What is the date (day, month, and year) at the beginning of The Pirates of Penzance?

Solution follows.

The Pirates of Penzance is the story of Frederic, a reluctant pirate’s apprentice. As soon as he turns 21 his apprenticeship is over and he promptly leaves his pirate comrades for a more virtuous life on land, where he meets and falls in love with the beautiful Mabel.

Later, the Pirate King, Frederic’s erstwhile captain, shows up to enforce the terms of his indenture, which (no one had realized) specify that Frederic is to serve not until his 21st year but until his 21st birthday. As the Pirate King points out (in song, natch):

For some ridiculous reason, to which however I’ve no desire to be disloyal
Some person in authority — I don’t know who, very likely the Astronomer Royal –
Has decided that, although for such a beastly month as February, twenty-eight days as a rule are plenty
One year in every four, his days shall be reckoned as nine and twenty
Through some singular coincidence (I shouldn’t be surprised if it were owing to the agency of an ill-natured fairy)
You are the victim of this clumsy arrangement, having been born in leap year on the twenty-ninth of February
And so, by a simple arithmetical process, you will easily discover
That, though you’ve lived twenty-one years, yet if we go by birthdays, you’re only five and a little bit over.

Frederic, an admitted “slave of duty,” reluctantly agrees that he is still bound to the pirates. Breaking the news to Mabel, he sings to her,

In 1940 I of age shall be,
I’ll then return and claim you, I declare it!

From that you might conclude that, if his twenty-first birthday is in 1940, then he’ll be 84 that year (21×4) and that he was therefore born in 1856, but you’d be wrong!

Everyone knows that leap year comes once every four years, but not as many people know that it’s more complicated than that. It’s a leap year if:

• The year is a multiple of four;
• But not if the year is a multiple of one hundred;
• Except when the year is a multiple of four hundred.

1996 was a multiple of four, and not a multiple of 100, and not a multiple of 400, so it was a leap year. 2000 was a multiple of four, so it would have been a leap year, except that it was also a multiple of 100, so it would not have been a leap year, except that it was also a multiple of 400, so it was a leap year — and because of that, it allowed some people who believed “every four years” is the complete rule to continue believing that.

So by these rules, 1900 was not a leap year. If Frederic had been born in 1856, then 1940 would have been only his twentieth birthday, not his twenty-first. For 1940 to be his twenty-first birthday he had to have been born in 1852.

So the action at the beginning of The Pirates of Penzance takes place on 1 March 1873 — the first day after Frederic’s twenty-first year. (Assuming Frederic was aware that 1900 would not be a leap year when he told Mabel “1940″ — which, being such a stickler for the rules, he surely was.)

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A lower price wasn’t the Fit’s only advantage over the Prius. While car shopping I rented a Prius for a one-day extended test drive, ending up with three specific complaints:

1. Visibility through the rear window is poor;
2. The console’s large, animated engine-performance display is dangerously distracting;
3. The keyless engine-start button is (a) unsafe with small children around but (b) too cool to disable with the child-safety lock.

More than a year ago I replaced my 1998 Honda Civic hatchback with a new Honda Fit. Fuel efficiency was a key decision criterion for me, and naturally I considered the Toyota Prius; but the Prius gets its best gas mileage in city driving, and at the time of my purchase most of my driving was on the highway, where the Fit’s efficiency was close to that of the Prius, at a much lower price.

I’ve been tracking my Fit’s fuel consumption on a spreadsheet for several months now and the trend is clear: its efficiency is consistently in the 35 MPG range. Nothing to sneeze at, especially given the dismal fuel economy of almost all other cars on the market; but disappointingly it falls short of the mileage I was getting with my Civic at the end, which occasionally exceeded 40 MPG — with the previous decade’s engine technology!

You can see the mileage I’m getting, fill-up-by-fill-up, in my Google Docs spreadsheet.

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I have a thousand songs on one of those thumb drives, and I always play them in “shuffle” mode. Yet it seems that there is always a lot of overlap between one listening session and another — the same songs that I heard yesterday are in today’s mix. You’d think that with a thousand songs to choose from, it would be a while before I hear the same song twice, unless there’s something not sufficiently random about the PlayStation’s song randomizer.

