From Patrick Mayor comes a question about something we’re doing these days to keep ourselves and others safe: social distancing.

You’re walking along the middle of a wide sidewalk when you see someone walking toward you from the other direction, also down the middle of the sidewalk, 12 feet away. Being responsible citizens, you pass each other while maintaining a distance of at least 6 feet at all times. By the time you reach each other’s original positions, you should be back in the middle of the sidewalk again.

You should assume that the other person follows the same path you do, but flipped around (since they’re walking in the opposite direction). For example, you could both walk 3 feet to the left, 12 feet forward and finally 3 feet back to the right, walking a total of 18 feet before swapping positions.

Being lazy (I mean, *efficient*), you’d like to know the *shortest *distance you and the other person could walk so that you can switch positions, all while staying at least 6 feet apart at all times. What is this distance?

*Extra credit:* Now suppose the person walking toward you has no intention of straying from the center of the sidewalk (sigh), and it’s entirely up to you to maintain a distance of at least 6 feet. In this case, what is the *shortest* distance you would have to walk to reach the other person’s original position?

From Phil Imming comes his favorite riddle, which he was the only student in his calculus class to solve back in 1965:

One morning, it starts snowing. The snow falls at a constant rate, and it continues the rest of the day.

At noon, a snowplow begins to clear the road. The more snow there is on the ground, the slower the plow moves. In fact, the plow’s speed is inversely proportional to the depth of the snow — if you were to double the amount of snow on the ground, the plow would move half as fast.

In its first hour on the road, the plow travels 2 miles. In the second hour, the plow travels only 1 mile.

When did it start snowing?

Congratulations to Chesson Yauk of Kansas City, Missouri, winner of last week’s Riddler Express.

Last week, you helped me get exactly 10 gloves out of a box. It was difficult to pull exactly two gloves out of the box at a time — sometimes I’d pull out two gloves, other times three and yet other times four. Somehow, I never pulled out any other number of gloves at a time.

How many distinct ways were there for me to remove all 10 gloves from the box? Note that the order mattered here — for example, pulling out two gloves, then four gloves and then another four gloves was distinct from pulling out four gloves, another four gloves and then two gloves.

One approach to this puzzle was to simply list out all the possibilities. Solver Andrew Heairet used a tree to work through each case:

Sure enough, there were **17** distinct ways to remove the gloves two, three or four at a time.

Many solvers, meanwhile, used a recursive approach. Suppose you already knew that there were *f*(*N*−4), *f*(*N*−3), *f*(*N*−2) and *f*(*N*−1) ways to remove *N*−4, *N*−3, *N*−2 and *N*−1 gloves, respectively. So then how many were there to remove *N* gloves? There were three distinct ways to reach *N*:

- There were
*f*(*N*−4) ways to remove*N*−4 gloves, after which you could remove four gloves. - There were
*f*(*N*−3) ways to remove*N*−3 gloves, after which you could remove three gloves. - There were
*f*(*N*−2) ways to remove*N*−2 gloves, after which you could remove two gloves.

If you ever found yourself having removed *N*−1 gloves, however, then there was no way to remove *N*, since you couldn’t remove just one glove. That meant *f*(*N*), the number of ways to remove *N* gloves, was equal to *f*(*N*−4) + *f*(*N*−3) + *f*(*N*−2). Before applying this formula, you had to work out a few small cases: There was one way to remove zero gloves (you simply don’t remove any), zero ways to remove one glove, one way to remove two gloves and one way to remove three gloves.

From there, you could use the recursive formula to find that there were two ways to remove four gloves, two ways to remove five gloves, four ways to remove six gloves, five ways to remove seven gloves, eight ways to remove eight gloves, 11 ways to remove nine gloves and, sure enough, 17 ways to remove 10 gloves.

Finally, solver Benjamin Dickman used an advanced technique known as generating functions, looking at expressions like (*x*^{2} + *x*^{3} + *x*^{4})^{5}. If you expand this expression, the coefficient of the *x*^{10} term winds up being the number of ways you can combine *five* twos, threes and fours to get a sum of 10. Similarly, the coefficient of the *x*^{10} term in (*x*^{2} + *x*^{3} + *x*^{4})^{4} tells you how many ways you can combine *four* twos, threes, and fours, and the coefficient of the *x*^{10} term in (*x*^{2} + *x*^{3} + *x*^{4})^{3} tells you how many ways you can combine *three* twos, threes, and fours. Combining these results, Benjamin once again found that there were 17 ways in total.

In the end, I was able to remove the gloves I needed — and I wasn’t stuck with one glove left over.

Congratulations to Kyle Tripp of Concord, CA, winner of last week’s Riddler Classic.

Last week, you started with a fair 6-sided die and rolled it six times, recording the results of each roll. You then wrote these numbers on the six faces of *another*, unlabeled fair die. For example, if your six rolls had been 3, 5, 3, 6, 1 and 2, then your second die wouldn’t have had a 4 on it; instead, it would have two 3s.

Next, you rolled this second die six times. You took those six numbers and wrote them on the faces of *yet another* fair die, and you continued this process of generating a new die from the previous one.

Eventually, you’d have a die with the same number on all six faces. What was the average number of rolls it would take to reach this state?

This puzzle got hairy quickly — after just one round, there were many possibilities to consider: You could still have all six numbers, or you could have five, four, three or two; there was even a small chance — 1 in 6^{5}, or 1 in 7,776 — that you’d roll the same number six times in a row and be done after just one round!

Many readers broke the problem down into cases. For example, if you knew the average number of rounds it would take when you had only *two* numbers left, you could work backwards to figure out how many rounds it would take when you had *three* numbers left. However, not all cases of having two numbers left were the same. While having one number on five faces (e.g., 1, 1, 1, 1, 1 and 2) gave you a whopping 33 percent chance of ending on the next round, having two numbers on three faces each (e.g., 1, 1, 1, 2, 2 and 2) provided a minuscule 3 percent chance of ending the next round. All that is to say — there were many cases to consider here.

Nevertheless, Riddler Nation powered through. Some turned to their computers, running countless simulations to arrive at an approximate answer. Angelos Tzelepis ran 2 million simulations, finding the average number of rounds was approximately 9.66.

Others were able to find an exact answer using Markov chains and transition matrices. Solver Allen Gu found that the average number of rounds was precisely **31,394,023/3,251,248, or about 9.656**. (The original question asked for the number of rolls. Since each round had six rolls, I also accepted answers that were six times larger, or approximately 58.)

Last week’s extra credit went beyond 6-sided dice, asking you to consider *N*-sided dice. Again using Markov chains, Allen identified the average number of rounds up through 10-sided dice. Beyond that, things got … dicey. (Sorry.)

Angela Zhou went even further, demonstrating that the average number of rounds needed for an *N*-sided die was approximately 2*N*:

So if you had a fair 1,000-sided die, it would take a little under 2,000 rounds on average until all the faces were the same number.

If ever there was a time when you expected a linear relationship to appear in a Riddler Classic, this was *not* it.

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!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>These days I always have a pack of latex gloves nearby. But it’s notoriously difficult to pull exactly two gloves out of the box at a time. Sometimes I’ll pull out two gloves, other times three, and yet other times four. Somehow, I never pull out any other number of gloves at a time.

This morning, I noticed that there are 10 gloves left in the box. How many distinct ways are there for me to remove all 10 gloves from the box? Note that the order matters here — for example, pulling out two gloves, then four gloves and then another four gloves is distinct from pulling out four gloves, another four gloves and then two gloves.

From Chris Nho comes a question of rolling (and re-rolling) a die:

You start with a fair 6-sided die and roll it six times, recording the results of each roll. You then write these numbers on the six faces of *another*, unlabeled fair die. For example, if your six rolls were 3, 5, 3, 6, 1 and 2, then your second die wouldn’t have a 4 on it; instead, it would have two 3s.

Next, you roll this second die six times. You take those six numbers and write them on the faces of *yet another* fair die, and you continue this process of generating a new die from the previous one.

Eventually, you’ll have a die with the same number on all six faces. What is the average number of rolls it will take to reach this state?

*Extra credit:* Instead of a standard 6-sided die, suppose you have an *N*-sided die, whose sides are numbered from 1 to *N*. What is the average number of rolls it would take until all *N* sides show the same number?

Congratulations to Lloyd Kvam of Lebanon, New Hampshire, winner of last week’s Riddler Express.

Last week, a manager instructed his team to hold a sale on their widgets every morning, reducing the price by 10 percent. Every afternoon, he had them increase the price by 10 percent from the sale price, with the (incorrect) idea that it would return it to the original price.

After *N* days, the manager walked through the store in the evening, horrified to see that the widgets had been marked more than 50 percent off of their original price. What was the smallest possible value of *N*?

Reducing the price by 10 percent was the same as multiplying it by 9/10, while increasing the price by 10 percent was the same as multiplying it by 11/10. So if the starting price on a given day was *x*, then the sale price was 0.9*x*, while the final price was 1.1(0.9*x*), or 0.99*x*. In other words, the price was in fact *decreasing* by 1 percent every day. Also, recall that the manager had seen that day’s final price from the afternoon, rather than the sale price in the morning.

So the question was essentially asking how many incremental 1 percent decreases resulted in an overall 50 percent decrease — that is, what’s the smallest value of *N* for which 0.99^{N} is less than 0.5?

Some solvers, like Jake Russo, listed or graphed the various prices over time. Meanwhile, solver Sarry Al-Turk took a more direct approach. You wanted to find *N* such that 0.99^{N} < 0.5. Taking the logarithm of both sides gives the inequality log(0.99^{N}) = *N*log(0.99) < log(0.5). Dividing both sides by log(0.99) — a negative number, which means we have to flip our inequality — gives* N* > log(0.5)/log(0.99), which is approximately 68.97. The smallest *N* was therefore **69**.

A few solvers gave the correct answer to a slightly different question: What was the first time the price *ever* dipped below 50 percent of the original price? This happened on the morning of the 60^{th} day of the sale, when the sale price was 0.9(0.99)^{59}, or about 0.497, times the original value.

Anyway, if the manager had better number sense, he would have asked his team to lower the price by 10 percent each morning, and then raise it by 11.11… percent each afternoon. On the bright side, at least he was able to purchase widgets half-off.

Congratulations to Eric Widdison of Kaysville, Utah, winner of last week’s Riddler Classic.

Last week’s Riddler Classic was all about the game SET.

In SET, there are 81 total cards, and each card has four attributes:

*Number:*Each card has one, two or three shapes on it.*Shape:*Each shape on a card is identical, and can be oval, diamond or “squiggle.”*Color:*Each shape on a card is identical in color, which can be red, green or purple.*Shading:*Each shape on a card is identical in shading, which can be solid, shaded or outlined.

Importantly, a “set” (not to be confused with the game’s title) of cards consists of three cards that are either all alike or all different in each attribute; if two of the cards have a common attribute that is not shared by the third, they cannot be a set.

Before we get to the specific questions from last week, it’s worth pointing out that SET* is a very popular game*, and, as a result, these questions have been asked (and answered) before. In particular, I’d like to send a shoutout to Gary Gordon and Liz McMahon, who not only submitted correct solutions but also literally wrote the book on SET’s deeper math puzzles with their daughters.

In solving this puzzle, it was also helpful to first think about the total number of sets there were in a deck of 81 cards. The key here was to realize that, given any two cards in the deck, there was always exactly one card that would complete a set. For any attributes the first two cards shared, the third card in the set would have to be the same, and for any attributes the first two cards didn’t share, the third would have to be different from both. So knowing any two cards in the set told you what the third card had to be.

So to count up all the sets, you could choose any two cards, and there were 81·80/2, or 3,240 ways to do this. However, for any given trio of cards, there were three ways to pick two, meaning we triple counted each set. The grand total number of sets in a deck was 3,240/3, or 1,080 sets.

That’s a lot of sets! Here they are, visualized:

Each of the 81 points represents a card in the deck, and the 1,080 randomly colored triangles connecting the points are the sets.

Solver Laurent Lessard wrote some code to efficiently scour the 81 cards and 1,080 sets, ultimately building up a maximal board of **20 cards** with zero sets among them. There were many ways to pick these 20 cards, and here’s one of them, courtesy of Laurent:

Laurent also found a board of 12 cards with the maximal number of **14 sets**:

Can you find them all?

To find the probability of having at least one set among a random board of 12 cards, most solvers again turned to their computers. Steve Loomis kindly shared his code, finding that this probability was approximately **96.8 percent.**

Without the aid of a computer, this was indeed a very challenging Riddler Classic. For example, proving that the maximum number of cards with no sets (also known as the largest “cap set”) was 20 was only done in 1971, and it involved rather advanced algebraic theory. For more on these relatively recent developments, I suggest checking out this 2016 article from Quanta Magazine, which describes a new upper bound on the size of cap sets.

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!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>A manager is trying to produce sales of his company’s widget, so he instructs his team to hold a sale every morning, lowering the price of the widget by 10 percent. However, he gives very specific instructions as to what should happen in the afternoon: Increase the price by 10 percent from the sale price, with the (incorrect) idea that it would return it to the original price. The team follows his instructions quite literally, lowering and then raising the price by 10 percent every day.

After *N* days, the manager walks through the store in the evening, horrified to see that the widgets are marked more than 50 percent off of their original price. What is the smallest possible value of *N*?

This week’s Riddler Classic features several puzzles, independently submitted by Tyler Barron and Angela Zhou, about the game SET.

In SET, there are 81 total cards, and each card contains has four attributes:

*Number:*Each card has one, two or three shapes on it.

*Shape:*Each shape on a card is identical, and can be oval, diamond or “squiggle.”

*Color:*Each shape on a card is identical in color, which can be red, green or purple.

*Shading:*Each shape on a card is identical in shading, which can be solid, shaded or outlined.

Here is a “board” of 12 such cards:

Importantly, a “set” (not to be confused with the game’s title) of cards consists of three cards that are either all alike or all different in each attribute; if two of the cards have a common attribute that is not shared by the third, they cannot be a set.

