# What are you trying to optimize in Wordle?

Here’s my solution to the Wordle puzzle from yesterday, Sunday, August 7, 2022:

This is a good example of the strategy I generally follow in Wordle:
– start with CRANE, because WordleBot says it’s the best first move. (I used to start with STERN, inspired by the Wheel of Fortune starting letters RSTLNE.)
– usually guess words that are consistent with the information received from previous words. This is what Wordle calls “hard mode” but I don’t actually turn on that setting.

In this case, the E, A, and R in EARTH have different positions from those in CRANE; similarly for RELAY. That in turn forces the position of the R and the E in the fourth word – by the fourth guess the answer must be ??EAR.

So what should my fourth guess be? WordleBot, the New York Times tool for analyzing Wordle play, says that after guessing RELAY, there are three possible solutions, SMEAR, SPEAR, and SWEAR. I believed these were three possible solutions but couldn’t be sure they were the only ones.

Three guesses left, three possible solutions left, so at this point I’m guaranteed to win.

But say I only had two guesses left, and I want to maximize my chance of winning. Then the optimal strategy is to guess a word that includes at least two of P, M, and W – let’s say WIMPY. WIMPY is a wimpy guess in that it’s guaranteed to be wrong, but one of W, M, and P will turn yellow, and this gives the information to get the next guess right. The strategy of guessing SMEAR, SPEAR, and SWEAR in turn has a two-thirds chance of winning.

On the other hand, say I only had one guess left. Then WIMPY has probability zero of winning; any of SMEAR, SPEAR, or SWEAR has two-thirds, so I may as well go for it.

If we ignore the arbitrary six-guess limit, and assume we’re playing to minimize the expected total number of guesses (say, because we have to pay for each guess), then it doesn’t matter what we do – either way the expected number of guesses needed is two. But Wordle collects statistics on how many times you’ve won, and doesn’t compute an average number of guesses, so the framing is really towards maximizing win probability.

There’s a sportsball analogy here. If there’s plenty of time left in the game, playing to maximize the expected number of points is probably the right move; but if there’s little time left, the strategy that maximizes the probability of winning may be different from the probability that maximizes the expected score. Examples include intentional walks in baseball, going for two-point conversions instead of extra points in American football, etc.

Also, picking Sarah Palin as your running mate. That was probably a negative-expectation move for John McCain, but he was already behind. A negative-expectation but high-variance strategy might have been the right one.

# Why is a euro close to a dollar?

As I write this post 1 US dollar = 0.9914 euros. For a moment on Thursday, July 14, 2022 the US dollar was slightly above to euro, where it hasn’t been in twenty years. The chart below is from xe.com (the URL gives a chart for the previous week). chart of USD to EUR conversion rate, July 9 to 16, 2022

This has been in the news (NPR, AP, Reuters, NYT). Of course the barrier is only psychological. But is there some reason the euro is basically in the neighborhood of 1 US dollar? This is astoundingly hard to search for, even before EUR/USD parity cam into the news.

But it turns out there is. The euro is the successor to the European currency unit (ECU), which was a currency basket used internally by the European economic community. The ECU was in turn a successor to the European Unit of Account (EUA), which was defined in 1950 to be 0.888671 grams of gold, and was redefined in terms of a basket of European currencies in the seventies.

Seems like a strange number, 0.888671 grams of gold.