I was all prepared to fire off an indignant letter to Sony’s customer support department when I decided I first needed to understand exactly how unlikely was the overlap I was encountering.

Figure that a “listening session” includes twenty songs. There are 339,482,811,302,457,603,895,512,614,793,686,020,778,700 (339 duodecillion) different ways to choose twenty songs from a collection of a thousand. This result is given by the combinatorial formula:

n! / k!(n-k)!

where n is the number of items to choose from (1,000, in this case), k is the number of items to choose (20), and “!” is the “factorial” operator that means “multiply the preceding number by every other number between it and 1.” Five factorial, for instance, is written “5!” and is equal to 5×4×3×2×1, which is 120.

The combinatorial formula above is sometimes abbreviated “nCk,” pronounced “n choose k.” The very very big number is the result of calculating 1000 C 20.

So there is a vast number of possible listening sessions. But in how many ways can one listening session overlap with another? Let’s consider a second listening session that doesn’t overlap at all with the first. The way to think about this is that the first listening session “used up” twenty of the available songs, leaving 980 to choose from — specifically, 980 from which to choose 20, or 980 C 20, which is 225,752,650,356,644,030,123,857,337,771,499,346,518,885 (225 duodecillion).

So of the 339 duodecillion ways to choose 20 songs from a thousand, 225 duodecillion, or 66%, do not overlap — but that means that 34% do overlap. There is a one-in-three chance that at least one song in the second session will be the same as one in the first.

This was a stunning result to me. I never expected the odds of an overlap to be so high.

That doesn’t mean that the PlayStation is working correctly, necessarily; it’s my impression that I’m getting multiple-song overlaps, and I’m getting them much more than one-third of the time, so the PlayStation still may not be adequately randomizing its playlist. But this result does send me back to the drawing board to gather objective data about just how much overlap I am getting.

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(Didn’t know that I had an adult stepdaughter? I haven’t mentioned her here before, but we added her to the cast a couple of seasons ago in a Cousin Oliver moment to boost our sagging ratings.)

Update: Oh drat, Pamela’s 26, not 27. What’s interesting about the relationship between her age and Archer’s now? Umm… Archer’s is the number of suits in a deck of playing cards, and Pamela’s is the number of red (or black) cards? Sorry, that’s the best I’ve got.

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Come to think of it, that’s only if you consider my mounting AdSense balance to be an arithmetic progression. On the other hand, ten dollars is a thousandfold increase over my balance the last time I reported it almost two years ago. With just two data samples it’s impossible to tell whether the progression is arithmetic or geometric. If the latter, then my balance has been growing at better than 1% per day. At that rate, by the time I retire you will all bow before benevolent supreme dictator Bob and his 8.7 duodecillion dollars, mwa ha ha ha ha ha! Even if the dollar collapses, that oughta be worth something on eBay. Thanks, AdSense!

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To goose my weight-loss regime, which appears to have stalled once again — though I am holding my own against the “Google 15” — I am adding a daily exercise regimen for the first time, inspired by hundredpushups.com: each day I will do one more push-up than the day before, starting with one today and culminating with a hundred push-ups by my birthday.

If I keep exactly to that plan, the total number of push-ups I’ll do is 5,050 — one today, two tomorrow, three on Wednesday, four on Thursday, and so on. The sum of the first N numbers from 1 through N is, in general,

(N+1) × N/2

an elegant intuitive proof of which is as follows. List the first N numbers, let’s say 6 for this example:

1 2 3 4 5 6

The sum of the two “outer” numbers is 7:

1 2 3 4 5 6

Removing those, the sum of the next two “outer” numbers is also 7:

2 3 4 5

Removing those, the sum of the final pair is also 7:

3 4

That sum — N+1 — is repeated N/2 times, giving rise to the formula

(N+1) × N/2

“Wait a minute,” I hear you say. “What about when N is odd? Then there’s one extra innermost number with no partner.” That’s true. In that case, the number of pairs that add up to N+1 isn’t N/2, it’s only (N-1)/2:

1 2 3 4 5 6 7

Here N is 7, and there are 3 pairs that add up to 8 — 1 and 7, 2 and 6, 3 and 5 — and 4 is all alone in middle. So the sum is:

(N+1) × (N-1)/2 + the middle number

But the middle number is always (N+1)/2, so this becomes:

(N+1) × (N-1)/2 + (N+1)/2

which is the same as

(N+1)/2 × (N-1) + (N+1)/2

which can be read as adding one more (N+1)/2 to a collection of N-1 of them, for a total of N (N+1)/2’s:

(N+1)/2 × N

which is the same as

(N+1) × N/2

which is the same as the original formula above whether N is odd or even. QED.

