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| XGID=------EBBBBB--------------:0:0:1:00:0:0:1:0:10 | |||||
Blue's objective | |||||
The object of this puzzle is to reach the above formation for Blue (five on 6pt, two on each of the five higher points).
Congratulations to Mike (Clapsadle), Rory and Scotty for solving Problems 1 and 2!
Scroll down for solutions to all four associated problems.
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| XGID=------E-C----E----------B-:0:0:1:00:0:0:1:0:10 | |||||
| Blue's starting position, without opposition | |||||
From his starting position, how can Blue reach the (previously shown) outer prime formation in four rolls?
First, some logic: Blue's starting position is the usual 167 pips, and his final position is 120 pips. He therefore moves 47 pips in the four rolls. If he uses three doublets and one non-doublet, the non-doublet must be 3, 7 or 11 pips (in order to have a multiple of 4 left over for the doublets). There are two non-doublet duos, though for one of them (21 62, which appears in two roll sets), the 62 can sometimes be treated as a 22 roll with fewer options. The other duo (41 64, which appears in only one roll set) will be discussed in the section for problem 2.
There are many solutions, and most of them happen to make use of the points that White would normally occupy (her midpoint, 8pt and 6pt, which are Blue's 12pt, 17pt and 19pt). Below are all possible roll sets:
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.....21 66 33 22.....61 66 22 22.....43 66 33 11
.....21 66 33 62.....61 66 33 11.....43 55 44 11
.....21 66 44 11.....61 55 44 11.....52 66 33 11
.....21 55 44 62.....61 55 33 22.....52 55 44 11
.....21 55 44 22.....61 44 44 22.....65 55 33 11
.....................................64 55 33 41
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For example, Mike's original submission (66b, 22d, 22d, 61P) played the 24pt checkers to the (usually occupied) 12pt, followed by 22 making the 11pt and 10pt, another 22 making the 9pt, and 61 making the bar point. Here's a different solution that makes use of both the 17pt and 12pt: 11B (23pt + 12pt), 66b (11pt), 61, 33. And here's one that uses both the 19pt and 12pt: 55R (14pt), 44, 61, 11. (Where the moves seemed sufficiently obvious, I listed only the rolls.)
For any of the sixteen roll sets listed, the rolls can be played in any order, and if those orders count separately there are 360 solutions. Moreover, if numbers landing on specific points is accounted for individually (relevant when pairs of rolls within the same roll set have redundant numbers: 44 44, 44 22, 61 66, 52 55, etc.), the number of solutions runs into the thousands. That is, it depends on how liberally one chooses to count. (Without the five-on-6pt condition, even more solutions would be possible by squandering partial rolls off the 6pt.)
My favorite problem-1 solution -- in part because the 19pt and 12pt must both be used regardless of roll order (though this is also true of four other roll sets) -- involves the only roll set that contains 65, and in the tricky order 11 65 33 55. Here's a film clip:
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| XGID=------EAB--AAC----------B-:0:0:1:65:0:0:3:0:10 | |||||
Blue played 11, and now rolls 65 | |||||
Blue played 11 as 13/12 13/11 8/7, and next rolls 65. Regardless of when 65 appears in the sequence, the 6 must be played 13/7, as shown in the next diagram. [Also, btw, at least one the five 5s must be played 24/19, so we can already be certain that the 65 55 33 11 roll set does not qualify for a problem 2 solution.]
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| XGID=------EBB--AAB-----A----A-:0:0:1:33:0:0:3:0:10 | |||||
Blue played 65 (24/19 13/7), and now rolls 33 | |||||
Blue played 65 (24/19 13/7), and now rolls 33.
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| XGID=------EBBABA-------A-A----:0:0:1:55:0:0:3:0:10 | |||||
Blue played 33 (24/21 13/10(2) 12/9), and now rolls 55 | |||||
I trust you can visualize Blue's prime-making move with 55 and we can skip the final diagram. If not, please review the very first diagram of this post. :)
Scroll down for solutions to problem 2.
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| XGID=-b----E-C---eE---c-e----B-:0:0:-1:00:0:0:1:0:10 | |||||
| Starting position of backgammon | |||||
Again, reach the outer prime formation (shown in the very first diagram), except the first of Blue's (four) rolls must be a non-doublet, and he cannot move any roll portion to his 19pt, 17pt or 12pt (i.e., the usual points White occupies as shown above -- and she does not move).
