                            |
| |||||||||||||||
| XGID=-b---cE-C-CB---AA-adb-ba--:0:0:-1:62:18:4:1:19:10 | ||||||||||||||||
 to play 62 | ||||||||||||||||
is Player 1 eXtreme Gammon Version: 1.11
Hi Nack
Thank you for replying to my earlier message and taking up your invitation to post positions I have posted the one above.
Based on what you have told us so far and with some guess work inspired by your Naccel 1 article I have had a go at some skating on my own. Be warned – it is not very elegant.
Here’s what I did
White
Firstly I understand I can ignore all the checkers on the 6 point (the Naccel 0-point)
My eye jumped to the two sets of 3 checkers on both the 8 and 10 points. This is equivalent to 6 checkers on the 9 point. In Naccel the 9 point is N3 and six checkers on the Naccel 3 point gives a count of 3.
I thought I would then mentally move the two checkers on the 11 point back 1 pip each to the 12 point. Two on the twelve point gives a count of 2.
To compensate for the shifting back of 2 pips I then thought I would need to move one of the remaining two checkers in black’s outer board up 2 pips. (I mentally moved the white checker on the 15 point to the 13 point).
I then mentally moved these two checkers still in Black’s outer board 5 pips (remembering one of them is notionally now on the 13 point) so they could join their imaginary friends on the 12 point. This gives a count of a further 2 (because 2 more checkers are on the 12 point) and there is the count of 5 pips they had to travel which has not been compensated for elsewhere.
This give me a total Naccel count for white of 7 (5)
Black
There are two black checkers on the 24 point so that gives a count of 2 X 3 = 6
I mentally shifted the 3 Black checkers on the 20 point forward to the 18 point – by virtue of being on the 18 point this gives a count of a further 6 ( 3 x2 ) plus there is the journey of 6 pips I have used to go forward
Going to Black’s Home board
I can ignore everything on the 6 point and I can ignore the checker on the seven point because I am also going to similarly ignore one of the checkers on the 5 point (these two checkers cancel each other out).
I mentally shifted the two checkers on the 3 point to the 6 point – this is a movement of 6 pips backwards and compensates for the earlier same pip movement in the opposite direction from the 20 point to the 18 point. Once on the 6 point the checkers can be ignored.
This leaves one unaccounted checker on the 5 point and one unaccounted checker on the 2 point.
The Naccel count is
On 24 point 2 x 3 = 6
Plus
On (notional) 18 point 3 x 2 = 6
But we need to factor in the remaining pips -
As the remaining pips need to be brought backwards to the Zero point these 5 pips need to be subtracted and a count of 12 less 5 pips is equivalent to 11 (1)
I therefore make the Naccel counts White 7 (5) and Black 11 (1)
I would obviously welcome any corrections and/or observations. There are no doubt many ways to make the counts easier.
Incidentally the position comes from the match you played against Joe Sylvester in the US Open in June 1989 - but you probably recognised that in anycase :-)
Best regards,
Jim