For example, in the image above, the single diamond in the left column and the top and bottom cards in the right column form a set. They each have different numbers (one, two and three), different shapes (diamond, squiggle and oval), the same color (red) and the same shading (shaded). If you look carefully, you might also find other sets within this board of 12 cards.

*Question 1:* What is the maximum number of cards you could have (from a single deck of 81 cards) such that there are *no* sets among them?

*Question 2: *What is the largest number of sets one can possibly find among 12 cards? You are free to pick any board of 12 cards you like — your goal is to maximize the number of sets the board contains.

*Question 3:* If you pick 12 cards at random (again, from a single deck of 81 cards), what’s the probability that they contain *at least one* set?

Congratulations to Lazar Ilic of Austin, Texas, winner of last week’s Riddler Express.

Last week, you were looking for powers of 2 that came very close to powers of 10. For example, 2^{10} equals 1,024, which is very close to 1,000, or 10^{3}. After 2^{10}, what was the next (whole number) power of 2 that came even closer to a power of 10? (To be clear, “closer” didn’t refer to the absolute difference — it meant your power of 2 needed to differ from a power of 10 by less than 2.4 percent.)

One way to find the answer was by brute force: calculating powers of 2 and checking how close they came to a power of 10. Solver Eugene Tsai did exactly this, with a lengthy spreadsheet to show for it.

Another approach made use of logarithms. You were looking for whole numbers *A* and *B* such that 2*A* ≈ 10*B*. Taking the base 10 logarithm of both sides and rearranging gave the expression *B*/*A* ≈ log_{10}2. In other words, you were looking for a fraction that approximated log_{10}2, which itself is about 0.301, and you wanted the denominator of your fraction, *A*, to be as small as possible.

So what were some fractions that were good approximations for 0.301? Well, 1/3 came to mind, which meant 2^{3} ≈ 10^{1}. I mean, sure, 8 is pretty close to 10. The next, better approximation was 3/10, which was *quite* close to 0.301. That meant 2^{10} ≈ 10^{3}. However, at this point, we’ve merely caught up to the original puzzle. So what was the next, better fractional approximation for log_{10}2 after 3/10?

Solver Tejas Guruswamy used a technique known as “continued fractions,” which do exactly what we want — they tell you the best fractional approximations for different numbers, including irrational numbers like log_{10}2. Meanwhile, solvers Matthew Miller and Dennis Okon both noted that there’s an OEIS sequence for what we’re looking for. (Isn’t there always?) Any which way, the next approximation for log_{10}2 was 28/93, which meant 2^{93} ≈ 10^{28}. In fact, 2^{93} is approximately 0.9904 × 10^{28}, only 0.96 percent less than 10^{28} — and so **2 ^{93}** was the correct solution.

Many solvers offered an answer of 2^{103}, which was 1.4 percent greater than 10^{31}. However, the original problem didn’t specify that the power of 2 had to be *greater* than its corresponding power of 10, so 2^{103} was not the correct answer.

A final approach worth mentioning was to graphically analyze the problem using polar coordinates, inspired by 3Blue1Brown’s related video on prime numbers.

In the animation above, each point represents a power of 2. The point’s distance from the origin is its power of 2 (so, for example, 2^{7} is a distance of 7 from the origin). The point’s angle, meanwhile, indicates how close that power of 2 comes to a power of 10, with an angle of zero degrees meaning it’s an exact power of 10. Here, we’re interested in the points that came closest to the zero-degree axis, labeled in red. Reading them off, they were: 3, 10, 93, 196, 485, 2,136, 13,301, 28,738, 42,039, and 70,777. These were exactly the powers of 2 that came progressively closer to powers of 10 in the aforementioned OEIS sequence!

In case you were curious, 2^{70,777} equals 1.00000716 × 10^{21,306}. Pretty darn close to a power of 10!

Congratulations to Donald M. of Dallas, Texas, winner of last week’s Riddler Classic.

Last week, you paid another visit to Tiffany’s barbershop. But this time, the riddle was stated slightly less ambiguously than it had been during your first visit.

This time, all of Riddler City decided to get a haircut at Tiffany’s, and everyone lined up at the entrance in a random order a few minutes before the shop opened at 8 a.m. After opening, the shop’s four barbers started cutting hair for their first customers at random times between 8 a.m. and 8:15 a.m. Each haircut then lasted exactly 15 minutes.

Also, each person in line had a 25 percent of preferring Tiffany. Whenever such a person was at the front of the line, if a barber *other *than Tiffany became available, they’d allow the person behind them to skip them in line … unless, of course, that person *also* preferred Tiffany, in which case the third person in line would skip both of them … unless, of course that person *also* *also* preferred Tiffany — you get the idea.

Sadly, you found yourself toward the back of this very, very long line. To pass the time while you waited, you spent a long time thinking about last week’s Riddler column, completely unaware of the passage of time. The next thing you knew, you were second in line, with one person waiting in front of you. At this point, how long should you expect to wait for your haircut from Tiffany?

Now, you might have thought that the person in front of you had a 25 percent chance of preferring Tiffany. Surprisingly, that wasn’t right. Over time, as more people got their hair cut, a backlog of Tiffany fans built up at the front of the line, while the non-Tiffany customers were weeded out.

Think of the line of customers as a string of *T*s (the Tiffany fans) and *U*s (the non-Tiffany customers). Over the course of each 15-minute cycle, Tiffany and the other three barbers each took turns picking a customer — Tiffany took the next in line, regardless of whether that person was a *T* or a *U*, while the other three barbers took the next available *U*. After a few cycles, all the early *U*s would have been taken by the other three barbers, leaving a whole bunch of *T*s at the head of the line. Over time, the number of *T*s at the front continued to grow with the total number of customers. (The average number of leading *T*s after *N *15-minute cycles of four haircuts looked suspiciously like the square root of *N*. But asking you to prove that would have been worthy of yet another Riddler!)

That meant, as solver Jason Shaw verified with code, that as the number of customers in front of you went to infinity, the probability the person in front of you was also a *T* approached 1. Since Riddler City had a very, very large population, it was safe to assume that the person in front of you was *definitely* a Tiffany fan. Thus, you had to wait an average of 7.5 minutes for Tiffany to finish her current haircut, and then another 15 minutes for the person in front of you, for a grand total of **22.5 minutes**.

An alternate, even more challenging interpretation of the problem some readers had was that, after your long wait outside the barbershop, you found yourself second in line, with *all four barbers currently *cutting hair. This was indeed a different scenario, since after a long time you would have expected the other three barbers to have serviced all the *U*s in the line, leaving nothing but *T*s (i.e., only Tiffany was left cutting hair).

Needless to say, Tiffany and her team of barbers clearly specialize in mathematical extensions. (See what I did there?)

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!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>Most people will tell you that a kilobyte is 1,000 bytes, that a megabyte is 1,000,000 bytes and that a gigabyte is 1,000,000,000 bytes. But that’s not quite right — a kilobyte is in fact 1,024, or 2^{10}, bytes. A megabyte is 1,048,576, or 2^{20}, bytes. A gigabyte is 1,073,741,824, or 2^{30}, bytes.

While these aren’t whole number powers of 10, they’re all pretty close. And that’s thanks to the happy coincidence that there’s a power of two that’s very close to 1,000. Working through the numbers, 2^{10} is a mere 2.4 percent more than 10^{3}.

But surely there are other, higher powers of 2 that are even *closer* to a power of 10. After 2^{10}, what’s the next (whole number) power of 2 that comes closer to a power of 10? (To be clear, “closer” doesn’t refer to the absolute difference — it means your power of 2 should differ from a power of 10 by less than 2.4 percent.)

This week’s Riddler Classic is a new take on a recent puzzle. Two weeks ago, you were waiting in line at a barbershop. There were four barbers working simultaneously, and each haircut took exactly 15 minutes. There were almost always one or more customers waiting their turn on a first-come, first-served basis.

Being a regular, you preferred to get your hair cut by the owner, Tiffany. If one of the other three chairs opened up, and it was your turn, you would have said, “No thanks, I’m waiting for Tiffany.” The person behind you in line would then be offered the open chair, and you’d remain at the front of the line until Tiffany became available.

Unfortunately, you weren’t alone in requesting Tiffany. In general, a quarter of the other customers were expected to hold out for Tiffany, while no one held out for any of the other barbers. The question was then, if you had one person in line ahead of you, and the four barbers were independently at random places in their respective haircuts, how long would you expect to wait until your haircut?

While last week’s solution explored some interesting mathematics, it made a critical assumption: that the person in front of you had a 25 percent chance of waiting for Tiffany. But as it turns out, this was not a reasonable deduction to make.^{5} And so, for this week’s Classic, you are being asked to analyze a more specific statement of the original problem.

Suppose all of Riddler City, with its incredibly vast population (25 percent of whom will wait for Tiffany), decides to get a haircut at this barbershop one fine morning, and everyone lines up at its entrance in a random order a few minutes before the shop opens at 8 a.m. After opening, the four barbers will start cutting hair for their first customers at random times between 8 a.m. and 8:15 a.m. Each haircut then lasts exactly 15 minutes.

Sadly, you find yourself toward the back of this very, very long line. To pass the time while you wait, you spend a long time thinking about this week’s Riddler column, completely unaware of the passage of time. The next thing you know, you’re second in line, with one person waiting in front of you — the exact conditions from the original puzzle. At this point, how long should you expect to wait for your haircut from Tiffany?

(Hint: Think about the probability that the person in front of you will request Tiffany. Is it still 25 percent? Also, keep in mind that if there are *multiple* Tiffany requesters at the front of the line, and a barber other than Tiffany becomes available, the next non-Tiffany requester will effectively jump the line.)

Congratulations to Veronica Pillar of Ithaca, NY, winner of last week’s Riddler Express.

Last week, you had a tic-tac-toe board with nine pieces you could place: five Xs and four Os. If you randomly placed all nine pieces in the nine slots on the board (with one piece in each slot), what was the probability that X won? That is, what was the probability that there would be at least one occurrence of three Xs in a row at the same time there were no occurrences of three Os in a row?

A good place to start was to figure out how many possible arrangements there were. If there are nine total slots, and you’re choosing where to place the five Xs (or, alternatively, the four Os), then there are 9 choose 4, or (9·8·7·6)/(4·3·2·1) = 126, possible arrangements. So among these 126 total ways to place the Xs and Os, how many resulted in three Xs in a row but not three Os in a row?

There were several different strategies for counting these up. One way was to look at different cases, depending on where the three winning Xs showed up in the grid.

- If the three winning Xs were arranged horizontally in a row, then as long as each of the other two horizontal rows had one X, there wouldn’t be three winning Os. There were three possible rows for the winning Xs, three possible columns for one of the Xs in another row and three possible columns for the X in the remaining row. This accounted for 3·3·3 = 27 winning arrangements.
- If the three winning Xs in a were arranged vertically in a column, then as long as each of the other two vertical columns had one X, there wouldn’t be three winning Os. Once again, this accounted for 3·3·3 = 27 winning arrangements.
- If the three Xs were arranged diagonally, then there was no way for there to also be three winning Os. There were two possible diagonals for the winning Xs, and then 6 choose 2, or 15 possible ways to arrange the remaining two Xs. That accounted for another 30 winning arrangements.

It would seem that there were a total of 27+27+30 = 84 winning arrangements. However, there was one slight flaw in the above reasoning — some winning arrangements were counted twice!

For example, it’s possible to arrange the five Xs so that they occupy *both* diagonals of the grid. This arrangement was counted once for one of the diagonals, and once again for the other diagonal. It was also possible for the five Xs to occupy both a winning row and a winning column at the same time, another arrangement that was double-counted — once for the row, and again for the column. Since there were three rows and three columns, there were 3·3, or nine arrangements like this. Finally, it was possible to have five Xs occupy both a winning diagonal as well as either a winning row or a winning column. Since there were two diagonals and six total rows and columns, there were 2·6, or twelve arrangements like this.

At the end of the day, there appeared to be 84 winning arrangements, but 1+9+12 = 22 were double-counted. That meant the actual number of winning arrangements was instead 84−22, or 62. And so the probability of placing the pieces in a winning arrangement was 62/126, or **31/63** — just a shade under 50 percent.

Solver Julian Gerez found the answer via computer simulation, and even extended the puzzle, finding the chances of having two sets of winning Xs and simultaneous winning sets of Xs and Os.

Jeff Harrison, meanwhile, took a more traditional approach. He sketched out all the arrangements:

Who needs a computer when you have graph paper?

Congratulations to Dan Rose of Hopkins, Minnesota, winner of last week’s Riddler Classic.

Last week, you looked at a modified version of the game “Guess Who,” in which each player first randomly (and independently of their opponent) selected one of *N* character tiles. (It was possible, albeit unlikely, that both players chose the same character.) Each of the four characters was distinct in appearance — for example, characters had different skin tones, hair color, hair length and accessories like hats or glasses.

Each player also had access to a board with images of all *N* characters. The players alternated taking turns, and during each turn a player had two options:

- Make a specific guess as to their opponent’s selected character. If correct, the player who made the guess immediately won. Otherwise, that player immediately lost.
- Ask a yes-or-no question about their opponent’s chosen character, in order to eliminate some of the candidates. Importantly, if only one possible character was left after the question, the player still had to wait until their
*next*turn to officially guess that character.

Both players were assumed to be highly skilled at choosing yes-or-no questions, so that they could always craft a question that identified any subset of characters, who would then be ruled in or out. Also, both were playing to maximize their own probability of winning.

When *N* was four, how likely was it that the player who went first would win?

You might have thought that splitting the candidates in half was the way to go. While that’s not a bad strategy, was it really the *best* strategy?

To find out, you could play a mock game in which both players use this strategy. First, Player 1 would ask a yes-or-no question that resulted in a “yes” for two of the candidates and a “no” for the other two. Either way, Player 1 would have narrowed the field from four candidates to two. Next, Player 2 would have done the same.

