A T1 counterexample
Josh Purinton
sprouts positions [1] is that o+(n+X) = o-(n+X) where n = 3k-1 for all
integers k >1. That is, "Any game which has a component consisting of
5,8,11,14,... initial spots is won by the same player in normal and
misere play". To date, no counter-example has been published. I
provide one here.
The position 5+3P0+2P0P0P1 is P in normal play (according to Glop) and
N in misere play (according to Aunt Beast). In Glop notation, this
position is "0.0.0.0.0.}0.0.0.2.}0.0.A.}0.B.}A.C.}B.C.}]!"
This position could be reached via 15 1(16)1[2-10] 2(17)2[3-5]
2(18@3)17 1(19)16[11] 12(20)12 12(21)20[13-15] 13(22)13
13(23)22[14-15]. (See attached diagram.) In the misere game, there is
only one winning move, namely: 6(24)6. But this loses in normal play.
15- ... 6(24)6 P 6(25)24 N 3(26)3[4] P
15+ ... 6(24)6 N 6(25)24 P 3(26)3[4] N
[1] Khorkov, Roman. (2007-08-05) "Theorems 1 and 2, Combined and
Corrected" <http://www.geocities.com/chessdp/theorems12cc.htm>
Roman
n>2:
Gz = (x^y + (3n - 1)spots)- = (x^y + (3n - 1)spots)+
The intuitive explanation:
k=2,5 are exception of the general rule
k=2: choice Gx (x>1), 2- 2(3)2 Gx or 2- 1(3)2 Gx
k=5: choice Gx (x>1), 5- 5(6)5 Gx
k=8: no choice Gx (x>1) (any move creates G1)
k=11: no choice Gx (x>1) (any move creates G0)
k=14: no choice Gx (x>1) (any move creates G1)
k=17: no choice Gx (x>1) (any move creates G0)
.
.
.
So
New Theorem:
n>2:
Gz = (x^y + (3n - 1)spots)- = (x^y + (3n - 1)spots)+
> ATTACHMENT: image/png (t1counterexamplegame.PNG)
>
Josh Purinton
> k=11: no choice Gx (x>1) (any move creates G0)
This turns out not to be the case. According to Aunt Beast and Glop,
G-(4L6) = G+(4L6) = 2.