I have this disk. It's shaped like a CD, and divided into twelve sectors. Some of those sectors are opaque, and some are transparent. But the opaque and transparent sectors are not arranged in the particular order shown above.
Actually, I have two disks like this. They're completely identical to each other.
I place the two disks on top of each other. You can see through some sectors and not through others. And then I begin to spin the top disk. Every time it rotates one twelfth of a full circle, the number of transparent sectors changes. It forms a sequence of numbers.
Here's part of the sequence:
..., 2, 3, 4, 4, 0, 4, ...
Which sectors of the disk are transparent, and which are not?
Solution posted
7 comments:
So just to be clear, the sequence of numbers is the number of transparent sectors at each 1/12 turn of the top disk?
Yes.
I found by brute force programming (without logic and cleverness):
oottoottotot
o = opaque
t = transparent
One disc has to be turned by 180° in 3D !
I'm sure that Squirell will do it by logic.
I don't really have time to do it right now... All i've figured out are that atleast 6 of the twelve sectors are opaque because of the 0.
Can one disk be flipped upside down? or are they both facing the same way so that theres a point where all the transparent are matched up and all the opaque are matched up?
I would have done it by brute force programming too. (Had I the time.) Generate-and-test.
Dartheye, you tell me. Can you flip a disk upside-down? Isn't that against the laws of physics or something? Last time I tried flipping a disk, it tore a hole in reality, and a bunch of my roomie's snacks got sucked in before it closed up again. Yes, that's where the snacks went. That's my story, and I'm sticking to it.
You have to flip one of the discs unless they're symmetrical, in which case flipping doesn't change anything. As Dartheye says, there must be at least six opaque sectors. The 4-0-4 sequence tells us that that there are four places where the sequence is -X and four where it is X-.
The 4-4 seq indicates that there are more than 4 transparent sectors. For each previously doubly-transparent sector blocked (going from 4 to 4) there must be a corresponding one created so there must be an even number of transparent sectors. Since we already know that there at at least 6 opaque sectors, there must be also 6 transparent ones.
Since the transparent sectors must be grouped into 4 contiguous areas (by 4-0-4 above), they can only be 3111, 2211, or 2121.
I couldn't think of a logical or clever reason for excluding 3111 but I didn't like it so I tested it and... no good. The fact that 3 doubly-transparent sectors crops up means that the arrangement of the sectors is not rotationally symmetrical, since that will only produce even numbers. That leaves 2211.
So, it's as Ed has it: --XX--XX-X-X
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