aa r X i v : . [ m a t h . C O ] M a y Crossing numbers of periodic graphs
Zdenˇek Dvoˇr´ak ∗ Bojan Mohar †‡ August 22, 2018
Abstract
A graph is periodic if it can be obtained by joining identical piecesin a cyclic fashion. It is shown that the limit crossing number of aperiodic graph is computable. This answers a question of Richter [1,Problem 4.2].
The asymptotic behavior of the crossing number of periodic graphs, i.e., thegraphs that can be obtained by joining identical pieces in a cyclic fashion,plays fundamental role in constructions of crossing-critical graphs [4] and inexplaining certain phenomena. The first systematic treatment of this area isdue to Pinontoan and Richter [5], who provided basic results and motivatedseveral questions. In this paper we provide a simplified, yet equivalent set-ting for considering periodic graphs and answer a question of Pinontoan andRichter.We allow graphs to have loops and parallel edges. A tile is a triple T =( G, A, B ), where G is a graph and A = ( a , a , . . . , a k ) and B = ( b , b , . . . , b k ) ∗ Computer Science Institute of Charles University, Prague, Czech Republic. E-mail: [email protected] . Supported by the Center of Excellence – Inst. for Theor.Comp. Sci., Prague (project P202/12/G061 of Czech Science Foundation). † Department of Mathematics, Simon Fraser University, Burnaby, B.C. V5A 1S6. E-mail: [email protected] . Supported in part by an NSERC Discovery Grant (Canada), by theCanada Research Chair program, and by the Research Grant P1–0297 of ARRS (Slovenia). ‡ On leave from: IMFM & FMF, Department of Mathematics, University of Ljubljana,Ljubljana, Slovenia. G of the same length k . Note that each vertexof G may appear several times in A and B . The length k of the sequences iscalled the width of the tile, and the graph G is the underlying graph of thetile. If T = ( G , A , B ) and T = ( G , A , B ) are tiles of the same width,we denote by T T the tile ( G, A , B ), where G is the graph obtained fromthe disjoint union of G and G by adding, for 1 ≤ i ≤ k , an edge betweenthe i -th vertices of B and A . By ⊙ ( T ), we denote the graph obtained from G by adding edges between the i -th vertices of A and B for i = 1 , . . . , k .If T is a tile, we define T = T and T n = T n − T for integers n >
1. We callthe edges of ⊙ ( T n ) that belong to the copies of G internal , and the edgesbetween the copies external .For a tile T , we would like to determine the asymptotic behavior of thecrossing number of ⊙ ( T n ). Let us remark that this can significantly differfrom the crossing number of T n (compare e.g. the cylindrical m × n grid withthe toroidal m × n grid). Let c n ( T ) = cr( ⊙ ( T n )). Pinontoan and Richterproved in [5] that the limit c ( T ) = lim n →∞ c n ( T ) n exists (they prove this only for tiles whose underlying graph is connected,however the claim holds for disconnected tiles as well, see Lemma 9). How-ever, their proof gives no way to determine or to estimate the limit. Ourmain result is a bound on the convergence rate of the limit, giving an algo-rithm to approximate c ( T ) within arbitrary precision. Theorem 1.
For every tile T and for every ε > , there exists a computableconstant N = O (1 /ε ) such that (cid:12)(cid:12)(cid:12)(cid:12) c t ( T ) t − c ( T ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ ε for every integer t ≥ N . Therefore, to estimate the limit within given precision ε , it suffices todetermine the constant N of Theorem 1 and compute the crossing number The abstract of [5] erroneously states an estimate on the convergence rate of the limit,but the results of the paper do not give this, as confirmed in a private communicationwith the authors. ⊙ ( T N ), which can be done using the algorithm of Chimani et al. [2]. Theresulting algorithm is exponential in a polynomial of ε .It seems intuitively obvious that for large n , there is an optimal drawing of ⊙ ( T n ) exhibiting some kind of periodic behavior, and thus c ( T ) is a rationalnumber. Nevertheless, we were not able to prove this, and we propose thefollowing conjecture. Conjecture 1.
