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Fifteen checkers on the bar, fewest moves -- solutions

Posted By: Nack Ballard
Date: Saturday, 12 June 2010, at 6:42 a.m.

The fifteen-checker puzzle (in two parts) was originally posted here.

From the starting position of backgammon, what is the fewest number of moves required to get 15 checkers on the bar?

Extra credit: What is the fewest number of moves required to get 15 checkers on the bar and make a six-point prime?

For the first part of the puzzle, Michael Nielsen found a 7-move solution. Bravo, Michael!

As a supplement, see Keene's interesting analysis of the puzzle and my proof that a solution with fewer than 7 moves is impossible, here.

For the "extra credit" part of the puzzle, hitting 15 checkers and making a six-point prime, Michael was able to find an 8-mover, while no other respondent found a solution with fewer than 9 moves.

Not counting transpositions and similar sequences, I found two 7-move solutions. Here is the first:

21: White will play 13/11 8/7

44: Blue will play 24/16 13/9(2)

33: White will play 24/21 13/10 6/3(2)

51: Blue will play 16/11 8/7

44: White will play 24/20 13/9 8/4 6/2

33: Blue will play 13/10(3) 10/7

11: White will play 13/12 6/5 3/1*

11: Blue will play Bar/24*/23*/22*/21*

From here, it's easy to see what happens. Blue rolls this set of double 1s and two more, and employs the hoover maneuver to pick up all twelve blots on the far side of the board, while White fans. Finally, Blue rolls a fourth 11 and plays 13/12*/11 6/5*/4. The position after Blue plays his seventh (final) move is:

White has fifteen checkers on the bar

A summary list of the 7 moves is:

(1) 21: 13/11 8/7... 44: 24/16 13/9(2)
(2) 33: 24/21 13/10 6/3(2)... 51: 16/11 8/7
(3) 44: 24/20 13/9 8/4 6/2... 33: 13/10(3) 10/7
(4) 11: 13/12 6/5 3/1*... 11: Bar/24*/23*/22*/21*
(5) 66: Fan... 11: 21/20*/19*/18*/17*
(6) 66: Fan... 11: 17/16*/15*/14*/13*
(7) 66: Fan... 11: 13/12*/11 6/5*/4*

None of the rolls in the above solution may be transposed or changed.

The second solution makes a different prime, and the player with 15 checkers on the bar gets hit 16 times. It looks like this:

21: White will play 13/11 8/7

11: Blue will play 8/7 6/5(3)

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

44: Blue will play 24/16* 13/9(2)

33: White will play Bar/22 24/21 13/10 6/3

33: Blue will play 13/10(3) 10/7

11: White will play 24/23 13/12 6/5 2/1*

11: Blue will play Bar/24*/23*/22*/21*

From here, Blue rolls double 1s four times, taking three moves to hit the outside twelve blots and one move to hit the three inside blots. He has an ace left over (to be optionally squandered with 17/16 or 12/11 or 2/1).

In list form, the 7 moves are:

21: 13/11 8/7... 11: 8/7 6/5(3)
44: 13/9 8/4 6/2(2)... 44: 24/16* 13/9(2)
33: Bar/22 24/21 13/10 6/3... 33: 13/10(3) 10/7
11: 24/23 13/12 6/5 2/1*... Bar/24*/23*/22*/21*
(5) 65: Fan... 11: 21/20*/19*/18*/17*
(6) 65: Fan... 11: 16/15*/14*/13*/12*
(7) 55: Fan... 11: 5/4*/3*/2* and a fourth ace

In this solution, the second player's first and third rolls (11 and 33) can be transposed. No other transpositions or changes of rolls work. (There are, of course, many orders in which the aces can be played in the last four rolls, in both solutions.)

Nack