Now assume that in the last round R did not play according to (d).
Then at the end of the game we will have the situation where all circles are
covered with the same amount of green and red points, all key positions are
taken (dN/Ke key positions on each circle) and R placed his points on key
positions only. Then there is a tie. ■
Our next result deals with the case where K | N where we show that
R has a tying strategy in this case as well. The strategy is simple and is
therefore defined directly in the proof of the theorem. We shall refer to it
as T* as well since its identity will be clear from whether K | N or not.
Theorem 3 Let hN, {Cj}jK=1i define a game on the family of disjoint circles
with K ≥ 2. If K | N, then R has a tying strategy (which will be also called
T*).
Proof. Observe that R can implement K/N red arcs of equal size on
each of the K circles by placing N points. To see this, consider the following
strategy of R (called T*): place exactly N/K equidistant points in round r
on circle Cr . Obviously since Cr is continuous, G can always do that.
Thus at the end of the game both R and G placed the same amount N of
points in K circles. Moreover R points take N/K key positions determined
by a red point and N/K in each of these circles. If there is no green interval
on the circles then there is a tie. Observe that if there is a green interval,
then its size is always ≤ K/N while each red interval has size K/N . Thus
if there are any green intervals, then by the fact 1 it cannot be that G won.
■
By Theroem 2 and Theroem 3 we have shown that if K ≥ 2, then R, the
first mover, has a tying strategy in the game and hence the second mover
advantage present under K = 1 disappears. The question that arises now
is: can the first mover do better? The answer is no as we show the existence
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