There exists a computable function f that assigns to everytile T a positive integer f ( T ) such that the number c ( T ) is rational withdenominator at most f ( T ) . If this were true, our result would give an algorithm to determine c ( T )exactly, by choosing ε = f ( T ) and rounding the result to the nearest ratio-nal number with denominator at most f ( T ). Note that c ( T ) is not alwaysintegral, as shown by Pinontoan [3].Let us remark that we define the cyclic composition ⊙ ( T n ) of copies of atile T = ( G, A, B ) in a way slightly different from the definition of Pinontoanand Richter [5]. They assume that the 2 k vertices appearing in A and B are all distinct and that for i, j = 1 , . . . , k , a i and a j are adjacent if andonly if b i and b j are adjacent. Then, instead of adding the edges a i b i joiningconsecutive copies of T , they identify a i and b i for i = 1 , . . . , k . Our choiceof a different definition turns out to be very convenient in our proofs, as wecan treat the external edges added between the tiles specially.It is easy to see that the same set of cyclic graphs can be constructedby both definitions. Given a tile T in our definition, we can add vertices a ′ i and b ′ i and edges a i a ′ i and b i b ′ i for i = 1 , . . . k , and after this change theconstruction of [5] gives the same result as our definition up to vertices ofdegree 2, which have no effect on the crossing number. On the other hand,given the cyclic composition H of n copies of T constructed as in [5], we canchoose an arbitrary edge-cut in T separating A from B and split H on thecopies of this edge-cut, resulting in n identical tiles which can be composedto form H using our definition of tile composition. Before starting with the main proofs, we show that it suffices to considertiles that are connected and linked. We say that a tile is connected if itsunderlying graph is connected. A tile (
G, A, B ) of width k is weakly linked if3here exist pairwise edge-disjoint paths P , . . . , P k in G and a permutation π of [ k ] such that for 1 ≤ i ≤ k , the path P i starts at the i -th element of A andends at the π ( i )-th element of B . A tile ( G, A, B ) of width k is linked if thereexist pairwise edge-disjoint paths P , . . . , P k in G such that for 1 ≤ i ≤ k ,the endvertices of P i are the i -th elements of A and B . Proposition 2.
For every tile T = ( G, A, B ) of width k there exists a weaklylinked tile T of width t ≤ k such that for every n ≥ , we have ⊙ ( T n ) = ⊙ ( T n ) .Proof. Let G ′ be the graph obtained from G by adding new vertices a and b and k edges between a and the elements of A and k edges between b andthe elements of B . Let us first consider the case that there exists a minimaledge-cut S = { s , s , . . . , s t } in G ′ separating a from b with t < k . For1 ≤ i ≤ t , let u i be the vertex incident with s i belonging to the componentof G ′ − S containing a , if this vertex is different from a . If s i is incident with a and the edge s i joins a with the j -th element of A , then let u i be the j -thelement of B . Symetrically, let v i be the other vertex of s i if it is not equalto b , and let v i be the corresponding element of A if s i is incident with b . Let T be the tile ( G ′′ , ( v , . . . , v t ) , ( u , . . . , u t )), where G ′′ is the graph obtainedfrom ⊙ ( T ) by removing the edges u v , u v , . . . , u t v t . It is easy to see that ⊙ ( T n ) = ⊙ ( T n ) for every n ≥ T is smaller than the width of T , we can perform thistransformation only a bounded number of times. Eventually, we obtain a tile T = ( G ′′ , ( v , . . . , v t ) , ( u , . . . , u t )) of width t ≤ k such that every edge-cutbetween the added vertices a and b has size at least t . By Menger’s theorem,there exist pairwise edge-disjoint paths P , . . . , P t in G ′′ and a permutation π : [ t ] → [ t ] such that for 1 ≤ i ≤ t , the endvertices of P i are v i and u π ( i ) . Asmentioned above, we also have ⊙ ( T n ) = ⊙ ( T n ) for every n ≥ T and T are weakly linked tiles and π and π are thecorresponding permutations, then T T is weakly linked for the permutation π ◦ π . Proposition 3.
