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Naccel 2  post #8 (midblot magic)
Posted By: Nack Ballard
Date: Thursday, 28 January 2010, at 8:33 a.m.
Welcome to Post #8. You can review earlier Naccel posts by clicking on the links below.
FAQ, Post #1, Post #2, Post #3
Post #3x, Post #4, Post #4x, Post #5
Post #5x, Post #6, Post #6x, Post #7I'll start this post by demonstrating a few combined integer counts.
When checkers occupy Supers, you can't go far wrong counting them as is. On the other hand, you will so often find one or two checkers on S3 (their starting location) that it is worth knowing the counts that combine S3's with other common integer formations.
For example, a commonly arising position after 51$42S43@ is shown below:
The S3 checker counts 3, and the nearby pair counts 5. This three backchecker formation is worth committing to memory as an instant count of 8.
S3 = 3, Pair = 5; three back checkers count 8
If you add an S3 checker to Blue's formation, the count of his double anchor is 11. The larger formation is less common but still occurs frequently enough that I made a point of learning it.
For a still larger example, see White's five back checkers in the eighth diagram here. The S3 point (count of 6) and the n14 triplet (count of 7) coexist often enough that over time I've learned their combined count of 13 without really trying.
Not all combined counts need be highsingle or doubledigited. In positions of moderate count, it is typically more desirable to combine a positive count with a negative count, generating a net integer count that is zero (best), small positive (second best) or small negative (third best).
Consider the second diagram here. The back pair counts 5, and the front pair counts 1; easy enough. But if you combine them as a mirror of 4, you don't have to subtract 1 from 5: you've saved a step. (If you don't know how mirrors work, review the third and fourth diagrams here.)
Having conveyed the concept, I will now show you some small combined integer counts that frequently arise.
Each of the formations below combine the S3 point (count of 6) with a sym of negative count in the inner board. A "sym" (short for sixsym) is any six checkers symmetrical around a point, though I have chosen to illustrate the most common sym, the threeprime. (If you don't know how to count a sym, review the second and third diagrams here.)
The four diagrams below are a series. Traditional point numbers are added in an extra row under each diagram (the purpose of which you'll see in a minute).
S3 point + t5 sym = 5 5
S3 point + t4 sym = 4 4
S3 point + t3 sym = 3 3
What is cute and memorable about this series is that the traditional point number ("t") on which the sym is centered is the count of the formation! In the above four diagrams, the sym is centered on t5, t4, t3 and t2, making the counts 5, 4, 3 and 2, respectively.
S3 point + t2 sym = 2 2
If there is a third checker on S3, add 3 to each of the above counts. If there is a mere blot on S3 (more likely), subtract 3; for example, you can count the secondtolast diagram here as t3  3 = 0 (among other ways to count it) if you haven't already learned it is a poof.
Next, I will cover "midblot" formations. To be clear, I'm not referring to a blot on the midpoint, but rather to a formation with some number of checkers on the midpoint combined with a blot elsewhere. There are quite a few midblot formations, but the counts are always easy to determine.Consider the basic formation above. There is ONE checker on the midpoint, and a blot on the FIRST point (n1) in the inner board. The combination of ONE and FIRST produces a count of ONE. (I left a bunch of checkers on S0 only to make clear that you skip the Superpoint  you start with the first point in the field.)
Midblot: One/first = 1 1
Compare to the basic formation below. Here the combination of TWO (midpoint checkers) and SECOND (the blot is on the second point of the field, n2) produce a count of TWO.
Extending the logic illustrated in the last two diagrams:
Midblot: Two/second = 2 2
Three checkers on the midpoint plus a blot on the third point (n3), a description we'll shorten to "three/third," counts 3.
Four/fourth counts 4. (Four on midpoint, blot on n4.)
Five/fifth counts 5. (Five on midpoint, blot on n5.)Let's build on that knowledge to create similar counts when the blot is in other quadrants. From the formation diagrammed above, hop the blot back one quadrant to the outer board, like this:
In this formation, the blot is in the quadrant where S1 (instead of S0) lives. Because of S1, you add another 1 to the count. Putting the pieces of the puzzle together, "two/second" (two on mid, with blot on second point) counts 2, plus 1 (for S1), gives you a count of 3. [White checkers are included in the diagram for orientation purposes only.]
