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| XGID=-a-C-bD-C--CbB-----bbb-bb-:0:0:-1:00:0:0:3:0:10 | |||||
White on roll | |||||
For the Mystery of the Big Stack, here were the two clues:
...(1) From here, both sides play (the checkers and cube) PERFECTLY, according to XGR++ evaluation.
...(2) Upon completion of his fifth turn (the end of the ten-roll sequence), Blue's 15 checkers are all STACKED on the same point.
What is a workable sequence?
...(3) (extra credit): Also, White has a closed board. How is this possible?
...(4) (extra extra credit): Also, White's first four (of her five) rolls are identical. How is this possible?
Nack
I'll fast forward to the (unique) "extra extra credit" solution, though I'll mention some simpler possibilities along the way.
Obviously, there is nowhere near enough time to get the three checkers on Blue's 3pt hit and recycled to the 6pt. Indeed, if you count roll portions it becomes clear there isn't even time to target the 2pt or 1pt. The super-stack target must be Blue's 3pt, where he already has three checkers.
The basic plan is this: White breaks the midpoint, leaving an ace shot, where Blue hits. White anchors on her 24pt, her two anchors working together to freeze Blue's 5s from his 6pt and 3s from his 8pt. So, Blue plays checkers safely to his 8pt, 6pt and 3pt. Then he rolls enough 3s to clear his 6pt, and enough 5s to clear his 8pt -- until all 15 checkers are on his 3pt.
Starting with four identical rolls (the extra-extra-credit stipulation) of 31 gets the job done for White. With the first 31, she leaves a blot on her midpoint (less undesirable than breaking her board). With her second 31, she comes off the roof and anchors on the 24pt -- perfect. Finally, her third and fourth 31's are just barely good enough to set up the closing of the board (the extra-credit stipulation), as you will see.
Below is the position after White kicks off the sequence by rolling her first 31:
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| XGID=-a-C-bD-C--CaB--a--bbb-bb-:0:0:1:61:0:0:3:0:10 | |||||
White played 31D, Blue rolls 61 | |||||
With any (non-doublet) ace, White was induced to leave a blot on the midpoint, so as not to break her board. (You can forget about a non-ace, where White brings two checkers down, White's lone 24pt checker can't defend itself, and the sequence is cooked.)
It is important to plan ahead the exact number of checkers needed on Blue's points. For four of his five turns, he'll need to move checkers off the 8pt and 6pt four checkers at a time, or he won't be able to transport all his checkers to the 3pt in time. There are currently three checkers on the 3pt, so it would be ideal to allocate the other twelve to multiples of four on each of the 8pt and 6pt. How can we achieve that?
Surveying the situation carefully, you'll see that Blue can either play 52 as 13/11 13/8 or play 43 52 or 61 as 13/7, followed by one set of double 5s, perfectly ending up with four checkers on his 8pt and eight on his 6pt. However, only one of those first-roll possibilities also hits White's blot, and that roll is 61.
[It is interesting that if White had instead rolled 21 (played 13/10), the extra pip she trails in the race is just enough to tempt Blue to play 61 conservatively -- to make his 2pt instead. Hence, any qualifying solution must start with a White roll of 31, 41, 51 or 61.]
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| XGID=aa-C-bE-C--C-A--a--bbb-bb-:0:0:-1:31:0:0:3:0:10 | |||||
Blue played 61T, White rolls 31 | |||||
Blue is set up to roll double 5s, except that we can't let him make his 1pt with it. Fortunately, just in time, White unleashes her second 31, using her ace to make the vital 24pt anchor (see diagram below).
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| XGID=-b-C-bE-C--C-A-----cbb-bb-:0:0:1:55:0:0:3:0:10 | |||||
White played 31C, Blue rolls 55 | |||||
As advertised, Blue roll double 5s, which clears his four rear checkers and creates stacks of four-multiples on his only two remaining "feed" points.
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| XGID=-b-C-bH-D----------cbb-bb-:0:0:-1:31:0:0:3:0:10 | |||||
Blue played 55O., White rolls 31 | |||||
Oddly enough, it is no longer necessary for White to hold her 24pt; she only needed it for that one Blue roll. Careful, though: we DO need White to maintain her 20pt anchor. With any roll that permits White to "leap" her 20pt checker over Blue's 8pt, she will indeed run a checker off her 20pt, Blue's 33 can be played 8/5*(4), and that will cook our goose.
White is therefore relegated to a roll of 11, 21, 31 or 33. Of these, 33 is the most flexible, played 24/12. This gives her several ways to close the board in two rolls (there is even a way to do it in one roll) -- though the board-close is only for the extra-credit bonus. That said, the only way to get extra extra credit is to go with an admittedly awkward roll of 31, which is played as shown below.
