The puzzle was originally posed here. Succinctly:
- From the starting position of backgammon, what is the fewest number of moves that must transpire for both players to vacate their originally occupied points and end up with a stack of five checkers on each of three new points?
Here is my solution:
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|---|---|
| 61: Blue plays 8/7 8/2 | |
| 11: White plays 8/7(2) 6/5(2) | |
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| 22: Blue plays 13/11(2) 6/4(2) | |
| 11: White plays 24/23* 8/7 6/5(2) | |
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| 22: Blue plays Bar/23 13/11 6/4(2) | |
| 66: White plays 24/18* 23/17* 13/7(2) | |
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| 22: Blue plays Bar/23(2) 13/11(2) | |
| 11: White plays 13/12(3) 6/5 | |
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| 11: Blue plays 24/23(2) 6/4 | |
| 65: White plays 18/12 17/12 | |
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| Final position | |
61: 8/7 8/2... 11: 8/7(2) 6/5(2)
22: 13/11(2) 6/4(2)... 11: 24/23* 8/7 6/5(2)
22: Bar/23 13/11 6/4(2)... 66: 24/18* 23/17* 13/7(2)
22: Bar/23(2) 13/11(2)... 11: 13/12(3) 6/5...
11: 24/23(2) 6/4... 65: 18/12 17/12
I thought my solution to be unique, but I now see my misstep in logic. Stein found a different solution.
While it is not feasible to mesh my Blue play with Stein's White play (both sides can't end up occupying White's 2pt), it IS possible to mesh my White play with his Blue play, producing a third solution:
31: 24/21 8/7... 11: 8/7 6/5(3)
33: 8/2 6/3(2)... 11: 24/23* 8/7 6/5(2)
44: Bar/21 13/9(3)... 66: 24/18* 23/17* 13/7(2)
44: Bar/21(2) 13/9(2)... 65: 18/12 17/12
33: 24/21 6/3(3)... 11: 13/12(3) 8/7...
(Note that I transposed the rolls and subplays a bit this time for variety.)
If/when I republish this problem, I may add a condition to the original phrasing of the problem so that one of these solutions is defined as "best."
Nack