World
Game
Of
Sprouts
Association

Strategy for N = 7, 13, …

Let's look at the game: n=6k+7, k=0,1,2... (7,13,19,25...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 2(n+4)2[n-3,n-2]

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

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]

16) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)3

16.1) n=7

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

16.2) n>7

3(n+4)n+3[4,5]

17) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)3[4-m] m=2(3s+2), s=0,1,2…

3(n+4)n+3[m+1]

18) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)3[4-m] m=6s+5, s=0,1,2…

3((n+4)@4)n+3

19) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)3[4-m] m=6s, s=1,2,3…

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

20) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)3[4-m] m=6s+1, s=1,2,3…

3(n+4)n+3[m+1]

21) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)3[4-m] m=2(3s+1), s=1,2,3…

3((n+4)@4)n+3

22) 1(n+1)2 1(n+2)1[3-(n-4)] 3(n+3)3[4-m] m=3(2s+1), s=1,2,3…

22.1) n=m+4

3((n+4)@n+2)n+3

22.2) n>m+4

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