Research problems

Discrete Mathematics 257 (2002) 599 – 624

www.elsevier.com/locate/disc

Research problems The research problems section presents unsolved problems in discrete mathematics. Older problems are acceptable if they are not as widely known as they should be or if the exposition features a new partial result. Concise de)nitions and commentary (such as motivation or known partial results) should be provided to make the problems accessible and interesting to a broad cross-section of the readership. Problems are solicited from all readers. Ideally, they should be presented in the style below, occupy at most one journal page, and be sent to Professor Douglas B. West Department of Mathematics University of Illinois 1409 West Green St. Urbana, IL 61801-2975, USA Comments and questions of a technical nature about a particular problem should be sent to the correspondent for that problem. Other comments and information about partial or full solutions should be sent to Professor West. The research problems in this volume are dedicated to Daniel J. Kleitman. They are based on problem sessions led by Professor Kleitman at the conference “Kleitman and Combinatorics” at MIT in August 1999 and at the regional AMS meeting at the University of South Carolina in March of 2001. In addition, other problems were submitted later in honor of Professor Kleitman. We have grouped these problems to start with discrete geometry, followed by graph theory, partially ordered sets, and other combinatorial structures. The problems touch on Professor Kleitman’s interests throughout his career. In the spirit of Professor Kleitman’s approach to mathematics, we note that open problems and problem-solving are the lifeblood of mathematics. We hope that this large group of problems will stimulate the community to contribute research problems to this forum on a continuing basis.

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Problem 362. Piercing families of planar convex sets Posed by GCeza TCoth. Correspondent: G. TCoth Renyi Institute Hungarian Academy of Science Budapest, Hungary [email protected] A family of sets has the (p; q)-Helly property if among any p sets in the family there are q sets with a common point. The transversal number or piercing number of a family of sets is the minimum number of points in a set that intersects every member of the family. Question. Is it true that every family of convex sets in the plane that has the (4,3)Helly property also has piercing number at most 3? Comment. In general, Hadwiger and Debrunner [3] conjectured that for families of closed convex sets in Rd that have the (p; q)-Helly property, there is a constant bound f(p; q; d) on the piercing number. This was proved by Alon and Kleitman [1,2]. The Alon–Kleitman argument is very elegant but gives very high bounds. For the )rst nontrivial case, their general bound plus some easy tricks yields about f(4; 3; 2)6200. Kleitman et al. [5] improved this to f(4; 3; 2)613. On the other hand, an easy construction by Danzer (see [4]) gives the best known lower bound f(4; 3; 2)¿3, which is conjectured to be tight by Kleitman and others. References [1] N. Alon, D.J. Kleitman, Piercing convex sets and the Hadwiger–Debrunner (p; q) problem, Adv. in Math. 96 (1992) 103–112. [2] N. Alon, D.J. Kleitman, A purely combinatorial proof of the Hadwiger–Debrunner (p; q) conjecture, The Wilf Festschrift, Philadelphia 1996, Electron. J. Combin. 4(2) (1997) Research paper 1. M [3] H. Hadwiger, H. Debrunner, Uber eine Variante zum Hellyschen Satz, Arch. Math. 8 (1957) 309–313. [4] H. Hadwiger, H. Debrunner, V. Klee, Combinatorial Geometry in the Plane, Holt, Rinehart and Winston, New York, 1964. [5] D.J. Kleitman, A. GyCarfCas, G. TCoth, Convex sets in the plane with three of every four meeting, Combinatorica 21 (2001) 221–232.

Problem 363. Saving Borsuk’s Problem Posed by Gil Kalai. Correspondent: G. Kalai Hebrew University Jerusalem Israel [email protected]

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In 1933, Borsuk asked whether every set of diameter one in Rd can be partitioned into d + 1 sets of diameter smaller than one. The conjectured answer “yes” became known as Borsuk’s Conjecture. It was disproved in high dimensions in 1993 by Kahn and Kalai [1]. Weaker versions of Borsuk’s Conjecture may still be true. The idea is to forbid the counterexamples, which use con)gurations of points that are coincidental. The version of coincidental presented below is similar to a condition called the “strong Arnold property” in the theory of Colin de Verdier’s invariants of graphs. Let Z be a )nite set subset of Rd with diameter 1. Let E be the subset of ( Z2 ) consisting of all pairs whose elements are distance one apart. Thus E is the set of edges of a graph G with vertex set Z. We ignore isolated vertices. We say that E is stress-free if there is no way
to assign weights w to the edges (not all zero) such that every point z ∈ Z satis)es x∈NG (z) w(z; x)(x − z) = 0. Conjecture (Finite case). If E is stress-free, then Z can be partitioned into d + 1 parts such that no part contains a pair in E. For the general formulation, we take a set Z in Rd of diameter 1 and the set E ⊆ Z×Z of pairs such that d(x; y) = 1 for (x; y) ∈ E. We say that E satis)es the strong Arnold Property if there exists ¿0 such that for every continuous function T : E → [1 − ; 1 + ] there is a continuous deformation D of Z to Z  such that if (x; y) ∈ E, then d(D(x); D(y)) = T (x; y). Conjecture (General case). If E satis)es the strong Arnold property, then Z can be covered by d + 1 closed sets such that no set contains a pair (x; y) ∈ E. References [1] J. Kahn, G. Kalai, A counterexample to Borsuk’s conjecture, Bull. Amer. Math. Soc. 29 (1993) 60 – 62.

Problem 364. Splitting lines for planar point sets Posed by Noga Alon. Correspondent: N. Alon Tel Aviv University Tel Aviv Israel [email protected] Let X be a set of n points in the plane, with n¿2. A c-splitting line for X is a line containing at least two points of X such that the numbers of points on the two sides of the line diRer by at most c. Question. Does there exist an absolute positive constant c such that for every X and n there is a c-splitting line?

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Comment. Pinchasi proved that setting √ c = O(log log n) suSces. This improves earlier bounds of O(log n) by Perles and O( n) by Alon, all unpublished. The problem is to )nd a constant bound. The problem originated with Kupitz [1], who conjectured that c = 1 suSces. Alon found in)nitely many counterexamples, but it remains unknown whether c = 2 suSces. Alon’s examples have 8k + 4 points for k ∈ N. The simplest starts with a triangle ABC, extends the edges to add points with subscripts 1 and 2 as shown, and )nally adds points A3 ; B3 ; C3 at the intersections of the lines A1 B, B1 C, and C1 A as shown. The larger examples start with 2k + 1 central points instead of 3.

References [1] Y.S. Kupitz, Extremal Problems in Combinatorial Geometry, Lecture Notes Series, Vol. 53, Matematisk Institut, Aarhus Universitet, 1979 (see MR81d:05018, R.C. Entringer).