It’s 1/35 troy ounce. The US dollar, under the gold standard, was convertible to gold at $35 per troy ounce. So basically the predecessor to the euro was defined to be worth one US dollar. # Rolling the dice From the March 11, 2022 “Riddler”: We’re playing a game where you have to pick four whole numbers. Then I will roll four fair dice. If any two of the dice add up to any one of the numbers you picked, then you win! Otherwise, you lose. For example, suppose you picked the numbers 2, 3, 4 and 12, and the four dice came up 1, 2, 4 and 5. Then you’d win, because two of the dice (1 and 2) add up to at least one of the numbers you picked (3). To maximize your chances of winning, which four numbers should you pick? And what are your chances of winning? Some first thoughts: • You want numbers that are common as the sums of two dice – middling numbers, numbers near seven. • The problem has a reflection symmetry. The dice values $x_1, x_2, x_3, x_4$ win with the target sums $y_1, y_2, y_3, y_4$ if and only if the dice values $7-x_1, 7-x_2, 7-x_3, 7-x_4$ win with the target sums $14-y_1, 14-y_2, 14-y_3, 14-y_4$. Putting these together, a symmetric set of middling numbers seems likely to be the best target set – something like 5, 6, 8, 9. This is a nasty case analysis, but it’s easy to do by brute force in R. library(tidyverse) dice_and_targets = expand.grid(d1 = 1:6, d2 = 1:6, d3 = 1:6, d4 = 1:6, t1 = 2:12, t2 = 2:12, t3 = 2:12, t4 = 2:12) %>% filter(t1 < t2 & t2 < t3 & t3 < t4) %>% mutate(s12 = d1 + d2, s13 = d1 + d3, s14 = d1 + d4, s23 = d2 + d3, s24 = d2 + d4, s34 = d3 + d4) The data frame dice_and_targets has a row for every possible combination of dice results (d1 … d4) and targets (t1 … t4), and the sums of the dice (s12 … s34). It’s a big data frame, with $6^4 \times {11 \choose 4} = 1296 \times 330 = 427680$ rows, one for each of the 1296 possible dice rolls and 330 choices of targets. Let’s take a look at a sample of this data frame, consisting of 10 randomly selected rows: set.seed(1) dice_and_targets$idx = sample(nrow(dice_and_targets))
dice_and_targets %>% filter(idx <= 10) %>% select(-idx)

d1 d2 d3 d4 t1 t2 t3 t4 s12 s13 s14 s23 s24 s34
1   1  2  3  5  2  3  5  8   3   4   6   5   7   8
2   4  1  6  5  2  3  4  9   5  10   9   7   6  11
3   2  3  4  5  3  6  7  9   5   6   7   7   8   9
4   2  1  5  6  3  5  8  9   3   7   8   6   7  11
5   1  6  4  6  2  6  9 10   7   5   7  10  12  10
6   3  4  4  6  3  6  7 11   7   7   9   8  10  10
7   2  2  4  2  4  8 10 11   4   6   4   6   4   6
8   1  2  3  5  2  3  4 12   3   4   6   5   7   8
9   2  6  6  1  2  6  8 12   8   8   3  12   7   7
10  1  6  3  6  6  8 11 12   7   4   7   9  12   9

Consider, for example, the first row. In this case we roll 1, 2, 3, and 5; the targets are 2, 3, 5, and 8; the pairwise sums are 1+2 = 3, 1+3 = 4, 1+5 = 6, 2+3 = 5, 2+5 = 7, and 2+6 = 8; and we win the game, in fact three times over, since the pairwise sums include three of the targets, namely 3, 5, and 8.

Next we can work out which rows win, leveraging some bitwise operations because how often do I get a chance to use these?

dice_and_targets = dice_and_targets %>% mutate(target_bits = bitwOr(bitwOr(2^t1, 2^t2), bitwOr(2^t3, 2^t4)))
dice_and_targets = dice_and_targets %>%
mutate(sum_bits = bitwOr(bitwOr(bitwOr(2^s12, 2^s13), bitwOr(2^s14, 2^s23)), bitwOr(2^s24, 2^s34)))
dice_and_targets = dice_and_targets %>%
mutate(win = bitwAnd(target_bits, sum_bits) > 0)

In this case target_bits has the bit corresponding to $2^t$ set if $t$ is one of the targets; sum_bits has the bit corresponding to $2^s$ set if $s$ is one of the pairwise sums. Then bitwAnd(target_bits, sum_bits) has a nonzero bit if and only if we have a winning combination.

Let’s look at those randomly selected rows, now with the wins figured out:

dice_and_targets %>% filter(idx <= 10) %>% select(-idx)

d1 d2 d3 d4 t1 t2 t3 t4 s12 s13 s14 s23 s24 s34 target_bits sum_bits  win
1   1  2  3  5  2  3  5  8   3   4   6   5   7   8         300      504 TRUE
2   4  1  6  5  2  3  4  9   5  10   9   7   6  11         540     3808 TRUE
3   2  3  4  5  3  6  7  9   5   6   7   7   8   9         712      992 TRUE
4   2  1  5  6  3  5  8  9   3   7   8   6   7  11         808     2504 TRUE
5   1  6  4  6  2  6  9 10   7   5   7  10  12  10        1604     5280 TRUE
6   3  4  4  6  3  6  7 11   7   7   9   8  10  10        2248     1920 TRUE
7   2  2  4  2  4  8 10 11   4   6   4   6   4   6        3344       80 TRUE
8   1  2  3  5  2  3  4 12   3   4   6   5   7   8        4124      504 TRUE
9   2  6  6  1  2  6  8 12   8   8   3  12   7   7        4420     4488 TRUE
10  1  6  3  6  6  8 11 12   7   4   7   9  12   9        6464     4752 TRUE


In the first row, here, target_bits is $2^8+2^5+2^3+2^2 = 100101100_2 = 300$ and sum_bits is $2^8+2^7+2^6+2^5+2^4+2^3 = 111111000_2 = 504$. And bitwAnd(target_bits, sum_bits) (not shown) is $2^8 + 2^5 + 2^3 = 100101000_2 = 296$. Since it’s greater than zero, that counts as a win.

