* Not the nimber 1, corresponding to the one-counter heap, but a half-tame game as 4 + 4 game has value 0^0. News : in 2007 it was shown both by Josh Purinton and Danny Purvis that the 4 + 4 + 4 game is a misere losing position. There are strong evidences that this pattern continues for any sum of 4 spots components.

If only one number is provided, it is for expected Normal Spague-Grundy value.
3^0(0^1,1^1,2^1)+3^3(0^1,1^0,2^2) 3^0(0^1,1^1,2^1)+2^2(0^1,1^0)=(3^0,2^1,0^4,2^2,3^3)=1^5 2^1(0^0,1^0)+1^1(0^0)=(1^1,0^0,2^2)=3^3
(3^3,2^2,1^5,3^0,2^1,1^5)=0^4         3^3(0^1,1^0,2^2)+2^1(0^0,1^0)=(3^3,2^2,2^1,3^0,0^4)=1^5 3^0(0^1,1¨1,2^1)+0^0(2^2)=(0^0,1^1,2^2,1^5)=3^3
3^0(0^1,1^1,2^1)+1^1(0^0)=(1^1,0^0,3^3,2^2)=4^4

3 spots = 1^0

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

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 a_is and the b_is : 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

Table An or Anago n is for position you get after first move in fianchetto (connecting a spot to another one). For example A2 is for the position reached after :
2 1(3)2. So the Loop Table combined with An table gives you all the values after Left first move.

  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 compoonent 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' componeent 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.