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]