Pip Counting

 Naccel 2—aN ACCELerated Pipcount Part 12:  Motion Poofs Nack BallardJanuary 2010

This is Part 12 of the Naccel 2 series.

I'll start by reintroducing an old friend:

The midpoof was first illustrated (and discussed) in Part 6. The midpoof shown in the left-hand diagram arises more often than the midpoof in the right-hand diagram (shown for reference).

Next, I'll rehash two basic squads:

The left-hand diagram shows the most commonly arising triplet, having a count of 1. The right-hand diagram shows a pair, also having a count of 1.

### Chairs

Combining the triplet and pair produces this squad combination:

The formation on the left, combining the pair and triplet, is called a "chair," and the count is 1 + 1 = 2.

If we replace the triplet 1 (on n2) with the triplet 2 (on n4), shown in the right-hand diagram, we also have a chair. The right-hand chair has a count of 3.

Moving an entire squad or squad combination one quadrant alters the count by the number of checkers in the formation. For example, let's look at the same chairs located a quadrant lower:

The chairs in the previous diagram pair counted +2 and +3; therefore, the chairs in this diagram pair (five checkers, one quadrant lower) count −3 and −2.

Chairs usually appear in the near-side quadrants, but they occasionally appear in the back quadrant:

These five-checker formations are two quadrants higher than the original chairs shown of +2 and +3, so their counts are ten higher: +12 and +13 (left-hand and right-hand diagrams, respectively).

To count any chair, take five times the nearest Super and add 2 (if behind the Super) or subtract 2 (if in front of the Super). For example, the chair above left is 2 × 5 + 2 = 12, and the chair above right is 3 × 5 − 2 = 13.

[Some alternative counting methods for the chair are: (a) Take five times the lower Super and add 2 (when close) or 3 (when not close); or (b) Take three times the nearest Super and add twice the other Super; or (c) Add your known triplet and pair counts together.]

It is easy enough to count a chair; nevertheless, I have found it worthwhile to commit them to memory as instant counts, particularly the near-side chairs of −3, −2, +2 and +3 (listed from left to right).

Chairs are useful counts on their own. They can also provide the basis for useful poofs; for example:

The S3 blot counts +3 and the chair counts −3. This six-checker formation is a "chair poof."

### Shift Poofs

Next, let's discuss "shift-poofs." You can review one example in Part 5. Below is another example:

For visualization purposes, the "roofer" (the checker on the roof) is actually on a point to the left of S3. Therefore, if you shift the roofer 1 pip to the right, it lands on S3. To keep the count the same, you can countershift one of the n−4 spares to n−3. By doing so, you achieve the poof in the right-hand diagram (same as the previous diagram).

So ... isn't the left-hand diagram already a poof? Yes, that's right! The only difference between a shift-poof and a poof is that the latter is a standard formation or one that you already recognize as a poof.

Consider the two poofs below. (And for variety, let's switch from Blue to White.)

As you know, a closed board (with or without checkers on S0) counts −5. Okay, now:  Any formation with a count of +5 can be poofed against a closed board.

In the left-hand diagram, White has a pair of +5. This pair is sometimes referred to as the "silver anchor" in backgammon. In combination with the closed board (count of −5), I call the entire formation the "silver poof."

In the right-hand diagram, White's midtriangle counts +5, and so it poofs against White's closed board of −5 (which is the same regardless of the presence or absence of checkers on S0).

Pop quiz: What is the count of the diagram below?

An easy way to count this is to shift the two n−4 spares to n−3, and two checkers on n0 to n−1, thereby "closing" the board. It doesn't matter that S0 ends up occupied by fewer than two checkers. Indeed, even if S0 started with no checkers on it, you can freely conjure them.

The result of shifting is essentially shown in the previous (left-hand) diagram.

Okay, let's try another (requiring more imagination). What is the count of the left-hand position below?

While there are other ways of counting, try this long-range shift: double-hop the roofer to the midpoint, and then (to offset) counter-hop the two bar point checkers back to the midpoint. This gives you the mid-triangle poof in the right-hand diagram.

### Motion

Another way you can achieve poofs is with "motion" (movement/shift without equal countermovement). For example:

There is more than one way to count the left-hand diagram, but a good, clear way is to move the trailing checker 9 pips, thereby covering n1 and creating a basic near-side poof. The movement of 9 pips can be thought of as a hop of +1, plus a slide of 3 pips, for a forward motion of 1(3).

The result of the 1(3) motion is shown in the right-hand diagram.

You can also use partially offsetting motion (i.e., both forward and back). For example:

From the left-hand diagram, there are many ways to use partially offsetting motion. I'll list three:

1. n7 to n2, and n1 (back) to n2
2. n7 to n2, and n−2 (back) to n−1
3. n7 to n1, and both n−2 checkers (back) to n−1.

