BGonline.org Forums

Posted By: Nack Ballard
Date: Friday, 13 June 2014, at 9:59 p.m.

Actually, having conditions or stipulations would make the problem even more challenging. For example, the checkers on the five point have to be the back checkers. Also, either of the back checkers cannot land on a point already occupied by its own checker(s). That is, each of the back checkers must clearly maintain its identity throughout the journey to the five point.

I have a feeling that Tim will find your added set of stipulations (to Bob's problem) less than elegant, but let's put that aside for now. I have a different quandary: I'm not sure exactly what stipulations you intend (though I agree in any case that they make the problem more challenging).

(1) By the "back checkers," do you mean the original back checkers? (Other checkers that are sent back become back checkers.)

(2) If that player enters with some other checker onto the 24pt while one or both of his original back checkers have not yet moved, can he play a roll of 21 as bar/24/22 and claim that the 24/22 part is with an original back checker and not with the checker that entered?

Assuming the least generous interpretation (original back checkers only reach the 5pt, and disallowing that player a 24pt entrance onto an original back checker, in addition to the other stated restrictions), here is an eight-roll sequence that works:

........63\$-33R-32K-11U-65r-64H-51R-55P

Key: R (Run 2011) = 24/21(2) 8/5 6/3, and "r" (run) = 24/18 20/15.

[If doublets are allowed on the opening roll, it can be done in seven rolls.]

Nack

Post Response

Subject:
Message: