Nack, do you have pip-counting times to report?
I'm a little puzzled why people aren't answering the question directly.
Sorry for the length of this answer, but I feel that it first deserves some context. Also, I have some related thoughts to share.
Here were the Naccel counts for Tenland's three positions (for Blue and White, respectively), six separate counts altogether, as they accumulated in my head:
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........8, 9, 16(3)
........8, 9
........6, 6(1)
........3, 5, 4
........3, 10, 11(5)
........4, 8(4)
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If you wish to translate the above six Naccel totals of l6(3), 9, 6(1), 4, 11(5) and 8(4), to trad (traditional) totals, then multiply by 6, and add 90 plus any spare pips (in parentheses). For reference, the trad counts are, respectively: 189, 144, 127, 114, 161 and 142.
While honing Naccel, I tried to get a feel for how many seconds it would cost me to convert the Naccel count to traditional count. I discovered something unexpected, though perhaps it won't surprise you. As I practiced the counts (typically in Naccel's 0 to +10 range, but as wide as –5 to +20), I was memorizing and converting the "6x + 90" part without trying at all. They became inseparably linked. "1" automatically means 96, or "2" is 102, "3" is 108, etc. All that is left is to add the spare pips (if any). For example, the brain registers the Naccel count of "11(5)" instantly as "156 + 5," the +5 addition costing only a fraction of a second.
To save even this fraction of a second, but mainly so that I wouldn't have to convey or support the above experience every time someone unfamiliar with Naccel expressed misgivings about the conversion, I set out to create a race formula for Naccel that would eradicate the need for conversion.
The result of my efforts was Nack58, or just "N58" for short, which can be used with any total-pipcount system. It is the only race formula to perfectly reproduce the GST (Gold Standard Table), except possibly at a leader count of 111 (not sure NeilKaz did a rollout for that one, but if wrong then it is slightly better to add 1 to point of last take). Trice62 misses by a pip at leader counts of 61, 69, 70, 78, 79, 80, 88, 89, 90, 99 and 100, and I don't know if there is another formula closer than that.
Here is the N58 formula: For traditional counts, subtract 32 from the leader and round down to the nearest square number; its root is the point of last take. (From that, subtract 4 for a borderline double, or 3 for a borderline redouble.) For example, suppose the leader has "114." Subtract 32 gives you 82, round down to 81, making 9 the point of last take.
Oddly, in shifting my attention to perfectly match the GST, I ended up eliminating only the "add 90" part of the Naccel-to-trad conversion. Oh well, there was only a fraction of a second at stake anyway. You still have to multiply by 6 (the counts for which become automatic with practice as they did before) and add the spare pips (if any); then instead of subtracting 32, you add 58. Hence one reason for the name of the N58 formula. The other reason (conveniently) is that for trad leader counts below 58, you switch to Trice's low-count rule (subtract 5, divide by 7, round down).
Using the same 114 example, which would be a Naccel count of "4" that becomes 24, add 58 to get 82 and round down to 81, making 9 the point of last take.
Okay, so Naccel still gets docked a fraction of a second when doing a straight-race cube calculation. However, as Stick and others have stated or implied, most of the time you don't need to do one of those even when you've done a two-total count. A significant majority of the time, either (a) the position is NOT a straight race (the difference in counts might be useful but you wouldn't apply a race formula), or (b) it IS a straight-race but the difference between the two pipcount totals is such that it obvious what to do without a race-formula calculation; e.g., 6 vs 9 in Naccel, which is 90 vs 108 in trad, is a no-brainer double/pass. So, we're really only talking about a fraction of a second, a fraction of the time!
Wait, back up. How does a Naccel counter know that a lead of 6 vs 9 is a double/pass? Answer: The same way a traditional counter knows that 90 vs 108 is a double/pass! The difference in count (3 in Naccel, or 18 in trad) is clearly too large a difference to take (except at total counts so high it couldn't even be a minimal-contact position, let alone a straight race). Suggesting that a veteran Naccel counter must convert 6 or 9 or whatever to a traditional count, is like suggesting to a native French speaker that he must translate his French essay into Esperanto. A Naccel counter knows exactly what his counts mean and what to with them just the way they are. (I'm just making this additional point for anyone who still dreads the idea of conversions.)
