aa r X i v : . [ m a t h . C O ] F e b Radon Numbers for Trees
Shoham Letzter ∗ October 11, 2018
Abstract
Many interesting problems are obtained by attempting to generalize classical results on con-vexity in Euclidean spaces to other convexity spaces, in particular to convexity spaces ongraphs. In this paper we consider P -convexity on graphs. A set U of vertices in a graph G is P -convex if every vertex not in U has at most one neighbour in U . More specifically, weconsider Radon numbers for P -convexity in trees.Tverberg’s theorem states that every set of ( k − d + 1) − points in R d can be partitionedinto k sets with intersecting convex hulls. As a special case of Eckhoff’s conjecture, we showthat a similar result holds for P -convexity in trees.A set U of vertices in a graph G is free , if no vertex of G has more than one neighbour in U .We prove an inequality relating the Radon number for P -convexity in trees with the size of amaximal free set. Radon’s classical lemma [8] states that every set of d + 2 points in R d can be partitioned into twosets whose convex hulls intersect. Tverberg [9] generalized this to partitions into more than twosets. Namely, every set of at least ( k − d + 1) + 1 points in R d can be partitioned into k setswhose convex hulls have a point in common.Inspired by this, Eckhoff conjectured in [3] that the situation is similar in general convexity spaces.A convexity space is a pair ( X, C ) where X is a set and C is a collection of subsets of X , called convexsets , such that ∅ and X are convex and the intersection of convex sets is convex. The convex hull of a set S ⊆ X , denoted by H C ( S ) , is the minimal convex set containing S , i.e. the intersectionof all convex sets containing S . For a set S , a k -Radon partition is a partition of S into k setswhose convex hulls have a point in common. A set is k -anti Radon (or k -a.r.) if it has no k -Radon partition. The k th Radon number of ( X, C ) is the minimal number (if it exists) r k ( C ) suchthat every set S ⊆ X of size at least r k ( C ) has a k -Radon partition. Eckhoff [3] conjectured that r k ( C ) ≤ ( k − r ( C ) −
1) + 1 holds in every convexity space. This conjecture has been provedin several convexity spaces including trees with geodesic convexity [6]. However, the generalconjecture has recently been disproved by Bukh [1]. ∗ Department of Pure Mathematics and Mathematical Statistics, Centre for Mathematical Sciences, Wilberforce Road,Cambridge, CB3 0WB, UK. Email: [email protected] k -a.r set can be generalized to multi-sets by considering partitions of multi-setsrather than sets. In this paper, we define ˜ r k ( C ) to be the size of the largest k -a.r. multi-set. Notethat ˜ r k ( C ) ≥ r k ( C ) − , with equality for k = 2 (as a -a.r. multi-set is a set, i.e. no element canappear more than once). When k = 2 , we often omit the prefix k , e.g. a -a.r. set may be called ana.r. set and ˜ r ( C ) = ˜ r ( C ) .In this paper we shall study P -convexity in trees. For a graph G , a set U of vertices of G is P -convex or, briefly, convex if every vertex not in U has at most one neighbour in U . Equiva-lently, U is convex if it contains all middle vertices in the paths of length between two verticesof U . P -convexity was first considered in the context of directed graphs and tournaments (see[4],[5],[7],[10]).Throughout this paper graphs are always finite, simple and undirected. For a graph G , let ˜ r k ( G ) denote the k th Radon number for P -convexity on G , and for a set U ⊆ V ( G ) , let H G ( U ) denotethe convex hull of U in G .As the first main result of our paper, we show that Eckhoff’s conjecture holds for P -convexity ontrees. Theorem 1.1.
Let T be a tree, k ≥ . Then ˜ r k ( T ) ≤ ( k − r ( T ) . Given a graph G , call a set A ⊆ V ( G ) free if every vertex of G has at most one neighbour in A .Note that every free set in a graph G is also convex and the converse does not hold in general. Let ˜ α ( G ) be the size of the largest free set in G . It follows that ˜ r ( G ) ≥ ˜ α ( G ) . Our second main theoremanswers a question posed by Dourado et al. [2]. Theorem 1.2.
Let T be a tree. Then ˜ r ( T ) ≤ α ( T ) . We shall show that this theorem is sharp in the sense that there are infinitely many trees for whichwe have equality.The last inequality is not true in general as shown by the graph G in Figure 1. Every two of theseven vertices of G have a common neighbour, hence ˜ α ( G ) = 1 . It is easy to check that theset A = { , , } of vertices of G is a.r. and that every set of vertices of G is not a.r., therefore ˜ r ( G ) = 3 .
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Figure 1: Graph G We prove Theorem 1.1 in Section 2 and Theorem 1.2 in Section 3.2
Proof of Theorem 1.1
Before proving this theorem, we introduce some notation. For a graph G and a vertex v ∈ V ( G ) ,define ˜ r ∗ k ( G, v ) = max {| R | : R is a k -a.r. multi-set and v / ∈ H G ( R ) } . (1) Proof of Theorem 1.1.