Midblot: Two/second + S1 = 3 3
To practice further, let's increase the number of checkers on the midpoint to FOUR, and therefore slide the blot to the fourth point. (Without this corresponding slide, the count won't remain an integer).
You have FOUR checkers on the midpoint and thus a blot on the fourth point. Add one for S1 being in the quadrant, and you get a count of 5.
Midblot: Four/fourth + S1 = 5 5
Getting the hang of it? Okay, let's backhop this fourthpoint blot to the next quadrant:
The Blue formation has increased again by one: it is now four/fourth plus S2, for a count of 6.
Midblot: Four/fourth + S2 = 6 6
Let's backhop the blot one more time:
Blue's count has increased yet again by one. It is now four/fourth plus S3, for a count of 7.
Midblot: Four/fourth + S3 = 7 7
Below, for reference, all twenty possible midblot formations are shown, in five blocks of four diagrams. (These include the midblot examples already diagrammed.) In each case, the caption reminds you how to count, and the count itself is repeated on the far right of the caption line.
If you start with the top left diagram of any fourgroup and go clockwise, the counts sequentially decrease by 1, or if you start with the bottom left diagram and go counterclockwise, the counts increase by 1. Alternatively, if you choose an orientation in the one/first group and then look at the same orientation in the two/second group, three/third group, etc., respectively, the counts will sequentially increase by 1. The more such counting patterns, connections and morphs of other formations you can spot, the better your understanding of Naccel and the more surefooted your counting will become.
[Note: Two of the formations are marked "uncommon." Blue seldom has a blot on the n2/t8 (trad 8pt) or n0/t6 (trad 6pt) of White, who occupies these points at the outset of the game.]
I should emphasize that these formations are not vital to counting in Naccel. However, they will help you imprint other formations you already know and they are additional tools to help you achieve lightning counts.
This complete set of Midblot reference diagrams doesn't end the post. At the bottom there is further discussion, followed by a fullboard position to count for Blue and White.
One/first + S3 = 4 (also "reflection") 4
One/first + S2 = 3 (also "wide") 3
Above: "One/first" Midblots
One/first + S0 = 1 (also "diag") 1
One/first + S1 = 2 (also "mirror") 2
Two/second + S3 = 5 5
Two/second + S2 = 4 (wedge) 4
Above: "Two/second" Midblots
Two/second + S0 = 2 2
Two/second + S1 = 3 3
Three/third + S3 = 6 6
Three/third + S2 = 5 (also "triangle") 5
Above: "Three/third" Midblots
Three/third + S0 = 3 3
Third/three + S1 = 4 4
Four/fourth + S3 = 7 7
Four/fourth + S2 = 6 (also "sock") 6
Above: "Four/fourth" Midblots
Four/fourth + S0 = 4 4
Four/fourth + S1 = 5 5
Five/fifth + S3 = 8 (uncommon) 8
Five/fifth + S2 = 7 (also "sixstack") 7
Above: "Five/fifth" Midblots
Five/fifth + S0 = 5 5
Five/fifth + S1 = 6 6
The formations in the "One/first" diagramblock have only one checker on the midpoint, but you can double or triple (etc.) their size. For example, in the top left "One/first" diagram, the count is 4. If you put two checkers on each of those points, the count is 8, etc.The midblot formations in the "Two/second" diagramblock can be doubled as well. For example, let's double the patterns in the upper left and lower left of that diagramblock:
Twice [Two/second + S3] = 10 10
The first of the above two formations was originally described by "two/second + S3" = 5. It is twice the size of that, so we count it as "twice [two/second + S3]" = 10. The formation immediately above is "twice [two/second + S0]" = 4.
Twice [two/second + S0] = 4 4
To become fluent in midblot (and twice midblot) counts, you need only practice the basic counting technique a little, and before long what seem now to be methodically plodding counts will become quicker, then gradually ingrained as instant counts.