OK, let’s get this regimen started. Rrrrrnnnnngghh — one. Whew.

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It comes down to a calculation of the kinetic energy of the projectile. Fortunately it’s very easy to approximate by making a few assumptions and by ignoring the effects of air resistance and the spray of cake batter.

The projectile was the mixer’s own removable endcap. As soon as it vibrated loose, fell into the bowl, and was struck by the spinning mixer blades, it was on a ballistic trajectory, arcing up, past my head, and then down behind me. Let’s assume that the highest point of the trajectory was about level with the top of my head, roughly 1.5 meters off the ground, and that this height was attained just as the projectile was passing me, meaning that once it did pass me, it had already started down.

We can decompose the motion of the projectile into a pair of vectors: the one pointing straight down to the floor and the one perpendicular to it, pointing horizontally past my head. The speed in that direction was constant until the endcap hit the floor. The speed in the floorward direction was increasing due to gravity.

Since we’ve assumed the endcap reached its apex as it passed my head, we know that its downward velocity at that moment was zero. We also know that the acceleration in that direction (due to gravity) is 9.8 meters per second per second. Finally we know from high school physics that:

distance = initial-velocity×time + acceleration×time²/2

and since we know initial-velocity is 0, we can rearrange this to say:

time = √distance/acceleration

And since we know “distance” is 1.5 meters and “acceleration” is 9.8 m/s², we know that it took about 0.4 seconds for the endcap to fall from the height of my head, regardless of its motion in the horizontal direction.

In that 0.4 seconds I estimate (based on where I later found the endcap) that it covered a horizontal distance of 3 meters, giving it a speed of 7.5 meters per second.

I haven’t weighed the endcap but I’m going to guess it’s about 0.25 kilograms (around half a pound). Again, high school physics tells us that kinetic energy is:

mass×velocity²/2

which means the endcap, if aimed just a bit differently, would have struck me with about 7 joules of energy.

How bad would that have been, compared to a bullet? Apparently even the wimpiest guns deliver hundreds of joules to their targets, so we’re not looking at a shearing-off-the-top-of-my-head scenario here. On the other hand, I was there, and I’m not exaggerating when I say that hunk of metal could have dealt me a grievous injury at the very least. If that was what seven joules looks like, I have a whole new appreciation for the stopping power of a bullet.

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Knowing the distances involved, it seems implausible that enough photons would be headed in my exact direction — of all the other places in the universe they could have gone! How implausible? Let’s figure it out.

We’ll start by estimating that the area of my dilated, night-sky-gazing pupils is about one square centimeter, or (to keep all measurements using the same units) one ten-thousandth of a square meter.

Thanks to Wikipedia I know that a single photon of visible light carries about 4×10^-19 joules of energy which is just enough to excite a photoreceptor in my eye. How many must arrive each second for me to perceive a continuous image? I know that movies, projected at 24 frames per second, are good enough to trigger the persistence-of-vision effect. Since a distant star has far less detail than a frame of movie film, I’m guessing that it can appear steady at even fewer than 24 “frames” per second. Let’s guess 10 photons per second must arrive in my eyes to perceive a steadily shining star. 10 photons delivering 4×10^-19 joules every second is 4×10^-18 watts (because one watt is equal to one joule per second).

Now, how far is that star? A bit of googling reveals that a typical faintly visible star is 200 light years away (though many are much closer and many are much farther). 200 light years is 1.89×10¹⁸ meters.

From the perspective of that star, what fraction of its “sky” is taken up by my eyes? If you imagine a gigantic spherical shell centered on and surrounding that star, with a radius of 200 light years and my eyes on the inner surface, then the total area of that inner surface — the star’s “sky” — is 4×π×200light-years², which is 503,000 square light-years, which is 4.5×10³⁷ square meters. My pupils, one ten-thousandth of a square meter, comprise 1/2.22×10⁴² of that “sky.” That’s two millionths of a trillionth of a trillionth of a trillionth — not like that’s going to make my big ego feel small or anything.