Let T be a weakly linked tile and let π be the permutationfrom the definition of weakly linked tiles. Let m be the least common multipleof the lengths of the cycles of the permutation π . Then, the tile T m is linked.Proof. As we observed, the tile T m is weakly linked for the permutation π m ,which is the identity permutation. Consequently, T m is linked.4he following proposition shows that the components of linked tiles arethemselves tiles (this is not necessarily the case in general; e.g., it is possiblethat a component contains different numbers of vertices from A and B ). Proposition 4.
Let T = ( G, A, B ) be a linked tile, where G is the disjointunion of graphs G and G . Let A and A be the sequences of vertices of A in G and G , respectively, in the same order as in A . Let B and B be thesequences of vertices of B in G and G , respectively, in the same order asin B . Then T = ( G , A , B ) and T = ( G , A , B ) are linked tiles and forevery n ≥ , ⊙ ( T n ) is the disjoint union of ⊙ ( T n ) and ⊙ ( T n ) .Proof. Let k be the width of T and consider some i ∈ { , . . . , k } . Let a i and b i be the i -th vertices of A and B , respectively, and let P i be the path fromthe definition of linkedness joining a i with b i . Note that either P i ⊆ G or P i ⊆ G , and in particular a i belongs to G if and only if b i belongs to G .We conclude that T and T are linked tiles. Furthermore, the edges betweenthe same vertices are added in the constructions of ⊙ ( T n ) and of ⊙ ( T n ) and ⊙ ( T n ), and thus ⊙ ( T n ) = ⊙ ( T n ) ∪ ⊙ ( T n ).Let T = ( G, A, B ) be a tile of width k . Let H be a graph with vertices v ,. . . , v k and no edges, and let Z be the tile ( H, ( v , . . . , v k ) , ( v , . . . , v k )). Let Z ′ be a copy of the tile Z with vertices v ′ , . . . , v ′ k . A tile drawing of T is a draw-ing of ZT Z ′ in a closed disk such that the vertices v , . . . , v k , v ′ k , v ′ k − , . . . , v ′ are drawn in the boundary of the disk in order.We define M ( T ) = (cid:0) | E ( G ) | +2 k (cid:1) . Note that there exists a tile drawing of T such that any two edges cross at most once; hence, this drawing has atmost M ( T ) crossings. By connecting n such tile drawings into a cycle, weconclude that c n ( T ) ≤ M ( T ) n for every n ≥
1. We often use variants of thefollowing useful observation (an analogous result with s = 1 was proved byPinontoan and Richter [5]). Lemma 5.
Let T be a tile of width k and let s ≥ an integer such that T s is connected. For every n ≥ s + 1 , there exists a tile drawing of T n with atmost c n ( T ) + (8 k + 1) M ( T ) s + (cid:0) k (cid:1) crossings.Proof. Let G be a drawing of ⊙ ( T n ) with c n ( T ) crossings. For 1 ≤ i ≤ n , let a i denote the number of crossings involving edges incident with the verticesof the i -th copy of T in G , and let a ′ i = P i + s − j = i a i , where a n + x = a x for x ≥ a i , and5hus P ni =1 a i ≤ c n ( T ) ≤ M ( T ) n and P ni =1 a ′ i ≤ M ( T ) ns . Hence, withoutloss of generality we have a ′ ≤ M ( T ) s .Let S be the set of external edges of G drawn between the first and thelast copy of T , and let S be the set of external edges of G drawn betweenthe s -th and the ( s + 1)-th copy of T . Let v be any vertex of the first tile,and for each e ∈ S ∪ S , let P e be a path starting with e and ending in v contained in the first s copies of T (which exists since T s is connected).Let G be the drawing of T n − s obtained from G by removing the first s copies of T . The drawing G has at most c n ( T ) crossings. Let G bethe drawing obtained from G by adding back the vertex v and for each e ∈ S ∪ S , adding an edge e ′ between v and the endvertex of e containedon G , such that e ′ is drawn along the path P e (perturbed slightly to avoidedges intersecting in infinite number of points, or three edges intersecting inone point). Note that for each e ∈ S ∪ S , every intersection of e ′ with anedge f of G apears next to an intersection of f with P e . Consequently, e ′ intersects G in at most a ′ points.By splitting the vertex v into 2 k vertices of degree one and shifting thevertices slightly, we can transform G into a tile drawing G of T n − s thatextends G , without creating any new intersections with G . Let S ′ = { e ′ : e ∈ S ∪ S } . If two edges of S ′ intersect more than once in the tile drawing G ,we can eliminate the two crossings by swapping the parts of the edges betweenthese crossings and shifting the edges at the crossings slightly (this does notaffect the crossings with other edges). Furthermore, we can eliminate thecrossings of edges of S ′ with themselves by removing parts of the edges. Thisway, we transform G into a tile drawing G such that no edge of S ′ intersectsitself and each two edges of S ′ intersect at most once. Hence, G has at most c n ( T ) + 2 ka ′ + (cid:0) k (cid:1) ≤ c n ( T ) + 8 kM ( T ) s + (cid:0) k (cid:1) crossings.By combining G with s tile drawings of T (each with at most M ( T )crossings), we obtain the required tile drawing of T n with at most c n ( T ) +(8 k + 1) M ( T ) s + (cid:0) k (cid:1) crossings.For an integer n ≥
1, let t n ( T ) denote the minimum number of crossingsin a tile drawing of T n . Pinontoan and Richter [5] observed that t n ( T ) issubadditive (i.e., for every n , n ≥
1, we have t n + n ( T ) ≤ t n ( T ) + t n ( T ),and by Fekete’s subadditive lemma, the limit lim n →∞ t n ( T ) /n exists. Notethat a tile drawing of T n can be turned into a drawing of ⊙ ( T n ) with thesame number of crossings by identifying the corresponding vertices in theboundary of the drawing and suppresing the resulting vertices of degree two.6ence, we have c n ( T ) ≤ t n ( T ), and Lemma 5 gives a rough converse. Hence,we obtain the following. Corollary 6.
Let T be a tile. If there exists an integer s ≥ such that T s is connected, then the limit c ( T ) = lim n →∞ c n ( T ) /n exists and is equal to lim n →∞ t n ( T ) /n . A simple consequence of Lemma 5 is the following relationship between c n ( T ) /n and c m ( T ) /m for m ≫ n . Lemma 7.
Let T = ( G, A, B ) be a connected tile of width k and let ε > bea real number. Let n = 2 (cid:0) (8 k + 1) M ( T ) + (cid:0) k (cid:1)(cid:1) /ε and a = 2 M ( T ) /ε . If n ≥ n and m ≥ a n , then c m ( T ) /m ≤ c n ( T ) /n + ε .Proof. By Lemma 5 applied with s = 1, there exists a tile drawing G of T n with at most c n ( T ) + (8 k + 1) M ( T ) + (cid:0) k (cid:1) crossings. Let G be a tiledrawing of T with at most M ( T ) crossings. Suppose that m = an + b , where a, b are nonnegative integers and b ≤ n −
1. Let G be the drawing of ⊙ ( T m )obtained by combining a copies of G and b copies of G . Then, G has at most a (cid:0) c n ( T ) + (8 k + 1) M ( T ) + (cid:0) k (cid:1)(cid:1) + bM ( T ) ≤ mn c n ( T ) + mn (cid:0) (8 k + 1) M ( T ) + (cid:0) k (cid:1)(cid:1) + nM ( T ) crossings. Consequently, c m ( T ) m ≤ c n ( T ) n + 1 n (cid:18) (8 k + 1) M ( T ) + (cid:18) k (cid:19)(cid:19) + nm M ( T ) ≤ c n ( T ) n + ε. Corollary 8.
Let T be a connected tile and let ε > be a real number. Let n be as in Lemma 7. If n ≥ n , then c ( T ) ≤ c n ( T ) /n + ε . Before we proceed with the proof of the main theorem, let us argue thatthe limit c ( T ) exists even if T is not connected. Lemma 9.
