Pip Counting

Naccel 2—aN ACCELerated Pipcount
Part 5:  Primes and Hopping
Nack Ballard
January 2010

This is Part 5 of the Naccel 2 series. As always, Naccel point numbers are labeled in white, Super (Super-point) numbers in black.


I'll start by showing you how to count a six-point prime:

 ' ' ' ' ' ' ' ' ' ' ' '

 '2O2O2O2O2O2O ' ' ' ' '


To count any six-prime (including a closed board), simply add together the endpoints. For the prime shown above, the endpoints are the (Naccel) −4pt and 1pt. Adding these together, −4 + 1 = −3.

[You can also choose to count this formation as two six-syms: one around the −3pt and the other around the 0pt. This essentially adds together the second and fifth points of the prime. Yet another option is to sum the middle points: −2 and −1 sum to −3.]

Let's apply that nugget of knowledge to the first of the two positions submitted by Jim:

 '2X1X2X '2X3X2X2O ' ' '

2X2O2O2O2O2O1O ' ' '1O1X

Naccel seems to get more than its fair share of luck. Here, you can kill two birds with one stone with the following 4-pip shift:

 ' ' ' ' ' '2O ' ' ' ' '

 '2O2O2O2O2O2O ' ' ' ' '


The count is −3 for the six-prime (as we just learned), +4 for the two checkers on S2, and +3(1) for the roofer. Total of 4(1).

[If you know (as I did) that an S3 blot (+3) offsets/poofs this six-prime (−3), the count is even shorter: 4 for S2, plus 1 pip to enter the roofer.]

On to White's position. It can be divided into two groups. The first is


 ' '1X2X '2X3X2X ' ' ' '

 ' ' ' ' ' ' ' ' ' ' ' '

Can you see this is a poof? Just move the blot and spare equally towards each other, filling in the vacant point. This means White has only five checkers (shown below) that you actually need to count.

 '2X ' ' ' ' ' ' ' ' ' '

2X ' ' ' ' ' ' ' ' ' '1X

I happen to know that the Naccel 18pt (S3) and the −4pt are equidistant from the handy midpoint (7pt), so I just shifted (more like swept) the left four checkers to join the fifth on the midpoint, for 5(5). I am impressed that Ian found this solution as well.

That "five-sym" mixed fraction count is advanced or hard to find, so instead I'll suggest to you this 2-pip shift:

Zag plus 5 pips5(5)

 ' '2X ' ' ' ' ' ' '1X '

2X ' ' ' ' ' ' ' ' ' ' '

Do you recognize the zag (zag mirror)? Take a look at this diagram (left), isolate just the left zag (of the double zag there), and notice that White's zag here is the upside-down image. Practice spotting mirror and zag-mirror formations for both sides—it gets a lot easier.

A zag counts 5, plus 5 pips for the blot; White's entire count is 5(5).

Recapping both sides: Blue is ahead 4(1) to 5(5), which is a lead of 1(4), or 10 pips.

Okay, now for Jim's second position:

1X ' ' ' ' ' ' ' ' ' ' '

1O3O2O2O3O3O '1O ' ' ' '

For Blue, a trivial 1-pip shift will do the trick:

 ' ' ' ' ' ' ' ' ' ' ' '

2O2O2O2O2O4O '1O ' ' ' '

Closed board plus 2 pips−5(+2)

A closed board is a standard count of −5 (or sum the end points: −5 + 0 = −5). Adding 2 pips for the outside blot yields a count of −5(+2).

Now, White's position:


1X ' ' ' ' ' ' ' ' ' ' '

 ' ' ' ' ' ' ' ' ' ' ' '

Simplest, perhaps, is to bear off 4(1) + (1) = 4(2) and subtract 15 (i.e., 1 for each borne off checker). Alternatively: enter the roofer 2 pips, creating a mirror of 2, that's 2(2) so far; then subtract the 13 borne off. Either way, White has a count of −11(+2).

Ian chose to view White's two checkers as symmetrical around the midpoint—equivalent to having them on the midpoint = 2(2). Then he subtracted the 13 borne off, achieving the same −11(+2) result.

