                        |
| ||||
| XGID=---BaBC-B---dB---A-jB---A-:0:0:1:00:0:0:0:9:10 | |||||
| Reach in 8 rolls (Blue starts) | |||||
"Reach the above position LEGALLY (no need to play well) in just eight rolls. Blue rolls first, White last. However, I'll accept any solution of twelve rolls or fewer."
There are many ways to legally arrive at the above position, even in as few as eight rolls, though they involve themes that are difficult to find. In order to limit the discussion to a manageable number of possibilities, (for the purposes of this post) I'll attach a new condition: The players receive only the minimum number of doublets needed to accomplish the task. That minimum turns out to be two doublets for Blue, and one doublet for White.
Blue can be assigned any of eleven opening moves. (I'll elaborate after Blue's next turn.) Let's give him opening 41$, as diagrammed below.
        |
| ||||
| XGID=-b---AD-CA--eD---c-e----B-:0:0:-1:22:0:0:3:0:10 | |||||
Blue played 41$, White rolls 22 | |||||
White has 22 to play, and given that it is her only doublet, she needs four crossovers. She must play O. (13/11 8/6(3)), as shown below.
   |
| ||||
| XGID=-b---AD-CA--dDa----h----B-:0:0:1:62:0:0:3:0:10 | |||||
White played 22O. (13/11 8/6(3)), Blue rolls 62 | |||||
The only roll we can give Blue now is 62, and he must double-slot with & (13/7 6/4) as shown below.
   |
| ||||
| XGID=-b--AACACA--dCa----h----B-:0:0:-1:65:0:0:3:0:10 | |||||
Blue played 62&, White rolls 65 | |||||
To this point, Blue's four roll portions -- 13/9, 13/7, 6/5 and 6/4 -- can be paired off in six different ways. However, 13/7 can be replaced by 13/10 or 13/11, which triples the permutations, minus two (the illegal 13/11 6/4 permutes twice). Hence, technically, this position and the two parallel ones can be reached in 16 different ways.
What's going to happen is that Blue will be hit twice, one blot per turn, but his 9pt and 5pt blots are to be left alone. (The reason will later become apparent.)
White now rolls 65. She hits with the 6, and has a 5 to play. She needs to end up with just three remaining crossovers for the checker on her 18pt and 11pt, and currently there are four (because the 18pt checker can't go to the 12pt). So, she must play 18/13 or 11/6. (13/8 also makes a crossover, but it oddly fails. White would have to lift 8/6 on the very next turn so as not to be hit there, and that would leave her with two 5s that cannot be played with one non-doublet.)
White's roll does not have to be 65. Instead it can be 63 or 64, with the 3 played either 18/15 or 13/10, or with the 4 played either 18/14 or 13/9. Also, looking at parallel solutions, if Blue's 7pt blot were on the 10pt, or 11pt, White's only play would be to hit one blot with 63r, or 64r, respectively.
That said, Blue's blot is actually on the 7pt, and White has rolled 65 (hitting with the 6). As explained, the 5 can go to either her midpoint or 6pt but I have chosen the latter (see diagram below).
    |
| ||||
| XGID=-a--AACaCA--dC-----i----BA:0:0:1:55:0:0:3:0:10 | |||||
White played 65C, Blue rolls 55 | |||||
Blue now springs the first of his two doublets. He enters and makes his 3pt with 55M_ (bar/20 13/8 8/3(2)), as shown below.
 |
| ||||
| XGID=-a-BAACaBA--dB-----iA---B-:0:0:-1:43:0:0:3:0:10 | |||||
Blue played 55M_, White rolls 43 | |||||
The critical part of White's current roll is the 3, hitting Blue on the 21pt. The other half of her roll can be a 6, 5 or 4 (not a 2, because 18/16 hits again, and 13/11 leaves two 5s), so that she will be reduced to two crossovers on the outside.
Looking parallel again, if White's 18pt checker were on her 15pt (with one of the 6pt checkers still back on her 11pt), her roll now could be only 63 or 53. Or if it were on her 14pt, her roll could be 53 or 43. Or if it were on her 13pt, her roll could be any of 64, 54 or 43.
