Some Favorite Problems

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Some Favorite Problems

• Dan Kleitman, M.I.T.

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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 .

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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 n2d, the diameter is at most nd.
• This implies a linear bound on diameter of one of
  a polytope and its dual.
  
  
  
  
  
  
  
  
  
  
  
  
  
• Can you prove that statement?

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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

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Robert Cowans Problem

• You want to choose a graph on n vertices which
  has at least t induced triangles, to maximize the
  number of induced K4s
• 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.

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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.

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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

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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

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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

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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 nx games then in the first type of
  excluded vector nx1 components are fixed to be
  1s and in the second n-x1 components are fxed
  to be 0s.

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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

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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.

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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 n2, 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.

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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 n1 to 2n of A beat n1 to 2n of B
• Players n of A beats 2 throught n1 of B
• Player 2n of A beats 1 and n2 to 2n of B.
• Then the exclusions of 10 players in A are enough
  to exceed all exclusions in the conjectured best

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Extension to general win pattern among A and B

• Requires no new ideas, just some dogwork.

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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

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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