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| Blue on roll | |||||
Objective: In the above DMP game, in spite of both sides playing perfectly (arbited by XGR++), Blue picks up a second checker and wins the game.
(a) How can Blue accomplish this?
(b) What is the fewest number of total rolls required for Blue to accomplish this?
Congratulations to Stick for demonstrating how a second checker can be picked up with perfect play. (He omitted Blue's subsequent bear-in and bear-off, presumably because he considered it a trivial matter of accounting.)
Mike Mannon's submission technically doesn't qualify as a solution, but if we change his opening 61D to the correctly played 13/7 2/1*, I'm sure he would get there, as he otherwise derived the same brute-force method that Stick did for picking up the second checker.
It is not Blue's objective to pick up a second checker, and it is not White's objective to either cooperate or run interference. Blue and White only play the best moves (according to XGR++ eval). Rather, given that constraint, it is the puzzle-solver's objective to create a roll sequence that causes Blue to pick up a second checker (and win) -- and ideally in the fewest rolls possible.
There are three main ways to get the second checker hit:
(1) Blue leaves at least THREE blots that White hits on her way around the board. Blue anchors on the 20pt, 21pt, 22pt or 23pt, and keeps one checker (or more) on the bar. White rolls a number with an ace such that she lands just in front of the anchor and bears the ace off (forced), thereby reducing the number of checkers on her 1pt from 14 to 13. Her next three rolls are 11, which leaves White two blots (one remaining on her 1pt and the other stuck in front of Blue's anchor). Blue comes off the bar and hits both blots (not necessarily on the same roll).
(2) White hits only TWO blots on her way home. Blue anchors on the 20pt, 21pt, 22pt or 23pt (no checker on the bar). As above, White lands just in front of the anchor and bears an ace off. Three double 1s then bear off 12 checkers, leaving a blot on her 1pt. Now Blue hits the other blot, White hits him on the way back around, then Blue enters and hits both blots.
(3) White hits only ONE blot on her way home. Blue stays on the bar or at some point might enter on the 23pt. White rolls a number that comes into the 6pt and bears an ace off (because it is her best play -- it isn't forced as in #1 or #2 above). She follows this with the thematic series of double 1s, though it has to be her best play all three times to bear off four checkers. Blue hits the non-1pt blot, White hits on the way back around, and Blue is able to hit the 1pt checker (and the other one).
Method 1 (that Stick and Mike used) is the most robust and easiest to implement. Blue's anchor freezes one blot while his checker on the bar waits to pounce on the second blot that poor White will expose on her 1pt. Let's examine Stick's sequence: "63H-61K-41-65H-41-21-F-11-F-11-F-11-41..." (Blue's first two 41's each entered one checker.) If you can follow this from the original position above, you will arrive at the position below with Blue to play 41.
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| Blue on roll | |||||
If Blue is assigned a roll of double 1s (instead of 41), he gains a crossover, but it turns out to make no difference. If White fans for the rest of the game and Blue rolls all double 6s and White fans for the rest of the game, only 6 + 6 rolls remain (either way), for a total of 25 rolls.
The advantage of Method 2 (where only two checkers are hit) is that Blue can advance his board, allowing him to save two rolls (one for Blue and one for White) in his bearoff. [Granted, he has to get re-hit to pick up White's 24pt blot, but that only costs two rolls, and he gets those two rolls back by virtue of having only two (instead of three) checkers to bring around the board.]
We can adapt Stick's Method 3 sequence to Method 2, starting with his 63H-61K-41, and now (instead of 65H) White plays 549 and Blue 64Z. White rolls 41, bearing in the 4 and taking off the ace. Now, instead of fanning with a third checker (while White bears off 12 checkers), Blue's three rolls (e.g., 44 33 22) can be used to obtain a speedboard, as diagrammed below.
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| Blue on roll | |||||
From here Blue hits say with 51, White hits back with 55, Blue hits both checkers with 41. White fans while Blue rolls all 66's and claims gin with four checkers on the 1pt (before White's next roll). This sequence requires only 23 rolls.
For my solution, I use Method 3, which saves rolls by bearing off several checkers before even being hit. I start with 55D (13/8(2) 8/3 6/1*) and two double 6s, then White enters/hits with 44, which leads to the position shown below.
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| Blue on roll | |||||
Here, Blue fans, White rolls 11 (or 31), taking one checker off, Blue fans again, White bears off four checkers with 11, Blue plays 21 as bar/23 2/1, White bears off four checkers with 11, and Blue plays 21 as 3/1 2/1, White bears off four checkers with 11 one last time, Blue hits with double 2s (only way to get hit back immediately), White hits with 55, and Blue hits twice with 41, reaching the diagram below.
[There are many traps. For example, if Blue instead starts with three double 6s, his inner board consists of only six checkers (two each on his 3pt, 2pt and 1pt). If he fans a third time, White's 11 is played with 6/5 (according to XGR++), disrupting the plan. Or if he enters with 21, it is played bar/22 (instead of the board-breaking bar/23 3/2) and he has no way of staying for another roll. The seventh inside checker justifies his staying with 21 twice, which is why I begin the sequence with a less "powerful" 55.]
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| Blue on roll, cube action? | |||||
Finally, the coup de grace, which saves a roll. This is a little puzzle in and of itself: end the game in 3 rolls (Blue-White-Blue). Hint: Double 6s is too strong. Answer: Instead, double 5s, White fans (or enters one), Blue rolls double 2s (!) and it's gin without another White roll.
Recapping the shortest solution (21 rolls): 55D-F-66-F-66-44-F-11-F-11-21P-11-21i-11-22R-55-41K-F-55-F-22 (gin).
Nack