World
Game
Of
Sprouts
Association

Strategy for N = 12, 18, …

Let's look at the game: n=6(k+1), k=1,2,3... (12,18,24,30...spots)

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

n+

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

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

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

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

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

6) 1(n+1)2 3(n+2)3[1,4] 5(n+3)6 3(n+4)n+2[5,7]

7) 1(n+1)2 3(n+2)3[1,4] 5(n+3)5 5(n+4)n+3[6,7]

8) 1(n+1)2 3(n+2)3[1,4] 5(n+3)5[6-m] m=6s, s=1,2,3…

8.1) n=m

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

8.2) n>m

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

9) 1(n+1)2 3(n+2)3[1,4] 5(n+3)5[6-m] m=6s+1, s=1,2,3…

5((n+4)@6)n+3

10) 1(n+1)2 3(n+2)3[1,4] 5(n+3)5[6-m] m=2(3s+1), s=1,2,3…

10.1) n=m+4

5(n+4)n+3[6]

10.2) n>m+4

5(n+4)n+3[(m+1)-(m+5)]

11) 1(n+1)2 3(n+2)3[1,4] 5(n+3)5[6-m] m=3(2s+1), s=1,2,3…

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

12) 1(n+1)2 3(n+2)3[1,4] 5(n+3)5[6-m] m=2(3s+2), s=1,2,3…

5((n+4)@6)n+3

13) 1(n+1)2 3(n+2)3[1,4] 5(n+3)5[6-m] m=6s+5, s=1,2,3…

13.1) n=m+1

5(n+4)n+3[6]

13.2) n>m+1

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