Lest you feel overwhelmed by the number of diagrams in this post, I'll reiterate the point I made earlier: Midblot formations are not crucial to making fast counts in Naccel. But they will help you count even faster.
Our choice of feature position in this post is inspired by a response post from Bill Madison. The position was originally posted here.Maik and Bill, in counting Blue's outfield, used a neutral shift, bringing down the four midpointers a pip, and backing up the blot four pips to compensate: all five checkers meet on S1 for a count of 5.
Petter also recognized Blue's fivechecker formation as a count of 5, though it's unclear to me what counting method he used: "...four on n7 and one on n2 is 5. I don't know if this formation has a name." Indeed, it does have a name  it's a midblot! The count is Four/fourth + S1 = 5, and after a few times running across this one it will become an instant 5 in your mind.
To reduce your scrolling, I've repeated the position for Blue below. We will now focus on the count of the left side of the diagram.
Maik, Bill and Petter counted the S3 point as 6, and the n2 sym as 2, netting a count of 4.
4
Ian counted the S3 point plus n3 point as a zag of 5, and the n1 sym as 1, also netting 4.
I don't deem Maik/Bill/Petter's solution to be better than Ian's: both are simple twopart counts and it comes down to whichever you see first. But it is worth noting that Maik/Bill/Petter excluded the (irrelevant) S0 point. Ian chose to include S0 as a visual aid, though he could have excluded it by counting the n2 and n1 points as a 1 block.)
[In case you're stumbling on terms or counts, review "sym" (i.e., sixsym) in the second and third diagrams here, "zag" in the ninth diagram here, and "block" here.]
Is there an even faster way to count Blue's left side? Yes. Review the third diagram of this post. When combined with the S3 point, a sym around t4 (the trad 4pt) yields a count of 4. Blue's left side is gone in the twinkle of an eye.
I recommend you make Blue's left side an instant "4" in your repertoire. Because S0 can always be included (even when it's vacant!), it may be easiest to remember it as: Best fourpoint board plus deepest anchor = 4.
In short, Blue's entire position can be lightningcounted as 4 + 5 = 9.
Daniel noted that clustercounting does well in this position: "For Blue, since one checker on the ace point and two checkers on the midpoint is 50 pips, two and four are 100. Since a full prime is 42, Blue's prime is 36. The odd man is 8. Total 144." Okay, well done; that's pretty fast for clustercounting. :)
Moving on, let's repeat the original position and count White:
Daniel and Garyo alertly pointed out that White has only moved 5 pips from the opening position and therefore the traditional count is 167 minus 5 pips = 162. Likewise for Naccel, 12(5) minus 5 pips = 12. If you have a convenient reference point, you may as well use it.
For practice, though, let's pretend we didn't get that fortunate. The fastest count without shifting appears to be S3 blot = 3, reflection = 4, midblot = 3, and triplet = 2; total of 12.
[In the above count, note that the reflection (of n17 + mid) uses up one midpoint spare, and the other three midpointers plus third point blot count 4. If you have trouble seeing it, refer to the top left of the one/first diagramblock and the bottom right of the three/third diagramblock in this post, the only difference being that the formations are for White instead of Blue.]
Faster counts can often be obtained with a little shift. My choice  also spotted by Maik and Petter  is the following 1pip shift:
The S3 point is 6, and the double zig is 6, for a total of 12.
S3 point + Double zig 12
Double zig? Yup. You'll see a zig in the third diagram here, and, as Maik and Petter found (or remembered), my third paragraph under that diagram states:
"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."
Recapping, then, for both colors, the count breaks down neatly between the left and right sides of the board:
12
9
Blue is 4 + 5 = 9, and (with the 1pip shift shown) White is 6 + 6 = 12.
Next position?Nack
 Naccel 2  post #8 (midblot magic)
Nack Ballard  Thursday, 28 January 2010, at 8:33 a.m.
 Naccel 2  How to make diagram with Naccel number *NM*
Claude Landry  Thursday, 28 January 2010, at 7:53 p.m.
 Naccel 2  How to make diagram with Naccel number
Nack Ballard  Thursday, 28 January 2010, at 8:53 p.m.
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