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No Pip Wastage

Posted By: Nack Ballard
Date: Saturday, 2 February 2019, at 7:27 p.m.

Typically, when I don't get a response to a post, I figure there is not enough interest or I've overreached in some way, and I don't bother following up with a rollout or answer. However, in this case I feel it's a shame to let such a gem die -- I'm inspired to try again.

		White is Player 2 score: 0 pip: 167 Unlimited Game Jacoby Beaver pip: 167 score: 0 Blue is Player 1
XGID=-b----E-C---eE---c-e----B-:0:0:1:00:0:0:3:0:10
Construct a no-pip-wastage game

Using standard backgammon rules, construct a legally played game such that

.....(1) One player bears off his last checker(s) when the other player has 13 or 14 checkers off.
.....(2) Neither side wastes any pips!

Can you do it?

Tie-breaker: Fewest number of rolls.

(For reference, in the original post, I listed a second tie-breaker.)

In truth, this puzzle is quite easy (and not particularly intriguing) without considering the tie-breaker. It's mainly a matter of giving one player exactly 167 pips (or more if we choose to have him get hit), bear his checkers around the board, and bear them off without wasting pips. (That is, in the bear-off stage, he takes only 6s off the 6pt, only 5s off the 5pt, etc. And meanwhile let the other player catch up (or slow down) so that s/he gets 13 checkers off before losing. You can construct that game with any rolls you want, whatever works. For example...

		White is Player 2 score: 0 pip: 140 Unlimited Game Jacoby Beaver pip: 123 score: 0 Blue is Player 1
XGID=------EBEa--dC--bc-e------:0:0:-1:00:0:0:3:0:10
White to play

To reach the above position, White ran with opening 65 and 44, while Black ran with 66 and 55. These plays happen to be the best plays on each side, but that is not a requirement. The moves merely need to be legal. Also, these rolls happen to be large, but that is not a requirement either.

Next, suppose we give White three double 6s and Blue two double 6s. That brings us to here, with Blue on roll:

		White is Player 2 score: 0 pip: 68 Unlimited Game Jacoby Beaver pip: 75 score: 0 Blue is Player 1
XGID=-BC---ECB--------abe-abbb-:0:0:1:00:0:0:3:0:10
Blue to play

Next, assume Blue rolls 44, 22 and 66, while White (in alternating turns) rolls 33, 66 and 41:

		White is Player 2 score: 0 pip: 27 Unlimited Game Jacoby Beaver pip: 27 score: 0 Blue is Player 1
XGID=-B-BBAA------------aababa-:0:0:1:00:0:0:3:0:10
Blue to play

Keep in mind that we can't just give Blue 66 or 55 here, as it would waste pips. For anyone who would care to visualize it, any other number is still okay, even 44 (if followed by, say 21, 31 and 31, or if White bears off first it doesn't matter if Blue gets stuck with more than one checker on the 1pt).

Let's give both players 22 and 33, which gets us to...

		White is Player 2 score: 0 pip: 7 Unlimited Game Jacoby Beaver pip: 7 score: 0 Blue is Player 1
XGID=-B---A---------------a-aa-:0:0:1:00:0:0:3:0:10
Blue to play

We must give Blue 31, 21 or 11 here. Anything larger will force him to waste pip(s) on the next roll. In response, we can assign White any non-doublet without a 6 or 5 (or 43), after which she will have (13 or) 14 checkers off at the moment Blue bears off his last checker(s), thereby fulfilling the conditions of the puzzle.

Note that if one player is getting too far ahead, we just slow him down until the other player catches up (so that he can bear off 13-14 checkers). There are countless ways to accomplish such a no-pip wastage game. (We can take a hundred rolls if we want, and the plays can be good or bad; doesn't matter.)

In this sample game, White and Blue rolled 12 times each, for a total of 24 rolls.

This brings us to the "tie-breaker." What is the fewest number of rolls such a game can last? (Hint: It is fewer than 24.)

		White is Player 2 score: 0 pip: 0 Unlimited Game Jacoby Beaver pip: 167 score: 0 Blue is Player 1
XGID=------E-C----E----------B-:0:0:1:00:0:0:3:0:10
Opening position for Blue's checkers only

Let's start with one side. Above is the opening position of Blue's checkers only. He must roll exactly 167 pips in order to bear off without any wastage. As 167 is not divisible by 4, at least one of his rolls must be a non-doublet.

Now, considering the crossovers, which I'll let you count for yourself, what is the minimum number of turns that it might take for Blue to bear off all his checkers?

Anyone?

Nack

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