At this point, Player 1 could guess Player 2’s selected character for a 50 percent chance at victory. Alternatively, Player 1 could have asked a question to narrow down the field to a single candidate. Then, to have any chance at winning, Player 2 would have to guess among their two remaining candidates. So Player 1’s chances of winning stand at 50 percent.

Surprisingly, Player 1 could improve their chances of winning by initially asking a question that would result in a “yes” for either one or three candidates. There would be a 25 percent chance of narrowing it down to a single candidate, forcing Player 2 to take a wild guess. In this case, Player 1 would win 75 percent of the time.

The other 75 percent of the time, when Player 1 was left with three candidates, keeping track of all the possibilities got pretty hairy. But it turned out that Player 1 *still* had a 50 percent of winning.

Putting it all together, 25 percent of the time Player 1 had a 75 percent chance of winning, while the remaining 75 percent of the time Player 1 had a 50 percent chance of winning. Player 1 would therefore win **9/16** of the time, or more than half the time. By splitting the field asymmetrically (i.e., not in half) Player 1 could better take advantage of having the first turn, and achieve better than even odds of winning.

While the case of *N* = 4 was certainly complicated, the extra credit problems (*N* = 24 and *N* = 14) were even more so. Fortunately, solvers turned to their computers for help, using either the technique of recursion (as written up by Liam Kirwin and Robert Sturrock) or dynamic programming. When *N* = 24, you might have thought splitting the field in half would be optimal, asking a question that would result in a “yes” for 12 candidates and a “no” for the other 12. However, the puzzle’s submitter, Andrew Lin, found via that as long as that first question generates a “yes” for between 8 and 16 candidates, Player 1 can still maximize their chances of winning, which turned out to be **5/9**. And in the case of *N* = 14, Player 1 could ask a question that generated a “yes” four, six, eight or 10 times (only the evens!), achieving a winning probability of **55/98**.

The extra credit may have stopped there, but Riddler Nation did not. Something was afoot — for example, notice how all those answers were slightly more than 50 percent? Will that always be the case for Player 1 for different values of *N*? That turns out to be the case, as shown by Angela Zhou:

Even as the number of characters get’s absurdly large (into the thousands), Player 1’s chances of winning take on a repeating pattern, bouncing between 55.5 and 56.3 percent.

Angela and Robert Spillner together also studied the optimal number of “yes” answers a player should seek, depending on how many candidates each player had left. The resulting graph is a thing of beauty:

It’s clear that the winning strategy for “Guess Who” goes far beyond splitting the field in half with every question. All that’s left is for someone to encode an AI that plays with this optimal strategy. I, for one, am ready to lose!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>A local cafe has board games on a shelf, designed to keep kids (and some adults) entertained while they wait on their food. One of the games is a tic-tac-toe board, which comes with nine pieces that you and your opponent can place: five Xs and four Os.

When I took my two-year-old with me, he wasn’t particularly interested in the game itself, but rather in the placement of the pieces.

If he randomly places all nine pieces in the nine slots on the tic-tac-toe board (with one piece in each slot), what’s the probability that X wins? That is, what’s the probability that there will be at least one occurrence of three Xs in a row at the same time there are no occurrences of three Os in a row?

Sticking with the board game theme, from Andrew Lin comes a closer examination of a classic game of reasoning and elimination:

In the game of “Guess Who,” each player first randomly (and independently of their opponent) selects one of *N* character tiles. While it’s unlikely, both players can choose the same character. Each of the *N* characters is distinct in appearance — for example, characters have different skin tones, hair color, hair length and accessories like hats or glasses.

Each player also has access to a board with images of all *N* characters. The players alternate taking turns, and during each turn a player has two options:

- Make a specific guess as to their opponent’s selected character. If correct, the player who made the guess immediately wins. Otherwise, that player immediately loses.
- Ask a yes-or-no question about their opponent’s chosen character, in order to eliminate some of the candidates. Importantly, if only one possible character is left after the question, the player must still wait until their
*next*turn to officially guess that character.

Assume both players are highly skilled at choosing yes-or-no questions, so that they can always craft a question to potentially rule out (or in) any desired number of candidates. Also, both are playing to maximize their own probability of winning.

Let’s keep things (relatively) simple, and suppose that *N* = 4. How likely is it that the player who goes first will win?

*Extra credit: *If *N* is instead 24 (the number of characters in the original “Guess Who” game), *now* how likely is it that the player who goes first will win?

*Extra extra credit:* If *N* is instead 14, *now* how likely is it that the player who goes first will win?

Congratulations to Greg Ebersole of Strongsville, Ohio, winner of last week’s Riddler Express.

Last week, you were in a dark room and seated at a table with a deck of 52 cards stacked into a single pile. You also happened to know that exactly 13 cards in the pile were face up, while the others were all face down.

While you couldn’t see the cards, you could feel them and move them around — but you couldn’t tell by touch which were face up and which were face down.

How could you make two piles of cards that were guaranteed to have the same number of face-up cards?

While this was not an original puzzle, its cleverness made it worth sharing here.

Some solvers, like Michael Campbell, interpreted the question as asking you to make a larger number of piles, such that two of them had the same number of face-up cards. In other words, you didn’t have to *know* which two piles had the same number of face-up cards, just as long as two such piles existed.

With this interpretation, there are many possible solutions, one of which is to take three cards off the top of the deck and make each one its own pile. By the pigeonhole principle, at least two of these “piles” (of a single card) must have the same number of face-up cards, whether that’s zero or one.

But honestly, that wasn’t a very interesting interpretation of the problem. Instead, what if you had to split all 52 cards into exactly two piles? Could you somehow guarantee that the two piles had the same number of face-up cards?

Surprisingly, the answer turned out to be yes. And the key was the fact that you knew 13 cards in the deck were face-up to begin with.

A winning strategy was to **take any 13 cards, move them to a separate pile, and flip that pile over**. Let’s check this: Suppose that, among the 13 cards you took, *x* cards were face-up. The meant that the remaining 13−*x* face-up cards were left in the pile of 39 cards. So at this point, you have a pile of 39 cards, of which 13−*x* are face-up, and a pile of 13 cards, of which *x* are face-up.

And now for the pièce de résistance — if you flipped that pile of 13 cards over, how many cards would now be face-up? Well, right before the flip, *x* cards were face-up, and the other 13−*x* cards were face-down. So after the flip, those two numbers were swapped: *x* cards were face-down, and 13−*x* cards were face up. At this point, *both* the pile with 39 cards and the pile with 13 cards each had 13−*x* face-up cards. You had no way of knowing what *x* was, and so you didn’t know *exactly* how many cards were face-up in each pile. But you *did* know that the two piles had the same number of face-up cards.

Perhaps the coolest thing about this problem was that if you were willing to give up on the certainty you expected to have — the precise number of face-up cards in each pile — you could find the certainty the problem was asking for. Good stuff.

Congratulations to Q P Liu of Santa Cruz, California, winner of last week’s Riddler Classic.

Last week, you were at your local barbershop, where there were four barbers working simultaneously. Each haircut took exactly 15 minutes, and there were almost always one or more customers waiting their turn on a first-come, first-served basis.

Being a regular, you preferred to get your hair cut by the owner, Tiffany. If one of the other three chairs opened up, and it was your turn, you would have said, “No thanks, I’m waiting for Tiffany.” The person behind you in line would then be offered the open chair, and you’d remain at the front of the line until Tiffany became available.

Unfortunately, you weren’t alone in requesting Tiffany — a quarter of the other customers held out for Tiffany, while no one held out for any of the other barbers.

One Friday morning, you arrived at the barber shop to see that all four barbers were cutting hair, and there was one customer waiting. You had no idea how far along any of the barbers were in their haircuts, and you didn’t know whether or not the customer in line would hold out for Tiffany.

What was your expected wait time for getting a haircut from Tiffany?

There were two cases to consider here: Either the customer in front of you was also waiting for Tiffany, with probability 1/4 (as stated in the problem), or they weren’t, with probability 3/4. If they *were* waiting for Tiffany, then your wait time was simply a matter of waiting out the current customer in her chair (which could take anywhere from zero to 15 minutes, for an average of 7.5 minutes), followed by that first customer in line (another 15 minutes). So one-fourth of the time, your average wait was 22.5 minutes.

But what about the other three-fourths of the time? Well, if Tiffany was the first of the four barbers to finish, then you were stuck waiting however long that took, plus *another* 15 minutes of the customer in line in front of you. But if Tiffany wasn’t the first to finish, then the customer in front of you would select a different barber, and as soon as Tiffany freed up you’d be getting your haircut.

Solver Benjamin Phillabaum reasoned (with some calculus that I’ll skip here) that if there were four barbers who were at random places within their respective 15-minute haircuts, on average, one barber would be done after three minutes, two barbers would be done after six minutes, three barbers would be done after nine minutes, and all four barbers would be done after 12 minutes. That meant that when Tiffany was the first to finish — which happened one-fourth of the time — you’d wait an average of 18 minutes (the three minutes it took her to finish, plus 15 minutes for the person in line ahead of you). And when Tiffany was not the first to finish — which happened the remaining three-fourths of the time — you’d wait an average of either six, nine or 12 minutes, which together average to nine minutes.

So when all was said and done, there was a one-in-four chance of waiting 22.5 minutes, a three-in-16 chance of waiting 18 minutes, and a nine-in-16 chance of waiting nine minutes. Rolling all these up meant, on average, you’d wait **14.0625 minutes**.

Many solvers, like this week’s winner, Q P Liu and Stephen Penrice, devised creative algebraic approaches that didn’t use any calculus.

Others, meanwhile, investigated the *distribution* of wait times. How likely were you to wait one minute, two minutes, three minutes, etc.? Allen Gu found that you could wait anywhere from zero to 30 minutes, and the resulting distribution had a rather unusual shape:

Quoc Tran and Angela Zhou ran with this idea, looking at how this distribution changed depending on the number of barbers.

That is one funky-looking probability distribution!

And if that wasn’t enough, Angela further extended the problem by looking for the *median* length of time you’d be waiting for your haircut, rather than the average, when there were four barbers. Here’s the exact expression she found:

This worked out to about 13.75 minutes.

Anyway, if you’re bummed about how long you’ll have to wait for Tiffany, just think about all the riddles you could solve in 14.0625 minutes.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>From Blue Taylor comes a puzzle of stacking cards that he has seen making the rounds on the internet (to avoid spoilers, I’ll reference original sources of the puzzle in next week’s column):

You’re in a dark room and seated at a table with a deck of 52 cards on it, stacked into a single pile. You also happen to know that currently, within that stack of cards, exactly 13 of them are face up, while all the others are face down.

While you can’t see the cards, you can feel them and move them around. But you can’t tell by touch which cards are face up and which are face down.

How can you make two piles of cards that are guaranteed to have the same number of face-up cards? (And yes, each of these two piles must have at least one card.)

From Dave Moran comes a question we’ve all faced at some point when waiting in line for a haircut:

At your local barbershop, there are always four barbers working simultaneously. Each haircut takes exactly 15 minutes, and there’s almost always one or more customers waiting their turn on a first-come, first-served basis.

Being a regular, you prefer to get your hair cut by the owner, Tiffany. If one of the other three chairs opens up, and it’s your turn, you’ll say, “No thanks, I’m waiting for Tiffany.” The person behind you in line will then be offered the open chair, and you’ll remain at the front of the line until Tiffany is available.

Unfortunately, you’re not alone in requesting Tiffany — a quarter of the other customers will hold out for Tiffany, while no one will hold out for any of the other barbers.

One Friday morning, you arrive at the barber shop to see that all four barbers are cutting hair, and there is one customer waiting. You have no idea how far along any of the barbers is in their haircuts, and you don’t know whether or not the customer in line will hold out for Tiffany.

What is the expected wait time for getting a haircut from Tiffany?

Congratulations to Marc Lachance of Munich, Germany, winner of last week’s Riddler Express.

Last week, it was a warm, sunny day. But when you toggled between the Fahrenheit and Celsius scales on your thermometer, you noticed something interesting: The digits of the temperature — when rounded to the nearest degree — had switched. For example, this would have worked for a temperature of 61 degrees Fahrenheit, which corresponds to a temperature of 16 degrees Celsius.

However, the temperature on that day was not 61 degrees Fahrenheit. What *was* the temperature?

Given the “warm, sunny day” description in the puzzle, it was safe to assume that the Fahrenheit temperature was a two-digit whole number (more on that later). One way to attack this puzzle was to then suppose the Fahrenheit temperature was *AB* — that is, it had the digit *A* in the tens place and the digit *B* in the ones place. Mathematically, that meant the value of the temperature was 10*A*+*B*.

As you may recall from science class, a Fahrenheit temperature *F* corresponds to a Celsius temperature *C* = 5/9(*F*−32). (If it’s been a while since you’ve seen this formula, you can check it with water’s freezing point, 0 C and 32 F, as well as its boiling point, 100 C and 212 F.) Plugging our Fahrenheit temperature, 10*A*+*B*, into this formula gave a Celsius temperature of 50/9*A*+5/9*B*−160/9.

In order for this new expression to be the same as the original temperature, but with its digits reversed, it needed to have a *B* in the tens place and an *A* in the ones place, meaning it had to equal 10*B*+*A*. In other words, you had to find solutions to the equation 50/9*A*+5/9*B*−160/9 = 10*B*+*A*. After some rearranging, this equation became *A* = (85*B*+160)/41.

At this point, you had to plug the digits from 0 to 9 in for *B*, and round the corresponding values of *A* in the above equation.

- When
*B*was 0,*A*rounded to 4. That means 40 F is equivalent to 04 C. However, there were two problems with this solution: It’s unclear whether the digits actually “flipped” here (04°C should really be written without the leading zero, as 4 C), and 40 F would*not*be the temperature of a “warm sunny day.” - When
*B*was 1,*A*rounded to 6. That means 61 F is equivalent to 16 C. But this was the temperature given in the original problem. - When
*B*was 2,*A*rounded to 8. That means 82 F is equivalent to 28 C. This looks like a winner… - For values of
*B*greater than 2, the corresponding value of*A*exceeded 10, meaning it could not have been a single digit.