We shall prove more than claimed in the statement of the theorem. Namely,we shall show that for every tree T • ˜ r ∗ k ( T, v ) ≤ ( k − r ∗ ( T, v ) holds for every v ∈ V ( T ) , • ˜ r k ( T ) ≤ ( k − r ( T ) .Our proof is by induction on n = | V ( T ) | . Both statements are clear for n ≤ .Let T be a tree with n ≥ vertices. The second statement follows easily by induction usingexpression 2 for ˜ r ∗ k below. For a vertex v ∈ V ( T ) , let v , . . . , v l be its neighbours, and for every i ∈ [ l ] let T i be the component of v i in T \ { v } . Then ˜ r ∗ k ( T, v ) = max i ∈ [ l ] ( X j = i ˜ r ∗ k ( T j , v j ) + ˜ r k ( T i )) . (2)We now prove that ˜ r k ( T ) ≤ ( k − r ( T ) . Let R be a k -a.r. multi-set of maximal size. If T hasendvertex v which is not in R , let T ′ = T \ { v } . Then by induction, | R | = ˜ r k ( T ′ ) ≤ ( k − r ( T ′ ) ≤ ( k − r ( T ) . Thus we may assume that R contains each endvertex of T at least once.In the rest of the proof we consider two possible cases which will be dealt with in different sub-sections. There is a longest path v , . . . , v m in T such that deg( v ) ≥ .Let z = v , y = v and x , . . . , x l be the neighbours of y other than z . Note l ≥ , and by the choiceof v , . . . , v m as a longest path, x , . . . , x l are all endvertices (see Figure 2).Denote by s i , i ∈ [ l ] , the number of appearances of x i in R and by t the number of appearances of y in R . By our assumption that R contains every endvertex at least once, s i ≥ for every i ∈ [ l ] .As R is k -a.r., s i , t ≤ k − . Let s = s + . . . + s l . We consider three cases according to the value of s .(a) s ≤ k − .Let τ = min { s, k − − s } , σ = ( s − τ ) / . Note that σ is an integer and τ + σ ≤ k − . Set T ′ = T \ { x , . . . , x l } (see Figure 2). Let R ′ be the multi-set obtained by adding σ copies of y to R ∩ V ( T ′ ) . Note | R | = | R ′ | + s − σ = | R ′ | + σ + τ ≤ | R ′ | + k − .3 x x l y z T ′ T ′′ Figure 2: Case 1
Claim 2.1. R ′ is k -a.r.. Proof.
We shall show the existence of sequences a , . . . , a σ , b , . . . , b σ satisfying • a j , b j ∈ [ l ] and a j = b j for every j ∈ [ σ ] , • |{ j ∈ [ σ ] : a j = i }| + |{ j ∈ [ σ ] : b j = i }| = s i for every i ∈ [ l ] .Note that the existence of such sequences completes the proof of this claim. Assume to thecontrary that R ′ = R ′ ∪ . . . ∪ R ′ k is a k -Radon partition of R ′ . Obtain R , . . . , R k by replacingeach of the σ new copies of y with a distinct pair x a j , x b j , j ∈ [ σ ] . By the choice of the a j ’s and b j ’s, R = R ∪ . . . ∪ R k is a partition of R . Clearly H T ( R l ) ∩ V ( T ′ ) = H T ′ ( R ′ l ) for every l ∈ [ k ] ,hence this partition is a k -Radon partition of R , contrary to the choice of R as a k -a.r. set.It thus remains to show the existence of such sequences. By induction on k , we show that if s = s + . . . + s l ≤ k − and s i ≤ k − for every i ∈ [ l ] we can find two sequences satisfyingthe above. We assume σ ≥ or equivalently s ≥ k , because otherwise there is nothing toprove. When k = 2 we thus have that without loss of generality s = s = 1 , and we set a = 1 , b = 2 . If k ≥ assume that s ≥ s ≥ . . . ≥ s l , and let a σ = 1 , b σ = 2 . Now set s ′ i = (cid:26) s i − i ∈ { , } s i otherwise . Note that s ′ = s ′ + . . . s ′ l ≤ k − and s ′ i ≤ k − (otherwise s ≥ k − and s + s + s ≥ k − > k − , a contradiction). Also σ ′ = 2 k − − s ′ = σ − . We can now continue byinduction.Using Claim 2.1 we conclude by induction that ˜ r k ( T ) = | R | ≤ | R ′ | + ( k − ≤ ˜ r k ( T ′ ) + ( k − ≤ ( k − r ( T ′ ) + 1) . The following claim completes the proof of Theorem 1.1 in Case 1a.
Claim 2.2. ˜ r ( T ) ≥ r ( T ′ ) + 1 . Proof.