You might get the idea that it’s impossible to lose from this sample. We got a little lucky here: mean(dice_and_targetswin) returns 0.874018. If you pick a random roll of four dice and four random targets out of 2, 3, …, 12, one of the pairwise sums will be in the target set 87% of the time. But we want to know which target set makes a win most likely. win_counts_by_target = dice_and_targets %>% group_by(t1, t2, t3, t4) %>% summarize(wins = sum(win)) %>% arrange(desc(wins)) head(win_counts_by_target) > head(win_counts_by_target) # A tibble: 6 x 5 # Groups: t1, t2, t3  t1 t2 t3 t4 wins <int> <int> <int> <int> <int> 1 4 6 8 10 1264 2 2 6 8 10 1246 3 4 6 8 12 1246 4 4 6 7 9 1238 5 5 7 8 10 1238 6 4 7 8 9 1236  There we go! And not a loop in sight. Once I have the 4, 6, 8, 10 target set it’s easy to come up with that number 1264. Consider just the dice that show even numbers – you win if there are at least two of these and they’re not all showing 6. Similarly you win if there are at least two odd dice and they’re not all showing 1. So the losing combinations are $(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 4), (1, 1, 1, 6), (1, 1, 6, 6), (1, 6, 6, 6), (3, 6, 6, 6), (5, 6, 6, 6), (6, 6, 6, 6)$ and their rearrangements, of which there are 32. Incidentally, my initial guess (5, 6, 8, 9) wasn’t bad – it wins 1228 times out of 1296, good enough for 14th place out of the 330 possible target sets. And the very worst target set? It’s (2, 3, 11, 12). No surprise there, although even that one wins 776 times out of 1296, nearly 60% of the time. # Expected number of all-Confederate World Series The World Series starts today. Atlanta vs. Houston. This is wrong for multiple reasons: • the Astros are cheaters • the Astros are a National League team • as a native Philadelphian, I’m obligated to hate the Braves, even though I moved to Atlanta • these are warm-weather teams and part of the fun of the ridiculously late postseason is that it’s not really baseball weather, but I just went for a walk and it’s pretty nice out. Nathaniel Rakich observed that this is the first-ever World Series between teams from the former Confederacy. This surprised me! But there are only five teams in the former Confederacy, out of 30 in MLB (29 of which are in the US). In chronological order of formation, they are • the Houston Astros (NL 1962-2012, AL 2013-present) • the Atlanta Cobb County Braves (NL 1966-present, moved from Boston) • the Texas (Dallas-area) Rangers (AL 1972-present, moved from Washington) • the Florida/Miami Marlins (NL 1993-present) • the Tampa Bay Devil Rays (AL 1998-present) In particular Missouri never seceded, which matters quite a bit here because the St. Louis Cardinals have been in the World Series the second-most of any team. First, a few words about Major League Baseball. There are currently two “leagues” comprising MLB, the National League and the American League. Each has 15 teams, of which one (the “pennant winner”) will make it to the World Series. Organized baseball got started in the late 19th century, and its “classic” alignment of 16 teams were all in northern cities, since there were few large southern cities at the time. From 1903 to 1952 the teams were located as follows: Boston x2, Brooklyn, Chicago x2, Cleveland, Cincinnati, Detroit, New York x2, Philadelphia x2, Pittsburgh, St. Louis x2, Washington. In 1953-1972 a bunch of teams moved but since then MLB has mostly grown via expansion. The former Confederacy is still is underrepresented in MLB – it has population of about 108 million, compared to the US population of 331 million, so it “ought” to have nine or ten teams. Or, if you’re going to argue that an MLB team has to be in a big city, nine of the thirty largest metropolitan areas are in the former Confederacy. (In order, they’re Dallas, Houston, Miami, Atlanta, Tampa, Orlando, Charlotte, San Antonio, and Austin. The first five have teams, and I believe the latter four have been thrown around as expansion candidates.) So although historically the country’s big cities may have been in the north, this is less true now. Given the historical locations of the teams, how many all-Confederate World Series would we expect? We start counting in 1972, when the American League got its first team in the former Confederacy. So for example, in each of 1972-76, the chances of both pennant winners coming from the former Confederacy were 2/12 x 1/12 = 2/144. With the current alignment it’s 2/15 x 3/15 = 6/225. The expected number of all-Confederate World Series is 5 x 2/12 x 1/12 + 16 x 2/12 x 1/14 + 5 x 3/14 x 1/14 + 15 x 3/16 x 2/14 + 9 x 2/15 x 3/15 = 0.978 which is honestly lower than I expected! But it’s only fairly recently that there have been an appreciable number of MLB teams in this part of the country, and the fact that you need teams from both leagues to get through really keeps this number down. # Which countries are better at the Winter Olympics than the Summer Olympics? From Reddit (posted by u/RoadyHouse): Which Olympic Games are these European countries