Pip Counting
 Naccel 2—aN ACCELerated Pipcount Part 3:  Mirrors and Triplets Nack BallardJanuary 2010

This is Part 3 of the Naccel 2 series. Thanks to Lucky Jim for his submission, shown below.

Jim informs us this position is from Ballard-Sylvester U.S. Open, June 1989 (though I'll have to confess my memory isn't what it used to be—sigh).

First, let's consider Blue's position. As always, the checkers on the Naccel 0pt (trad 6pt) are invisible. You need count only Blue's outfield checkers. These come in two groups, which I'll address separately.

Six-Syms

Six-sym3

The first group (shown above) is a six-sym (six checkers symmetrically arranged around a point—in this case, the 3pt), for a count of 3. Jim and Ian nailed this one, perhaps remembering the second and third diagrams of Part 2.

For the other Blue group (see previous diagram, two checkers furthest right plus two checkers on far side), Ian and Jim shifted the remaining near-side point one point to the right, and a far side blot 2 pips to the right to compensate, then moved both far side checkers to S1 to get a four-stack on S1 plus the 5 pips adjusted. That works, but I'll show you a faster way.

Mirrors

As a prelude, allow me to present the "mirror."

You could easily count this as two checkers on S2 (Super2), i.e., 2 × 2 = 4, and ignore the two zero-count checkers on S0 (Super0, same as the 0pt).

Alternatively, you can count this as a mirror. A "mirror" can be composed of blots, points, or even stacks of three or more checkers, as long as they are of equal size. Mirrors count the same as the total number of checkers used; here, for example, there are four checkers, so the count is 4.

The only wrinkle to remember is that the near-side member of a mirror is always 1 pip closer to the player's own bear-off tray if one drops the far-side member straight down. (It may help to note that the colors of the points match.)

If we slide the above mirror to the right a few pips, we get the commonly arising mirror below:

Again, there are four checkers in the mirror, so the count is 4. (To verify the count, shift both points 1 final pip towards each other so that they meet on S1. Four checkers on S1 count 4.)

You can probably see where I'm going with this. In the game position, two of the checkers are already on the near-side point of the above diagram. You need only move the far-side checkers 5 pips forward to get the midpoint mirror.

In other words, Blue's entire count has two components. The first is a six-sym around the 3pt (left diagram below) = 3. The second is a midpoint mirror plus 5 pips (right diagram below) = 4(5).

Six-sym3

Midpoint mirror plus 5 pips 4(5)

In short, Blue's count is 3 + 4(5) = 7(5).

Triplets

Before we count White, I'll reveal another trick. Just as six checkers on a point count the same as the point number, three checkers on a point count half the point number.

For example, consider each of White's triplets (three-stacks) below.

The trad 8pt is the Naccel 2-point. White's triplet on her 2pt counts half of 2, which is 1.

The trad 20pt is the Naccel 14-point. White's triplet on her 14pt counts half of 14, which is 7.

After you've seen these triplets a few more times, you'll just remember them as "1" and "7" without even having to divide the point number by 2.

Taking special note of White's latter triplet (count of 7), let's go back to the main position:

White's two checkers on S3 (Super3) count 6. Her triplet (as just explained) is 7. Adding those together, White's back checkers count 6 + 7 = 13. [I've seen this five-checker formation enough times that to me it is simply "13" (no addition needed).]

For White's near side, remove three of her irrelevant 0pt checkers, leaving one. Ignoring the outside blot on the (Naccel) 1pt for now, you have ...

This is a six-checker symmetry around the −2pt, for a count of −2.

Review

Now let's revisit Blue's position. I'll remove his 0pt checkers (the ones that you will learn to ignore), and you pretend that you're counting Blue for the first time but knowing what you know now:

Blue's count is a six-sym around the 3pt (count of 3) plus a (midpoint) mirror+ of 4(5), for a total of 7(5).

Now I'll remove all but one of White's 0pt checkers (below), and you count White for the "first" time, knowing what you know now.

White's count is 6 + 7 = 13 in the back, and −2 for the six-sym in her inner board. Add 1 pip for the outside checker, for a total of 11(1).

Once you "get" how it works, count both colors in exactly this way a couple/few more times so that you can get a sense of how fast Naccel counts are. It's just a matter of knowing a few tricks and spotting formations.

The difference between 7(5) and 11(1) is 4(−4), or 3(2). That is, Blue leads by a couple pips more than 3 supes (super-pips)—20 pips, to be exact.

(If for some reason you want to convert to trad, the counts (with the 0pt checkers added back to the board, of course) are 7 × 6 + 5 + 90 = 137, and 11 × 6 + 1 + 90 = 157.)

[For White's position (after counting S3 × 2), Jim and Ian moved the three-checker anchor out to S2 (count of 6); to compensate for that 6-pip adjustment, they moved the two checkers on the −3pt back to the 0pt (trad 6pt), causing it to disappear. Then they canceled a checker on the 1pt and −1pt, leaving only blots on the −1pt and −4pt left to count. This achieves the correct count but is, of course, more work than counting the three-checker anchor where it is.]

Go on to Part 4, or read the Further Discussion below.

Further Discussion
For reference, this diagram is repeated from above. Please note the similarity between White's formation below and Blue's six-sym in Part 2.

