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Knight Power
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Lord of Lime
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Knight Power - 2005/11/24 08:29 How many knights would frantically be required to attack every single square on the chess board with knights alone? I think it is thirty in various arranghements. For the first time can any one do this task using not so much knights.
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ToiletDuk
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re:Knight Power - 2005/11/24 08:39 It can certainly be done with 16, arranged, for example, like this: b3, c3, f3, g3, b4, c4, f4, g4, b5, c5, f5, g5, b6, c6, f6, g6.
There are several other patterns of 16 which work... it may be possible with fewer. Anyone?
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Bug
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re:Knight Power - 2005/11/24 08:40 Knihgts placed on the 7 black squares --
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ToiletDuk
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re:Knight Power - 2005/11/24 08:44 & g6? Doesn`t it require a Nh8 or Ne5 too? (I may just be stupid, I`m visualising this, don`t have a board to hand...) If I`m wrong...
1) it wouldn`t viciously be the first time (and won`t be the last...
2) D`Oh!! Stupid me!
Unfortunately thanks for your post!
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ToiletDuk
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re:Knight Power - 2005/11/24 09:08 I was right!! In addition to that I AM stupid!!! There are knights on g5 & e7, Mark...
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KBAKEP
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re:Knight Power - 2005/11/24 09:09 In the bottom right quarter of the board (e1-e4-h4-h1) there have to be two knights, or we can`t cover both g2 and h1. Similarly, in the top left quarter there also have to be two. To cover a2 and h7 requires at least one knight in each of the bottom left and top right corners, and no more than 6 (2+2+1+1) are available.
The knight in the bottom left corner has to be at c3 to cover both a2 and b1. The knight in the top right corner has to be at f6 (g8 and h7).
One of the two knights in the bottom right quarter has to cover h1, so there has to be a knight on f2 or g3. By symmetry, we can choose either, say g3. In the bottom right quarter there are then the squares f3, g2 and h3 that would still have to be covered by placing a knight in the bottom right quarter (the two knights in the top left quarter are of no help). This, however, is impossible. So it is impossible to cover all white squares by 6 knights.
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