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| XGID=-c----D-C---fAAAAAAf----B-:0:0:1:00:0:0:0:0:10 | |||||
With perfect play (using any match score), | |||||
PROBLEM 1: Beginning from the opening position (with either player going first), both players play perfectly (according to XGR++ evaluation). You may use any match score (or unlimited, Jacoby or not) that you like, as long as it is consistent throughout. Is it possible for Blue to end up with one blot on each of the six points in White's OUTER board, and if so how?
[What is relevant is the upper right quadrant of the diagram. The other 24 checkers (9 Blue, 15 White) are not necessarily on the points shown. I placed them on starting point locations just for orientation purposes.]
PROBLEM 2: What is the fewest number of moves that accomplish this feat?
PROBLEM 3: Same, but the cube does NOT remain centered.
When first pondering a six-blot formation in the opponent's outer board, I thought the fewest number of rolls to achieve it could be well over 20 (i.e., 10 moves for each player). However, more resources are available than I imagined.
My original idea was to give Blue his 18pt, 16pt and 14pt, break his midpoint, have White hit loose there, then give White a roll (53 or 64) that points on her 5pt or 4pt, leaving behind blots on her 10pt and 8pt as well. Blue then finishes with double 1s, entering and hitting thrice, turning the three points he owns in White's outer board into six blots. That scheme, however, requires that Blue get hit (sent back) at least five times (almost certainly more), use several roll portions to spring back checkers into the outfield, and of course double all that to account for White making a move every other roll.
To reduce the amount of Blue recycling, I streamlined to the archetypal setup exemplified below. White is on roll.
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| XGID=---a-BBBA---cAAcBcBe-A-A--:0:0:-1:66:0:0:3:0:10 | |||||
63S-63H-33M_64H-63@-62R-66B-33Z- | |||||
In the position above, White blitzes with 66. In response, Blue rolls 11, entering both checkers and hitting on the 17pt and 15pt, thereby achieving the outside six-blot formation.
The captioned sequence (perfectly played for Money) reports how the exact position arose. Variations: the three-roll snippet 31I-21H-42H midway through the sequence can be replaced by 21I-21H-43H (though it doesn't affect the final position) and/or White can hit with 63H (near the end, instead of 21H, adjusting her position 22/16). The length of these sequences is 17 rolls (i.e., 9 Blue, 8 White).
[Other final diagrams (resulting from slightly longer sequences) put Blue's rear blots on his 22pt + 20pt (instead of 23pt + 21pt) with White's last roll being 55 (instead of 66), or White also owns her 3pt or 1pt (or 4pt or 2pt in the parallel version) and/or Blue might have different offensive points, and/or both players' spare distributions can be altered (except for that of White's 10pt and 8pt).]
Subsequent to that bout, I looked for an early-game sequence that puts blots (of either/both colors) in White's outer board, aiming to create an urgent hit exchange that gives neither player time to clean up the outfield spray before Blue can six-ify it for himself. It turns out that such a sequence does exist and it allows us to save another roll.
In fact, I discovered that three rolls can be saved except for a single .014 error (according to XGR++). Hoping to perfect this sequence, I resorted to alternative match scores. This was an arduous task, because plugging the best-move leak inevitably caused leak(s) to appear somewhere else in the sequence. As you will see, the coveted 14-roll solution did ultimately emerge victorious, though I almost gave up before testing some pretty far-away scores.
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| XGID=-b----E-C---cEAaac-e----A-:0:0:-1:66:0:0:0:0:10 | |||||
White played 43D, Blue played 64R, White rolls 66 | |||||
For Money, White's 43D and Blue's 64R is the only perfectly-played opening-plus-reply combination to put three blots in White's outer board.
Mostly when White is ahead in the match, she will switch to 43Z (reverse split, 24/21 13/9) and/or Blue will switch to 43D-64S (Split, 24/18 13/9). Hence, when I circled back to try to make my near-perfect Money sequence match-perfect, I focused on scores with Blue leading, though most post-Crawford scores were also cooked by 43Z (all except p2, p6, p8, p10 and p12 -- I checked out to p20).
