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| XGID=-b----E-C---eE---c-e----B-:0:0:1:00:0:0:3:0:10 | |||||
Opening position | |||||
From the opening position, exactly 167 pips are required to bear off without wastage. As outlined in the previous post (with "Hints" in the subject header), there are three main ways with doublets + one non-doublet to sum 167 pips in 9 turns:
.....(a) 156 + 11, or
.....(b) 160 + 7, or
.....(c) 164 + 3.
... at least in theory. Which of these are possible in practice?
For each of these (a, b, c), we'll consider two cases: Bearing off for one side (i.e., with opponent's checkers conveniently off the board); and interactive (i.e., both sides playing).
[For any non-doublet I will address, it mechanically works to insert it into the middle of the sequence. However, the conditions of the problem confine it to either the beginning or end. The beginning, because according to the rules of backgammon a doublet cannot be played on the opening roll -- that slot must be reserved for a non-doublet. Or the end, because the losing side bears off 13 checkers having necessarily played all doublets and thus is left with the odd-pip roll remaining to be borne off. Hence, I confine my analysis to starting or finishing with the non-doublet.]
.....(a) The non-doublet is 21.
In order to traverse 34 crossovers with 8 doublets and 1 non-doublet, every roll portion must be a crossover. Thus, if the 21 is the opening roll (and we'll always assume Blue goes first), the ace must be played 13/12. As this point is occupied by White (her midpoint), an interactive solution is impossible with 21 at the beginning.
How about a one-sided solution? This is possible, though we must be careful. After 13/12 8/6 (13/11 doesn't work as the 11pt blot is an irreconcilable loose end), one 66 is automatically played 24/12(2). Now, if four checkers on the midpoint robotically take two rolls to end up in the inner board, we reach this position:
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| XGID=-----DF-B---C-------------:0:0:1:00:0:0:3:0:10 | |||||
Blue played 13/12 8/6, 24/12(2) and 13/5(2) twice. Now what? | |||||
I arbitrarily chose to put the midpoint-sourced checkers on the 5pt (reachable via 66 22 or 55 33 or 44 44 in either order), but the conundrum is the same regardless on which of the six inside points this four-stack is placed. If Blue were allowed to play 8/off(2) with 44, or 12/6(3) 6/off with 66, then there would be no problem, but the rules of backgammon require all checkers to be borne in before any can be borne off.
The mistake was bearing all the midpoint checkers in to early. Logically, it is better to play only the "forced" moves and see what remains. Seven 6s must be played to the 6pt, then play the eighth 6 down to the 7pt (because doublets come in groups of four, and 8/2 clearly doesn't work). This leaves:
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| XGID=------IAB----C------------:0:0:1:00:0:0:3:0:10 | |||||
The right way | |||||
Now it is possible. Blue can either play 33 as 13/10(3) 7/4 followed by 44 as 10/6(3) 8/4 and then 44 bears off cleanly, or he can play 44 as 13/9(2) 8/4(2) followed by 33 as 13/10 9/6(2) 7/4 and then 66 is on track. Either way, a three-big-set bearoff is achieved.
Okay, we looked at 21 first. Now let's look at 21 last. We'll examine it from the White side, as this represents White rolling doublets for her first eight turns, leaving blots on her 2pt and 1pt.
Six 6s must be played 24/6(2), so we'll start with that. And the only way to get blots on the 2pt and 1pt is 13/1 8/2 with three more 6s. That's nine 6s so far, and in order to round off a multiple of 4 the other three are essentially 13/7(3) [or conceivably 8/2 but only by transposition of 13/8/2 instead of 13/7/2], as shown below.
By the way, it is worth considering holding back the last 66 (keeping at least one 12/6 outside) for the option of splitting it between bear-in and bear-off, but it turns out to be unnecessary.
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| XGID=------------a----bcg---aa-:0:0:1:00:0:0:3:0:10 | |||||
Goal: end with blots on the 2pt and 1pt | |||||
Here, White's only path is 33 as 13/10 7/4(3) followed by 44. [If the last 66 had been held back, as mentioned above the diagram, then 33 and 44 could be played in either order.]
The secret (with 21, anyway) is that Blue or White (whichever is being analyzed in isolation) must put at least one checker on the 10pt at some point (i.e., by 13/10), so that either a follow-up 4 will go to the 6pt or a follow-up 6 will go to the 4pt, and either 44 or 66 can be split between bear-in and bear-off. This also explains why only 44 and 33 can accompany 66 66 66 (early) + 66 66 44 (bearing off); other doublets (11 22 55) won't work.
