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Fifteen checkers on the bar, fewest moves
Posted By: Nack Ballard In Response To: Fifteen checkers on the bar, fewest moves (Keene)
Date: Thursday, 10 June 2010, at 12:18 a.m.
Since it will take 11 distinct checker plays to turn the starting position into 15 blots, thats technically 3 sets to do that. Once all checkers are blots, opponent requires 15 checker plays to hit all of them - 4 sets of doubles, so you are looking at 7 move pairings to complete.
Let's assume Blue is the player that will end up with 15 checkers on the bar. At a minimum, White must play four times after Blue plays three times. That suggests that a 6.5 move solution may be possible except for the fact that the player going first must roll a non-doublet. I'm not sure you considered that aspect, but you did reach the right conclusion.
Even if Blue is allowed a doublet going first, it turns out he needs four rolls to create 15 blots. Three doublets aren't enough because of those two pesky five-stacks, even ignoring interrelated landing problems. For example, say the rolls are double a's, b's and c's. Blue can turn the two-stack and the three-stack into five blots with 3 (of the 12) roll subparts, and create six more blots (from the five-stacks) with 6 additional subparts. That leaves two two-stacks with 3 subparts to play. As 2 subparts will sadly cover a point that has been started by a, b, or c, it requires 2 subparts beyond that to acquire all fifteen blots. That's one more roll subpart (13) than Blue has (12) in only three rolls.
In other words, even if the starting position has been re-reached and therefore a doublet for the initial roll is allowed, it is not possible to construct a solution shorter than 7 full moves (i.e., 14 rolls, 7 by both sides).
Obviously, I havent figured out what they are, but doubtless its a bunch of 3s,4s,5s or 6s to split them all up, and probably a large number of 1s, 2s or 3s to collect them all.
For a 7-move solution, Blue (the player going first) has 14 roll subparts -- from one non-doublet and three doublets) with which to complete his 15-blot mission. As you said, 11 (eleven) subparts are strictly necessary. That means that no more than 3 "squanders" are allowed. Moving a blot, covering a blot, or moving to a stack is a squander.
Also note that moving more than one checker into the inner board is an automatic squander. Blue can't create six blots there out of seven checkers, so he would have to be hit there early and spend a subpart entering: that is also a squander. By the same token, bringing down a 6 is a squander because it forces a second 8pt checker inside, one of which must reenter. Bringing down a 5 is double-squander because it also commits the sin of stacking. (And double 5s is such an awkward roll for blotting purposes that it will put Blue over his limit of 3 squanders even if he can split to a vacated 19pt!).
Michael's solution (7 moves) uses exactly 3 squanders. Playing a second non-doublet gains non-covering flexibility over a doublet but it expends 2 squanders, and sliding 6/5 onto another blot before slotting the 4pt with his fourth ace accounts for the (allowed) third squander.
Note that if one could make 54 (6/2 6/1) Blue's first (instead of next) move in Michael's sequence and replace 42 with double 2s (played 13/11 13/9 6/4), that would be only 1 squander. After Blue's 33 (played the same way, 24/21 13/10 8/5 6/3), then his 11 could be played 13/12 8/7, creating 15 blots and Blue doesn't need the other two aces (though being legally required to play them he would squander them with his back checkers).
However, there's a problem with that adjustment: 6/2 6/1 cannot be legally played first (White owns Blue's ace point -- the trickiest point for blots to reach), so Michael temporized by adding another non-doublet (even though it cost 2 squanders compared to a doublet). Going in reverse he likely set up the 33 and 11, noticed that 42 and 54 would slotfully fill in the vacant points, then made sure one of White's checkers escaped.
Actually, after Blue's initial 42, White's blot left behind can end up on any of Blue's near-side vacant points: if on the 1pt or 2pt then 54 is the last of Blue's four-roll sequence; if on the 3pt, 5pt, or 10pt then 33 is last; if on the 4pt or 7pt then 11 is last. (Some solutions involve White being hit twice.)
Nack
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