Playing Sprouts with Misere Grundy numbers
• # The Tables

 i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 G(i) 0/1 0/1 0/0 1/0 1/0* 1/1 0/1 0/0 0/0 1/0 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 Ppi 0/1 3/3 3/0 1/2 4/4 2/1 0/3 3/3 3/3 1/4 1/1 0/0 0/0 3/3 3/3 1/1 1/1 0/0 0/0 3/3 3/3 1/1 1/1 0/0 0/0 3/3

* Not the nimber 1, corresponding to the one-counter heap, but a half-tame game as 4 + 4 game has value 0^0

 Loop 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 0 1/0 2/2 2/1 0/2 2/2 2/2 2/2 2/1 2/2 3 3 2 2 2 2 3 3 2 2 2 2 3 3 2 2 2 2 1 2/2 0/1 0/3 2/0 1/0 1/2 2 2 0/2 2 2 1 2 2 0/0 2 2 1 2 2 0/0 2 2 1 2 2 0/0 2 2/1 0/3 0/0 2 1/1 1/1 2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 3 0/2 2/0 2 2 2 2 2/0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2/2 1/0 1/1 2 0/1 0/2 2 2 2 2 2 0/0 2 2 2 2 2 0/0 2 2 2 2 2 0/0 2 2 2 5 2/2 1/2 1/1 2 0/2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 6 2/2 2 2 2/0 2 2 2 2 2/2 2/0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 7 2/1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 8 2/2 0/2 0/0 2 2 1/1 2/2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 9 3 2 2 2 2 2 2/0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 10 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 11 2 1 1/1 2 0/0 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1 2 2 0/0 2 2 1/1

Outside the "wild" blue-grey square, games are firm of value a^a

 Pivot 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 0 0/1 1/0 1/0 1/1 0/1 0/0 1/0 1/0 1/0 1/1 0/0 0/0 1/1 1/1 1/1 1/1 0/0 0/0 1/1 1/1 1/1 1/1 0/0 0/0 1/1 1/1 1/1 1 1/0 1/0 1/1 0/1 0/1 0/0 1/0 1/1 1/1 0/1 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 2 1/0 1/1 0/1 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 3 1/1 0/1 0/0 0/0 1/0 1/1 0/1 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 4 0/1 0/1 0/0 1/0 1/0 1/1 0/1 0/0 0/0 1/0 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 5 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 6 1/0 1/0 1/1 0/1 0/1 0/0 1/0 1/0 1/1 0/1 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 7 1/0 1/1 1/1 0/0 0/0 0/0 1/0 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 0/0 0/0 0/0 1/1 1/1 1/1 8 1/0 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 9 1/1 0/1 0/0 0/0 1/0 1/1 0/1 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0 0/0 1/1 1/1 0/0 0/0 0/0

You may remark there is no value of 2 in this Pivot Table ! This is compliant with the numerous 2 valued games in the Loop Table which do not have 2 option.
Any real explanation of this fact is welcome.
• # Table Explanation

Explanation of first table G(i) : it gives the estimated Sprague-Grundy value and Misere Grundy value of a position of just i spots :
for example, the position below with three spots is a win on normal game G(3+)=1 with a possible move leading to a G0 : 1(4)1[2]   a loop with one spot inside and one spot outside 1L1 which value can be found in Loop Table 0^1. It is a lost in misere game G(3-)=0 hence the notation 1/0 or 1^0 which is the start of genus sequence notation 1^031... with every subsequent digit being the misere value of the game when you add a new G2 position to it (a G2 or 2^2 is a game with options of either a win or a lost at both normal and misere game : 0^1 and 1^0).
Another way of seeing it is that a 0^1 is a position with same parity of cannibals and survivors : 4 cannibals and 2 survivors in the example beneath ; whereas a 1^0 is a position leading opposite parity of cannibals and survivors.

 3 spots = 1^0 1L1, 4 cannibals give 2 survivors, same parity : 0^1

Explanation of first table PPi : after a looping move, when connecting the two spots produced on the loop, you get to a position with one spot belonging to two regions, it is called a pivot, as in the example bellow left where you get a pivot belonging to the external region with 0 spots and to the region with 3 spots : 0P3 of value 1^1 ; read in the 'Pivot table' (see article of Roman about this type of configuration).
When you connect this pivot to a spot in one of the two regions, this region becomes what I call a PPi (
Pivot Played to i spots region).
In the following example, the pivot is played to the 3 spots region which becomes a PP3.
You should remark you get a position with the same life force as an i spots region, namely 3xi liberties: two for the ghost, one for the pier spot and three for each novice cannibal. But the Grundy value given in second row of first table is often different of the one of the position of just i spots : for example, this position PP3 is a 3^0 : G(PP3+) has three options to 0^1, 1^1, 2^2  and none of them is a misere lost x^0, so it is a lost in misere game G(PP3-)=0 hence the notation 3/0 or 3^0.

 a 0P3 = 1^1 a PP3 = 3^0

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.

• # The Game

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).

 I play 13(32)31[23-27] Danny plays  1(33)30[28,29] I play  23(34)32 value 5^5 Roman's right move

Danny going to simplicity decides to transform this PP5 which is a 4^4 into a 1^0 by 23(35)23[24-26].
He leaves 4 biospheres :
- 11 spots = 1^1
- 9 spots =1^0
- 3 spots (1^0) + two pivots and a spots with option 0P1 which is a 1^0 hence a 0^1 : 1^0+0^1=1^0 (0^1 is the neutral element to this addition)
- a 0P2 = 1^0
1^1+1^0+1^0+1^0 = 0^0 because of the firm component composed by the sphere with 11 spots, the misere game has same value as the normal game.
I must try to break this firm component without making a UT:
2(36)2[3], this creates a 1L9 which is a 2^2, but I hope that Danny will not notice the disappearance of the firm component and will continue to play normal sprouts...

 Danny plays  23(35)23[24-26] I play  2(36)2[3]

But by 2(37)36[4], Danny makes a 1P8 which is a 1^1 because of the strong influence of 8 which is a 0^0 :
1^1+1^0+1^0+1^0 = 0^0 again.
I will transform it to a PP8 which like PP2 is a 3^3 and test again Danny's knowledge.
My hope is that by having a big grundy number I get more chances to get in return a 1^0 or a 2^2... : 5(38)37

 Danny plays  2(37)36[4] I play 5(38)37

Nothing to scare my opponent : 5(39)5[6-12] always trying to get a firm component, 7 spots being a firm 0^0, the extra-move from the pivot to the 38 spot changing it to a 1^1.
Would the PP7 have been worth trying? I should ask Danny what he thinks about it. News : Danny showed me PP7 is not a 0^1 because of its option
6(41)6[7-11] which is a winning 1^1. So update in the table PP7 to 0^3...
But nevertheless I played a 3L3 trying to complexify the position with the external living spot : 6(40)6[7-9] ?

 Danny plays   5(39)5[6-12] I badly try to shake things by  6(40)6[7-9]

And then the game proceeded 6(41)40[10,39] 39(42)41 34(43@27)35 11(44)12 11(45)11 II with no hope of coming back.

Thanks to Peter Alfeld for his nice Sprouts java applet used for these drawings :  http://www.math.utah.edu/~alfeld/Sprouts/
Thanks to Roman and Danny for their patience and some useful corrections,
Thanks to Josh for giving me exact normal grundy values for PPi up to PP5 and Pivots and Loops up to three spots, computed by his AuntBeast program.