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Naccel 2  post #10 (leftover counts)
Posted By: Nack Ballard
Date: Wednesday, 17 February 2010, at 1:00 a.m.
Welcome to post #10. You can review previous Naccel posts by clicking below.FAQ, Post #1, Post #2, Post #3
Post #3x, Post #4, Post #4x, Post #5
Post #5x, Post #6, Post #6x, Post #7
Post #8, Post #8x, Post #9, N57 Race Formula
In this post, I will discuss the concept of "leftover counts."With Naccel, as you have seen, you can freqently count the entire position in one or two formations, and, unless the count is a multiple of 6 pips, you have some sort of "leftover count." Below is a trivial example.
The n5 blot requires 5 pips to move/bear in to S0; thus, its count is 5 pips. If, say, your integer count is 3, and you have this n5 blot left over, your total count becomes 3(5), meaning 3 supes (superpips) plus 5 pips.
Leftover blot 0(5)
Here is a slightly more sophisticated example:
One way to count this leftover blot is to divide the n17 point number by 6, which is 17/6 = 2 5/6, meaning 2(5). More likely, however, you currently gravitate towards either
Leftover blot 2(5)
(a) hopping (+2) the blot forward to reach the previous diagram of 0(5), or
(b) moving the blot 5 pips out to S2, so that S2 + 5 pips movement = 2(5).Before long, you will see this blot as an instant 2(5) count; you'll no longer stutter with (a) or (b). Likewise for a blot anywhere on the board.
Two blots, or more commonly two checkers on a point, can also constitute a leftover count. A good example can be found in the sixth diagram here (and it is also instructive to see how the 2pip leftover count was stripped away and isolated by reading from the beginning of that post). A slightly trickier example is shown below.
To count this leftover twochecker point, you can add 4 + 4 = 8 pips and convert that to 1(2). Or you can move both checkers 2  2 = 4 pips backwards to get two checkers to S1, and count it as 2(4), same thing. Or you can divide the point number of n4 by 3 to get 1 1/3, which again is 1(2).
Leftover point 1(2)
However, the way I recommend counting this point is to move both checkers 1 pip forward, creating a standard pair on n3. The count of that pair is 1, and adding the 2 pips of motion gives you a total of 1(2). This method is visually clear, and its advantage in simplicity becomes more obvious the further away from S0 you move the point. For example...
Here, you can move the anchor forward 2 + 2 = 4 pips, to get a standard pair of 5. Therefore, the deeper anchor as diagrammed counts 5(4).
Leftover point 5(4)
You can handle two blots within a quadrant field in much the same way. Here are two examples:
In the lefthand diagram, you can move the back blot 2 pips and the front blot 1 pip (total of 3 pips forward) to get to a pair of 5. Thus the formation as it stands is 5(3).
Leftover blots 5(3)
Leftover blots 5(1)
For a somewhat more sophisticated example of the above pipmotion finale, look at the sixth diagram here. By moving Blue's far side blots a total of 5 pips to the midpoint, a mirror of 4 is formed, which means that the count of the formation diagrammed there is 4(5).
The righthand diagram above is 5(1). There are a number of ways to arrive at this count, but I find it easiest to move the back checker forward 1 pip to get a "split" of 5, which is a variant of the pair of 5. Or, if you don't know/remember the split (see the logic outlined in the first several diagrams here), you can move the back blot forward 2 pips and the front blot back 1 pip, netting +1 pips in acquiring the pair of 5. It is preferable to move in only one direction rather than two, but you can only apply what you know.
Occasionally, you might end up with blots in different quadrants. In such cases, you can hop and shift to a pair (or split or wide). Better yet, you can shift to a mirror (or diagonal mirror variant) or reflection. For example...
In the lefthand diagram, you can move the midpoint blot forward 2 pips to n5; this creates a (nearside reflection) poof with the blot on n5. Therefore, the count of the twochecker formation as it stands (premotion) is 0(2).
Leftover blots 0(2)
Leftover blots 2(3)
You can utilize similar motion on the far side  moving one blot (or point for that matter) to create a farside reflection (around S2) with another blot (or point), except that it is a count of 2 (instead of 0) per checker.
