Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Topic: Discrete math with IMO.
Replies: 5   Last Post: Feb 2, 2013 4:51 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 2,142
Registered: 12/13/04
Re: Discrete math with IMO.
Posted: Feb 2, 2013 4:39 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

> 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.
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.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.