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Naccel -- post #6 (midpoint formations)

Posted By: Nack Ballard <nack2000@sbcglobal.net>
Date: Thursday, 21 January 2010, at 6:57 p.m.

Welcome to "Post #6," the ninth of the Naccel 2 series.

For review, click on the following:

FAQ, post #1, post #2,
post #3, post #3x, post #4,
post #4x, post #5, post #5x.

The basic squads you have learned thus far are the six-stack (post #2, second diagram), triplet and pair (post #4x).

There are also shift-variants of the above squads, some of which you've seen (various six-syms and the wedge), but we'll put those aside for now.

Today, we'll look at another basic squad called the "block." Whereas the others mentioned above use six, three or two checkers, this squad uses four:

 '2O2O ' '5X '3X ' ' '2O

2X ' ' ' '5O2O2O ' ' '5X

Position with two Blocks14(2)

The block just to the right of the Naccel 0pt (trad 6pt) arises after Blue plays a standard opening 61.

To count a block -- or indeed any two points on the board -- add together the point numbers and divide by 3. The near-side block occupies the (Naccel) 1pt and 2pt, so: 1 + 2 = 3, divided by 3, gives you a count of 1.

The block on the other side of the board, formed by the four back checkers, is on the 16pt and 17pt. Summing these point numbers gives you 33, and dividing by 3 gives you 11. (Btw, blocks always generate odd-numbered counts.)

[An alternate counting method for blocks: Count three checkers as being on the nearest Super and the other as being on the second nearest Super. So, here, the near block counts (0 x 3) + 1 = 1, and the far block counts (3 x 3) + 2 = 11.]

As always, the 0pt checkers are invisible. Blue's entire position counts 11 + 1 for the blocks, plus 2(2) for the mid, for a total of 14(2).

Next, let's have a look at some useful ways to combine two checkers on the midpoint with a near side point:

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

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


You recognize this formation as a "mirror." (If not, review the third and fourth diagrams here.) The number of total checkers in a mirror gives you the count, so this mirror counts 4.

To reach the next formation, we will "zig" (move forward) the near-side point 3 pips. Two checkers times 3 pips is a total of 6 pips, or one supe, which means the overall count of 4 is reduced below by 1, to 3:

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

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

Zig mirror3

This formation is called a "zig mirror" or "zig" for short. A zig counts 3, one less than a regular mirror. [Zig is both a noun that is short for zig mirror, and a verb that means to move a point 3 pips forward.] For reference, if you drop the far-side point of a zig mirror straight down, the near-side point always leads it by 4 pips.

You can also count a zig in the same way you count a block (or any two points): add the point numbers and divide by 3. So, here (7+2)/3 = 3.

This zig is a combination of a stripped midpoint and stripped 2pt (trad 8pt). Also, as early as the opening roll, a double zig occurs when an opening 5 is brought down, creating a stack of four checkers on each point; the entire right side then counts 6.

Note that if you "hop" the two midpoint checkers down to the (Naccel) 1pt, you end up with the near-side block in the first diagram of this post. That block counts 1, and adding the 2 hops imagined here gives you a count of 3.

This hop example, and the series of zigs we're performing, help illustrate that all Naccel formations can be built and rebuilt upon board symmetries and simple, logical techniques. (To review hopping, see the twelfth diagram here.)

Let's zig the front point again, thereby further reducing the count by 1:

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

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


This position is called a "diagonal mirror," or "diag" for short. A diag counts 2, which is 1 less than a zig (previous diagram), or 2 less than a mirror (two diagrams back).

An easy way to reconcile this count of 2 is to hop the two midpoint checkers (+2) to the bar point, creating a little poof. Another way is to shift the midpoint to S1 (count of 2) and the -1pt to S0 (count of 0).

If you're good at visualizing, try this: After playing 31 (making the above near-side point) from the opening position, and ignoring the two back checkers on S3, you basically end up with a combination of the last two diagrams. The -1pt (trad 5pt) and 2pt (trad 8pt) each have two checkers; these can be coupled off with four (of the five) checkers on the midpoint, for a +2 diag and a +3 zig.

