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New proof reveals that graphs with no pentagons are fundamentally different

quantamagazine.org

31–40 of 116 posts

Re: New proof reveals that graphs with no pentagons are fundamentally different

#31
post #25

It would help if the article included at least one real world application in layman's terms.

Because you are finding the secrets of the Universe? Because math is fun? Not all work has to have a real world application ($$$). It would help if the you included at least one real world application of writing your comment.

Even when math does have profound real world applications they aren't always immediately apparent. Think of how many millennia work was done on primes before they became a huge part of cryptography, though they did have some other uses along the way.

Re: New proof reveals that graphs with no pentagons are fundamentally different

#33
post #23

Earlier quoted context omitted.

and an application ... what can I use it for?

I can't possibly imagine why you would want an application for recently developed mathematics. I would have thought that by now mathematics would have proven its worth enough to not have to justify itself.

Because most people here are engineers, not mathematicians. The newness of mathematics has little to do with its utility, so it is reasonable to ask if there are any relevant applications.

Re: New proof reveals that graphs with no pentagons are fundamentally different

#34

Earlier quoted context omitted.

You misread. It's either 3 people who know each other or 3 who have never met. Otherwise it's trivial with a 6-cycle.

3 people who know each other = T 3 people who don't know each other = F T || F = T Isn't that just True by default? Maybe I'm missing the context.

If I know Dan but not Dave, then Dan, Dave and I are not 3 people who know each other, but we are not 3 people who don't know each other either.

Re: New proof reveals that graphs with no pentagons are fundamentally different

#35

can I get an ELI5?

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.

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.

Re: New proof reveals that graphs with no pentagons are fundamentally different

#36
post #23

can I get an ELI5?

and an application ... what can I use it for?

Perhaps not much now but a use for them might come along and it will be good that this will be there ready to fill the need. It's my understanding that quaternions were mostly a curiosity when first described in the mid 1800s but now are used extensively in computing. From wikipedia "quaternions are used in computer graphics, computer vision, robotics, control theory, signal processing, attitude control, physics, bioinformatics, molecular dynamics, computer simulations, and orbital mechanics." When Sir William Rowan Hamilton first described them, he had no concept of most of the above, much less that quaternions would be important for them. Good for us that he still pushed forward with them.

Re: New proof reveals that graphs with no pentagons are fundamentally different

#37

can I get an ELI5?

There's a conjecture about graphs, and the conjecture was proven true for a new specific case. So we still don't know whether the conjecture is true in general.

For understanding the conjecture itself, I think the mathematical description is the easier to comprehend than any analogy. See the intro of the Erdős–Hajnal conjecture Wikipedia article[1]. Graphs are pretty intuitive mathematical structures and the description of the conjecture only uses basic graph structures. Although maybe I'm just not clever enough to come up with a suitable analogy.

[1]: https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Hajnal_conj...

Re: New proof reveals that graphs with no pentagons are fundamentally different

#38
post #3
post #2

I'm stuck on the beginning example. That in a group of at least six people, there are three people who all know each other. I think I can violate that one. I get someone I know from work and someone I know from one of my hobbies that I know don't know each other. To each of them, I have them get someone from their circle that I've never met. Then I get my wife to get someone from her circle I don't know. Then you put…

The article (and the theorem) says "there’s EITHER a group of three who all know each other, OR a group of three who have never met"

The way it's written this seems like it's exclusive. But it appears that both cases are possible at the same time. Maybe I'm reading the "OR" too much from the colloquial sense instead of the mathematic sense compared to "XOR".

Re: New proof reveals that graphs with no pentagons are fundamentally different

#39
post #35

can I get an ELI5?

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…

exponential -> polynomial

Re: New proof reveals that graphs with no pentagons are fundamentally different

#40

Earlier quoted context omitted.

You misread. It's either 3 people who know each other or 3 who have never met. Otherwise it's trivial with a 6-cycle.

Why even bother adding an edge?

Honest answer: because I could think of the term for cycle more quickly than path.
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