Pip Counting

 Naccel 2—aN ACCELerated Pipcount Part 8:  Combined Counts Nack BallardJanuary 2010

This is Part 8 of the Naccel 2 series. I'll start this post by demonstrating a few combined integer counts.

### Combined Counts

When checkers occupy Supers, you can't go far wrong counting them as is. On the other hand, you will so often find one or two checkers on S3 (their starting location) that it is worth knowing the counts that combine S3's with other common integer formations.

For example, a commonly arising position after 51\$-42S-43@ is shown below:

The S3 checker counts 3, and the nearby pair counts 5. This three back-checker formation is worth committing to memory as an instant count of 8.

If you add an S3 checker to Blue's formation, the count of his double anchor is 11. The larger formation is less common but still occurs frequently enough that I made a point of learning it.

For a still larger example, see White's five back checkers in this diagram of Part 3. The S3 point (count of 6) and the n14 triplet (count of 7) coexist often enough that over time I've learned their combined count of 13 without really trying.

Not all combined counts need be high-single- or double-digited. In positions of moderate count, it is typically more desirable to combine a positive count with a negative count, generating a net integer count that is zero (best), small positive (second best) or small negative (third best).

Consider this diagram in Part 4. The back pair counts 5, and the front pair counts −1; easy enough. But if you combine them as a mirror of 4, you don't have to subtract 1 from 5: you've saved a step.

Having conveyed the concept, I will now show you some small combined integer counts that frequently arise.

Each of the formations below combines the S3 point (count of 6) with a sym of negative count in the inner board. A "sym" (short for six-sym) is any six checkers symmetrical around a point, though I have chosen to illustrate the most common sym, the three-prime. (If you don't know how to count a sym, review the second and third diagrams of Part 2.)

The four diagrams below are a series. Traditional point numbers (the purpose of which you'll see in a minute) are added in black under each diagram.

What is cute and memorable about this series is that the traditional point number ("t") on which the sym is centered is the count of the formation! In the above four diagrams, the sym is centered on t5, t4, t3 and t2, making the counts 5, 4, 3 and 2, respectively.

If there is a third checker on S3, add 3 to each of the above counts. If there is a mere blot on S3 (more likely), subtract 3; for example, you can count this diagram of Part 7 as t3 − 3 = 0 (among other ways to count it) if you haven't already learned it is a poof.

### Midblot Formations

Next, I will cover "midblot" formations. I'm not referring to a blot on the midpoint, but rather to a formation with some number of checkers on the midpoint combined with a blot elsewhere. There are quite a few midblot formations, but the counts are always easy to determine.

Consider this basic formation (which also happens to be a blot diag). There is one checker on the midpoint, and a blot on the first point (n−1) in the inner board. [I left a bunch of checkers on S0 to clarify that you skip the Super—you start adjacent to it: at the first (highest) point in the field.] The combination of one and first produces a count of one.

Compare to the basic formation below. Here the two (midpoint checkers) and second (the blot is on the second highest point of the field, n−2) combine to produce a count of two.

Extending the logic illustrated in the last two diagrams:

Three checkers on the midpoint plus a blot on the third point (n−3), a description we'll shorten to "three/third," counts 3.

Four/fourth counts 4. (Four on midpoint, blot on n−4.)

Five/fifth counts 5. (Five on midpoint, blot on n−5.)

Let's build on that knowledge to create similar counts when the blot is in other quadrants. From the formation diagrammed above, hop the blot back one quadrant to the outer board, like this:

In this formation, the blot is in the quadrant where S1 (instead of S0) lives. Because of S1, you add 1 to the count. Putting the pieces of the puzzle together, "two/second" (two on mid, with blot on second point) counts 2, plus 1 (for S1), gives you a count of 3.

To practice further, let's increase the number of checkers on the midpoint to four, and therefore slide the blot to the fourth point. (Without this corresponding slide, the count won't remain an integer).

You have four checkers on the midpoint and thus a blot on the fourth point. Add one for S1 being in the quadrant, and you get a count of 5.

Getting the hang of it? Okay, let's back-hop this fourth-point blot to the next quadrant:

The blot is still on the fourth point of its quadrant (counting from the S2 Super). The Blue formation has increased again by one: it is now four/fourth plus S2, for a count of 6.

Let's back-hop the blot one more time:

Blue's count has increased yet again by one. It is now four/fourth plus S3, for a count of 7.