Let's zig the front point forward one last time:

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

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


The previous formation counts 2, so this one counts 1. If you like, you can call this formation a diagonal zig mirror, or dizig for short. However, the only other formation with the same relationship, one space to the left, rarely arises; so rather than fill your head with the term dizig (or even diag for the mirror relationship in the previous diagram), it might be easier just to remember this specific midpoint formation as 1 (and the previous one as 2).

Okay, now let's add a specially selected "pair" to the above formation. (If you are not sure what a pair is or how to count it, review the second diagram here.)

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

 '2O2O ' ' ' ' ' ' ' ' '


And here it is. Midpoint plus -4pt counts +1, and the pair on the -3pt counts -1. This important six-checker formation counts zero: I call it a midpoint poof or "midpoof" for short.

If you shift the two low points outwards, so that Blue instead owns the -5pt (trad 1pt) and -2pt (trad 4pt), that is also a midpoof, though it doesn't tend to arise as frequently in practice.

Commit the diagrammed midpoof to memory, and you will encounter it often in your Naccel adventures. We'll even have a chance to use the midpoof later in this post.

Now we'll look at midpoint formations on the far side of the board.

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

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

Far-side Reflection8

This "far-side reflection" was introduced here. It counts the same as if you were to stack all four checkers on S2 (the point around which they reflect). In other words, the count is double the number of checkers you see, which is 8.

Okay, let's zig (move forward) the back point, reducing the count by 1:

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

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


I call this formation "midgold" because it combines the midpoint with the golden anchor. It counts 1 less than the far-side reflection of 8, which makes it 7.

There are many other ways to reconcile midgold as a count of 7. To name a few: (a) Add the point numbers 14 + 7 and divide by 3. (b) Shift the golden anchor back a point to get a pair (count of 5) and the midpoint forward to S1 (count of 2); (c) Shift one midpoint checker forward 7 pips to the 0pt and the other back 7 pips to the golden anchor, creating a 7-count triplet there. (You can review triplets and pairs here.)

Midgold comes up a lot, so you would do well to remember that its count is 7. We'll even have a chance to apply midgold later in this post.

Let's zig the back checkers forward again, but we'll skip over the next spot because Opp typically occupies her 2pt (trad 8pt). Instead we'll zig twice, reducing the count from 7 to 5. Can you visualize the resulting formation, and what is another way to count it (other than 7 - 2)? Answer below.

 ' ' ' ' ' ' ' ' ' '2O2O

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


Having hopped (or zigged twice) the gold part of midgold, you now have a "block." If you don't remember exactly what a block is, review the first diagram of this post.

I call this block "midblock," because it contains the midpoint. To count it, sum its point numbers 8 + 7 and divide by 3, for a count of 5.

When you have four (or more) checkers on the midpoint and you need a back 2-pip countershift, the midblock is very handy.

Let's look at Lucky Jim's most recent submission:

1X1X2X '2O3X '4X2X ' '2O

 ' '2O2O '2O1O3O1O ' '2X

Poof, and Midgold7

First, let's count Blue.

Maik quickly spotted Blue's near-side poof: simply move the 2pt spare back, covering the 3pt, and lift the 1pt blot to the 0pt. Poof! (If you don't see the poof, please review the last two diagrams here, the last diagram here (especially), the second diagram here, and the fourth diagram here.) There's nothing on the near side to "count."

That leaves the far side. Aha, "midgold" = 7; that's Blue's entire count!

Let's repeat the diagram with point numbering from White's perspective:

1X1X2X '2O3X '4X2X ' '2O

 ' '2O2O '2O1O3O1O ' '2X

For the White count, I recommend this simple 1-pip shift:

Poof, and Six-stack plus 1 pip2(1)

 '2X2X ' '3X '5X1X ' ' '

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

Do you see the midpoof? It is the six-checker formation composed of the midpoint and the four low-point checkers: count of zero. And of course the (invisible) 0pt checkers also count zero.

You need count only the six checkers on the top right. This is simply a six-stack on the (Naccel) 2pt plus 1 pip. White's entire count is 2(1).

(If you don't know what a six-stack is or how to count it, review the second diagram here. If you don't understand the midpoof used in White's count or the midgold used in Blue's count, scroll back a few diagrams in this post.)

Next position?


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