Let S ′ be an a.r. set in T ′ of maximal size. Set S = (cid:26) S ′ ∪ { x } y / ∈ S ′ ( S ′ \ { y } ) ∪ { x , x } y ∈ S ′ . Note that | S | = | S ′ | + 1 . We shall show that S is a.r, thus proving the claim. Assume to thecontrary that there exists a partition S = A ∪ B with H T ( A ) ∩ H T ( B ) = ∅ . We assume x ∈ A .Consider the following three possibilities. 4 y / ∈ S ′ .Set A ′ = A \ { x } . Note – S ′ = A ′ ∪ B is a partition of S ′ . – H T ( A ) = (cid:26) H T ′ ( A ′ ) ∪ { x } z / ∈ H T ′ ( A ′ ) H T ′ ( A ′ ) ∪ { x , y } z ∈ H T ′ ( A ′ ) . – H T ( B ) = H T ′ ( B ) , and y / ∈ H T ( B ) .Therefore H T ( A ) ∩ H T ( B ) = H T ′ ( A ′ ) ∩ H T ′ ( B ) = ∅ , a contradiction. • y ∈ S ′ and x , x ∈ A .Let A ′ = ( A \ { x , x } ) ∪ { y } . Then H T ( A ) = H T ′ ( A ′ ) ∪ { x , x } , and H T ( B ) = H T ′ ( B ) .As before we reach a contradiction. • y ∈ S ′ and x ∈ A , x ∈ B .As S ′ is a.r. and y ∈ S ′ , z / ∈ H T ′ ( A \ { x } ) ∩ H T ′ ( B \ { x } ) . Without loss of generality, z / ∈ H T ′ ( B \ { x } ) . Set A ′ = ( A \ { x } ) ∪ { y } . As before H T ( A ) ⊆ H T ′ ( A ′ ) ∪ { x , y } and H T ( B ) = H T ′ ( B ) ∪ { x } . This leads to a contradiction to S ′ being a.r..(b) s = 2 k − .Define T ′ as before, and let R ′ be the union of R ∩ V ( T ′ ) with a copy of x and k − copies of y . Claim 2.3. R ′ is k -a.r.. Proof.
Replacing s by s − returns us to the setting of Claim 2.1. Following the same argu-ments we obtain this claim.Set T ′′ = T ′ \ { y } , R ′′ = R ′ ∩ V ( T ′′ ) (see Figure 2). Then z / ∈ H T ′′ ( R ′′ ) as otherwise we canpartition R ′ into k parts, k − of which contain y , and the last contains both x and z . Thispartition is such that y is in the convex hull of all parts, contradicting the fact that R ′ is k -a.r..Thus ˜ r k ( T ) = | R | = 2 k − | R ′′ | < k −
1) + ˜ r ∗ k ( T ′′ , z ) ≤ ( k − r ∗ ( T ′′ , z )) . The following claim completes the proof of Theorem 1.1 in Case 1b.
Claim 2.4. ˜ r ( T ) ≥ r ∗ ( T ′′ , z ) . Proof.
Let S ′′ be an a.r. set of T ′′ satisfying z / ∈ H T ′′ ( S ′′ ) . We shall show that S = S ′′ ∪{ x , x , x } is a.r. thus proving the claim (note that l ≥ , so S is well defined). Assumethat we have a Radon partition S = A ∪ B . Without loss of generality x , x ∈ A . Set A ′ =( A ∩ V ( T ′ )) ∪ { y } , B ′ = B ∩ V ( T ′ ) . Then H T ( A ) ∩ H T ( B ) = H T ′ ( A ′ ) ∩ H T ′ ( B ′ ) . Claim 2.4follows from the following claim. Claim 2.5.
Let T be a tree, v ∈ V ( T ) and S an a.r. set in T satisfying v / ∈ H T ( S ) . Let T v ← u denotethe tree obtained from T by adding a new vertex u and connecting it to v . Then S ∪ { u } is a.r. in T v ← u . roof. We prove the claim by induction on | V ( T ) | . The claim is clear when T has at most onevertex. Let T ′ = T v ← u and S = A ∪ B a partition of S . We show H T ′ ( A ∪ { u } ) ∩ H T ′ ( B ) = ∅ .Let v , . . . , v l be the neighbours of v in T . Denote by T i the component of v i in T \ { v } and S i = S ∩ V ( T i ) , A i = A ∩ V ( T i ) , B i = B ∩ V ( T i ) . If for every i ∈ [ l ] , v i / ∈ H T i ( A i ) , then H T ′ ( A ∪ { u } ) = H T ′ ( A ) ∪ { u } and H T ′ ( B ) = H T ( B ) . Thus, as S is a.r., H T ′ ( A ∪ { u } ) ∩ H T ′ ( B ) = H T ( A ) ∩ H T ( B ) = ∅ .So we can assume v ∈ H T ( A ) . As v / ∈ H T ( S ) , this means that v i / ∈ H T i ( S i ) for every i ≥ .As S is a.r., v / ∈ H T ( B ) . Thus H T ′ ( A ∪ { u } ) = H T ( A ) ∪ { u } ∪ ( [ j ≥ H ( T i ) vi ← v ( A i ∪ { v } )) H T ′ ( B ) = [ i ≥ H T i ( B i ) = H T ( B ) ∪ ( [ i ≥ H ( T i ) vi ← v B i ) . Therefore H T ′ ( A ∪ { u } ) ∩ H T ′ ( B ) = [ i ≥ ( H ( T i ) vi ← v ( A i ∪ { v } ) ∩ H ( T i ) vi ← v ( B i )) The proof follows using the induction hypothesis with T i , i ≥ .(c) s ≥ k .Similarly to Claim 2.1, we can conclude that the multi-set obtained by adding k copies of y to R ∩ V ( T ′ ) is k -a.r. , which is obviously a contradiction. In every longest path v , . . . , v m of T , deg( v ) = 2 .Fix a longest path v , . . . , v m in T . Denote v = z , and note that each of its neighbours other than v is either an endvertex or has degree and is adjacent to an endvertex (by the choice of the longestpath and the definition of Case 2). Let y , . . . , y p be the neighbours of z other than v which havedegree , and y p +1 , . . . , y q the neighbours of z other than v which are endvertices. Let x , . . . , x p be the neighbours of y , . . . , y p which are endvertices respectively (see Figure 3a).Denote by s i , i ∈ [ p ] , the number of appearances of x i in R ; t i , i ∈ [ q ] , the number of appearancesof y i in R and u the number of appearances of z in R . Let t = t + . . . + t q . As in the previous case,we conclude from the fact that R is k -a.r. that t ≤ k − . Consider the following three cases.(a) q = 1 .Then t + min { s , u } ≤ k − (otherwise obtain a k -Radon partition of R by putting a copy of y in t sets, and a copy of x and z in the other k − t sets). Thus s + t + u = t + min { s , u } + max { s , u } ≤ k − . zy y p y p +1 y q x p x T ′ T ′′ (a) Case 2 x y z v T ′ (b) Case 2a Figure 3: Case 2Let T ′ = T \ { x , y , z } , R ′ = R ∩ V ( T ′ ) (see Figure 3b). Then ˜ r k ( T ) = | R | ≤ | R ′ | + 2( k − ≤ ˜ r k ( T ′ ) + 2( k − ≤ ( k − r ( T ′ ) + 2) . The proof of Theorem 1.1 in Case 2a follows from the following claim.