In the righthand diagram, you can move the farside blot forward 3 pips, thereby achieving a mirror with a count of 2 with the nearside blot. Therefore, as it stands, the diagrammed formation counts 2(3).
Alternatively, you can view the (righthand) twoblot formation as half of a (point) zag; half of 5 is 2 1/2, which is 2(3). [The "zag" is explained starting with the eighth diagram here.] While for the most part during the main count you should confine yourself to integer counts, you have more leeway in handling leftover counts.
It is even possible to have a leftover count with three checkers. Typically, an easy way to deal with them is to move to the nearest triplet formation. For example...
This threechecker stack is on an oddnumbered point; that's the bad news. The good news is that you can either move the three checkers 1 pip (each) forward to form a triplet on n14 (count of 7) or backward to form a triplet on n16 (count of 8). These two methods are the most consistent with mainstream counting. However, if you prefer, without motion, you can forcefully divide the n15 point number by 2. Any way you cut it, the count is 7(3).
Leftover threestack 7(3)
Alternatively, you can count two of the checkers as a pair 5 and the (peeled off) spare as 2(3), which adds to 7(3). But (and we'll put aside the motion alternatives for a moment) it is probably faster to divide n15 by 2. This example demonstrates that sometimes larger leftover counts can be a speedier resolution than peeling off all the integer counts available before dealing with leftovers.
Allow me to further expand your concept of leftover counts. Perhaps the most useful leftover formation that is composed of more than five checkers is the "tween sym." Below are two examples.
In the lefthand diagram, White has sixchecker sym around an imaginary point halfway between n4 and n3. The count, therefore, is halfway between 4 and 3, which is 3(3). Because the 3 1/2 point center of symmetry is in between two real points, this formation is called a "tween sym."
Tween Sym 3(3)
Tween Sym 12(3)
In the righthand diagram, White has a tween sym centered around her imaginary n12 1/2 point (where the bar is), located halfway between her n15 and n10 points. This tween sym counts 12 1/2, which is 12(3).
The most practical leftover counts involve checkers on the midpoint, both because it is frequently occupied and because it is located just one pip higher than a Super. Leftover counts of 1(1), 2(2), 3(3), 4(4) and 5(5) are very easy to tack onto your main integer count.
Consider this example:
From the lefthand diagram, you have two ways to quickly reach an integer count:
5(1) or 4(5)
Leftover midpointers 4(4)
(a) You can move the blot backwards 1 pip to create a midtriangle of 5, as shown in the lefthand diagram below. The count of the lefthand diagram above is therefore 5(1).
(b) You can move a midpoint spare backwards 1 pip to create a midblock, as shown in the righthand diagram below. Again the count of 5 minus the 1pip motion demonstrates that the count of the formation in the lefthand diagram above is 5(1).
If you are still at or near the beginning of your count and choose the midpoint corner first for whatever reason, employing (a) or (b) with a 1pip countershift somewhere else is an optimal strategy. But if you've reached the end and this is a leftover count, you don't have to worry about carrying around baby pips. Under these commonplace circumstances, I find the leftover midpoint counts so routine that instead I tend to move the blot forward, as shown in the righthand diagram above. The 4(4) count there plus the 1 pip of motion means that the formation in the lefthand diagram counts 4(5). [This is the same as the 5(1) counted by the backmotion methods.]
Many examples of leftover counting are illustrated in this post, but I stress now that you should only apply these techniques to the final formation (i.e., at the end of the count), so that you don't have to carry forward mixed counts only to make another pip adjustment later  a process that will slow down your counts and might confuse you. It may be difficult to resist the temptation of a clever count like a tween sym, but you'll generally do better if you allow such gems to fall into your lap as leftovers.
Midtriangle 5
Midblock 5
Integer counts are always available, at most one shift away; there's just no reason to compromise your speed by getting bogged down with fractional counts and fractional movements midexecution. If you spend a little extra time to find large integer count formations and sweep them away, the nurturing of good habits will reward you with lightning counts before long.
