Presentation on theme: "Some Graph Problems. LINIAL’S CONJECTURE Backgound: In a partially ordered set we have Dilworth’s Theorem; The largest size of an independent set (completely."— Presentation transcript:
Some Graph Problems
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?
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.
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).
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?
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.
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.
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!)
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.
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.
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.
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?