Problem 365. Intersection graphs of convex sets Posed by JCanos Pach. Correspondent: J. Pach RCenyi Institute Hungarian Academi of Science Budapest Hungary [email protected] Given a family P of convex sets in the plane, let f(P) denote the maximum size of a subfamily that is pairwise intersecting or pairwise disjoint. Let f(n) be the minimum of f(P) over all families consisting of n convex sets. The determination of f(n) is equivalent to the following question.

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Question. What is the maximum m such that every family of n convex sets in the plane has m pairwise intersecting members or m pairwise disjoint members? Comment. The question asks for a large guaranteed independent set or complete subgraph in the intersection graph of the sets. √ The classical √ bounds on Ramsey numbers yield f(n)¿c log n.√A family consisting of n sets of n pairwise-crossing segments shows that f(n)6 n. The best bounds known (see [1,2]) are n:25 6f(n)6n:41 . References [1] G. KCarolyi, J. Pach, and G. TCoth, Ramsey-type results for geometric graphs, I, Discrete Comput. Geom. 18 (1997) 247–255. [2] D. Larman, J. MatouTsek, J. Pach, J. TMor˝ocsik, A Ramsey-type result for planar convex sets, Bull. London Math. Soc. 26 (1994) 132–136.

Problem 366. The incidence graph between points and lines Posed by Dominique de Caen and LCaszlCo SzCekely. Correspondent: L.A. SzCekely University of South Carolina Columbia, SC, USA [email protected] A family of n points and m straight lines through these points generates a bipartite incidence graph in which point x is adjacent to line l if x lies on l. SzemerCedi and Trotter [2] proved that the number of edges in the incidence graph is bounded by O((mn)2=3 + m + n). de Caen and Szekely [1] proved that the number of copies of the 4-vertex path P4 in G is bounded by O(mn). This strengthens the SzemerCedi–Trotter Theorem, but the proof uses it many times and also uses the second moment method. Conjecture. The number of 6-cycles in the incidence graph of n points and m straight lines in the plane is bounded by O(mn). Comment. Such cycles in the incidence graph correspond to ordinary triangles in the con)guration of points and lines, where the vertices of the triangle lie in the point set and its sides are on lines in the line set. The conjecture would immediately imply the result of [1], perhaps via an easier proof. The authors believe that the number of 6-cycles is always less than mn. Eric Moorehouse oRered $50 for a counterexample to the O(mn) conjecture. References [1] D. de Caen, L.A. SzCekely, On dense bipartite graphs of girth eight and upper bounds for certain con)gurations in planar point-line systems, J. Combin. Theory Ser. A 77 (1997) 268–278. [2] E. SzemerCedi, W.T. Trotter, Extremal problems in discrete geometry, Combinatorica 3 (1983) 381–392.

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Problem 367. What is the true crossing number? Posed by JCanos Pach and GCeza TCoth. Correspondent: G. TCoth RCenyi Institute Hungarian Academi Science Budapest, Hungary [email protected] A drawing of a graph G is a representation of G in the plane such that its vertices are (represented by) distinct points and its edges are (represented by) simple continuous curves so that the incidence relation between the curves and the points is the same as the incidence relation between edges and vertices in G. In particular, (a) no edge passes through any vertex other than its endpoints. For simplicity, we also assume that (b) if two edges have a common interior point, then at this point they properly cross each other, and (c) no three edges cross at the same point. A common interior point of two edges is a crossing point or, in short, a crossing. The crossing number of a graph G is usually de)ned as “the minimum number of edge crossings in any drawing of G in the plane” [1]. The term “edge crossings” has been interpreted in diRerent ways. What is usually meant by the crossing number cr(G) is the minimum number of crossing points. This is always achievable by a drawing in which every two edges cross at most once. However, it may be possible to draw G with fewer pairs of crossing edges (at the cost of more crossing points) if we allow edges to cross more than once. Let pcr(G) denote the minimum number of pairs of edges that cross at least once, over all drawings of G; this is the pairwise crossing number of G. As we have observed, pcr(G)6cr(G) for every graph G. Question. Is it true that pcr(G) = cr(G) for every graph G? Comment. Checking this equality seems hard even for relatively small graphs, because determining cr(G) and pcr(G) are both NP-hard problems [2,4]. Nevertheless, cr(G)62(pcr(G))2 does hold for every graph G [4]. It is not hard to check that pcr(G) = cr(G) for every graph G with pcr(G)63. Question. Is it true that pcr(G) = cr(G) for every graph G with pcr(G) = 4? Comment. Even the crossing numbers of the complete graphs are not known. The best lower bound is due to Kleitman [3]. Even the following very special question is unanswered: Question. Does the limit lim

n→∞

pcr(Kn ) cr(Kn )

exist, and, if so, does it equal 1?

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References [1] S.N. Bhatt, F.T. Leighton, A framework for solving VLSI graph layout problems, J. Comput. System Sci. 28 (1984) 300 –343. [2] M. R. Garey and D. S. Johnson, Crossing number is NP-complete, SIAM J. Algebraic Discrete Methods 4 (1983) 312–316. [3] D. J. Kleitman, The crossing number of K5; n , J. Combin. Theory 9 (1970) 315–323. [4] J. Pach, G. TCoth, Which crossing number is it anyway?, J. Combin. Theory Ser. B 80 (2000) 225–246.

Problem 368. The convex crossing number Posed by Farhad Shahrokhi. Correspondent: F. Shahrokhi University of North Texas Denton TX USA [email protected] There are many models for drawing graphs in the plane; a survey appears in [3]. One model is to place all the vertices at integer points on the horizontal axis and then draw each edge as a semicircle in the upper half-plane (equivalently, one could place the vertices on a circle and draw each edge as a chord). The convex crossing number of a graph G is the minimum number of edge crossings over all such drawings of G. Of course, the number of edge crossings in such a drawing is determined by the ordering of the vertices; two edges cross if and only if their endpoints alternate in the linear order on the vertices. The linear arrangement number is the minimum sum of the distances between the endpoints of edges when the vertices are placed at integer points on the horizontal axis. Let Gk denote the
class of bipartite graphs with maximum degree at most k. For a graph G, let D(G) = v∈V (G) (d(v))2 , where d(v) is the degree of vertex v. Let n and m denote the numbers of vertices and edges in a graph G. Conjecture. There is a constant ck such that for G ∈ Gk , the convex crossing number plus D(G) is always at least ck = log n times the linear arrangement number. Comment. The conjecture is true over a subclass of these drawings obtained by restricting the vertex order. If the vertices of one partite set all precede the vertices of the other on the horizontal axis, then the number of crossings is always at least "(G)=36 times the linear arrangement number, where "(G) is the minimum vertex degree [4]. Shahrokhi posed a stronger version of the conjecture for ordinary crossing number of arbitrary graphs, but this was proved false by a construction of SzCekely. When m (the number of edges) exceeds 4n, every drawing of G has at least m3 =(64n2 ) crossings [1,2]. Therefore, when m¿4n, the crossing number for graphs in Gk grows asymptotically as fast as D(G), and the additive term in the conjecture is not signi)cant. If the conjecture is true, then the linear arrangement number approximates