the best? This is a map which shades countries: • blue if they’ve won more gold medals in the Winter Olympics than the Summer Olympics • yellow if they’ve won more gold medals in the Summer Olympics than the Winter Olympics • red if they’ve won no gold medals The only blue countries are Norway, Austria, Switzerland, and Liechtenstein. These certainly seem like a wintry set of countries (one of the big tourist attractions in Oslo is the ski jumping hill) but surely, say, Sweden should be on here? Or the Dutch with the speed skating? Or Canada? Do they even have summer there? (A side note about that ski jumping hill – you can take the subway to it. But then you have to climb up a hill to get there! This is obvious in retrospect – of course the ski jumping hill would be on a hill! – but it was still exhausting. Also, they don’t really explain why ski jumping is a thing. I assume it involves young men and alcohol.) The answer is that there are just a lot more events in the Summer Olympics than the Winter Olympics (and the Summer Olympics have been going on longer). So there have been 5,121 gold medals awarded all-time in the Summer Olympics but only 1,062 in the Winter Olympics, according to the all-time medal table at Wikipedia. Consider for example Sweden. They’ve won 148 summer gold medals out of the total of 5,121, or 2.89% of all summer gold medals. They’ve won 57 out of the 1,062 winter gold medals, or 5.36% of all winter gold medals. So it’s reasonable to say that Sweden is better at the winter Olympics than the summer Olympics. If you wanted to put a number on it, 5.36%/2.89% = 185% so you could say they’re 85% better at winter than summer. If I’ve done it right, the list of countries that are better at the winter Olympics than the summer Olympics are, in order from most: Liechtenstein (their only gold medals ever are in winter), Austria, Norway, Switzerland, Canada, Belarus, Czech Republic, Netherlands, Germany, Finland, Estonia, Sweden, South Korea, Russia, Slovakia, Croatia, East Germany, Slovenia, Latvia. Do you like maps? Here that is as a map. I’m not surprised that this list is so Eurocentric. The Soviet Union, West Germany, Italy, and France just barely miss it. (I haven’t made any effort to merge together the various Germanies, or deal with the Soviet Union and its various successor states.). Many Winter Olympic sports have a high barrier to entry just in terms of what facilities are available – to take an extreme example there are only fifteen luge tracks in the world – and lots of countries just don’t have enough winter to have winter sports. So this is essentially a map of rich, cold countries. As one reporter put it during the 2018 Olympics, “From a sports perspective, Norway is rich as shit“. Mountains help too – Denmark has 48 summer gold medalists but no winter gold medalists. Maybe the Danes should take up speed skating like the Dutch. (This post originated as a Reddit comment. Map made using mapchart.net.) # Products equaling sums From The Riddler, June 25, 2021, a puzzle from Matt Enlow (which was originally published in Math Horizons, although I can’t find the original): A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is exactly 20 percent. How many marbles of each color are in the bag? It doesn’t seem like there’s enough information, but there is. I recommend not trying to solve this puzzle in your head while navigating the day care parking lot, though. Let’s say the numbers of the three colors are x, y and z. Then the number of sets of three marbles which contain one of each color is just xyz. And the number of total sets of three marbles is ${100 \choose 3}$. So we need to find positive integers x, y, z such that x + y + z = 100 and $5xyz = {100 \choose 3}.$ It’ll help to find the prime factorization of ${100 \choose 3}$, so we expand it: ${100 \choose 3} = {100 \times 99 \times 98 \over 3 \times 2 \times 1} = {2^2 \times 5^2 \times 3^2 \times 11 \times 2 \times 7^2 \over 3 \times 2} = 2^2 \times 3 \times 5^2 \times 7^2 \times 11.$ Dividing out the 5, we need $xyz = 2^2 \times 3 \times 5 \times 7^2 \times 11.$. The three big factors – the two 7s and the 11 – look like they’ll be worrisome. So let’s spread them out: let x and y both be multiples of 7, and let z be a multiple of 11. To get them to add up to 100, we need to find a multiple of 7 and a multiple of 11 that add up to 100; it’s easy to find 44 and 56. So z = 44. That leaves $xy = 3 \times 5 \times 7^2$, so x and y must be 21 and 35. Alternately, we can write some code to solve it, but what fun is that? n = 100 target = (n * (n-1) * (n-2))/6 for x in range(1, n+1): for y in range(x, int((n-x)/2)+1): z = n-x-y; if target == 5*x*y*z: print(x, y, z)  which prints out 21 35 44 as expected. But now that we have some code, is there something special about 100? Let’s iterate over the first 1000 integers: for n in range(1, 1000): target = (n * (n-1) * (n-2))/6 for x in range(1, n+1): for y in range(x, int((n-x)/2)+1): z = n-x-y; if target == 5*x*y*z: print(n, x, y, z)  returns 