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Re: Discrete math with IMO.
Posted:
Feb 2, 2013 4:39 AM
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> 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.
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