Two main aspects distinguish problem 2 from problem 1. One is how the back checkers are handled. Deprived of the 12pt landing spot, they cannot move with 66 first, else 13/7(2) will dillify the 8pt spare. And they cannot move with 55 first, nor 22 then 55 (obviously, because the 19pt and 17pts are occupied). If 55 is included, the back checkers must be moved with 33 or 44 first. Or if 66 is included, at least one back checker must be moved with a 2, 3 or 4 first.
The second aspect that distinguishes problem 2 from problem 1 is that double 1s cannot be included (though it takes all three White-occupied points to exclude it!). It would still be possible to thread the needle with 33 (24/21 13/10(3), 11 (24/22 10/9 8/7), 66 (22/16 21/9 13/7) and 52 (16/11 13/11), except for the additional condition that Blue must START with a non-doublet. In short, roll sets with 11 do not pass muster for levels higher than problem 1.
The upshot is that half of the roll sets (that qualified for problem 1) are eliminated. Here are the ones that work for problem 2:
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.....21 66 33 22.....61 66 22 22
.....21 66 33 62.....61 55 33 22
.....21 55 44 62.....61 44 44 22
.....21 55 44 22.....64 55 33 41
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Let's start with Rory's solution: 61P, 22B, 66B, 22D. As it happens, this is a variant of Mike's problem 1 submission listed earlier (66b, 22d, 22d, 61P). Rory played Mike's last roll first and his third roll last, which doesn't matter. What does matter is that Rory reversed the order of the other 22 and the 66, which allowed him to bypass the 12pt (i.e., on the way to the 10pt, his back checkers land on the 22pt instead of the 12pt), thereby qualifying his sequence as a Problem-2 solution.
For his own problem-2 submission -- 21d, 44b, 55B, 22D -- Mike radically altered his original roll set in order to solve the problem with his back checkers: This time they go to the 20pt (with half of 44) before going to the 10pt (with 55).
As only one roll set employs 41 -- and 64 for that matter, I'll feature it here. Although any roll order that starts with non-doublet 64 or 41 and in which 33 precedes 55 is legitimate, each of the four rolls can be played in only one way. I will describe one such roll order, hoping that the aid of one diagram is all you'll need to follow the entire sequence.
Blue starts with 41d (13/9 8/7), then covers the two blots with 64D. The midway position is shown below.
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| XGID=-b----EBBB--eB---c-e----B-:0:0:1:33:0:0:3:0:10 | |||||
Blue played 41d and 64D (He'll finish with 33B and 55B) | |||||
From here, Blue plays 33B (24/21(2) 13/10(2), and finally 55B (21/11(2)). It's a clean, straightforward solution.
[With most problem-2 move combinations, and for at least some of the roll orders, it is possible to insert White moves in between -- assuming the conditions specified in the next paragraph are met -- thereby promoting them to problem-3 solutions. For example, the one I just described could be expanded to 62S-41H-32@-64D-41P-33B-xxx-55B (where 41P can optionally be 64P, and for xxx White rolls anything except 44).]
Blue reaches the same outer-prime formation from the opening position, but it's interactive: WHITE starts, each side plays FOUR moves, and all checker plays and cube actions are perfect (according to XGR++ evaluation, money game).
As Mike and Rory realized, a solution that works for problem 2 is automatically viable for problem 1. By the same token, Scotty pointed out that any problem 3 solution (even though his initial effort did not meet 3's perfect-play requirement) works for problems 1 and 2. Hence, he did not bother submitting separate solutions for 1 and 2 -- any problem-3 solution is a subset.
Scotty does not have XG. I should point out, though, that anyone with access to Snowie or GnuBG can check that bot and typically get the same answers. (And even if a solution happens to depend on a very close play, you can always ask.) So, while Scotty got further than anyone else (kudos), blanket-assigning White "fan" to her last three moves seems less than a stellar effort. On the other hand, I think I see Scotty's motivation: he cleverly managed to partially qualify for a problem 4 solution as well, because in his final position, Blue leads by 46 pips ("more than 45" being a stated problem-4 requirement). He just hoped that -- somehow ("gee, how would I know, I don't have XG") -- Blue would never have had a better checker play nor proper cube along the way!