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the convex crossing number of these graphs up to a factor of log(n), since it is easy to see that the linear arrangement number is at most a constant multiple of the convex crossing number. This would improve, by a factor of log n, the quality of approximation algorithms for the crossing number of bipartite graphs with bounded degree. References [1] M. Ajtai, V. ChvCatal, M.M. Newborn, E. SzemerCedi, Crossing-free subgraphs, in Theory and Practice of Combinatorics, Ann. Discrete Math. 12 (1982) 9–12. [2] F.T. Leighton, Complexity Issues in VLSI: Optimal Layouts for the ShuWe-Exchange Graph and Other Networks, Foundations of Computing, MIT Press, Cambridge, MA, 1983, p. 96. [3] F. Shahrokhi, O. SCykora, L.A. SzCekely, I. VrCto, Crossing number problems: bounds and applications, in: I. BCarCany, K. BMorMoczky (Eds.), Intuitive Geometry Bolyai Society Math. Studies 6 (1997) 179–206. [4] F. Shahrokhi, O. SCykora, L.A. SzCekely, I. VrCto, On bipartite drawings and the linear arrangement problem, SIAM J. Comput. 30 (2001) 1773–1789.

Problem 369. Drawings of Qk Posed by Pavel Valtr. Correspondent: P. Valtr Charles University Prague Czech Republic [email protected] Let Qk denote the graph of the k-dimensional hypercube, having 2k vertices that are binary k-tuples and k2k−1 edges joining pairs of k-tuples that diRer in one position. A drawing of Qk in the plane represents its vertices by points and its edges by curves joining the points representing their endpoints. Question. Is it possible for all k to draw Qk in the plane so that the maximum number of pairwise crossing edges is bounded by a constant? Comment. It is known that there is a constant cr such that every drawing of a graph with n vertices and at least cr n log n edges has r pairwise crossing edges [1]. The hypercube with n vertices has 12 n log2 n edges. If the answer to the question above is “yes”, then the threshold obtained in [1] is sharp in its order of growth in n; this motivates the question. References [1] P. Valtr, On geometric graphs with no k pairwise parallel edges, Discrete Comput. Geom. 19 (1998) 461– 469.

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Problem 370. Acyclic edge-coloring of graphs Posed by Noga Alon. Correspondent: N. Alon Tel Aviv University Tel Aviv Israel [email protected] A proper coloring of the edges of a simple graph G is acyclic if no cycle in G is 2-colored. The acyclic edge-chromatic number of G, denoted by a (G), is the least number of colors in an acyclic edge-coloring of G. Question. Does a (G)6%(G) + 2 hold for every graph G, where %(G) denotes the maximum vertex degree in G? Comment. Examples are known that require %(G) + 2 colors. It is known that a (G)6 16%(G) for every simple graph G [1,3]. It is also known that there is a constant c such that a (G)6%(G) + 2 for every simple graph G with girth at least c%(G) log %(G), and that a (G)6% + 2 for almost all %-regular graphs [2]. References [1] N. Alon, C.J.H. McDiarmid, B.A. Reed, Acyclic coloring of graphs, Random Structures Algorithms 2 (1991) 277–288. [2] N. Alon, B. Sudakov, A. Zaks, Acyclic edge-colorings of graphs, J. Graph Theory 37 (3) (2001) 157–167. [3] M. Molloy, B.A. Reed, Further algorithmic aspects of the local lemma, Proceedings of the 30th ACM Symposium Theory of Computation, May 1998, pp. 524 –529.

Problem 371. Graph coloring with required adjacencies Posed by Steve Hedetniemi and Renu Laskar. Correspondent: S.T. Hedetniemi Clemson University Clemson, SC USA [email protected] A proper coloring is a fall coloring if every vertex has at least one neighbor in each other color class. It is NP-hard to test whether a graph has a fall coloring [1]. When a fall coloring of a graph G exists, the minimum number of colors in such a coloring is the fall chromatic number &f (G). Question. Among graphs that have fall colorings, can &f (G) − &(G) be arbitrarily large?

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A restriction on proper colorings has the interpolation property if whenever a graph has this type of coloring using i colors and using k colors, it also has this type of coloring using j colors for every j between i and k. A partial Grundy k-coloring [3] of a graph G is a proper coloring with colors 1; : : : ; k such that for each i, some vertex with color i has neighbors in each color class j with j¡i. A colorful coloring [1] is a proper coloring such that every color class has at least one vertex whose neighborhood intersects every color class other than its own. Conjecture. The interpolation property holds for partial Grundy colorings and for colorful colorings. A partition of the vertex set of a graph G is a complete partition if for each choice of two distinct parts there is a vertex of one adjacent to a vertex of the other. The pseudo-achromatic number of a graph G, denoted p (G), is the maximum number of parts in a complete partition of V (G), while the achromatic number (G) of G is the maximum number of parts in a complete partition of V (G) such that every Vi is an independent set. Conjecture. If T is a tree, then

(T )6 p (T )6 (T ) + 1.

Comment. Edwards [2] constructed a tree with pseudo-achromatic number one larger than its achromatic number, disproving Hedetniemi’s original conjecture of equality. References [1] J.E. Dunbar, S.M. Hedetniemi, S.T. Hedetniemi, D.P. Jacobs, J. Knisely, R.C. Laskar, D.F. Rall, Fall colorings of graphs, J. Combin. Math. Combin. Comput. 33 (2000) 257–273. [2] K. Edwards, Achromatic number versus pseudoachromatic number: a counterexample to a conjecture of Hedetniemi, Discrete Math. 219 (2000) 271–274. [3] P. Erd˝os, S.T. Hedetniemi, R.C. Laskar, G. Prins, On the equality of the partial Grundy and upper ochromatic numbers of graphs, submitted for publication.

Problem 372. List coloring and maximum degree Posed by Noga Alon. Correspondent: N. Alon Tel Aviv University Tel Aviv Israel [email protected] Let L be a function that assigns to each vertex of a graph G a list of k integers. The graph G is k-choosable if for every such L there is a proper coloring c of G such that c(v) ∈ L(v) for all v ∈ V (G). The choice number of G is the minimum k such that G isk-choosable.

