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| XGID=--B-BBB------C---eee--AAB-:0:0:1:00:0:0:3:0:10 | |||||
Problem 1: ..Reach this position in seven rolls | |||||
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| XGID=--B-BBB------C---eee-AA-B-:0:0:1:00:0:0:3:0:10 | |||||
Problem 2: ..Reach this position in seven rolls | |||||
Treat these as separate problems. Hint: The player who starts in one position is different from the player who starts in the other position.
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Above is the pair of seven-roll retro-problems originally posted here.
We know that in one of the problems White must finish in three rolls. Her pipcount has been reduced by 62 pips (from a starting 167 to 105), and if she rolls her biggest non-doublet of 65, not even two double 6s suffice. Therefore, she must roll three doublets totalling at least 64 pips (the next highest multiple of 4).
It doesn't take long to see that there is no way that White can reach her final position with 66 66 66 or 66 66 44 or 66 55 55 no matter where she is hit. That means that she rolled 66 66 55 (total 68) and lost 6 pips along the way by being hit (or by being blocked, but that can be nixed pretty easily with a little investigation).
So, we have deduced that White lost 6 pips by being hit. There is only one point on which that can happen: Blue's 6pt! For this to happen, of course, Blue must break his 6pt, hit White on his 6pt, and later remake his 6pt.
Given White's known roll combination of 66 66 55, Blue's two recycled checkers can only be hit on his 6pt. (Not his 5pt, because White cannot play 24/19, get hit, then play bar/20* on separate rolls; and not his 11pt or 12pt because Blue's two checkers cannot be hit at the same time -- he needs to enter each checker separately with doublets in order to use enough portions/submoves to reach his forward position.)
Blue needs to move four checkers off his 6pt, two checkers off his midpoint, five checkers from outer board to inner board, and enter two checkers from the roof. That's 13 "portions" (submoves). As a non-doublet and three doublets add up to 14 portions, he can afford to spend only one non-crossover portion.
Looking at the problem 2 position, Blue's recycled checkers would have to have entered with 44 and 33, or 33 and 22, and the (only) other doublet would have had to have occurred before he was hit on the 6pt. With these restrictions, it does not take long to see that a late double 4s is too awkward, and that squandering Blue's last portion on 23/21 does not leave enough flexibility to rebuild the 6pt.
As problem 2 does not work for White achieving her position in only three rolls, it must be problem 1 in which Blue goes first. Moreover, the information accumulated just prior to the previous paragraph works for problem 1. Continuing to apply logic will make it possible to solve thoroughly, but trial and error should be manageable enough (and more fun) from here that you can find a solution. Here, again is problem 1:
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| XGID=--B-BBB------C---eee-AA-B-:0:0:1:00:0:0:3:0:10 | |||||
Problem 2: ..Reach this position in seven rolls | |||||
Feel free to try to solve this problem with the clues you have been given. Otherwise, read on.
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My solution is illustrated in a series of diagrams below, with captions. If you widen your window (or use Ctrl- to zoom out), you will see the diagrams side by side, with Blue's plays on the left and White's plays on the right. Otherwise, just follow the progression in one vertical column.
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| XGID=-b--AAC-C---eE---c-e----B-:0:0:1:00:0:0:3:0:10 | |||||
Blue rolled 21 and played 6/5 6/4 | |||||
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| XGID=-b--AAC-C---aE---cde----B-:0:0:1:00:0:0:3:0:10 | |||||
White rolled 66 and played 13/7(4) | |||||
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| XGID=-bB-BAA-BA--aD---cde----B-:0:0:1:00:0:0:3:0:10 | |||||
Blue rolled 44 and played 13/9 8/4 6/2(2) | |||||
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| XGID=--B-BAa-BA-a-D---dde----BA:0:0:1:00:0:0:3:0:10 | |||||
White rolled 55 and played 24/19*/14 24/19 13/8 | |||||
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| XGID=a-B-BBA-A-Aa-C---dde--A-B-:0:0:1:00:0:0:3:0:10 | |||||
Blue rolled 33 and played bar/22 13/10 9/6* 8/5 | |||||
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| XGID=--B-BB--A-A--C---eee--A-BA:0:0:1:00:0:0:3:0:10 | |||||
White rolled 66 and played bar/19*/7 14/8 | |||||
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| XGID=--B-BBB------C---eee--AAB-:0:0:1:00:0:0:3:0:10 | |||||
Blue rolled 22 and played bar/23 10/6 8/6 | |||||
Unfortunately, I was too lazy (mostly) to spend the necessary time proofing for uniqueness. I naively thought something like, "This fits together so neatly, how could there be another solution?"
Then, when I saw Stein's solution, I realized that I had overlooked two aspects. One was simple. For White's first turn, she can play one of her 6s out to the 18pt (which she follows by playing a slightly different 55 and 66, to reconcile).
The other aspect is a tricky transposition for Blue, with one of the four 4s being played differently, and then reversing my order of 33 and 22. This (Stein's) solution, which is equally clever, is listed below:
........B: 21: 6/5 6/4....................W: 66: 24/18 13/7(3)
........B: 44: 13/9(2) 6/2(2).........W: 55: 24/19* 18/8 13/8
........B: 22: b/23 8/6* 8/4..........W: 66: b/19*/7 13/7
........B: 33: b/22 8/5 9/6(2)
Hence, there are technically four solutions to problem 1. Mine, Stein's, and each of ours with the other's initial fourth 6 (kicking off a slightly different White sequence). Nactation is designed more for practical backgammon play than for puzzle solutions, but for compactness I'll list all four sequences below.
........21&-66D-44i-55s-33M-66B-22P
........21&-66Z-44i-55b-33M-66B-22P
........21&-66D-44N-55s-22M-66B-33C.
........21&-66Z-44N-55b-22M-66B-33C.
I'll post my solution to problem 2 later -- the one where White goes first. (There's still time if you want to work on it.)
Nack