Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
Riddler Express
From Dan Levin comes a hazardous riddle for the ironclad and eagle-eyed:
The U.S. Open concluded last weekend, with physics major Bryson DeChambeau emerging victorious. Seeing his favorite golfer win his first major got Dan thinking about the precision needed to be a professional at the sport.
A typical hole is about 400 yards long, while the cup measures a mere 4.25 inches in diameter. Suppose that, with every swing, you hit the ball X percent closer to the center of the hole. For example, if X were 75 percent, then with every swing the ball would be four times closer to the hole than it was previously.
For a 400-yard hole, assuming there are no hazards (water, sand or otherwise) in the way, what is the minimum value of X so that you’ll shoot par, meaning you’ll hit the ball into the cup in exactly four strokes?
Riddler Classic
From Dean Ballard comes a pernicious pizza puzzle:
Dean ordered a personal pizza that was precisely 10 inches in diameter. He ate half of it, and he wants to save the remaining semicircle of pizza in his refrigerator. He has circular plates of all different sizes, so to save space in his fridge, he’ll place the pizza on the smallest plate that holds the entire semicircle (i.e., with no pizza hanging off the plate).
Unfortunately, the smallest plate that can hold half of a 10-inch pizza is just a circle with a 10-inch diameter. So much for saving space.
But Dean has a thought: If he cuts the pizza, he can squeeze both of the resulting pieces onto a smaller circular plate — again, with no pizza hanging off the plate and without the pieces lying on top of each other.
If Dean makes a single straight slice, what is the diameter of the smallest circular plate onto which he can fit the two resulting pieces?
Extra credit: Dean wants to save even more space in his fridge. So instead of one straight slice, he will now make two straight slices. First, he will cut the semicircular pizza into two pieces. Then, he will take one of those pieces and make his second slice, giving him a total of three pieces. What is the diameter of the smallest circular plate onto which he can fit all three pieces?
Solution to last week’s Riddler Express
Congratulations to 👏 Graham E. McGrath 👏 of Boston, Massachusetts, winner of last week’s Riddler Express. (This problem might have hit a little too close to home for Graham, who is a lab technician and uses centrifuges for a living!)
Last week, you were doing your best not to break a microcentrifuge, a piece of equipment that separates components of a liquid by spinning around very rapidly. Liquid samples were pipetted into small tubes, which were then placed in one of the microcentrifuge’s 12 slots evenly spaced in a circle.
For the microcentrifuge to work properly, each tube had to hold the same amount of liquid. Also, importantly, the center of mass of the samples had to be at the very center of the circle — otherwise, the microcentrifuge would not be balanced and likely break.
You needed to spin exactly seven samples. In which slots (numbered 1 through 12, as in the diagram above) could you have placed them so that the centrifuge was balanced?
Balancing centrifuges is certainly not a new problem, and was previously addressed on the Numberphile YouTube channel, with Holly Krieger explaining the solution.
In the case of 12 slots, it was always possible to balance the centrifuge for any even number of tubes. All you had to do was arrange them into pairs and place each pair on diametrically opposite ends of the circle. The center of mass of each pair was at the center of the circle, which meant the total center of mass was also at the center.
Alas, the problem asked how you could place seven tubes, and seven is not an even number. Several readers suggested adding an eighth tube, which was against the spirit of the puzzle.
As for odd numbers of tubes, balancing a single tube was not possible. However, it was indeed possible to balance three tubes by arranging them evenly around the circle so they were the vertices of an equilateral triangle (e.g., in positions 1, 5 and 9 in the diagram above). It was further possible to balance five tubes by again arranging three so they formed an equilateral triangle (e.g., in positions 1, 5 and 9) and another two on opposite sides (e.g., in positions 2 and 8). Again, because the centers of mass for both the trio and the pair were at the center of the circle, their combined center of mass was also at the center of the circle.
