# A T1 counterexample

2 views

### Josh Purinton

Jan 4, 2009, 2:09:29 PM1/4/09
One of Roman Khorkov's primary heuristics for evaluating misere
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>

t1counterexamplegame.PNG

### Roman

Jan 5, 2009, 3:24:42 AM1/5/09
Correction T1:

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)
>