That said, awareness of how to handle leftover counts is important, and knowing what types of formations you would prefer to save for leftover counts is a valuable intuition to cultivate. Your time reading this post has been well spent.
Last time (twothirds of the way into the post), I discussed asymmetrical squad poofs. The most commonly arising tend to involve blocks, pairs, and triplets.Below, I've diagrammed a few more asymmetrical squad poofs (one of which will shortly appear in a fullboard position in this post) in order to broaden your horizons. S0 checkers have been arbitrarily added. Even if these exact situations aren't the ones that arise pip for pip, or even if you only use one side of the squad/squad formation at times, the more familiarity you have with squads and their variants, the more shifting options you'll have.
Triangle/Wide Poof 0
Sock/Layer Poof 0
Of course, as you have seen, poofs, asymmetrical or otherwise, need not involve squads. Poofs are very efficient formations regardless. Below are a few examples of subsquad poofs.
Triplet/Split Poof 0
Wedge/Stack Poof 0
0
0
The formation in the lefthand diagram immediately above is very handy to remember, because it is also an "asym"  an asymmetrical sixchecker formation around a point (in this case n0). You can slide this formation (not including the invisible n0 checkers) to the left or right, and/or reverse it left/right and it is always an integer.
0
0
In other words, any formation composed of a point with four checkers plus a point 3 pips away with two checkers is an integer asym.
It's time to practice counting full board positions. Petter supplies our first:The key in finding the fastest count for Blue is to attack the biggest problem area(s) first (typically the checkers that are furthest from S0) and turn it to advantage. For this purpose, the "midpoof" is one of my favorite formations; you can review it in the sixth diagram here.
Having spotted the midpoof (two checkers each on n7, n3 and n4), it is not terribly hard to find Petter's dualpurpose 2pip shift (n7 spare forward 2 pips, n1 spare back 2 pips):
Two formations can now be easily identified. One is the midpoof (now stripped of its distracting spare), and the other is a tight (symmetrical) eightchecker poof around n0. After removing those two poofs from the board, the only checker you actually have to count is the n5 blot for a count of 0(5)!
Midpoof, small poof, and +5 pips 0(5)
Let's repeat the position, with pointnumbering from White's perspective:
Again, Petter found the best solution, IMO. Without shifting at all, this position can be broken into two formations, shown below.
Triplet/Split Poof 0
The first of the above two formations is a triplet/split poof diagrammed earlier in this post (with an extra n0 spare). The second formation, and the only one you need actually count, is a sym around n2, for a total of 2.
Sym 2
Finally, let's repeat the main position again, and apply what we know of the race to determine correct cube action. We'll assume that Blue is on roll:
Blue has a count of 0(5) or 5 pips. White has a count of 2 (supes) or 12 pips. To get the trailer's (White's) last take point, we can apply the Nack 57 formula (N57), which I'll restate here:
2
No Double 0(5)
Add 57 to the leader count, double it, and find the nearest square root.
So, 5 + 57 = 62, doubled is 124. The nearest square root is 11, so that's the take point.
The minimum doubling point is always 3 pips fewer (than the take point) and the minimum redoubling point is 2 pips fewer.
Blue's (minimum) double point is therefore 8. But he leads by only 7. (That is, Blue has a count of 5 pips to White's 12). Hence, he should not double.
The second position is also kindly provided by Petter:For Blue: Petter, Ian and Jim shifted the n0 spare forward 2 pips, closing the board. Ian and Peter countershifted a checker in the outer board (though not the same one) back 2 pips. For the easier of the two countershift choices, this creates a count of 5 +1(3) +S1 = 3(+3).
Instead of a 2pip countershift, Jim moved the midpoint blot 1 pip forward. His count is therefore: Closed board 5, S2*2 = 2, and motion of +3 pips, for a net of 3(+3).
[For negative supe counts, I recommend always prefacing the number of pips with either a plus or a minus. Otherwise, "3(3)" could easily be mistaken as being halfway between 3 and 4 supes (much in the way that 3 1/2 or 3.5 is interpreted) instead of halfway between 3 and 2 supes. Indeed, the respondent's usages were inconsistent with each other.]
