aa r X i v : . [ m a t h . C O ] S e p Monotone Paths in Dense Edge-Ordered Graphs
Kevin G. Milans ∗ July 8, 2018
Abstract
The altitude of a graph G , denoted f ( G ), is the largest integer k such that under each orderingof E ( G ), there exists a path of length k which traverses edges in increasing order. In 1971,Chv´atal and Koml´os asked for f ( K n ), where K n is the complete graph on n vertices. In 1973,Graham and Kleitman proved that f ( K n ) ≥ p n − / − / f ( K n ) ≤ ( + o (1)) n . We show that f ( K n ) ≥ ( − o (1))( n/ lg n ) / . A totally ordered graph is a graph G that is associated with a total ordering of its vertex set V ( G )and a total ordering of its edge set E ( G ). We use T ( G ) and T ′ ( G ) to denote the total orderings of V ( G ) and E ( G ) respectively. When only the vertices or only the edges of G are totally ordered,we call G an ordered graph or an edge-ordered graph , respectively. An ordering , edge-ordering , or total ordering of a graph G is an ordered, edge-ordered, or totally ordered graph whose underlyinggraph is G .In an edge-ordered graph G , a monotone path is a path which traverses edges in increasing orderwith respect to T ′ ( G ). A monotone trail is similar, except that a trail is allowed to revisit vertices.The altitude of a graph G , denoted f ( G ), is the maximum integer k such that every edge-orderingof G contains a monotone path of length k . Also, let f ⋆ ( G ) be the maximum integer k such thatevery edge-ordering of G contains a monotone trail of length k .In 1971, Chv´atal and Koml´os [4] asked for f ( K n ) and f ⋆ ( K n ), where K n denotes the completegraph on n vertices. Citing private communication, Chv´atal and Koml´os noted in their 1971 paperthat Graham and Kleitman had already proved Ω( n / ) ≤ f ( K n ) < ( + ε ) n and established f ⋆ ( K n )exactly: f ⋆ ( K n ) = n − n ∈ { , } , in which case f ⋆ ( K n ) = n .To show f ⋆ ( K n ) ≥ n −
1, Graham and Kleitman [6] proved that if G has average degree d , then f ⋆ ( G ) ≥ d . Friedgut communicated to Winkler [13] an elegant formulation of their proof, known asthe pedestrian argument . For an n -vertex edge-ordered graph G , the pedestrian argument involves n pedestrians, with one starting at each vertex in G . An announcer calls out the names of the edgesin order according to T ′ ( G ). When e is called, both pedestrians at the endpoints of e traverse e ,trading places. Since each pedestrian travels along a monotone trail and each edge is traversed bytwo pedestrians, the average length of a pedestrian’s monotone trail is 2 | E ( G ) | /n , which equals d .The pedestrian argument has recently been modified to produce monotone paths (see [8] and [5]). ∗ West Virginia University, Morgantown, WV 26505, ( [email protected] ). f ( K n ) and f ⋆ ( G ). In particular, they proved that p n − / − / ≤ f ( K n ) < n/
4, and they conjectured that f ( K n ) is closer to their upper boundthan their lower bound. They also commented that, with additional effort, their lower bound couldbe improved to f ( K n ) ≥ ( c − o (1)) √ n for some c >
1. In his Master’s thesis from the same year,R¨odl [11] proved that if G has average degree d , then f ( G ) ≥ (1 − o (1)) √ d ; for G = K n , R¨odl’sresult matches the Graham–Kleitman lower bound asymptotically. R¨odl also noticed that the ideasin the Graham–Kleitman upper bound can be combined with results in design theory to prove f ( K n ) ≤ ( + o (1)) n . Alspach, Heinrich, and Graham (unpublished, see [3]) further improved theupper bound to f ( G ) ≤ ( + o (1)) n . In 1984, Calderbank, Chung, and Sturtevant [3] obtainedthe best known upper bound: f ( K n ) ≤ ( + o (1)) n . After 1984, explicit progress on determining f ( K n ) slowed (but see [2] for exact values for n ≤ f ( G ) ≤ G is planar and showed that f ( C n ∨ K ) ≥ n ≥
99, where C n ∨ K is the planar graph obtained by joining the n -vertexcycle C n and a pair of non-adjacent vertices. Consequently, the maximum of altitude of a planargraph is between 5 and 9.Clearly, f ( G ) ≤ f ⋆ ( G ). The edge-chromatic number of G , denoted χ ′ ( G ), is the minimum k such that E ( G ) is the union of k matchings. Ordering E ( G ) so that each matching is an intervalshows that f ⋆ ( G ) ≤ χ ′ ( G ). Vizing’s theorem [12] states that χ ′ ( G ) ≤ ∆( G ) + 1, where ∆( G ) is themaximum degree of G . It follows that f ( G ) ≤ ∆( G ) + 1.Improving a result of Yuster [14], Alon [1] gave a short proof that there exist k -regular graphs G with f ( G ) ≥ k , as follows. The girth of G is the length of a shortest cycle in G . If G has girth g ,then every trail of length less than g is a path. Therefore f ( G ) ≥ min { g − , f ⋆ ( G ) } ≥ min { g − , d } ,where d is the average degree of G . In particular, if G is k -regular and has girth larger than k ,then f ( G ) ≥ k . For k = 3, better constructions are known. Mynhardt, Burger, Clark, Falvai1, andHenderson [9] characterized the 3-regular graphs with girth at least 5 and altitude 3, and then usedthe characterization to show that the flower snarks are examples of 3-regular graphs with altitude 4.For k ≥