Claim 2.6. ˜ r ( T ) ≥ ˜ r ( T ′ ) + 2 . Proof.
Let S ′ be an a.r. set in T ′ . Set S = S ∪ { x , y } . It is easy to verify that S is a.r..(b) t ≤ k − .As before, set τ = min { t, k − − t } , σ = ( t − τ ) / (then t − σ ≤ k − ). Let T ′ = T \{ x , . . . , x p , y , . . . , y q } (see Figure 3a) and let R ′ be the multi-set obtained by adding σ copiesof z to R ′ ∩ V ( T ′ ) . Then as in Claim 2.1, R ′ is k -a.r. and thus ˜ r k ( T ) = | R | = | R ′ | + s + t − σ ≤ ˜ r k ( T ′ ) + ( p + 1)( k − ≤ ( k − r ( T ′ ) + p + 1) . The proof of Theorem 1.1 in this case follows from the following claim.
Claim 2.7. ˜ r ( T ) ≥ ˜ r ( T ′ ) + p + 1 . Proof.
Let S ′ be a.r. in T ′ . Set S = (cid:26) S ′ ∪ { x , . . . , x p , y } z / ∈ S ′ ( S ′ \ { z } ) ∪ { x , . . . , x p , y , y } z ∈ S ′ Showing that S is a.r. is similar to the proof of Claim 2.2.(c) t = 2 k − .Let T ′ = T \ { x , . . . , x p , y , . . . , y q } , T ′′ = T ′ \ { z } (see Figure 3a). Let R ′′ = R ∩ V ( T ′′ ) and let R ′ be the multi-set obtained by adding k − copies of z to R ′′ and a copy of y . As in Case 1b, R ′ is k -a.r. in T ′ and R ′′ is k -a.r. in T ′′ with v / ∈ H T ′′ ( R ′′ ) . Thus ˜ r k ( T ) = | R ′′ | + s + 2 k − ≤ ˜ r ∗ k ( T ′′ , v ) + s + 2 k − ≤ ( k − r ∗ ( T ′′ , v ) + s + 2 k − . (3)As s i ≤ k − for every i ∈ [ p ] , s ≤ ( k − p . If s < ( k − p , we obtain ˜ r k ( T ) ≤ ( k − r ∗ ( T ′′ , v ) + p + 2) . And the proof of Theorem 1.1 in this case follows from the claim below.7 laim 2.8. ˜ r ( T ) ≥ ˜ r ( T ′′ , v ) + p + 2 . Proof. If S ′′ is a.r. in T ′′ with v / ∈ S ′′ then similarly to the proof of Claim 2.4, S = S ′′ ∪{ x , . . . , x p , y , y } is a.r. in T .Thus we may assume s = ( k − p i.e. s = . . . = s p = k − . Claim 2.9. t = . . . = t p = 0 . Proof.
Assume otherwise, then without loss of generality t ≥ . Let φ = k − t . Similarly tothe proof of Claim 2.1 we will show the existence of a , . . . , a φ , b , . . . , b φ such that • a j , b j ∈ [2 , q ] and a j = b j for every j ∈ [ φ ] , • |{ j ∈ [ φ ] : a j = i }| + |{ j ∈ [ φ ] : b j = i }| ≤ t i for every i ∈ [2 , q ] .This leads to a contradiction as we can then obtain a k -Radon partition of R by putting a copyof y in t of the sets, and putting a copy of x and a pair y a l , y b l in each of the other k − t sets. y will be in the intersection of the convex hulls of the sets (here we use the assumption that s = k − so this is indeed possible).If t i ≤ k − t − for every i ∈ [2 , q ] , we proceed as in Claim 2.1 to prove the existence of suchsequences. Otherwise, let i be such that t i ≥ k − t . Note that X j =1 ,i t i = t − t i − t ≥ k − − ( k − − t = k − t . Thus in this case we can choose a = . . . = a φ = i and b , . . . , b φ ∈ [2 , q ] \ { i } to satisfy therequirements.Using Claim 2.9 it follows that k − t = t p +1 + . . . + t q . As t i ≤ k − for every i ∈ [ q ] , q − p ≥ . Claim 2.10. ˜ r ( T ) ≥ ˜ r ∗ ( T ′′ , v ) + 3 + p . Proof.
