Forum Archive : Puzzles

Priming puzzle

From:   Gregg Cattanach
Address:   gcattanach@prodigy.net
Date:   9 May 2005
Subject:   Priming puzzle
Forum:   GammOnLine

From the opening position, with X moving three times consecutively,
form a full 6 prime in front of O's anchor. (O doesn't move, doubles
are OK on the first roll.)

In other words, reach this position after three rolls:

     24  23  22  21  20  19      18  17  16  15  14  13
    | X                   O |   |     O               X |
    | X                   O |   |     O                 |
    |                     O |   |     O                 |
    |                     O |   |                       |
    |                     O |   |                       |
    |                       |   |                       |
    |                       |   |                     O |
    |                       |   |                     O |
    |                       |   |                     O |
    | O   X   X   X   X   X |   | X                   O |
    | O   X   X   X   X   X |   | X                   O |
      1   2   3   4   5   6       7   8   9  10  11  12

How do you do it?

Řystein Johansen  writes:

This is the Bill Davis "Great Prime" puzzle. It was published in an
online backgammon magazine in December 1999.

How to start the thinking progress. Trying and failing won't take you
anywhere. The first thing to think about is: How many pips do you need?
Which dice combinations of 3 rolls can give me this number?


TarHeelFan  writes:

There are two "tricks" to solving this sort of problem. The first is to
determine which dice rolls could possibly get you to the position. For
this one, we need 52 pips in 3 rolls. It's very unlikely that using the
same set of doubles twice will be usefull, so I'll ignore sets like 55,
55, 33 and 55, 44, 44. That leaves 66, 55, 22 and 66, 44, 33 as the only
combos I see that give the correct number of pips. Since 5's don't allow
us to diversify our checkers, we'll start with 66, 44, 33 as the most
likely rolls.

The second trick is to forget about how you would normally play
checkers. (I learned this trick from a similar chess problem where you
had to put your queen in a position to be captured to solve it)


  - We have to more 3 checkers from the 6, and we have to use 2 rolls to
    do it. That means either 2 fours and a three or 2 threes and a four.

  - We have the same situalion with the 3 checkers on the 8 point (we
    can't play 8/5(2), 8/2 with 33 since that wouldn't leave a 3 for the
    6 point checkers)

  - Obviously, any solution would include 13/7(2), but there is no
    solution involving 8/2(2), since that would leave no 4 to play from
    the 6 point

So far, we have 13/7(2), 8/5, 8/4, 6/3, 6/2 as our "forced" plays,
leaving 2 6's, 2 3's and 2 4's:

     24  23  22  21  20  19      18  17  16  15  14  13
    | X                   O |   |     O               X |
    | X                   O |   |     O               X |
    |                     O |   |     O               X |
    |                     O |   |                       |
    |                     O |   |                       |
    |                       |   |                       |
    |                       |   |                     O |
    |                       |   |                     O |
    |                     X |   |                     O |
    | O                   X |   | X                   O |
    | O   X   X   X   X   X |   | X   X               O |
      1   2   3   4   5   6       7   8   9  10  11  12

  - We still have 2 checkers on the mid, which cannot play 6's. We can
    quickly see that playing both of these checkers with the same number
    does not work, so we need to play 13/10, 13/9

  - The remaining plays are trivial: 10/4, 9/3, 8/5, 6/2

So, the plays are (the order of the 44 and 33 doesn't matter):

44: 13/9, 8/4, 6/2(2)
33: 13/10, 8/5(2), 6/3
66: 13/7(2), 10/4, 9/3

I haven't looked for a solution using 66, 55, 22 or repeating rolls, but
I find it highly unlikely that such a solution exists.

Gregg Cattanach  writes:

It appears there are at least two different ways to do this, considering
transpositions as not being a separate solution:

Here's the way I've always done it.

66: 13/7(3), 8/2
44: 13/5, 8/4, 6/2
33: 8/5, 7/4, 6/3(2)

Cool! I didn't know there was a 2nd way to do it!


TarHeelFan  writes:

This is something of a transposition, but:

33: 13/10, 8/5(2), 6/3
66: 13/7(3), 10/4
44: 8/4, 7/3, 6/2(2)

My error was when I said "We still have 2 checkers on the mid, which
cannot play 6's..." I had no logical backing for that, and the solution
branches at this point.
Did you find the information in this article useful?          

Do you have any comments you'd like to add?     



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