Leslie Hogben. Iowa State University and American Institute of Mathematics. University of Puerto Rico, Rio Piedras Campus February 25, 2015

Transcription

1 s Iowa State University and American Institute of Mathematics University of Puerto Rico, Rio Piedras Campus February 5, 015 s Joint work with E. Gethner, B. Lidický, F. Pfender, A. Ruiz, M. Young.

2 Definitions and classical problems Zarankiewicz Z(n 1, n ) and Hill H(n) of the complete tripartite graph crossing bound Upper bounds by drawings Lower bounds by counting of cr( r K of n) CR( r as n K n) Random geodesic spherical drawings s

3 s All graphs are simple and undirected. A good drawing of a graph is a plane drawing where at most edges intersect at any point not a vertex (and no other degeneracies). The crossing cr(d) of a good drawing D is the of non-vertex edge intersections in D. The crossing of a graph G is cr(g) := min{cr(d) : D is a good drawing of G}. G is planar if and only if cr(g) = 0.

4 Classical problems: Brick Factory Problem cr(k n1,n ) s Turán considered the crossing of the complete bipartite graph K n1,n in a brick factory in Budapest during World War II. 3 kilns each were connected to 3 storage areas by rails and rail crossings were disruptive to transport. Brick Factory Problem: What is cr(k n1,n )? After the war Turán posed the Brick Factory Problem to other mathematicians, including in lectures in Poland in 195.

5 Zarankiewicz Z(n 1, n ) s The Brick Factory Problem cr(k n1,n ) sparked the interest of several other mathematicians including Kazimierz Urbanik and Kazimierz Zarankiewicz. Zarankiewicz published his solution in Fundamenta Mathematicae. He claimed what is now called Zarankiewicz Conjecture: cr(k n1,n ) = Z(n 1, n ) := n1 n 1 1 n n 1. Zarankiewicz proved cr(k n1,n ) Z(n 1, n ).

6 for Z(n 1, n ) Zarankiewicz showed cr(k n1,n ) Z(n 1, n ) by producing a line drawing of K n1,n that has Z(n 1, n ) crossings. s Z(5, 4) := = 8

7 Classical problems: cr(k n ) s Anthony Hill was a British artist who discovered good drawings of K n and other combinarotrial results through art in about He wrote papers on crossing s with Frank Harary, Richard Guy, and others. Hill s Conjecture: cr(k n ) = H(n) := 1 4 n n 1 n n 3. Blažek and Koman proved cr(k n ) H(n).

8 Cylindrical drawings of K n Hill introduced a cylindrical drawing of K n that has H(n) crossings. s H(7) = = 3 3 = 9 4

9 s A good drawing of G is rectilinear if every edge is drawn as a straight line segment. The rectilinear crossing of a graph G is cr(g) := min{cr(d) : D is a rectilinear drawing of G}. cr(g) cr(g). Fáry showed in 1948 that any plane graph has a rectilinear drawing without crossings, i.e., cr(g) = 0 implies cr(g) = 0.

10 Rectilinear drawings for K n1,n Zarankiewicz in fact showed that cr(k n1,n ) Z(n 1, n ) because his line drawing of K n1,n is rectilinear and has Z(n 1, n ) crossings. s

11 cr(k n ) > cr(k n ) s Harary and Hill conjectured in a 196 paper that cr(k n ) = H(n) and cr(k 8 ) = 19 > 18 = H(8) = cr(k 8 ). They presented a drawing for K 8 with 18 crossings and a rectilinear drawing for K 8 with 19 drawings. The values cr(k 8 ) = 18 and cr(k 8 ) = 19 were later established, and it is known that cr(k n ) < cr(k n ) for n 10.

12 graphs s Much less is known about crossing s tripartite graphs K n1,n,n 3. cr(k 1,3,n ) = n n 1 + n [Asano 86]. cr(k,3,n ) = 4 n n 1 + n [Asano 86]. cr(k 1,4,n ) = n(n 1) [Huang & Zhao 08] cr(k,4,n ) = 6 n n 1 + n [Ho 13].