That meant the temperature had to be **82 degrees Fahrenheit, or 28 degrees Celsius**.

While this algebraic approach was all well and good, many solvers turned to their computers. That included a first-time solver in 8-year-old Aaron Vaughn, who found the answer using a Python script.

Welcome to Riddler Nation, Aaron!

Some solvers had even more fun with this puzzle, extending it beyond two digits. They searched for three-, four- and even five-digit temperatures such that the digits were reversed when written in Fahrenheit and Celsius. Solver Jan Hattendorf identified the next few temperatures that worked after 82 F:

- 4,862 F, which equals 2,684 C
- 5,082 F, which equals 2,805 C
- 96,635 F, which equals 53,669 C

Angela Zhou found an even higher temperature — an absurdly hot 9,663,413,658,635 F, which equals 5,368,563,143,669 C.

While the problem was slightly ambiguous, it was fair to assume that the “warm, sunny day” was in fact on the Earth’s surface, and so temperatures like 96,635 F were not the expected answer. But if any members of Riddler Nation solve these puzzles in or around the sun’s corona, I’d love to hear about it.

Congratulations to Tom Bassine of Farmington, Connecticut, winner of last week’s Riddler Classic.

Last week, you played a game with two fair coins, labeled A and B. When you flipped coin A, you got 1 point if it came up heads, but you lost 1 point if it came up tails. Coin B was worth twice as much — when you flipped coin B, you got 2 points if it comes up heads, but you lost 2 points if it came up tails.

To play the game, you made a total of 100 flips. For each flip, you chose either coin, and you knew the outcomes of all the previous flips. In order to win, you had to finish with a *positive* total score. (Finishing with 2 points was just as good as finishing with 200 points, and finishing with 0 or −2 points was just as bad as finishing with −200 points.)

If you optimized your strategy, what percentage of games would you win?

First, a shoutout to Angela Zhou, who noticed the similarity between this puzzle and a 2017 Riddler Classic from Guy Moore about miniature football. For long-standing members of Riddler Nation, this Classic was like greeting an old friend.

One way to solve this puzzle was with dynamic programming — that is, recursively breaking the problem up into smaller problems, solving those and putting them back together. For example, if you were on your 100th and final coin flip, should you flip A or B? If your points were well into positive or negative territory, it didn’t make a difference.

But what if your score was close to zero? If you had −1 points, then your only chance of finishing with a positive score and winning was to flip coin B (worth 2 points). If you had 0 or 1 points, it again didn’t make a difference which coin you picked, as your chances of winning were 50 percent. And if you had 2 points, you could guarantee a win if you played it safe, flipping coin A (worth 1 point).

From there, you could work your way back to your 99th, penultimate flip, knowing full well what your chances of winning would be after the flip (e.g., if you had 2 points after your 99th flip, you’d still be guaranteed a victory). And then back further to your 98th coin flip, and so on. Eventually, you had a map of your chances of winning as a function of your current score, and how many flips you had left, much like the one generated by solver Jason Ash:

When you worked your way back to your very first coin flip, you found that your chances of winning this game stood at about **64 percent**.

At first, it might have seemed counterintuitive that your chances of winning could be anything other than 50 percent. After all, each coin flip had an equal chance of increasing or decreasing your score by the same amount. On average, you’d expect to finish the game with zero points. So then how could you possibly expect to win more games than you lost?

The answer had to do with the optimal strategy. While dynamic programming gave the correct answer, you needed further analysis to see what was really going on. It turned out that you should only have flipped coin A when your score was positive. It’s analogous to basketball strategy in the fourth quarter of a close game: When you’re ahead, you want to milk the clock, but when you’re behind, you want to speed up the pace of play.^{8}

When your score was positive, you’d flip coin A, so that your score wouldn’t move around as quickly, and you’d make it to the end of your 100 flips with your lead intact. But if your score was negative, you wanted to score as many points as you could to give yourself a chance of winning at all. In other words, you’d flip coin B.

In the end, you were more likely to have a small, positive score, and less likely to have a larger, negative score. Your average final score still worked out to be zero, but your chances of winning did indeed exceed 50 percent.

After finding the answer, many solvers extended the problem in new and fascinating directions. Josh Silverman wondered what would happen if, instead of 100 coin flips, the game called for 1,000 flips, or 1,000,000 flips. As the number of coin flips approached infinity, what happened to your chances of winning the game? Shockingly, the answer turned out to be 2/3. To see an explanation for this, check out Josh’s writeup.

Solver Laurent Lessard also took on the problem’s extra credit, which asked what would happen to your chances of winning when coin A wasn’t fair. When coin A always came up tails (with probability 1), then your best bet was to stick with coin B until coin A guaranteed victory. And when coin A always came up heads, then you could guarantee a victory by only flipping coin A. In between, the math got hairier, resulting the the following nonlinear function:

When coin A was fair, you can see your chances of winning were close to 2/3. But if coin A strayed slightly from fairness in either direction, your chances of winning would rapidly rise or fall.

Laurent even looked at when *both* coins were unfair:

Anyway, if this sort of riddle ever makes a *third* appearance in this column, it will likely require an Elam Ending to make the endgame strategy more interesting.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

**CORRECTION (Feb. 28, 1:02 p.m.):** An earlier version of the Riddler Classic solution said that after coin A comes up tails, you want to stick with coin B for the entire game. But the correct solution sticks with coin B for most of the game, only until coin A guarantees a victory.

From Nick Harper comes a question of tempered temperatures:

On a warm, sunny day, Nick glanced at a thermometer, and noticed something quite interesting. When he toggled between the Fahrenheit and Celsius scales, the digits of the temperature — when rounded to the nearest degree — had switched. For example, this works for a temperature of 61 degrees Fahrenheit, which corresponds to a temperature of 16 degrees Celsius.

However, the temperature that day was not 61 degrees Fahrenheit. What *was* the temperature?

From Abijith Krishnan comes a game of coin flipping madness:

You have two fair coins, labeled A and B. When you flip coin A, you get 1 point if it comes up heads, but you lose 1 point if it comes up tails. Coin B is worth twice as much — when you flip coin B, you get 2 points if it comes up heads, but you lose 2 points if it comes up tails.

To play the game, you make a total of 100 flips. For each flip, you can choose either coin, and you know the outcomes of all the previous flips. In order to win, you must finish with a positive total score. In your eyes, finishing with 2 points is just as good as finishing with 200 points — any positive score is a win. (By the same token, finishing with 0 or −2 points is just as bad as finishing with −200 points.)

If you optimize your strategy, what percentage of games will you win? (Remember, one game consists of 100 coin flips.)

*Extra credit:* What if coin A isn’t fair (but coin B is still fair)? That is, if coin A comes up heads with probability *p* and you optimize your strategy, what percentage of games will you win?

Congratulations to Steve Schaefer of Carlsbad, California, winner of last week’s Riddler Express.

Last week, you undertook an urban planning challenge in Riddler City, which was a large, circular metropolis, with countless square city blocks that each had a side length of 1 km. At the very center of the city was Riddler City Hall, whose many employees all walked to and from work, and their homes were evenly scattered across the city. The sidewalks they walked along had always been adjacent to the streets.

But recently, several employees requested that the sidewalks instead cut diagonally across the city blocks, connecting nearby street intersections. These were represented by the thicker blue lines in the diagram below, which showed a *small section* (some readers missed this!) of the city:

What fraction of City Hall employees would have had a shorter walk home (that is, to the street intersection nearest to their home) if the city replaced its traditional sidewalks with these diagonal sidewalks?

First off, many readers pointed out that for employees whose front doors were on a street and halfway down a block, these new sidewalks wouldn’t even get them to their front door. Indeed, last week’s puzzle was ambiguous about exactly *how* an employee would enter their home once they had reached their block. Because it was stated that Riddler City was very, very large, you could safely assume that virtually all of each employee’s commute was spent just trying to reach their block. They’d then get into their home some way or another.

Okay, back to the problem. Most solvers used a coordinate grid in thinking about their answer, where City Hall was located at the point (0, 0), and an employee’s home (i.e., the intersection nearest their home) was located at (*x*, *y*). Here, *x* represented how many blocks east of City Hall their home was located, and *y* represented the number of blocks north of City Hall. So if *x* was negative, then the employee lived *west* of City Hall; if *y* was negative, the employee lived south of City Hall.

Under the original system of sidewalks, the employee would have had to walk a total of |*x*| blocks east or west, and a total of |*y*| blocks north or south. (We took the absolute values of *x* and *y* so that the number of blocks walked would be positive — you can’t walk a negative number of blocks!)

But what about when the sidewalks were diagonal, as the puzzle stated? It helps to look at a specific point close to City Hall, like (5, 3), as illustrated in the diagram below.

There are several ways to travel from City Hall to the home at (5, 3) using the diagonal sidewalks, and one such way is shown in red. First, you’d walk northeast to the point (4, 4), a distance of 4√2, and from there you’d walk an additional √2 southeast to reach the point (5, 3). In general, to reach the point (*x*, *y*), the employee would have to walk a distance of |x+y|/√2 northeast (or southwest), and then a distance |x−y|/√2 southeast (or northwest).

So when was this new commute shorter than the original? As solver Kimberly Powell noted, the original sidewalks resulted in shorter commutes for homes closer to the *x* and *y* axes, while the diagonal sidewalks resulted in shorter commutes to homes closer the lines *y* = *x* and *y* = −*x* — that is, along lines that made 45 degree angles with the axes.

With a little trigonometry, solver Mike Strong found that Riddler City was precisely carved in half. That is to say, **half** the employees would have a shorter commute with the new sidewalks.

If you’re still not convinced, here’s an animation showing the original and diagonal sidewalk paths to each block. When it’s a shorter walk via the original sidewalks, the intersection is shown as a gray dot; when it’s faster via diagonal sidewalks, the intersection is a blue dot. Sure enough, the diagonal sidewalks resulted in a shorter commute for half the blocks.

So when it came to diagonal sidewalks, the employees were evenly split. No wonder the measure didn’t pass.

Congratulations to Mark Jackson of Rochester, New York, winner of last week’s Riddler Classic.

Moving beyond diagonal sidewalks, you were next asked to consider octagonal sidewalks (similar to those found in Barcelona), as shown in the diagram below.

If the city replaced its traditional sidewalks with these octagonal sidewalks, *now *what fraction of employees had a shorter walk home?

Once again, many readers pointed out that some front doors might be inaccessible via these new sidewalks, or even that entire homes might have been bulldozed to make way for these rhomboidal intersections! And once again, these concerns could be treated as negligible due to the vastness of Riddler City.

To solve this puzzle, it was helpful to have a bathroom floor with octagonal tiles.

If you didn’t have one of those, it was also helpful to start with a diagram of an employee who just happened to live very close to City Hall, which we said could be represented by the point (0, 0). For example, consider an employee who lived near the intersection (5, 3):

The red path represents this employee’s shortest path home. How long is it? Well, they started at (0, 0) in the lower left corner and first walked up toward (3, 3), represented by the black point. During this part of the trip, walking along the diagonal sidewalks *shaved off* some distance compared to the original sidewalks. For homes like this, where *x* and *y* were positive and *x* > *y*, the distance saved turned out to be 2*yd*(√2−1), where *d* was the length of the short diagonal segments on each block.

Upon reaching (3, 3), the path straightened out, heading due east toward home. But not *exactly* due east, thanks to the diagonal sidewalks at the intersections. During this part of the trip, the diagonals *added* some distance, since it would have been a straight line to home with the original sidewalks. Again, for homes where *x* and *y* were positive and *x* > *y*, the additional distance turns out to be (*x*−*y*)*d*(2−√2).

Thus, the problem became a balancing act: When did the distance saved exceed the distance gained? When this happened, an employee would prefer the octagonal sidewalks; otherwise, they’d prefer the original grid.

Combining these shortening and lengthening effects meant the octagonal sidewalks wouldn’t affect the commute when 2*yd*(√2−1) — the reduction in distance — *equaled* (*x*−*y*)d(2−√2) — the additional distance. After rearranging some equations (amazingly, the length of the diagonal segments of the sidewalk didn’t matter), these effects canceled out when *y* = (√2−1)*x*. Employees who lived above this line, and closer to the line *y* = *x*, preferred the octagonal sidewalks, while employees who lived below this line, and closer to the *x* intercept, preferred the original sidewalks.

But that was just for the case when *x* and *y* were both positive, and *x* > *y* — that is, below the line *y* = *x*. Similar results could be worked out for all the other cases (in all four quadrants). And for those of you who worked through the trigonometry of last week’s Riddler Express, you might have noticed that the Riddler Classic gave you the *exact same result!* That is, in the case of a very, very large Riddler City (as stated in the puzzle), **half** of the employees preferred octagonal sidewalks, while the other half preferred to keep the original sidewalks. This symmetry did not go unnoticed by Riddler Nation.

After all this analysis, it would appear that Riddler City is stuck in gridlock when it comes to sidewalk legislation.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>Riddler City is a large circular metropolis, with countless square city blocks that each have a side length of 1 km. A small section of the city, composed of 36 blocks, is shown in the diagram below:

At the very center of the city lies Riddler City Hall. Its many employees all walk to and from work, and their homes are evenly scattered across the city. The sidewalks they walk along have always been adjacent to the streets — but that may be changing.

Recently, several city hall employees submitted a petition, requesting that the sidewalks should no longer lie alongside the streets. Instead, they want the sidewalks to cut diagonally across the city, connecting nearby street intersections. These proposed sidewalks are represented by the thicker blue lines in the diagram below:

The mayor of Riddler City has tasked you with resolving this dispute in a mathematical manner. She would like you to answer the following question: What fraction of the city hall employees would have a *shorter* walk home (that is, to the street intersection nearest to their home) if the city replaced its traditional sidewalks with these diagonal sidewalks?

From David Lewis comes an additional, original twist on Riddler City’s urban planning:

The mayor ultimately decided not to pursue diagonal sidewalks, but the petitioners haven’t given up yet. One of them recently visited Barcelona and was inspired by its octagonal city blocks.