These motions are 5 − 1, 5 − 1, and 6 − 2 pips, respectively, for a net of +4 pips in all three cases. The result of example (A) is shown in the right-hand diagram.

It is also possible to combine shifting with motion. For example (from the left-hand diagram), you can:

1. Move the midpointer down 4 pips and shift together the two blots.
2. Shift the n7 and n1 blots a pip towards each other and move the trailing blot 4 pips (making the n2 point).
3. Shift apart the n−2 checkers by a pip, and move the midpointer down 4 pips (in either order).

Either (a) or (b) achieves the right-hand diagram.

Of these three options, (a) seems most natural to me. Also, the advantage of (a) over (A) is that as you gain experience, you will see n−2 n−2 n1 n3 as a small (shift-) poof (equivalent to the right-hand diagram) after the one-way motion of 4 pips faster than you'll see and account for the two-way motion of 5 − 1 = 4 pips. In other words, (a)'s little shift step will gradually decrease in mental impact and ultimately you'll transcend (skip) it.

I'll finish off here with my usual reminder: Motion (i.e., non-offsetting or partially offsetting) is best used for leftover counts or to create a position for which you can see that the remainder is exclusively integer counts (preferably poof(s)). Don't get in the habit of using non-integer motion or other leftover counting techniques more than once during a count.

If you find yourself using a lot of motion or multiple shifts to count a single position, you're straining unnecessarily; I recommend that you practice looking for Naccel formations that are more readily at hand. Allow me to modify a line from a Moody Blues song: "They're all around you if you can but perceive."

### Problem 1

It's time for full board counting. Petter kindly supplies our first position:

All participants wisely chose to poof the n1 and n−1 blots and (of course) the n0 point. For the remaining eleven checkers, Lucky Jim not only found a fast way to count them, but he also summarized his count efficiently: "m2+3 + Sym−1 + pair−1 = 3." (Bravo.) Those three formations are shown separately below.

Jim's first diagram is a straightforward midblot formation. "m2" means two midpointers plus a blot on the second point after a Super. The "+3" part identifies the relevant Super (in this case, S3).

Jim's second diagram is a sym around n−1, and his third diagram is a simple pair.

Sym−1

Pair−1

(Ian came up with the same correct count, except he subdivided the sym−1 (second diagram) into triplet−2 and triplet1.)

Let's have another look at Blue's position:

There is another counting method that is at least as fast. Remember the midpoof? It is composed of two checkers on each of n−4, n−3 and the midpoint.

If we remove the midpoof (as well as the small four-checker poof everyone else did), we are left with only five checkers to count:

This simplifies the count to: Mirror 2 + triplet 1 = 3.

Now let's repeat the position for White:

You can combine eleven checkers right away by shifting the two n−4 checkers back a pip and two n0 checkers forward a pip. That closes the board (noting that it makes no difference whatsoever how many checkers are or are not on n0) for a count of 5. Now that you've seen that White's inner board is 5, you just look to create a +5 formation to offset it.

In achieving this offset, there are numerous solutions. The one that occurred to me immediately is shifting the roofer 1 pip forward (to S3) and the midpointer back 1 pip, like this:

Here, S3 + mirror = 5 offsets (shifted) closed board = −5. Poof! That leaves only the n8 blot, for a count of 1(2).

Note that in the left-hand diagram I didn't actually illustrate the inside shift to close the board, in order to more closely simulate over-the-board practice. (After all, once you know the inside formation is 5, you don't have to keep revisualizing it as a closed board.) Nevertheless, I have bluntly added the right-hand (post-shift) diagram, for reference.

Let's revisit White's main position:

Another good solution is this motion poof:

Double-hop (+2) the roofer to the midpoint, and back up the n3 checker to the midpoint (−4 pips). This motion of 2(−4) is the entire count because you end up with

The midtriangle (also summarized as m3+2), has a count of 5. It therefore offsets the −5 count in the inner board. Poof!

In the right-hand diagram, the inner board is shifted and closed, for further clarity. This midtriangle poof was diagrammed earlier.

Again, the entire count of White's original position is the 2(−4) motion that was used to create the midtriangle poof. [This count agrees with the 1(2) leftover count generated by the (1-pip) shift-poof method.]

Below is another motion-poof solution, this time with one-way motion. From White's original position, the roofer hops just once (+1) and the midpointer moves +2 pips, which gives us

White's freshly moved n13 and n5 blots are a rev 3, while her n9 and n3 blots are (still) a mirror 2. Together, their +5 offsets the −5 in the inner board. Poof! (Again, the right-hand diagram includes the inside shift as well, for complete reference.)

The entire count is the forward motion of 1(2).