Below, for reference, all twenty possible mid-blot formations are shown, in five blocks of four diagrams. (These include the midblot examples already diagrammed.) In each case, the caption reminds you how to count, and the count itself is repeated on the far right of the caption line.

If you start with the top left diagram of any four-group and go clockwise, the counts sequentially decrease by 1, or if you start with the bottom left diagram and go counterclockwise, the counts increase by 1. Alternatively, if you choose an orientation in the one/first group and then look at the same orientation in the two/second group, three/third group, etc., respectively, the counts will sequentially increase by 1. The more such counting patterns, connections and morphs of other formations you can spot, the better your understanding of Naccel and the more sure-footed your counting will become.

[Note: Two of the formations are marked "uncommon." Blue seldom has a blot on the n2 (trad 8pt) or n0 (trad 6pt) of White, who occupies these points at the outset of the game.]

I should emphasize that these formations are not vital to counting in Naccel. However, they will help you imprint other formations you already know and they are additional tools to help you achieve lightning counts.

This complete set of Midblot reference diagrams doesn't end the post. At the bottom there is further discussion, followed by a full-board position to count for Blue and White.

One/first Midblots

One/first + S3 = 4 (also "reflection")4

One/first + S2 = 3 (also "wide")3

One/first + S0 = 1 (also "diag")1

One/first + S1 = 2 (also "mirror")2

Two/second Midblots

Two/second + S3 = 55

Two/second + S2 = 4 (also "wedge")4

Two/second + S0 = 22

Two/second + S1 = 33

Three/third Midblots

Three/third + S3 = 66

Three/third + S2 = 5 (also "triangle")5

Three/third + S0 = 33

Three/third + S1 = 44

Four/fourth Midblots

Four/fourth + S3 = 77

Four/fourth + S2 = 6 (also "sock")6

Four/fourth + S0 = 44

Four/fourth + S1 = 55

Five/fifth Midblots

Five/fifth + S3 = 8 (uncommon)8

Five/fifth + S2 = 7 (also "six-stack")7

Five/fifth + S0 = 55

Five/fifth + S1 = 66

The formations in the One/first diagram-block have only one checker on the midpoint, but you can double or triple (etc.) their size. For example, in the top left "One/first" diagram, the count is 4. If you put two checkers on each of those points, the count is 8, etc.

The midblot formations in the Two/second diagram-block can be doubled as well. For example, let's double the patterns in the upper left and lower left of that diagram-block:

The first of the above two formations, at half size, was originally described as "two/second + S3" = 5. At full size, we count it as "twice [two/second + S3]" = 10. In a similar vein, the formation immediately above doubles the simpler "two/second + 0" = 2 and becomes "twice [two/second + S0]" = 4.

To become fluent in midblot (and twice midblot) counts, you need only practice the basic counting technique a little, and before long what seem now to be methodically plodding counts will become quicker, then gradually ingrained as instant counts.

Lest you feel overwhelmed by the number of diagrams in this post, I'll reiterate the point I made earlier: Midblot formations are not crucial to making fast counts in Naccel. But they will help you count even faster.

### Problem

Our choice of feature position in this post is inspired by a response post from Bill Madison. The position was originally posted here.

Maik and Bill, in counting Blue's outfield, used a neutral shift, bringing down the four midpointers a pip, and backing up the blot four pips to compensate: all five checkers meet on S1 for a count of 5.

Petter also recognized Blue's five-checker formation as a count of 5, though it's unclear to me what counting method he used: "... four on n7 and one on n2 is 5. I don't know if this formation has a name."

Indeed, it does have a name—it's a midblot! The count is Four/fourth + S1 = 5, and after a few times running across this one it will become an instant 5 in your mind.

To reduce your scrolling, I've repeated the position for Blue below. We will now focus on the count of the left side of the diagram.

Maik, Bill and Petter counted the S3 point as 6, and the n−2 sym as −2, netting a count of 4.

Ian counted the S3 point plus n−3 point as a zag of 5, and the n−1 sym as −1, also netting 4.

I don't deem Maik/Bill/Petter's solution to be better than Ian's: both are simple two-part counts and it comes down to whichever you see first. But it is worth noting that Maik/Bill/Petter excluded the (irrelevant) S0 point. Ian chose to include S0 as a visual aid, though he could have excluded it by counting the n−2 and n−1 points as a −1 block.)

