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bonanova

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Everything posted by bonanova

  1. bonanova

    The Most Wonderful Prince

    The King has decreed that his daughter the Princess shall marry the most wonderful Prince in all the land. One hundred suitors have been selected from their written applications, and on a certain day the King arranges for them, in turn, to interview the Princess. Each suitor must either be chosen or eliminated on the spot. If the Princess does not choose any of them, she will marry the last Prince to speak with her. You have been chosen as the Royal Advisor to the Princess and tasked with implementing her best strategy to choose the Most Wonderful Prince of the realm. You devise an evaluation scheme by which the princess can assign a unique "wonder number" to each prince as she meets him. The strategy then is to have the Princess reject, but record the highest score of, the first N princes that she meets. The Princess will then choose the first Prince that she subsequently interviews whose score exceeds that recorded score. That's it. The puzzle is basically solved. Except, of course, to decide on the optimal value of N. It requires some thought. If N to too high, the most wonderful prince is likely to be eliminated at the outset, and she ends up with the last guy. If N is too small, the Princess will likely settle for a fairly undistinguished prince. What value of N optimally balances these two risks? What is the probability that the Most Wonderful Prince will be chosen? Disclaimer: I recall this puzzle being posted before, with different flavor text. And it's somewhat of a classic. To give it a fair play here, I'll ask not to post any links and not to just give the answer if you know it, at least not without "showing your work."
  2. bonanova

    Salesman in a Square

  3. bonanova

    Salesman in a Square

    Al made sales calls at a number of cabins, which lay in a square field, one mile on a side. He drove his car to the first cabin, then visited the remaining cabins and returned to his car on foot, walking several miles in the process. If Al had had a map and perhaps a computer, he could have picked the shortest route to take, (traveling salesman problem). Lacking these amenities, Al simply chose to visit (after the first one) the (un-visited) cabin closest to his current location. If there were 6 cabins in all, how might they be placed, so that using Al's nearest-neighbor algorithm, and selecting the worst initial cabin, Al would be forced to walk the greatest distance; and what is that distance? Examples: 2 cabins: diagonal corners, starting at either: 2 x sqrt(2) = 2.828... miles. 3 cabins: any three corners, starting at any of them: 2 + sqrt(2) = 3.414... miles. Check out n=4 and n=5 as a warm-up.
  4. bonanova

    The Most Wonderful Prince

    @plasmid Not sure why derivative failed, but you're right about the result. Nice solve.
  5. bonanova

    Random passengers

    These cases are the ones I calculated as well, and led me to the right conclusion. (I had to solve it as it's not mine originally and I did not receive the solution.) It took a little insight to get me thinking along productive lines. See the spoilers in my April 1 post - btw not an April Fools post - to get there.
  6. bonanova

    Salesman in a Square

  7. bonanova

    The Most Wonderful Prince

    Great job! Want to look at the reciprocal of that number and guess the exact result?
  8. bonanova

    Fettered random walk

    @Molly Mae blazed the trail and @plainglazed nailed it.
  9. bonanova

    Fettered random walk

    Consider a random walk in the plane where each step is taken, beginning at the origin, in either in the positive x or positive y direction, i.e. either east or north, each choice being made by the flip of a fair coin. The length of each step is 1/2 the length of the previous step, and the first step has length √2. After infinitely many steps have been taken, what is your expected distance from the origin? Edit: Ignore the original text in pink. Instead, What is the distance to the origin of the centroid of the possible termination points? You find the centroid of a set of points by averaging respectively their x- and y- coordinates. First correct answer wins, but style points will be awarded as well.
  10. bonanova

    Five rings

    You are given 5 circles, A, B, C, D, and E, whose radii are, respectively, 5", 4", 2", 2", and 1". Can you find a way to overlap circle A with portions of some or all of the other four circles so that the un-overlapped portion of A has the same area as the sum of the unoverlapped portions of the other four circles? That is, the red area is equal to the sum of the green areas. Circles B, C, D and E may overlap portions of each other as well as a portion of A.
  11. bonanova

    Fettered random walk

    You are both on the right track, but I realize now that I mis-stated the OP. I didn't ask for what I wanted. What I wanted to get at was the average location, that is the average of all the possible ending location coordinates, more precisely, their centroid, and its distance from the origin. That's not the same as the expected distance of the ending points -- which does take sort-of serious math. My bad. I edited the OP.
  12. bonanova

    Five rings

  13. bonanova

    I might be a memory aid

    Sir, I bear a rhyme excelling In mystic force and magic spelling Celestial sprites elucidate All my own striving can't relate. See, I have a rhyme assisting My feeble brain, its task ofttimes resisting.
  14. bonanova

    Who finished 2nd?

    Al, Bert, and Charlie competed in a track and field event in which points were awarded for 1st, 2nd, and 3rd, place only. At the end of the day, Al had accumulated 22 points, while Bert and Charlie each garnered only 9 points. No other competitor earned points. Bert was 1st in the shot put. Who finished 2nd in the javelin throw? This is a Gold star puzzle.
  15. bonanova

    Who finished 2nd?

    Yes and yes.
  16. bonanova

    Fettered random walk

    Here’s a challenge: I think this puzzle can be solved with almost no math at all. Like, think of a single “what if” question to ask that changes the conditions slightly.
  17. Six line segments serve to connect pairwise any four points in the plane, no three of which are collinear. It's clear that no placement of the points permits all six to have the same length. How many unique placements permit the segments to have only two distinct lengths? Example:
  18. bonanova

    Limit of a shrinking function

    I have a strong feeling, and I'm working on a proof, that
  19. bonanova

    Limit of a shrinking function

    Which way should we evaluate this? [ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ].... or x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ]
  20. bonanova

    Random passengers

    One approach, among several, is to and
  21. bonanova

    Squirrel

    I guess I've always been a little confused about what is being asked. Sharing is not permitted, yet an "average" result is requested. If average values are used, then two holes is enough. If there is no sharing or averaging, survival is assured only by the (very unlikely) worst case of 100 holes. If the question is what is the expected number of holes that together yields at least 100 nuts, we have an answer from simulation. Is there a way to say precisely what else might be needed?
  22. bonanova

    100 white marbles

    @harey, some guidance about completing a formula would be appreciated. I'm interested in the "surprise."
  23. bonanova

    The holy mans walk up the mountain

    Yes you cand do that. The elevation of the two men as they ascend and descend are continuous functions of time. If at one time one is greater and at a later time the other is greater there must be a time when they are equal. Since they both stay on the path, and if the the slope of the path never changes sign, then they meet at that time. If the slope does change sign, then multiple points on the path will have the same elevation and there can be multiple times when the elevation of the two men are the same. They meet at one of those times.
  24. bonanova

    Shortest set of lines in a square, revisited

    Good start -- That's shorter than 3, which comes from any three of the square's edges.
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