A cube with size 10 * 10 * 10 consists of 1000 unit cubes, all colored white. A and B play a game on this cube. A chooses some pillars with size 1 * 10 * 10 such that no two pillars share a vertex or side, and turns all chosen unit cubes to black. B is allowed to choose some unit cubes and ask A their colors. How many unit cubes, at least, that B need to choose so that for any answer from A, B can always determine the black unit cubes?
Setting aside some possible quibbles about the meaning of 'share a vertex or side' and 'pillar', choosing the cubes {1,1,1},{2,2,2},...{10,10,10} will disclose a black cube from each of the 'pillars' but not their orientation.
Assume that, say, {5,5,5} is confirmed to be black. Then either {6,5,5} or {5,6,5} or {5,5,6} is black too, but since only 3 orientations are possible, it suffices to check 2 of these to confirm the orientation of the 'pillar'. Say{6,5,5} is confirmed to be black and call this orientation A.
Assume that another triplet {x,x,x} is confirmed to be black. Orientation A is now excluded by the terms of the problem, for any other 'pillar' in that orientation would share sides with it. Accordingly, say the remaining possible orientation B would include (x,x,(x+1)} and C, (x,(x+1),x}. Only one of these needs to be checked for the same reason as given above.
Assume that another triplet {y,y,y} is confirmed to be black. Orientations A and B are now excluded by the terms of the problem, so the third 'pillar' must have the remaining orientation, C.
There cannot be more than 3 'pillars' since the 3 already discovered must by the terms of the problem touch every side.
So B need choose no more than 13 cubes to determine all the unit cubes that are black.
Edited on July 30, 2021, 8:25 am
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Posted by broll
on 2021-07-30 08:14:55 |