Pip Counting

Naccel 2—aN ACCELerated Pipcount
Part 6:  Midpoint Combinations
Nack Ballard
January 2010

This is Part 6 of the Naccel 2 series.

The basic squads you have learned thus far are the six-stack, triplet, and pair. There are also shift-variants of these squads, some of which you've seen (various six-syms and the wedge), but we'll put those aside for now.


Today, we'll look at another basic squad called the "block." Whereas the others mentioned above use six, three or two checkers, this squad uses four:

 '2O2O ' '5X '3X ' ' '2O

2X ' ' ' '5O2O2O ' ' '5X

Position with two Blocks14(2)

The block just to the right of the Naccel 0pt (trad 6pt) arises after Blue plays a standard opening 61.

To count a block—or indeed any two points on the board—add together the point numbers and divide by 3. The near-side block occupies the (Naccel) 1pt and 2pt, so: 1 + 2 = 3, divided by 3, gives you a count of 1.

The block on the other side of the board, formed by the four back checkers, is on the 16pt and 17pt. Summing these point numbers gives you 33, and dividing by 3 gives you 11. (By the way, blocks always generate odd-numbered counts.)

[An alternate counting method for blocks: Count three checkers as being on the nearest Super and the other as being on the second nearest Super. So, here, the near block counts (0 × 3) + 1 = 1, and the far block counts (3 × 3) + 2 = 11.]

As always, the 0pt checkers are invisible. Blue's entire position counts 11 + 1 for the blocks, plus 2(2) for the mid, for a total of 14(2).

Zig Mirrors

Next, let's have a look at some useful ways to combine two checkers on the midpoint with a near side point:

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

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


You recognize this formation as a mirror. The number of total checkers in a mirror gives you the count, so this mirror counts 4.

To reach the next formation, we will "zig" (move forward) the near-side point 3 pips. Two checkers times 3 pips is a total of 6 pips, or one supe, which means the overall count of 4 is reduced below by 1, to 3:

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

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

Zig mirror3

This formation is called a "zig mirror" or "zig" for short. A zig counts 3, one less than a regular mirror. [Zig is both a noun that is short for zig mirror, and a verb that means to move a point 3 pips forward.] For reference, if you drop the far-side point of a zig mirror straight down, the near-side point always leads it by 4 pips.

You can also count a zig in the same way you count a block (or any two points): add the point numbers and divide by 3. So, here (7 + 2) ÷ 3 = 3.

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.

Note that if you hop the two midpoint checkers down to the (Naccel) 1pt, you end up with the near-side block in the first diagram of this post. That block counts 1, and adding the 2 hops imagined here is another way to achieve the count of 3 for the diagram directly above.

This hop example, and the series of zigs we're performing, help illustrate that all Naccel formations can be built and rebuilt upon board symmetries and simple, logical techniques.

Diag Mirrors

Let's zig the front point again, thereby further reducing the count by 1:

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

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

Diag mirror2

This position is called a "diagonal mirror," or "diag" for short. To count a diag, divide the number of checkers in the formation by 2. Here there are four checkers, so the diag is 2. This is 1 less than a zig (previous diagram) or 2 less than a point mirror (two diagrams back).

An easy way to reconcile this count of 2 is to hop the two midpoint checkers (+2) to the bar point, creating a little poof. Another way is to shift the midpoint to S1 (count of 2) and the −1pt to S0 (count of 0).

As your pattern recognition improves, you will spot (or create) larger groups that are combined from smaller ones. For example ...

 ' ' ' ' ' ' ' ' ' ' '4O

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

Combined formation5


 ' ' ' '3O5X '3X '2X '4O

1X ' ' '2O4O '2O ' ' '4X


The left-hand diagram is a combination of the previous two diagrams. Two "mids" (midpoint checkers) plus the −1pt are a diag (+2), and two mids plus the 2pt are a zig (+3). If you see this formation enough times, the eight checkers (plus any that are on S0) will jump out as an instant +5.

The right-hand diagram puts the formation in a context. Blue's entire position counts +5 (midpoint and near side) + 7 (triplet) = 12.

Diag Zig Mirrors

Continuing from the Diag diagram, letís zig the front point forward one last time:

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

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


The previous formation counts 2, so this one counts 1. If you like, you can call this formation a diagonal "zig mirror," or "dizig" for short. However, the only other formation with the same relationship, one space to the left, rarely arises; so rather than fill your head with the term "dizig," it might be easier just to remember this specific midpoint formation as 1.


Okay, now let's add a specially selected pair to the above formation.

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

 '2O2O ' ' ' ' ' ' ' ' '


And here it is. Midpoint plus −4pt counts +1, and the pair on the −3pt counts −1. This important six-checker formation counts zero: I call it a midpoint poof or "midpoof" for short.

If you shift the two low points outwards, so that Blue instead owns the −5pt (trad 1pt) and −2pt (trad 4pt), that is also a midpoof, though it arises less frequently in practice.

Commit the diagrammed midpoof to memory, and you will encounter it often in your Naccel adventures. We'll even have a chance to use the midpoof later in this post.


