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 the next 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.
Tonight marks the sixth night of Hanukkah, which means it’s time for some more Menorah Math!
I have a most peculiar menorah. Like most menorahs, it has nine total candles — a central candle, called the shamash, four to the left of the shamash and another four to the right. But unlike most menorahs, the eight candles on either side of the shamash are numbered. The two candles adjacent to the shamash are both “1,” the next two candles out from the shamash are “2,” the next pair are “3,” and the outermost pair are “4.”
The shamash is always lit. How many ways are there to light the remaining eight candles so that sums on either side of the menorah are “balanced”? (For example, one such way is to light candles 1 and 4 on one side and candles 2 and 3 on the other side. In this case, the sums on both sides are 5, so the menorah is balanced.)
The solution to this Riddler Express can be found in the following column.
Winter is almost upon us, at least for those of us in the Northern Hemisphere. Official ice master and deliverer Kristoff and his trusty pal Sven are preparing ice cubes that will be sent to the hot springs, where Kristoff’s extended family lives.
At the moment, they are studying a cube whose side lengths measure 1 meter, which means its surface-area-to-volume ratio is 6. (Kristoff doesn’t particularly care about the units here, but knows they are inverse meters.) Sven is concerned that much of the ice will melt before they reach the hot springs, so he suggests that Kristoff cut the ice along a single plane to minimize the surface-area-to-volume ratio of the larger resulting piece.
Along which plane should Kristoff cut the ice? And what will the resulting surface-area-to-volume ratio be?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Derrick Butler 👏 of Lexington, Kentucky, winner of last week’s Riddler Express.
Last week, you were on the Food Network’s latest game show, Cranberries or Bust, where you had a choice between two doors: A and B. One door had a lifetime supply of cranberry sauce behind it, while the other door had absolutely nothing behind it. And boy, did you love cranberry sauce.
Of course, there was a twist. The host presented you with a coin with two sides, marked A and B, which corresponded to each door. The host told you that the coin was weighted in favor of the cranberry door — without telling you which door that was — and that door’s letter would turn up 60 percent of the time. For example, if the sauce had been behind door A, then the coin would have turned up A 60 percent of the time and B the remaining 40 percent of the time.
You could flip the coin twice, after which you had to make your selection. Assuming you optimized your strategy, what were your chances of choosing the door with the cranberry sauce?
Before working through this scenario with two coins, let’s take a step back and look at one coin. Since the coin was slightly weighted in favor of the cranberry door, your best strategy was to choose whichever door was indicated by the coin. Then you’d be right — and win your delectable cranberry sauce — 60 percent of the time.
Surely, you had better odds of winning with two flips instead of one. Right?
Wrong. To see why that was, solver Rebecca Harbison looked closer at the possible results of the two flips. Instead of A and B, let’s relabel the coin C (for “Cranberry sauce is behind this door”) and D (for “Dangit, no cranberry sauce behind this door”). As stated by the problem, the coin had a 60 percent chance of landing on C and a 40 percent chance of landing on D.
With two flips, there were four possible outcomes:
- The first flip was C and the second flip was also C, which occurred with probability (0.6)(0.6), or 0.36.
- The first flip was C and the second flip was D, which occurred with probability (0.6)(0.4), or 0.24.
- The first flip was D and the second flip was C, which occurred with probability (0.4)(0.6), or 0.24.
- The first flip was D and the second flip was D, which occurred with probability (0.4)(0.4), or 0.16.
Putting these together, there was a 48 percent chance that the two flips were different. When this happened, you had absolutely no information how the coin was weighted, since the coin came up either side an equal number of times. In other words, you had to guess, which meant you’d be right half the time.
The other 52 percent of the time the two flips were the same. The majority of the time (i.e., with probability 36/52, or 9/13) that meant both flips were C. So when the two flips were the same, your best move was to guess the door both flips corresponded to.
So then what were your chances of winning the cranberry sauce? Well, 48 percent of the time you had a one-half chance of winning, while the other 52 percent of the time you had a 9/13 chance of winning. Numerically, these combined to (12/25)(1/2) + (13/25)(9/13), which simplified to 6/25 + 9/25, or 15/25 — in other words, 60 percent.
Shockingly (to me, at least), that second flip didn’t improve your chances one bit. You might as well have simply flipped the coin once and chosen the resulting door.