So if 4×10^-18 watts of visible light is delivered to 1/2.22×10⁴² of the star’s sky, then the total power delivered to its full sky is that first number divided by that second, which is 1.8×10²⁴ watts, or 1.8 trillion terawatts. Is that a plausible number?

Yes it is. Our own sun has a visible-light output of 3.8×10²⁶ watts. Which means, incidentally, that (using my guesses from above) it would be visible as a steadily shining star to a distance of 2.7×10¹⁹ meters, or about 2,900 light-years. Future criminals exiled from our galaxy will lose sight of their home star while barely a thousandth of the way to their new home in Andromeda.

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Of course abortion remains a fantastically polarizing subject. It seems like the country is divided fairly evenly in favor of choice and opposed to it (and not necessarily along traditional party or ideological lines); and that most opinions are strongly held.

I for one believe that Bill Clinton got it exactly right when he said that abortion should be “safe, legal, and rare,” but why do I believe that? No opinion is worth having if it can’t be examined, challenged, and defended, so here is my defense.

First we must determine whether it’s possible to attack or defend abortion dispassionately, without appeal to emotion. Most people who oppose abortion do so because they believe that it is murder, a naturally emotional subject. We could try to remove or distort the emotional component (e.g., by confining ourselves to a discussion of costs and benefits — as routinely happens in cases of state-sponsored murder such as executions and wars), but I think there’s another way that sidesteps the question of murder altogether.

Whether or not abortion is murder depends on whether or not a life exists to terminate. Indeed that’s what it came down to in the Roe v. Wade decision:

We need not resolve the difficult question of when life begins. When those trained in the respective disciplines of medicine, philosophy, and theology are unable to arrive at any consensus, the judiciary, at this point in the development of man’s knowledge, is not in a position to speculate as to the answer.

Nearly everyone agrees that life does not exist prior to conception. Nearly everyone agrees that life does exist upon birth. So where, in between those two events, is the switch flipped? We don’t need medicine, philosophy, or theology to decide; just simple math.

Life may not exist the entire time, but what does exist is an ever-increasing probability of life — more precisely, a probability of being born alive. The appearance of a fetal heartbeat at about eight weeks after conception improves the odds. “Viability” (a minimal ability to survive outside the womb) at about 22 weeks improves it some more. The chances of being born alive continue slowly to climb as the fetus develops, until finally it reaches 100% at the moment the baby takes its first breath.

Can we scientifically choose a probability threshold before which we say abortion is OK and after which we say it isn’t? Sort of. A threshold very close to 100% would be unacceptable to most people, even pro-choice advocates, for emotional reasons (although this has not been true at all times or in all cultures, some of which routinely disposed of unwanted infants simply by exposing them to the elements). It makes no sense to talk about a threshold of 0% as some extreme anti-abortionists might prefer it, because the probability is higher than zero even before conception — especially if Barry White is playing and the lights are turned low. Any other choice between 0% and 100% would be arbitrary, so the best we can do is to choose the least arbitrary number in that interval: 50%. As it happens, in modern America a 50% chance of a live birth appears to be reached, on average, between the 22nd and 28th week of pregnancy.

So that’s my position: abortion should be legal (and safe and rare) before about 22 weeks, and illegal (with the usual pragmatic exceptions — rape, incest, health of the mother) otherwise. Your emotions aside, I think I’ve shown that no other position on this subject is more rational than that one.

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The year 2000 had 7 prime factors (2×2×2×2×5×5×5), which was the most since 1984, which also had 7 (2×2×2×2×2×2×31). 1944 had 8 (2×2×2×3×3×3×3×3), which we won’t see again until 2016 (2×2×2×2×2×3×3×7). 1920 had 9! (2×2×2×2×2×2×2×3×5.)

Those of us who can make it until 2048 will get to see an 11-factor year (2×2×2×2×2×2×2×2×2×2×2) and the first year since 1024 that’s a power of two, which is a big deal to computer nerds like me.

The most recent prime-number year was 2003 and the next is 2011. The most recent twin-prime years were 1997 and 1999, and the next are 2027 and 2029.

What is the significance of all this? Not a damn thing. But you read it all with interest, didn’t you?

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