4, it remains open to decide whether there are graphs G with ∆( G ) = k and f ( G ) = k + 1.A Hamiltonian path in a graph is a path containing all of its vertices. Katreniˇc and andSemaniˇsin [7] proved that deciding whether a given edge-ordered graph contains a Hamiltonianmonotone path is NP-complete. Although it seems likely that computing the altitude of a givengraph is NP-hard or worse, we note that the result of Katreniˇc and Semaniˇsin does not directlyimply this.Lavrov and Loh [8] investigated the maximum length of a monotone path in a random edge-ordering of K n . They showed that with probability tending to 1, a random edge-ordering of K n contains a monotone path of length at least 0 . n . Consequently, edge-orderings of K n that givesublinear upper bounds on f ( K n ), if they exist, are rare. They also proved that with probabilityat least 1 /e − o (1), a random edge-ordering of K n contains a Hamiltonian monotone path. Thecommon strengthening of these results leads to a natural and beautiful conjecture. Conjecture 1.1 (Lavrov–Loh [8]) . With probability tending to 1, a random edge-ordering of K n contains a Hamiltonian monotone path.Recently, De Silva, Molla, Pfender, Retter, and Tait [5] proved that f ( Q n ) ≥ n/ lg n where Q n is the n -dimensional hypercube and lg denotes the base-2 logarithm. They also showed that if2 ( n ) → ∞ and p ≤ ( ω ( n ) ln n ) /n / , then with probability tending to 1 the Erd˝os–R´enyi randomgraph G ( n, p ) has altitude at least (1 − o (1)) npω ( n ) ln n . Consequently, there are graphs with averagedegree √ n (ln n ) and altitude at least (1 − o (1)) √ n . These graphs are sparse and yet the lowerbound on their altitude asymptotically matches the lower bound on f ( K n ) due to Graham andKleitman.In this paper, we improve R¨odl’s result for sufficiently dense graphs. We show that if G is an n -vertex graph with average degree d and s /d → s = Θ( n / (log n ) / ), then f ( G ) ≥ (1 − o (1)) d s . For G = K n , we obtain f ( K n ) ≥ ( − o (1))( n/ lg n ) / . Our proof is based on asimple algorithm to extend monotone paths. In his Master’s thesis, R¨odl [11] gave a simple and elegant argument that f ( G ) ≥ (1 − o (1)) √ d where d is the average degree of G , which we outline as follows. Let G be an edge-ordered graphwith average degree d and suppose that k is an integer with d ≥ (cid:0) k +12 (cid:1) = 2(1 + · · · + k ). Obtain G ′ from G by marking at each vertex v the k largest edges incident to v (or all edges incident to v if d ( v ) < k ) and then removing all marked edges. Since G ′ has average degree at least d − k ,by induction G ′ contains a monotone path x . . . x k − of length k −
1. Since x k − is not isolated in G ′ , it follows that x k − is incident to at least k edges in E ( G ) − E ( G ′ ), and one of these extends x . . . x k − to a monotone path of length k . R¨odl’s idea of reserving large edges at each vertex forpath extension plays a key role in our approach. We make a slight change in that we require thevertices to have disjoint sets of reserved edges. We organize the edges in a table.Let G be a totally ordered graph. The height table of G is an array A whose columns areindexed by V ( G ) and rows are indexed by the positive integers. Each cell in A is empty or containsan edge in G . For u ∈ V ( G ) and a positive integer i , we use A ( i, u ) to denote the contents of thecell in A located in row i and column u . We order the cells of A so that A ( i, u ) precedes A ( i ′ , u ′ )if and only if i < i ′ or i = i ′ and u precedes u ′ in T ( G ). We define A iteratively. Given that thecontents of all preceding cells have been defined, let A ( i, u ) be the largest edge (relative to T ′ ( G ))incident to u not appearing in a preceding cell; if no such edge exists, then A ( i, u ) is empty. Notethat each edge appears in exactly one cell in A . We define the height of e in G , denoted h G ( e ), tobe the index of the row in A containing e .Extending a given monotone path is a key step in our algorithm. The height of a nontrivialmonotone path x . . . x k is the height of its last edge x k − x k . Lemma 2.1.
