I apologize for being so negative, but sprouts really needs a first class (or maybe even a second class (or maybe even a third class)) analyst, which I am not. No, the sprouts world for me is a slapstick, unstable place, like a work area inhabited by the Three Stooges. But all I can do, without quitting entirely, is keep plugging away, consoling myself that, after all, I am by necessity limiting my analytical work to a couple of hours a week. There is also some comfort in the knowledge that as an analyst, as opposed to a competitor, I can be assured that my sprouts world, despite appearances, is ultimately Augustinean, in Norbert Wiener’s sense. My best efforts should tend to accumulate. Wiener sees the competitor’s world as Manichaean; an active opponent seizes upon lapses, causing success or failure to be linked to one’s worst, rather than best, moments. All of this I am saying to explain the quotation marks in my title. Maybe what follows is wrong. But someday, someday, if I keep at it, if I live long enough and keep my minor wits long enough, perhaps I will be rewarded with a trivial, but shimmering island of order.
After 8+ 1(9)2 1(10)1[2,3] 4(11)10 I now see the sphere S(2) as a switch of Sprague-Grundy number 3. My revisions to the “Game of the Week” for 1/27 were apparently in error. (I will leave that apparently faulty article out there for a while, so that anyone can see what I am talking about.) The original game notes were probably correct as far as they went. Right can continue, as in that game, 4(12)11 with at least a cogent plan.
This plan is rather unique in my experience. Right tries to treat the single biosphere S(4) as two separate biospheres! The hope is that spots 4 and 12 will behave like a switch of Sprague-Grundy number 2. According to this theory, these two spots will produce zero or one survivor, depending upon whether the hollow sphere H(4,12) gets filled or not. The remaining spots, spots 5 through 8, Right hopes will behave like an inverter of Sprague-Grundy number 1. Four original spots normally produce three survivors, and even if an extra liberty is supplied by spots 4 and 12, three survivors are likely. Right hopes the total position will behave as if it consists of a Sprague-Grundy 3 (Sphere S(2)), a Sprague-Grundy 2 (the spots 4 and 12), and a Sprague-Grundy 1 (spots 5-8), in which case Right will win. Right knows, of course, that Left has tries such as 4(13)5 which immediately disrupt plans to treat the single biosphere as two biospheres, but he hopes these tries will not work. Left can only win in this fashion if he finds a move which leaves the sphere S(4) a biosphere of Sprague-Grundy number 3, assuming that my assessment of sphere S(2) is correct.
The move I saw as refuting Right’s strategy in my revised notes is not a clear refutation. Probably when I played the original game I saw Right’s proper response and then forgot it. After 8+ 1(9)2 1(10)1[2,3] 4(11)10 4(12)11 2(13)3, Right can play 3(14)13. The sphere S<3> has been returned to Sprague-Grundy number 3 and Right’s plan still is in play.
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