Can 21-last be interactive? No. If White ends up with a 21 bearoff, she must have played 13/7/1 at some point to meet her crossover obligation; yet Blue must have played 24/18/12 at some point to meet his obligation (true regardless of his non-doublet opener), and the two are in direct conflict. In short, Blue and White can't get by each other; a 21-last game is impossible.
.....(b) The non-doublet is 43.
In a similar vein, I believe 43 is the only 7-pip non-doublet that fits, and even so it is the trickiest find. Because there are only 160 pips remaining (instead of 164), it is necessary to shave 4 pips from one of the doublets, and the only one that appears to work is changing one of the 66s to 55 (i.e., 66 66 66 66 55 44 44 33).
Let's put aside the 6s that bear off from the 24pt and 6pt, which always use thirteen sixes (three double 6s plus an extra 6). What is flexible is the numbers coming from the 8pt and 6pt, and here are breakdowns:
When the non-doublet is 21: Four of the 13pt checkers each use 6+4+3 to bear off and the other 6+6+1. Two of the 8pt checkers use 4+4 to bearoff and the other 6+2 (total of seven 6s, eight 4s and four 3s, plus 21).
By contrast, when the non-doublet is 43: Three of the 13pt checkers use 6+4+3 and two use 5+4+4. One of the 8pt checkers uses 4+4 and the other two use 5+3 (total of three 6s, four 5s, nine 4s and five 3s, the non-doublet accounting for the fifth 4 and fifth 3).
The reorganization of the roll portions helps explain why 43 can work for a non-doublet but neither 52 nor 61 qualilfy. The reallocation of the 4s and 3s (in order to meld with the 5s instead of 6s) barely holds together as it is.
For the opening 43, the 3 may be played 13/10 or 8/5, and the 4 may be played 13/9 or 8/4, or 13/6 in conjunction with each other. As both 44 (twice) and 33 will be subsequently rolled, most transpositions work. For simplicity, I've chosen 13/6 (i.e., 13/10 + 13/6). Then adding 24/6(2) 13/7(2) with two double 6s reaches this position:
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| XGID=------HBC----B------------:0:0:1:00:0:0:3:0:10 | |||||
What combination is successful? | |||||
From here, 33J (8/5(2) 7/4(2)), 44O (13/5(2)), and 440 (8/4 4/off(3)) does the trick, pristinely leaving 66 66 55 to bear off.
What about 43 at the end? As with the last position for Blue, we'll assume two sets of 66 played 24/6(2) 13/7(2), reaching the same position as White, minus 13/6:
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| XGID=------------c----cbg------:0:0:1:00:0:0:3:0:10 | |||||
Aim to end with 4pt+3pt blots, reached here by clever slotting | |||||
White next rolls 44 and plays 13/9(2) 9/5 7/3, then 33 as 13/10 8/5(2) 7/4, then bears off a checker with 44 (i.e., 10/6 7/3 8/4/off), which leaves 66 66 55 to bear off. Challenging to visualize, but you'll get a chance to see it more directly at the end of the next sequence.
Can the opening-43 and end-43 sequences be interactive? Yes, if the double 6s are delayed until the opponent's midpoints are cleared, though it's a narrow feat. The moves look different, but the roll portions end up in the same place.
For example, Blue opens with 43T (13/6), White plays 44O (13/9(3) 9/5), Blue plays 33O (13/10(2) 8/5(2)) and now:
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| XGID=-b---BF-A-B-bB--bc-ea---B-:0:0:-1:00:0:0:3:0:10 | |||||
Start of interactive 43 vs 43 game | |||||
Here, White plays 33O (13/10(2) 8/5(2)), Blue plays 66C (24/18(2) 10/4(2)) and now:
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| XGID=-b---BE-A-B--B-bba-ec---B-:0:0:1:66:0:0:3:0:10 | |||||
Blue to play 66 | |||||
Now, White rolls 66 and double slots with 24/18(2) 10/4 9/3!), then Blue rolls 44 and plays 9/5(2). Both sides then roll 66 and play 18/6(2), and...
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| XGID=----BDH-A------aaa-gcaa---:0:0:-1:00:0:0:3:0:10 | |||||
Endgame conversion to 4pt+3pt blots | |||||
Here, White simply rolls 44 and takes a checker off (while rounding off the 6pt and 5pt stacks), then Blue neatly takes three checkers off with his 44. This leaves eight checkers on the 6pt and four on the 5pt for both sides, plus the (intentional) 4pt and 3pt blots that White will still have when the game ends in three more turns (each).
.....(b) The non-doublet is 65.