Let G be a totally ordered graph. For ≤ k < r , each monotone path of length k andheight r extends to a monotone path of length k + 1 and height at least r − k .Proof. Let A be the height table of G , and let x . . . x k be a monotone path of length k and height r . Since x k − x k appears in row r in A , this edge did not already appear when A ( i, x k ) is definedfor i < r . It follows that for i < r , the cell A ( i, x k ) contains an edge incident to x k which is largerthan x k − x k in T ′ ( G ). Let S = { A ( i, x k ) : r − k ≤ i ≤ r − } . Since | S | = k and x k − x k S , someedge in S joins x k with a vertex outside { x , . . . , x k − } and extends the path as claimed.Starting with a single edge and iterating Lemma 2.1, we obtain the following. Lemma 2.2.
Let G be a totally ordered graph and let x x be an edge in G of height r . If t is apositive integer and (cid:0) t (cid:1) < r , then G contains monotone path x x . . . x t of height at least r − (cid:0) t (cid:1) . roof. By induction on t . The lemma is clear when t = 1. For t >
1, the inductive hypothesisimplies that G contains a monotone path x x . . . x t − of height at least r − (cid:0) t − (cid:1) . With k = t − x . . . x t with height at least ( r − (cid:0) t − (cid:1) ) − ( t − r − (cid:0) t (cid:1) .Using Lemma 2.2, we match R¨odl’s bound f ( G ) ≥ (1 − o (1)) √ d asymptotically. We include theshort proof for completeness. Theorem 2.3. If G has average degree d , then f ( G ) ≥ ⌊ / √ d ⌋ .Proof. Let H be a total ordering of G , and let x x be an edge of maximum height r . Since eachrow of the height table contains n cells, it follows that r ≥ | E ( G ) | /n = d/
2. If t is a positive integerand (cid:0) t (cid:1) < d/
2, then we may apply Lemma 2.2 to extend x x to a monotone path of length t in H . Hence, (cid:0) t (cid:1) < d/ f ( G ) ≥ t . With t = ⌊ / √ d ⌋ , we have that (cid:0) t (cid:1) < d/ f ( G ) ≥ ⌊ / √ d ⌋ .Let G be a totally ordered graph and let x . . . x k be a monotone path in G . Viewing heightas a resource, extending x . . . x k becomes more expensive as k grows. When extending becomestoo expensive, we delete { x , . . . , x k − } from G to form a new totally ordered graph G ′ (whichinherits the orderings of V ( G ) and E ( G )), and we extend x k − x k to a monotone path in G ′ . Forthis to work, we must show that the height of x k − x k does not decrease too much when we delete { x , . . . , x k − } from G . Definition 2.4.
Let G be a totally ordered graph. For S ⊆ V ( G ) and an edge e in G − S , we definedrop( G, S, e ) to be h G ( e ) − h G − S ( e ). For s ≤ n −
2, let g ( n, s ) be the maximum of drop( G, S, e )over all n -vertex totally ordered graphs G , all sets S of s vertices in G , and all edges e ∈ E ( G − S ).Note that g ( n, s ) is monotonic in n , since adding isolated vertices to a totally ordered graph G and inserting them arbitrarily into the vertex ordering gives a larger totally ordered graph G ′ suchthat drop( G, S, e ) = drop( G ′ , S, e ) for all S ⊆ V ( G ) and e ∈ E ( G − S ). Lemma 2.5.