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Question. Is there an absolute constant c so that the choice number of every bipartite graph with maximum degree d is at most c log2 d? Comment. With probability tending to 1 as np tends to in)nity, the choice number of a random subgraph of Kn; n with independent edge probability p is (1 + o(1)) log2 (np) [2]. It may even be true that the choice number of every d-regular bipartite graph diRers from log2 d by o(log2 d) (as d tends to in)nity). On the other hand, the choice number of every graph with average degree d is at least ( 12 − o(1)) log2 d [1]. References [1] N. Alon, Degrees and choice numbers, Random Structures Algorithms 16 (2000) 364 –368. [2] N. Alon, M. Krivelevich, The choice number of random bipartite graphs, Ann. of Combin. 2 (1998) 291–297.

Problem 373. A list version of the Hajnal–Szemer8edi Theorem Posed by Alexandr Kostochka. Correspondent: A.V. Kostochka University of Illinois Urbana, IL USA [email protected] An equitable k-coloring of a graph G is a proper k-coloring in which each of the k colors appears on n=k or n=k vertices. Hajnal and SzemerCedi [1] proved that if k¿%(G) + 1, then G has an equitable k-coloring. A slightly weaker but perhaps more applicable restriction is to require that each color be used at most n=k times. This permits generalizing the question to list colorings. Each vertex is assigned a list of available colors, but diRerent vertices may have diRerent lists. Question. Given a graph G and an integer k with k¿%(G) + 1, suppose that every vertex of G is assigned a list of k available colors. Does there always exist a proper coloring of G such that the color on each vertex belongs to its list and such that each color is used at most n=k times? Comment. For %(G)62, such a coloring can be found by a greedy procedure. For %(G) = 3, the question has been answered aSrmatively by Pelsmajer [2]. Pelsmajer also gave aSrmative answers for outerplanar graphs and for interval graphs, improving the latter to k¿%(G). References [1] A. Hajnal, E. SzemerCedi, Proof of a conjecture of P. Erd˝os, in: Combinatorial Theory and its Applications, II (Proceedings of the Colloquium BalatonfMured, 1969), North-Holland, Amsterdam, 1970, pp. 601– 623. [2] M. Pelsmajer, Equitable list coloring for % = 3, preprint.

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Problem 374. Paired-list coloring Posed by AndrCe KMundgen and Radhika Ramamurthi. Correspondent: A. KMundgen California State University San Marcos, CA USA [email protected] Suppose that each vertex of a graph G is assigned a list of four colors from a set of colors. The colors come in pairs of the form i; i such that a vertex has i in its list if and only if it has i in its list. Such an assignment is a paired-list assignment of colors. Question. Is it true that, for every planar graph G and every paired-list assignment of four colors to each vertex, there is a proper coloring of G in which the color on each vertex comes from its list? Comment. The question is raised in [1]. An aSrmative answer immediately implies the Four Color Theorem, since we can view a )xed set of four colors as two pairs of colors. The statement is true for every plane graph whose dual is Hamiltonian [2]. References [1] A. KMundgen, R. Ramamurthi, Coloring face-hypergraphs of graphs on surfaces, J. Combin. Theory Ser. B 85 (2002) 307–337. [2] R. Ramamurthi, Coloring problems on graphs and hypergraphs, Ph.D. Thesis, University of Illinois, 2001.

Problem 375. Iterated domination Posed by Stephen T. Hedetniemi and Renu C. Laskar. Correspondent: S.T. Hedetniemi Clemson University Clemson, SC USA [email protected] We iteratively partition the vertices of a graph G. Iteratively removing a maximal independent set from the remaining graph produces a proper coloring of G. The minimum number of steps in such an iteration is &(G). Every maximal independent set S in a graph is a minimal dominating set, since every vertex outside S has a neighbor in S, and deleting a vertex from S would leave that vertex with no neighbor in S. Iteratively deleting a minimal dominating set of the remaining graph produces a domination shelling of G [1]. As observed above, the minimum number of steps in a domination shelling of G is at most &(G).

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Question. Is it possible to prove without the Four Color Theorem that every planar graph has a domination shelling in at most four steps? References [1] S.M. Hedetniemi, S.T. Hedetniemi, A.A. McRae, D. Parks, J.A. Telle, Iterated colorings of graphs, submitted for publication.

Problem 376. Separator theorems and small hereditary classes Posed by Gil Kalai and Moshe Litvin. Correspondent: G. Kalai Hebrew University Jerusalem Israel [email protected] A class G of graphs is small if the number of n-vertex isomorphism classes in G is bounded by C n for some constant C. The classes of trees and planar graphs are small, but the class of cubic graphs is not. A class G is hereditary if every subgraph of a graph in G is also in G. An (a; b)-separator for an n-vertex graph G is a set A ⊆ V (G) such that |A|6a and G −A has no component with more than bn vertices. With a expressed as a function of n, a class G has a (a; b)-separator theorem if for all n and all G ∈ G with n vertices, G has an (a(n); b)-separator. Question. For which functions a is it true that every class having an (a; 2=3)-separator theorem is small? Comment. Let  be a positive real number. If G is a hereditary class having an (a; 2=3)separator theorem with a(n) = cn=(log n)2+ , then G is small [1]. Furthermore, this is no longer true when (log n)2+ is replaced with (log n)1− [1]. Hence the problem is to close the gap between log n and (log n)2 . For the upper bound on a(n), consider graphs obtained by replacing each edge of a cubic n-vertex graph with a path of length (log n)1− . For a sketch of the lower bound on a(n), let f(n) be the number of isomorphism classes of n-vertex graphs in G, and let g(n) = log f(n)=n. We seek f(n)6C n , and hence it suSces to show that g(n) is bounded from above by a constant when G has a (a; 2=3)-separator theorem with a(n) = cn=(log n)2+ . Every G ∈ G with n vertices can be built by choosing G1 ; G2 ∈ G of orders k and n − k + a(n), respectively (where n=36k6n=2), and identifying a(n) vertices of G2 with vertices of G1 . At most f(k)· f(n − k + a(n)) · n2a(n) graphs can be created in this way when G1 has k vertices. Thus f(n)6n

max

n=36k6n=2

f(k) · f(n − k + a(n)) · n2a(n) :

This relation implies the desired upper bound for g(n).

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References [1] M. Litvin, Counting graphs, counting edges and separation properties, M.Sc Thesis, Jerusalem 1996.