However, the sequence chosen has White's checker on her 18pt and the roll is 43. White could play the non-hitting half to the 14pt, but she will actually play it to her 9pt, as shown below.
  |
| ||||
| XGID=---BaACaBA--cB--a--iA---BA:0:0:1:44:0:0:3:0:10 | |||||
White played 43Z, Blue rolls 44 | |||||
Blue now unleashes the second of his doublets. Three of his 4s (bar/21/17 24/20) exactly make his position on the far side, and the reason for Blue having single checkers on each of the 9pt and 5pt becomes clear. He needs a fourth 4 to exactly make his position on the near side as well. Blue's position after playing the 44 is shown below.
 |
| ||||
| XGID=---BaBCaB---cB--aA-iB---A-:0:0:-1:53:0:0:3:0:10 | |||||
Blue played 44C_, White rolls 53 | |||||
White now completes his last two "crossovers" with 18/13 9/6.
Looking parallel one final time, in the outfield White might have had two blots in her own outer board in various ways, or one in each outer board in various ways (this being one example), or just one checker on her 15pt, 14pt or 13pt. (These are all two-crossover situations.) My parallellism notes demonstrate that there are a large number of specific solutions (even with the two-plus-one non-doublet restriction I spontaneously attached), but they all rise out of one theme.
Anyway, in this universe, White played 53c (18/13 9/6), reaching the target position (shown at the beginning of this post and repeated below).
|
| ||||
| XGID=---BaBC-B---dB---A-jB---A-:0:0:1:00:0:0:3:0:10 | |||||
41$-22O.62&-65C-55M_43Z-44C_53c.....Final Position | |||||
As before, reach the above position legally in just eight rolls. (Blue rolls first.) However, all of White's rolls are doublets and in ascending order of pips!
When we know that the roll possibilities for a player are restricted (e.g., White's rolls get bigger each turn), it can often make our job easier. On the other hand, if the puzzle has a limited number of solutions (or only one) and there are a lot of roll portions (as there are with doublets), our job is definitely harder on balance. A further testament to this is that (as it turns out) we must give Blue a third doublet of his own to make it all work. (We can't give Blue a fourth doublet because he is rolling first, and the rules of backgammon require the game start with a non-doublet.) There is only one workable theme.
First, some logic: (a) Obviously, White's last roll cannot be 66 to reach this position. Yet, we know that she rolled all doublets and the number of pips increases with each roll. Therefore, she rolled four out of five rolls in the order 11 22 33 44 55 (i.e., with one of those excluded), which also means the number of total pips rolled is 40, 44, 48, 52 or 56. Yet (b) We know White gains only (167 - 133 = ) 34 pips to reach the target position. She must be hit at some point(s).
Can White's roll set be 11 33 44 55? If it were, the only way to get all three checkers off her 8pt would be to play 8/6(2) with her 11, then have Blue hit the third, which is on his 17pt. As White must be sent back exactly 52 - 34 = 18 pips with this roll set, Blue must also hit her on his 1pt (because 17 + 1 = 18), but White already spent her aces with which she could hit back from the roof and Blue has no checker on his 1pt in the final position. Therefore, White must use at least one deuce to play 8/6 -- her roll set canNOT be 11 33 44 55.
So, White must get sent back (not 18, but) either 6, 10, 14 or 22 pips -- by getting hit one or more times during the sequence. Clearly, Blue lacks enough roll portions to break and remake his 6pt; therefore, if White is hit once it will be on Blue's 10pt, 14pt or 22pt. Keeping in mind that roll portions are precious for Blue, if White is hit twice (for 6, 10, 14 or 22 pips), it is more likely than not to happen on points that Blue ends up occupying, so we should be aware of hit combinations on the 3pt + 3pt, 5pt + 5pt, 17pt + 5pt, or at least one of the two points being end-occupied -- e.g., 9pt + 5pt, 11pt + 3pt. (Not 20pt + 2pt, because white can never get to Blue's 2pt if her first roll and smallest doublet is 22.)
