, 3 , then B's answer depends on 3 mod 3. If 3 ? 0 mod 3, B may reply by taking 1 vertex in S , so as to leave P 3?1 + S 1,1,2,3 . Indeed, from Lemma 24, we know that G(S 1,1,2,3 ) = 2 = G(P 3?1 ), thus P 3?1 + S 1,1,2,3 is a P-position. If 3 ? 1 mod 3, B may reply by taking 2 vertices in S

S. -if, 3+3 , then B removes 4 vertices from S , so as to leave P 3?1 + P 3?1, vol.1

P. 3+2,

, B can reply by taking 1 vertex in S , so as to leave P 3 + S 1,2,2,3 , which is a P-position. Indeed, from Lemma 27, we know that G(S 1,1,2,3 ) = G(P 3 ) = 3 mod 3. If S = S 1,2,3+3 , then B may reply by removing 1 vertex from S , so as to leave P 3 + P 3+3 , which is a P-position. If S = S 1,5,3 , then B may reply by removing 1 vertex from S , so as to leave P 3 + P 3+6, vol.2

;. Case, , p.4

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