Let G be an n -vertex totally ordered graph and let x x be an edge of height r . If s is a positive integer and s ≤ n − , then G contains a monotone path extending x x of length atleast sk + 1 , where k = ⌊ ( r − / ( (cid:0) s +12 (cid:1) + g ( n, s )) ⌋ .Proof. By induction on n . If k = 0, then the lemma is clear. Otherwise, r − ≥ (cid:0) s +12 (cid:1) + g ( n, s ) andwe may apply Lemma 2.2 to obtain a monotone path x . . . x s +1 of height at least r − (cid:0) s +12 (cid:1) . Let S = { x , . . . , x s − } and let G ′ = G − S . We have that h G ′ ( x s x s +1 ) = h G ( x s x s +1 ) − drop( G, S, x s x s +1 ) ≥ r − (cid:0) s +12 (cid:1) − g ( n, s ).Applying the inductive hypothesis to G ′ and x s x s +1 , we obtain a monotone path P ′ in G ′ extending x s x s +1 of length at least sk ′ + 1, where k ′ = $ r − (cid:0) s +12 (cid:1) − g ( n, s ) − (cid:0) s +12 (cid:1) + g ( n − s, s ) % ≥ $ r − (cid:0) s +12 (cid:1) − g ( n, s ) − (cid:0) s +12 (cid:1) + g ( n, s ) % = k − . Prepending x . . . x s to P ′ produces a monotone path in G of length at least s + sk ′ + 1, and s + sk ′ + 1 ≥ sk + 1. 4 The Token Game
Our goal is to prove an upper bound on g ( n, s ). Let G be an n -vertex totally ordered graph, andlet S be a set of s vertices of G . We analyze an iterative process which obtains the height table of G − S from the height table of G . Let G ′ = G − S , let A be the array obtained from the heighttable of G by deleting columns indexed by vertices in S , and let A ′ be the height table of G ′ . Notethat the cells of both A and A ′ are indexed by Z , where Z = { , , , . . . } × V ( G ′ ). We order Z in the same order as the corresponding cells in A ′ are defined; that is, ( i, u ) ≤ ( i ′ , v ) if and onlyif i < i ′ or i = i ′ and u ≤ v in T ( G ′ ). For β ∈ Z , the open down-set of β , denoted D ( β ) is { γ ∈ Z : γ < β } and the closed up-set of β , denoted U [ β ] is { γ ∈ Z : γ ≥ β } . Similarly, the interval [ β, γ ] is { δ ∈ Z : β ≤ δ ≤ γ } .We produce a sequence of arrays { A β : β ∈ Z } which initially resemble A and later resemble A ′ . For β ∈ Z , the cells of A β are indexed by Z and are partitioned into a lower part indexed by D ( β ) and an upper part indexed by U [ β ].For β ∈ Z , each cell in A β is either empty, contains an edge in G ′ , or contains an object calleda hole . Moreover, each edge in G ′ appears in one cell in A β . Each A β also has a critical interval [( i, u ) , ( j, u )], where β = ( i, u ) and j is the least integer such that j ≥ i and A β ( j, u ) does notcontain a hole. Lemma 3.1.
There is a sequence of arrays { A β : β ∈ Z } such that each column in the initial arrayhas at most s holes, and for each β ∈ Z the following hold.1. If δ < β , then A β ( δ ) = A ′ ( δ ) .2. If δ ≥ β and A β ( δ ) does not contain a hole, then A β ( δ ) = A ( δ ) .3. If γ is the successor of β in Z , then A γ is obtained from A β by swapping A β ( β ) and A β ( δ ) ,where δ is in the critical interval of A β . Moreover, if β and δ index cells in distinct columns u and v , then A β ( δ ) = uv .Proof. Recall that A is obtained from the height table of G by deleting columns indexed by verticesin S . Note that A omits every edge with both endpoints in S and contains every edge in G ′ . Anedge uv ∈ [ S, S ] with u S and v ∈ S appears in A if and only if uv is in column u in the heighttable of G . Let α be the minimum element in Z , and let A α be the array obtained from A byreplacing edges in [ S, S ] with holes. If u indexes a column in A α , then each hole in column u replaces an edge uv in G with v ∈ S , and therefore each column in A α contains at most s holes.Clearly, every edge in G ′ appears once in A α and A α satisfies properties (1) and (2).We obtain other arrays iteratively. Let β = ( i, u ), let γ be the successor of β , and suppose that A β has been previously defined but A γ is not yet defined. Analogously to A β , we partition of thecells of A ′ into a lower part indexed by D ( β ) and an upper part indexed by U [ β ]. Since A β and A ′ contain the same set of edges and agree on their lower parts, it follows that the upper parts of A β and A ′ contain the same edges (possibly in a different order). We consider two cases, depending onwhether A ′ ( β ) is empty or contains an edge in G ′ . Case 1 : A ′ ( β ) is not empty. Let e = A ′ ( β ), and let δ be the index of the cell in A β containing e . We claim that δ is in the critical interval [( i, u ) , ( j, u )] of A β . Since e is in the upper partof A ′ , it follows that e is in the upper part of A β and so δ ≥ β = ( i, u ). Since δ, ( j, u ) ∈ U [ β ]and neither A β ( δ ) nor A β ( j, u ) contains a hole, it follows from (2) that A ( δ ) = A β ( δ ) = e and A ( j, u ) = A β ( j, u ). Suppose for a contradiction that δ > ( j, u ). Note that e is available for A ( j, u )5hen building the height table of G , and so A ( j, u ) = e ′ for some edge e ′ incident to u such that e ′ > e in T ′ ( G ). Since A β ( j, u ) = A ( j, u ) = e ′ , it follows that both e and e ′ appear in the upperpart of A β and hence in the upper part of A ′ also. Therefore both e and e ′ are available for A ′ ( β )when building the height table of G ′ . The selection of e over e ′ for A ′ ( β ) implies that e > e ′ in T ′ ( G ′ ), contradicting that e ′ > e in T ′ ( G ). Therefore δ ≤ ( j, u ) and δ is in the critical interval of A β as claimed. Obtain A γ from A β by swapping the contents of cells A β ( β ) and A β ( δ ) (if β = δ ,then A γ = A β ). Note that if δ indexes a cell in column v and v = u , then A ′ ( β ) = e and A ( δ ) = e imply that e is incident to both u and v , so that A β ( δ ) = e = uv , satisfying (3).We check that A γ satisfies (1) and (2). Since γ is the successor of β and A γ ( β ) = A β ( δ ) = e = A ′ ( β ), it follows that A γ satisfies (1). If the critical interval [( i, u ) , ( j, u )] of A β has size 1, then β = ( i, u ) = δ = ( j, u ) and A γ = A β , implying that A γ satisfies (2). Otherwise j > i and A β ( β )contains a hole. Relative to A β , the only change in the upper part of A γ is that A γ ( δ ) becomes ahole after swapping A β ( β ) and A β ( δ ), and so A γ satisfies (2). Case 2 : A ′ ( β ) is empty. This implies that the upper part of A ′ contains no edge incident to u ,and so the upper part of A β also contains no edge incident to u . In particular, A β ( j, u ) is empty,where [( i, u ) , ( j, u )] is the critical interval of A β . We obtain A γ from A β by swapping the contentsof cells A β ( i, u ) and A β ( j, u ), satisfying (3). Since A γ ( β ) and A ′ ( β ) are both empty, A γ satisfies(1). Relative to A β , the upper part of A γ is either unchanged or contains a new hole at A γ ( j, u ).It follows that A γ also satisfies (2).Given the sequence of arrays { A β : β ∈ Z } from Lemma 3.1, we obtain a useful upper boundon drop( G, S, e ). Lemma 3.2.
Let e be an edge in G ′ and choose β ∈ Z so that A ′ ( β ) = e . If [( i, u ) , ( j, u )] is thecritical interval of A β , then drop( G, S, e ) ≤ j − i .Proof. Since β = ( i, u ) and e appears in row i of the height table of G ′ , it follows that h G ′ ( e ) = i .Let δ index the cell in A β containing e . Since the successor A γ of A β satisfies A γ ( β ) = A ′ ( β ) = e ,it follows that A γ is obtained from A β by swapping A β ( β ) with A β ( δ ). By (3), we have that δ is inthe critical interval [( i, u ) , ( j, u )] of A β , and so δ = ( ℓ, v ) where i ≤ ℓ ≤ j . Since δ ≥ β and A β ( δ ) isnot a hole, by (2) we have that e = A β ( δ ) = A ( δ ). Therefore e appears in row ℓ of the height tableof G and so h G ( e ) = ℓ . We conclude drop( G, S, e ) = ℓ − i ≤ j − i .We define the height of a critical interval [( i, u ) , ( j, u )] to be j − i . Note that the height of thecritical interval of A β is at most the number of holes in column u of A β . Also, by property (1) ofLemma 3.1, all holes of A β are contained in the upper part of A β . Analyzing the movement of theholes as β increases in Z naturally leads to a single player game.A token game is a game played on an array B with rows indexed by the positive integers andcolumns indexed by a finite list. Let B ( i, u ) denote the cell in row i and column u . Each cell in B isempty or contains a token . A token in cell B ( i, u ) is grounded if all cells in column u below B ( i, u )contain tokens; a token which is not grounded is ungrounded . One of the columns is distinguishedas the active column .A step in a token game modifies B to produce a new array B ′ , subject to certain rules. Let u be the active column. If column u contains grounded tokens, then the player may optionally movethe highest grounded token in column u from its cell B ( i, u ) to an empty cell B ( i ′ , v ), provided that i ′ ≤ i and no prior step in the game moved a token between columns u and v . Next, all ungroundedtokens in column u shift down by one cell, and the active column advances cyclically. A step in6hich a token moves between columns is a transfer step . The list of arrays produced in a tokengame is its transcript .An ( n, s ) -token game is a token game with n columns, each of which initially contains at most s tokens. Let ˆ g ( n, s ) be the maximum number of tokens that can be placed in a single column inan ( n, s )-token game. The following gives the connection between g ( n, s ) and ˆ g ( n, s ). Lemma 3.3. g ( n, s ) ≤ ˆ g ( n − s, s ) Proof.
