Some Favorite Problems

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

2. 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 .

3. 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?

4. 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

5. 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 K4‘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.

6. 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.

7. 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:

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

9. 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

10. 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.

11. 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

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.

13. First Step Idea Example • If there are seven players in Asuch 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.

14. 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

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

16. 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

17. 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