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]