The non-doublet of 65 is in the vast majority of variations (mainly because 13/8 creates two four-stacks). All one-sided solutions can also be interactive, and work for both first and last roll, with any of these doublet combinations:
66 66 66 66 66 66 22 11
66 66 66 66 66 55 33 11
66 66 66 66 66 55 22 22
66 66 66 66 66 44 33 22
66 66 66 66 66 55 33 11
66 66 66 66 66 55 22 22
66 66 66 66 66 44 44 11
66 66 66 66 66 44 33 22
66 66 66 55 55 55 33 22
66 66 66 55 55 44 44 33
66 66 66 55 44 44 44 44
66 66 66 55 55 55 33 33
In the original post, I listed a second tie-breaker, which is to create a game in which the positions reached (along the way) have as high a beaver percentage as possible. What follows is a game in which all the positions are beavers.
(If the added tier of complexity doesn't interest you, just ignore the beaver-related comments and follow it like the other sequences -- as a cooperative game sequence that starts with 65 and finishes in the minimum of 17 rolls.)
Blue opens with 652 (13/2); an odd play for sure. However, he needs to be assigned a weaker play than S (24/18 13/8) or D (13/8 13/7), else after White's upcoming play she won't have a beaver.
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| XGID=-bA---E-C---eD---c-e----B-:0:0:-1:22:0:0:3:0:10 | |||||
White to play 22 | |||||
White plays 22O (13/11(3) 8/6), which has a well-chosen error size. A general (beaver-irrelevant) solution can be easily pieced together with other 22 plays. However, if she plays O (13/11(2) 8/6(2)), she will be a hair too weak to beaver; and if she plays 13/11(4), then Blue will be (slightly) too weak to correctly beaver after his upcoming play.
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| XGID=-bA---E-C---bDc--b-f----B-:0:0:1:22:0:0:3:0:10 | |||||
Blue to play 22 | |||||
Blue plays 22d (13/11(4)), vacating the midpoint with the minimum number of pips. If he comes down with 33 or 44, his race/position will be too favorable three rolls from now for White to beaver (as you will see).
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| XGID=-bA---E-C--Db-c--b-f----B-:0:0:-1:66:0:0:3:0:10 | |||||
White to play 66 | |||||
White plays 66b (makes the 12pt), and obviously Blue will not beaver. Next...
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| XGID=--A---E-C--Dbbc--b-f----B-:0:0:1:66:0:0:3:0:10 | |||||
White to play 66 | |||||
Blue fights back with 66C (24/18(2) 6/5(2)). He needs to make the 18pt to be able to beaver, unless he splits while making his strongest board (24/18 11/5(2) 8/2), in which case White's upcoming beaver will fail (details on request).
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| XGID=--A--BE-C--Bbbc--bBf------:0:0:-1:55:0:0:3:0:10 | |||||
White to play 55 | |||||
White plays 55O (13/8(2) 11/6(2)). She needs to clear her midpoint, and to guarantee all-doublet parity she also needs to create a four-stack. She can do so without leaving a shot by playing 22O (13/11(2) 8/6(2)) [the extra 11pt checker will become the 5pt blot in the end either way], but it would lead to a less balanced leapfrogging situation in the race, and meanwhile White has enough equity to beaver even with the shot being left (which, incidentally, wouldn't be the case if Blue had rolled 33 or 44 instead of 22 three rolls earlier).
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| XGID=--A--BE-C--B-ba--dBh------:0:0:1:66:0:0:3:0:10 | |||||
Blue to play 66 | |||||
Alternating between Blue being ahead by 15 and down by 9 (all positions therefore being beavers), the two players exchange three pairs of double 6s, reaching this position:
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| XGID=--D--DD------------ea--d--:0:0:1:22:0:0:3:0:10 | |||||
Blue to play 22 | |||||
Here, locking in the rest of the beavers requires care. If Blue starts with 66, White will keep pace with 66 and we'll lose one beaver: either Blue rolls 22 and won't have a beaver; or he rolls 55 and after White's 22 she won't have a beaver. We are in a similar pickle if Blue starts with 55. Beavers remain 100% only if he rolls 22 here.
White will respond with his own 22, Blue will roll 66 or 55, White will roll 66, and Blue will bear off with whichever of 66 or 55 is left. White will be left with one checker on his 6pt and one on his 5pt, as planned.
Recapped sequence: 652-22O22d-66B-66C-55O-66T-66J-66J-66-66-66-220-220-66-66-55
In summary, the conditions of the puzzle can be met in the minimum number of rolls (9 + 8 = 17) in a variety of ways. It is accomplished with the least difficulty if the non-doublet is 65 (which dovetails with a dozen doublet combinations), 21 is harder, 43 is hardest. All three work as the first roll or the last roll (unrequited bearoff), and against themselves or any of the others, except that 21 does not interact (it is one-sided only).
The no-beaver game is an added level and not intended to be the main thrust of the problem.
Nack