Pip Counting
 Naccel 2—aN ACCELerated Pipcount Part 4:  Reflections, Zags, and Wedges Nack BallardJanuary 2010

This is Part 4 of the Naccel 2 series. For this post, Lucky Jim submitted two positions.

### Far-Side Reflections

You know about reflections on the near side (see poofs). Cancellations around S0 (trad 6pt) count zero, which means there are a lot of checkers you don't have to count at all.

There are also reflections on the far side. Because they reflect around S2 (opponent's bar point) instead of S0, they count 2 per checker—that is, count the number of checkers you see and double it.

With the midpoint featured, this is the most commonly arising far-side reflection:

Doubling the number of checkers, 4, gives us a count of 8. (If these were blots instead of points, the count would be only twice 2 = 4.)

Let's apply that knowledge to the Blue side of Jim's first position:

Blue's checkers on the near side reflect around the 0pt and therefore cancel out except for the 2pt checker (which counts 2 pips). So ...

8 (reflection), plus 9 for S3, plus 2 pips for the blot, gives a count of 17(2).

[Jim and Ian counted similarly, except they didn't yet know about far-side reflection. Instead, they shifted to S3 and S1, which is fine but it adds a step of visualization.]

White's count is even easier (as Jim and Ian found as well), so it's hardly worth a new diagram, but I'll spend one anyway so that the point numbering is from White's point of view.

White's entire near side goes poof (reflection/cancellation around the 0pt), so the entire count is 3(3).

The midpoint is nice for leftover counts because it's trivial to multiply 1(1) for each checker, and if it has as many as six checkers it just counts 7. (Naturally—a six-stack on the 7pt = 7.)

In short, the count is 17(2) to 3(3), a difference of about 14 supes (super-pips).

[Round supes (which can be thought of as crossovers) are a good way to assess Blue's timing. As 12 of the 14 supes are accounted for in bringing home the vital-to-keep anchors, that only leaves 2 spare supes, which are roughly offset when White's 6s then 5s are killed bearing in later. And the S3 spare might not escape and get its 3 supes. Yikes, looks like a monster pass.]

If you are unclear as to why Blue's near side counts only 2 pips and White's near side is a giant poof that doesn't need to be counted at all, review the fourth and fifth diagrams in Part 1.

Below is Jim's second position. There are many ways to count Blue (I'll show you my count later) but I'll give you a way that works well with what you know so far.

For Blue, try this easy mental shift:

Now, here's the trick that you need: Conjure a new checker on the 0pt. Remember, the 0pt is not only a black hole but also a white hole: even when the 0pt looks vacant, phantom checkers always exist there for your convenience. Now, lift that conjured 0pt checker onto the −1pt, which is a 1 pip adjustment.

So, your count: Four on S3 counts 12, six-sym (around the −3pt) is −3, the engineered six-stack is −1, and add the 1-pip lift needed to get that stack. That's a net of 8(1).

Now let's count White. Her position again is ...

Try this mental shift (2 pips forward, 2 pips back):

The stack is −3, the six-sym (three-prime) is −1, and the three spares on the −1pt are −3 pips. Total of −4(−3). In case you want to convert, −4 × 6 − 3 + 90 = 63.

Do you see a similarity with what we did in Blue's inner board and in White's inner board? Six-stacks and six-syms (which are often three-primes) are powerful weapons.

When both Blue and White checkers are home, cluster-counting is relatively at its best. With only one quadrant to count, Naccel's edge disappears. But if you want to become lightning fast, it pays to practice counting all positions with Naccel (instead of lazily falling back), even though at first it may be slow and even confusing in certain situations. That's to be expected—you're still tying your skate laces at this point.

### Zag Mirrors

Finally, I promised to show you the way I counted Blue (noting that the alternative method suggested above is perfectly legit). To understand my count, you'll need two key pieces of information.

First, let's review the mirror, introduced in Part 3. Below is another example:

For a regular mirror, the near-side point is (as you can see) a pip closer to the bear-off tray (than the far-side point would be if it were dropped straight down). The number of checkers = 4, and thus the count = 4.

For advanced anchors (or blots), you can often use regular mirrors to offset checkers in the home board. But for deep anchors, you will tend to use "zag-mirrors" (or "zags" for short).