Now, there’s a second petition on the mayor’s desk, asking that the grid layout of the city’s sidewalks be replaced with an octagonal pattern, represented by the thicker blue lines in the diagram below:

Under this second proposal, *now* what fraction of the employees would have a shorter walk home if the city replaced its traditional sidewalks with these new sidewalks?

Congratulations to Andrew Yuan of Brisbane, Australia, winner of last week’s recent Riddler Express.

Last week, you were asked to find how many more palindrome dates (in the MM/DD/YYYY format) there would be this century. The most recent occurrence was this past Groundhog Day, 02/02/2020.

As many solvers noted, the fact that the date had to occur in *this* century meant that it had to have the form MM/DD/20YY. Then, in order to be a palindrome — meaning it reads the same forwards and backwards, if we ignore the slashes — the form becomes MM/02/20YY.

At this point, the two last digits of the year are the same as the digits of the month, but flipped. And so Andrew went through the 12 months of the year to see which resulted in palindromes that hadn’t yet occurred. The 12 palindrome dates are:

- 01/02/2010 (already passed)
- 02/02/2020 (
*just*passed) - 03/02/2030
- 04/02/2040
- 05/02/2050
- 06/02/2060
- 07/02/2070
- 08/02/2080
- 09/02/2090
- 10/02/2001 (already passed)
- 11/02/2011 (already passed)
- 12/02/2021

Of the 12 possible palindrome dates, four have already passed. That means there are **eight** palindrome dates remaining this century.

Solver Sami from London extended the riddle, looking also at palindrome dates in the DD/MM/YYYY format (which, appropriately enough, is used in London). Following a similar approach, Sami found 23 such palindrome dates. Interestingly, one of these is Feb. 29, 2092 — a palindrome *leap* day!

Back in the American format, the next palindrome date will occur on Dec. 2, 2021, which sounds like the perfect time to ask this riddle again. (Spoiler alert: The answer will be seven.)

Congratulations to Bram Carlson of Fort Lauderdale, Florida, winner of last week’s Riddler Classic.

Last week, you were introduced to an ambiguous mathematical expression with absolute values. Offered as an example, the expression |−1|−2|−3| had *two* possible interpretations, and as a result, two corresponding values:

- The two left bars were a pair, and the two right bars were a pair. In this case, we had 1−2·3 = 1−6 = −5.
- The two outer bars were a pair, and the two inner bars were a pair. In this case, we had |−1·2−3| = |−2−3| = |−5| = 5.

That was all well and good. But instead of analyzing just two cases, you had to find how many different values the expression |−1|−2|−3|−4|−5|−6|−7|−8|−9| could have. Yikes!

As you might have expected, some solvers used pencil and paper, while others turned to their computers. While this riddle seemed imposing, pencil and paper did an excellent job at providing an *upper bound* for the answer.

For the expression |−1|−2|−3|, there were *two* ways to pair the absolute value bars, and it turned out there were *two* possible values — one for each pairing. Similarly, finding the total number of ways to pair the absolute value bars in the expression |−1|−2|−3|−4|−5|−6|−7|−8|−9| gives us all possible values for the expression. (However, not all of these values may be *unique* — some may be duplicates of each other. But we can cross that bridge when we get to it … in a few paragraphs).

To count up the number of pairings of absolute value bars, we could imagine each pair consisting of an “open” bar and a “close” bar — just like parentheses. In other words, the number of pairings of the 10 bars in |−1|−2|−3|−4|−5|−6|−7|−8|−9| was the same as the number of ways to write out five pairs of — or 10 total — parentheses. Many solvers pointed out that this related problem is quite famous. Given *N* pairs of parentheses, the total number of valid sequences is known as the *N*^{th} Catalan number.

Catalan numbers show up all over the place in discrete mathematics and combinatorics. Don’t believe me? Well, try figuring out how many ways you can divide up a convex polygon into non-overlapping triangles by connecting the polygon’s internal diagonals. Go ahead, I’ll wait. What’s that? The answer turns out to be a Catalan number? You don’t say.

Since there were *five* pairs of absolute value bars in the original expression, |−1|−2|−3|−4|−5|−6|−7|−8|−9|, that meant there were 42 (the *fifth* Catalan number) total ways to pair up the bars. A few solvers stopped here, satisfied with an answer of 42. Of course.

But, as we said earlier, 42 is just the *upper bound* for our answer. There can’t be more than 42 unique values, but if some pairings of the absolute value bars result in duplicate values, the answer will be less than 42.

Indeed, there were several such duplicates. Ian Rhile evaluated all 42 interpretations of the original expression and plotted them on a number line:

Of the 42, three were duplicates, meaning there were only **39** unique values for the expression. For anyone who wants to know, this week’s winner, Bram, identified those pesky duplicates: −375, −25 and 105.

A few brave solvers tried their hands at evaluating even longer expressions, like |−1|−2|−3|−4|−5|…|−37|, which Angela Zhou found to have 664,540,593 unique values. (Her poor computer.)

While the number of unique values closely matches the Catalan numbers for shorter expressions, Angela found that they drop off — |−1|−2|−3|−4|−5|…|−33| is the first such expression whose number of unique values isn’t even half of its corresponding Catalan number.

Finally, there’s a tradition here at The Riddler of referencing The On-Line Encyclopedia of Integer Sequences (OEIS). There seems to be a known sequence out there that gives you the answer almost every week. Solver Tyler Barron was delighted to report that no one has yet submitted this sequence — the number of unique values for |−1|−2|−3|−4|−5|…|−(2*N*−1)| — to OEIS.

And in true Riddler fashion, Tyler called dibs.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>I’ve spent the last 10 years studying the history of presidential books for my own book, “Author in Chief.” But along the way, I discovered a strange little data set that helps explain how we went from an age when writing your autobiography was seen as risky and vain, to one when we’re constantly updating our autobiographies — one Instagram story at a time.

It all starts with a reference librarian named Louis Kaplan. In 1946, Kaplan began working on a deceptively ambitious project — a bibliography that would list every autobiography that had been published in America to that point. Kaplan studied old periodicals. He spent months on the road, visiting rare book libraries around the country. He befriended private collectors. And he got lots of help, especially when it came time to review the Library of Congress’s massive catalog.

After 14 years, Kaplan and his colleagues finally had their “Bibliography of American Autobiographies.” They ended up identifying 6,377 autobiographies published in the U.S. between 1675 and the 1940s, and their data shows the genre’s notable growth. Between 1800 and 1809, Americans published a total of just 27 autobiographies. One century later, during the decade between 1900 and 1909, that number had exploded to 569, easily outpacing population growth.

Kaplan’s data had flaws, of course. It was easier to miss older books, because they were older and because the prohibitive cost of manufacturing books meant fewer copies existed in the first place.^{11} More recent scholars have also critiqued Kaplan’s relatively narrow definition of “autobiography” and his undercounting of books written by women and people of color. But Kaplan’s book, along with Mary Louise Briscoe’s follow-up, remains the best resource available. As Jeffrey Levin and Thomas Cole put it in a 1996 journal article, “These limitations pale before the great wealth of historical material contained in Kaplan’s ‘Bibliography.’”

Some of the best analysis of Kaplan’s data set has come from Diane Bjorklund, a sociologist at Illinois State University who coded Kaplan’s entries for her own book, “Interpreting the Self.” Bjorklund spent about a month sorting Kaplan’s thousands of autobiographies by their authors’ professions — soldier, farmer, scientist, and so on. This sorting was both blunt and subjective; two farmers could write two very different books. “My coding can only provide a rough guide,” Bjorklund said in an email.

Still, that rough guide contains some fascinating hints about the historical anxieties surrounding autobiography. This anxiety has always shown up in anecdotes. One early American reader, for example, dismissed Jean-Jacques Rousseau’s classic “Confessions” as “an unnatural compound of vanity, meanness, and contemptible self-love.” It shows up in the data, too. According to Bjorklund’s coding, in the first half of the 19th century, more than half of all autobiographies published in U.S. came from one of two professions: religious figures and criminals. The explanation seems simple: One group had divine authority to tell their life stories. The other had nothing left to lose.

Slowly, the stigma against publishing autobiographies dissipated. “I’d point to Benjamin Franklin and his autobiography as the turning point,” said Susan Clair Imbarrato, an English professor at Minnesota State University-Moorhead who’s written extensively about the genre.

Bjorklund and Kaplan’s data reveals that as autobiography became more popular, it became more diverse. New varieties could rise and fall. During the 1840s and 1850s, there was a 600-percent increase in slave narratives, or autobiographies written by fugitive slaves that captured the brutality of life in the South. The nation’s real-time obsession with the Civil War also made it easier for generals and politicians to do what John Adams could not: publish their memoirs during their own lifetimes. In the 1860s, 50 military figures published their autobiographies — more than in the previous six decades combined. And during the Gilded Age, there was a boom in business autobiographies, with 48 appearing between 1880 and 1899 — more than had appeared in the previous *eight* decades combined.

That’s one of the most striking things about this data: The autobiographies that thrived during any particular moment said something about the desires of America and its readers. By the Roaring Twenties, the clergy-and-criminal bookshelf had dropped to less than a combined 20 percent of all autobiographies — the same era that saw a spike in memoirs written by a new kind of celebrity, the entertainer.^{12}

In the 1940s, the last decade covered by Kaplan’s “Bibliography,” memoirs by entertainers proved one of the most popular categories. And yet, even those accounted for only 5.6 percent of the 1,043 memoirs that appeared in America during that span. That’s the other striking thing about these autobiographical numbers: Once autobiography truly caught on, no one category could dominate because readers had a huge variety of authors and styles to choose from.

This abundance feels like a good description of our current moment — not just on the print bestseller lists, but in social media, as well. And yet the data makes clear that, once the anxiety over autobiography faded, the genre’s range became an essential part of its appeal. Readers have been curious about all sorts of experiences for a while now. Back in 1909, William Dean Howells, perhaps the most influential critic in the U.S. at the time, wrote an essay championing the genre. While he praised Franklin’s book, Howells seemed most excited about the diversity among autobiographies. “All autobiographies are good for one reason or other,” Howells wrote, before saying that autobiographies aren’t restricted to authors of “any age or sex, creed, class, or color.”

The data largely supports Howells’s assertion. While researching that same decade — 1900 to 1909 — Bjorklund counted 31 different categories that contained at least two new autobiographies. A significant number of people reading, sharing and responding is not a byproduct of Instagram. That’s a byproduct of autobiography itself.

]]>From James Anderson comes a palindromic puzzle of calendars:

This past Sunday was Groundhog Day. Also, there was a football game. But to top it all off, the date, 02/02/2020, was palindromic, meaning it reads the same forwards and backwards (if you ignore the slashes).

If we write out dates in the American format of MM/DD/YYYY (i.e., the two digits of the month, followed by the two digits of the day, followed by the four digits of the year), how many more palindromic dates will there be this century?

Also on Super Bowl Sunday, math professor Jim Propp made a rather interesting observation:

At first glance, this might look like one of those annoying memes about order of operations that goes viral every few years — but it’s not.

When you write lengthy mathematical expressions using parentheses, it’s always clear which “open” parenthesis corresponds to which “close” parenthesis. For example, in the expression (1+2(3−4)+5), the closing parenthesis after the 4 pairs with the opening parenthesis before the 3, and *not* with the opening parenthesis before the 1.

But pairings of other mathematical symbols can be more ambiguous. Take the absolute value symbols in Jim’s example, which are vertical bars, regardless of whether they mark the opening or closing of the absolute value. As Jim points out, |−1|−2|−3| has *two* possible interpretations:

- The two left bars are a pair and the two right bars are a pair. In this case, we have 1−2·3 = 1−6 = −5.
- The two outer bars are a pair and the two inner bars are a pair. In this case, we have |−1·2−3| = |−2−3| = |−5| = 5.

Of course, if we gave each pair of bars a different height (as is done in mathematical typesetting), this wouldn’t be an issue. But for the purposes of this problem, assume the bars are indistinguishable.

How many different values can the expression |−1|−2|−3|−4|−5|−6|−7|−8|−9| have?

Congratulations to Stephen Kloder of Berkeley, California, winner of last week’s recent Riddler Express.

Last week’s Riddler Express was inspired by Nick Brett’s roll at the recent World Indoor Bowls Championships in Great Yarmouth, England:

You were asked to imagine yourself in Nick’s shoes, trying to split two of your opponent’s bowls. Each bowl was a sphere with a radius of 1, and your opponent’s two red bowls were separated by a distance of 3 — that is, their centers were separated by a distance of 5. If the angle between the path your bowl was on and the line connecting your opponent’s bowls was 75 degrees, then you could indeed split your opponent’s bowls, as shown in the animation below.

What was the *minimum* such angle that would have allowed your bowl to split your opponents’ bowls without hitting them?

Stephen reasoned that, when the path of the green bowl made the minimum angle, it would be tangent to the two red bowls as it passed by them, meaning it just barely grazes each of them at a single point.

Solver Thomas Stone sketched it out and applied some trigonometry.

Triangle ABC in Thomas’s sketch is a right triangle, since a circle’s radius is always perpendicular to the tangent it intersects. That means the sine of the angle in question was equal to 2 (twice the radius of each bowl) divided by 2.5 (half the distance between the centers of your opponent’s two bowls). The angle itself turns out to be approximately **53.13 degrees**.

And if you’re still not convinced, here’s an animation of that perfect roll:

Nick Brett may be a great professional indoor bowler, but no one can match the precision of trigonometry.

Congratulations to Daniel Hennessy of West Chester, Ohio, winner of last week’s Riddler Classic.

Last week you looked at Magna-Tiles that were isosceles triangles with one 30 degree angle and two 75 degree angles. If you arranged 12 of these tiles with their 30 degree angles in the center, they would lay flat and form a regular dodecagon. If you put fewer (between three and 11) of those tiles together in a similar way, they would form a pyramid whose base was a regular polygon.

To maximize the volume contained within the resulting pyramid, how many tiles should you use?