Problem 377. Two questions on long cycles Posed by Stephen Locke. Correspondent: S.C. Locke Florida Atlantic University Boca Raton, FL USA [email protected] Bondy and Locke [1,2] proved that every 3-connected 3-regular graph having a path of length m contains a cycle of length at least 2m=3. Question. What is the largest constant c such that every 4-connected 4-regular graph having a path of length m contains a cycle of length at least cm? Comment. Locke [4] proved that c¿1=2. Question. If G is a 3-connected graph with minimum degree d and X is a set of four vertices on a cycle in G, must G have a cycle through X with length at least min{2d; |V (G)|}? Comment. Locke and Zhang [5] proved that if G is a 2-connected graph with minimum degree d and X is a set of three vertices on a cycle in G, then G has a cycle through X with length at least min{2d; |V (G)|}. Egawa et al. [3] proved that if G is a k-connected graph with minimum degree d and X is a set of k vertices in G, then G has a cycle through X with length at least min{2d; |V (G)|}. References [1] J.A. Bondy, S.C. Locke, Relative lengths of paths and cycles in k-connected graphs, Proceedings of the First Canada–France Combinatorial Colloquium, Ann. Discrete Math. 8 (1980). [2] J.A. Bondy, S.C. Locke, Relative lengths of paths and cycles in 3-connected graphs, Discrete Math. 33 (1981) 111–112. [3] Y. Egawa, R. Glas, S.C. Locke, Cycles and paths through speci)ed vertices in k-connected graphs, J. Combin. Theory Ser. B 52 (1991) 20 –29. [4] S.C. Locke, Relative lengths of paths and cycles in k-connected graphs, J. Combin. Theory Ser. B 32 (1982) 206–222. [5] S.C. Locke, C.-Q. Zhang, Cycles through three vertices in 2-connected graphs, Graphs and Combin. 7 (1991) 265-269.

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Problem 378. A linear version of k-ordered graphs Posed by Aradhana Narula Tam and Philip J. Lin. Correspondent: Timothy Chow Tellabs Research Center [email protected] A graph is k-trail-ordered if for every list v0 ; : : : ; vk of k + 1 distinct vertices, there is a trail that visits v0 ; : : : ; vk in order (among a possibly longer trail). Question. Is every k-edge-connected graph also k-trail-ordered? If not, then what is the least edge-connectivity f(k) such that every f(k)-edge-connected graph is k-trailordered? Comment. Note that in the desired trail, each vi may appear repeatedly, as long as the list of vertices visited has v0 ; : : : ; vk as a sublist in order. The problem arose in the context of a wavelength assignment problem in a WDM optical network. Whitney’s Theorem on edge-disjoint paths yields f(2) = 2, as desired. In [2] it was proved that f(3) = 3. Florian Pfender points out that a result of Okamura [3] yields a proof that f(k) = k when k is even, and consequently f(k)6k + 1 when k is odd. Therefore, it remains to decide between k and k + 1 when k is odd. A graph satisfying the analogous condition in which every list of k distinct vertices is visited in order by some cycle is a k-ordered graph (see [1] for results on these graphs). Every k +1-ordered graph is k-trail-ordered, but the connectivity conditions for k + 1-ordered graphs are stronger than needed for k-trail-ordered graphs. In particular, the graph obtained from the 8-cycle by adding the eight chords joining vertices at distance 2 is 4-connected but not 4-ordered. References [1] J.R. Faudree, R.J. Faudree, R.J. Gould, M.S. Jacobson, L. Lesniak, On k-ordered graphs, J. Graph Theory 35 (2000) 69–82. [2] A. Narula-Tam, P. J. Lin, E. Modiano, EScient routing and wavelength assignment for recon)gurable WDM networks, IEEE J. Selected Areas Comm. 20 (2002) 75–88. [3] H. Okamura, Paths in k-edge-connected graphs, J. Combin. Theory Ser. B 45 (1988) 345–355.

Problem 379. Long paths and the cycle space of a graph Posed by Stephen Locke. Correspondent: S.C. Locke Florida Atlantic University Boca Raton, FL USA [email protected]

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A graph is k-path-connected if every two vertices are connected by a path of length at least k. The cycle space of a graph is the binary vector space spanned by the incidence vectors of its cycles. Question. How fast can a function f be allowed to grow such that in every k-pathconnected graph, the cycles of length at least f(k) generate the cycle space? Is it possible that f can grow linearly? Comment. If the cycles of length at least k generate the cycle space of a 2-connected graph G, then G is k=2-path-connected [1]. Also, if G is k-path-connected, then √ the cycles of length at least m generate the cycle space of G, where m is about k [1]. Thus f can grow at least as fast as the square root function and no faster than linearly. One cannot guarantee f(k)¿k, since f(k)617k=18 for some examples based on the dodecahedron. References [1] S.C. Locke, Long paths and the cycle space of a graph, Ars Combin. 33 (1992) 77–85.

Problem 380. Representations of graphs modulo n Posed by Darren Narayan. Correspondent: D.A. Narayan Rochester Institute of Technology Rochester, NY [email protected] A graph G is representable modulo k if there is an injective map f : V (G)→Zk such that u and v are adjacent if and only if |f(u)−f(v)| is relatively prime to k. Erd˝os and Evans [1] showed that every )nite graph is representable modulo some positive integer. The representation number rep(G) is the smallest k such that G is representable modulo k. Representation numbers for many classes of graphs were determined in [2,3]. The representation number is easy to compute for complete graphs and for independent sets: rep(Kn ) is the smallest prime number that is at least n, and rep(mK1 ) = 2m. Representation numbers for paths, most cycles, and matchings are found by combining known lower bounds from [4] with known upper bounds from [2]. For example, rep(Pn ) is the product of the smallest log2 (n − 1) primes, rep(C2n+1 ) is the product of the smallest log2 n primes after 2 when n is at least 4 and not a power of 2, and rep(nK2 ) is the product of the smallest 1 + log2 n primes. Evans et al. [3] proved that rep(mKn ) is the product of the m smallest primes that are at least n if and only if there is a family of m − 1 mutually orthogonal Latin squares of order n. Question. What is rep(C2k +1 ) when k¿3? What is rep(G) when G is a complete multipartite graph or a disjoint union of complete graphs of arbitrary orders? What is the behavior of the representation number for random graphs?

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Comment. The only cycles whose representation number is unknown are those whose length is one more than a power of 2 (except for C3 and C5 ). Besides the results on complete graphs and bounds for stars in [3], little is known about representations of complete multipartite graphs. Using the Chinese Remainder Theorem and the fact that the product dimension of an n-vertex graph is at most n − 1 [4], Narayan [5] proved that when n¿4, the representation number of an n-vertex graph other than Kn is at most the product of the n − 1 smallest primes that are at least n − 1. Furthermore, this holds with equality for K n1 + K 1 . References [1] P. Erd˝os, A.B. Evans, Representations of graphs and orthogonal Latin square graphs, J. Graph Theory 13 (1983) 593–595. [2] A.B. Evans, G.H. Fricke, C.C. Maneri, T.A. McKee, M. Perkel, Representations of graphs modulo n, J. Graph Theory 18 (1994) 801–815. [3] A.B. Evans, G. Isaak, D.A. Narayan, Representations of graphs modulo n, Discrete Math. 223 (2000) 109–123. [4] L. LovCasz, J. NeTsetTril, A. Pultr, On a product dimension of graphs, J. Combin. Theory Ser. B 29 (1980) 47– 67. [5] D. Narayan, An upper bound for representations of graphs modulo n, preprint.

