New proof reveals that graphs with no pentagons are fundamentally different
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 us all in a room.I know two people, they know two people, there are two people who only know one other person. And then there's the person my wife invited who knows no one.There are certain potential violations that can happen. Let's label everyone, I'm A, my guests are B and F, their guests are C and E respectively, my wife's guest is D. Our graph is essentially E-F-A-B-C and D. I know F-B can't happen because I've deliberately chosen people in that manner. And no one other than F and B can connect to A as the instructions were to invite people I don't know.C-E-F-C, B-C-E-B, D-E-F-D, B-C-D-B, and C-D-E-C are all possible graphs however. But it's also possible that they're not. I'm pretty sure I can engineer it so that it won't be.But this kind of feels like it violates the spirit of the theory as it's not a natural group, it's a contrived group.