Can I prove Naccel is the fastest total-pipcount system (that currently exists) in a real-life demonstration? Well, I suppose if a world pip-counting championship were to be held, I could coach some bright young backgammon player (say between the age of 15 and 30, who has lost way fewer brain cells) for a couple of months, and enter him in the competition. However, even if he (or I) won the event, some would argue that Naccel isn't fastest, rather that the contest was won on general pip-counting talent or mental agility. Besides, no longer being young and ambitious, I'm very content with the way I'm living my life; I doubt anyone would offer me enough motivation to put myself through that.
Bear with me. I haven't forgotten about your question.
Regarding Tenland's three positions, I breezed through five of the six counts. However, I confess that I handled the Blue side of the third position poorly. I counted 20pt+13pt as midgold (+7), which is a fine start, but then I stalled for a bit (sensing I was missing something) before moving the roof checker forward 3 pips to create a mirror (+2) with Blue's 2pt blot. That left an 8pt triplet (+1), and 8 final pips on the 10pt, for a total count of 10(11) [which, by the way, is fine the way it is]. I'd guess that count took me about 17 seconds (ugh), though years ago it would have taken about half that time even with the balk.
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Right afterwards (hand slapping forehead), I noticed that the five-checker formation in the outer board is a sym (six-checker group symmetrical around a point) with one checker missing, and all I have to do is move the 2pt blot back to the 10pt! The obvious 8-pip offset is to move the roof checker forward (25/17), which becomes a leftover count of 1(5). Adding the midgold of +7, Sym of +3, and the 1(5) is a total of 11(5). I'd guess that more efficient count (which also unfairly assumes no balk) would have taken me about 4 or 5 seconds, or at my peak (years ago) about 2 seconds.
For an aspiring Naccel counter, it is important to develop good habits. If the counter is set on clunking his way around the board, back to front, he might report his count of the position this way:
"Okay, I'll start by moving the roof checker 5 pips to make a triplet on the 20pt -- oops, I mean the Naccel 14pt, which I divide by 2 for a count of +7... that's 7(5). Then I can hop the midpoint checkers down to the bar point (that's +2) and 2 more pips will bring them into the 6pt / Naccel 0pt, where they vanish along with the others there. That adds 2(2) to... what was the count... oh yeah 7(5), making it 9(7) so far -- remember that! What's left here is the triplet of +1, and the other two checkers count 8 pips, so that comes to 10(15). Finally, subtract 4 pips for the blot on the negative 4pt, for a final count of 10(11)."
Okay, our guy got the count right (not a given), but he dragged around a mixed count the whole way and threw most of the strengths of Naccel out the window. We see that he knows how to count triplets (though not with instant recall, yet) and he knows how to "hop" (the technique of moving a checker or checkers a total of 6 pips or multiple thereof). As cancelling around the 6pt (Naccel 0pt or N0) is relatively basic, we can assume he knows that as well. Having gleaned all that, we can analyze his count and explain to him how faster counts are already within his grasp. We could say to him:
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(a) "Your start of making a triplet by moving the roof checker 5 pips forward was commendable. However, your counting will speed up when you learn to offset by moving the near-side blot back the same distance, so as not to carry around 5 baby pips -- a bad habit that comes from thinking in linear fashion. Ideally, baby pips (if any), should be added just once, at the end. But let's proceed with just the 25/20 part that you moved.
(b) Next, it's better to hop the two midpoint checkers just 3 pips each (count of +1), rewarding you with another triplet -- a formation you clearly like. Then you can count triplets of +1 and +2 there (which you will eventually learn how to count even faster as one Sym). You were too set on eliminating the midpoint checkers by essentially counting them as 2(2), which is generally reserved for remainder counts. Instead, you should look for a way to combine or integrate them -- for example the +1 hop just mentioned. Another example is 13/11(2) offset by 2/6, and yet another is 13/9(2) offset by 2/10.