For every tile T = ( G, A, B ) of width k , the limit c ( T ) =lim n →∞ c n ( T ) /n exists. Furthermore, there exist integers m ≤ k ! and r ≤ m | V ( G ) | and connected linked tiles T , . . . , T r with at most m | V ( G ) | verticessuch that c ( T ) = ( c ( T ) + . . . + c ( T r )) /m .Proof. By Proposition 2, we can assume that T is weakly linked. Let m beas in Proposition 3 and let T , . . . , T r be the maximal connected subtiles7f T m . By Proposition 4, these subtiles are linked. Furthermore, the limits c ( T ), . . . , c ( T r ) exist by Corollary 6.For 1 ≤ i ≤ r , let T ′ i be the subtile of T i contained in the first copy of T in T m . There exists a permutation π of [ r ] such that T i = T ′ i T ′ π ( i ) T ′ π ( i ) . . . T ′ π m − ( i ) .Let t be the number of cycles of π , and for 1 ≤ i ≤ t , let ℓ i be the length ofthe i -th cycle of π , let b i an arbitrary element of the i -th cycle of π , and let S i = T ′ b i T ′ π ( b i ) T ′ π ( b i ) . . . T ′ π ℓi − ( b i ) . Then, for every n ≥ m , ⊙ ( T n ) is the disjointunion consisting of gcd( n, ℓ i ) copies of ⊙ (cid:16) S n/ gcd( n,ℓ i ) i (cid:17) for i = 1 , . . . , t . Notethat S m/ℓ i i = T b i is connected. Hence, by Corollary 6, the limit c ( S i ) exists.Consider any ε >
0, and let n ≥ m be large enough that | c n ( S i ) /n − c ( S i ) | ≤ ε/t for every n ≥ n and 1 ≤ i ≤ t . Let us remark that ℓ i divides m , and thus ℓ i ≤ m . Hence, for any n ≥ n m , we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c n ( T ) n − t X i =1 c ( S i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t X i =1 gcd( n, ℓ i ) n c n/ gcd( n,ℓ i ) ( S i ) − c ( S i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ t X i =1 (cid:12)(cid:12)(cid:12)(cid:12) c n/ gcd( n,ℓ i ) ( S i ) n/ gcd( n, ℓ i ) − c ( S i ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ ε. We conclude that c ( T ) = lim n →∞ c n ( T ) /n exists (and is equal to P ti =1 c ( S i )).Clearly, c ( T ) = c ( T m ) /m , and by Proposition 4, we have c ( T m ) = P ri =1 c ( T i ).Hence, c ( T ) = ( c ( T ) + . . . + c ( T r )) /m as required. Let T = ( G, A, B ) be a tile and n an integer. Consider a drawing G of ⊙ ( T n )and let β > weight of a crossing between twointernal edges is 1 + 2 β , between an internal and an external edge is 1 + β and between two external edges is 1. Let cr β ( G ) be the sum of the weightsof the crossings in the drawing G . The reason for introducing this weightedcrossing number is the following lemma. Lemma 10.
Let T = ( G, A, B ) be a linked tile of width k , let n be an integerand let α ≥ , Q and β > be real numbers. Let n ≥ be an integer and let G be a drawing of ⊙ ( T n ) such that cr β ( G ) ≤ α n + Q . If the internal edges of ome copy of T in the drawing G participate in at least β (cid:0)(cid:0) k (cid:1) + α (cid:1) crossings,then there exists a drawing G ′ of ⊙ ( T n − ) with cr β ( G ′ ) ≤ α ( n −
1) + Q .Proof. Let c be the number of crossings in G involving the internal edges ofa copy T ′ of T in G , and suppose that c ≥ β (cid:0)(cid:0) k (cid:1) + α (cid:1) . Let P , . . . , P k be thedrawings of the paths from the definition of the linkedness of T ′ accordingto G . Let G ′ be the drawing of ⊙ ( T n − ) obtained from G by removing T ′ and by connecting the corresponding external edges incident with T ′ alongthe paths P , . . . , P k . In case that more than two of the paths intersect in avertex of the underlying graph of T ′ , we shift the new edges slightly so thatat most two edges intersect in each point. Let S be the set of the resultingedges. Note that each of the crossings with the internal edges of T ′ in G either disappears or becomes a crossing with an external edge of S in G ′ , andthus its weight is decreased at least by β . On the other hand, by drawingthe edges along the paths P , . . . , P k , we could introduce crossings betweenthe edges of S ; let s be the number of these new crossings. This shows thatcr β ( G ′ ) ≤ cr β ( G ) − βc + s .If two edges of S intersect more than once, we can eliminate the twocrossings by swapping the parts of the edges between these crossings andshifting the edges at the crossings slightly (this does not affect the crossingswith other edges). Furthermore, we can eliminate the crossings of edges of S with themselves by removing parts of the edges. In this way, we transformthe drawing G ′ to a drawing G ′ , where the number of crossings among theedges of S is at most (cid:0) k (cid:1) . Thus, we have cr β ( G ′ ) ≤ cr β ( G ) − βc + (cid:0) k (cid:1) ≤ cr β ( G ) − α ≤ α ( n −
1) + Q , as required.For a tile T and two edges e and e of ⊙ ( T n ), the cyclic tile distance between e and e is the minimum number of distinct tiles that a path in ⊙ ( T n ) between the two edges must intersect. Lemma 11.
