In the third stage, whenever player R creates a red interval he is gaining
an advantage of the size of this interval. If G breaks the interval, the game
is in a tie again. Otherwise in the next round (if there is a next round) R
breaks the green interval, regaining his advantage. Now assume the game
is in its last round. Assume the length of the bichromatic interval within
which R created his last red interval is l. If G is to win, he must create
a green interval within a bichromatic interval of the size > l - ε and the
created interval cannot be bigger than the red one by more than a margin
< ε (as a maximal remaining bichromatic interval has size ≤ l). ■
3.1.1 Nash equilibrium of the ε-adjusted game
Since G wins for sure in the one-by-one variant of the game, that points
are be placed on continuous curves and no single point on a circle can be
served by more than one loctions, R does not really have an optimal strat-
egy. Hence, a Nash equilibrium in this version of the game does not exist.
However, strategy Y 0 becomes a dominant strategy for player R if we restrict
attention further to one-by-one games where R is not allowed to place his
points within a distance smaller than 1/dN/Ke ε > 0 to a green point.
This is because, as follows from proof of Theroem 7, just before the last
round R is not loosing and in the last round G can create an interval of
the size greater than the interval created by R, by at most ε. With this
observation, the following theorem is immediate.
Theorem 8 Let hN, {Cj}jK=1i define a game on the family of disjoint circles
with N > K ≥ 2 and assume that players face a very strict resource mobil-
ity constraint so that they are allowed to place exactly one point at a time.
Suppose also that R, the first mover, is not allowed to place his points within
a distance smaller than 1/dN/Ke ε > 0 to a green point. Then strat-
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