Let G be an n -vertex totally ordered graph and let S be a set of s vertices in G such thatdrop( G, S, e ) = g ( n, s ) for some edge e in G − S . Let G ′ = G − S , let A ′ be the height table of G ′ ,obtain A from the height table of G by deleting columns indexed by S , and apply Lemma 3.1 toobtain the sequence of arrays { A β : β ∈ Z } . We use this sequence to play the ( n − s, s )-token gameso that at least g ( n, s ) tokens are placed in some column.Construct a sequence { B β : β ∈ Z } of token arrays as follows. Let β = ( i, u ). We put a tokenin B β ( j, v ) if and only if A β ( k, v ) contains a hole, where k = j + i if v < u in T ( G ′ ) and k = j + i − B β from A β by removing all edges so that only holes and emptycells remain, shifting cells down to discard the lower part of A β , and replacing holes with tokens.We claim that the sequence { B β : β ∈ Z } is the transcript of an ( n − s, s )-token game in whichthe active column of B β is the second coordinate in β . Let α be the minimum element in Z , andnote that each column in A α contains at most s holes by Lemma 3.1. It follows that each columnin B α contains at most s tokens, satisfying the initial condition of an ( n − s, s )-token game.Let β = ( i, u ) and let γ be the successor of β . From property (3) of Lemma 3.1, we have that A γ is obtained from A β by swapping A β ( β ) and A β ( δ ) for some δ in the critical interval [( i, u ) , ( j, u )]of A β . If the critical interval has size 1, then A γ = A β and column u of B β contains no groundedtokens. We obtain B γ from B β by allowing the tokens in column u to shift down by 1 cell. Theactive column advances, completing a legal move in the token game.Otherwise j > i . Recall that the cells of B β correspond to the upper part of A β . The cellsindexed by the critical interval [( i, u ) , ( j, u )] of A β correspond to the cells in B β of height at most j − i , except that the last cell A β ( j, u ) corresponds to B β ( j − i + 1 , u ) which has height j − i + 1.Since A β ( ℓ, u ) contains a hole for i ≤ ℓ < j , it follows that B β ( ℓ, u ) contains a grounded tokenfor 1 ≤ ℓ ≤ j − i . Since A β ( β ) contains a hole and A β ( δ ) does not, it follows that δ > β andwe obtain A γ from A β by swapping the contents of distinct cells A β ( β ) and A β ( δ ). Therefore weobtain B γ from B β by firstly moving the grounded token in B β (1 , u ) to an empty cell of height atmost j − i or to B β ( j − i + 1 , u ) and secondly shifting the contents of all cells in column u down by1 cell. Equivalently, we obtain B γ from B β by optionally moving the highest grounded token from B β ( j − i, u ) to an empty cell of height at most j − i and shifting the ungrounded tokens in column u down by 1 cell. This is allowed in a token game provided that we have not executed a transferstep between a pair of columns more than once.Suppose that the transition from B β to B γ represents the first transfer step between distinctcolumns u and v ; we may assume without loss of generality that a token is moved from column u in B β to column v in B γ . It follows that a hole in A β ( β ) is swapped with the contents of A β ( δ ) toform A γ , where β and δ index cells in columns u and v respectively. By property (3) of Lemma 3.1,we have that A β ( δ ) = uv . Since δ ≥ β , the edge uv is in the upper part of A β . On the otherhand, we have β < γ and A γ ( β ) = A β ( δ ) = uv , and so uv is in the lower part of A γ . In fact, A γ ′ ( β ) = A γ ( β ) = uv for γ ′ ≥ γ , and so uv is in the lower part of A γ ′ for all γ ′ ≥ γ . It follows thatthere are no subsequent transfer steps between columns u and v .7herefore { B γ : γ ∈ Z } is the sequence of arrays in an ( n − s, s )-token game. Let e be anedge in G ′ with drop( G, S, e ) = g ( n, s ), and let β be the index of the cell in A ′ containing e . ByLemma 3.2, we have that g ( n, s ) = drop( G, S, e ) ≤ j − i , where [( i, u ) , ( j, u )] is the critical intervalof A β . Since A β ( ℓ, u ) contains a hole for i ≤ ℓ < j , it follows that B β ( ℓ, u ) contains a groundedtoken for 1 ≤ ℓ ≤ j − i . Hence, it is possible to place at least j − i tokens in some column in an( n − s, s )-token game and so ˆ g ( n − s, s ) ≥ j − i .It remains to analyze the ( n, s )-token game. Our main tool is to show that in an ( n, s )-tokengame in which the number of tokens in a particular column grows substantially, it is possible to finda subgame with half the number of transfer steps in which a column gains a substantial number oftokens. Eventually, we obtain a contradiction since the number of tokens in a column cannot growby more than the total number of transfer steps. Lemma 3.4.
