I think
rsl12 is the only person who understands the question.
No, you may not move the card as a form of communication. The idea is that there is only one
bit of information that can be transfered at a time. You can assume the camp counsellors (or whatever they're called) will always centre the card on the table after you've flipped it (for neatness)...
ChickenNinja, perhaps it's pointless to say this but your scheme is both violent and useless. You still can't solve the problem using your "solution" but, at least, you'll have a bunch of kids with black eyes, if that's your cup of tea...
Quote:
Originally posted by rsl12
When he flips the card on Day X (i'll assume it's a boys camp), he knows that at least X-1 people (the people who saw black, including himself) will have entered the room by day 9. The people who saw black have nothing else to do--they've already been counted by the information gatherer. The other people (those who saw white or didn't enter the room at all) need to be counted now. Starting with the person who enters on Day 9, here's the procedure:
...
The information gatherer will count the number of times he sees the card is black after day 9. If he sees the card is black 9-(x-1) days, then he knows that everyone has been in the room!
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I don't understand the significance of day 9 (not to mention that there are ten kids, but that's arbitrary and unimportant). Suppose the the first kid to see the room twice is on the third day. I'm confused about how to proceed using your algorithm.
Other than that, it sounds like you understand the problem and, in fact, are on your way to one of the more efficient methods of doing so. There are two others, more effecient, but neither are terribly obvious...