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Naccel 2  post #4 (improved)
Posted By: Nack Ballard
Date: Monday, 18 January 2010, at 9:13 a.m.
This is post #4 of the Naccel 2 series. If you would like to review or get caught up, here are post #1, post #2 and post #3.For this post (#4), Lucky Jim submitted two positions.
You know about reflections on the near side. (If not, click on "post #1" above and examine the fourth and fifth diagrams.) Cancellations around S0 (trad 6pt) count zero, which means there are a lot of checkers you don't have to count at all.
There are also reflections on the far side. Because they reflect around S2 (Opp's bar point) instead of S0, they count 2 per checker  that is, count the number of checkers you see and double it.
With the midpoint featured, this is the most commonly arising farside reflection:
Doubling the number of checkers, 4, gives us a count of 8. (If these were blots instead of points, the count would be only twice 2 = 4.)
Farside reflection 8
Let's apply that knowledge to the Blue side of Jim's first position:
Blue's checkers on the near side reflect around the 0pt and therefore cancel out except for the 2pt checker (which counts 2 pips). So...
17(2)
8 (reflection), plus 9 for S3, plus 2 pips for the blot, gives a count of 17(2).
[Jim and Ian counted similarly, except they didn't yet know about farside reflection. Instead, they shifted to S3 and S1, which is fine but it adds a step of visualization.]
White's count is even easier (as Jim and Ian found as well), so it's hardly worth a new diagram, but I'll spend one anyway so that the point numbering is from White's point of view.
White's entire near side goes poof (reflection/cancellation around the 0pt), so the entire count is 3(3).
3(3)
The midpoint is nice for leftover counts because it's trivial to multiply 1(1) for each checker, and if it has as many as six checkers it just counts 7. (Naturally  a sixstack on the 7pt = 7.)
In short, the count is 17(2) to 3(3), a difference of about 14 supes (superpips). [Round supes (which can be thought of as crossovers) are a good way to assess Blue's timing. As 12 of the 14 supes are accounted for in bringing home the vitaltokeep anchors, that only leaves 2 spare supes, which are roughly offset when White's 6s then 5s are killed bearing in later. And the S3 spare might not escape and get its 3 supes. Yikes, looks like a monster pass.]
If you are unclear as to why Blue's near side counts only 2 pips and White's near side is a giant poof that doesn't need to be counted at all, review the fourth and fifth diagrams here.
Below is Jim's second position. There are many ways to count Blue (I'll show you my count later) but I'll give you a way that works well with what you know so far.For Blue, try this easy mental shift:
Now, here's the trick that you need: Conjure a new checker on the 0pt. Remember, the 0pt is not only a black hole but also a white hole: even when the 0pt looks vacant, phantom checkers always exist there for your convenience. Now, LIFT that conjured 0pt checker onto the 1pt, which is a 1 pip adjustment.
8(1)
So, your count: Four on S3 counts 12, sixsym (around the 3pt) is 3, the engineered sixstack is 1, and add the 1pip lift needed to get that stack. That's a net of 8(1).
Now let's count White. Her position again is...
Try this mental shift (2 pips forward, 2 pips back):
The stack is 3, the sixsym (threeprime) is 1, and the three spares on the 1pt are 3 pips. Total of 4(3). In case you want to convert, 4*6 3 +90 = 63.
4(3)
Do you see a similarity with what we did in Blue's inner board and in White's inner board? Sixstacks, and sixsyms (which are often threeprimes), are powerful weapons.
When both Blue and White checkers are home, clustercounting is relatively at its best. With only one quadrant to count, Naccel's edge disappears. But if you want to become lightning fast, it pays to practice counting all positions with Naccel (instead of lazily falling back), even though at first it may be slow and even confusing in certain situations. That's to be expected  you're still tying your skatelaces at this point.
Finally, I promised to show you the way I counted Blue (noting that the alternative method suggested above is perfectly legit). To understand my count, you'll need two key pieces of information.First, let's review the "mirror." It was introduced here (third and fourth diagrams). Below is another example:
For a regular mirror, the nearside point is (as you can see) a pip closer to the bearoff tray (than the farside point would be if it were dropped straight down). The number of checkers = 4, and thus the count = 4.
Mirror 4
For advanced anchors (or blots), you can often use regular mirrors to offset checkers in the home board. But for deep anchors, you will tend to use "zagmirrors" (or "zags" for short).
Starting with the mirrored point shown above (count of 4), if you zag (move) back either point back 3 pips, you increase the count by 6 pips or 1 supe, to a count of 5, and you have yourself a zag mirror.
If it is the farside point that is zagged, we get the zag mirror shown below.
Whereas with regular mirrors, the nearside point is 1 pip closer to the bearoff tray; with zag mirrors, the nearside point is 2 pips further away.
Zag Mirror 5
As stated, zagmirrors count 5. The occupied points are opposite colors, which helps remind you that the count is odd.
Now for the second part of the explanation:
A triplet (threestack), which was introduced here (see seventh diagram), is a basic squad (i.e., a group of checkers within a quadrant field). An example is shown below.
As was explained in Post #3, you count a triplet by dividing the number of the point on which it resides by 2. This triplet is on the 2pt, so the count is 1.
Triplet 1
The triplet has a variant squad, which can be obtained by moving two of the checkers 1 pip towards the nearest Super, and the third checker 2 pips in the other direction, to compensate. It looks like this:
This squad is called a "wedge" (it is shaped like a door wedge). To count a wedge, take the point number that is in between the blot and point but closer to the latter (or just shift to it, thereby reenacting a triplet on the appropriate point, if you need the visual help), and divide it by 2.
Wedge 1
Here, the inbetween point is the 2pt. Dividing by 2 gives you this wedge's count of 1.
You can also double the height of the wedge which gives you a "double wedge," and this alsocommonly seen formation counts the same as the inbetween point number (e.g., if you double the height of the wedge shown above, the count is 2). Both the wedge and the double wedge can be great counting tools.
Now that you know what a zag (zag mirror) and a wedge are (or if you don't remember, then review), I'll show you how I counted Blue in Jim's position, repeated below.
To count Blue, all I did was to move one pip forward, from the 0pt to the 1pt, and sum the count of two formations:
Double Zag 10
That's 10 for the double zag, 2 for the double wedge, plus the 1pip lift: my count is 8(1).
Double Wedge 2
The checker left on the 0pt, is, as usual, invisible.
Next position?Nack

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