Let S ′′ be an a.r. set in T ′′ with v / ∈ H T ′′ ( S ′′ ) . Let S = S ′′ ∪ { x , . . . , x p , y p +1 , y p +2 , y p +3 } .It is easy to see that S is a.r. in T .Recalling inequality 3, we obtain ˜ r k ( T ) ≤ s + 2 k − k − r ∗ ( T ′′ , v ) ≤ ( k − p + 3 + ˜ r ∗ ( T ′′ , v )) ≤ ( k − r ( T ) . And the proof of Theorem 1.1 is complete. 8
Proof of Theorem 1.2
We need the following definition for the proof of Theorem 1.2. Let G be a graph, v ∈ V ( G ) . Define ˜ α ∗ ( T, v ) , max {| A | : A is free and x / ∈ A } . Proof of Theorem 1.2.
We prove a stronger statement than what is claimed in this theorem. Weshall show that for every tree T • ˜ r ∗ ( T, v ) ≤ α ∗ ( T, v ) for every vertex v ∈ V ( T ) , • ˜ r ( T ) ≤ α ( T ) .We prove these statements by induction on n = | V ( T ) | . Both statements are clear for n ≤ .To prove the first statement, let v ∈ V ( T ) , and denote by v , . . . , v l its neighbours. For every i ∈ [ l ] ,let T i be the connected component of v i in T \ { v } . It is easy to see that ˜ α ∗ ( T, v ) = max j ∈ [ k ] { X i = j ˜ α ∗ ( T i , v i ) + ˜ α ( T j ) } . Note the similarity to expression 2 from the previous section. It thus follows by induction that ˜ r ∗ ( T, v ) ≤ α ∗ ( T, v ) .We now proceed to proving that ˜ r ( T ) ≤ α ( T ) . Let R be an a.r. set of maximal size in T . As inthe proof of Theorem 1.1, we can assume that R contains all endvertices of T . The brothers of anendvertex v are the endvertices in distance from v . Then in particular, every endvertex has atmost brothers, as no vertex of T can have more than neighbours in R .The following claim will be useful in the rest of the proof. Claim 3.1.
Let T be a tree. There exists a free set A ⊆ V ( T ) of size ˜ α ( T ) satisfying that for every endvertex v ∈ V ( T ) either v ∈ A or one of its brothers is in A . Proof.
Let A be a free set in T of maximal size, v an endvertex in T and u its only neighbour.If v ∈ A we are done. Otherwise, by the maximality of A , A ∪ { v } is not free. As u is the onlyneighbour of v , there is a neighbour w = v of u which is contained in A . If w is an endvertex,we are done. Otherwise, set A ′ = ( A \{ w } ) ∪ { v } . Then A ′ contains v and is free of size ˜ α ( T ) .Continuing similarly will result in a free set of size ˜ α ( T ) with the property that for each endvertexeither it or one of its brothers is in the set.We consider three cases concerning longest paths in T . Note that the theorem can be easily verifiedif the longest path in T has at most vertices, thus we assume that a longest path in T contains atleast vertices. We devote a separate subsection for each case. There is a longest path v , . . . , v m such that the component of v in T \ { v } has no endvertex indistance from v with brothers. 9 v v T ′ (a) Case 1a v v v (b) Case 1b In this figure and the following ones:a black vertex is in R , a white one is not in R ,and for a grey vertex it is unknown if it is in R . Figure 4: Cases 1a, 1b . We consider six cases.(a) v , v ∈ R .Set T ′ = T \ { v , v } (see Figure 4a), R ′ = R ∩ V ( T ′ ) . R ′ is a.r. in T ′ and v / ∈ H T ′ ( R ′ ) . Thus, by induction, ˜ r ( T ) = | R ′ | + 2 ≤ ˜ r ∗ ( T ′ , v ) + 2 ≤
2( ˜ α ∗ ( T ′ , v ) + 1) . Note that ˜ α ∗ ( T ′ , v ) + 1 ≤ ˜ α ( T ) , because if A ′ ⊆ V ( T ′ ) \ { v } is free, then A ′ ∪ { v } is free in T .Therefore ˜ r ( T ) ≤ α ( T ) in this case.(b) v , v ∈ R (see Figure 4b).Set R ′ = ( R \{ v } ) ∪ { v } . It is easy to see that R ′ is a.r. and it follows from Case 1a that ˜ r ( T ) ≤ α ( T ) .We can now assume that the above two cases do not occur. Consider the neighbours of v otherthan v . Each such neighbour either is an endvertex, or has degree and its other neighbour is anendvertex (using the fact v , . . . , v m is a longest path and that we are in Case 1). Let S i , i ∈ { , } ,be the set of neighbours of v other then v with degree i . Note that v ∈ S , and by our previousassumptions: S ⊆ R , S ∩ R = ∅ . In particular | S | ≤ . Consider the remaining four cases.(c) | S | ≥ .Let T ′ be the component of v in T \ ( S \ { v } ) (see Figure 5a). Then ˜ r ( T ′ ) ≥ ˜ r ( T ) − ( | S | − , as R ∩ V ( T ′ ) is a.r. in T ′ and R \ V ( T ′ ) contains only the endvertices which are neighbours ofvertices in S \ { v } . Furthermore ˜ α ( T ) ≥ ˜ α ( T ′ ) + ( | S | − . To see this, let A ′ ⊆ V ( T ′ ) be a maximum sized free set in T ′ containing v (recall Claim 3.1).Then v / ∈ A ′ and the set obtained by adding the endvertices which are neighbours of thevertices in S to A ′ is free in T . Hence, by induction, ˜ r ( T ) ≤ ˜ r ( T ′ ) + | S | − <