For extra credit, you looked at what happened when you were allowed three or four flips, instead of just two. For three flips, the optimal strategy was to choose whichever door was indicated by a majority of the flips. All three flips came up C with probability (0.6)3, or 0.216. Meanwhile, the probability of two Cs and one D was 3(0.6)2(0.4), or 0.432. Adding these together meant you’d win the cranberry sauce 64.8 percent of the time — an improvement over your chances with just one or two flips.
But just as two flips were no better than one, four flips were no better than three. That fourth flip either confirmed your decision based on the first three flips, or changed things so that now there were two flips for one door and two flips for another (opening up twice as many possibilities, and forcing you to take a random guess).
Solvers Jake Gacuan and Emily Boyajian both found a general formula for your chances of winning the cranberry sauce as a function of the number of flips N. Sure enough, having an even number of flips was no better than having the preceding odd number of flips. And as N increased, your chances of choosing the correct door approached 100 percent.
Solution to last week’s Riddler Classic
Congratulations to 👏 Izumihara Ryoma 👏 of Toyooka, Japan, winner of last week’s Riddler Classic.
Last week, you were introduced to Trig the turkey, who could not fly. Instead of flying, he decided to jump as far as he could with a running start up one of the famed Sinusoidal Hills, all of which were precisely the same size.
Trig knew that he simply could not jump a horizontal distance of more than two hills. Also, he prefered a smooth takeoff and landing. That is, when he took off from the ground and landed on it again, the slope of his parabolic trajectory through the air had to perfectly match the instantaneous slope of the ground beneath him.
The animation below shows a jump where the takeoff and landing were precisely 1.2 hills apart.
What was the greatest horizontal distance Trig could have jumped, such that his takeoff and landing were smooth?
At first glance, it was tempting to think the answer was 1.5 hills. Trig would take off just as the concavity of one hill was flipping from positive to negative. Then, Trig would land at a similar inflection point 1.5 hills away, as the concavity was flipping from negative to positive. Indeed, several readers believed this was the answer.
But there was a slight problem with this trajectory. Zooming in to Trig’s launch (or landing) revealed that he would have needed to pass through the ground in order to match the instantaneous slope of the ground at takeoff and landing, as shown below. Clearly, there was more to this problem than first met the eye, and closer inspection was necessary.
This puzzle could be reframed as two mathematical conditions. Given a sine wave, you were looking for a downward-facing parabola such that (1) it was tangent to the sine wave at two distinct locations, and (2) the parabola was always greater than the sine wave between those two points of tangency.
Before going further, it was also worth noting that the parabola representing the longest possible jump had an axis of symmetry that passed through a trough of the sine wave (as shown in the animations above). As solver Arvind Hariharan explained, if that weren’t the case, then Trig’s jump could be further extended by making the parabola symmetric.
Now for the calculus. We can represent the hills as h(x) = −cos(x). Meanwhile, we can represent the parabola as p(x) = −ax2 + c. (There was no linear term in the parabola due to its aforementioned symmetry.) If we suppose these two functions were equal at x0 — where Trig launched (and symmetrically landed) — then h(x0) = p(x0) and h’(x0) = p’(x0).
Mathematically, that meant −cos(x0) = −ax02 + c, and sin(x0) = −2ax0. You could solve for a in the second equation: a = −sin(x0)/(2x0). Plugging that value into the first equation gave c = −cos(x0) − x0sin(x0)/2.
Having solved for the parabola’s coefficients, the final step was to ensure that the parabola never fell below the sine wave between the points of tangency. There was a transition between the parabola dipping below the sine wave and always staying above — a transition that occurred when the second derivatives were also equal, i.e., h’’(x0) = p’’(x0), which simplified to tan(x0) = x0. Solving this transcendental equation gave you the largest possible value of x0.
The smallest nonzero solution was when x0 was approximately 4.493. The total distance Trig jumped was actually twice this, or about 8.987. Finally, if the Sinusoidal Hills were represented by the function h(x) = −cos(x), then the width of each hill was 2𝜋. So in terms of hill-widths, Trig’s longest jump was approximately 1.43 hills.
Here’s an infinite loop of that longest jump, courtesy of Maxim Wang:
Trig may not have flown, but this past week he certainly soared into our hearts.
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 firstname.lastname@example.org.