Is there an even faster way to count Blue's left side? Yes. Review the S3 point + t4 sym diagram from earlier in this post. When combined with the S3 point, a sym around t4 (the trad 4pt) yields a count of 4. Blue's left side is gone in the twinkle of an eye.

I recommend you make Blue's left side an instant "4" in your repertoire. Because S0 can always be included (even when it's vacant!), it may be easiest to remember it as: Best four-point board plus deepest anchor = 4.

In short, Blue's entire position can be lightning-counted as 4 + 5 = 9.

Daniel noted that cluster-counting does well in this position: "For Blue, since one checker on the ace point and two checkers on the midpoint is 50 pips, two and four are 100. Since a full prime is 42, Blue's prime is 36. The odd man is 8. Total 144."

Okay, well done; that's pretty fast for cluster-counting. :)

Moving on, let's repeat the original position and count White:

Daniel and Garyo alertly pointed out that White has only moved 5 pips from the opening position and therefore the traditional count is 167 minus 5 pips = 162. Likewise for Naccel, 12(5) minus 5 pips = 12. If you have a convenient reference point, you may as well use it.

For practice, though, let's pretend we didn't get that fortunate. The fastest count without shifting appears to be S3 blot = 3, reflection = 4, midblot = 3, and triplet = 2; total of 12.

[In the above count, note that the reflection (of n17 + mid) uses up one midpoint spare, and the other three midpointers plus third point blot count 4. If you have trouble seeing it, refer to the top left of the one/first diagram-block and the bottom right of the three/third diagram-block in this post, the only difference being that the formations are for White instead of Blue.]

Faster counts can often be obtained with a little shift. My choice—also spotted by Maik and Petter—is the following 1-pip shift:

The S3 point is 6, and the double zig is 6, for a total of 12.

Double zig? Yup. As Maik and Petter found (or remembered), my third paragraph under this diagram in Part 6 states:

"This zig is a combination of a stripped midpoint and stripped 2pt (trad 8pt). Also, as early as the opening roll, a double zig occurs when an opening 5 is brought down, creating a stack of four checkers on each point; the entire right side then counts 6."

Recapping, then, for both colors, the count breaks down neatly between the left and right sides of the board:

Blue is 4 + 5 = 9, and (with the 1-pip shift shown) White is 6 + 6 = 12.

Go on to Part 9, or read more below.

This is a clarification in response to a recent post from Lucky Jim.

 I would be grateful for confirmation and views on the following few points—Mirrors, Zigs and Zags. My understanding is that the Mirror applies in Naccel when you have two sets of checkers on opposite sides of the board. For the "standard" Mirror, the set on the side with the home board for that colour's checkers is always one pip closer to the bearoff tray.

Correct. From this mirror, you can march both points as many pips equally leftward or rightward as you like and the count remains 4 (the number of checkers in the mirror).

### Zigging and Zagging

 If you move a set of 2 checkers backwards 3 pips (zag) the count goes up by 1 (two checkers × 3 pips = 6 pips). If you move a set of 2 checkers forward 3 pips (zig) the count goes down by 1 (two checkers × 3 pips = 6 pips).

Right, and right again. A mirror (of two points) counts 4. A zag counts 5, and a zig counts 3.

 Am I correct that provided you keep them on opposite sides of the board you can move the set on either side of the board the three pips forward or backwards for the purposes of doing a count?

Yes, though I'd like to clarify that your goal is to instantly see (by the angle) these other-mirror formations as zig = 3 and zag = 5, rather than every time thinking "it would be a mirror of 4 if it were here, but since it is slid forward 3 pips it is a zig mirror of 3" (or "... since it is slid back 3 pips here it is a zag mirror of 5"). I realize this instant vision doesn't come all at once.

Points of a formation need to be on opposite sides of the board to be called a mirror (such as the one diagrammed below). However, to be clear, there is no restriction on where you zig or zag any more than there is a restriction on where you hop a single checker. Indeed, zigging or zagging is really just another form of hopping: two checkers 3 pips each (instead of one checker 6 pips).

In other words, you can freely cross sides of the board when hopping, zigging or zagging. For example ...

This is your friend the mirror—count of 4. If you zig the front point forward 3 pips you end up with a zig. If you zig the back point forward you also get a count of 3, but you end up with ...

