                      |
| ||||
| XGID=-b--CBCBD----A-a--bbbbbb--:0:0:1:00:0:0:3:0:10 | |||||
Final Position | |||||
I looked into a number of hitting possibilities and am pretty sure that they don't help, but the complications of hitting were the main reason I felt too lazy to construct an actual proof of uniqueness.
Good point. I convinced myself early in the creation of the puzzle that it does not work for either player to be hit but then forgot to mention it in my solution post.
Blue cannot be hit. Playing (say) 24/20 and being hit will cost not only a pip (24pt to 25pt) but also an entire turn, and there is no way to make up the loss (no legal way to roll 69 pips in three turns).
White also cannot be hit, due to a shortage of resources. I believe I saw a simpler proof, but I'll lay out the one that occurs to me now, using reductio ad absurdum. That is, I'll start by assuming White CAN be hit, which leads to a contradiction that proves she CANNOT be hit.
Under "descriptive solution" (paragraph 3), I proved at the outset that White needs at least 15 move portions. Add 2 more now that we know White has to remake the 6pt, and it comes to 17. An opening non-doublet plus four doublets is 18 portions, so there is one portion to spare.
With this spare portion, White must play 24/xx, get hit, and enter with double 2s (the non-doublet will have been used up). With no portion left to waste after playing bar/23 6/4(2), at least one deuce is played 7/5. (If it is played 13/11, it will block Blue's later 55 unless it is moved with 11/5, a scenario that transposes to 13/7 + 7/5.)
With no portions left to spare, two portions must go directly from White's midpoint to her 7pt, plus a third 13/7 there to set up 7/5. (The 8pt checkers must go directly to the inner board, hence 8/7 is out.)
With 7/5 and 6/4(2) spoken for, there are only five spots (5pt 3pt 3pt 2pt 2pt) left for the six checkers remaining on White's 8pt and 6pt. This situation is irreconcilable. Therefore, the premise that White can be hit is false.
Nack