Some Favorite Problems Dan Kleitman, M.I.T.

The Hirsch Conjecture 1. How large can the diameter of a bounded polytope defined by n linear constraints in d dimensions be? HC claims n-d. (one step along an edge between two vertices of the polytope is distance 1) a vertex (assume no degeneracy) is characterized by the d facets which meet at it. A polynomial upper bound in n or d is not now known.

Relatively new results on HC Suppose only condition is one can get from any vertex in a facet to any other staying in it. Then there is an old upper bound to maximum diameter, and new lower bound that is almost quadratic. (ask Gil Kalai for reference) Suppose also ub(n,d) is at least ub(n,d-1)-1. Then if n=2d, the diameter is at most n+d. This implies a linear bound on diameter of one of a polytope and its dual. Can you prove that statement?

Simple Subset Union Problem Consider subsets of a 2n element set whose sizes are either n or n-2 How many can you have if the union of two of the smaller ones is not one of the bigger ones? Easier problem: how many sets of size n-2 can you have if no 2 have union of size n? (does Frankl Wilson answer this?) Obviously there are many similar questions

Robert Cowan’s Problem You want to choose a graph on n vertices which has at least t induced triangles, to maximize the number of induced K 4 ‘s There are some partial results, conjectures and generalizations, too numerous to mention I have been about to write a paper on this for many years but have never gotten around to it.

Partitioning a girth 5 Planar Graph into A Forest and ? The edges of Girth 6 planar graph can be partitioned into a forest and a graph of maximum degree 2. this statement is very tight Can the edges of a Girth 5 planar graph be partitioned into a forest and a graph of maximum degree 3? (would not be tight; should be true and not so hard to prove) Other one question: if girth is instead 8, can there be partition into forest and matching? True for 9 (Kostochka et al.) Strangely a tight 8 subcase is easy.

Maximum size of Diameter 2 Tripartite Tournament of size (2n,2n,?) Problem raised many years ago by Petrovic et al. Conjectured Solution: among 2n size parts A and B: nnnn

Conjectured Solution Players in Third part C win half of their games each. Present Result: True for sufficiently large n. Possible improvements: 1.How large is sufficiently large? 2.Better argument

How big is C in Conjectured Solution? Some facts: Diameter 2 means every edge is in a directed triangle and every non-edge is a diagonal of a directed 4-cycle, The second of these statements implies that C is an anti-chain in the sense we now describe: Denote each player in C by its 0-1 win vector, components corresponding to members of A and B

What the Directed Triangle Condition Implies A player in C cannot defeat a player X in A (or B) and also everyone X beats in B. Also it cannot lose to a player X in A and everyone X loses to in B. This excludes vectors having form: (…,1, …,1,1,1,1,1,1,1,…) and (0,0,0, 0,... ) If X wins n+x games then in the first type of excluded vector n+x+1 components are fixed to be 1’s; and in the second n-x+1 components are fxed to be 0’s.

Max Size of C in conjectured Solution In Same, all players in A and B win half their games, so that each excludes 2C(3n-1,2n) vectors of weight 2n. All n corresponding to one single vertex of the 4-cycle exclude 2(C(3n,2n)-C(2n,n)). The overlap among these exclusions for different 4-cycle vertices is 8C(2n,n)-12, which gives a total maximum size for C of C(4n,2n) – 8C(3n,2n) + 16C(2n,n) - 12

Method of Proof Show first that conjectured Solution is Best among tournaments in which all players in A vs B win half their games, and all players in C do so as well. This means C players will correspond to all weight 2n vectors except those excluded; so we need look only at exclusions rather than look at anti- chains Key idea in proving this: exclusions from 10 or fewer vertices in 0all other possible tournaments exceed those from the conjectured best tournaments.

First Step Idea Example If there are seven players in A such that the union of each of their 21 pairs of win-sets is of size at least n+2, then together their exclusions of vectors of weight 2n from representing players of C exceed those of all players of A in the conjectured solution, by a finite fraction. We find 5 exhaustive if statements like this for which the same conclusion follows. All single players exclude alike, the further their win- sets are from one another the smaller these exclusions overlap and the greater their total exclusion.

The hardest case Occurs when A and B is as close as possible to the Conjectured Best but slightly Different: Players 1 to n-1 of A beat 1 to n of B Players n+1 to 2n of A beat n+1 to 2n of B Players n of A beats 2 throught n+1 of B Player 2n of A beats 1 and n+2 to 2n of B. Then the exclusions of 10 players in A are enough to exceed all exclusions in the conjectured best

Extension to general win pattern among A and B Requires no new ideas, just some dogwork.

Extension to general anti-chain for C of n-n-n-n 4-cycle A-B graph If any part C vectors have weight strictly greater than 2n you can replace the top weight vectors by at least as many vectors of weight one less. Argument is pretty, uses special properties of n-n-n-n 4-cycle exclusions

General Extension to General Tournament Uses fact that proof for A-B win patterns other than n-n-n-n 4-cycle give too many exclusions from only at most 10 players in A or in B. This makes the argument easy and fun. And that is the end of the story