Starting with the mirrored point shown above (count of 4), if you zag (move back) either point 3 pips, you increase the count by 6 pips or 1 supe, to a count of 5, and you have yourself a zag mirror.

If it is the far-side point that is zagged, we get the zag mirror shown below.

Whereas with regular mirrors, the near-side point is 1 pip closer to the bearoff tray; with zag mirrors, the near-side point is 2 pips further away.

As stated, zag-mirrors count 5. The occupied points are opposite colors, which helps remind you that the count is odd.

### Wedges

Now for the second part of the explanation:

A triplet (three-stack), which was introduced in Part 3, is a basic squad (i.e., a group of checkers within a quadrant field). An example is shown below.

As was explained in Part 3, you count a triplet by dividing the number of the point on which it resides by 2. This triplet is on the −2pt, so the count is −1.

The triplet has a variant squad, which can be obtained by moving two of the checkers 1 pip towards the nearest Super, and the third checker 2 pips in the other direction, to compensate. It looks like this:

This squad is called a "wedge" (it is shaped like a door wedge). To count a wedge, take the point number that is in between the blot and point but closer to the latter (or just shift to it, thereby reenacting a triplet on the appropriate point, if you need the visual help), and divide it by 2.

Here, the in-between point is the −2pt. Dividing by 2 gives you this wedge's count of −1.

You can also double the height of the wedge which gives you a "double wedge," and this also-commonly seen formation counts the same as the in-between point number (e.g., if you double the height of the wedge shown above, the count is −2). Both the wedge and the double wedge can be great counting tools.

Now that you know what a zag (zag mirror) and a wedge are, I'll show you how I counted Blue in Jim's position, repeated below.

To count Blue, all I did was to move one pip forward, from the 0pt to the −1pt, and sum the count of two formations:

Double Zag10

Double Wedge-2

That's 10 for the double zag, −2 for the double wedge, plus the 1-pip lift: my count is 8(1).

The checker left on the 0pt, is, as usual, invisible.

Go on to Part 5, or read more below.

This bonus post contains illustrations and basic explanations of two basic squads: the "triplet" (review) and the "pair" (new).

As always, (Naccel) point numbers are labeled in white and Super (Super-point) numbers in black.

### Triplets

Triplet (two examples)7 and 1

A "triplet" can appear on any even-numbered point. To count a triplet, simply divide its point number by 2.

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. For another example, this triplet is on the −2pt and therefore counts −1.

(By dividing by 2, you are in essence counting the number of two-point (third-quadrant) steps to the Naccel 0pt. The 14pt triplet counts 7 because it takes 14 ÷ 2 = 7 two-point steps to journey it to the 0pt.)

[Scholarly note: A more obscure method is to count the closest Super twice and the other flanking Super once. It essentially shifts two checkers 2 pips each and the other checker 4 pips in the opposite direction. For example, the far side of the above diagram can morph into the far side of this diagram, which is 3 + 2 + 2 = 7.]

### Pairs

Pair (two examples)5 and −1

A "pair" can appear on any third point of the board; either on a Super (though then it is commonly counted as twice the Super) or, as here, in the middle of any field. A "field" (or "squad field") is the five-point area composed of all the points in a quadrant minus its Super.

There are two ways you can count a pair. The first way (related 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" or "neg 3 point") and therefore counts −1.

(By dividing by 3, you are in essence counting the number of 3-pip or half-quadrant steps to the Naccel 0pt. The 15pt pair counts 5 because it takes 15 ÷ 3 = 5 half-quadrant steps 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 like you can equally shift one checker to each). The far-side pair sits between S2 and S3 and therefore counts 2 + 3 = 5. The near-side pair sits between S−1 (the minus 1 Super, the bearoff tray) and S0 and 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 a pair or triplet (or any other squad or formation) for which you've already learned the count. Once you know, for example, that the far side pair is 5, and the far side triplet in the previous diagram is 7, a method to achieve that count becomes academic. Nothing beats an instant count.

9(3)

In this position, White has opened with 42 and Blue has replied with double 5s. Both players made an inside point.

Blue has a basic squad in each of his 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; a 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:

White has an even more basic poof on her near side. The points with two checkers on either side of her 0pt are symmetrical. So, White's entire count is 6 (for two on S3), plus 5(5) for five on her midpoint, making 11(5).

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.

Go on to Part 5.

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