13 s := i=1,,3 {j,k}={1,,3}\{i} ( nj + ni nj 1 ni 1 Conjecture cr(k n1,n,n 3 ) = cr(k n1,n,n 3 ) =. nk nk 1 ) nj n k. coincides for known results for cr(k 1,3,n ), cr(k,3,n ), cr(k 1,4,n ), cr(k,4,n ).

14 Main results for complete tripartite s Theorem (GHLPRY, 014+) For all n 1, n, n 3 1, cr(k n1,n,n 3 ) cr(k n1,n,n 3 ). Theorem (GHLPRY, 014+) For n large enough, 0.666A(n, n, n) cr(k n,n,n ). Theorem (GHLPRY, 014+) For n large enough, 0.973A(n, n, n) cr(k n,n,n ).

15 s

16 Alternating 3-line drawing of K n1,n,n 3 Draw rays r 1, r, r 3 that originate from one point (the center) with an angle of 10 between each pair of rays. For every i {1,, 3}, draw a ray r i the opposite direction of r i. r i and r i form line l i. For i = 1,, 3: and bi := n i 1. a i := n i. On 1 r i, place a i points at distances from the center in a i +1, a i a i +1 a i +1,... from the center. 3. On r i, place b i points at distances 3, 4,..., b i + from the center. For each pair of points not on the same line l i, draw the line segment between the points. s

17 Alternating 3-line drawing of K n,n,n has A(n, n, n) crossings s Proof sketch: Count the of crossings in alternating 3-line drawing of K n,n,n. Two types of pairs of points that can result in crossings, (,) and (,1,1), depending on how the endpoints of edges are partitioned. 3 ( n ) ( + 6 n )( n ) ( + 3 n ) crossings of type (,). 6 ( n ) n n ( + 6 n ) n n crossings of type (,1,1).

18 Proof sketch for K n,n,n continued s n 1 n ( n 1 n ) A(n, n, n) = 3 + n ) ( n )( n ) n ) = 3( ( + n ) n 6( n n ) n + 6( n Theorem (GHLPRY, 014+) cr(k n,n,n ) cr(k n,n,n ) A(n, n, n).

19 Lower bound cr(k n,n,n ) Theorem (GHLPRY, 014+) For n large enough, 0.666A(n, n, n) cr(k n,n,n ). Proof sketch cr(k,3,n ) = 4 n n 1 + n n [Asano 86]. Each copy of K,3,n in K n,n,n has approximately n crossings. The of copies of K,3,n in K n,n,n is 6 ( n ) ( n ) ( 6 n n3 6 = 1 n7 crossings (counting each crossing multiple times). )( n 3)( n n). The of times a crossing gets counted varies with whether the endpoints are partitioned of type (,) or type (,1,1). s

20 Proof sketch contiinued A crossing of type (,) is counted 4n4 3 times. A crossing of type (,1,1) is counted n 4 times. 1 n7 4n 3 3 = 3 8 n4 = 3 ( 9 16 n4 ) = A(n, n, n) > 0.666A(n, n, n) 3 The counting method used in this proof has a structural limitation. We use the count for a (,) partition as the of times a K,3,n is counted. Asymptotically /3 of the crossings in an alternating 3-line drawing of K n,n,n are of type (,1,1). Even assuming cr(k n,n,n ) = A(n, n, n), this method cannot produce a lower bound of ca(n, n, n) with c close to 1. s

21 Lower bound cr(k n,n,n ) s Theorem (GHLPRY, 014+) For n large enough, 0.973A(n, n, n) cr(k n,n,n ). Proof uses counting with flag algebras. Norin and Zwols used flag algebras to show that 0.905Z(m, n) cr(k m,n ).

22 Complete s A complete multipartite graph is balanced if the partite sets all have the same cardinality. The r-partite graph K n,...,n will be denoted by r K n. r K n is the join of r copies of the complement of K n. K n = K n,n, 3 K n = K n,n,n, and n K 1 = K n.