Let T = ( G, A, B ) be a connected linked tile of width k andlet β, ε > and α ≥ be real numbers. Let c = β (cid:0)(cid:0) k (cid:1) + α (cid:1) and Q =2 k (2 | E ( G ) | + 2 c + 4 k )(1 + β ) + 4 k + 2 α . Let n ≥ n ≥ be integers andlet Q ≥ Q be a real number, and suppose that n is the smallest integer suchthat n ≥ n and there exists a drawing G of ⊙ ( T n ) with cr β ( G ) ≤ αn + Q .Furthermore, assume that • G is chosen among the drawings of ⊙ ( T n ) so that cr β ( G ) is the smallestpossible, and for n ≤ m < n , there is no drawing G ′ of ⊙ ( T m ) with cr β ( G ′ ) ≤ α m + Q .If n ≥ n + 2 , then the cyclic tile distance between any two crossing edgesof G is at most n + 1 .Proof. Since n > n , the minimality of n and Lemma 10 imply that everycopy of G in G is intersected at most c times.Suppose that G contains two crossing edges e and e with cyclic tiledistance at least n + 2. Let G and G be the subdrawings of G consistingof the tiles on the two paths between e and e in the cycle forming ⊙ ( T n );in case that e or e is an internal edge of a tile, this tile is included neitherin G nor in G . For i ∈ { , } , G i is a drawing of T k i , where k and k areintegers such that k + k = n if both e and e are external, k + k = n − e and e is internal and k + k = n − e and e are internal.Furthermore, we have k , k ≥ n . Since n ≥ n + 2, we can also assumethat k ≥ n .Suppose that e is an external edge, and let S be the set of edges betweenthe two copies of G that are joined by e . Let x be the sum of weights ofcrossings of e in G . Consider an edge e ′ ∈ S and let y be the sum of weightsof crossings of e ′ in G . Since T is connected, we can redraw e ′ to follow apath in the tiles with which it is incident to come close to the ends of e and then follow the drawing of e instead of its current drawing. Since everycopy of G is intersected at most c times, the crossings on the redrawn edge e ′ would have weight at most (2 | E ( G ) | + 2 c + 4 k )(1 + β ) + x . By the minimalityof cr β ( G ), we have (2 | E ( G ) | + 2 c + 4 k )(1 + β ) + x ≥ y , and by symmetry, | x − y | ≤ (2 | E ( G ) | + 2 c + 4 k )(1 + β ). Therefore, we can redraw all edges of S along e and increase cr β ( G ) by at most k (2 | E ( G ) | + 2 c + 4 k )(1 + β ). If e is external, we perform a similar transformation for e as well; this mayincur an additional penalty of at most 4 k for the intersections between thererouted edges. By allowing this penalty, we can achieve any specified orderof the edges in S close to the crossing of e with e . In case that e or e areinternal, we can perform the same transformation after first eliminating thecopies of G containing e or e similarly to the proof of Lemma 10.Finally, we can match the rerouted edges in the vicinity of the crossing of e and e and obtain drawings G ′ and G ′ of ⊙ ( T k ) and ⊙ ( T k ), respectively,such that cr β ( G ′ ) + cr β ( G ′ ) ≤ cr β ( G ) + 2 k (2 | E ( G ) | + 2 c + 4 k )(1 + β ) + 4 k =cr β ( G ) + Q − α . Since n > k ≥ n , the minimality of n implies thatcr β ( G ′ ) > αk + Q . Furthermore, note that cr β ( G ′ ) > αk + Q (this follows10rom the assumptions of the lemma if k < n , and by the minimality of n otherwise). Therefore, cr β ( G ′ ) + cr β ( G ′ ) > α ( k + k ) + Q + Q ≥ αn + Q + Q − α . Therefore, we have cr β ( G ) > αn + Q , which is a contradiction.The main approximation result follows straightforwardly from the nextlemma. Lemma 12.