What follows is Scotty's problem 3 "solution" (which was partial, but I'm completing it):
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| XGID=aa----EAB--AdDa--c-e----B-:0:0:-1:00:0:0:3:0:10 | |||||
62S-21H, and now White is on roll | |||||
Here, Scotty gave White a "fan" (her first of three), then had Blue cover his third and fourth point in the outer prime with 62D. In a return post, I clarified that White's opening 62S and Blue's 21H reply were on a good track, but if White now fans Blue's 62D becomes a whopper. In the form of a hint I asked, "What White move can you find that will prevent (or discourage) Blue from making his 5pt with 62?" It doesn't look like Scotty or anyone else found the answer (or perhaps he hasn't seen the post yet, or is still working on the follow-up).
Perhaps because my puzzles don't tend to be trivial, readers assume it is not worth looking for an easy answer to any question I pose. But easy it is, in this case. To prevent Blue from making his 5pt, White simply anchors there herself! Her play is 54@, as shown below.
[Subtler solutions are possible from the above diagram. Technically, Scotty was still alive at the 3-level even after delegating White her first fan, though it makes the path to glory more difficult to find. For example (following 62S-21H-F) -44B-43@-55B-66O_22D -- or instead of 66O_ any roll from 8 to 16 pips (without a 5) is sufficient, though in that case Blue will ship the cube before finishing with 22D.]
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| XGID=-----bEAB--AdDa--c-e----B-:0:0:1:62:0:0:3:0:10 | |||||
62S-21H-54@, now Blue rolls 62 | |||||
Blue is now free to roll 62 and correctly play D (Down, 13/11 13/7). It is interesting that if White's 11pt blot is on her 10pt (having played an opening 63S instead of 62S), Blue's best XGR++ play (beating D by a mere .003) is S (Split, 24/18 13/11). [Some of you may be familiar with the theory that it is more efficient to split to the 18pt against a 10pt builder than against other builders.] That is, a solution starting with 63S-21D-54@-62D is XGR++ imperfect, though I would accept it. As it is, White's builder is on her 11pt, so Blue's 62D, as played below, is definitely best.
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| XGID=-----bEBB--BdBa--c-e----B-:0:0:-1:11:0:0:3:0:10 | |||||
62S-21H-54@-62D, now White rolls 11 | |||||
The trick here is to induce Blue into playing 44B or 33B (half anchor, half down) rather than playing her back checkers out all the way. Let's discuss 44 first. As it is not possible for White to make her 9pt (a roll of 42 or 22 would be better spent), her only successful resource is to overwhelm pip-wise with 66D (Down). Blue then hedges with (the half-and-half) 44B, White rolls anything (except 55), and Blue snuggles into the missing point of the prime with 55B.
However, I chose a variation that caters to Blue's upcoming 33 (instead of 44). White makes her bar point (to stop Blue from anchoring there) with 64, 41 or 11. Any of those White rolls do the trick. I chose 11(N) for this example, to which (as advertised) Blue responds with 33B. Both moves are added to the board below:
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| XGID=-----bEBB-BBd-a--abcbB----:0:0:-1:00:0:0:3:0:10 | |||||
62S-21H-54@-62D-11N-33B, White on roll | |||||
Here, as long as White rolls anything other than 44, 33, 31 or 11 (with which she would cook the solution by making her 16pt or 10pt), Blue will complete the coveted prime with 66B.
Incidentally, it is possible to convert Mike's problem-2 submission (21d, 44b, 55B, 22D) to a problem-3 solution that begins with the same opening + reply sequence (62S-21H) Scotty employed. For example, 62S-21H-32@-44b-66D-55B-11@-22D. (44b here translates to the nifty 24/20(2) 13/9(2) -- Blue delays covering the 7pt until the end of the sequence.)
Again, Blue reaches the (initially diagrammed) outer-prime formation from the opening position interactively (White starts, with four perfectly played moves each), but in addition: It is never proper to turn the cube, and in the final position one side leads the other by more than 55 pips.
Unfortunately, I wrote "more than 45 pips" in the original post. After being pleased to find a solution with a final pipcount as lopsided as 120 to 176, I subtracted the smaller number from the larger number and got 46. That may seem pathetic, but the explanation is simple enough. When you reach my age, arithmetic is more challenging.
With the pip-lead condition as intended -- more than 55 -- I believe there is only one sequence that works, and it's all the more fitting that it employs the only roll set for Blue that contains two double 4s. Here is a film of that unique solution:
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| XGID=----b-EBB---dD--ac-e----B-:0:0:1:44:0:0:3:0:10 | |||||
41S-61P-32@, Blue rolls 44 | |||||
It may seem that 43Z-61P-21@ and 64S-61P-43@ are acceptable alternate routes to the above position. However, XGR++ evaluation (the arbiter of these puzzles) ranks opening 43D and opening 64P (rather than 43Z and/or 64S) as best. Hence, the uniqueness of the sequence is still intact.