Solver Sam Koehn used design software to mock up the different pyramids and measure their volumes directly. Here are his results for pyramids of between four and 11 tiles:

Plugging in a side length of 3 inches for each tile’s shortest side, Sam found that the volume was maximized when **10 tiles** were used, for a total volume of about 73 cubic inches.

It was, of course, possible to calculate this without visually rendering the pyramids (but now that you’ve seen them, that’s definitely the cooler way to solve the problem). Many solvers set pencil to paper and slogged through the messy trigonometry, which was apparently last week’s theme.

Solver David Zimmerman found the volume as a precise function of the number of tiles. For any given number of tiles, he first measured the dimensions of the polygonal base they would form, and then used the Pythagorean theorem to find the resulting pyramid’s height. At this point, the volume can be found by multiplying the area of the base by the height and dividing by three. I’ll spare you the exact formula, which has cotangents and cosecants to spare.

The puzzle’s author, Rob Berger, took a similar approach, and graphed the area of the base, the height, and the volume for each of the possible pyramids. As you increase the number of tiles, the area of the base increases and the height decreases. Sure enough, the product of the base and height is maximized when there are 10 tiles.

In this problem, the greatest number of tiles you could put together was 12, which resulted in a flat pancake shape. Several solvers were interested in what would happen if this number of tiles, 12, was increased. In other words, if you have some large number *N* of congruent isosceles triangles that fit together to form a pancake, what number of triangles are needed to form a pyramid with maximal area?

By taking limits and setting derivatives equal to zero, solver Hector Pefo found that the volume is maximized when you use slightly more than 80 percent of the tiles, or, more precisely, a fraction of them that’s close to √(2/3).

So if you ever find yourself in a position where you have lots of skinny triangles arranged in a circular formation, and you need to build a container out of them, be sure to get rid of a fifth of them before you break out the glue.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>At the recent World Indoor Bowls Championships in Great Yarmouth, England, one of the rolls by Nick Brett went viral. Here it is in all its glory:

In order for Nick’s green bowl to split the two red bowls, he needed expert precision in both the speed of the roll and its final angle of approach.

Suppose you were standing in Nick’s shoes, and you wanted to split two of your opponent’s bowls. Let’s simplify the math a little, and say that each bowl is a sphere with a radius of 1. Let’s further suppose that your opponent’s two red bowls are separated by a distance of 3 — that is, the centers of the red bowls are separated by a distance of 5. Define *ɸ* as the angle between the path your bowl is on and the line connecting your opponent’s bowls.

For example, here’s how you could split your opponent’s bowls when *ɸ* is 75 degrees:

What is the *minimum* value of *ɸ* that will allow your bowl to split your opponents’ bowls without hitting them?

From Robert Berger comes a question of maximizing magnetic volume:

Robert’s daughter has a set of Magna-Tiles, which, as their name implies, are tiles with magnets on the edges that can be used to build various polygons and polyhedra. Some of the tiles are identical isosceles triangles with one 30 degree angle and two 75 degree angles. If you were to arrange 12 of these tiles with their 30 degree angles in the center, they would lay flat and form a regular dodecagon. If you were to put fewer (between three and 11) of those tiles together in a similar way, they would form a pyramid whose base is a regular polygon. Robert has graciously provided a photo of the resulting pyramids when three and 11 tiles are used:

If Robert wanted to maximize the volume contained within the resulting pyramid (presumably to store as much candy for his daughter as possible), how many tiles should he use?

Congratulations to Tom Lauwers of Pittsburgh, Pennsylvania, winner of last week’s recent Riddler Express.

Last week, you analyzed the voting for the baseball Hall of Fame when there were 20 players on the ballot and 400 voters, each of whom could select up to 10 players for induction without voting for any given player more than once. To gain entry, a player must have been selected on at least 75 percent of the ballots.

Under these circumstances, what was the *maximum* number of players that could be inducted into the Hall of Fame?

Tom reasoned that to get as many players into the Hall of Fame as possible, each of the 400 voters should have selected the maximum allowable number of 10 players, meaning a total of 4000 votes would have been cast. Meanwhile, to gain entry, a player needed to appear on 75 percent of the ballots, meaning they needed to receive at least 300 votes. The greatest number of players who could have received 300 votes out of 4000 total votes was 4000 ÷ 300, which is slightly greater than **13**. That meant the number of inductees was *at most* 13 — but was it possible for exactly 13 players to be voted in?

Indeed it was. Solver Ramsey Jade showed one possible way by first labeling the 13 players with the letters from A through M, and dividing the 400 voters into four groups of 100.

- The first 100 voters would select players A, B, C, D, E, F, G, H, I and J.
- The next 100 voters would select D, E, F, G, H, I, J, K, L and M.
- The third set of 100 voters would select A, B, C, G, H, I, J, K, L and M.
- The final set of 100 voters would select A, B, C, D, E, F, J, K, L and M.

Twelve of the 13 players received 300 votes, and player J apparently joined Mariano Rivera as the only unanimous inductees.

Interestingly, this answer didn’t depend on the number of eligible players or the number of voters — just the maximum number of players on a ballot and the percentage needed to get inducted.

Some solvers wondered about the minimum possible number of inductees, which was, unsurprisingly, zero. The voters weren’t obligated to list 10 players on their ballots, so they could have simply cast empty ballots. And even if they had each voted for the maximum 10 players, it would still be possible for the 20 candidates to split the votes so that no one made it into the Hall of Fame.

Solver Ravi Chandrasekaran further studied how much more crowded Cooperstown would be if voters colluded to maximize the number of inductees. According to Ravi’s analysis, there would be more than seven times as many players enshrined.

Fortunately, this is not something that Larry Walker will have to worry about anymore.

Congratulations to Sion Verschraege of Ghent, Belgium, winner of the last week’s Riddler Classic.

Last week, I invited you to play a game of “Pinching Pennies,” which started off with between 20 and 30 pennies. First, I divided the pennies into two piles any way I liked. Then we alternated taking turns, with you first, until someone won the game. For each turn, a player took any number of pennies he or she liked from either pile, or instead took the same number of pennies from both piles. Each player also had to take at least 1 penny every turn. The winner of the game was the one who took the last penny.

If we both played optimally, what starting numbers of pennies (again, between 20 and 30) guaranteed that *you* could always win the game?

Solver Jason Ash tackled this puzzle by first looking at potential endgame strategies. For example, if it were your turn and you encountered two piles that each had 1 penny, which we can indicate with the ordered pair (1, 1), then you could win (by taking both coins). Similarly, any ordered pair of the form (*n*, 0), (0, *n*), and (*n*, *n*) would result in a one-move victory for you.

But what about the ordered pair (1, 2), meaning one pile had 1 penny and the other had 2? Here are your four possible moves and what I (your opponent) would do next:

- Remove 1 penny from the first pile, leaving (0, 2). I would then take both pennies from the second pile, and you’d lose.
- Remove 1 penny from the second pile, leaving (1, 1). I would then take one penny from each pile, and you’d lose.
- Remove 2 pennies from the second pile, leaving (1, 0). I would then take the last penny from the first pile, and you’d lose.
- Remove 1 penny from each pile, leaving (0, 1). I would then take the last penny from the second pile, and you’d lose.

As you can see, (1, 2) was a state of the game you’d prefer to avoid, as it represented a no-win scenario for you. Meanwhile, any pile arrangements of the form (1, *n*+2) would guarantee a victory for you, since you could remove the *n* pennies from the second pile and then *I’d* be faced with (1, 2).

This week’s winner, Sion, continued working up to larger and larger numbers, finding the game states that we’d each prefer to avoid, and an interesting pattern began to emerge:

Those red cells in Sion’s grid show the states we both want to avoid: (1, 2), (3, 5), (4, 7), (6, 10), (8, 13), (9, 15), (11, 18), etc. They appear to fall along two symmetric lines.

But I digress. The answer to the original question, which was asking for the initial numbers of pennies that guaranteed a victory for you, was **20, 22, 23, 25, 26, 27, 28 and 30**. Had we instead played with the *other* numbers between 20 and 30, I would have won: I would have split 21 coins into piles of 8 and 13, 24 coins into piles of 9 and 15 and 29 coins into piles of 11 and 18. (Note: I also accepted answers that omitted 20 and 30 — a different interpretation of “between 20 and 30” — as well as answers of 21, 24 and 29, which were winning numbers for *me*.)

Now, back to the pattern in Sion’s grid. As it turns out, this game pre-dates last week’s column and is better known as Wythoff’s Game, named for the Dutch mathematician and game theorist. The game has been well studied, and those losing positions are related to the golden ratio — notice that the second number in each of the above pairs is approximately 1.6 times greater than the first, very close to the golden ratio. Another famous sequence of numbers with close ties to the golden ratio is the Fibonacci sequence, whose numbers like 5, 8 and 13 you’ll also see lurking in Sion’s grid.

For more on Wythoff’s Game, check out his original paper from 1907. (Special thanks to solver Laurent Lessard for finding this gem!)

Whether it’s 1907 or 2020, a good riddle is timeless.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>**Video/motion graphics intern.**This person will work on videos in politics, sports and more. The intern will help conceive of and research topics for FiveThirtyEight’s YouTube channel and the ABC video player; animate, edit and produce; and help package the videos for online and social media distribution**Designer.**This intern will work with the art director and editors to decide how to make our stories visual, including photo research, cropping and captioning, and making your own illustrations for daily and feature stories.**Podcast producer**. This person will help produce our award-winning podcasts, including doing topic research, setting up the recording studio and editing. Candidates should have basic audio-editing skills and an interest in data journalism.**Politics data reporter**. This intern will help our politics team as it covers the 2020 election. The internship will be a mix of data-mining and analysis, writing, research, reporting and fact-checking.**Sports data reporter.**This intern will write, conduct research and contribute to longer features. Basic knowledge of the major sports is essential, and expertise in topics outside the traditional areas of coverage is a plus.**Copy editor****.**This intern will assist our copy desk in editing stories for accuracy, numeracy, grammar and style, with a particular focus on fact-checking and research.**Data visualization intern****.**This intern will design, edit and produce static graphics to accompany articles. The ideal candidate will have experience making data graphics using design tools such as Illustrator or programming languages such as JavaScript, Python and R. General comfort with numbers is essential, as is experience with spreadsheet software such as Excel.

For more information and to apply, please use the following links:

]]>From Stephanie Thompson comes a question of ballot optimization:

Derek Jeter and Larry Walker were just elected to the Baseball Hall of Fame! That got Stephanie thinking. Suppose there are 20 players on the ballot and 400 voters in a given year. Each voter can select up to 10 players for induction without voting for any given player more than once. To gain entry, a player must have been selected on at least 75 percent of the ballots.

Under these circumstances, what is the *maximum* number of players that can be inducted into the Hall of Fame?

From Dean Ballard comes another coin-related challenge — a game of “Pinching Pennies”:

The game starts with somewhere between 20 and 30 pennies, which I then divide into two piles any way I like. Then we alternate taking turns, with you first, until someone wins the game. For each turn, a player may take any number of pennies he or she likes from either pile, or instead take the same number of pennies from both piles. Each player must also take at least one penny every turn. The winner of the game is the one who takes the last penny.

If we both play optimally, what starting numbers of pennies (again, between 20 and 30) guarantee that you can win the game?

Congratulations to Kealan Vasquez of Ave Maria, Florida, winner of last week’s recent Riddler Express.

Last week, you were a coach in the Riddler Football League, and you had devised a new strategy for scoring after a touchdown. Your team would line up 2 yards away from the goal line in such a way that it could attempt either a 1-point conversion or a 2-point conversion. Your opponent could only properly defend against one of those two possibilities, so they’d have to guess.

If you attempted a 1-point conversion that was properly defended, you’d score 90 percent of the time; otherwise, you’d score 100 percent of the time. If you instead attempted a 2-point conversion that was properly defended, you’d score 40 percent of the time; otherwise, you’d score 60 percent of the time.

It was up to you to communicate to your team’s captain the probability with which they should attempt each. However, given all the spying that occurs in the League these days, you could assume that your message was overheard by your opponent, who also knew the probability of you scoring in each of the four scenarios listed above.

With all that said, what were the best offensive and defensive strategies here, and how many points would you score, on average, after each touchdown?

This was an example of a zero-sum game, meaning points were just as good for you as they were bad for your opponent. Since you and your opponent both had complete information about the probabilities, you’d pursue strategies that would result in a Nash equilibrium — that is, an effective stalemate where neither you nor your opponent would benefit from switching strategies.

Solver Ravi Chandrasekaran explored what would happen if you and your opponent tried different probabilities between always going for (or defending) 1 point vs. 2 points. Both Ravi and Kealan found that your best strategy was to go for 1 point 80 percent of the time and go for 2 points 20 percent of the time. To see why that is, suppose your opponent defends against 1 point with probability *p* and against 2 points with probability 1−*p*. How many points would you score on average? Let’s look at the four possible cases:

- You’d go for 1, and they’d defend it with probability 0.8
*p*, earning 0.9 points on average (90 percent of 1 point). - You’d go for 1, and they wouldn’t defend it with probability 0.8(1−
*p*), earning 1 point on average (100 percent of 1 point). - You’d go for 2, and they’d defend it with probability 0.2(1−
*p*), earning 0.8 points on average (40 percent of 2 points). - You’d go for 2, and they wouldn’t defend it with probability 0.2
*p*, earning 1.2 points on average (60 percent of 2 points).

Combining all these cases, the average number of points you’d get will be 0.8*p *· 0.9 + 0.8(1−*p*) · 1 + 0.2(1−*p*) · 0.8 + 0.2*p* · 1.2 = 0.96 points. Notice how all the *p*’s cancel out? That means when you go for 1 point 80 percent of the time, it doesn’t matter *what* your opponent does — you’ll score an average of 0.96 points after each touchdown. If you were to veer away from these probabilities (say, by going for 2 more often), your opponent could take advantage (by defending against the 2 more often) and lower your average number of points.

