Right is to move. He counts five cannibals in the sphere S(2). (Please note that spot 10 belongs to the sphere S(2) as well as to the sphere S(3)). He therefore needs the sphere S(2) to produce an even number of survivors. Since this sphere has a life force of 15 (add the liberties of each spot) certainly four survivors must be his goal. Well, four survivors seem rather a lot. In the old days I played several interesting games starting with 5(12)6, thinking by analogy with the two spot game that this move tends to increase fecundity. But I was an absolute ignoramus not to immediately consider, instead, 9(12)10[2,5], creating a second trap. Here is a good rule of thumb: the player needing fecundity should always consider creating a trap, if that is possible. Here is an even more general, and infallible rule: any time it is possible to create a trap, that line must be considered. This move, in fact, wins immediately for Right. Left is now faced with a losing cannibal war. The sphere S(2) has two cannibals and will produce two survivors. Left will eventually be forced to move to one of the traps, letting Right have his way. (Right will use this opportunity of force Left to eventually move to the second trap.) Crashing the universal trap S(7) by playing 7(13)12 does not create double switches; the sphere S(2) still produces two survivors. The answer to the Polar Bear puzzle, therefore, is 9(12)10[2,5].
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