Date: Feb 2, 2013 4:39 AM
Author: mina_world
Subject: Re: Discrete math with IMO.
> 2n ambassadors are invited to a banquet.

>

> Every ambassador has at most (n-1) enemies.

>

> Prove that the ambassadors can be seated around a round table,

>

> so that nobody sits next to an enemy.

>

>

>

> ------------------------------------------------------------------------------------------------------------

>

> Solution)

>

> First, we seat the ambassadors in any way.

>

> Let H be the number of neighboring hostile couples.

>

> We must find an algorithm which reduces this number whenever H > 0.

>

>

>

> Let (A, B) be a hostile couple with ? sitting to the right of A(Fig.

> 1.3).

>

> We must separate them so as to cause as little disturbance as possible.

>

> This will be achieved if we reverse some arc ? A' getting Fig. 1.4.

>

> H will be reduced(***) if (A, A') and (?, ?') in Fig. 1.4 are friendly

> couples.

I know that..

It's really reduced.

Because, reverse some arc ? A'.

In fact, I thought only B and A' (reverse).

Arc is a important key in this problem.

> It remains to be shown that

>

> such a couple always exists with B' sitting to the right of A'. (***)

Yes, it's true.

Because,

there is 2n chairs around the table.

Fix A and B. (A, B) as given problem.

There is at least n friends of A.

We start in A and go around the table counterclockwise.

A, B, A_1, A_2, ... , A_n (counterclockwise)

so, n+2

remainder chairs (2n)-(n+2) = n-2

And there is at least n friends of A.

B_1, B_2, ... , B_(n-1), B_n

Since remainder chairs (n-2) < Friends of B (n),

there is at least two as A_i = B_j , A_k = B_m

It means that always exists with B' sitting to the right of A'

Hm, no problem ?

I think that lower explanation is wrong.

> We start in A and go around the table counterclockwise.

>

> We will encounter at least n friends of A.

>

> To their right, there are at least n seats.

>

> They cannot all be occupied by enemies of ? since ? has at most (n-1)

> enemies.

>

> Thus, there is a friend A' of A with right neighbor B', a friend of B.