(Ian's choice relates to the method described for bundling five White checkers, four diagrams back. The shift is easy to see here if you "hop" the roofer +2 and back-hop the other checker −2, so that they meet on the midpoint. Hopping is a valuable visual aid that will be explained later in this post.)

[When the set of checkers as a whole is closer to S−1 (the bearoff tray) than to S0, a reasonable option is to bump up the Supers by 1 so that S0 is the bearoff tray). However, I strongly advise against adding this wrinkle until your regular S0 counts feel confident and natural.]

This time, we have a third position, submitted by Ian.

2X '3X3X3X3X ' '1O ' ' '

3O5O5O ' ' ' '1O ' ' ' '

For Blue, again a 1-pip shift is all that's needed:

2X '3X3X3X3X ' '1O ' ' '

4O4O4O1O ' ' '1O ' ' ' '

Mirror plus twelve-sym, minus 2 pips−6(−2)

The mirror outside is 2, the two six-syms (twelve-sym) around the −4pt count −8, and the inside blot is (−2), yielding a count of −6(−2). If you don't see how two six-syms are symmetrical around the −4pt, picture two three-primes on top of each other (for example).

[Alternatively, you can poof the 2pt and −2pt blots, count the twelve-sym as −8, and add back in 1(4) for the far-side checker, resulting in −7(+4). That agrees with the −6(−2) count above.]


Before we count White's position, I'll introduce you to a technique called "hopping."

Visualize three hops3

 ' ' ' '1X ' ' ' ' ' ' '

 ' ' ' ' ' ' ' ' ' ' ' '

There are (at least) three ways to count this two-checker formation as "3." One is to learn it as an automatic 3 count. Another is to recognize it as a mirror variant I haven't shown you yet (and complicated by having to visualize the roofer where it really is—one point to the left of S3).

The third way is to "hop." A hop is a visualized movement of one supe (6 pips), counting +1 for each forward hop or −1 for each backward hop. Having seen and played a lot of double 6s over the years, we all hop naturally.

By hopping around the board in the above diagram, White's roofer will land on her own bar point (Naccel 1pt). How convenient! This creates a little poof, the checker on the 1pt and −1pt cancelling each other out. (I trust you can visualize the position after White's triple-hop without the aid of an after-diagram.)

Returning to the main position:

2X '3X3X3X3X ' '1O ' ' '

3O5O5O ' ' ' '1O ' ' ' '

Here, you can reach an easy count in two movements (totalling 3(3)): one is to move White's roofer to her bar point in 3 hops. The other is to close her board with 3 pips. Like this:

Closed board (and a tiny poof)-5

2X2X2X2X3X3X1X ' ' ' ' '

 ' ' ' ' ' ' ' ' ' ' ' '

As explained earlier in this post, a closed board counts −5 (and the extra three checkers on the 1pt, 0pt and −1pt are a tiny poof). Accounting for the movement of 3(3) it took to get here gives you a net count of −2(+3).

Hopping will open whole new worlds of understanding and speed counting to you. In this case, if you know/remember that roofer plus −1pt checker = 3, great. If you don't know or forget, then hop, and you'll be more likely to remember the instant 3 count of that little formation for next time.

Recapping, Blue's count is −6(−2) and White's count is −2(+3). Blue leads by 4(5).

Go on to Part 6, or read more below.

Further Reading
Jim has been especially courageous in laying out his early attempts at Naccel counting for all to see. Here, as a reward, is an unplanned post, to hopefully benefit him and anyone else who takes the time to read through.

We will walk through Jim's count of a position in detail. I hope he can forgive any critique that seems less constructive than it is fully intended to be. We can all use help in improving our counting habits.

For this post, I felt that including diagrams other than the main position would interrupt the flow. While not essential, it might help to set up the position below and follow along on a board.

For reference, a faster count for Blue in this position is illustrated in this diagram above.

Counting Habits

It is very helpful to see your and Ian's Naccel counts. It gives me a better idea what issues to address, and how quickly.

