Extreme problems

[Bill Daly]

On occasion, when someone publishes a chess or bridge problem, it turns
out to be impossible, in the sense that the problem position cannot
possibly be created by any legal sequence of plays. This would appear to
be less of an issue with backgammon, since there are fewer constraints
on the kinds of positions that can be created. Nevertheless, there are
certainly positions which cannot legally arise, for example all 30
checkers on the bar. I have been ruminating about this somewhat
inconclusively, but my focus is currently on two questions. 1) Assuming
that both players cooperate and are free to choose the rolls of the
dice, how can they play so as to put all 15 of X's checkers on the bar?
Can they do this in such a way that O also has at least one checker on
the bar? And 2) what is the maximum number of checkers (on both sides)
which can be on the bar at the same time?

Perhaps these questions would be better suited to one of the puzzles
newsgroups, but I'm not sure that backgammon is well enough understood
elsewhere to make this interesting.

Regards,

Bill

[Kevin Cline]

Bill Daly <bill...@mail.interport.net> wrote:

>1) Assuming
>that both players cooperate and are free to choose the rolls of the
>dice, how can they play so as to put all 15 of X's checkers on the bar?

It is just possible for O to put all 15 of X's checkers on the bar while O's
checkers return to their original position.

[Tommy K.]

Bill Daly wrote:

"On occasion, when someone publishes a chess or bridge problem, it turns
out to be impossible, in the sense that the problem position cannot
possibly be created by any legal sequence of plays. This would appear to
be less of an issue with backgammon, since there are fewer constraints
on the kinds of positions that can be created. Nevertheless, there are
certainly positions which cannot legally arise, for example all 30
checkers on the bar. I have been ruminating about this somewhat

inconclusively, but my focus is currently on two questions. 1) Assuming

that both players cooperate and are free to choose the rolls of the
dice, how can they play so as to put all 15 of X's checkers on the bar?

Can they do this in such a way that O also has at least one checker on
the bar? And 2) what is the maximum number of checkers (on both sides)
which can be on the bar at the same time?

Perhaps these questions would be better suited to one of the puzzles
newsgroups, but I'm not sure that backgammon is well enough understood
elsewhere to make this interesting.

Regards,

Bill"

Tommy K. Replies:

It is a simple matter for a player [X] to spread out his 15 checkers on
different points and get them all hit. The opposing player [O] cannot
simultaneously have a checker on the bar, however. Once [X] has 1-4 men on
the bar (a necessary transitional stage) [O] can only hit at most 6 more
men before getting all of his men in to hit [X]'s checkers outside the home
board. If [X] hits a man when most of his guys are already up, [O] will
need to get all of his guys in in order to his the entering [X] checker
back, so it is impossible for one player to have 15 men up and the other to
have at least one up.

As for the total number of checkers that can be on the bar, that is also
15, by similar arguments. Once [X] has even one man on the bar, he can
only hit [O] by bringing a checker in at the same time, so the total number
of checkers on the bar doesn't increase.

The most checkers that the lesser of the two players can have up is 6,
which can occur with 7 opposing checkers up. This can occur by [X] leaving
13 blots exposed, with one innerboard point closed. [O] then picks up the
13 blots, and slots his own inner board. The final slot consists of
breaking [O]'s last inner board point and leaving one blot on each point.
[X] then rolls to hit all six blots in the next three rolls, bringing in 6
guys of his own. [O] meanwhile can bounce because [X] had one inner board
point closed. After the third roll [X] has 7 men on the bar, and [O] has
6.

Tom Keisler