Similarly, on the defensive end, when your opponent defends against 1 point 40 percent of the time and against 2 points 60 percent of the time, you’ll again score an average of 0.96 points, no matter what your offensive strategy is. At the end of the day, you could expect to score **0.96 points** after each touchdown.

I am pleased to note that the many game theorists of Riddler Nation pointed out that, due to the defense overhearing what the offensive strategy was going to be (which was to go for 1 point 80 percent of the time), it technically didn’t matter *what* the defensive strategy was. It’s a great point! (Or was it a great 2 points?)

Congratulations to Adam Richardson of Old Hickory, Tennessee, winner of the last week’s Riddler Classic.

Last week, two delirious ducks were having a difficult time finding each other in their pond, which contained a 3×3 grid of rocks.

Every minute, each duck randomly swam from one rock to a neighboring rock in the grid — up, down, left or right but *not* diagonally. So if a duck was at the middle rock, it would next swim to one of the four side rocks with probability 1/4. From a side rock, it would swim to one of the two adjacent corner rocks or back to the middle rock, each with probability 1/3. And from a corner rock, it would swim to one of the two adjacent side rocks with probability 1/2.

If the ducks both started at the middle rock, then on average, how long would it take until they’re at the same rock again?

If the ducks were to make the same first move, they would have found each other after one minute. On the other hand, they could have gone around the rocks countless times without ever finding each other. A key insight was to realize that if you knew where the ducks were at some time, you could find a precise probability distribution for where they’d be one minute later. And because of the symmetry of the rocks, solver Jason Ash noted there were only five possible arrangements you had to keep track of:

For example, suppose after some time the ducks happen to be in the “outer across” arrangement in the above diagram. One minute later, there’s a 1/4 chance they’ll meet (if the left duck moves right and the right duck moves left), there’s a 1/4 chance they’ll switch to the “middle across” arrangement (if both ducks move down) and a 1/2 chance they’ll switch to the “diagonal outer” arrangement (if one duck moves down and the other moves horizontally).

By determining how likely the ducks transitioned between each of these different arrangements, solver Vikrant Kulkarni set up a system of equations relating the average times it took the ducks to meet from each arrangement, including the initial arrangement where both ducks were at the middle rock.

For problems like this, with random transitions between a fixed number of arrangements, many solvers, including Adam and Jason, used a technique knowns as Markov chains, which solve the very same equations Vikrant came up with. On average, the ducks will meet after **363/74**, or about 4.905, minutes.

Solver Angelos Tzelepis went one step further, finding (after 3 million simulations) the probabilities of *where* the ducks would ultimately find each other. The most likely meeting point was where they started — the rock in the middle:

Despite his mathematical prowess, Angelos did make one critical error: He incorrectly referred to the ducks as “drunk.” They were merely *delirious*.

For extra credit, you were asked to consider the case of three or more ducks: If they all started in the middle rock, on average, how long would it take until they were all at the same rock again?

Again using Markov chains, solver Laurent Lessard was able to find exact values for the cases of three, four, five and six ducks, which would respectively take on average 18.4, 66.7, 237.4 and 825.3 minutes to all meet up. As the number of ducks increased, the time it took for them to meet appeared to grow exponentially, and the computation required to find the exact result became increasingly difficult.

To find approximate solutions for the general case where there were *N* ducks, solvers Guy D. Moore and Hector Pefo used the fact that after an *odd* number of minutes the ducks would all be on one of the four sides, and after an *even* number of minutes the ducks would all be in the middle or one of the four corners.

After an odd number of minutes — when the ducks were on the sides — no side was more likely to be where they met than any of the others. That meant the probability the ducks were all together at any given minute was 1/4^{N−1}, where the minus one in the exponent came from the fact that the first duck could have been on any of the four sides, while the other *N*−1 ducks had to be on that same side. By performing a similar analysis after an even number of minutes, Guy and Hector found that *N* ducks would all meet up after approximately 2·3^{N} minutes.

That meant it would take a dozen delirious ducks approximately two years to find each other. Meanwhile, it would take a *dozen* dozen delirious ducks far longer than the age of the universe to find each other.

Those poor dozen dozen ducks.

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>The Riddler Football League (RFL) playoffs are upon us! As the coach, you’ve devised a new strategy for scoring after a touchdown. Your team will line up 2 yards away from the goal line in such a way that it could attempt either a 1-point conversion or a 2-point conversion. (Unlike other football leagues, the distance is the same for both types of conversion, and you need not announce which you’ll be attempting.) Your opponent can only properly defend against one of those two possibilities, so they’ll have to guess.

If you attempt a 1-point conversion and the other team defends against it properly, you’ll score 90 percent of the time. If they don’t defend it properly, you’ll score 100 percent of the time.

If you instead attempt a 2-point conversion and the other team defends against it properly, you’ll score 40 percent of the time. If they don’t defend it properly, you’ll score 60 percent of the time.

To tell your team which they should attempt, your team’s offensive coordinator will communicate to your team’s captain the probability with which they should attempt each. For example, the coordinator might say: “Go for 1 with a 51 percent chance, and go for 2 with a 49 percent chance.” Using a random number generator, the captain will then ultimately decide to go for 1 point or 2 points. (Naturally, every athlete in the RFL has a random number generator handy.)

However, given all the spying that occurs in the RFL these days, you can assume that the offensive coordinator’s message will also be heard by your opponent — that means the defense knows the exact probability with which you’ll attempt either conversion. Your opponent also knows the probability of you scoring in each of the four scenarios listed above.

With all that said, what strategy will maximize the average number of points you’ll score (i.e., how often should your team go for 1 or 2)? What should your opponent’s defensive strategy be? How many points will you score, on average, after each touchdown?

After a long night of frivolous quackery, two delirious ducks are having a difficult time finding each other in their pond. The pond happens to contain a 3×3 grid of rocks.

Every minute, each duck randomly swims, independently of the other duck, from one rock to a neighboring rock in the 3×3 grid — up, down, left or right, but *not* diagonally. So if a duck is at the middle rock, it will next swim to one of the four side rocks with probability 1/4. From a side rock, it will swim to one of the two adjacent corner rocks or back to the middle rock, each with probability 1/3. And from a corner rock, it will swim to one of the two adjacent side rocks with probability 1/2.

If the ducks both start at the middle rock, then on average, how long will it take until they’re at the same rock again? (Of course, there’s a 1/4 chance that they’ll swim in the same direction after the first minute, in which case it would only take one minute for them to be at the same rock again. But it could take much longer, if they happen to keep missing each other.)

*Extra credit:* What if there are three or more ducks? If they all start in the middle rock, on average, how long will it take until they are all at the same rock again?

Congratulations to Nathan Holmes-King of Fremont, California, winner of last week’s recent Riddler Express.

Last week, you were asked to find a fraction (with a whole number numerator and denominator) that was greater than 1/2020, less than 1/2019 and with the smallest possible denominator.

Solver Amy Leblang used an algebraic approach, looking for a fraction *a*/*b* (where *a* and *b* are whole numbers) such that 1/2020 < *a*/*b* < 1/2019. Flipping all the fractions, you can rewrite this inequality as 2019 < *b*/*a* < 2020. Finally, multiplying through by *a* gives us 2019*a* < *b* < 2020*a*.

Again, our goal is to find the denominator *b* that’s as small as possible, and that will happen when *a* is also small. If we let *a* = 1, then there’s no whole number *b* that sits between 2019 and 2020, so that won’t work. But if *a* = 2, then we’re looking for a value of *b* between 4038 and 4040, which means *b* = 4039. Larger values of *a* will produce larger values of *b*, which this riddle wasn’t asking about. In other words, **2/4039** is the correct answer.

Many solvers, like Angela Zhou, observed that this question was straightforward if you knew a thing or two about Farey sequences, which are ordered sets of fractions between 0 and 1. For the Farey sequence in which all the denominators are at most 2020, the fractions 1/2020 and 1/2019 are “Farey neighbors,” meaning they’re next to right next to each other in the sequence. This riddle is effectively asking you to identify the first Farey sequence where 1/2020 and 1/2019 are *no longer* neighbors — that is, one where there’s another fraction between them. That fraction will be what’s called the mediant of the fractions on either side, generated by adding the numerators and the denominators: (1+1)/(2020+2019) = 2/4039.

Just to be extra sure that’s the right answer, solver Nolan Gannage wrote code to search all the fractions between 1/2020 and 1/2019 whose denominators were also 50,000 or less. Sure enough, the one with the smallest denominator was 2/4039.

That’s one of the Riddler maxims: “When in doubt, code it out.”

Congratulations to Austin Calico of Ashland, Kentucky, winner of the last week’s Riddler Classic.

Last week, you looked at an alphanumeric code inspired by “Gematria,” where words were assigned numerical values based on their letters. Each A was worth 1 point, each B was worth 2 points, and so on. The value of a word was then the sum of the values of its letters. For example, RIDDLER had an alphanumeric value of 70, since R + I + D + D + L + E + R became 18 + 9 + 4 + 4 + 12 + 5 + 18 = 70.

But what about the values of different numbers themselves, spelled out as words? The number 1 (ONE) had an alphanumeric value of 15 + 14 + 5 = 34, and 2 (TWO) had an alphanumeric value of 20 + 23 + 15 = 58. Both of these values were *bigger* than the original numbers.

Meanwhile, if we looked at larger numbers, 1,417 (ONE THOUSAND FOUR HUNDRED SEVENTEEN) had an alphanumeric value of 379, while 3,140,275 (THREE MILLION ONE HUNDRED FORTY THOUSAND TWO HUNDRED SEVENTY FIVE) had an alphanumeric value of 718. These values were much *smaller* than the original numbers.

If we considered all the whole numbers that were *less than* their alphanumeric value, what was the largest of these numbers?

First off, this question was a little ambiguous. The intent was to find the largest number *N* that was less than its “Gematria score” of *N*, which we’ll call *G*(*N*) — that is, *N* < *G*(*N*). However, you could also have read the question as asking for the largest value of *G*(*N*) among numbers *N* where *N* < *G*(*N*). The majority of readers answered the first question, but we’ll address both here.

Almost all solvers wrote code for this one, with the general strategy of (1) systematically describing how numbers are codified as words in English, and (2) scoring those words. A few solvers, like Cameron Shelton, took the time to work it all out by hand. In the words of Cameron, this “gives me a better feel for the problem and because I enjoy it more.” Bravo, Cameron!

Either way, the answer turned out to be **279**, or TWO HUNDRED SEVENTY NINE, which had a Gematria score of 284. To confirm this result, here’s a graph from solver Jason Ash showing the scores for the numbers from 1 to 500.

Sure enough, 279 is the last number above the dotted line, meaning it’s the greatest number to exceed its Gematria score. Interestingly, 80 is the smallest number that’s *less than* its Gematria score: EIGHTY is only worth 74 points. And for those of you who had the alternate interpretation of the original question, the number above the dotted line that’s worth the most points is 277, or TWO HUNDRED SEVENTY SEVEN, which is worth a whopping 307 points.

If you’re curious to see the scores of Gematria scores beyond those of the first 500 numbers, then solver Quoc Tran has an animation for you, showing the scores from ONE to ONE HUNDRED THOUSAND:

Finally, James Chapman took this riddle even further, solving it for multiple languages (not just English). James found that Finnish, French, and Polish each had answers just shy of 400. Dan Miller even went on to suggest using Roman numerals — maybe next time!

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>To celebrate the new year, here’s a quick puzzle about the number 2020. Of all the fractions out there that are greater than 1/2020 but less than 1/2019, one has the smallest denominator. Which fraction is it?

(Before you ask, by “fraction” I mean that both the numerator and denominator should be whole numbers.)

From Leonard Cohen comes a puzzle at the intersection of language and mathematics:

In Jewish study, “Gematria” is an alphanumeric code where words are assigned numerical values based on their letters. We can do the same in English, assigning 1 to the letter A, 2 to the letter B, and so on, up to 26 for the letter Z. The value of a word is then the sum of the values of its letters. For example, RIDDLER has an alphanumeric value of 70, since R + I + D + D + L + E + R becomes 18 + 9 + 4 + 4 + 12 + 5 + 18 = 70.

But what about the values of different numbers themselves, spelled out as words? The number 1 (ONE) has an alphanumeric value of 15 + 14 + 5 = 34, and 2 (TWO) has an alphanumeric value of 20 + 23 + 15 = 58. Both of these values are *bigger* than the numbers themselves.

Meanwhile, if we look at larger numbers, 1,417 (ONE THOUSAND FOUR HUNDRED SEVENTEEN) has an alphanumeric value of 379, while 3,140,275 (THREE MILLION ONE HUNDRED FORTY THOUSAND TWO HUNDRED SEVENTY FIVE) has an alphanumeric value of 718. These values are much *smaller* than the numbers themselves.

If we consider all the whole numbers that are *less than* their alphanumeric value, what is the largest of these numbers?

Congratulations to Quinn Rose of Des Moines, Iowa, Mike Cromwell of Novi, Michigan, and Mark Ritchie of Cleveland, Ohio, winners of last week’s recent Riddler Express.

Last week, you were presented with three images, in which different nations’ flags had their pixels randomly rearranged. You were tasked with figuring out which flag was which.

The first flag was an even mix of red, white and blue…

…or was it blue, white and red? Quinn correctly identified it as the flag of **France**:

While there are many national flags that make use of this color palette, about 40 percent of solvers correctly identified this as France’s flag. Solver Stew Schrieffer checked this by comparing the number of pixels of each color. There were approximately an equal number of blue, white and red pixels, and the precise colors perfectly matched France’s flag. The most popular (wrong) submissions were the United Kingdom, the United States of America and Russia, each at about 15 percent.

The second flag was mostly green, followed by a mix of yellow and blue, with a pinch of white.

Mike correctly identified this as the flag of **Brazil**:

I’m happy to report that most solvers got this right. It turns out that Brazil has a very identifiable color scheme. Who knew?

The third and final flag had the greatest variety of colors and had a relatively even mix of those colors. This made it more difficult to identify.