On to the third-roll play. If you didn't know how to play 66 for White in the above position, you might save yourself from a blunder in the future. According to XGR++ eval, the knee-jerk play of B (Both, 24/18(2) 13/7(2)) is second best but an error of .060 for Money (and wrong at all match scores).
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| XGID=------EbC---cEAaaa-e---bA-:0:0:1:63:0:0:0:0:10 | |||||
White played 66c, Blue rolls 63 | |||||
After White makes the best play of 66c (cross, 24/18(2) 8/2(2)), there are already four blots in her outer board.
Blue now rolls 63 and makes the obvious play...
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| XGID=a-----EbC---cEAAaa-e---b--:0:0:-1:65:0:0:0:0:10 | |||||
Blue played 63H, White rolls 65 | |||||
After Blue hit with 63H (as shown), there are still four blots in White's outer board, but now a second one has changed from White to Blue (the color that we eventually want them all to be).
Now how does White play her roll of 65?
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| XGID=-----aEbC---bEAAaaae---b--:0:0:1:00:0:0:0:0:10 | |||||
White played 65Z, Blue on roll | |||||
If you guessed that White should play Z (reverse split, bar/20 13/7), then (a) you show fine backgammon judgment [as the choice over C (Cross, bar/20 9/3) is very close] and/or (b) you demonstrate acute puzzle awareness (13/7 puts yet another blot into White's outer table)! White can afford to be blotty in that area because Blue's back checkers have both leapt past.
We now have two objectives for Blue: (1) To diminish his midpoint to one checker, and (2) To get at least three of his checkers sent back so that they can hit White on her 7pt, 8pt and 9pt. Ideally, we want to accomplish (1) and (2) simultaneously via a hit exchange (i.e., hit back, hit back, hit back...) somewhere in Blue's outer board -- in direct range of his midpoint.
For example, Blue could hit loose on his 5pt -- or on his 4pt or 3pt if White entered there instead, White hits and runs to Blue's 11pt, 10pt or 9pt, and the primary hit exchange can occur there. But not just any loose hit will do. If Blue hits from the 14pt or 15pt, extra rolls will be wasted resurrecting that outfield blot later. Blue must hit from the 13pt, or play 13pt-checker down and hit from the 6pt or 8pt. Unfortunately, Blue prefers to play safe or semi-safe, leaving at most one outfield blot. An interesting twist is that at Gammon-Go-ish scores some of those loose hits become correct, but in that case White is symmetrically Gammon-Save-ish and would not have opened with 43D (nor Blue 43D-64R). In short, we can eliminate the loose inside hit here at all scores.
In order to minimize your scrolling, I'll repeat the previous diagram. Keep in mind that the discussion at this point is a bit broader, covering situations where White has entered with any of 62Z, 63Z, 64Z or 65Z, or fanned. In other words, sometimes you want to visualize White's back checker blot on Blue's 2pt, 3pt or 4pt (instead of his 5pt), or in some cases on the bar (with White's 13/7 slot to be played when she enters).
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| XGID=-----aEbC---bEAAaaae---b--:0:0:1:00:0:0:0:0:10 | |||||
White played 65Z, Blue on roll | |||||
Fortunately, there is another approach. Blue can roll doublets -- pointing on White -- with 33A (Attack, 8/5*(2) 6/3*(2)) or 44A (8/4*(2) 6/2*(2)), depending upon where White has entered, and either of these board-making plays leave a Blue blot on the 8pt. It is there, on the 8pt, where the subsequent hit exchange will take place.