The above formation counts 3, same as the zig; however, it is not called a "zig" (because the points are on the same side of the board and they are adjacent). It has morphed into a block.

The zig and the block are based on the same underlying design: two checkers each on the fourth and fifth points of a quadrant field—those points just happen to be in the same field now.

By the same token, starting from the mirror in the previous diagram (count of 4), you can zag the back point backwards, achieving a zag (count of 5), or you can zag the front point backwards (also getting a count of 5), which looks like this:

Again, you have crossed sides of the board, brought your checkers together and created a block. This one is called the midblock (count of 5), first introduced in Part 6.

You can just as easily morph the blocks (of the last two diagrams) into zig mirrors and zag mirrors by employing the same technique in reverse. Just remember that when you zig (forward) you subtract 1 from the count, and when you zag (backward) you add 1 to the count.

Each time you make a movement (zig, zag, hop, shift, whatever), I recommend that you use it as an opportunity to learn or reinforce the counts of both the pre-move formation and the post-move formation; this will bring you closer to recalling them as instant counts the next time you see them.

 For the position [above] the blue count was a Mirror for 4 (n−3 and n15). The rest of the checkers in the homeboard and the n1 and n2 are a poof which just leaves moving the checker on n14 needing to go two pips forward to the 2 Super for a count of 2(2) giving a total of 6(2).

Very good! For succinctness, you might consider using the term "near side" in your explanation, for example:

"n−3 and n15 is a mirror of 4. The rest of the near side is a five-prime poof, and the blot counts 2(2). Total of 6(2)."

In my opinion, a slightly faster way is to mirror all three back checkers against the front, for a count of 6, and add the 2 pips necessary to create a poof. Total of 6(2).

For clarity, I've removed the mirrored checkers below (that gave the count of 6). Playing +2 pips (either n2 to n0, or n0 to n−2) restores the poof.

 In terms of white's checkers I wasn't sure of the best way to count this and would be interested in how others would tackle this.

Petter spotted that by moving the roofer 1 pip onto S3, it can be poofed against the six-stack on n−3. This useful poof is shown in this diagram of Part 7, the only difference being that the sym is a three-prime instead of a six-stack.

Petter then added this +1 pip movement (i.e., getting roofer to S3) to n−3 to n0 (−3 pips) and four checkers on n1 to n0 (+4 pips) for a net of +2 pips. He added that to the triplet remaining on n2, which counts 1, for a total of 1(2).

Petter's alternative count also involved +1 pip for the roofer (poofing against the n−3 stack), then moved the n−3 spare 5 pips back to n2 (net of −4 pips) and counted 2 for the double block. That's a total of 2(−4), which agrees with his 1(2) count above.

I try to make my shifts two-sided (shift and countershift) whenever possible, and it's amazing how often there is a dual purpose shift available. I chose this one:

S3 and six checkers on n−3 is a poof, and the seventh checker on n−3 poofs with the n3 blot. For counting, that leaves only six checkers, composed of a block of 1 plus the 2 pips on n1. Count of 1(2).

 Final query in this post. In practical terms could Nack confirm how far up the 6 times table a Naccel counter should know for the calculations likely to be performed over the board?

You already know this part but I'll review for others: To convert a Naccel count to a traditional count, multiply by 6, add (or subtract) the leftover pips and add 90. For example Blue's count of 6(2) here converts to 6 × 6 + 2 + 90 = 128, and White's count of 1(2) converts to 1 × 6 + 2 + 90 = 98.

When you practice converting Naccel to traditional, start by memorizing just:

 −5 = 60 0 = 90 5 = 120 10 = 150

That way when your count is "1" you'll see it's 90 + 6 = 96, when your count is "9" you'll see its 150 − 6 = 144, etc., and before long you'll know all the 6-counts (so that the conversion process will be nothing more than adding or subtracting the baby pips).

That said, aside from counting practice (to confirm your Naccel counts with trad/decimal totals, which are currently the only way that books and bots display pipcounts), the only reason to convert to trad is that you're in a straight race and the difference between totals is around 10 pips (so that you might need to apply a race formula). In that case, note that you need convert only one side to a trad total (you already know the difference in pips).

Later, I'll coach you on the Naccel race formula, and then you won't be converting to trad at all (unless you want to).

Go on to Part 9.

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