23 Maximum crossing The maximum crossing of a graph G is CR(G) := max{cr(d) : D is a good drawing of G}. CR( r K n ) can be computed as the of choices of 4 endpoints that can produce a crossing. CR( r K n ) can be realized by a rectilinear drawing with vertices evenly spaced on a circle and vertices in the same partite set consecutive. CR(K n ) = ( n 4). CR(K n,n ) = ( n. ) ) ( + r r 1 )( n )( n ) ( 1 + r )( n ) 4, 4 1 CR( r K n ) = ( r n )( obtained by choosing points partitioned among the partite sets as (,), (,1,1), and (1,1,1,1). s

24 of crossing as fraction of the maximum: bipartite, complete s Theorem (Richter, Thomassen, 1997) and lim n lim n cr(k n ) CR(K n ) lim n cr(k n,n ) CR(K n,n ) lim n H(n) CR(K n ) = 3 8 Z(n, n) CR(K n,n ) = 1 4. Theorem (GHLPRY, 014+) lim sup n cr(k n,n,n ) CR(K n,n,n ) lim n A(n, n, n) CR(K n,n,n ) = 1 4.

25 of crossing as fraction of the maximum: r-partite s ζ(r) := 3(r r) 8(r +r 3). ζ() = 1 4 = lim n ζ(3) = 1 4 = lim n lim r ζ(r) = 3 8. Z(n,n) CR(K n,n). A(n,n,n) CR(K n,n,n). Theorem (GHLPRY, 014+) lim sup n cr( r K n ) CR( r K n ) ζ(r). Conjecture lim n cr( r K n ) CR( r K n ) = ζ(r).

26 Random geodesic spherical drawings A geodesic spherical drawing of G is a good drawing of G obtained by placing the vertices of G on a sphere, drawing edges as geodesics, and projecting onto the plane. In their 196 paper, Harary and Hill gave a drawing of K 8 with 18 = cr(k 8 ) drawings that they interpret as a geodesic spherical drawing. In a random geodesic spherical drawing, the vertices are placed randomly on the sphere. For r and n 1, s(r, n) is the expected of crossings in a random geodesic spherical drawing of r K n. s Theorem (GHLPRY, 014+) s(r, n) For r, lim n CR( r K n ) = ζ(r).

27 Random geodesic spherical drawing lemma Lemma In a random geodesic spherical drawing of a pair of disjoint edges, the probability that the pair crosses is 1 8. Proof sketch A pair of edges is determined by two sets of endpoints. Each set of two endpoints determines a great circle and two great circles intersect in two antipodal points, which are the potential crossing points. A crossing occurs if and only if both edges include the same antipodal point. The probability that the geodesic between a pair of points includes one of the two antipodal points is 1. The probability that both edges include an antipodal point is 1 4. Half the time these are the same antipodal point. s

28 Proof that lim n s(r,n) CR( r K n ) = ζ(r) Assume n large. The probability of getting a crossing among four points in a geodesic spherical drawing of r K n depends on how the points are partitioned among the partite sets, because different partitions of four points have different s of pairs of disjoint edges. Type A 0 pairs: The four points are partitioned among partite sets as (4) or (3,1). Let α r denote the probability that four randomly chosen points in r K n are of this type. Type B pairs: The four points are partitioned among partite sets as (,) or (,1,1). Let β r denote the probability that four randomly chosen points in r K n are of this type. Type C 3 pairs: The four points are partitioned among partite sets as (1,1,1,1). Let γ r denote the probability that four randomly chosen points in r K n are of this type. s

29 Proof that lim n s(r,n) CR( r K n ) = ζ(r) continued For Type C we must choose four distinct partite sets, so γ r = r(r 1)(r )(r 3) = (r 1)(r )(r 3). r 4 r 3 For Type A there are two choices. For partition (4) the probability is 1 4r(r 1) r 4 = 4(r 1) r 3 β r = 1 α r γ r.. For partition (3,1) the probability is r 3. Thus α r = 4(r 1) + 1 = 4r 3. r 3 r 3 r 3 Let q be the of 4-tuples of points. The of pairs of disjoint edges is (3γ r +β r )q = (3γ r +(1 α r γ r ))q = (+γ r α r )q, s The expected of crossings in a geodesic spherical drawing is 1 8 the of pairs of disjoint edges, so s(r, n) = 1 8 (γ r + β r )q = 1 8 ( + γ r α r )q.