What can we conclude about Blue? Well, he has a lot of work to do -- he has to play a lot of roll portions to reach his final position. It seems clear he must roll at least one 55 and probably a second 55, in order to have two of his checkers hit and make the 20pt, strip his midpoint and make the 3pt. It is useful to take that into account in deciding where White will most likely be hit (as discussed in the previous paragraph).
This gives you an idea of the sort of deductions and observations that can be worthwhile making for the most challenging problems. They can help you eliminate impossibilities -- and focus on the most likely possibilities that remain -- as you try to retrofit a sequence. With all that said, I'll fast forward to the solution.
Blue must start with a 51 or 42. If the roll is 51, the 5 must be 13/8, though the ace can be 24/23, 8/7 or 6/5. If the roll is 42, the play must be 13/7. I assigned Blue 51S (as shown below), as I may as well let him play the best move. :)
  |
| ||||
| XGID=-b----E-D---eD---c-e---AA-:0:0:-1:11:0:0:3:0:10 | |||||
Blue played 51S (24/23 13/8), White rolls 11 | |||||
White rolls 11. As the checkers on her midpoint are blocked (else 13/11, along with 8/6 or 24/22 or even 11/9, could dovetail with the solution), White must play 8/6 with half her roll. The other half can be played either 8/6 or 24/22. I chose the latter, and added it to the board below.
   |
| ||||
| XGID=-a-a--E-D---eD---b-f---AA-:0:0:1:11:0:0:3:0:10 | |||||
White played 11C (24/22 8/6), Blue rolls 11 | |||||
Blue rolls a 11 of his own. With the 23/22 8/7 6/5(2) play, he sets up everything perfectly for his brilliant move next turn.
|
| ||||
| XGID=-a-a-BCAC---eD---b-f--A-A-:0:0:-1:22:0:0:3:0:10 | |||||
Blue played 11Y_ (23/22 8/7 6/5(2)), White rolls 22 | |||||
White rolls 22. This is her last chance to play 8/6, so that takes care of two deuces. Her back checkers are right where she wants them, so she plays 13/9 with her other two deuces, which gives her a useful 3 to play with her next roll. White's move is reflected in the diagram below.
|
| ||||
| XGID=-a-a-BCAC---dD--a--h--A-A-:0:0:1:55:0:0:3:0:10 | |||||
White played 22o (13/9 8/6(2)), Blue rolls 55 | |||||
Blue rolls 55. He hits loose (rather than pointing on White) because he needs two checkers sent back. His move 55G_ (22/17 13/8(2) 8/3*) is shown below.
  |
| ||||
| XGID=aa-A-BCAD---dB--aA-h----A-:0:0:-1:33:0:0:3:0:10 | |||||
Blue played 55G_ (22/17 13/8(2) 8/3*), White rolls 33 | |||||
White rolls 33, and can hardly make her play fast enough. She hits two checkers and stacks the last 3 with 9/6. Whew -- another checker safe! The after-position appears below.
 |
| ||||
| XGID=---a-BCaD---dB---A-i----AB:0:0:1:55:0:0:3:0:10 | |||||
White played 33C. (bar/22* 24/18* 9/6), Blue rolls 55 | |||||
Blue rolls another 55, executing the second of two hits on his 3pt. He is happy covering, as there is no need to get hit back this time. Besides, it is the last move he will play, and he has reached his final position. In short, Blue's move is 55C (bar/20(2) 8/3*(2)) -- shown below, the second of two backgammon-best plays he made.
 |
| ||||
| XGID=a--B-BCaB---dB---A-iB---A-:0:0:-1:44:0:0:3:0:10 | |||||
Blue played 55C (bar/20(2) 8/3*(2)), White rolls 44 | |||||
White rolls 44 and couldn't be more delighted. In unlikely fashion, she is able to super-stack yet another checker. Her move is 44F_ (bar/21 18/6), diagrammed below.
|
| ||||
| XGID=---BaBC-B---dB---A-jB---A-:0:0:1:00:0:0:3:0:10 | |||||
51S-11C-11Y_22o-55G_33C.55C-44F_.....Final Position | |||||
White has tied Igor's record for having ten checkers on the 6pt even before the middle game is in full swing.
Nack