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BGonline.org Forums
Sixth solution?
Posted By: Timothy Chow In Response To: Peever vs Kenji Cube Action (Bob Koca)
Date: Tuesday, 16 March 2010, at 3:54 p.m.
I think there are only five solutions. I emailed Bob and he said he now sees only five.
The argument is as follows. First we simplify matters by observing that Blue's checkers on the 23-point, 18-point, and 13-point could not have just moved. Then we ask, on which points were White's blots? The only possible points on which White could have been hit are points that are currently occupied by Blue checkers, or points that are multiples of 4 earlier than those (provided they are not "blocked" by White checkers): i.e., points 4, 5, 6, 9, 10, 11, 14, and 21.
If a blot was hit on the 4-point, then Blue must have moved 8/4*(2) as part of his move; this move involves a "wasted" 4, by which I mean a checker that moves without hitting. Since 3 checkers were hit in total, we can afford at most one unit of wastage. By similar reasoning, if a blot was hit on the 11-point, there would have been one unit of wastage.
If we consider the 5-point and the 9-point collectively, we see that it is not possible for blots to have been hit on both of these points because the only possible move would have been 13/9*/5*(2), which wastes two 4's. So at most one of the 5-points and 9-points had a blot, and in either case there must have been one unit of wastage.
Now consider the 6, 10, and 14 points collectively. It's not possible for both the 14-point and the 10-point to have had blots because Blue would have had to move seven 4's to hit them and arrive at the current position. Nor could there have been blots on both the 6-point and the 10-point because that would waste two 4's. The possibilities are: (a) a blot on the 10-point alone, involving one unit of wastage; (b) a blot on the 6-point alone, involving one unit of wastage; (c) a blot on the 14-point alone, involving no wastage; (d) blots on both the 14-point and 6-point, involving one unit of wastage. (Or, of course, no blots on any of these points.)
Collecting this information, we see that out of the 4, 5, 6, 9, 10, and 11 points, we can pick at most one, since each one of these involves one unit of wastage. Therefore there must have been blots on the 14-point and 21-point since there were three blots in total. The blot on the 14-point precludes a blot on the 10-point, as we explained above. So there are at most five solutions: we must pick exactly one from the set {4, 5, 6, 9, 11}. It is easily checked that all five of these options are feasible.
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