In the interest of being efficient, for the most part, I'll be sticking to the general Naccel posts rather than addressing individual counts. Today, I'm going to make an exception because I have noticed some counting pitfalls worth discussing.

Your notes start below the position diagram, in italics.

 '2X1X2X '2X3X2X2O ' ' '

2X2O2O2O2O2O1O ' ' '1O1X


With this position the approach I took was as follows:

Blue: For counting purposes the man on the bar needs to go on the 24 point so that is one pip forward to the 24 point for a count of 3(1).

A poor start. Leading off with a fractional count slows down the count and reduces your options. Not only do you have to carry a separate baby pip-count around, but you can no longer use the roofer as a shifter or as an offset checker (in case that comes in handy)—he's gone. Maybe you'll get lucky and snag integer counts the rest of the way, but the best you'll do is break even.

If you must lead off with a single checker, make it part of an integer formation or at least hop it (i.e., 6 or a multiple of 6 pips) onto another blot or to an existing point to consolidate the board position (minimize confusion).

The midpoint is often a good place to hop a roofer (a movement of +2) because I have created special formations involving the midpoint and it is a great collector of leftover counts. (It's easy to multiply 1(1) by the number of checkers there.)

In the actual position, the midpoint is vacant, but the bar point (Naccel 1pt) turns out to be a magnificent landing spot. You can wipe out twelve checkers with a single blow (and any integer count it happened to be would be fine—the fact that the −3 count and +3 hop exactly offset is a further bonus). You could then finish with a dual purpose 4-pip shift: back up the far-side point to S2 (= 2 × 2), and slide the lower right blot forward to the 1pt, for a count of 4(1).

Oh, one more thing. Please refer to points by Naccel numbering (trad minus 6), which is intimately connected to the counts of the formations: get in the habit immediately. For example, the highest point on the board is "S3" or the "18pt," or you can say "Naccel 18pt" or "n18" as an occasional reminder). If you want to reference trad for some reason, "trad 24pt" or "t24" works.

For the remainder of this post, I will convert your point numbers from trad to Naccel. The square brackets designate my edit of your original text.

I moved the man on the blue [(Naccel) 1pt] 4 pips backwards to go on top of the checker on the [5pt] so as to create a zag for 5(−4).

Okay, you didn't know/see the six-prime, but the zag is a nice consolation prize (and shows you've paid attention). However, instead of carrying around another 4 pips in baggage, you should be looking for a way to countershift 4 pips profitably (or at least neutrally). The only place left is in the inner board, so that's where you do the deed.

Probably the best 4-pip countershift is lifting the −1pt checkers to the −3pt. That separates the irrelevant 0pt from the rest of the pack so it is easy to see what you have left, and, more importantly, it creates a pair (−1) atop a three-prime (−3). Indeed, that would complete the count.

This means we are running with a net (−3 pips).

Right (if you are referring solely to the baby pips and not the supes also accumulated thus far): the +1 pip from the 3(1) leadoff, and the −4 pips from the one-sided shift to the zag. The questionable technique of counting the 4 pips instead of countershifting aside, note there was no advantage in leading off with the roofer. Even if you find no way to use it, counting the extra pip last (or at least as late as possible) is simpler.

The six-sym around n−3 gives a count of −3.

Okay, good, that's the best you have now. However (if you can pardon me driving the point home) note that if you still had the roofer, you could either: (a) hop it +3 to the (now vacant) bar point and wipe out the −3 three-prime so that you don't have to alter your running supe count; or (b) cancel the +3 against the −3 and alter your count by only the leftover pip. Of these two choices, (a) is usually better technique, and in a moment you'll see the benefit.

There are two pips to move back from the n−1 point to the n0 point so add those to the existing −3 pips gives a total of −5 pips.

If you had still had the roofer and chose method (a) above, you could cancel a 1pt pip against one of the two −1pt pips, and you'd only now have to subtract 1 pip from your baby-pip count instead of having added 1 pip earlier and now subtracting 2.