6 1 1 4 10 2 2 6 22 4 7 11 26 5 8 13 27 5 9 13 35 7 11 17 36 7 12 17 40 8 13 19 46 11 12 23 56 11 21 24 65 13 24 28 76 19 20 37 100 21 35 44 126 31 35 60 330 82 94 154 345 86 98 161 352 78 120 154 352 91 96 165 406 101 116 189 436 109 124 203 497 124 142 231 512 128 146 238 737 161 268 308  The sequence 6, 10, 22, 26, 27, 35… isn’t in the OEIS as of this writing, although some other Riddler puzzles have made it; it may be there later. These are all numbers where not just n has a lot of factors, but also n-1 and n-2 – for example 345 = 3 x 5 x 23, 344 = 2 x 2 x 2 x 43, 343 = 7 x 7 x 7. Thus ${n \choose 3}/5$ will have many factors, and many ways to be written as a product of three integers, making it more likely that one of those will have the right sum. I had been expecting an algebraically cleaner solution, because this reminded me of another puzzle: a bag has r red and b blue balls in it. Find r, b such that the probability that two balls picked at random have different colors is exactly one-half. In this case similar code returns the solution 4 1 3 9 3 6 16 6 10 25 10 15 36 15 21 49 21 28 64 28 36 81 36 45 100 45 55 from which we can guess that the solutions are $r = (n^2-n)/2, b = (n^2+n)/2$. # Two dice problems Nick Berry asks: you roll a pair of six-sided dice and sum that to get a total. Your good friend does the same. What are the chances that you both get the same total? My instinct is to say that it should be somewhere between 1/6 and 1/11 – you’d get 1/6 if the question were about one die and 1/11 if it were about picking uniformly a number from 2, 3, …, 12. Let’s generalize. What if you roll two n-sided dice? In that case, the probabilities of getting 2, 3, …, n+1 are $1/n^2, 2/n^2, \ldots, n/n^2$, and the probabilities of getting n+2, n+3, …, 2n are $(n-1)/n^2, (n-2)/n^2, \cdots, 1/n^2$ – generalizing the well-known triangular distribution. The probability that you and your friend both roll 2, then, is $\displaystyle \left( {1 \over n^2} \right)^2 + \left( {2 \over n^2} \right)^2 + \cdots + \left( {n-1 \over n^2} \right)^2 + \left( {n \over n^2} \right)^2 + \left( {n-1 \over n^2} \right)^2 + \cdots + \left( {1 \over n^2} \right)^2$ or, putting everything over a common denominator, $\displaystyle {1^2 + 2^2 + \cdots + (n-1)^2 + n^2 + (n-1)^2 + \cdots + 1^2 \over n^4}$ At this point we can already see that there are order-of-n terms in the numerator, of order $n^2$, so we should get a result of order 1/n. We can peservere and do the algebra and we get $\displaystyle {2n^3 + n \over 3n^4}$ or just a hair over 2/3n. Berry also asks what happens if you have more dice. For k six-sided dice he gives the results of a Monte Carlo simulation which has roughly $1/{6\sqrt{k}}$. He writes that “the percentage chance of a match score falls off a lot slower than I would have predicted.” Why does it fall off so slowly? The mean result from a single die is 7/2, with variance 35/12. So the mean result from k dice is 7k/2, with variance 35k/12 – so only values within a few multiples of $\sqrt{35k/12}$ are possible with any appreciable frequency. There are on the order of a few square roots of k of these, so the answer should be $1/\sqrt{Ck}$ for some $k$. Like Berry, I am too lazy to find the constant $C$. Edited to add: Matthew Aldridge gives in the comments an argument (for approximating that for $k$ six-sided dice, withk$large, the probability of both getting the same total is approximately $1/\sqrt{pi \times 35k/3}$, by approximating both sums as normal and applying the central limit theorem. This is about $1/(6.054 \sqrt {k})$. If we bear in mind that the variance of a single roll of an n-sided die is $(n^2-1)/12$, then we get $1/\sqrt{pi/3 \times (n^2-1)k}$ for rolling k n-sided dice. This approximation is surprisingly good even for k = 1, where it has no business being good. It gives $1/\sqrt{pi/3 \times (n^2-1)}$. This reduces to 1/n, the correct answer, if you let $\pi = 3$ and ignore the -1. # Rational approximations of logs of small primes Let’s say for some reason we need to approximate decimal logarithms of some small primes. We know $2^{10} \approx 10^3$, so clearly $log 2 \approx 3/10$. (All logarithms in this post are to base 10.) Can we approximate the logs of any other primes this way? When I run through powers of small primes, the target that jumps out at me is $7^4 = 2401 ~ 2400 = 2^5 3^1 5^2$. Taking logs of both sides gives $4 L_7 \approx 5 L_2 + L_3 + 2 L_5$ where $L_x = \log_{10} x$. And of course we have $L_2 + L_5 = 1$, so this relation becomes $4L_7 \approx 2 + 3 L_2 + L_3$. Are there other such relations? It turns out that there are finitely many numbers n such that n and n+1 have all their prime factors < 7 – apparently a conseuqence of the abc conjecture – so I just take the largest ones and derive relations like these. So I can derive similar relations from $\log 224 \approx \log 225$ and$\log 4374 \approx \log 4375, namely $7 L_2 + L_7 \approx 2 L_3 + 2$ and $5 L_2 + 7 L_3 \approx 4 + L_7$ This is a system of three (approximate) equations