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BGonline.org Forums
Criss-Cross Count
Posted By: Nack Ballard In Response To: Criss-Cross Count (Michael)
Date: Wednesday, 26 September 2018, at 8:51 p.m.
The final step (shifting) is the part of any half-crossover-system count (e.g., Coconut, Urquhart or Criss-cross) that takes the longest (except possibly Urquhart's weighted crossover-difference step, IMO). It requires practice to make it go quickly.
[For anyone that still doesn't "get" it, the final step is to shift to the center point of each "slice" (three-point area). I think this will become clear if you review my Criss-Cross example (especially around the final diagram), Max's explanation of the same position (under his seventh diagram), and my explanation and list of tips in the original Coconut post (starting with "Step 4" down to the dotted line).]
To be clear, the board is divided into eight three-point areas, which can also be called slices, zones, triples, triads, half-quadrants, etc. (You erroneously referred to "...eight quadrants"). One more name for these three-point (one-eighth-board) areas is "octants," though not as many people are familiar with that word. You are basically shifting to the second and fifth point of each quadrant, except you must never cross the invisible zone-edge line between either player's 3pt/4pt or 15pt/16pt. Thus, these shifts are all 1 pip (except when it is 2 pips from the bar or bear-off tray).
You mentioned being confused over seeing Michael's plus and minus signs (scroll down to seventh diagram). The checker on Black's 1pt there is marked "–1" because it is being shifted to the 2pt. Here are two ways to reconcile: (a) The 1pt-to-2pt shift increases Black's pipcount by 1, so to compensate it is counted as –1. Or (b) the 1pt checker is 1 pip lower than the 2pt, so it is counted as –1. [There are equally reasonable ways to reconcile seeing the + and – the other way around in an inverted system, but I won't mention them here.]
The "+2" on Black's 6pt is because two checkers are circled, and they are each shifted 1 pip lower (therefore are +, to compensate) or can be thought of as two checkers each 1 pip higher than the 5pt (center of slice) and thus counted +2. Note that the 4pt-5pt-6pt define a different slice/zone/octant than that of the 1pt-2pt-3pt.
Similarly, you will notice (in Max's same diagram) that all checkers 1 pip lower than the center point of their slice are marked with a minus (–), and all checkers 1 pip higher than the center point of their slice are marked with a plus (+).
In this way, you can work your way around the board, netting out all the plusses and minus until you're left with a small plus or minus number (or zero). However, I find it considerably faster to cancel as many of these plusses versus minuses as you can, and there are many efficient ways and tricks to do this. (It comes with practice.) I encourage you to reread my analysis under the final diagram here and keep asking questions until you fully understand it. That will boost your comprehension and over-the-board speed substantially.
As I've mentioned before, it is often unnecessary to do that final step (shifting), when the relative count isn't close enough to a number that would make you wonder about your decision. For example, if, prior to the final step, you have accumulated a relative count of 21 in a (medium-length, say) straight race, you won't bother with the final step because the chance of reducing the count into the low teens is extremely unlikely; you should just turn the cube or pass the cube being offered without spending more time on the clock.
Hope that helps.
Nack
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