Let B , . . . , B k be the transcript of a token game with a total of m tokens and at most ℓ transfer steps. Suppose that some column initially contains a tokens in B and ends with b tokensin B k . If a ′ and r are integers such that m < ( a ′ + 1) r/ , then some subinterval of B , . . . , B k is thetranscript of a token game with m tokens and at most ℓ − transfer steps, in which some columninitially has at most a ′ tokens and ends with at least b ′ tokens, where b ′ = b − a − r + 1 .Proof. Choose j so that both B , . . . , B j and B j , . . . , B k are transcripts of token games with atmost 2 ℓ − transfer steps.Let u index a column which initially has a tokens in B and ends with b tokens in B k , andlet R be the set of tokens which end in column u but were not always in column u . Clearly, | R | ≥ b − a . Let { t , . . . , t r } be the tokens in R which are in the r highest positions in B k , and let R = { t , . . . , t r } . Each token in R has height at least b ′ in B k . Moreover, since the height of atoken is non-increasing throughout the game, it follows that each token in R has height at least b ′ in every array.Since each t i ∈ R is moved to u during the token game, we may choose columns v , . . . , v r andindices ℓ , . . . , ℓ r such that t i is moved from v i in B ℓ i to u in B ℓ i +1 . Since a token game forbids morethan one transfer between a pair of columns, v , . . . , v r are distinct. Choose I ∈ { [0 , j ] , [ j, k ] } sothat | R | ≥ r/
2, where R = { t i ∈ R : { ℓ i , ℓ i + 1 } ⊆ I } . Since t i has height at least b ′ throughoutthe game, column v i in B ℓ i has at least b ′ grounded tokens.Note that it is not possible for each of the columns in { v i : t i ∈ R } to begin the subgame { B i : i ∈ I } with more than a ′ tokens, since ( a ′ + 1) | R | ≥ ( a ′ + 1)( r/ > m . It follows thatsome column v i has at most a ′ tokens in the first array of { B i : i ∈ I } but has at least b ′ tokens in B ℓ i .Iterating Lemma 3.4 gives the following. Lemma 3.5.
In a token game with m tokens and at most ℓ transfer steps, each column gains anet of at most ℓ ⌈√ m ⌉ tokens.Proof. If ℓ = 0, then the lemma is clear. For larger ℓ , suppose that there is a token game B , . . . , B k with at most 2 ℓ transfer steps in which some column begins with a tokens and ends with b tokens.We iterate Lemma 3.4 with a ′ = r = ⌈√ m ⌉ to obtain, for each 1 ≤ t ≤ ℓ , a subgame with m tokens and at most 2 ℓ − t transfer steps in which some column begins with at most a ′ tokens andends with at least ( b − a ) − (2 t − r tokens. 8ith ℓ = t , we obtain a token game with at most 1 transfer step in which some columnbegins with at most a ′ tokens and ends with at least ( b − a ) − (2 ℓ − r tokens. We conclude( b − a ) − (2 ℓ − r − a ′ ≤
1, which implies b − a ≤ ℓr . Corollary 3.6.
Always ˆ g ( n, s ) ≤ n ( √ ns + 1) + s + 1 = O ( s + √ ns lg n ) . In particular, if n ≥ max { , s } , then ˆ g ( n, s ) ≤ √ ns lg n .Proof. Consider an ( n, s )-token game in which some column starts with at most s tokens and endswith ˆ g ( n, s ) tokens. Let b = ˆ g ( n, s ). Note that an ( n, s )-token game contains at most 2 ℓ transfersteps provided that 2 ℓ ≥ (cid:0) n (cid:1) ; it suffices to choose ℓ = ⌊ n ⌋ . Let m be the number of tokens inour ( n, s )-token game; clearly m ≤ ns . It now follows from Lemma 3.5 that b − s ≤ ℓ ⌈√ m ⌉ ≤ n ( √ ns +1). When n ≥ max { , s } , algebra gives the simpler bound ˆ g ( n, s ) ≤ √ ns lg n .Improvements to Corollary 3.6 directly translate to improved bounds on f ( G ) via Lemma 2.5and Lemma 3.3. Unfortunately, our next theorem shows that there is not much room to improveCorollary 3.6. Theorem 3.7.