Let T = ( G, A, B ) be a connected linked tile of width k andlet ε ≤ be a positive real number. Let α = c ( T ) + ε , β = ε/ (8 α ) , c = β (cid:0)(cid:0) k (cid:1) + α (cid:1) , Q = 2 k (2 | E ( G ) | + 2 c + 4 k )(1 + β ) + 4 k + 2 α , n = ⌈ Q /ε ⌉ , Q = 8 c ( n + 1)(1 + β ) + 4 k ( n + 2) + 2 (cid:0) k (cid:1) and n = ⌈ Q/ε ⌉ . Then, thereexists n such that n ≤ n ≤ n + 1 and c n ( T ) /n ≤ c ( T ) + ε .Proof. Since lim t →∞ c t ( T ) t = c ( T ), for every sufficiently large t , there existsa drawing G of ⊙ ( T t ) with cr( G ) ≤ ( c ( T ) + ε/ (4(1 + 2 β ))) t . Note thatcr β ( G ) ≤ (1 + 2 β ) cr( G ), and thus cr β ( G ) ≤ ( c ( T ) + 2 βc ( T ) + ε ) t ≤ αt forsuch an integer t . Hence, there exists the smallest integer n such that n ≥ n and for some drawing G of ⊙ ( T n ), we have cr β ( G ) ≤ αn + Q . Let G be sucha drawing with the smallest possible cr β ( G ).If there exists m such that n ≤ m < n and there is a drawing G ′ of ⊙ ( T m ) with cr β ( G ′ ) ≤ αm + Q , then the claim of the lemma holds, sincecr( G ′ ) ≤ cr β ( G ′ ) ≤ αm + Q = ( c ( T )+ ε + Q /m ) m ≤ ( c ( T )+ ε + Q /n ) m ≤ ( c ( T ) + ε ) m . Therefore, assume that there is no such m .Similarly, if n ≤ n + 1, then the claim holds, since cr( G ) ≤ cr β ( G ) ≤ αn + Q = ( c ( T ) + ε + Q/n ) n ≤ ( c ( T ) + ε + Q/n ) n ≤ ( c ( T ) + ε ) n . Thus,we may assume that n ≥ n + 2 and we can apply Lemma 11 to concludethat any two crossing edges in G have cyclic tile distance at most n + 1.Let k and k be integers such that k + k = n and k , k ≥ n . Thedrawing G can be decomposed to drawings G and G of T k and T k . Considerthe drawing G , and let A and B be the vertices of the first and the lasttile of G , respectively, incident with the external edges of G \ G . Let S bethe set of edges of G \ G incident with A ∪ B . Let Z be the set of edgesof G that are at cyclic tile distance at least n + 2 from the edges of S andlet R be the edges of G at cyclic distance at most n + 1 from S . Since T islinked, there exist pairwise edge-disjoint paths P , . . . , P k in G ∪ S joiningthe corresponding vertices of A and B . For i = 1 , . . . , k , let us add an edgeto G between the i -th vertices of A and B drawn along P i ; let G ′ be theresulting drawing and let S ′ be the set of newly added edges.11y Lemma 11, the edges of S ′ do not intersect Z . Let us estimate theweight of crossings between S ′ and R . Note that each such crossing corre-sponds to a crossing in G between an edge of R and of G . The internal edgesof R belong to at most 2( n + 1) distinct tiles of G , and thus by Lemma 10,they are intersected at most 2 c ( n + 1) times. Consider the external edgesof R . By Lemma 11, they can only intersect the edges of G at cyclic tiledistance at most n + 1 from S . By Lemma 10, it follows that there are atmost 2 c ( n + 1) intersections between external edges of R and internal edgesof G . Finally, note that each two external edges of G intersect at most once,as otherwise we could decrease cr β ( G ). Therefore, the number of crossingsbetween external edges of R and G is bounded by 2 k ( n + 2) .As usual, we can assume that each two edges of S ′ intersect at mostonce in G ′ by redrawing them if necessary. We conclude that cr β ( G ′ ) ≤ cr β ( G ) + 4 c ( n + 1)(1 + β ) + 2 k ( n + 2) + (cid:0) k (cid:1) = cr β ( G ) + Q . Similarly, wecan obtain a drawing G ′ of ⊙ ( T k ) with cr β ( G ′ ) ≤ cr β ( G ) + Q . We havecr β ( G ) + cr β ( G ) ≤ cr β ( G ), and thus cr β ( G ′ ) + cr β ( G ′ ) ≤ cr β ( G ) + Q .However, since n ≤ k , k < n , the minimality of n implies that cr β ( G ′ ) >αk + Q and cr β ( G ′ ) > αk + Q . Therefore, cr β ( G ) > αn + Q , which is acontradiction.We are ready to give the proof of the main theorem. Proof of Theorem 1.
By Lemma 9, we can assume that the tile T = ( G, A, B )is connected and linked (otherwise, we consider each connected subtile of T m separately). Let k be the width of T . Let ε be a constant to be chosen laterand let us consider the following quantities: • α d = ε and α u = M ( T ) + ε • β d = ε / (8 α u ) and β u = ε / (8 α d ) • c d = β u (cid:0)(cid:0) k (cid:1) + α d (cid:1) and c u = β d (cid:0)(cid:0) k (cid:1) + α u (cid:1) • Q ,d = 2 k (2 | E ( G ) | + 2 c d + 4 k )(1 + β d ) + 4 k + 2 α d and Q ,u = 2 k (2 | E ( G ) | + 2 c u + 4 k )(1 + β u ) + 4 k + 2 α u • n ,d = ⌈ Q ,d /ε ⌉ and n ,u = ⌈ Q ,u /ε ⌉• Q u = 8 c u ( n ,u + 1)(1 + β u ) + 4 k ( n ,u + 2) + 2 (cid:0) k (cid:1) and • n ,u = ⌈ Q u /ε ⌉ . 12hese numbers are chosen in such a way that they are related to the valuesused in Lemma 12. More precisely, α d , β d , c d , . . . give lower bounds and α u , β u , c u , . . . give upper bounds on the corresponding quantities in Lemma12, used with ε in the role of ε . In particular, if n and n are the constantsof Lemma 12 for T and ε , then n ≥ n ,d and n ≤ n ,u . Furthermore, n ,d = Θ(1 /ε ) and n ,u = Θ(1 /ε ) are computable, given T . (Here andbelow, all constants involved in the Θ-notation depend only on T .)Let n = Θ(1 /ε ) and a = Θ(1 /ε ) be the constants of Lemma 7 appliedfor ε/
2. Let us now choose ε ≤ ε/ n ,d ≥ n and ε = Θ( ε ). Let N = a n ,u and note that N = Θ(1 /ε ).By Lemma 12, there exists n such that n ,d ≤ n ≤ n ,u and c n ( T ) /n ≤ c ( T ) + ε ≤ c ( T ) + ε/
2. By Lemma 7, we have c t ( T ) /t ≤ c n ( T ) /n + ε/ ≤ c ( T ) + ε for every t ≥ N . Conversely, Corollary 8 implies that c t ( T ) /t ≥ c ( T ) − ε . Acknowledgement
The authors are grateful to BIRS for providing research environment duringa workshop on crossing numbers [1] where the results of this paper haveemerged. The authors also acknowledge helpful discussions with severalworkshop participants within a work group where they presented their initialideas.
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