Had you done (b), your top spare on the trad 10pt would exactly cancel against the inside blot. (If it didn't, you could move them towards each other until one reaches the 6pt (Naccel 0pt) and count the leftover pip(s) of whichever checker remains."
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This counter shows promise in that he makes an honest effort. He doesn't overreach or expect too much. Mainly, he has to break out of his rigid habit of back-to-front (at the cost of summing together mixed counts). The joy of Naccel is not merely explained by the small numbers -- units of 6 shrunk to "1" and cancellations around its 0pt. The geometry of the backgammon board visually basks in these multiples of 6, by virtue of each quadrant containing six points. Patterns in one quadrant are replicated in every other quadrant, giving you a 4-to-1 bang for the buck for learning local groups. For global groups, the board is set up for vertical, horizontal and even diagonal symmetries, all of which a budding Naccel-er can eventually exploit. Formations are everywhere, just pick.
By contrast, some people who try counting with Naccel expect miracles, that they will achieve lightning counts in short order, and give up in frustration when that doesn't happen. It doesn't work that way. In fact, early on you should even expect a few miscounts. If you only want a pip-counting method you can apply at a reasonable speed in games immediately, give up on being the fastest counter you can be and find another system. I respect your choice.
Finally, Tim, in answer to your question: My times have doubled compared to what they used to be. I stopped working at it years ago, and when I did involve myself with Naccel or other systems it was in order to teach other people how to count. I believe I could get back to my peak or a bit better if I desired to put in the work. Back then, I estimate my average time to count the total for one side was 4 seconds. Most fell under that, with an occasional blooper of 10–15 seconds (as you might expect the curve to be).
It is interesting to observe how someone -- call him person "A" -- might patiently explain how he is counting a position (Stick's fine presentation comes to mind, though I'm going to use the position above as our example), while person "B" might say:
"25, 65, 91, 111, 135, 159, 161 for Blue, and 40, 92, 113, 131, 139, 142 for White. See how easy and fast that is?"
Indeed, for the same position, I list my counts with Naccel as
"3, 10, 11(5) for Blue, and 4, 8(4) for White."
Is Stick's count slower because he took space to explain his method in detail? Not necessarily! Is my count faster than B's count because it looks shorter? Not necessarily!
You can't judge the list of progressive accumulative counts by the length of the character string alone. Granted, the Naccel numbers are smaller, and there are fewer of them; that string certainly makes it look like a faster count. However, there are a couple of aspects to consider:
(1) For the Naccel count, I glossed over the 5-pip offset I made for Blue (which is unlike White's going from 4 to 8(4), merely a midpoint-remainder count.) Time is added due to such an offset, both in the finding of it and the care taken not to make a visualization error. On the other hand, B's count glosses over the fact that each step (five out of seven for Blue, and five out of six for White) involves multiplication (as well as the addition obviously implied).
(2) Person B's method is the basic traditional one, counting from the rear to the front. He always knows exactly what he is going to count next. By contrast, Naccel is a group-finding system that sometimes resorts to shifting to create such groups. (Likewise with Cluster.) Using a "finding and shifting" system effectively means applying your memory and/or some basic visualization skills. There may be stutters as the counter figures out what to do next (though that mostly goes away with practice).
The more you count with Naccel -- once you've crossed a threshold fairly early on -- the more fascinating and fun it becomes (albeit judging by my own experience). For those interested in giving it a try: After studying some basic ideas and groups for Naccel, I recommend looking at just one position. Treat it as a puzzle, and write down different ways of counting with the Naccel knowledge you have acquired to that point. Then pick what you believe is the fastest of those ways and move on to the next position. You are even invited to use me as a resource. Post your fastest Naccel count for a position, and the next time I visit this site I'll let you know what I think.
Nack