# Arrange people at round table so that everyone knows the two people next to them

Each of the guests know: a) more than half of the guests b) at least half of the guests. Prove that in both of these cases it is possible to arrange them to sit around a round table so that everyone knows the two people next to them.

I believe that if we prove b) then we have at the same time proven a) as well. Can anyone give me a hint? I've tried drawing, but I'm not sure how to formally prove it. I was considering relationship properties, such as symmetry and transition, but couldn't work it out. Thanks in advance.

Construct a graph with $$n$$ vertices representing the people and connect two vertices if the two people they represent know each other. For a), the degree of each vertex if greater than $$\frac{n}{2}$$, By dirac's theorem, there is a Hamiltonian cycle. And this implies we can arrange the people in a circle so that each person knows the ones sitting next to them. Same for b). I think.
Or consider first a random arrangement of the people around the table, Suppose a neighboring pair $$(A,B)$$ is a hostile couple with $$B$$ sitting to the right of $$A$$, then if we can find a neighboring pair $$(A'B')$$ with $$B'$$ sitting to the right of $$A'$$ and $$B'$$ is friend with $$B$$ and $$A'$$ is a friend of $$A$$. we can then swap $$B$$ with $$A'$$ and that will reduce the number of hostile neighboring couples. So it remains to show $$(A'B')$$ exists. Well $$A$$ has at least $$n$$ friends sitting to his right, and there are $$n$$ sits to the right of frinds of $$A$$. $$B$$ has at most $$n-1$$ enemy, So there is a friend of $$A$$,$$A'$$, with $$B'$$ sitting right to him, a friend of $$B$$. Done?