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Paging Matt Ryder!!!

Posted By: Matt Ryder
Date: Monday, 11 January 2010, at 5:38 a.m.

In Response To: Paging Matt Ryder!!! (Nack Ballard)

Sorry, I think I had my pager switched to silent! :-)

I may have been a little overzealous in my editing of Nack's HTML. Apologies to IE users. I'll try again here with reformating Post #3.

I've tried to do some cross-browser testing, but let me know if the diagrams in Nack's edited post below DO NOT look okay. (They should hopefully now be corrected for IE, Firefox and Safari; anybody use Chrome?)

Nack, the trick to fixing these turns out to be quite simple. Search and replace both src="../Images/n_high.gif" and src="../Images/n_low.gif" in your HTML with an empty string (in other words, remove all these references to the upper and lower pip numbers entirely).

Nack's Post #3
-----------------------------------------

For reference, here are Naccel 2's post #1 and post #2. The current post is Naccel 2 -- post #3. I appreciate all the bgonliner feedback and participation.

Thanks to Petter and Matt R. for making me aware that my diagrams aren't properly viewable with Firefox or Safari, and sorry about that. Please use (Windows) Internet Explorer when reading Naccel posts #1, #2 and #3.

Meanwhile, Matt is figuring out which html tags to amend so that I can accommodate other browsers. [Matt, you're not quite there: the White bearoff tray image is missing (at least it is in IE), causing the top half of the board to be off-pip. When you've solved that, please e-mail me the Gnu-export html for the opening position of backgammon with point numbers and all extraneous stuff removed so I can compare to mine.]

Thanks to Lucky Jim for his submission. In the future, it would be better for the readers (who might want to take a pre-crack at counting) if you post your position in Blue/White or at least dark-near / far-light, and/or have the inner boards on the left, though if you can't or it's inconvenient then please submit anyway; I will always rediagram the position in my post.

Below is Jim's position (which he informs us is from Ballard-Sylvester U.S. Open June 1989, though I'll have to confess my memory isn't what it used to be -- sigh).

 '1X2X '2X4X1X '1O1O ' '

2X ' ' '3X5O '3O '3O2O '


First, let's consider Blue's position. As always, the checkers on n0 (the Naccel zero point, trad 6pt) are invisible. You need count only Blue's outfield checkers. These come in two groups, which I'll address separately.

 ' ' ' ' ' ' ' ' ' ' ' '

 ' ' ' ' ' ' '3O '3O ' '
Six-sym . . . . . . . . n0 . . . . . . . n33


The first group (shown above) is a "six-sym" (six checkers symmetrically arranged around n3), for a count of 3. Jim and Ian nailed this one, perhaps remembering the second and third diagrams here.

For the second group, Ian and Jim shifted the remaining near-side point one point to the right, and a far side blot 2 pips to the right to compensate, then moved both far side checkers to S1 to get a four-stack on S1 plus the 5 pips adjusted. That works, but let me show you a faster way.

As a prelude, allow me to present the "mirror."

 ' ' ' ' ' '2O ' ' ' ' '

 ' ' ' ' '2O ' ' ' ' ' '
Mirror4


You could easily count this as two checkers on S2 (Super-2), i.e., 2 x 2 = 4, and ignore the two zero-count checkers on S0 (Super-0, same point as n0, the Naccel zero point).

Alternatively, you can count this as a mirror. A "mirror" can be composed of blots, points, or even stacks of three or more checkers, as long as they are of equal size. Mirrors count the same as the total number of checkers used; here, for example, there are four checkers, so the count is 4.

The only wrinkle to remember is that the near-side member of a mirror is always 1 pip visually closer to the bear-off tray. (It may help to note that the colors of the points match.)

If we slide the above mirror to the right a few pips, we get the commonly arising mirror below:

 ' ' ' ' ' ' ' ' ' ' '2O

 ' ' ' ' ' ' ' ' ' '2O '
Midpoint Mirror4


Again, there are four checkers in the mirror, so the count is 4. (Btw, these points can be shifted 1 pip towards each other to S1, creating a four-stack there.)