2( ˜ α ( T ′ ) + | S | − ≤ α ( T ) . v v v T ′ (a) Case 1c v v v v T ′ (b) Case 1d v v v v T ′ (c) Case 1e Figure 5: Case 1c, 1d, 1e v v v v T ′ (a) v v v v v T ′ T ′′ (b) Figure 6: Case 1fWe can now assume that S = { v } .(d) | S | ≤ .Let T ′ be the connected component of v in T \ { v } (see Figure 5b). Then ˜ r ( T ′ ) ≥ ˜ r ( T ) − , as R ∩ V ( T ′ ) is a.r. in T ′ , and v , v / ∈ R (otherwise consider Cases 1a,1b). Also ˜ α ( T ) ≥ ˜ α ( T ′ ) + 1 ,because if A ′ ⊆ V ( T ′ ) is free in T ′ then A ′ ∪ { v } is free in T . We obtain ˜ r ( T ) ≤ ˜ r ( T ′ ) + 2 ≤
2( ˜ α ( T ′ ) + 1) ≤ α ( T ) . (e) | S | = 3 .Let T ′ be as in the previous case (see Figure 5c) and set R ′ = R ∩ V ( T ′ ) . Then v / ∈ H T ′ ( R ′ ) and | R | = | R ′ | + 4 . Also ˜ α ( T ) ≥ ˜ α ∗ ( T, v ) + 2 , because if A ′ ⊆ V ( T ′ ) \ { v } is free, then A ′ ∪ { v , v } is free in T . Hence ˜ r ( T ) ≤ ˜ r ∗ ( T ′ , v ) + 4 ≤
2( ˜ α ∗ ( T, v ) + 2) ≤ α ( T ) . (f) | S | = 2 .Set T ′ = T \ { v , v } (see Figure 6a). If T ′ contains a free set of maximal size A ′ such that v / ∈ A ′ , then A ′ ∪ { v } is free, so in this case ˜ α ( T ) ≥ α ( T ′ ) and ˜ r ( T ) ≤ r ( T ′ ) < α ( T ′ )) ≤ α ( T ) . Therefore we may assume that v is contained in every maximum sized free set of T ′ . Let S bethe set of neighbours of v other than v and for v ∈ S let T v be the connected component of v in T \ { v } .We need the following claim. 11 laim 3.2. T v has depth as a tree rooted in v for every v ∈ S . Proof.
Let v ∈ S , v = v (note that the claim is clear if v = v ). By the choice of v , . . . , v m asa longest path in T , T v has depth at most . We now show that T v has depth at least , i.e. v has neighbours which are not endverties. Let A ′ be a free set of maximal size in T ′ . If v has noneighbour in T v which is not an endvertex, then ( A ′ \ { v } ) ∪ { v } is also a free set of the samesize in T ′ , contradicting our previous assumption.If for some v ∈ S , T v is not isomorphic to T v (as rooted trees at v , v respectively), by changingthe selected longest path to go through v instead of v , we go back to one of the previous cases.Thus we may assume that the trees T v , v ∈ S , are all isomorphic to T v .Let T ′′ be the component of v in T \ { v } (see Figure 6b), R ′′ = R ∩ V ( T ′′ ) . Then | R \ R ′′ | ≤ | S | + 1 as for each v ∈ S , | R ∩ V ( T v ) | = 3 and possibly v in R . Note also that ˜ α ( T ) ≥ ˜ α ( T ′′ ) + 2 | S | , because the union of a free set of T ′′ with the endvertices in distance from v and their neighbours is free in T . Therefore ˜ r ( T ) ≤ ˜ r ( T ′′ ) + 3 | S | + 1 ≤ α ( T ′′ ) + 4 | S | ≤ α ( T ) . Case 1 does not hold, and there exists a longest path v , . . . , v m such that the component of v in T \ { v } has no endvertices in distance from v with more than one brother.Choose the longest path such that v has exactly one brother v ′ . Then v , v ′ ∈ R and as R is a.r., v / ∈ R . We consider the neighbours of v other than v . Note that they can be of degrees , or only and that if they have degree or the other neighbours are endvertices. Let S i , i ∈ { , , } ,be the set of neighbours of v other than v with degree i . Consider the following six cases.(a) S = ∅ .Let T ′ be the component of v in T \ S (i.e. remove all neighbours of v of degree , see Figure7a). Then, by induction ˜ r ( T ) ≤ ˜ r ( T ′ ) + 2 | S | ≤
2( ˜ α ( T ′ ) + | S | ) ≤ α ( T ) . The last inequality follows from the fact that a maximum free set in T ′ can be assumed tocontain v (see Claim 3.1), so it does not contain v and we can add the | S | endvertices thatwere discarded to obtain a free set in T .We now assume S = ∅ .(b) | S | ≥ .Set T ′ to be the component of v in T \ ( S \ { v } ) (see Figure 7b). R contains | S | − of thediscarded vertices, thus, as in the previous case, ˜ r ( T ) ≤ ˜ r ( T ′ ) + 2( | S | − ≤