Yet another motion poof was produced by Petter. From the original position, he moved the roofer 1 pip to S3 and the midpointer 1(1) to n0, for a total of 1(2) motion, attaining this position:

[Note that this is effectively the same position reached in my solution. The difference is that I chose a 1-pip shift that isolated a blot having a leftover count of 1(2), whereas Petter used motion of 0(1) + 1(1) to actually put the blot onto n0.]

The checker on S3 plus the outside mirror 2 cancels against the inner board of −5. Poof! (If you scroll back, you'll see that's the way I counted it.)

It is interesting that (apparently) none of the participants noticed that the inner board could be closed with a simple shift (whereas in the past they've exhibited an awareness, eagerness even, to dunk this easy −5 count). This suggests a subconscious resistance to breaking the traditional 6pt (which in this case started with three checkers), but don't worry; that mental block will evaporate before long.

So, how did Petter count his post-motion position above? Answer: He poofed the S3 checker with the block −3 (consisting of two n−5 and two n−4 checkers), and he poofed the remaining double-split −2 inside with the mirror +2 outside. Nothing wrong with that! Petter's entire count is his 1(2) motion explained above the diagram.

### Problem 2

Jim brings us our second position, shown below:

We'll start with Blue. When the count is this low, it is reasonable to count forward to the bear-off tray: Motion of 3(5) gets the roofer to the 2pt, plus 2 for the six-stack yields a count of 5(5).

However, counting Blue to the bear-off tray means you either have to subtract 15 (which means you may as well count the normal way to begin with) or count White to the bear-off tray as well (which wouldn't be efficient). The dilemma exists only because it is so unusual to count a position where one side has a lead this enormous, but we'll play along anyway.

Counting the regular way (to n0), you can still start with motion of 3(5) but then subtract the six-stack of −4, to get a difference of −1 pip. Then subtract the 9 checkers borne off, for a count of −9(−1). This is pretty much how Petter did it.

It is also possible to count this position with minimal movement; just a 1-pip shift and a 1-pip (backwards) motion, in either order. Suppose we start with the shift: enter the roofer to S3, and move one of the n−4 spares to n−3; that gives us

From the left-hand diagram, −1 pip of motion gives you the chair poof in the right-hand diagram. (The chair −3 poofs with the S3 checker.)

The −1p of motion, and −9 for the borne off checkers (not shown here), yields a count of −9(1).

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

Ian found a great solution: he shifted the n15 blot 2 pips to the left and the n8 blot 2 pips to the right, as shown in the left-hand diagram below:

The n17 blot and n−5 spare create a mirror +2, plus the S3 checker gives a count of 5, which poofs with the closed board of −5. (As always, the presence or absence of checkers on n0 is irrelevant.) The only checkers that contribute to Ian's count are the two on S1, for a total count of 2.

It happens that the method I chose, shown in the right-hand diagram, was nearly identical to Ian's. The difference: I countershifted S1 to n4 (instead of n8 to S1), creating a right-side mirror of 2. Since all else poofs (as before), the total count is 2.

Below is another shift that works well. From the original position, the S3 and n15 checkers move a pip closer to each other:

In the left-hand diagram, the just-shifted blot on n16 and the blot on n8 form a reflection (around S2) with a count of 4. Add the S1 checker for 5, poofing with the closed board of −5. All that remains is the mirror (formed by the new n17 blot and n−5 spare), the entire count of 2.

Incidentally, it is equally good to countershift n8 instead of n15, as illustrated in the variant (right-hand) diagram. That creates a reflection 4 composed of n15 + n9 (instead of n16 + n8).

Note the marvelous efficacy of dual-purpose shifts. In the two (similar) cases above, the (1-pip) shift yields a mirror and the (1-pip) countershift yields a reflection.

Petter's shift involved more movement, but he attained an even cleaner position: He shifted S3 to n15 and n8 to S1—a total of 5 pips, and countershifted n−5 to n0, thereby reaching

On the left is a silver poof, which is a cancellation of a pair +5 with a closed board −5. The two checkers on S1 are the entire count of 2.

Finally, I want to reiterate my thanks to Matt Ryder for his excellent work on Naccel-numbered diagrams (which anyone can now create), and to Ian Shaw for showing me the ropes for formatting side-by-side diagrams. These two elements have made it possible to offer much more lucid explanations and presentations.

This is in response to Lucky Jim's recent post.

 The following is position 460 from Robertie's 501 Essential Backgammon Problems.

 Blue: Most of it is a poof except one spare on n2 and two checkers on n4. The checker on n2 can go forward two pips for a count of 0 and one of the checkers om n4 can go back two pips for a count of 1 that just leaves the remaining checker on n4 needing to go 4 pips forward for a total count of 1(4)

Good. You poofed the five-prime (plus two n0 spares), leaving only the three checkers below (left-hand diagram) to count:

The blot plus one of the n4 checkers is a split, for a count of 1. For further reference/imprinting, I have isolated the split in the right-hand diagram.

Split 1, plus the 4 pips (from n4), produces a count of 1(4).