Problem 381. Large triangle-free graphs Posed by Fan Chung Graham and Tom Trotter. Correspondent: Stephen C. Locke [email protected] It is a well-known exercise that every graph has a bipartite subgraph with at least half its edges. The kth power of a graph G is the graph G  with the same vertex set as G such that xy ∈ E(G  ) if and only if the distance between x and y in G is at most k. Question. What is the largest triangle-free subgraph in the kth power of a long cycle? Comment. Chung and Trotter [2] observed that if j¿k=2, then one can obtain a bipartite subgraph by alternating j vertices of one partite set with j vertices of the other (occasionally using j +1 instead of √ j to reach around the cycle). One can then optimize j; this yields a lower bound of 2 − 2 for the edge density, which they suggested may be best √ possible. Their upper bound for the edge density of a triangle-free subgraph is (5 + 3)=11. For k624, Bondy and Locke showed that this construction is best possible for suSciently long cycles [1,3]. Is this also true for all k? Chung and Trotter [2] were interested in triangle-free n-vertex graphs with bandwidth m. The bandwidth of a graph is the minimum, over all labelings of the vertices with

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distinct integers, of the maximum diRerence between the labels on√adjacent vertices. Their upper and lower bounds for the question below are (2 − 2)nm and ((5 + √ 3)=11)nm. Question. What is the maximum number of edges in a triangle-free n-vertex graph with bandwidth m? References [1] J.A. Bondy, S.C. Locke, Triangle-free subgraphs of powers of cycles, Graphs Combin. 8 (1992) 109–118. [2] F.R.K. Chung, W.T. Trotter, Triangle-free graphs with restricted bandwidth, Progress in graph theory (Proceedings Waterloo, Ontaria, 1982), Academic Press, Toronto, 1984, pp. 175–190. [3] S.C. Locke, Further notes on: largest triangle-free subgraphs in powers of cycles, Ars Combin. 49 (1998) 65–77.

Problem 382. Greedy vertex partition of a graph Posed by Lenore Cowen and Douglas B. West. Correspondent: L.J. Cowen Tufts University Medford, MA USA [email protected] Let G be a simple graph with n vertices and m edges, and consider a partition of V (G) into two sets. A vertex is greedy with respect to a partition if it has at least as many neighors in the other set as in its own set. A partition of V (G) into two sets is greedy if every vertex is greedy with respect to it. Given a non-greedy partition, the operation of moving a non-greedy vertex to the opposite set is a ,ip, and a list of Zips performed successively is a ,ip sequence. Question. Among all graphs with n vertices and m edges, and all starting partitions of the vertex set into two sets, what is the maximum length of a Zip sequence? Comment. LovCasz [1] observed that every Zip increases the number of edges crossing the partition, and hence every sequence of Zips is )nite. A maximal Zip sequence produces a greedy partition, and by the degree-sum formula a greedy partition captures at least half the edges of the graph across the cut. Since each Zip increases the number of edges crossing the partition, every Zip sequence has length at most m. The star K1; n−1 achieves equality, but what is the best √ bound in terms of n? For n = k 2 and m = 12 (n − 1) n, we have a graph Gk with a Zip sequence of length m (G4 appears below). The )rst question is whether it is possible for an n-vertex graph to have a Zip sequence of length more than (1=2)n3=2 . Also interesting is the minimum number of Zips needed to obtain a greedy partition. Let f(G) denote the minimum length of a Zip sequence that transforms the trivial partition (∅; V (G)) into a greedy partition. If every vertex of G has odd degree, then

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f(G)6n=2. (For Kn , every Zip sequence has length at most n=2.) The greedy algorithm for Zip sequences always Zips a vertex that produces the greatest increase in the number of edges crossing the cut. Question. Is f(G) always bounded by n=2? If so, is there a fast algorithm that produces a greedy partition in at most n=2 Zips? What is the worst-case length of the Zip sequence produced by the greedy algorithm? References [1] L. LovCasz, Combinatorial Problems and Exercises, North-Holland, Amsterdam, 1979.

Problem 383. Linear discrepancy of posets Posed by Tom Trotter. Correspondent: W.T. Trotter Arizona State University Tempe, AZ USA [email protected] A linear extension of a partially ordered set is a linear order L on its elements such that x¡y in P implies x¡y in L. We seek linear extensions that keep incomparable pairs of elements close together. The linear discrepancy l(P) of a poset P is the minimum, over all linear extensions of P, of the maximum diRerence in height on the extension between two incomparable elements of P. When P has k pairwise incomparable elements, the linear discrepancy must be at least k − 1. Such a set of elements becomes a clique in the incomparability graph of P, and in general the linear discrepancy of P is at least the bandwidth of the incomparability graph of P. Fishburn et al. [1] proved that equality always holds. (The bandwidth of a graph is the minimum, over all labelings of the vertices with distinct integers, of the maximum diRerence between the labels on adjacent vertices.) The maximum number of pairwise incomparable elements in P is its width w(P); we have observed that l(P)¿w(P) − 1. A well-known result of Hiraguchi states that the dimension of P is bounded by the width. Hence always dim P6l(P) + 1. (The

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dimension dim P is the minimum k such that P embeds in Rk under the coordinate-wise order.) Question. For which posets P is it true that dim P = l(P) + 1? Comment. Equality can hold for some posets of dimension at most 4; below we show the diagrams of two such posets. The one on the right is the only such poset known where the dimension is 4. Further results, motivation, and background about linear discrepancy appears in [2]. Conjecture. If P is a poset with dimension at least 5, then dim P6l(P).

References [1] P.C. Fishburn, P.J. Tanenbaum, A.N. Trenk, Linear discrepancy and bandwidth, Order 18 (2001) 237– 245. [2] P.J. Tanenbaum, A.N. Trenk, P.C. Fishburn, Linear discrepancy and weak discrepancy of partially ordered sets, Order 18 (2001) 201–225.

Problem 384. The Flattened Antichain Conjecture Posed by Paulette Lieby. Correspondent: Jerrold R. Griggs [email protected] Let Bn denote the collection of subsets of [n] ordered by inclusion. An antichain in Bn is a set of pairwise unrelated elements of Bn (none contains another). Conjecture. For every antichain F in Bn , there is an antichain F  such that F  has the same number of elements as F and the same sum of sizes of elements as F, and the members of F  diRer in size by at most 1. Comment. Lieby [2] proved the conjecture for antichains whose sets diRer in size by at most three. During the time since the problem was presented at the M.I.T. meeting in honor of Kleitman, KisvMolcsey [1] presented a proof of the conjecture.

