Download presentation

Presentation is loading. Please wait.

Published byLeonel Millison Modified over 3 years ago

1
Some Graph Problems

2
LINIAL’S CONJECTURE Backgound: In a partially ordered set we have Dilworth’s Theorem; The largest size of an independent set (completely unordered set) is the smallest number of blocks in a partition of the elements into completely ordered sets. (Since each element of an independent set must lie in a separate block from any other, the latter number is obviously larger than the former.) There is a generalization: The largest size of the union of k independent sets is the smallest of the sum of the lesser of k and the size of the block over all blocks in any such partition. (The latter number is similarly obviously larger than the former.) Question: What of this generalizes to directed graphs with blocks consisting of simple paths, and independent sets being independent sets of vertices?

3
Linial’s Conjecture: For directed graphs the largest size of the union of k independent sets is always at least the smallest possible size of the sum of the lesser of k and the cardinality of the path over all paths in a partition of the vertices into simple paths. (The inequality that was obvious for orders is no longer true in general for directed graphs: Let G be a path of vertex-length 3. Then G is only one path while the independence number is 2) The conjecture is that the non-obvious inequality in the order case is still true in general.

4
Known Results: The conjecture is: true if G is acyclic (I think some proofs of harder part for orders work for all directed graphs.) true for k=1 (old result) or k=2 (newer) Let K be length of longest simple path in G; Then the conjecture is true for k=K (old) or k=K-1 (newer).

5
2. A conjecture by two people whose names I have forgotten. Suppose G is a complete tripartite graph, with each edge directed in one direction. If two parts (A and B) each have 2n vertices, and G has diameter at most 2, how many vertices can the third part (C) have?

6
The conjecture is: The largest possible such G has the edges between A and B directed as a “cloned” 4-cycle: all edges from n vertices in A directed toward the same n in B whose other edges are directed toward the other n vertices of A, whose other edges are directed toward the other edges of B whose other edges are directed toward the first n in A. Also every vertex of C has exactly 2n out-edges. For such G it is easy to compute the maximum size of C.

7
Some comments on this conjecture The condition that G has diameter two means: Every edge is in a directed triangle, and Every pair of vertices within any part are diagonally opposite in some directed 4-cycle. Suppose we describe the directions of the edges containing v in C by a 4n-vector, with component 1 when the corresponding edge goes out from v and 0 otherwise. Then these vectors must form an anti-chain in the usual order on such vectors.

8
The first condition above tells us that v cannot have an outward edge to w and also to all the vertices that w has outward edges to. This means that any vertex w of A or B of outward degree d to B or A excludes C(2n+d-1,2n)+C(4n-d-1,2n) vectors of weight 2n from representing vertices of C, the former from in edges to w and to 2n-d others, the latter from edges to w and d others. For the conjectured optimum this yields an upper bound of C(4n,2n)-8C(3n,2n)+a bit more. (work out the answer yourself!)

9
3 Excluded Triangles Consider graphs G. The vertices are all chosen from among n and n-2 element subsets of a 2n element set. The edges are the containment edges and also edges between each pair of n-2 element subsets that have Hamming distance 4 between them. (so that their union has cardinality n.) Also, G contains no triangle having an n element vertex. How large can G be? Conjecture: G can have no more than C(2n,n)(1+1/n) vertices.

10
4.A somewhat more general problem Consider graphs G. The vertices are all chosen from among n and n-k element subsets of a 2n element set. The edges are the containment edges and also edges between each pair of n-k element subsets that have Hamming distance 2k between them. (so that their union has cardinality n.) Also, G contains no triangle having an n element vertex. How large can G be? Conjecture: G can have no more than C(2n,n)(1+1/n) vertices.

11
5. A Counting Problem Let G be the graph whose vertices are the n, n+1 and n-1 element subsets of a 2n element set, with edges between each pair that are ordered by inclusion. Question: How many independent sets are there in G? Obviously there are 2^C(2n,n) independent sets that consist entirely of sets of size n, but there are lots more than that. If you choose a typical one of these of cardinality roughly C(2n,n)/2, each of the other sizes has a 2^(-(n+1)/2) chance of being addable to it.

12
Conjecture: Most independent sets here consist of lots of sets of cardinality n, (roughly half of them or a bit less) along with roughly half of those of the other two sizes not adjacent to them. Question: Is there some other profile of independent sets of which there are more?

Similar presentations

Presentation is loading. Please wait....

OK

CS 336 March 19, 2012 Tandy Warnow.

CS 336 March 19, 2012 Tandy Warnow.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on cultural heritage of india Ppt on isobars and isotopes of elements Ppt on centering prayer Ppt on world book day games Ppt on weapons of mass destruction 2016 Ppt on blood and its components Ppt on technical analysis of stocks Ppt on hydrogen fuel cell Ppt on statistics in maths what is median Ppt on chromosomes and genes are actually replicated