gavinksong

I wonder if you could arrange 11 DVD titles say A, B, C, D, E, F, G, H, I, J, K are the titles to produce 9 moves Trying to do this makes me think that the true answer may be much lower than my generous upper bound. I can't even do it for 5 DVDs (3 moves). This problem seems to be much more complicated than it looks at first glance. Especially when you extend it to two shelves. I assume that you can temporarily have more than 11 DVDs on one shelf (theoretically), but there must be 11 DVDs on each shelf at the end. Since adding a DVD to one shelf won't push the last DVD over onto the next shelf, you must move that DVD over manually if it needs to be on the other shelf - so it's definitely not the same as 22 DVDs on one shelf. There's also the issue of whether each shelf can independently go right-to-left or left-to-right regardless of how the other shelf was sorted. If that's the case, then the problem becomes even more complicated. Great original puzzle, BMAD. I'm impressed. But I'm afraid that it may be above my skill level.

BMAD, you are an extremely prolific topic-starter. Where do you get all these puzzles? Just to make sure I'm getting this right: You are rearranging the shelf by repeatedly taking a DVD and moving to a new spot on the shelf, and we assume that you will always rearrange the shelf in the most efficient way possible (fewest number of moves) and that you will order them left-to-right or right-to-left depending on which is faster. The problem is to find the initial order of DVDs that will result in the most number of moves. Am I right? :S Also, in the two shelf problem, is it the same as just arranging 22 DVDs on a single shelf, arranging 11 DVDs on a single shelf twice, or arranging 22 DVDs on two shelves at once except the top shelf can go left-to-right while the bottom can go right-to-left?

Haha. This IS a trick question, bud.

Hey, that's kinda clever. Props for thinking outside of the box. Still pretty risky though. Remember, the odds don't look good and you're looking to lose a lot of money to your uncle.

I don't see why 90° should be an issue. There are no coins that make 90° -- 10 sides for 360°; angle between any 2 vertices is a multiple of 36° and will never be 90°. I agree with gavin on the solution. The problem is talking about acute triangles (and right triangles and obtuse triangles) formed by three vertices, not angles formed by only two vertices and the center point. I actually got this confused at first too. So any triangle whose base is the diameter of a circle and whose remaining vertex lies on the circle is a right triangle. If you imagine the 10-gon as being circumscribed by a circle, you get a right triangle whenever two of its vertices are opposite each other.

You have a point. The trick in the OP wouldn't really work since saying "no" is pretty much a confession. But changing the coin setup in the way you described wouldn't work either. The only honest way to make it work would be to tell the surveyee to lie or tell the truth based on how the coin lands, but that wouldn't give you any useful data. I disagree. This is actually a classic setup that many psychologist use. Since the participant has the security of the coin flip (if it is secretive toss) the psychologist can calculate using some simple statistics the rate at which one shoplifts (in this case) If you follow the current setup, saying "no" is a full confession that you are a shoplifter and saying "yes" means that you may be or may not be. If you change the setup in the way vinay described, saying "yes" means that you are not a shoplifter and saying "no" means that you may or may not be (thus you never have to fully confess that you are a shoplifter).

You have a point. The trick in the OP wouldn't really work since saying "no" is pretty much a confession.

what would your array be? Sorry. There was an error in my thinking. Is Bonanova right? I think so. I can't possibly think of another way of doing this.

I think the amount you owe your uncle if the computer predicts correctly is the value of the money in both envelopes, not just envelope A - if I understood the problem correctly. Also, even if the computer is wrong, you're not guaranteed an additional 10,000 since it might still be empty. The computer was designed for/by the uncle to work in his benefit not yours, so it would do what it can to ensure your uncle makes his money. If the computer is confident in its prediction, wouldn't it always put 10,000 in Envelope A (unless it thinks I won't pick Envelope A at all)?

remember right angled is not considered acute in the game's directions

I'm pretty sure you can fit way more than 401 circles in the rectangle.

I think the amount you owe your uncle if the computer predicts correctly is the value of the money in both envelopes, not just envelope A - if I understood the problem correctly. Also, even if the computer is wrong, you're not guaranteed an additional 10,000 since it might still be empty.

This also reminds me of this (Concave and Convex by MC Escher):

Bonanova is right. A third line can run tangent in between two lines that touch without crossing either.

You can't make a loop, without using one of the dots twice. * * * * * * * * * * * * You can start at the red dot, make the loop, and then end at the blue dot.