The Y region (the also called A5) is a 3^0. (normal grundy value:3, misere:0 )
If therefore you find a child of the Z region with value *0 you are sure to win in misere as
3^0 + *0 = 3^0
You can possibly do the same with a 0^1 as
(usually) 3^0 + 0^1 = 3^0
But if the child is found that is G0 in normal play but not a 0^1
lets say a *2+*2 or a *3+*3) therefore a 0^0
then (usually) 3^0 +0^0 = 3^3 , which is not a misere win for the player that plays this child.
(Note: A5=3^0 but in this case 3^0+0^0 = A5 + (*2+*2) = 3^2 (still not a misere win for the player that plays this).
In conclusion, you are correct, finding a G0 is not enough.
One has to find a child which is normal G0 and misere G1. (and stiil not be sure to win unless it's a confirmed *0)
On Fri, Dec 12, 2008 at 3:06 AM, Yper Cube <yper...@gmail.com> wrote:
The Y region (the also called A5) is a 3^0. (normal grundy value:3, misere:0 )
If therefore you find a child of the Z region with value *0 you are sure to win in misere as
3^0 + *0 = 3^0What does "with value *0" mean? Does it mean just that G+(Z) = 0 and G-(Z) = 1? Or are there further conditions?
I'm theorizing that his choice in would be to convert Y to either a shallow odd or to a deep even. (Now, this would be plenty of options if Z is deep. But if Z is shallow, either both options are good or both options are bad, depending on Z's survivor parity.) [...] Note two key lines:
1(26)1[3-5] shallow odd1(26)1[3,4] deep even
In rot13, the winning move is: gjragl gjraglfvk frira rapybfvat gjraglguerr.
> Roman wrote: "Let Y be the area which contains points 1,2,16,3,4,5. Let Z be
> the area which contains all other points 11,7... If you find a move which
> creates a normal-play G0 in region Z, then you have won."
For what it's worth, the winning move leaves the sphere Z with a
normal-play Grundy number of 3.