World
Game
Of
Sprouts
Association

Strategy for N = 8, 14, …

Let's look at the game: n=2(3k+4), k=0,1,2... (8,14,20,26...spots)

The first player's best chance is 1(n+1)2.

n+

1) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)n+2 3(n+4)n+3

2) 1(n+1)2 1(n+2)1[3-(n-4)] 2(n+3)n+1 2((n+4)@n-3)n+1

3) 1(n+1)2 1(n+2)1[3-(n-4)] n-3(n+3)n+2 n+1(n+4)n+3[2]

4) 1(n+1)2 1(n+2)1[3-(n-4)] n-3(n+3)n-2 n+1(n+4)n+2[2,n-3]

5) 1(n+1)2 1(n+2)1[3-(n-4)] 2(n+3)n-3 n+1(n+4)n+2[n-3,n-2]

6) 1(n+1)2 1(n+2)1[3-(n-4)] n-3(n+3)n+1 n+2(n+4)n+3[2]

7) 1(n+1)2 1(n+2)1[3-(n-4)] 2(n+3)2 n+1(n+4)n+2[n+3]

8) 1(n+1)2 1(n+2)1[3-(n-4)] 2(n+3)2[n-3] n+1(n+4)n+3[n-2,n-1]

9) 1(n+1)2 1(n+2)1[3-(n-4)] 2(n+3)2[n-3,n-2] n+1(n+4)n+3[n-1]

10) 1(n+1)2 1(n+2)1[3-(n-4)] 2(n+3)2[(n-3)-(n-1)] n+1(n+4)n+2[n+3]

11) 1(n+1)2 1(n+2)1[3-(n-4)] 2(n+3)2[(n-3)-n] n+2(n+4)n+3

12) 1(n+1)2 1(n+2)1[3-(n-4)] n-3(n+3)n-3 n-3(n+4)n+3

13) 1(n+1)2 1(n+2)1[3-(n-4)] n-3(n+3)n-3[n-2] n-3(n+4)n+3[2]

14) 1(n+1)2 1(n+2)1[3-(n-4)] n-3(n+3)n-3[n-2,n-1] n-3(n+4)n+3[2]

15) 1(n+1)2 1(n+2)1[3-(n-4)] n-3(n+3)n-3[(n-2)-n] n-3(n+4)n+3[n-2]

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