1
$\begingroup$

There are six delegates on the world festival of youth. Between three randomly chosen delegates, there are two of them that can communicate on some language. Prove that there exists a delegate triplet where communication is possible between any pair of delegates.

We can represent delegates as vertices of a graph.

Then, between three randomly picked vertices, two of them are connected.

How can we prove that there exist a delegate triplet where communication is possible between any pair of delegates?

  • 0
    Is connectivity between 3 vertices enough to satisfy communication between all three? Or do we need a cyclebetween 3 people to say that they can communicate? – 2017-01-23
  • 0
    @Lelouch, That is not specified, but I think that we don't need a cycle. – 2017-01-23

2 Answers 2

1

In the following we consider that connectivity between 3 vertices is sufficient for communication between them. Just in case, we define a set of vertices $(v_i)$ to be connected iff for every $v_x,v_y; x \ne y,\in (v_i)$, it is possible to travel from $v_x to v_y$ along the edges connecting the different $(v_i)$.(This was a semi formal definition). So now suppose the six vertices are a,b,c,d,e,f. Consider the triplets a,b,c, and suppose a and b communicate. Similarly among c,d,e suppose c and communicate. And among e,f,a, e and f commnicate. Convince yourself that any othercase would lead a triple i,j,k which is connected(trivial case). So now we have the graph as

$a-b, c-d,e-f$. Where $-$ denotes an edge. Next consider the triplet b,c,e. If you conectany two,supose b,c the graph looks like:

$a-b-c-d,e-f$ clearly, there exists a connected triplet and hence proved. Try it yourself with the necessary condition being a cycle.

1

This is a variation on a very famous problem. see here.

Build a complete graph in which there are $6$ vertices representing the persons. Color an edge betweent two persons red if the two persons can communicate.

By the theorem in the video there must be a red triangle or a blue triangle. But you are told by hypothesis that there is no blue triangle. Therefore we conclude a red triangle exists.

  • 0
    Yes this is actually the shortest and most elegant proof. I had missed out on this beautiful property. – 2017-01-23