bonanova

What happens at the boundaries? Do you prohibit moves that would cross them, or like a real E-a-S, just ignore the turning of the knob?

No shenanigans. I don't think the real numbers have to all be different. OP just says each box contains a real number. The probability of any two being the same is zero, but it could happen, and still solve the problem. Edit: When the OP says "countably" it has to mean "countably infinite." There must be a box in each room for every counting number for the strategy to work.

DeGe, No. To answer plasmid's question about communicating, let's frame the question this way: When a mathematician enters his room, the door locks behind him. Consider the room to be an information black hole, except that each box communicates to a central computer. Each box has a keypad capable of transmitting any real number (along with its box number) to the computer. If and only if the computer receives correct (real number - box number) pairs from 99 or 100 rooms, the doors unlock themselves. Otherwise, ... the world loses 100 mathematicians. That is, it's not like the 100 prisoners with red or blue hats: they do hear each other. Here, the strategy is so cool that they don't need to know each other's guesses. [spoiler=A similar strategy will also solve this puzzle]An infinite number of people will receive hats bearing a distinct real number unseen to the wearer but seen by all the others. When all the hats have been placed, each will shout a number, to guess the one on his own particular hat. There may be incorrect guesses but, to win, only finitely many. Before they receive their hats they craft a strategy that may require infinite resources. After that, they may not communicate.

Yes. Congratulations! Can you get the rest of the clues, and decode the quotation itself? The coveted bonanova Gold Star will be the prize.

Yeah, I know that. She still made The Movie.

Look at the matrix of situations in this link. http://www.sorting-algorithms.com/ There are eight columns, representing eight sorting algorithms. There are four rows, representing four different initial orderings. Tap the green arrows in the upper left corner. Each cell shows a sort process that requires a certain number of moves. If we hypothetically insert a row for EVERY initial ordering, We would get a row of eight numbers for each of them. The number in which cell represents the answer to the OP?

"Best moves" and "worst disarray" seem interrelated. The worst disarray for one algorithm might be duck soup for another. (Duck soup means easy - an Americanism.)

OP asks, what is the greatest number of moves? The number of moves depends on the original degree of disorder and the choice of algorithm. So it sounds like asking, for any starting order, what is the least efficient algorithm? Or, what starting order is the most challenging for the best choice of algorithm? I don't have an answer, but I found this page very interesting while thinking about it: http://www.sorting-algorithms.com/

Any thoughts on the clues themselves? I'll answer un-spoilered guesses/questions.

Since we're allowed facetious answers (but when did we ever require permission) then 3 is perfectly clear:

Look backwards. If the second point were in the middle empty area, where does that put the first point? Maybe this logic can be extended to the other smaller empty areas as well... I can't look backwards from the initial point. If the initial point were in the central area, the second point would not also be there.

Explanation: The prohibited area comprises a series of triangles of doubled linear scale proceeding from the vertices. That means if you start outside this area you can't get inside it by going half-distances toward a vertex. But I'm not clear what happens if your initial point were inside.

I think uncountably countably is all that's needed.

Nope. Or is that a cryptic way of saying you found the answer?

This puzzle begins with a familiar ring: you have a bag of visually indistinguishable, but not identical, coins. They are variously gold, silver and bronze, and therefore they are of three distinct weights. The King asks you to give him a gold coin. You consider just drawing a coin at random, but you fear (correctly) that you will lose your head if the coin you give to the King is not gold. Being of above average intelligence, you cast about for a better approach. You search the BrainDen archives for a balance scale. Luckily, you find one. Unluckily, it's damaged: Instead of giving the usual three outcomes, { <, +, > } it can discern only whether the two sides of the scale are in balance, or not. Whatever you place in the two pans can only be determined to have equal weight, or not. But weight, there's more. The scale has a further idiosyncrasy. It will allow an object to be weighed only two times. If an object is weighed for a third time, the scale will magically cause the object to disappear. That's it. The King is tapping his fingers impatiently on the table. There is no time to find another scale. If it heldps, I just looked in the bag and counted 1521 coins. I threw them all into a mass spectrometer that I just happened to have with me and determined that 827 of the coins are gold. It's algorithm time. Can you fulfill the king's wishes using this really weird scale?

Ah, got it. Very nice.

The quote is English, actually, and, although quite well known, the answer might require some calculation.

120? 5?

120?

Nothing yet, but the responses earn