"Zwischenzug" (in between move) is a term borrowed from chess. Imagine that White is still on the bar (has not yet rolled, 13/7 not slotted either). Further suppose she fans on this turn or the next, and on the other turn enters with a non-ace (and 13/7 with the other half of the move). Blue can build a four-point board with (a) 44P (13/5(2)) or 55P (13/3(2)), combined with (b) 44A (8/4(2) 6/2(2)). Either (a) or (b) points on White, and the other move is the zwischenzug. Puzzle-wise, reducing the midpoint by two checkers has equivalent value to a single-checker reduction plus getting hit.
[Caveats: White can enter with a 4 before or after her fan. However, she cannot enter with a 2 before she fans because Blue's 44N (Near, 13/9*(2) 6/2*(2) is better than 44A, and Blue leaves no near-side blot. And she cannot enter with a 3 or 5 (i.e., 63Z or 65Z) after she fans because 44A will be the move Blue has not yet played.]
Some zwischenzug variations fail when the midpoint reduction occurs first; e.g., F-55P-64Z-44A is okay, but 64Z-44A-F-55 is not -- Blue will (cookingly) play 15/10 with the fourth 5 rather than restart the 8pt. Other zwischenzugs fail due to a cube timing problem; e.g., 64Z-44A-F-44P-53H is no double before the 44P but a cash after 53H. An example of a sound variation is 63Z-55P-F-C-44A-53H- (the "C" sneaking in by a mere .005), and for those curious it can be finished on the 16th roll with -51H-53H-51H-53H-44K-1-42H.
For a while, I thought 16 rolls to be the fewest achievable. However, there is one needle-threading method that saves an additional two rolls (and even then only at select match scores) and it can only stem from an immediate 8pt-hit exchange (i.e., without a zwischenzug).
Thus, the featured sequence continues with 65Z-33A, as diagrammed below. A couple of notes, though:
(1) 63Z-33A would reach the identical position, except Blue plays 33N (Near, 13/10(2) 6/3*(2)) instead. Thus 65Z-33A.
(2) With 65Z-33, while A (Attack, 8/5*(2) 6/3(2)) is best for Money, there are many match scores at which O (Outer, 14/5* 8/5) evaluates better. The scores at which all other moves in the feature sequence work -- that fail due solely to 33O beating 33A, are either Crawford (c8) or Post-crawford (px, where "x" is an odd number).
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| XGID=a--B-BCbA---bEAAaaae---b--:0:0:-1:62:0:0:0:0:10 | |||||
Blue played 33A, White rolls 62 | |||||
After Blue makes his 33A play, White begins the hit exchange on the 8pt...
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| XGID=a--B-BCbA---bDAAaaae---bA-:0:0:-1:11:0:0:0:0:10 | |||||
White played 62H, Blue played 51H, White rolls 11 | |||||
Two plays have been added. White hit on Blue's 8pt with a 62, and Blue hit back there with 51. This diagram is the same as the previous diagram, except that one of Blue's midpoint spares has been transported to the 24pt.
If this simple exchange of hits (with 62H-51H) repeats three more times, Blue's midpoint will finally be diminished to one checker (so far, so good) but his 24pt will be stacked with four checkers (overkill). From there, even if F-C-33V-1-54K were legit (it fails by .004 when Blue plays 13/10 with one of his 3s), that solution would require 18 rolls.
A nice improvement is to give Blue a timely double 4s when he is down to exactly two checkers on his midpoint and three on his 24pt (with White's position being unaltered). Blue will hit twice with bar/17* 24/16*, White will enter both (so that she can take the cube), then Blue can finish off with 65K (24/18* 13/8*) and achieve the six-blot formation on the 16th roll. This entire sequence (43D-64R-66c-63H-65Z-33A-62H-51H-62H-51H-62H-51H-62H-44K-41-C-65K) is perfectly played for Money and a wide range of scores.
The only chance to improve further is to somehow advance Blue's 24pt checkers into direct range of more White outer-board blots. But how? We might try making Blue's hit-back number 53 or 54 (rather than always 51) so that some of his entrances can be on the 22pt or 21pt (i.e., not all on the 24pt). The problem is that Blue's best play will be bar/17* or bar/16*, which puts a premature halt to the near-side hit exchange (before enough midpoint checkers have been recycled).