Now we'll look at midpoint formations on the far side of the board.

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

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

Far-side Reflection8

This far-side reflection counts the same as if you were to stack all four checkers on S2 (the point around which they reflect). In other words, double the numbers of checkers you see: that gives you a count of 8.

Okay, let's zig (move forward) the back point, reducing the count by 1:

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

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


I call this formation "midgold" because it combines the midpoint with the golden anchor. It counts 1 less than the far-side reflection of 8, which makes it 7.

There are many other ways to reconcile midgold as a count of 7. To name a few:

  1. Add the point numbers 14 + 7 and divide by 3.
  2. Shift the anchor back a point to get a 5-count pair and the midpoint forward to S1 (count of 2);
  3. Shift one midpoint checker forward 7 pips to the 0pt and the other back 7 pips to the anchor, creating a 7-count triplet there.

Midgold occurs with regularity, so you would do well to remember that its count is 7. We'll even have a chance to apply midgold later in this post.


Let's zig the back checkers forward again, but we'll skip over the next spot because Opponent typically occupies her 2pt (trad 8pt). Instead we'll zig twice, reducing the count from 7 to 5. Can you visualize the resulting formation, and what is another way to count it (other than 7 − 2)? Answer below.

 ' ' ' ' ' ' ' ' ' '2O2O

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


Having hopped (or zigged twice) from the gold part of midgold, you now have a block. I call this block "midblock," because it contains the midpoint. To count it, sum its point numbers 8 + 7 and divide by 3, for a count of 5.

When you have four (or more) checkers on the midpoint and you need a back 2-pip countershift, the midblock is very handy.


Let's look at Lucky Jim's most recent submission:

1X1X2X '2O3X '4X2X ' '2O

 ' '2O2O '2O1O3O1O ' '2X

Poof, and Midgold7

First, let's count Blue.

Maik quickly spotted Blue's near-side poof: simply move the 2pt spare back, covering the 3pt, and lift the 1pt blot to the 0pt. Poof! (If you don't see the poof, please review the last two diagrams of Part 1, the last diagram of Part 2 (especially), the second diagram of Part 4, and the fourth diagram of Part 5.) There's nothing on the near side to "count."

That leaves the far side. Aha, midgold = 7; that's Blue's entire count!

Let's repeat the diagram with point numbering from White's perspective:

1X1X2X '2O3X '4X2X ' '2O

 ' '2O2O '2O1O3O1O ' '2X

For the White count, I recommend this simple 1-pip shift:

Poof, and Six-stack plus 1 pip2(1)

 '2X2X ' '3X '5X1X ' ' '

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

Do you see the midpoof? It is the six-checker formation composed of the midpoint and the four low-point checkers: count of zero. And of course the (invisible) 0pt checkers also count zero.

You need count only the six checkers on the top right. This is simply a six-stack on the (Naccel) 2pt plus 1 pip. White's entire count is 2(1).

Go on to Part 7, or read more below.

Further Reading
In this extra post, I'm combining two clarifications. The first is a recap of the Blue and White counts in the submitted position discussed above. The second is a clarification of the location of "blocks."

Midpoof, and six-stack plus 1 pip2(1)

1X1X2X '2O3X '4X2X ' '2O

 ' '2O2O '2O1O3O1O ' '2X

Poof, and Midgold7

White's count is 2(1) and Blue's count is 7, so White leads by nearly 5 supes (29 pips, to be exact). It therefore looks like a strong money double or redouble, though with the existing midpoint contact Blue can still take.

The remainder of this post is a question-and-answer discussion of the block. Jim said:

Nack, can I just check that I am looking at this correctly? It looks to me as if the block only applies to certain sets of points, namely those which divide exactly by 3 (4 sets of points on one side and 4 on the other). Had, in your first example, the four blue checkers in white's home board been on the n16 and n15 points, you would not have been able to use the block but would have done something else?

Astute of you to notice.

Strictly speaking, you can count any two points on the board (in the same quadrant or in different quadrants) by adding the point numbers together and dividing by 3. However, if you choose those points at random, there is only a 33% chance you will obtain an integer count. (In the case of two consecutive points in a quadrant, your chances improve to 40%; and if you additionally exclude the Super, you're 50%.)

Your example of n16 + n15 gives you a count of 31 ÷ 3 = 10(2). If the other eleven checkers had already been counted (or canceled), this would actually be a good way to obtain the leftover count. Otherwise, it leaves you with the burden of a baby-pip count to carry forward.

A proper block is one that yields an integer count.

A six-point quadrant, from high point to low, is composed of the Super, the high block points, the pair point, and the low block points. Each will visually stand out, and even jump off the board, the more you practice.

In short, the block points flank the pair point. In your example, n15 is a pair point and therefore cannot be part of a block. Likewise, with n3 in the diagram below: Blue's blocks exist on either side of the vacant pair point (the only point in that quadrant where Blue can have a pair).


2O ' ' ' '5X '3X ' ' ' '

2X ' ' ' '5O2O2O '2O2O5X

Blue's two Blocks flank the Pair point10

Go on to Part 7.

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