2( ˜ α ( T ′ ) + ( | S | − ≤ α ( T ) . v v v v ′ T ′ (a) Case 2a v v v v v ′ T ′ (b) Case 2b Figure 7: Cases 2a, 2b v v v v v ′ T ′ (a) Case 2c v v v v v ′ T ′ (b) Case 2d v v v v v ′ T ′ (c) Case 2e Figure 8: Cases 2c, 2d, 2eHence we can assume | S | = 1 . Note that | S | ≤ , since R is a.r..(c) | S | = 2 .Set T ′ the component of v in T \ { v } (see Figure 8a). Then ˜ r ( T ) ≤ r ∗ ( T ′ , v ) ≤ α ∗ ( T ′ , v )) ≤ α ( T ) , since v / ∈ H T ′ ( R ∩ V ( T ′ )) ( R is a.r.) and the union of a free set in V ( T ′ ) \ { v } with { v , v } remains free.(d) S = ∅ and v ∈ R .Choose T ′ as in the previous case (see Figure 8b). Again v / ∈ H T ′ ( R ∩ V ( T ′ )) and similarly ˜ r ( T ) ≤ r ∗ ( T ′ , v ) < α ∗ ( T ′ , v )) ≤ α ( T ) . (e) S = ∅ and v / ∈ R .Again set T ′ as before (see Figure 8c). Here R contains only of the discarded vertices and wecan add v to any free set of T ′ to obtain a free set of T . Thus ˜ r ( T ) ≤ ˜ r ( T ′ ) + 2 ≤
2( ˜ α ( T ′ ) + 1) ≤ α ( T ) . (f) All previous cases do not hold, i.e. | S | = | S | = 1 and S = ∅ .Set T ′ = T \ { v , v ′ , v } (see Figure 9a). If T ′ contains a maximum free set A ′ with v / ∈ A ′ , then A ′ ∪ { v } is free in T and thus ˜ r ( T ) ≤ ˜ r ( T ′ ) + 2 ≤ α ( T ′ )) ≤ α ( T ) . Therefore, we may assume that every maximum free set of T ′ contains v . As in Case 1f, let S be the set of neighbours of v different from v and for v ∈ S define T v to be the component of v in T \ { v } . Similarly to Claim 3.2, T v has depth as a tree rooted at v for every v ∈ S . If T v is13 v v v v ′ T ′ (a) v v v v ′ v v T ′ T ′′ (b) Figure 9: Case 2fnot isomorphic to T v or to the graph in Case 1f (as rooted trees), by changing the longest pathto go through v , we can continue as before. (Note that in all but the present case and Case 1f,we did not consider other neighbours of v ).Set T ′′ to be the component of v in T \ { v } (see Figure 9b). R contains three vertices of T v for every v ∈ S and possibly it contains v as well. Also ˜ α ( T ) ≥ ˜ α ( T ′′ ) + 2 | S | , as we can twovertices from each T v to a free set of T ′′ to obtain a free set. Thus ˜ r ( T ) ≤ ˜ r ( T ′′ ) + 3 | S | + 1 ≤
2( ˜ α ( T ′′ ) + 2 | S | ) ≤ α ( T ) . For every choice of a longest path v , . . . , v m the connected component of v in T \ { v } has anendvertex with brothers in distance from v .We choose the longest path such that v has brothers v ′ and v ′′ . Then v , v ′ , v ′′ ∈ R , so v , v / ∈ R .Similarly to Cases 2a,2b, we can assume that all the neighbours of v are endvertices (except formaybe v ) having one or two endvertices as neighbours. By the choice of v , . . . , v m as a longestpath, the neighbours of v other than v are either endvertices or have degree and are neighboursto endvertices. Thus, as R is a.r., v can have degree or only. Set T ′ to be the component of v in T \ { v } . We consider seven possible cases.(a) v has degree with the only neighbour other than v and v being an endvertex (see Figure10a).Then v / ∈ H T ′ ( R ∩ V ( T ′ )) . Note that ˜ α ( T ) ≥ ˜ α ∗ ( T ′ , v ) + 2 , as we can add v and v to a freeset in V ( T ′ ) \ { v } to obtain a free set. Thus ˜ r ( T ) ≤ ˜ r ∗ ( T ′ , v ) + 4 ≤
2( ˜ α ∗ ( T ′ , v ) + 2) ≤ α ( T ) . (b) v has degree with the only neighbour other than v and v , u , having three neighbourswhich are endvertices (see Figure 10b).Then again v / ∈ H T ′ ( R ∩ V ( T ′ )) , and ˜ α ( T ) ≥ α ∗ ( T ′ , v ) , as we can add v , v and anendvertex which is a neighbour of u to a free set of T ′ . Thus ˜ r ( T ) ≤ r ∗ ( T ′ , v ) ≤ α ∗ ( T ′ , v )) ≤ α ∗ ( T ) . v v v ′ v ′′ v T ′ (a) Case 3a v v v v ′ v ′′ v u T ′ (b) Case 3b v v v v ′ v ′′ v T ′ (c) Case 3c Figure 10: Cases 3a, 3b, 3c v v v v ′ v ′′ v v T ′ T ′′ (a) Case 3d v v v v ′ v ′′ v v T ′′ T ′ (b) Case 3e Figure 11: Cases 3d, 3eIn the remaining cases we assume that v has degree . Let T ′′ be the component of v in T \ { v } .(c) T ′ has a maximum free set A ′ with v / ∈ A ′ (see Figure 10c).Then A ′ ∪ { v , v } is free and ˜ r ( T ) ≤ r ( T ′ ) <
2( ˜ α ( T ′ ) + 2) ≤ α ( T ) . We may now assume that every maximum free set of T ′ contains v . Let S be the set of neighboursof v other than v and v . Claim 3.3.