That brings us to the nifty six-move combination that starts with White rolling double 1s (from the diagram above). There is a play that for Money XG 3-ply evaluates as third best (¡V.057) yet XGR++ evaluates it as +.050 (easily best)! Can you guess that play? It has been added to the diagram below.
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| XGID=-a-B-BCaa---bDAAaaae----bB:0:0:1:33:0:0:0:0:10 | |||||
White played 11e, Blue rolls 33 | |||||
White's play of 11e (each, bar/24 18/17* 2/1*(2)), shifting from her 2pt to 1pt, has left only 4 return shots (22 33 44 55). When both players have a bunch of blots lying around, the first player to roll doublets is a huge beneficiary, often icing an advantage or turning around a disadvantage. Typically, the idea is that after hitting two (or more) blots, which is unlikely to meet with a strong retaliation, the next turn can be used to continue the attack or consolidate. If Blue himself had fewer blots, White would have played the more structural E (bar/24 18/17* 6/5(2)) or F_ (bar/24 8/7 6/5(2)), depending on score.
Now or on earlier turns White would not have risked bar/23 18/17* (with 21) or bar/21 18/17* (with 41). The hitting of a second checker (with 11) offers enough protection to justify breaking her anchor.
In terms of the puzzle, 11e is a rejuvenating resource. It creates a choice of points (Blue's 7pt and 8pt, both of which get used) on which to continue the hit exchange, and makes it possible to advance two back checkers to the 22pt (in direct range of all three White outfield blots), given Blue's current roll of double 3s.
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| XGID=aa-B-BCAa---bCAAaaae--B-b-:0:0:-1:62:0:0:0:0:10 | |||||
Blue played 337, White rolls 62 | |||||
In response to White's 11e play, Blue's just-played 337 (24/22(2) 13/7*) resembles a snapback in the game of Go.
[Back at the 65Z-33 diagram, if instead White had entered with a 4 and Blue played 44A, he would now own his 4pt and 2pt instead of his 5pt and 3pt, and Blue's best play with the 33 rolled here would have been to hit on the 8pt instead of on the 7pt, costing us the 14pt blot.]
Now White can continue the hit exchange, this time on Blue's 7pt. If she does so with 61 (or 52 or 43, same thing), she will be cubed out for Money, but not necessarily at other scores. Therefore, it seems the roll we assign White might still be any of 64, 62 or 61. The reason we are choosing 62 will be explained in a moment.
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| XGID=--aB-BCaa---bCAAaaae--B-bA:0:0:1:55:0:0:0:0:10 | |||||
White played 62H, Blue rolls 55 | |||||
Now that we know Blue's next roll (55), it is easy to see why White was not assigned 64H -- yielding the same position but with the rear blot on her 21pt (Blue's 4pt) instead of her 23pt. In that scenario, Blue (after entering and using his second 5 to hit 13/8* no matter what) would hit with 14/4*, and we'd lose the 14pt blot.
The refutation of White's 61H (same position but with the rear blot on her 24pt instead of 23pt) is much subtler. White's 23pt blot is in direct range of Blue's 8pt, and this gives Blue slightly more incentive to lock up the 8pt. Without that incentive, Blue is more willing to risk being hit there with (an unlikely) 44 and will diversify his builders with 15/10 instead of the stiffer 13/8. Therefore, with 61H rejected in this paragraph, and 64H rejected in the previous paragraph, our only remaining chance (to procure a 14-roll solution) is to assign White 62H.
Even so, this 8pt-securing factor rarely makes a pivotal difference. For one thing, it will never be enough if the cube has been turned, and for Money (prior to the 55) it is a D/T. It is at this point we can finally conclude that for Money no solution of 14 rolls is possible (fewest is 16). Indeed, other scores with Blue trailing can be likewise eliminated -- the cube action will go D/T or D/P just after the 62H (if not sooner).