After studying the color profile of the pixels, Mark found that there were five colors: blue and green in “equal(ish)” amounts, as well as red, white and yellow in lesser amounts. The flag that best fit this bill was **Namibia**:

About a third of respondents correctly identified Namibia, with about 20 percent guessing South Africa, 15 percent guessing the Seychelles and 10 percent guessing Turkmenistan.

Overall, this was some excellent vexillological sleuthing by Riddler Nation. Sheldon Cooper would be proud!

Congratulations to Peter Ji of Madison, Wisconsin, winner of the last week’s Riddler Classic.

Last week, you analyzed the Spelling Bee word game from The New York Times. In Spelling Bee, seven letters are arranged in a honeycomb lattice, with one letter in the center. The goal is to identify as many words that meet the following criteria:

- The word must be at least four letters long.
- The word must include the central letter.
- The word cannot include any letter beyond the seven given letters.

Note that letters can be repeated. Four-letter words are worth 1 point each, while five-letter words are worth 5 points, six-letter words are worth 6 points, seven-letter words are worth 7 points, etc. Words that use all of the seven letters in the honeycomb are known as “pangrams” and earn 7 bonus points (in addition to the points for the length of the word).

Your task was to find the seven-letter honeycomb that resulted in the highest possible game score. To be a valid choice of seven letters, no letter could be repeated, it couldn’t contain the letter S (that would have been too easy) and there had to be at least one pangram. For consistency, you were asked to use this word list (courtesy of computer scientist Peter Norvig) to check your game score.

The highest scoring honeycomb had an R in the center, with the letters AEGINT around it. This arrangement scored a whopping 3,898 points. If you don’t believe me, here’s a visualization showing each word in the honeycomb, with the letters in each word highlighted in pink:

There are some particularly high-scoring words contained in this honeycomb, like REAGGREGATING and REINTEGRATING (each worth 20 points).

As if finding this highest-scoring honeycomb wasn’t challenging enough, several members of Riddler Nation set out to write increasingly efficient algorithms for solving this puzzle.

At first, you might think there are countless letter combinations that you (that is, your computer) would have to sift through to find the best honeycomb. Out of 25 letters (remember, we excluded the S), there are more than 3 million possible ways to pick a central letter and then six other letters around it. Yikes! Solver Tyler Barron estimated that it would have taken his computer 584 days to crank through all those letter combinations.

But Peter Ji, our winner, didn’t try all 3 million honeycombs. He narrowed his search to letters that showed up most frequently in the word list. This approach didn’t *guarantee* the right answer, but it worked.

Meanwhile, solver Laurent Lessard definitively proved this was the best honeycomb by taking advantage of the fact that there had to be a pangram — that is, a word with exactly seven unique letters. In the given word list, there were only 14,741 pangrams, which corresponded to 55,902 possible honeycombs. It’s still a pretty big number, but it’s way smaller than 3 million. Along the way, Laurent also found the highest scoring honeycomb that *did *include an S (with an E in the center and AINRST around it, worth a cool 8,681 points) and the lowest scoring honeycomb (with an X in the center and CINOPR around it, worth a pathetic 14 points). Solver David Robinson wrote about a similar approach that also leveraged matrix operations.

Finally, Peter Norvig (the provider of the word list!) documented his incredibly efficient approach. By looking at the possible subsets of letters within a honeycomb, the final version of his program found the right answer in 2 seconds. Wow!

Solver Hector Pefo suggested that The New York Times use this highest-scoring honeycomb for the April Fools’ Day edition of Spelling Bee. It would surely drive the Spelling Bee community (also known as the #HiveMind) bonkers. How awesome would that be?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

]]>Today’s number is 14.82 million, the number of people who tuned in to watch the first night of “Jeopardy: The Greatest of All Time” tournament on Wednesday. Champions Ken Jennings, Brad Rutter and James Holzhauer are scheduled to conclude the three-game tournament tonight at 8 p.m.

Even small increases in earnings can have huge health outcomes. A study published in the Journal of Epidemiology & Community Health this week found that states that raised their minimum wages by $1, reduced the suicide rates by 3.4 percent to 5.9 percent among adults between the ages of 18 and 64, whose highest educational attainment was a high school diploma or less. Researchers analyzed 25 years of monthly data from all 50 states and the District of Columbia. [Washington Post]

A new study from the National Center for Health Statistics shows drinking is rising in popularity among more Americans, resulting in more alcohol-related deaths. Researchers looked at almost 20 years of data and found nearly 73,000 people died in the U.S because of liver disease and other alcohol-related illnesses in 2017, more than twice as many people who died from the same health reasons in 1999. Researchers found that some of the greatest increases were among women and those who were 50 and older. [National Public Radio]

Chris Paul’s performance with the Oklahoma City Thunder this season has been so remarkable, FiveThirtyEight’s Chris Herring says he might “legitimately be the NBA’s best closer.” Through his scoring help, the team had won 10 of out of their last 12 games going into Thursday’s match against the Houston Rockets, and as Herring noted they are now tied for the second-most wins in the league. Paul’s ability to score in the fourth quarter has set him apart from the rest of the league, with his total of 103 points in the clutch as of Thursday afternoon, 20 more than Zach LaVine, the next closest player. [FiveThirtyEight]

The Russian Academy of Sciences appointed a commission to investigate unethical publication practices in an industry that has been known for low standards and thousands of cases of plagiarism as well as “questionable authorship,” according to Science Magazine, and now more than 800 papers from various Russian journals will be retracted. One former staffer at the U.S. National Science Foundation called the preliminary report from the RAS “a bombshell” for documenting the breadth of the problem. [Science]

On Thursday, Mexican environmental authorities announced that a red tide algae bloom was responsible for the deaths of 292 sea turtles from the country’s southern Pacific coast. The toxic algae had caused many of the turtles to stop breathing or keep their heads up. Volunteers, researchers and other authorities were able to save 27 of the Pacific Green sea turtles, though. [Associated Press]

**Educational debt is usually something that can’t be included when filing bankruptcy,** but one case might soon help more people drowning in student loans. Judge Cecelia G. Morris of the U.S. Bankruptcy Court in Poughkeepsie, N.Y., ruled that a U.S. Navy veteran didn’t have to pay $220,000 in student loan debt because “satisfying his law school debt in full would impose an undue hardship.” [Wall Street Journal]

]]>

After a magnitude 6.4 earthquake struck Puerto Rico on Tuesday, the country is still struggling to restore power to the island. About one million people still had no power of Wednesday, partly because the island’s major power plant Costa Sur power plant, was seriously damaged in the quake. Thousands of people have slept outside their homes due to concerns further tremors could cause other buildings to collapse. [New York Times]

The World Health Organization says more than 6,000 people have died from a measles epidemic in the Democratic Republic of Congo, and there have been approximately 310,000 suspected measles cases reported since the beginning of last year. BBC News reports an emergency vaccination program was launched by the Congolese government and WHO last September, but the epidemic is currently the world’s largest. More than 18 million children under the age of 5 have been vaccinated, but the WHO says it would cost an extra $40 million to also vaccinate Congolese children between the ages of six and 14 to strengthen response to the out break. [BBC News]

Of the 176 passengers and crew who died in the crash of Flight PS752, which left Iran early Wednesday morning, 63 people were Canadians and almost half of them were from the city of Edmonton. A large number of the Canadian victims were doctors, scientists, researchers, and graduate students. The Ukraine International Airlines flight was on its way to Kyiv when it crashed only a few minutes after taking off from Tehran’s main airport. [CBC News]

The ecological damage of the Australian bushfires in Victoria and New South Wales is now estimated to be at least 800 million animals, as well as “hundreds of billions of insects.” The Sydney Morning Herald reports more than 6 million hectares have been destroyed so far, “including rainforest normally considered too wet to burn.” The damage is so extensive that “at least one species is feared extinct.” [The Sydney Morning Herald]

]]>

IKEA, the popular Swedish furniture company, will pay $46 million to a California couple whose two-year-old son died from his injuries from a Malm dresser that tipped over and crushed him. In a statement, the couple’s lawyers said the financial amount from IKEA “is the largest wrongful death settlement related to one child in U.S. history.” In 2016, IKEA settled with families of three other children who were killed by the same line of dressers for a total of $50 million. [CNN Business]

At least 32 people are dead and dozens more have been wounded during a stampede at the funeral for Qassem Soleimani, one of Iran’s most powerful military leaders, who was killed last week in a targeted airstrike attack by the U.S. Iranian television said the stampede happened in Soleimani’s hometown, where he is to be buried. [National Public Radio]

You’re not supposed to get better with age as a defenseman in pro hockey, but John Carlson is proving to be a great exception to the rule. Even though the Washington Capitals player is turning 30 this week, Carlson is currently his team’s leader in scoring with 52 points (13 goals, 39 assists). FiveThirtyEight’s Terrence Doyle and Neil Payne also note that Carlson is so good, he’s “currently enjoying the ninth-best defenseman scoring season since 1943.” [FiveThirtyEight]

If you’ve ever wondered if the woman or nonwhite person you saw in an advertisement or on a dating app was real, artificial intelligence start-ups might make it much harder to figure that out. The Washington Post reports companies like the design firm Icons8 are now selling digital images of computer-generated faces that “look like the real thing” to marketing companies and dating apps. The sales pitch of the AI software is how it can quickly generate 1 million images in a single day, allowing companies to “increase diversity” without the costly process of finding real people. [Washington Post]

There were a lot of great movies that highlighted a female perspective or represented the diverse demographics of the United States, but you wouldn’t know it looking at who is nominated in the 2020 BAFTA film awards. This year, all of the 20 main acting nominations went to white actors and no women were nominated for best director. Marc Samuelson, chair of BAFTA’s film committee, told Variety, “It’s just a frustration that the industry is not moving as fast as certainly the whole BAFTA team would like it to be.” [TIME]

Many rare languages are at risk of disappearing, and Seke, which is spoken in just five villages in Nepal has only approximately 700 speakers left in the world, according to a recent study by the Endangered Language Alliance. The organization estimates there are roughly 100 Seke-speakers living in New York, and 50 of them live in one building in Flatbush, Brooklyn. One of the youngest residents there, Rasmina Gurung, has several relatives in the building, and is helping the Endangered Language Alliance compile a Seke-English dictionary. “I feel so much pressure,” she told the New York Times. “I need to get as much knowledge as possible. And fast.” [New York Times]

]]>

LSU quarterback Joe Burrow has had a pretty remarkable turnaround, becoming what FiveThirtyEight’s Josh Planos describes as a “potential future No. 1 draft pick.” Burrow has nabbed 8 out of the 65 best single-game QBR performances this season. And during last weekend’s Chick-fil-A Peach Bowl, Burrow passed for 493 yards and scored eight touchdowns, an unprecedented accomplishment. His other stats include throwing 5,208 yards, 55 touchdowns, and completing more than 77 percent of his throws, which Planos notes puts Burrow on track to break the all-time record. [FiveThirtyEight]

A 5.8-magnitude earthquake happened on the American territory of Puerto Rico shortly before the sun rose on Monday, causing landslides, power outages as well as damage to at least 34 homes. Several more tremors followed in the hours after the initial quake, including a 5.0-magnitude one. The Associated Press says power lines were shaken and five of the damaged homes had collapsed, but no casualties or injuries have been reported. [Associated Press]

The editorial staff of Sports Illustrated announced their plan to form a union on Monday after the legacy magazine saw drastic layoffs in October following the acquisition of the magazine by Maven, a Seattle-based tech and media company. Roughly 80 staffers in print, digital and video are represented by the union. [CNN]

A new report from the National Bureau of Economic Research says the cost of President Trump’s trade war has been paid almost entirely by American businesses and consumers, not China. Experts and economists from the Federal Reserve Bank of New York, Columbia University and Princeton said analysis of tax levies found “approximately 100 percent” of import taxes fell on Americans, despite the president’s assertion the country was “taxing the hell out of China.” Some of the implemented tariffs on Chinese goods are as high as 25 percent. [New York Times]

]]>

The heat from wildfires across southeastern Australia is so intense that clouds are raining down fire, not water. There are approximately 161 fires raging across New South Wales and Victoria. Temperatures have exceeded 100 degrees Fahrenheit and the blazes have already destroyed more than 22,000 square miles. [Wall Street Journal]

A massive crash involving two tractor trailers, a tour bus and several other vehicles has left at least five people dead and approximately 60 people sent to local hospitals. The crash, which happened early Sunday morning on the Pennsylvania Turnpike around 3:30 a.m., shut down 86 miles of the highway in both directions. Stephen Limani, a spokesperson from the Pennsylvania State Police, told USA Today, “It was kind of a chain-reaction crash” after the tour bus was struck by two tractor-trailers, followed by a collision with another truck and a passenger vehicle. [USA Today]

During the opening weekend of the NFL playoffs, the New England Patriots were dispatched by the Tennessee Titans, 20-13, bringing an end to an incredible era of post-season dominance. It is “their earliest elimination from the NFL playoffs since 2009” and dashed quarterback Tom Brady hopes for a seventh Super Bowl with the team. The 42-year-old quarterback has said he will not retire, but there are lingering questions about whether he will play with the Patriots next season or join another team. [NBC News]

Demand for theater is certainly alive in Chicago. A touring production of Lin-Manuel Miranda’s show “Hamilton” ended there this weekend after generating $400 million in sales over its three year stint in the city, making it one of the most financially successful shows in the city’s history. The Chicago Tribune reports the production sold “most of its weekend center orchestra seats for premium prices,” there were very few unsold seats, and “it spent very little on advertising, only mounting a significant marketing campaign in its final year.” [Chicago Tribune]

The health commission in Wuhan, a major city in Central China, says a viral pneumonia outbreak that has affected 59 people as of Sunday is not connected to the the flu-like and highly contagious Sars virus. Sars, which stands for severe acute respiratory syndrome, was responsible for the deaths of 349 people in mainland China, as well as 299 people in Hong Kong in 2003. The comments from the health commission were in response to false information published online speculating about the medical cases possibly indicating a resurgence of the virus. [AFP]

]]>