in three unknowns, and solving it in the usual way gives $L_2 = 72/239 \approx 0.30125, L_3 = 114/239 \approx 0.47699, L_7 = 202/239 \approx 0.84519$. while the true values are 0.30103, 0.47712, and 0.84510 – so this is good enough for three-place accuracy. In practice, who wants to have a rational with 239 in the denominator as an approximation? More practical is the following: • remember $2^{10} \approx 10^3$ (from familiar powers of two), or, taking 120th roots, $2^{1/12} \approx 10^{1/40}$. • remember $2^{19} \approx 3^{12}$ (a fundamental fact of musical temperament). That gives $\log 2 \approx 3/10$ right away, $\log 3 \approx 19/12 \log 2 = 19/40$, and then $\log 2400 = 2 + 3 \log 2 + \log 3 = 3.375$ so $\log 7 \approx 3.375/4 = 0.84375$. This is essentially I. J. Good’s singing logarithms, which exploits the fact that a lot of rational numbers with small numerator and denominator can be approximated as powers of $2^{1/12}$, the fact on which our usual musical tuning system is based: $2^{7/12} \approx 3/2, 2^{4/12} \approx 5/4$, and a bit less accurately $2^{10/12} \approx 7/4$. (Musically these are the facts that the perfect fifth, major third, and minor seventh are the ratios 3/2, 5/4, and 7/4.). From the last we can derive $2^{34/12} \approx 7$, which gives $\log 7 \approx 34/40 = 0.85$. This is a less good approximation than those of $\log 3$ and $\log 5$, just as the seventh harmonic isn’t really a note in the equal-tempered scale – the harmonic or “barbershop” seventh is noticeably flat compared to the minor seventh. # Objects in my house with icosahedral symmetry Somehow toys just show up in this house, a phenomenon which I think is familiar to many parents. Some of them have icosahedral symmetry. Like this one. Hard plastic icosahedron ball, with stars for vertices. You can see that the designer was outlining the vertex of an icosahedron with that five-pointed star shape, but they couldn’t quite commit – the star points don’t actually point to the next vertex! (And no, they don’t turn.) You can get a better sense of the stars corresponding to the vertices of an icosahedron here: There’s also this one, which is softer and has a couple of distinguished vertices antipodal to each other: Soft icosahedron, with rattly bits at two antipodal vertices. And this one which I bought for myself years ago, presumably in some sort of store that sold housewares. (Remember stores?) Skeleton of an icosahedron. The nice thing about this one is that you can see through it, which makes for some interesting photographic possibilities, such as this view where two antipodal vertices are aligned: Skeleton of an icosahedron, photographed down an axis through two vertices. and this view with that emphasizes a threefold rotational symmetry: Skeleton of an icosahedron, emphasizing threefold symmetry. Of course we have a soccer ball somewhere. You know what a soccer ball looks like, I’m not taking a picture. Finally, the Twitter feed for this blog has as its icon an icosahedron, which I got from Wikimedia Commons: I also use this as an avatar in various work systems that need one – these generally require small pictures and a face wouldn’t show up well, and it’s easier to pick out than the default in a lot of these systems which is just someone’s initials in a circle. I do not yet have an icosahedron as a tattoo, but I’ve liked it for a while. If I were to get a tattoo it would be either an icosahedron or the diagram from Byrne’s rendition of Euclid’s proof of the Pythagorean theorem. (The link goes to Nicolas Rougeux’s interactive enhancement of the same.) # STOP POTS POST. SPOT OPTS TOPS. I had thought, for a few decades now, that STOP was the four-letter word with the most anagrams, with six: STOP itself, POST, POTS, TOPS, OPTS, SPOT. So of course when Josh Millard put out these STOP permutations signs, I had to buy one. It’s a limited edition of 24 stop sign prints, one for each permutation. (I opted for OPTS. As of this writing there are 18 still available; the 6 that have been bought are the five anagrams of STOP other than STOP itself, and SOTP.) But then I had to check that claim. Peter Norvig has, meant to accompany a chapter on NLP, some word lists, of which I’ve used the enable1.txt list before for word puzzles. (I’m not sure who compiled this list.) We can put words into a canonical form by alphabetizing the letters – for example michael becomes acehilm, and stop becomes opst. Scrabble players call this an alphagram. Then to find the four-letter word with the most anagrams is just a matter of counting. library(tidyverse) words = read_csv(url('https://norvig.com/ngrams/enable1.txt'), col_names = FALSE) colnames(words) = 'word' alphabetize_word = function(w){paste(sort(strsplit(w, '')[]), collapse = '')} wordsalphagram = sapply(words$word, alphabetize_word) words$len = nchar(words\$alphagram)
alphagram_counts = words %>% group_by(alphagram, len) %>%
summarize(n = n(), anagrams = paste0(word, collapse = ', '))
alphagram_counts %>% filter(len == 4) %>% arrange(desc(n))