The closed board is a nice count, but IMO it can be improved upon. I got rid of the polarized checkers with this 3pip shift:
The n5 and n5 points poof. The sym around n2 plus the spare on n3 counts 2(3).
Poof, sym 2, and 3 pips 2(3)
The advantage of this shift is that less counting is required and/or in only one direction. As with the midpoof solution for Blue in the previous position, poofing the extreme lowcount and highcount checkers simplifies the count.
Note that other 3pip back shifts are workable (you might like to experiment) but I chose this one specifically because it moves one checker 1 pip and another checker 2 pips (mimicking the forward shift in the outer board), it separates the other checkers from the (poofed) n5 point, and it creates a clear inside count.
Here is the position repeated with pointnumbering from White's perspective.
It is tempting to close the board with a few pips (5, in this case), but once again poofing is faster. The two outer board checkers are pretty close to n0, but you can still get rid of all four n1 checkers by applying the asym poof that Petter found and was diagrammed earlier in this post. After shedding those (as well as the irrelevant n0 checkers), these six checkers remain:
Here, Petter applied the leftover tactic of the "tween sym" diagrammed earlier in this post. These six checkers are symmetrical around an imaginary point halfway between n4 and n3. The count is 3 1/2 supes, which is 3(3). Brilliant.
Tween Sym 3(3)
If you don't know/remember the tween sym, you can still obtain a threeprime sym around n3 by moving the n5 blot back 3 pips, for a count of 3(3). Or, you can similarly create a threeprime around n4 by moving the n2 blot forward, for a count of 4(+3).
Here is the main position again:
For bonus credit, see if you can visualize this shift for White: An n3 checker forward 2 pips (covering n5), and the n0 spare back 2 pips (putting a spare on n2). This creates a sock/triplet poof around n0, and the five checkers on the left are a block of 3 plus the n3 blot: again, that's a count of 3(3).
3(3)
2(3)
Ian took the position a step further, assumed White was on roll, and figured out the proper cube action. To do so, he converted the leader count to pips (3*6  3 = 21) and applied the N57 formula, as described in my analysis of the previous position (or you can click here.
So, 21 + 57 = 36, twice that is 72. That's closer to 64 than to 81, so the take point is 8. Blue (the trailer), with 2(3), is 6 pips behind White's 3(3). Ian's assessment is that the potential wastage from Blue's second n5 checker costs him another halfpip or so, adjusting the practical difference to about 6.5 pips. As 6 is already sufficient, White has a clear redouble. Meanwhile, Blue, who can take even if he is behind 8 pips, has an easy take being down only about 6.5.
In short, White has a clear redouble and Blue has a clear take. Nice assessment, Ian. I agree.
The third position (White checkers only) is brought to us by Lucky Jim:This time, Claude is our star. He found this simple but elegant twochecker 2pip shift:
POOF! The count is zero.
Poof! 0
0
I saw two similar shiftpoof solutions (though I missed Claude's), shown below:
POOF! The count is zero in all three diagrams (left, right, and previous).
Poof! 0
Poof! 0
For practice, I'm repeating Jim's original position below. See if you can visualize all three shiftpoof solutions. (There are more, but they're less straightforward.)
Jim said (I paraphrase): "I could see immediately that White has two checkers on n5 and four on n4 for an inner board count of 14 pips, and she has four on n3 and two on n1 for an offsetting +14 pips. Poof. Does Nack have any comments this method?" [Jim also offered a second solution that involves offsetting 12 pips while leaving two checkers on n1 and two on n1.]
It may depend on your exact meaning of "immediately." It appears to me, Jim, that you're still somewhat (though less so) mired in checkerbychecker counting, though in positions where the checkers are close together it's not likely to make much difference.
IMO, your second solution (12p vs +12p) is a slight improvement on your first (14p vs +14p) because you visually (rather than countingly) cancelled a 2p versus a +2p portion. That's a step in the right direction.
I definitely find largescale assaults faster: I search for formations (poofs especially) with integer counts that involve many checkers or at least shed extreme ones. I recommend that you practice looking for dualpurpose shifts and large poofs, and soon you'll be counting faster than ever.
Next position(s)?
Nack

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