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| XGID=-b-C-bH-D----------bbabbb-:0:0:1:33:0:0:3:0:10 | |||||
White played 31P, Blue rolls 33 | |||||
Note that even with White owning her 24pt, we can't give Blue a 55 here (or on the next roll), because his subsequent 3s wouldn't be forced to come off the 3pt; they would bear off instead! Thus, Blue is now assigned his first of two double 3s...
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| XGID=-b-G-bD-D----------bbabbb-:0:0:-1:31:0:0:3:0:10 | |||||
Blue played 33, White rolls 31 | |||||
Again, we can't let White come off her 20pt, and given the delicate state of her board she would now even do so with a roll of 21. We can give her 11 (though it costs the board-closing extra credit), or better we can give her 33, which gives her 6 numbers (all 8s) to close her board. OR, we can assign White her fourth 31 (the best play being 24/20), to keep alive the extra extra credit. Will that work?
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| XGID=-a-G-cD-D----------bbabbb-:0:0:1:33:0:0:3:0:10 | |||||
White played 31U, Blue rolls 33 | |||||
Although Blue leads by 45 pips, he does not have an initial double (not that the legitimacy of the sequence was at stake -- there was no keep-cube-in-the-center stipulation). Anyway, Blue rolls his second 33 here, and...
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| XGID=-a-K-c--D----------bbabbb-:0:0:-1:44:0:0:3:0:10 | |||||
Blue played 33, White rolls 44 | |||||
Aha -- White has a board-covering roll up her sleeve (20/4). [By the way, she had suddenly become a favorite but didn't have a cube.]
Next, Blue rolls 55, and we reach the final position:
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| XGID=-a-O-b-------------bbbbbb-:0:0:-1:00:0:0:3:0:10 | |||||
31D-61T-31C-55O.31P-33-31U-33-44P-55 | |||||
Bob Koca perceived this to be a variant of my puzzle, while I called it "Bob's Puzzle." (I guess we won't be fighting over a patent.) I'm now filing it under "Big Stack."
Initially, Bob had failed to realize that my intent was to start with the position repeated at the top of this post (originally posted by Stick but with Blue on roll and having a specific number to play). However, in the process of correcting himself he suggested that stacking 15 checkers from the opening position might be a good alternate puzzle. Arguably, it is better, because it stems from the most common backgammon position that exists -- the opening position.
As you might expect (from the opening position), stacking 15 checkers can be accomplished most quickly if the 6pt is the object point. However, I was still in 3pt mode from the original puzzle, so I targeted the 3pt for my initial attempt here. My solution stacks on Blue's 8th turn (15 rolls altogether); below is an abbreviated account.
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| XGID=-b-C--D-D---eB---b-db---B-:0:0:-1:65:0:0:3:0:10 | |||||
Blue plays 53P, White plays 31P, Blue rolls 55O., White rolls 65 | |||||
The sequence reaching this position is captioned above, though there can be substitutions for White: Instead of opening 31P, we can give her 42P, 53P or 64P, and instead of the current 65 roll she can have 64 or 63. All eight combinations work (plus a ninth: 53P-53P-55O.62).
Bringing two checkers down is standard for the third-roll position 31P-55P-65. It is even more correct when Blue's 53P has preceded the sequence.
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| XGID=ab-C--D-D---cC---c-db---A-:0:0:-1:62:0:0:3:0:10 | |||||
White played 65D, Blue played 65R, White rolls 62 | |||||
After White came down with 65, Blue hit and ran to the midpoint with his own 65, and now White rolls 62...
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| XGID=-b-C-bD-D---bD---c-db-----:0:0:1:00:0:0:3:0:10 | |||||
White played 62Z, Blue played 65R, White played 53@ | |||||
Trailing by even more in the race, White (after entering with a 2) naturally came down with the 6 again; then Blue hit and ran with another 65 -- this time with his last back checker. White then entered and anchored on the 20pt with 53, yielding the position shown above.
Note how similar this is to the third diagram of the original problem's solution film (after White entered with an ace and made her second anchor), and the ending is much the same. Here, Blue brings four checkers down with 55, and his remaining moves are forced when he super-stacks with 55, 33 and 55. (Blue is pretty frail just prior to his final 55 joker, but if you want to keep the cube centered just keep the status quo -- don't make White's board too strong with her two interim rolls.)
[With this particular 3pt-target theme (used in both problems), you can get by with fewer than four (or eight) checkers on the 6pt, because there will simply be a 3 (or 3s) that cannot be played. However, the 8pt must have a four-multiple; otherwise, when Blue rolls his final 55, he will bear any leftover 5 (or 5s) off the 3pt, and the stack won't have the prescribed 15 checkers "upon the completion of Blue's fifth turn."]
After finding the above 3pt-stack sequence, I tried targeting the 6pt instead and was able to reduce the length (from 15) to 13 rolls. Before long (possibly by the time you read this paragraph), I'll give that solution in a nearby post (which I'll entitle something like"Big Stack -- 6pt SOLUTION").
Nack