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]