Naccel -- post #7 (shortcuts)
Posted By: Nack Ballard <email@example.com>
Date: Monday, 25 January 2010, at 12:09 a.m.
Welcome to post #7 of the Naccel 2 teaching series.
For review, click on the following:
FAQ, Post#1, Post#2, Post#3, Post#3x,
Post#4, Post#4x, Post#5, Post#5x, Post#6, Post#6x.
The current post (#7) is in response to Matt Ryder's post entitled Learning Naccel - Problem 1.
I applaud your courage, Matt, and the fine example you set :)
Here is your position:
For Blue, you shifted the front spare (back to S0) and the blot (to S1) for a count of 1, poofed the five-prime, and added 1 for the pair (on n3), for a total of 2. Great! That beats non-Naccel methods of counting the position hands down.
That said, I want to suggest an even faster idea: "Hop" the blot down to n2, as shown below. (To review hopping, see the twelfth diagram here.
There are two ways to count this, both of which start with counting +1 for the hop already executed. One way is to poof everything but the pair on n3 (adding another +1).
Big poof + Pair, or Small poof + Six-prime 1
The other way is to poof only the spares, and count the six-prime: sum the endpoints (i.e., -2 + 3 = 1), or, alternatively, double the high point and subtract 5 (i.e., 3*2 - 5 = 1). Either way, you add that 1 to the +1 hop, achieving the total of 2.
For you, hopping is best (assuming you notice the thirteen-checker poof), and then it mostly comes down to which of the pair or six-prime you can count faster, or spot first. (In time, this pair and six-prime will each be an obvious, instant count of 1 to you, so it won't matter.)
A supplemental lesson here is that when you see a poof, quickly check to see if there might be an even bigger one.
Now I'll share with you the way I counted Blue (first with a prep diagram).
Compare the above formation to the one in the fourth diagram here. It's slid over 1 pip to the left, but it's still a "diag" (diagonal mirror). Note that I had said that a diag counts 2, but in context I was referring to a formation of two points (i.e., a point diag counts 2). Actually a diag's count equals half the number of checkers in the formation; here, there are a total of two checkers, so the count is 1. What I'm getting at is that a blot diag can be counted as 1 directly without shifting to the Supers (as you did).
Blot Diag 1
Let's repeat the full position for Blue:
If you know the relevant formations, there is no need to even shift or hop. You have a six-prime and a blot diag, for a count of 1 + 1 = 2, as I see it. The only way to count faster is to instantly recognize the entire formation as 2 (and if it comes up again soon I will)!
Six-prime and Blot Diag 2
To be clear, I don't expect you to spot the diag (instead of shifting or hopping) or to count the six-prime as an instant 1 until you've had more practice: I'm just giving you a glimpse of two of the many corners you will eventually cut.
Let's repeat the position below and count for White.
You chose a 1-pip shift of S3 to n17 and counter-shift of S-1 (conjured) to n-5, like this:
(Gnu won't let me put sixteen checkers on the board, so I stole one from S0. You can imagine that there are still two checkers on S0, if you like, but it is irrelevant to the count.)
Block and Closed board 6
That's clever, Matt! The block counts (16 + 17) / 3 = 11, and the closed board (counted like any other six-prime) is -5, for a net count of 6.
The only problem, as Petter said, is that you forgot to add 1 for the checker you conjured from S-1. (If you like, think of it this way: either (a) you "hopped" a checker from S0 to S-1, adding 1 to the count, then performed your 1-pip shift; or (b) your countershift was from n0 to n1 and then you hopped forward, closing the board.) Had you done so, you would have gotten the correct count of 7.
If you don't want to deal with the hop, you can do this shift instead:
Block, Triplet and Sym 7
For clarity, I made the (irrelevant) S0 checkers invisible. The block is still 11, the triplet is -2 (half of its point number, as illustrated in the seventh diagram here) and the six-sym around n-2 is -2, for a count of 11 - 2 - 2 = 7.
Btw, I call White's nine-checker squad a "truck" (which counts the point number of its cab plus thrice the Super adjacent to its rear). I know this truck as -4, so I don't need to separately add the triplet and six-sym. My count would be a bit shorter: 11 - 4 = 7.
Petter's count is cool, too. From the main position (three diagrams back), he counted the six-sym around n-2 as -2, and n16 plus n-4 as a point mirror of 4. However, he did more work than necessary when he shifted n17 and n-5 to the Supers; these blots are already mirrored. Instead, (after counting the six-sym of -2), it is better to count the S3 blot as it is, leaving the six-checker mirror formation shown below:
A mirror of six checkers counts 6. Adding in 3 for S3 and -2 for the six-sym in the main diagram, that nets a count of 7.
Six-checker mirror 6
Jim -- as you suspect, spotting mirror-family formations for White is just a matter of practice. I don't stutter in the slightest when going back and forth from Blue counts to White counts.
Matt -- negative numbers are your friend. That is what allows you to poof so many checkers and that is why all the numbers in Naccel are so small. Don't be overly concerned with left-over counts; once you're counting right you'll get down to having one left at the end (positive or negative). Then again, I think you already know that, intuitively :)
Finally, I'll show you how I chose to count White.
I find the following to be a handy poof:
You already have the ability to easily count this diagram in two stages: as 3 for the S3 blot, and -3 for the six-sym around n-3. But the more intimate you become with these two formations, the more quickly you'll see them as a +3 and a -3 that cancel each other out. Poof!
Poof (S3 and -3 sym) 0
After poofing the above (to me, standard) seven-checker formation (and of course S0 is poofed too), and therefore having had to count nothing so far, I was left with these six checkers:
The blots are a mirror, and the points are a zag, for a count of 2 + 5 = 7.
Blot mirror, and zag 7
To review mirrors, see the third and fourth diagrams here. To review zags, see the eighth and ninth diagrams here or the sixth diagram here (zag for White).
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