Blue now rolls 44. He should grab the 16pt (hitting there). The after-position is diagrammed below.
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| XGID=a---b-EBB---dD--Bc-e------:0:0:-1:22:0:0:3:0:10 | |||||
41S-61P-32@-44R, White rolls 22 | |||||
As a puzzle creator, I am fascinated with the situation that has arisen. In order to maximize Blue's race lead in the final position, I sought a way to jar loose another White checker that Blue can hit with his pending double 2s. Originally, I had devised the opening sequence 32S-61P-43Z-44R, which reaches the same position diagrammed above except with White having played an extra 13/11 (a "pre-jarred" checker, if you will). That's exactly where I want White's next blot to be! Moreover, there are several entering White rolls in that pre-jarred position (e.g., 31, 11) with which she will keep the blot on the 11pt. However, in all those cases, in responding with 22, Blue (or rather XGR++) refuses to hit and make his 11pt (as I would like); instead he hits and aggressively makes his 5pt (or 4pt).
Fortunately, I stumbled on the concept of altering the opening sequence (to the starker 41S-61P-32@) in conjunction with giving White double 2s (from the roof) here, so that the (absent) 11pt blot re-materializes. After all, White's 11pt checker is a bigger asset as a builder than it is a liability as a blot, so she wants it there anyway. The key is that this is the only way to also give White an extra point in her board. [I was unable to profitably acquire an extra point in the position where the blot started on the 11pt, because White's (entering) 22 makes the 4pt and covers the 11pt; with 11 she makes the 20pt rather than the 5pt; and with 33 and 44 she advances the 11pt blot. The only roll that leaves the blot and sufficiently encourages Blue to make his own 11pt is White's 55 (played bar/5), but the largeness of the roll negates Blue's pip-gain from one of the hits.]
The position after White plays her 22 is shown below.
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| XGID=--a-b-EBB---cDa-Bc-c-b----:0:0:1:22:0:0:3:0:10 | |||||
41S-61P-32@-44R-22Y_, Blue rolls 22 | |||||
The pivotal effect of White having the extra point in her board (which she doesn't if the sequence is, for example, 32S-61P-43Z-44R-31C) is that Blue is discouraged from playing as tactically as 16/14*(2) 7/5(2) with his double 2s. He now settles for the more positional 16/14*(2) 13/11(2). Mission accomplished!
The position after Blue plays his 22 is shown below.
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| XGID=a-a-b-EBB--BcBB--c-c-b----:0:0:-1:21:0:0:3:0:10 | |||||
41S-61P-32@-44R-22Y_22B, White rolls 21 | |||||
Even if White is allotted the minimum-sized roll, her best move (or for that matter all but one of her legal moves!) is able to stave off the cube -- good news. Accordingly (after looking ahead and proofing it for the desired 44 response from Blue), I gave White a roll of 21, which she plays as shown below.
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| XGID=--b-b-EBB--BcBB--c-bab----:0:0:1:44:0:0:3:0:10 | |||||
41S-61P-32@-44R-22Y_22B-21$, Blue rolls 44 | |||||
As you have no doubt anticipated, Blue's final roll is 44, with which he locks up the prime (though that play beats 13/5(2) by only .003). This move is reflected in the final diagram below.
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| XGID=--b-b-EBBBBBc----c-bab----:0:0:-1:00:0:0:3:0:10 | |||||
41S-61P-32@-44R-22Y_22B-21$-44D, Final Position | |||||
Carefully subtracting one pipcount from the other reveals Blue's lead to be 56 ("more than 55," which will be the stated condition if/when I ever reprint this puzzle). I accidentally asked only for "more than 45" in the original post, and therefore I would have accepted any answer with a pip lead exceeding 45, and several are possible. The solutions ranging from 46- to 55-pip leads are variations of the same sequence with White's final roll altered (from 21) to 11, 32, 42, 52, 22 or 53. (The second anchor is necessary to ensure Blue primes with 44, rather than making or pointing on an inside point, though it turns out not to quite work for 31 or 62.) The only other sequence I know of that falls into this range is 63S-21H-32@-44b-11@-55B-11Y_62D, which yields a 47-pip lead.
In short, there are (at least) seven solutions in the 46- to 55-pip-lead range. The sequence filmed above is the only solution that surpasses 55.
Nack