The vertices in S are endvertices in T . Proof.
Let v ∈ S , and T v the component of v in T \ { v } . Then T v has depth at most as a treerooted in v (by the choice of v , . . . , v m as a longest path). Let A ′ be a free set of maximal size in T ′ , then v ∈ A ′ . If v has a neighbour in T v , u , it is either an endevertex, or all of its neighboursexcept for v are endvertices. Then ( A ′ \ { v } ) ∪ { u } is free in T ′ , a contradiction. Thus v has noneighbours in T v , i.e. it is an endvertex in T .Clearly, the claim implies | S | ≤ .(d) S = ∅ (see Figure 11a).Then | R ∩ ( V ( T ) \ V ( T ′′ )) | ≤ and we can add v , v to a free set of T ′′ . Thus ˜ r ( T ) ≤ r ( T ′′ ) ≤ α ( T ′′ )) ≤ α ( T ) . We can assume now that S = ∅ . 15 v v v ′ v ′′ v v v T ′′ T ′′′ T ′ (a) Case 3f v v v v ′ v ′′ v v v u T ′′ T ′ T ∗ (b) Case 3g Figure 12: Cases 3f, 3g(e) There is a free set A ′′ of maximal size in T ′′ with v / ∈ A ′′ (see Figure 11b).Then | R ∩ ( V ( T ) \ V ( T ′′ )) | ≤ and the union of A ′′ with v , v and an endvertex from S is free,thus ˜ r ( T ) ≤ r ( T ′′ ) ≤ α ( T ′′ )) ≤ α ( T ) . Thus we can assume that every maximal free set in T ′′ contains v .(f) v has degree in T .Let T ′′′ be the component of v in T \ { v } (see Figure 12a). Then, as in the previous case, ˜ r ( T ) ≤ r ( T ′′′ ) ≤ α ( T ′′′ )) ≤ α ( T ) . (g) v has a neighbour u = v , v .Let T ∗ be the component of u in T \ { v } (see Figure 12b). The following claim can be provedsimilarly to the proofs of Claims 3.2, 3.3, using the above assumptions. Claim 3.4. T ∗ has depth as a tree rooted in u . By considering a longest path going through u instead of v , we can assume that the compo-nent of u in T \ { v } satisfies the same conditions as the component of v . However, in thiscase T ′′ has an endvertex in distance from v , a contradiction to the assumption that everymaximum free set of T ′′ contains v .The following example shows that Theorem 1.2 is sharp.This is sequence of trees T m , m ≥ , with m vertices. ˜ r ( T ) ≥ m (the set of all endvertices isa.r.). Let A be a free set of T m with maximal size. We can assume that A contains the endvertices x , . . . , x m , y , . . . , y m . Thus u , . . . , u m / ∈ A . Also, A contains at most one of the neighbours of u i for each i ∈ [ m ] . Hence { x , . . . , x m } ∪ { y , . . . , y m } ∪ { w , . . . , w m } is a free set of maximal size, so ˜ α ( T ) = 3 m . By Theorem 1.2, ˜ r ( T ) ≤ α ( T ) = 6 m . Thus ˜ r ( T ) = 6 m = 2 ˜ α ( T ) . In this paper we proved two results about the Radon number for P -convexity in graphs. It maybe interesting to consider these problems for general graphs. Regarding Theorem 1.1, it is still an16 u w x y v u w x y v m u m w m x m y m Figure 13: Sharpness of Theorem 1.2open problem to determine whether Eckhoff’s conjecture holds for P -convexity in all graphs. Weshowed that the inequality ˜ r ( G ) ≤ α ( G ) from Theorem 1.2, does not hold for all graphs G , but itmay still be the case that a similar but weaker inequality holds in general. Furthermore, for bothresults, it would be interesting to characterize the trees for which the results hold with equality. References [1] B. Bukh. Radon partitions in convexity spaces, 2010. arXiv:1009.2384v1 [math.CO].[2] M. C. Dourado, D. Rautenbach, V. Fernandes dos Santos, P. M. Sch¨afer, J. L. Szwarcfiter, andA. Toman. An upper bound on the P -Radon number. Discrete Math. , 312:2433–2437, 2012.[3] J. Eckhoff. The partition conjecture.
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