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The Erdős–Hajnal conjecture says that for any graph `H`, then every graph in the set of graphs `F_H` that does not contain `H` as an induced subgraph will eithe
by mFixman 5y ago
The Erdős–Hajnal conjecture says that for any graph `H`, then every graph in the set of graphs `F_H` that does not contain `H` as an induced subgraph will either have a polynomial amount of cliques or a polymomial amount of independent sets. The growth rate of the exponent depends on the size of each graph in `F_H` and on some properties `H`.
Like with most simple questions in Ramsay theory no proof or contradiction has been found for the general conjecture, but there has been some progress for some `H`.
The latest paper proves that the conjecture is true if `H` is a cycle of 5 graphs. Since proving this was so hard and provided so much insight, there's hope that the general conjecture can be proved without a lot more work.
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Here's an actual ELI5:
There's a party where some pairs of guests know each other and some don't. A genie told you that there isn't any group of 5 people that only know two people from that group in a way that those relationships form a loop (1 — 2 — 3 — 4 — 5 — 1).
Knowing that, you reason that this cannot be a regular party. Instead, it has to be one of these two:
⒈ A class reunion, where there's a large group of people that all know each other.
⒉ A group blind date, where there's a large group of people where nobody knows anyone.
Being a smart 5-year-old the genie pushes you to publish a 19-page paper proving this, but you find out in Hacker News that some academics beat you and published it this February.
- Hfjfjdjfjceijfj 5y agoexponential -> polynomial
- mFixman 5y agoThanks, edited. I first understood the conjecture as bounding the exponential function on the size of the graph of an element of `F_H`, but apparently the odd thing is the exponent depends solely on `H`.
- ball_of_lint 5y agoI think I'm misunderstanding or you left out an important condition. For your example where H forms a loop, I imagine an element of F_H which is a larger loop. Then it could be arbitrarily large, have no cliques above size 2, a linear number of cliques of size 2, and have only one independent set.
- Dagonfly 5y agoI'm not an expert, but I think you are confused about what 'independent set' means. It's basically 'no vertex pair v,w in the set has a edge (v,w)' So in your loop example, you can take every other vertex in the loop and those form an independent set (of size n/2 ish). And n/2 is in Ω(n).
- luxuryballs 5y agoaaand I just realized I’m only 3