This is Part 7 of the Naccel 2 series. It 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.
There are two ways to count this post-hop diagram:
- Poof the five-prime (from −2pt to 2pt) and the right side pair is +1. Or
- Poof the spares, and count the six-prime: either sum the endpoints (i.e., −2 + 3 = 1) or
double the high point and subtract 5 (i.e., 3 × 2 − 5 = 1).
In time, this pair and six-prime will each be an obvious count of 1 to you, so it won’t matter which of (a) or (b) you choose. Either way, you add that 1 to the original +1 hop, achieving a count of 2.
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).
Compared to the midpoint diag, the diag here is slid 1 pip to the left and it is composed of two blots instead of two points, but it still counts as half the number of checkers in the formation. A blot diag has two checkers and therefore counts 1.
What I'm getting at is that, like any mirror or mirror variant (once you have acquired the familiarity), a diag can be counted directly without shifting to the Supers (as you did).
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)!
I don't expect you to spot the diag (instead of shifting or
hopping) or to count the six-prime as an instant 1
without practice: I'm just giving you a glimpse or two of the many corners
you will eventually cut.
Let's repeat the position 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:
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.
Think of it this way: either (a) you hopped (+1) a checker from S0 to S−1, then performed your 1-pip shift; or (b) your countershift was from n0 to n1 and then you hopped forward (+1), 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:
I made the (irrelevant) S0 checkers invisible. The block is still 11,
the triplet is −2 (half of its point number) and the six-sym around n−2 is −2, for a count of 11 − 2 − 2 = 7.
By the way,
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, 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.
Jimas 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.
Mattnegative 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 leftover counts;
you'll get down to just one at the end, once you're counting right.
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.
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.
Go on to Part 8.