Always ˆ g ( n, s ) ≥ max { s, √ ns − s/ } . Consequently, ˆ g ( n, s ) ≥ Ω( s + √ ns ) .Proof. Clearly, ˆ g ( n, s ) ≥ s . Let k be the largest integer such that s (cid:0) k +12 (cid:1) ≤ n and note k ≥⌊ p n/s − / ⌋ . Using that n ≥ s (1 + 2 + · · · + k ), we let M , . . . , M k be disjoint sets of columnssuch that | M j | = sj for each j . Let u j, , . . . , u j,sj be the columns in M j . For each j , we constructa triangular pattern of tokens in M j so that for 1 ≤ i ≤ sj , the column u j,i contains i groundedtokens. We assume that the initial positions of all tokens are sufficiently high so that they fall intoplace as needed.For M , we initialize the board so that for 1 ≤ i ≤ s , the column u ,i starts with i tokens. For j ≥
2, we assume that we have played the token game so that for 1 ≤ i ≤ s ( j − u j − ,i in M j − contains i grounded tokens. We use the tokens in M j − to construct the desiredpattern in M j . We move the highest grounded token from each column in u j − , , . . . , u j − ,s ( j − to u j,sj in order. Since M j − has s ( j −
1) columns, this puts s ( j −
1) tokens in u j,sj and leaves asmaller triangular pattern in M j − where u j − ,i contains i − u j − , , . . . , u j − ,s ( j − to u j,sj − in order; this places s ( j − − u j,sj − . Iterating this play, we move all tokens in M j − to M j . We completethe triangular pattern by allowing min { i, s } tokens whose initial positions were high in column u j,i to fall into place.Note that we require at most s tokens in each column initially. Moreover, since M , . . . , M k arepairwise disjoint, no pair of columns is involved in more than one transfer step. After all steps,column u k,sk in M k contains sk tokens, implying that ˆ g ( n, s ) ≥ sk ≥ s ( p n/s − / g ( n, s )up to a logarithmic factor, it would still be interesting to obtain the exact order of growth. Ifˆ g ( n, s ) = O ( s + √ ns ) as we suspect, then the log term in the lower bound in Corollary 4.2 can beremoved. We do not know how sharp the inequality in Lemma 3.3 is; there may be room to makemore substantial improvements to our upper bound on g ( n, s ).9 Dense Graphs
Theorem 4.1.
Let G be an n -vertex graph, let s = n / (11 lg n ) / , and suppose that n is sufficientlylarge so that s ≤ n − . If G has average degree d and d > , then f ( G ) > d s (1 − d )(1 − s )(1 − s d − ) .Proof. Let H be a total ordering of G . Since H has nd/ x x hasheight at least d/
2. With s ′ = ⌊ s ⌋ , we apply Lemma 2.5 to obtain a monotone path of length atleast s ′ ⌊ d/ − ( s ′ +12 ) + g ( n,s ′ ) ⌋ + 1. Using Lemma 3.3, Corollary 3.6, and monotonicity of ˆ g ( n, s ), we have g ( n, s ′ ) ≤ ˆ g ( n − s ′ , s ′ ) ≤ ˆ g ( n, s ′ ) ≤ √ ns ′ lg n . We compute s ′ $ d/ − (cid:0) s ′ +12 (cid:1) + g ( n, s ′ ) % + 1 > s ′ $ d/ − (cid:0) s ′ +12 (cid:1) + 11 √ ns ′ lg n % ≥ ( s − $ d/ − (cid:0) s +12 (cid:1) + 11 √ ns lg n % = ( s − $ d/ − (cid:0) s +12 (cid:1) + s % ≥ ( s − (cid:18) d/ − s − (cid:19) = d s (cid:18) − d (cid:19) (cid:18) − s (cid:19) (cid:18) − s d − (cid:19) When the average degree d grows faster than s , Theorem 4.1 improves R¨odl’s result f ( G ) ≥ (1 − o (1)) √ d . In terms of n , an n -vertex graph must have average degree at least Cn / (lg n ) / for some constant C in order for Theorem 4.1 to offer an improvement. Corollary 4.2. f ( K n ) ≥ ( − o (1))( n lg n ) / Proof.
By Theorem 4.1, we have f ( K n ) ≥ n − n / (11 lg n ) / (1 − o (1)) ≥ n / n ) / (1 − o (1)).We make no attempt to optimize the constant 1 /
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