The eight spot normal game is difficult, but the nine spot normal game is rudimentary. We know, by AJS, that Left wins. Left needs an even number of survivors. The most natural try is to make the symmetrical enclosing move.
This try succeeds overwhelmingly. The move 1(11)2 is defeated by symmetry. Any move from spot 1 to spot 10 will clearly leave three survivors in the affected sphere. (The formation in sphere S<11> after 1(11)10[2,3] is well-known to produce three survivors. (This formation is called F(2,P,2) or simply 2P2.)) And the other sphere, the sphere consisting of four spots, produces three survivors.
Any enclosing move in sphere S<2> is transparently hopeless, so there is only one real try for Right.
But now Left has the simple task of insuring that three survivors will come from each region. There are multiple solutions, but perhaps most forceful is as follows.
Sphere<2> has a life force of nine, so it is no surprise that this sphere will produce at least two survivors. Sample lines include: (a) 2(13)5 2(14)11, and (b) 5(13)12 2(14)2 (but not 5(14)5, because of 2(15)3, when the sphere S<4,14> produces four survivors). Left achieves his goal.
9+, what a simple opening!
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