* Not the nimber 1, corresponding to the one-counter heap, but a
half-tame game as 4 + 4 game has value 0^0
Here is a
novelty, the option 1^1 which is a win at both normal and misere game
because of the Universal Trap with value 0^0. With the G(i) and Ppi
table we can check two options : 7(9)8 which is the value of 2 spots
game : 0^0, and playing the pivot 8 to make a PP2 : 3^3. To get the
value of a position a^b from its options, you use the mex rule (minimum
excludent) both on the ais and the bis : mex (0,3)=1,
mex (0,3)=1. I leave you to check the other option connecting the two
spots is not a x^1 nor a 1^x but is the sum of a 2^2 with a 1^0...
 a
position of value 1^1 because of its options right. The pier spots are
the magenta spots 7 and 8.
|
7(8)9, a position of value 0^0 (Universal Trap)
|
 a
position with one external survivor and a PP2 of value 3^3
|
Now just a few rules on how to calculate a
sum of positions to use these tables.
Despite normal impartial games where the sum of two positions of Grundy
numbers a and b is a XOR b (bitwise exclusive or), this is usually not
true in misere impartial games where there are a few cases depending on
the genus sequence, at least for tame games (nimbers of value 0^1, 1^0
and a^a) as is
explained in 'Winning Ways Vol.II':
- if one of the components is a firm component a^a equivalent to a heap
of 'a' counters in the game of Nim, you can use normal nim arithmetic,
a^b+c^d=(a XOR c)^(a XOR c), the misere
value is not taken into account. It is the case in most games where a
universal trap is present because of its value 0^0.
- if every component is a 'fickle' component 0^1 (an empty nim heap) or
1^0 (a nim heap of just one counter) :
a^b+c^d=(a XOR c)^not(a XOR c).
You can think of the position as a sum of nim heaps of one counter
each,
for example three heaps of one counter each is a
win for next player in normal game but a losing position for next
player in misere game (NP) because 1^0+1^0+1^0=(1 XOR 1 XOR 1)^not(1
XOR 1 XOR 1)=1^0.
Now lets take
the
example of a real game which I lost against Danny in our 2005 WGOSA
Sprouts tournament :
29- (Peltier-Purvis*)
A serve from Danny which pleased me because despite the high number of
spots, it is considered a simple position of value 1^1, with many
choices of good looping moves : the diagonal in the 'Loop Table' going
from (0,28) to (28,0) shows a 0/0 for looping moves 2L26, 5L23, 8L20,
11L17, 14L14... which are positions considered to be loosing for both
normal and misere play, hence of value 0^0.
I chose 11L17 with the move 1(30)1[2-12] : 11 spots inside, 17 spots
outside and 2 pivots.
You can check in the pivot table that every trapeze move joining the
two pivots in the inside region will give a biosphere of 17 spots (1^1
firm component) and a pivot iP(11-i) which is the diagonal in the
'Pivot table' going from (0,11) to (11,0) and of value 0^x : 1^1+0^x=(1
XOR 0)^(1 XOR 0)=1^1 a winning game for me.
The same is true of every trapeze move on the outside 17: 1^1+0^x=1^1.
So of course Danny did not play a trapeze move but a looping move in
the outside Sphere 13(31)13[14-22]
 |
|
| I
play 11L17 which
is a 0^0 :
1(30)1[2-12] |
Danny
plays 13(31)13[14-22] |
Now I really do not know what is the best move (1(32)13 seems
right to me now…).
The one I played 13(32)31[23-27] is dubious. Here were my reasons : a
5P2 is a 0^0 trying not to go to an 'all fickle' position that Danny
was heading to with his 9 sphere : 1^0.
Now every trapeze move in the 11 sphere will lead a 0^x hence a
position with value 0^0+1^0+0^x=1^1 and I was wrongly thinking that
after a looping move in the sphere of two spots that 0P2 or 1P1 being a
1^0, I could play a PP5 I thought was a 1^1 getting a
1^0+1^0+1^1+1^1=0^0, but Danny was not afraid of that knowing already
there was no escape in a PP5 which is a 2^2 (correction thanks to Josh, PP5 is a G4 !) as
we'll see...
So Danny played 1(33)30[28,29] and now you can see I am hooked with
modified value in the table, no option gives a x^0 :
- joining the two pivots gives 4 biospheres with: 11 spots (1^1), 9
spots (1^0), 5 spots (1^1) and 2 spots (0^0) : 1^1+1^0+1^1+0^0=1^1
- playing the PP2 (3^3) and a 0P5 (0^0) gives 1^1+1^0+0^0+3^3=3^3
- and the PP5 I played : 23(34)32 gives 1^1+1^0+4^4+1^0=1^1+4^4=5^5
- Latest News:
this position occurs in Roman's opening catalogue 6n+5, 1.6.2, he
explained me the right answer is 28(34)28[29], so I could still have
won (see 4th position).
3 spots = 1^0
1L1, 4 cannibals give 2 survivors, same parity : 0^1
a
0P3 = 1^1 
a
PP3 = 3^0