In other words, while you should have countershifted the 4 pips instead of tracking them, the roofer leadoff exacerbated the problem—it squandered a later chance to cancel both the −3 three-prime and one of the near-side pips at the end.

This gives 3 + 5 − 3 = 5 and −5 pips.

Correct. The full tracking of your count is 3 + 5 − 3 = 5 supes, and +1 − 4 − 2 = − 5 pips. Without the roofer leadoff, your count could have been 5 − 3 + 3 = 5 supes, and −4 − 1 = −5 pips. Had you countershifted the 4 pips as well, you could have had 5 − 1 − 3 + 3 = 4 supes, and +1 pip. And with the best countershift (six-prime and two-to-S2, see this diagram above), the count was 4 − 3 + 3 = 4 supes + 1 pip.

In these comparisons, the "−3 + 3" looks like two separate counts but the consecutiveness of the offset helps. The more adept you become the more it feels like a 0 count or not counting at all.

5 is the same as 4(6) so 4(6) less 5 pips is 4(1).

Naccel count for Blue is 4(1).

True, but in many or most cases, this is an unnecessary step. It is fine to leave it as 5(−5).

The main position is repeated below, with the point numbers labeled from White's perspective.

 '2X1X2X '2X3X2X2O ' ' '

2X2O2O2O2O2O1O ' ' '1O1X


White: Two men on the [18pt] gives a count of 6.

Okay, at least you're starting with an integer. Still, it is better to first find a problem and fix it. The S3 checkers can be added in anytime, and it might end up better to include them in a larger formation later. (I mentioned two such examples in the fifth and sixth diagrams above.)

Move the checker from the [7pt to the 6pt] for a count of 1(1).

There you go with the fractional counts again. Rather than count this checker as 1(1) you would be better off counter-shifting the −3pt blot and one of the −4pt checkers back to the 0pt—i.e., getting them off the board and keeping your zero count ... well, I mean your 6 count that you started prematurely.

You seem to have adopted a methodology of starting with the back checkers first and working your way around the board. The reason you have developed this habit is that for traditional counting it is easier to count the big numbers first, and you also fear you'll forget something.

In Naccel, all the numbers are small, and there are a lot more checker-combining resources available through board symmetry. You are squandering most of these resources if you just clunk around, back to front. Instead, you should be looking to blow away as many of your checkers as you can with your first formation, anywhere you can.

The advantage of your approach is that you pretty much know what area of the board you're counting in next. But you're giving away too much in the process. Fixing this habit may take you out of your comfort zone (if it can be called that) and further slow your counts at first, but that will be a temporary effect and your counts will gather a surprising amount of speed before long.

The checkers on the white [−2pt and 2pt] cancel each other out.

This is okay, though it seems more natural to me to wipe out the entire five-prime with the +2/−2 pip shift I pointed out earlier.

If we move the checker forward one pip from the [−3pt to the −4pt] then we have a triplet on n−4 so that gives a count of −2 (plus 1 more pip).

The triplet is a good idea. A better execution, though, instead of your last step of removing the (Naccel) 2pt and −2pt, would have been to shift the 1pt spare back to the 2pt, and at the same time the −3pt blot to the −4pt, creating two triplets. These triplets could either be counted separately (+1 − 2 = −1) or viewed as a six-sym around the −1pt (faster).

In general, it is better to shift checkers than to adjust a running baby-pip count. Shifting will develop better habits; before long you'll be spotting bigger formations, more of your shifts will be dual purpose and you'll need fewer of them.

This leaves a final 3 checkers on the [1pt to move to the 0pt] (3 more pips).

The count excluding pips is 6 + 1 − 2 = 5 and there are 5 pips to add so a Naccel count for white of 5(5).

Right. A count of 6 + 1 − 2 = 5 supes, and 1 + 1 + 3 = 5 pips.

You're improving. And, to be fair, you did better in the count of your second position of the same post; that's why I elected to comment on this one: I figured you'd learn more.

I hope that helps. You'll get there—stick with it!

Go on to Part 6.

Nack Ballard is a top international backgammon player.  He is a coauthor of Backgammon
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