v ) ≥ k + 2, then v is not contained inany maximum-sized independent set since α ( G ) = k . Hence G \ { v } is (1 , | E ( G \ { v } ) | ≤ k by what we just proved, hence | E ( G ) | ≤ k + deg( v ) ≤ k + 2 k = | E ( K n ) | . G is of degree at most k + 1, hence | E ( G ) | ≤
12 (2 k + 1)( k + 1) ≤ k + 2 k = | E ( K n ) | . The proof is done by combining both cases.
1. Is the odd cycle C n the only n -vertex tight (2 , n ≥
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