Six vs six on the bar, fewest moves -- Solution

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Six vs six on the bar, fewest moves -- Solution

Posted By: Nack Ballard
Date: Friday, 18 June 2010, at 6:56 p.m.

This question was originally posed here:
From the starting position of backgammon, what is the fewest number of moves required to reach a position in which the two players each have six checkers (12 total) on the bar?
The shortest solution is 7.5 moves and is accomplished as follows:

32: Blue plays 13/11 13/10

33: White plays 13/10 8/5 6/3(2)

33: Blue plays 13/7 8/5 6/3

44: White plays 13/9(2) 8/4 6/2

44: Blue plays 24/20* 13/9 8/4 6/2

22: White plays Bar/23*/21* 24/22* 3/1*

Blue rolls 66, fanning. And then...
44: White plays 24/20*/16 22/18* 21/17*

Blue rolls 66, fanning again. And then...
11: White plays 16/15*/14* 13/12* 6/5*

(Blue has 8 checkers on the bar)

(Blue has 12 checkers on the bar)

21: Blue plays Bar/24* Bar/23*

White fans. And then...
(Blue has 10 checkers on the bar)

43: Blue plays Bar/22* Bar/21*

White fans again. And then...
(Blue has 8 checkers on the bar)

65: Blue plays Bar/20* Bar/19*

(Blue and White each have 6 checkers on the bar)

Final position of sequence

Below is the (7.5 move) solution in list form (Blue...White):
(1) 32: 13/11 13/10... 33: 13/10 8/5 6/3(2)
(2) 33: 13/7 8/5 6/3... 44: 13/9(2) 8/4 6/2
(3) 44: 24/20* 13/9 8/4 6/2... 22: Bar/23*/21* 24/22* 3/1*
(4) 66: Fans... 44: 24/20*/16 22/18* 21/17*
(5) 66: Fans... 11: 16/15*/14* 13/12* 6/5*
(6) 21: Bar/24* Bar/23*... 66: Fans
(7) 43: Bar/22* Bar/21*... 66: Fans
(8) 65: Bar/20* Bar/19*...
White's 33 and 44 can be transposed, and Blue's last three rolls can (of course) be any permutation of the 6 numbers. With those caveats, I believe the solution is unique.
Nack