Ian said:

 Could you clarify the Symmetry in White's inner board, please? You seem to use the 1st man on the six point as part of the symmetrical pattern, while at the same time saying it counts zero because it's on S0. I can't see how it logically can count zero and also be used as part of a cluster, which implicitly makes it a contributor to part of a non-zero count.

This zero-count checker does indeed contribute to a non-zero count. However, its contribution is only as a visual aid; otherwise its existence is irrelevant.

The beauty of the 0pt (trad 6pt) checkers is that you can use them or not in a count, as convenient. Most of the time, it is best to remove/ignore all of them. But it can sometimes help to retain or even manufacture one (or more); neither alters the count. (Secrets of the modern universe: n0 is both a black hole and a white hole.)

Here, again, is White's complete near side position:

As you know, you can remove the four 0pt checkers without altering the count:

To get this position to a −2 count, you need to slide a checker one pip to the left. Any checker will do, though some patterns are more easily recognizable than others. You chose to slide the far-left checker further to the left, like this:

And that works fine: you have achieved a six-checker symmetry around the −2pt, for a −2 count.

There are two aspects to be aware of, however:

• The distance between the most polarized checkers of the formation is 6 pips; it is a little easier to spot symmetry when they are closer together.

• The bar buffers the n1 checker from the rest of the formation. That can be an impediment to perceiving symmetry clearly, especially on these diagrams where the bar is so wide (though, as with any counting system, bar distortion becomes much less noticeable as you practice with that system).

For those reasons, many people would find it easier/clearer to make this one-pip shift instead:

This is also a six-sym around the −2pt yielding a count of −2. Is it a problem that one of the checkers is on the 0pt? No. That would be like saying that (−5 + 1) = −4, but that (−4 + 0) can't be quantified. (The analogy refers to it being okay to have the −5pt and 1pt checkers reflect around the −2pt but somehow that it wouldn't be okay to have the −4pt and 0pt checkers reflect around the −2pt.)

Tandems

If you like (having accounted for the 1-pip adjustment and working from the diagram above), you can remove the 0pt checker, like this:

Since removing the 0pt checker has no effect, the count is still −2.

This five-checker formation happens to be a "tandem," because from White's perspective it looks like the top of a tandem bicycle with the blot being the handle bars and the points being the two riders. In White's own inner board location and facing this direction her tandem counts −2.

I put another tandem (bicycle) on the side of the board facing us so that you can see it right-side up. Here, the tandem counts 13 (15 more than −2, because it is a five-checker squad three quadrants away). By an eerie coincidence, if you shift White's back checker points outward a pip, you get the 13-count formation that exists in Jim's original position.

But I'm just having fun now; complex squads will come much later, if you need to learn them at all. The main point is that the six-sym (shown in the previous diagram) is a basic count, whereas the tandem (shown just above), even though it uses one fewer checker, is an advanced count. Hence, having that checker on the 0pt proves useful even though it contributes zero to the count.

Back to the original formation before the 1 pip was moved:

For my explanation involving this diagram above, rather than (a) remove all four of the 0pt checkers, "move" the 1pt checker to the 0pt, count the six-sym and add 1 pip for the earlier movement, I chose (b) to remove only three of the 0pt checkers, (leaving one 0pt-er still on the point), count the six-sym, and then count the checker outside as 1 pip.

It comes to the same thing, except when possible I slightly prefer (b), which counts extra pip(s) in the outer board last rather than (a), which requires accounting for pip movement(s) before the complete supe count has been obtained. (If you don't follow this part, it's okay—it's an unimportant detail.)

Go on to Part 4.

Nack Ballard is a top international backgammon player.  He is a coauthor of Backgammon
Openings
and the inventor of Nackgammon and Nactation.  His website is www.nackbg.com.

Naccel 2 Series
 Part 1: Introduction Supers • Poofs Part 2: Symmetry Six-Stacks • Six-Symmetry Part 3: Mirrors and Triplets Six-Syms • Mirrors • Triplets • Review • Tandems Part 4: Reflections, Zags, and Wedges Far-Side Reflections • Zag Mirrors • Wedges • Triplets • Pairs • Squad Poofs Part 5: Primes and Hopping Six-Primes • Hopping • Counting Habits Part 6: Midpoint Combinations Blocks • Zig Mirrors • Diag Mirrors • Diag Zig Mirrors • Midpoofs • Midgold • Midblock • Problem Part 7: Diags and Mirrors Hopping • Blot Diags • Truck • Mirror

 Part 8: Combined Counts Combined Counts • Midblot Formations • Problem • Zigging and Zagging Part 9: Squad Variants Pair • Split • Wide • Triplet • Layer • Wedge • Block • Triangle • Sock • Squad Poofs • Problem 1 • Problem 2 Part 10: Leftover Counts Leftover Counts • Tweensyms • Midpoint Leftover Counts • Squad Poofs • Problem 1 • Nack 57 • Problem 2 • Problem 3 Part 11: Midblot Refinement Midblot Counts • Revs • Leftover Counts • Problem 1 • Problem 2 Part 12: Motion Poofs Chairs • Shift Poofs • Motion • Problem 1 • Problem 2 • Nack 57 • Handy Count

 See: Other articles on pip counting Other articles by Nack Ballard Return to: Backgammon Galore