Thanks, Matt. I find your explanation and information immensely helpful.
To show that I'm paying attention, I'll discuss two basic squads with the help of your finely designed Naccel point-numbered diagrams below. [Please let me know if I made any diagramming errors.]
Point numbers are labeled in white, Super (Super-point) numbers in black.
| |
|---|---|
| Triplet (two examples) | 7 and 1 |
Above, the far-side triplet is on the (Naccel) 14pt and therefore counts 7. The near-side triplet is on the (Naccel) 2pt and therefore counts 1.
(By dividing by 2, you are in essence counting the number of third-quadrant (two-point) steps to the Naccel 0pt. The 14pt triplet counts 7 because it takes 14/2 = 7 third-quadrant steps to journey it to the 0pt.)
[Scholarly note: A more obscure method is to count the nearest Super twice and the second-nearest Super once. Hence, in the above cases, 2 + 2 + 3 = 7, and 0 + 0 + 1 = 1. In essence, you are shifting two checkers 2 pips each and the other checker 4 pips in the opposite direction and counting the occupied Supers.]
| |
|---|---|
| Pair (two examples) | 5 and -1 |
There are two ways you can count a pair. The first way (similar to the divide-triplet-by-two logic) is: Divide the point number by 3. Above, the far-side pair sits on the (Naccel) 15pt and therefore counts 5. The near-side pair sits on the -3pt (read as the "minus 3 point") and therefore counts -1.
(By dividing by 3, you are in essence counting the number of half-quadrants to the Naccel 0pt. The 15pt pair counts 5 because it takes 15/3 = 5 half quadrants to journey it to the 0pt.)
The second way to count a pair you may find even easier: Sum the Supers flanking it (because if you want to you can equally shift one checker to each). The far-side pair sits between S2 and S3 and it therefore counts 2 + 3 = 5. The near-side pair sits between S-1 (the minus 1 Super, the bearoff tray) and S0 and it therefore counts -1 + 0 = -1.
While these methods are great for helping you to build and remember counts for larger groups, you will no longer need to apply a method at all to pairs and triplets (and other squads) for which you've already learned the count. That is, once you know that the above pairs are 5 and -1, and the triplets (shown in the previous diagram) are 7 and 1, respectively, a counting method is academic. Nothing beats an instant count.
| 11(5) | |
|---|---|
| |
| 9(3) | |
Blue has a basic squad in each of her near-side quadrants. His pair counts -1 and his triplet counts +1. These two squads offset/reflect around n0, and this near-side formation (no matter how many checkers are on the irrelevant 0pt) is a standard poof; i.e., count of zero.
Blue's entire count is 6 (for two on S3), plus 3(3) for the midpoint checkers, making 9(3).
Let's repeat the diagram, and this time count for White:
| 11(5) | |
|---|---|
| |
| 9(3) | |
The difference between White's 11(5) and Blue's 9(3) is 2(2). That's how far Blue is ahead in the race after playing his double 5s.
Nack