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However, this proof relies on the dissertation of Lieby, which is a massive piece of work involving extensive computation. Thus, the problem of )nding a shorter and simpler proof remains. References C KisvMolcsey, Flattening antichains, March 2001, preprint. [1] A. [2] P. Lieby, Extremal Problems in Finite Sets, Ph.D. Dissertation, Northern Territory University, Australia, 1999.

Problem 385. Width of cutsets in truncated boolean lattices Posed by BCela Bajnok. Correspondent: B. Bajnok Gettysburg College Gettysburg, PA USA [email protected] For 06m6l6n, the truncated Boolean lattice Bn (m; l) is the set of all subsets of {1; : : : ; n} with size at least m and at most l, ordered by inclusion. By symmetry, we consider only the case m6l6n − m. Every maximal chain in Bn (m; l) has l − m + 1 elements. A cutset in a poset is a family of elements that intersects all maximal chains. The width of a family is the maximum number of pairwise incomparable elements in it. FMuredi et al. [3] showed that the largest minimal cutset in the full Boolean lattice contains almost all elements. We address a somewhat opposite question: what is the minimum width of a cutset? Conjecture. Let the integers n, m, and l satisfy 06m6l6n−m, and set c = l−m+1. If n is suSciently large in terms of m, then the minimum width of a cutset in Bn (m; l) is           n n n n n − + − + m m−1 m−c m−c−1 m − 2c   n − + ::: : m − 2c − 1 Comment. This was proved for m = 1 in [2] and for the “short” case c63 in [1]. The “tall” case is c¿m (equivalently, l¿2m). Here, the conjectured answer is     n n − m m−1

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independent of l. This states that when n is suSciently large, the minimum width of a cutset in Bn (m; 2m) is the same as in the much taller poset Bn (m; n − m) that contains it. In [1], using Griggs’s recursive construction [4] that covers the Boolean lattice using chains of size at most c, the minimum width of a cutset in Bn (m; l) was shown to be n ) when l¿2m. at least ( mn ) − ( m−1 The opposite inequality for l¿2m and the cases m + 36l62m − 1 remain open. References [1] B. Bajnok, On the minimum width of a cutset in the truncated Boolean lattice, Congr. Numer. 130 (1998) 77–81. [2] B. Bajnok, S. Shahriari, Long symmetric chains in the Boolean lattice, J. Combin. Theory Ser. A 75 (1996) 44 –54. [3] Z. FMuredi, J.R. Griggs, D.J. Kleitman, A minimum cutset of the Boolean lattice with almost all members, Graphs Combin. 5 (1989) 327–332. [4] J. R. Griggs, Saturated chains of subsets and a random walk, J. Combin. Theory Ser. A 47 (1988) 262–283.

Problem 386. Chain decompositions of LYM orders Posed by Jerry Griggs. Correspondent: J.R. Griggs University of South Carolina Columbia, SC USA [email protected] A graded poset is a poset in which all maximal chains have the same size, and the rank r(x) of an element x in a graded poset is its height above the minimum on the maximal chains containing it. Let Nk denote the number of elements with rank k. A graded
poset has the LYM property if every antichain A satis)es the LYM Inequality, x∈A 1=Nr(x) 61. A graded poset has a regular covering by chains if there is a list of maximal chains so that in each rank every element appears in the same number of chains in the list. A graded poset has the normalized matching property if for each rank k and each set A of elements with rank k, the set A∗ of elements with rank k that are related to elements of A satis)es |A∗ |=Nk+1 ¿|A|=Nk . Kleitman [3] proved that the LYM property, the existence of a regular covering by chains, and the normalized matching property are all equivalent, de)ning LYM orders. Let dk (P) denote the maximum size of a subposet of P containing no chain with more than k elements; d1 (P) is the maximum size of an antichain in P. Let C be a partition
of P into chains. Considering the contribution from each chain shows that dk (P)6 c∈C min{k; |C|}; the partition C is k-saturated if equality holds. Greene and Kleitman [1] proved that for every poset and every k, there is a chain partition that is both k-saturated and k + 1-saturated.

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Question. Does every LYM order have a chain partition that is k-saturated for every k? Comment. Such a partition has been called a completely saturated partition or a nested chain decomposition. The question was originally asked in [2], where it is aSrmed for LYM orders such that the sequence N0 ; N1 ; : : : is symmetric and unimodal. Recently, Logan [4] showed that deleting any interval from the inclusion lattice of subsets of a )nite set leaves a poset that is an LYM order and has a nested chain decomposition. References [1] C. Greene, D.J. Kleitman, The structure of Sperner k-families, J. Combin. Theory Ser. A 20 (1976) 41– 68. [2] J.R. Griggs, SuScient conditions for a symmetric chain order, SIAM J. Appl. Math. 32 (1977) 807–809. [3] D.J. Kleitman, On an extremal property of antichains in partial orders: the LYM property and some of its implications and applications, in: Combinatorics (Proceedings of the NATO Advanced Study Institute, Breukelen, 1974), Math. Centre Tracts 56 (Part 2), Math. Centrum, Amsterdam, 1974, pp. 77–90. [4] M.J. Logan, Sperner theory in a diRerence of Boolean lattices, Discrete Math. 257 (2002) 257 (2002) 501–512.

Problem 387. The Sperner problem for integer partitions Posed by Rod Can)eld. Correspondent: E.R. Can)eld University of Georgia Athens, GA USA [email protected] A partition 1 of the positive integer n is a nonincreasing list of positive integers with sum n. The indexed elements 11 ; 12 ; : : : of 1 are its parts (parts may be repeated). Among partitions of n, we say that 1 is a re2nement of 2, denoted 162, if 1 is obtained from 2 by replacing each part 2j with some partition of 2j and then reordering. The re)nement relation makes the set of partitions of n into a partial order Pn . The bottom element of Pn has n parts equal to 1, and the top element has a single part n. The partitions with a given number of parts are pairwise incomparable and form a rank. An antichain in a partial order is a family of pairwise incomparable elements. Question. How large is the largest antichain in the partition poset Pn ? What is the asymptotic relationship between the sizes of the largest antichain and the largest rank? Comment. Can)eld and Engel [1] showed that the maximum size of an antichain is no more than e(1 + o(1)) times the size of the largest rank. Computation through n = 45

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reveals that up to this value Pn has the Sperner property, meaning that the largest rank is an antichain of maximum size. References [1] E. R. Can)eld, K. Engel, An upper bound for the size of the largest antichain in the poset of partitions of an integer, Discrete Appl. Math. 95 (1999) 169–180.

