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Naccel 2 -- post #5 (improved)

Posted By: Nack Ballard
Date: Monday, 18 January 2010, at 4:45 p.m.

Welcome to Naccel 2 -- post #5.

If you would like to catch up or review earlier material:

Post #1, Post #2, Post #3,
Post #3x, Post #4, Post #4x.


As always, Naccel point numbers are labeled in white, Super (Super-point) numbers in black.

Iíll start by showing you how to count a six-point prime:


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

 '2O2O2O2O2O2O ' ' ' ' '

Six-prime-3


To count any six-prime (including a closed board), simply add together the endpoints. For the prime shown above, the endpoints are the (Naccel) -4pt and 1pt. Adding these together, -4 + 1 = -3.

[You can also choose to count this formation as two six-syms: one around the -3pt and the other around the 0pt. This essentially adds together the second and fifth points of the prime. Yet another option is to sum the middle points: -2 and -1 sum to -3.]

Letís apply that nugget of knowledge to the first of the two positions submitted by Jim:


 '2X1X2X '2X3X2X2O ' ' '

2X2O2O2O2O2O1O ' ' '1O1X


Naccel seems to get more than its fair share of luck. Here, you can kill two birds with one stone with the following 4-pip shift:


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

 '2O2O2O2O2O2O ' ' ' ' '

4(1)


The count is -3 for the six-prime (as we just learned), +4 for the two checkers on S2, and +3(1) for the roofer. Total of 4(1).

[If you know (as I did) that an S3 blot (+3) offsets/poofs this six-prime (-3), the count is even shorter: 4 for S2, plus 1 pip to enter the roofer.]

On to Whiteís position. It can be divided into two groups. The first is

Poof0


 ' '1X2X '2X3X2X ' ' ' '

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


Can you see this is a poof? Just move the blot and spare equally towards each other, filling in the vacant point. If you donít recognize the resulting five-prime as a poof (zero count), review the fourth and fifth diagrams here and the second diagram here.

This means White has only five checkers (shown below) that you actually need to count.


 '2X ' ' ' ' ' ' ' ' ' '

2X ' ' ' ' ' ' ' ' ' '1X


I happen to know that the Naccel 18pt (S3) and the -4pt are equidistant from the handy midpoint (7pt), so I just shifted (more like swept) the left four checkers to join the fifth on the midpoint, for 5(5). I am impressed that Ian found this solution as well.

That "five-sym" mixed fraction count is either advanced or hard to find, so instead I'll suggest to you this 2-pip shift:

Zag plus 5 pips5(5)


 ' '2X ' ' ' ' ' ' '1X '

2X ' ' ' ' ' ' ' ' ' ' '


Do you recognize the "zag" (zag mirror)? Take a look at the second to last diagram here, isolate just the left zag (of the double zag there), and notice that White's zag here is the upside-down image. Practice spotting mirror and zag-mirror formations for both sides -- it gets a lot easier.

A zag counts 5, plus 5 pips for the blot; White's entire count is 5(5).

Recapping both sides: Blue is ahead 4(1) to 5(5), which is a lead of 1(4), or 10 pips.


Okay, now for Jimís second position:


1X ' ' ' ' ' ' ' ' ' ' '

1O3O2O2O3O3O '1O ' ' ' '


For Blue, a trivial 1-pip shift will do the trick:


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

2O2O2O2O2O4O '1O ' ' ' '

Closed board plus 2 pips-5(+2)


A closed board is a standard count of -5 (or sum the end points: -5 + 0 = -5). Adding 2 pips for the outside blot yields a count of -5(+2).

Now, Whiteís position:
-11(+2)


1X ' ' ' ' ' ' ' ' ' ' '

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


Simplest, perhaps, is to bear off 4(1) + (1) = 4(2) and subtract 15 (i.e., 1 for each borne off checker). Alternatively: enter the roofer 2 pips, creating a mirror of 2, thatís 2(2) so far; then subtract the 13 borne off. Either way, White has a count of -11(+2).

Ian chose to view Whiteís two checkers as symmetrical around the midpoint -- equivalent to having them on the midpoint = 2(2). Then he subtracted the 13 borne off, achieving the same -11(+2) result.

(Ian's choice relates to the method described for bundling five White checkers, four diagrams back. The shift is easy to see here if you "hop" the roofer +2 and back-hop the other checker -2, so that they meet on the midpoint. Hopping is a valuable visual aid that will be explained later in this post.)

[When the set of checkers as a whole is closer to S-1 (the bearoff tray) than to S0, a reasonable option is to bump up the Supers by 1 so that S0 is the bearoff tray). However, I strongly advise against adding this wrinkle until your regular S0 counts feel confident and natural.]


This time, we have a third position, submitted by Ian.


2X '3X3X3X3X ' '1O ' ' '

3O5O5O ' ' ' '1O ' ' ' '


For Blue, again a 1-pip shift is all thatís needed:


2X '3X3X3X3X ' '1O ' ' '

4O4O4O1O ' ' '1O ' ' ' '

Mirror plus twelve-sym, minus 2 pips-6(-2)


The mirror outside is 2, the two six-syms (twelve-sym) around the -4pt count -8, and the inside blot is (-2), yielding a count of -6(-2).

If you donít see how two six-syms are symmetrical around the -4pt, picture two three-primes on top of each other (for example). If you donít understand why a three-prime is a six-sym or donít know how to identify or count a six-sym, review the second and third diagrams here, the second diagram here, and the fifth diagram here. If you don't know how to identify or count a mirror, review the third and fourth diagrams here and the eighth diagram here.

[Alternatively, you can poof the 2pt and -2pt blots, count the twelve-sym as -8, and add back in 1(4) for the far-side checker, resulting in -7(+4). That agrees with the -6(-2) count above.]

Before we count Whiteís position, Iíll introduce you to a technique called ďhopping.Ē

Visualize three hops3


 ' ' ' '1X ' ' ' ' ' ' '

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


There are (at least) three ways to count this two-checker formation as "3." One is to learn it as an automatic 3 count. Another is to recognize it as a diagonal mirror I haven't shown you yet (and complicated by having to visualize the roofer where it really is -- one point to the left of S3).

The third way is to "hop." A hop is a visualized movement of one supe (6 pips), counting +1 for each forward hop or -1 for each backward hop. Having seen and played a lot of double 6s over the years, we all hop naturally.

By hopping around the board in the above diagram, White's roofer will land on her own bar point (Naccel 1pt). How convenient! This creates a little poof, the checker on the 1pt and -1pt cancelling each other out. (I trust you can visualize the position after White's triple-hop without the aid of an after-diagram.)

Returning to the main position:


2X '3X3X3X3X ' '1O ' ' '

3O5O5O ' ' ' '1O ' ' ' '


Here, you can reach an easy count in two movements (totalling 3(3)): one is to move White's roofer to her bar point in 3 hops. The other is to close her board with 3 pips. Like this:

Closed board (and a tiny poof)-5


2X2X2X2X3X3X1X ' ' ' ' '

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


As explained earlier in this post, a closed board counts -5 (and the extra three checkers on the 1pt, 0pt and -1pt are a tiny poof). Accounting for the movement of 3(3) it took to get here gives you a net count of -2(+3).

Hopping will open whole new worlds of understanding and speed counting to you. In this case, if you know/remember that roofer + -1pt checker = 3, great. If you don't know or forget, then hop, and you'll be more likely to remember the instant 3 count of that little formation for next time.

Recapping, Blueís count is -6(-2) and Whiteís count is -2(+3). Blue leads by 4(5).


Next position?

Nack

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