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.<p>I think I can violate that one.<p>I get someone I know from work and someone I know from one of my hobbies that I know don't know each other.<p>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.<p>Then you put us all in a room.<p>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.<p>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.<p>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.<p>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.