Problem 388. Matchings in the symmetric group Posed by Jonathan Farley. Correspondent: J. Farley Vanderbilt University Nashville, TN USA [email protected] In its word form, a permutation 4 of [n] is a list 41 ; : : : ; 4n of the numbers 1 through n in some order. A descent or ascent occurs at position i if 4i+1 ¡4i or if 4i+1 ¿4i , respectively. In each permutation of [n], the number of descents plus the number of ascents is n − 1. Let Dk and Ak denote the sets of permutations of [n] with k descents and with k ascents, respectively. Reversal of the word form of a permutation is a bijection from Dk to Ak . An inversion in a permutation 4 is a pair of elements r; s such that r¿s but element r appears at an earlier position than element s. Let I (4) denote the set of inversions in 4. Question. For k¡(n − 1)=2, does there always exist a bijection 7 : Dk → Ak such that I (4) ⊆ I (7(4)) for all 4 ∈ Dk ? Comment. Edelman [1] constructed such a bijection for k = 1. The existence of such a bijection would yield the partial unimodality of the f-vector of a )nite distributive lattice [2]. Here the f-vector is the sequence of numbers with fi being the number of chains with i + 1 elements, for i¿−1. The full unimodality is the chief consequence of the Stanley–Neggers Conjecture (see [3–5]). References [1] P. Edelman, personal communication, 2000. [2] A. BjMorner, J.D. Farley, in progress. [3] V. Gasharov, On the Neggers–Stanley conjecture and the Eulerian polynomials, J. Combin. Theory Ser. A 82 (1998) 134 –146. [4] F. Brenti, Log-concavity and combinatorial properties of Fibonacci lattices, European J. Combin. 12 (1991) 6 459– 476 [5] J. Neggers, Representations of )nite partially ordered sets, J. Combin. Inform. System Sci. 3(3) (1978) 113–133.

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Problem 389. Graceful permutations Posed by Herbert S. Wilf. Correspondent: H.S. Wilf University of Pennsylvania Philadelphia PA USA [email protected] A permutation of [n] is graceful if the n − 1 diRerences between adjacent elements are all distinct. Let f(n) be the number of graceful permutations of [n]. Conjecture. lim f(n)1=n exists. Comment. A construction due to Deturck [1] shows that lim inf f(n)(1=n) ¿23=10 . This problem was originally posed in [2], but the lower bound is new. References [1] D. Deturck, private communication. [2] H. Wilf, in: Discrete Algorithms and Complexity, Proceedings of Japan–US Joint Seminar, Kyoto 1986, Academic Press, New York, 1987, p. 349.0

Problem 390. A problem in combinatorial algebra Posed by Alexander Sapozhenko. Correspondent: A. Sapozhenko Moscow Lomonosov State University Moscow Russia [email protected] A subset A in an additively written group G is sum-free if a + b ∈= A for all a; b ∈ A. Cameron, ErdMos, Alon, Kleitman, Calkin and other authors proved bounds on the size or the number of sum-free sets in )nite groups and in the set of integers (see [1–5]). The asymptotics for the maximum number of sum-free sets in )nite abelian groups of order 2k were obtained by Lev, Luczak, and Shoen [7] and independently by Sapozhenko [8], both presented at the meeting on “Finite and In)nite Combinatorics” in Budapest in January, 2001. The proof of the last author is based on the )nite case of a well known theorem of Kneser. Given sets A and B, let A + B = {a + b: a ∈ A; b ∈ B}. Kneser’s Theorem [6] states that given subsets A and B of an abelian group G such that |A+B|6|A|+|B|−1, there is a subgroup H of G such that A + B + H = A + B and that |A + B|¿|A + H | + |B + H | − |H |. An analogue of Kneser’s Theorem for all )nite groups would help determine the asymptotics for the number of sum-free sets in groups of odd order and perhaps have other applications. For the general case, we ask for a slightly weaker conclusion.

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Question. Given subsets A and B of a )nite group G such that |A + B|6|A| + |B| − 1, must there be a subgroup H of G such that A + B + H = A + B and that |A + B|¿ |A| + |B| − |H |? References [1] N. Alon, Independent sets in regular graphs and sum-free subsets of )nite groups. Israel J. Math. 73 (1991) 247–256. [2] N. Alon, D.J. Kleitman, Sum-free subsets, in: A. Baker, B. BollobCas, A. Hajnal, (Eds.), A Tribute to Paul Erd˝os, Cambridge University Press, Cambridge, 1990, pp. 13–26. [3] N. Calkin, On the number of sum-free sets, Bull. London Math. Soc. 22 (1990) 141–144. [4] P.J. Cameron, On the structure of random sum-free sets, Probab. Theory Related Fields 76 (1987) 523–531. [5] P.J. Cameron, P. Erd˝os, On the number of sets of integers with various properties, in: R.A. Molin (Ed.), Number Theory, de Gruyter, Berlin, 1990, pp. 61–79. [6] M. Kneser, Ein Satz uM ber abelischen Gruppen mit Anwendungen auf die Geometrie der Zahlen, Math. Zeit. 61 (1955) 429– 434. [7] V.F. Lev, T. Luczak, T. Shoen, Sum-free sets in abelian groups, Israel J. Math. 125 (2001) 347–367. [8] A.A. Sapozhenko, On the number of sum-free sets in abelian groups, Vestnik Moskovskogo Universiteta (in Russian), in press.

Problem 391. Combinatorics of orthogonal polynomials Posed by Steve Fisk. Correspondent: S. Fisk Bowdoin College Brunswick, ME USA )[email protected] A nonnegative weight function w determines a sequence of orthogonal polynomials in the following way: set p0 (x) =1, and for n¿0 set pn to be the unique monic polynomial of degree n such that pn (x)pr (x)w(x) dx = 0 for 06r6n − 1. Each of the polynomials in the resulting sequence has all real roots (see [1]). We ask whether analogous results hold for polynomials in two variables. We cannot expect linear factors, but we can ask for real roots when we )x the value of one variable. Given a nonnegative weight function w in two variables, the inductive de)nition of a family of orthogonal polynomials is like that for one variable. Let p0;0 = 1. For m and n nonnegative but not both 0, inductively de)ne pm; n to be the unique  polynomial with leading term xm yn such that pm; n (x; y)pr; s (x; y)w(x; y) dx dy = 0 whenever r6n, s6m, and (r; s) = (n; m). Conjecture. The polynomials in x that are coeScients of the powers of y in pn; m have all real roots. Furthermore, for every real number 9 and all positive integers m and n, the polynomial in x obtained from pn; m (x; y) by setting y = 9 has all real roots. References [1] G. Szeg˝o, Orthogonal Polynomials, 4th Edition, AMS Colloquium Publication, 23, American Mathematical Society, Providence, RI, 1981.