Returning to the original puzzle, it was possible to balance seven tubes by arranging three in an equilateral triangle, another two on opposite sides and the last two on a different pair of opposite sides. (Alternatively, some solvers noted that placing seven tubes was equivalent to choosing the five slots that didn’t have tubes, which was essentially the same problem as placing five tubes.) There were several correct answers here — more on that in a second — one of which was slots 1, 2, 5, 6, 8, 9 and 12, found by solver Alicia Zamudio.
For extra credit, you had to find the total number balanced arrangements for seven tubes. It turned out that there were exactly 12 arrangements, all rotations of each other. Solver Reece Goiffon verified this by finding all the ways seven vectors on a unit circle could add up to zero:
Meanwhile, in what I believe is a first for The Riddler, Chris Sears wrote code in Commodore 64 BASIC, checking all 792 (i.e., 12 choose 7) ways to place seven tubes in 12 slots.
The general version of this problem, with N slots and k tubes, was addressed in the aforementioned Numberphile video and was further proven by T. Y. Lam and K. H. Leung back in 2000. As for how many arrangements there are for given values of N and k, there’s an OEIS sequence for that! (Thanks to solver Eric Thompson-Martin for spotting it.)
When N = 12 and k increased from 0 to 12, the number of balanced arrangements was 1, 0, 6, 4, 15, 12, 24, 12, 15, 4, 6, 0 and 1. If you look closely, you’ll see that the sequence is symmetric, since placing k tubes was exactly the same problem as choosing where not to place k tubes.
I encourage readers to verify these results for themselves. But not the grad students or technicians out there. I don’t need angry lab directors emailing me about broken centrifuges.
Solution to last week’s Riddler Classic
Congratulations to 👏 Lowell Vaughn 👏 of Bellevue, Washington, winner of last week’s Riddler Classic. (Lowell is finally joining his son in the winner’s circle.)
Last week, you were introduced to an online game, Guess My Word, in which you tried to guess a secret word by typing in other words. After each guess, you were told whether the secret word was alphabetically before or after your guess. The game stopped and congratulated you once you had guessed the secret word.
The secret word was randomly chosen from a dictionary with exactly 267,751 entries. If you had this dictionary memorized and played the game as efficiently as possible, how many guesses should you have expected to make to uncover the secret word?
Most solvers recognized that the most efficient strategy was a binary search, where you first guessed the middle word in the dictionary. If the secret word came later, you then guessed the middle word in the second half of the dictionary. But if the secret word came earlier, you guessed the middle word in the first half of the dictionary. With each guess, you narrowed the possibilities by approximately half, and there was also a chance that the guess itself was the secret word.
The sequence of guesses could therefore be structured as a complete binary tree whose top node was your first guess (the exact middle word in the dictionary). If the secret word came earlier in the dictionary, you moved to the right; if the secret word came later in the dictionary, you moved to the left. Eventually, you’d find the secret word.
Now that you had the optimal guessing strategy, all that was left was calculating the average number of guesses. In your binary tree, one word — the middle word in the dictionary — would be guessed on the very first attempt. Two words required two guesses, four words required three guesses, eight words required four guesses, 16 words required five guesses, and so on. For each additional guess, there were twice as many words.
It turned out that the number of words in the dictionary, 267,751, was 1 + 21 + 22 + 23 + … + 216 + 217 + 5608. And each term in the sum required one more guess than the term before it. To find the average number of guesses, you had to divide each of these terms by 267,751 and then multiply by the corresponding number of guesses. In the end, the average was approximately 17.042 guesses.
Several solvers, like Michael Branicky, Josh Silverman and Rajeev Pakalapati, generalized this method, finding a formula for the expected number of guesses for a dictionary with any number of words. Meanwhile, Emma Knight went in a completely different direction — recursion — still arriving at the same solution.
Thanks again to Oliver Roeder for introducing Riddler Nation to this fun word game. (If only it weren’t so darn addictive.)
Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at riddlercolumn@gmail.com
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