And here are the four-letter words with the most anagrams:

alphagram_counts %>% filter(len == 4) %>% select(-len) %>% arrange(desc(n))

# A tibble: 2,655 x 3
alphagram     n anagrams
<chr>     <int> <chr>
1 aest          8 ates, east, eats, etas, sate, seat, seta, teas
2 aers          7 ares, arse, ears, eras, rase, sear, sera
3 ailr          7 aril, lair, lari, liar, lira, rail, rial
4 astw          6 staw, swat, taws, twas, wast, wats
5 opst          6 opts, post, pots, spot, stop, tops
6 ostw          6 stow, swot, tows, twos, wost, wots
7 aeht          5 eath, haet, hate, heat, thae
8 aels          5 ales, lase, leas, sale, seal
9 aelt          5 late, tael, tale, teal, tela
10 aelv          5 lave, leva, vale, veal, vela
# … with 2,645 more rows

No! Child me was wrong!

But wait! What is “seta”? Is “ates” really a thing – you can’t pluralize a verb like that! (“ate” appears to be Tagalog for “older sister”.) Perhaps the aers set, with seven anagrams, wins, but “sera” is technical (plural of serum), and as an American I have trouble recognizing “rase” as a legitimate spelling of “raze”. “lari” is a unit of money in Georgia (Tbilisi, not Atlanta) which I was unfamiliar with. And so on.

Fortunately Norvig also has a list of word frequencies (count_1w.txt), of the 332,202 most common words in a trillion-word corpus. (One of the perks of working at Google, I assume.) So we can read that in.

freqs = read_delim(url('https://norvig.com/ngrams/count_1w.txt'),delim = '\t', col_names = FALSE)
colnames(freqs) = c('word', 'freq')


The most common words are the ones you’d expect. (2.3% of words are “the”.)

> head(freqs)
# A tibble: 6 x 2
word         freq
<chr>       <dbl>
1 the   23135851162
2 of    13151942776
3 and   12997637966
4 to    12136980858
5 a      9081174698
6 in     8469404971

And the least common words are… barely words. (I don’t know the full story behind this dataset.) So it seems reasonable that all “real” words will be here.

> tail(freqs)
# A tibble: 6 x 2
word     freq
<chr>   <dbl>
1 goofel  12711
2 gooek   12711
3 gooddg  12711
4 gooblle 12711
5 gollgo  12711
6 golgw   12711

Now we can attach frequencies to the words. There are too many words in the sets for a table to be nice, so we switch to plots.

words %>% left_join(alphagram_counts)  %>%
filter(len ==  4 & n >= 6)  %>%
left_join(freqs) %>% arrange(alphagram, desc(freq)) %>%
select(alphagram,  word, freq) %>% group_by(alphagram) %>%
mutate(rk = rank(desc(freq))) %>%
ggplot() + geom_line(aes(x=rk,  y=log(freq/10^12, 10), group = alphagram, color = alphagram)) +
scale_x_continuous('rank within alphagram set', breaks  = 1:8, minor_breaks = c()) +
scale_y_continuous('log_10 of word frequency', breaks = -8:-3, minor_breaks  = c()) +
theme_minimal() + geom_text(aes(x=rk, y=log(freq/10^12, 10), color = alphagram, label = word)) +
ggtitle('Frequency of four-letter words with six or more anagrams')


And if we plot the frequency of each word against its rank in its own anagram set

then we can see that the STOP set consists of much more common words than any of the others. (STOP isn’t even the most common of its own anagrams, which surprises me – that honor goes to POST. But when I was a small child STOP seemed much more common, because of the signs.) I’m surprised to see SERA so high; this is either an extremely technical corpus or (more likely) contamination from Spanish.

And here’s a similar plot for five letters. Here I’d thought the word with the most anagrams was LEAST (among “common” words, 6: TALES, STEAL, SLATE, TESLA, STALE) but it looks like SPARE wins with room to spare, even if you don’t buy that APRES is an English word.