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Can You Crack The Case Of The Crescent Moon?

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.

Riddler Express

From Curtis Karnow comes a puzzle that boldly goes where no human has gone before:

You are creating a variation of a Romulan pixmit deck. Each card is an equilateral triangle, with one of the digits 0 through 9 (written in Romulan, of course) at the base of each side of the card. No number appears more than once on each card. Furthermore, every card in the deck is unique, meaning no card can be rotated so that it matches (i.e., can be superimposed on) any other card.

What is the greatest number of cards your pixmit deck can have?

Extra credit: Suppose you allow numbers to appear two or three times on a given card. Once again, no card can be rotated so that it matches any other card. Now what is the greatest number of cards your pixmit deck can have?

The solution to this Riddler Express can be found in the following column.

Riddler Classic

This past Monday marked the beginning of Ramadan in the U.S. The month-long observance traditionally begins with the sighting of a crescent moon, or hilal, after a new moon. For this week’s Classic, let’s take a closer look at a waxing crescent moon.

After a new moon, the crescent appears to grow slowly at first. At some point, the moon will be one-sixth full by area, then one-quarter full, and so on. Eventually, it becomes a half-moon, at which point its growth begins to slow down. The animation below provides some insight into what’s happening here:

Animation showing the phases of the moon, including waxing/waning and new/crescent/half/gibbous/full.

How many times faster is the area of the illuminated moon growing when it is a half-moon versus a one-sixth moon?

(Some simplifying assumptions you might make for this problem are that the moon is a perfect sphere, that its orbit around Earth is a perfect circle, that the moon orbits the Earth much faster than the Earth orbits the sun and that the sun is very, very far away. If you make additional assumptions, feel free to include them in your response.)

The solution to this Riddler Classic can be found in the following column.

Solution to last week’s Riddler Express

Congratulations to рџ‘ЏDavid Olsho рџ‘Џ of Seattle, Washington, winner of last week’s Riddler Express.

Last week, you had two 16-ounce cups — cup A and cup B. Both cups initially had 8 ounces of water in them.

You took half of the water in cup A and poured it into cup B. Then, you took half of the water in cup B and poured it back into cup A. You did this again. And again. And again. And then many, many, many more times — always pouring half the contents of A into B, and then half of B back into A.

When you finally paused for a breather, what fraction of the total water was in cup A?

A good first step was to work out what was happening with the first few pours. Writing the volumes of the cups as the ordered pair (A, B), the initial state was (8, 8). After the first pour from A to B, it was (4, 12); pouring half from B back to A resulted in (10, 6). Another pour from A to B resulted in (5, 11), and then back from B to A gave you (10.5, 5.5).

If you worked out a few more iterations, you noticed that with every subsequent pour from A into B, the volume of cup B appeared to be twice that of cup A. Similarly, with every pour from B into A, the volume of cup A appeared to be twice that of cup B. After many, many pours, with a final pour from B to A, cup A had two-thirds of the total water volume.

For extra credit, you had to consider a more general version of the problem, when both cups initially had somewhere between 0 and 8 ounces of water in them. You didn’t know the precise amount in each cup, but you knew that both cups were not empty. Again, you repeated the process of pouring half the water from cup A into cup B, and then half from cup B back to A. When you paused for a breather, what fraction of the total water was in cup A?

Surprisingly, no matter what the initial volumes in A and B were, the answer was always the same. The following animation shows 15 different starting ratios between the volumes in A and B. After just eight pours (i.e., four from A to B and another four from B to A), all of these cases were nearly the same, with two-thirds of the total volume in cup A.

15 distinct animations, showing 15 pairs of cups. Each pair has a different initial ratio of volume in cup A to cup B. After four pairs of pourings back and forth, all pairs have approximately the same 2:1 ratio of volume.

Up to this point, we have merely observed that cup A appears to have two-thirds of the total volume. A number of solvers went a step further and proved this fact using a variety of different approaches, with many venturing into linear algebra, eigenvalues/eigenvectors and stationary points.

Meanwhile, solver David Kravitz offered an elegant little proof. Initially, suppose the initial fraction of water in cup A was 2/3+x, which meant the fraction in cup B was 1/3−x. Note that every ratio between the volumes in A and B can be described using some value of x, which could take on any value between −2/3 and 1/3.

As an ordered pair, the initial volumes were (2/3+x, 1/3−x). After the first pour from A to B, you had (1/3+x/2, 2/3−x/2). Then, after pouring from B back to A, you had (2/3+x/4, 1/3−x/4).

What was the significance of this? With every pair of pours, the difference between the volume in cup A and two-thirds of the total volume decayed by a factor of four, as did the difference between the volume in cup B and one-third of the total volume. So no matter the initial state of the cups, their volumes exponentially approached a 2:1 ratio.

I’ll drink to that!

Solution to last week’s Riddler Classic

Congratulations to рџ‘ЏMike Strong рџ‘Џ of Mechanicsburg, Pennsylvania, winner of last week’s Riddler Classic

Last week, in an effort to make Riddler City friendlier for pedestrians and cyclists, the mayor decreed that all streets should be one-way. Meanwhile, the civil engineer overseeing this transition was not particularly invested in the project and was going to randomly assign every block of each street a particular direction.

For your daily commute to work, you drove a car two blocks east and two blocks south, as shown in the diagram below. What was the probability that, after each block was randomly assigned a one-way direction, there would still be a way for you to commute to work while staying within this two-by-two block region (i.e., sticking to the 12 streets you see in the diagram)? Here was one such arrangement of one-way streets that let you commute to work:

2 by 2 block grid, with each block surrounded by 1-way streets. A looped path is drawn that follows one-way streets, starting from the top left corner and finishing in the bottom right corner.

And no, you couldn’t get out of your car to hop on a bike or walk. I mean, you could have, but not in this puzzle.

Now, because this puzzle had 12 one-way streets that could all be one of two states — either left/right or up/down — that meant there were a total of 212, or 4,096, cases to consider in this problem. Yikes!

Solver Rohan Lewis realized that all viable paths had to be either 4, 6 or 8 blocks long. However, these paths were not mutually exclusive — some of the 4,096 cases included multiple paths, making the counting exercise even more difficult.

In the end, virtually everyone who solved this turned to their computers for some assistance. (A few brave souls nevertheless did their work by hand. I see you, Guy D. Moore and Ethan Rubin!) Of the 4,096 cases of one-way streets we have been discussing, 1,135 resulted in a viable commute to work. That meant your probability of being able to commute was 1,135/4,096, or about 27.71 percent.

A few solvers went even further. Jezlyn Schaa and Peter Ji both determined that the probability you could make a round trip (i.e., to work and back) along the one-way streets was lower: 170/4,096, or about 4.15 percent. Meanwhile, others looked beyond the two-by-two block scenario, at three-by-three (~20 percent), four-by-four (~15 percent), five-by-five (~12 percent), etc.

If you work even farther from home, you might want to consult this handy chart from Maxim Wang:

Grid showing the probability of a one-way trip for different rectangular dimensions of the commute to work. As work becomes more distant, the probability there is a viable path decreases.

Finally, it’s worth mentioning that Vikash Gayah, a professor of civil and environmental engineering at Penn State University, pointed out that “two-way streets are in fact safer for pedestrians and cyclists, and are even preferred for non-motorized roads too.” I’ll be sure to pass along this message to the Riddler City mayor!

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


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.