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  1. Today
  2. Cubicle Stack #2

  3. You've just found a neat way to place points uniformly randomly inside a unit circle: simply place points at random inside a circumscribed square -- x and y uniformly chosen on [-1, 1] -- and ignore the points near the square's corners that are outside the circle. There are other ways, but this works, and it's simple to do. And why are you excited about this? The reason is that you've often wondered about the expected size of randomly drawn triangles inside a unit circle. And now you can find out. You sequentially place a million sets of three random points in the circle, calculating (and then averaging) the areas of the million triangles they define. And you find something pretty amazing: the million triangles had an average area that is only ~ 7.4% of the circle's area. You also note that the median area was ~ 5.4%. OK, so that's a fairly long set-up for a pretty short puzzle. Read on. You tell a friend about how amazingly small random triangles constrained by a circle are, and he replies with a question of his own: "That's cool," he says, "but I wonder what fraction of those triangles cover the circle's center?" You admit that was a piece of information that you did not take note of. "Oh, that's OK," your friend replies, "I think I can tell you." What answer did your friend (correctly) come up with?
  4. Green and Yellow hats

    Hmmm. OP does not rule out movement. But it does rule out communicating. So let's say that if the prisoners want to be at some preferred location in the room, that's permissible. But their chosen location can't be in any way influenced by hat color -- i.e., all movement must occur before the hats are placed.
  5. Green and Yellow hats

    Are they allowed to move at all, like to sort themselves without communication? Or do they start off in their positions in the room with hats placed on them, only allowed to say one word?
  6. In a long hallway, 100 prisoners are given red or blue hats, whose color only the other prisoners can see. At a signal given by the warden the prisoners must walk single file through a door and take their places inside a large room. The room is circular and its wall, ceiling and floor are featureless. Nothing is said, nor are any gestures made to prisoners as they enter the room and take their place. When the last prisoner has taken his place the warden inspects the configuration of their hat colors. If the colors form two monotonic groups separable, say, by some straight line, then all the prisoners are freed. If their hat colors instead are intermingled, they are all executed. Prisoners are allowed to discuss strategy before receiving their hats. What is their fate? Let's see, what else? Oh ya, they can't just pass their hats around. They're super-glued on their heads. Ouch. And no one has a magic marker to ... uh ... you know, make a line ... or anything like that.
  7. Green and Yellow hats

    Nope. They act on what they see. Sorry.
  8. Green and Yellow hats

    As always, we hope for some communication. The prisoners can see each other. It’s not clear they can hear each other (after all, if they shout simultaneously, they can’t benefit from hearing the others). Are they allowed to turn their bodies to face in a variety of directions, or some such thing? You said “no communication”, and I fear you mean it, but just askin’...
  9. Green and Yellow hats

    Here's a toughie. A room full of prisoners is given hats, whose color only the others can see. And just to be different, let's say they are yellow or green. No communication is permitted. At a signal, given by the warden, the prisoners must simultaneously shout out the color of their own hat. Those who guess wrong are subsequently executed. Beforehand, the prisoners meet to determine a strategy -- a set of rules, not necessarily the same for each prisoner -- that will guarantee the greatest number of survivors. As an added wrinkle, the warden may attend the meeting and then use his knowledge of their strategy when he chooses the colors of their hats. If there are 100 prisoners, how many can be assured of surviving?
  10. Cubicle Stack #2

    New number for EEE. But I'm still getting the same result for CEE and CEM.
  11. Whodunit?

    I understand now. I interpreted it as making the triangle from the lines you get from trisecting. I didn't know you could connect the intersecting points. Thanks for clarifying.
  12. Yesterday
  13. This is another puzzle where precise wording is important -- I'll try to get it right, but if anything is unclear, please ask ... I'll start out by saying that all the circles in this puzzle have the same radius, the aspect ratio of the rectangle is not specified and does not matter, and its size, relative to the size of the circles is only indirectly implied. Only the constraints stated in the puzzle should be assumed. I've drawn 17 circles that at least partially overlap a rectangle. Their centers all lie within the rectangle. None of the circles overlap or even touch any of the other circles. There is no room for an 18th circle to be added to the group. That is, the circles are drawn in such a way that even though there is space between them, it is impossible to draw another circle whose center lies within the rectangle that does not at least partially overlap one of the first 17 circles. That is all you know about the relative sizes of things. And it is enough information to answer the following question: First, let's erase the circles that I drew. Then I will paint the rectangle red and give you a large supply of opaque white circles. What is the smallest number of circles you will need to completely cover the rectangle? (so that no red will be showing.) The centers of the circles, again, must lie within the rectangle, but now, of course, the circles can overlap each other.
  14. Building cars

    So when OP says "build exactly one car a day (no more, no less)" it means you can build any number of cars in the interval [1 2) because "and to be clear a partial car is as good as not building a car." So if you built 1 1/3 cars on the first day it would count as "exactly 1 car," because it would pass 1/3 of a car to the second day, when you would then have to build any number of cars in the interval [2/3, 1 2/3)? In general, is it correct to believe that at the end of every nth day you must have built [n, n+1) cars, except for n=7, after which you must have exactly 7 cars?
  15. Born on a Wednesday

    This is absolutely something I should have been able to reason myself into. D'oh.
  16. Building cars

    The only day where you cannot have a partial car built is the 7th day. The other days must have a whole car built to meet your quota but you can have part of a car as long as you don't make two in a given day. You must build something each shift.
  17. Building cars

    I'm not understanding something about two shifts building exactly one car. A whole car (in one shift) or two half-cars (in two shifts) seem to be the only cases. Maybe spoiler one other possibility as a means of explaining? Thanks.
  18. Jelly beans join the clean plate club

    It must be symmetric about the NW-SE diagonal, so your figure show all the cases you computed. Nice, btw. Hint
  19. Building cars

    You are in charge of building cars, you are tasked to build exactly one car a day (no more, no less) and to be clear a partial car is as good as not building a car. Your shift is separated into two parts in which you could either build a whole, half, third, fourth, or fifth of a car in a given shift. By the end of the week you are to have built 7 cars with no partial cars left over. How many ways can this be done assuming a 7-day work week?
  20. Jelly beans join the clean plate club

    The program isn't a proof, it's just an application of a greedy algorithm that starts from states with an empty plate and works backwards to see whether every state could eventually reach one with an empty plate. It covered every possible state for up to 500 jelly beans, but it doesn't prove that a plate can always be cleared if you have something like 8x1012 jelly beans.
  21. Whodunit?

  22. Jelly beans join the clean plate club

    @plasmid Does the program imply a proof that it can always be done? Or is it a statement that no counterexample has yet been found? A proof could be a repeated procedure which after each application reduces the smallest number of beans on a plate. Does your algorithm always reduce the number on place C?
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