30 Proof that lim n s(r,n) CR( r K n ) = ζ(r) continued In the drawing that maximizes the of crossings (points on a circle with vertices in each partite set consecutive), every 4-tuple of Type B and C produces one crossing. There are (β r + γ r )q such 4-tuples. Therefore CR ( r K n ) = (γr + β r )q = (1 α r )q. lim n s(r, n) CR( r K n ) = 1 8 ( + γ r α r )q (1 α r )q = r 3 + (r 1)(r )(r 3) (4r 3) 8 (r 3 (4r 3)) 3(r r) = 8 (r + r 3) = ζ(r). s

31 vs. crossing Question It seems that cr(k n1,n ) = cr(k n1,n ) and cr(k n1,n,n 3 ) = cr(k n1,n,n 3 ) but we know that cr(k n ) < cr(k n ) for large n. As we increase the of partite sets from to n ( n K 1 = K n ), at what stage do we lose the equality for rectilinear crossing and crossing? One fundamental change happens between and 3: partitions can be (,1,1) as well as (,). Another fundamental change happens between 3 and 4: partitions can be (1,1,1,1). Look for a complete 4-partite graph where rectilinear crossing is less than crossing? s

32 4-partite s Question Find good drawings for the complete 4-partite graph.

33 3-partite s Christian, Richter, and Salazar shown in 013 that for each positive integer m, there exists an integer n(m) such that: If cr(k m,n ) = Z(m, n) for all n n(m), then cr(k m,n ) = Z(m, n) for every n. Question Find an analogous result for complete tripartite graphs? Probably with two of the three part sizes restricted.

34 Acknowledgement s This work began the workshop Exact Crossing Numbers held at the American Institute of Mathematics (AIM) during the week April 8-May, 014, attended by all six authors. It was completed while three of the authors were general members of the Institute for Mathematics and its Applications (IMA). The authors thank AIM and IMA both for financial support (from NSF funds) and for creating wonderful collaborative research environments.

35 K. Asano. The crossing of K 1,3,n and K,3,n. Journal of Graph Theory, 10: 1 8, L. Beineke and R. Wilson. The early history of the brick factory problem. Mathematical Intelligencer, 3: 41 48, 010. J. Blažek and M. Koman. A minimal problem concerning complete plane graphs. In Theory of Graphs and Its Applications (ed. M. Fiedler), Czechoslovak Academy of Sciences, 1964, pp I. Fáry. On straight line representation of planar graphs. Acta Univ. Szeged, 11: 9 33, s

36 E. Gethner, L. Hogben, B. Lidický, F. Pfender, A. Ruiz, M. Young. s. F. Harary and A. Hill. On the of crossings in a complete graph. Proc. Edinburgh Math. Soc. () 13: , 196/1963. P. T. Ho. The crossing of K,4,n. Ars Combinatorica, 109: , 013. Y. Huang and T. Zhao. The crossing of K 1,4,n. Discrete Mathematics, 308: , 008. s

37 s S. Norin. Turán s brickyard problem and flag algebras. BIRS, 13w5091, Oct. 1, 013. R. B. Richter and C. Thomassen. Relations Between Crossing Numbers of Complete and Complete Bipartite Graphs. The American Mathematical Monthly, 104: , P. Turán. A note of welcome. Journal of GraphTheory, 1: 7 9, K. Zarankiewicz. On a problem of P. Turan concerning graphs. Fundamenta Mathematicae 41: , 1954.