Scores at which the 8pt-making play (bar/20 22/17* 13/8*(2)) is best mainly include: p3+ (all post-Crawford scores with opponent needing 3 or more), and c4, c6, c8 (etc.). However, at these scores 33O is better than 33A back on Blue's third turn (roll #6).
There are four surviving scores, ones that still dovetail with the featured sequence. They are: ¡V5¡V15, ¡V9¡V14, ¡V9¡V16 and ¡V10¡V16, as listed to the right of the diagram below.
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| XGID=b-aB-BCaB---bAAAaAaeA-A-b-:0:0:-1:42:0:0:0:0:10 | |||||
Blue played 55B, White rolls 42 | |||||
With the midpoint depleted to just one checker, all that remains is for Blue to hit on the 16pt and 18pt with his rearmost checkers. That can be done with a roll of 62 -- sounds simple enough. The catch is that Blue will instead hit on his 7pt with the 6, to keep White from anchoring there and/or to help build his own 4pt (or 7pt).
Fortunately, two 4s (instead of a 6 and a 2) can perform the desired double-hit. This can only happen with a roll of double 4s, but with the other two 4s making the 4pt is so compelling we don't have to worry about 4s within the outfield being better. However, given the chance, Blue will use three 4s to make the 4pt and hit loose on his 2pt -- or his entire roll to point on the 2pt (trumping one or both far-side hits). Therefore, White must enter/anchor with a 2 to prevent that. The question is whether White enters with a 4, or with a 1, or fans, with her other roofed checker.
Below is a table elucidating the three 14-roll solutions. The sequence (top row) is identical for all four surviving scores, except for White's "42," with which she can instead enter with a 2 or with 21, where noted in parentheses (far right). In all but one case, Blue is TG (too good) to double prior to the final 44. Assuming a solution of only 14 rolls, White entering with 42 at the ¡V5¡V15 score uniquely satisfies the "Problem 3" (extra credit) condition that the cube does NOT remain centered. That is, Blue should double only the exact situation depicted in the NEXT diagram.
-------------------------------------------------------------------------------------------------- ...-5-15........13..18..........0.3..8......................0.3....C.........(or 2 TG)
...Score........43D-64R-66c-63H-65Z-33A-62H-51H-11e-337-62H-55B-42-C-44C.....Alternatives to 42
...-9-14.........7..10..........5...........................2......TG........(or 2 TG or 21 TG)
...-9-16........10..13..........4...19......................0.3....TG........(or 2 TG or 21 TG)
..-10-16.........7..10..........4...........................0.0....TG........(or 2 TG or 21 TG)
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The cells of the table designate how many thousandths of a point a given move in the sequence beats the second best move. For example, at a score of ¡V5¡V15, 43D is better than its closest rival by .013. For another example, at the ¡V5¡V15 and ¡V9¡V16 scores, 55B defeats the second best move by .0003 (about a third of one thousandth of a point). If there is no number, that means that the sequence move bests the second best move by more than 20 (i.e., .020).
I've posted the above table in case future bot evaluations (or rollouts) overturn one or more solutions, in which case at least one 14-roll solution will probably still exist. The score at which the sequence(s) are least likely to be overturned is ¡V9¡V14.
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| XGID=--bBaBCaB---bAAAaAaeA-A-b-:0:0:1:00:10:0:0:15:10 | |||||
White entered with 42, Blue now Cubes and rolls 44 | |||||
At this score of ¡V5¡V15, Blue cubes (giving White a tiny take of -.996) and rolls 44 (dice not shown).
The final position, after Blue makes his best play of C (Cross, 22/18* 20/16* 8/4*(2)), is shown below.
Nack
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| XGID=c-bBBBCa----bAAAAAAe----b-:1:-1:-1:00:10:0:0:15:10 | |||||
43D-64R-66c-63H-65Z-33A-62H-51H-11e-337-62H-55B-42-C-44C | |||||