You can probably see where I'm going with this. In the game position, two of the checkers are already on the near-side point of the above diagram. You need only move the far-side checkers 5 pips forward to get the midpoint mirror.

In other words, Blue's entire count has two components. The first is a six-sym around n3 (first diagram below) = 3. The second a the midpoint mirror plus 5 pips (second diagram below) = 4(5).

 ' ' ' ' ' ' ' ' ' ' ' '

 ' ' ' ' ' ' '3O '3O ' '
Six-sym3


 ' ' ' ' ' ' ' '1O1O ' '

 ' ' ' ' ' ' ' ' ' '2O '
Midpoint mirror + 5 pips 4(5)


In short, Blue's count is 3 + 4(5) = 7(5).


Before we count White, I'll reveal another trick. Just as six checkers on a point count the same as the point number, THREE checkers on a point count HALF the point number.

For example, consider each of White's triplets (three-stacks) below.

. . . . . . . . . . . . . . . n0 . . . . .n2 (count of 1)
 ' ' ' ' ' ' '3X ' ' ' '

 ' ' ' '3X ' ' ' ' ' ' '
. . . . . . . . . . . .n14 (count of 7)


For reference, I have marked the relevant Naccel ("n") point numbers. You can either count up from n0, or subtract 6 from the trad point numbers you already know until Naccel point numbers become second nature to you.

The trad 8pt is n2 (the Naccel 2-point). White's TRIPLET on her n2 counts HALF of 2, which is 1.

The trad 20pt is n14 (the Naccel 14-point). White's TRIPLET on n14 counts HALF of 14, which is 7.

After you've seen these triplets a couple/few more times, you'll just remember them as "1" and "7" without even having to divide the Naccel point number by 2.

Taking special note of White's latter triplet (count of 7), let's go back to the main position:

 '1X2X '2X4X1X '1O1O ' '

2X ' ' '3X5O '3O '3O2O '


White's two checkers on S3 (Super-point 3) count 6. Her triplet (as just explained) is 7. Adding those together, White's back checkers count 6 + 7 = 13. [I've seen this five-checker formation enough times that to me it is simply "13" (no addition needed).]

For White's near side, remove three of her irrelevant n0 (trad 6pt) checkers, leaving one. Ignoring the outside blot on n1 for now, you have...

. . . . . . . . . . n-2
 '1X2X '2X1X ' ' ' ' ' '

 ' ' ' ' ' ' ' ' ' ' ' '
Six-sym


This is a six-checker symmetry around n-2 (the Naccel minus-2 point), for a count of -2. If the symmetry is less than obvious to you, review the explanation starting at the second diagram of post #2.

Now, I'll remove the Blue and White checkers we don't need on n0 (trad 6pt), and you pretend that you're counting the position for the first time. Count White and then Blue, knowing what you know now:

11(1)
 '1X2X '2X1X1X '1O1O ' '

2X ' ' '3X ' '3O '3O2O '
7(5)


White's count is 6 + 7 = 13 in the back, and -2 for the six-sym in her inner board. Add the n1 pip, for a total of 11(1).

Blue's count is a six-sym around n3 (count of 3) plus a (midpoint) mirror+ of 4(5), for a total of 7(5).

Once you "get" how it works, count both colors in exactly this way a couple/few more times so that you can get a sense of how fast Naccel counts are. It's just a matter of knowing the tricks and spotting the formations.

The difference between 11(1) and 7(5) is 3(2). That is, Blue leads by a couple pips more than 3 supes (super-pips); 20 pips, to be exact.

(If for some reason you want to convert to trad, the counts (with the 6pt checkers added back to the board, of course) are 11*6 + 1 + 90 = 157, and 7*6 + 5 + 90 = 137.)

[For White's position (after counting S3 x 2), Jim and Ian moved the three-checker anchor out to S2 (count of 6); to compensate that 6-pip adjustment, they moved n-3 (trad 3pt) back to n0 (trad 6pt), causing it to disappear. Then they canceled a checker on n1 and n-1, leaving only a blot on n-1 and n-4 left to count. This achieves the correct count but is, of course, more work than counting the three-checker anchor where it is.]

Next position?

Nack

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