You understood the essence of the split when you shifted (2 pips outwards) to the Supers. I'm just making the point that shifting is an unnecessary step once you know/remember the split.

 From White's point of view:

 White: I may not have counted this as well as I could as I have two separate mini pip counts. The checkers in the outer board are a six sym around the n3 for a count of 3.(I have to confess that I didn't see this immediately). I moved one checker on n−1 to n−2 to create a block for a count of −1 and then moved the checker on n7 one pip forward for an offsetting count of 1. This gives a total count of 3(2).

Great. You counted White's sym3 (i.e., six checkers around n3), then used a motion poof for the rest: n7 to S1 and n−1 to n−2 is a total of 2 pips of forward motion that creates a (block−1 vs S1) poof.

(Combining a motion from two areas isn't ideal but it's fine as long as it's in a single effort.)

I'll assume that the other readers can see the sym3 in White's outer board now that you've pointed it out, so I've removed it to more clearly show off your block/S1 poof (after the 2-pip forward motion)—see the left-hand diagram below. The four checkers on S0 are ignored, as always.

You can poof a blot on the midpoint with any formation of −7 pips in the inner board, and over time you may learn some as instant counts. The most common and easily spotted are those with a squad of −1 count plus an extra checker on n−1.

Starting from the main position (as you did), with a motion of 2 pips from n0 to n−2, I created a (blot) midpoof in the right-hand diagram. (Again, the sym3 has been removed.) If you happen to know this midpoof (as I did), you gain the further advantage of localizing your 2-pip motion instead of splitting it up (1 pip inside and 1 pip outside). This is just an advanced detail, though the glimpse might interest you.

Summary: Blue is 1(4), and White is 3(2). Blue leads by 10 pips. I'll repeat the main position:

 Clearly Blue is leading the race because Blue has a lower score to add to 90 if you were trying to generate a traditional pip count.

Blue leads 1(4) to 3(2). Forget about the 90; you are not burdened with it when you use Naccel!

### Nack 57

 What I am not sure about is how you would then use Nack 57 to decide if Blue should double and if White should take if doubled. Would Nack mind going very slowly through the approach you would take over the board if you were either blue considering whether to double and if you were white deciding whether or not to accept a double using Nack 57?

Chase's answer is exactly correct. To the leader's count, you:

1. Convert to pips
3. Double
4. Find nearest square root.

1. Blue's count of 1(4) = 10 pips.
2. Add 57, comes to 67.
3. Double; that comes to 134.
4. Find nearest sqrt, which is 12 (because 134 is closer to 144 than to 121).

In summary: 1(4) ... 10 ... 67 ... 134 ... 12.

Subtract 3 to get the minimum doubling point, or subtract 2 to get the minimum redoubling point. So, here, the take point of 12 means that Blue should double with a lead of 9 or redouble with a lead of 10.

[For more on the Nack57 formula, which works for both traditional pipcount and Naccel, review here.]

Blue leads 1(4) to White's 3(2), a difference of 10 pips. In short, Blue has the necessary 10-pip lead to redouble, and for White (who can take even when down 12 pips) it is an easy take, down only 10 pips.

 More generally, when counting, should you count the side you think is behind or ahead first or just count your own position first?

I don't feel strongly there is a best way, but consistency minimizes confusion. Personally, I count myself, then the opponent.

### Handy Count

 Do you have tips for retaining one score as you then count the other?

Try this: Keep the supe count on your left hand: One finger = 1 (or 6), two fingers = 2 (or 7), etc. Keep the baby pips on your right hand (works well for Naccel as these can only be 0, 1, 2, 3, 4, or 5). For a negative number, turn your hand over. (If, for example, resting your hand on your knee is your chosen "natural" position, do that when the count is positive and turn your hand facing up when the count is negative.)

So, here, Blue's count is 1(4). Rest 1 finger of your left hand on your left knee (with the others clenched into a fist) and rest 4 fingers of your right hand on your right knee. With that safety net in place, count White.

If, over the board, you prefer to keep the count on only one hand, for the supe count you can either subdivide your thigh into a dozen or so different areas or you can use the board itself. For example, resting 4 fingers in front of your n1 point means 1(4).

The faster that your counting becomes, the less you will need the "handy count" technique. The reason you are losing track now is that you are expending so much time and energy on the mechanics of the second player's count. Once you quickly and easily spot poofs and syms rather than working to find them, your second count will follow your first at such a rapid pace that you will wonder why you ever had trouble retaining the first count.

 Are there any other tips for tackling this other than obviously practice, practice and practice!

Can't think of anything else at the moment. There aren't many steps; just like anything else you have to practice enough so that you know by second nature what comes next. Simply reading the Naccel posts without practicing while continuing to count in traditional/Cluster ("see ... this still works ...") is a hopeless recipe for improving one's speed.

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