Three gaps, part 4

Okay, we’ve shown that when you generate a scale using an interval X, that scale has no more than three different gap sizes. Now let’s learn about the relationship between those sizes.

Let’s suppose we have three different gap sizes. For instance, the 8 note scale:Here the first note, the note we started with, is F, and the last note added is F♯. That means the Type I rigid gaps are FF♯ and F♯G (the yellow gap on the left, and the purple one immediately to its right, surrounding the last note). AB (the blue gap closest to the center) is a Type II rigid gap, because shifting it by X makes it coincide with EF♯, which isn’t a gap, it’s the two gaps surrounding the first note (EF and FF♯). Clearly that means the size of gap AB is the size of EF (which is the same as the size of rigid gap F♯G) plus the size of FF♯. So in this case the size of the largest gap is the sum of the sizes of the two smaller gaps.

But is that always true? Can you get a Type II gap that isn’t the same size as the sum of the gaps surrounding the first point? Or can you get a scale in which the two gaps surrounding the first point are both the size of one or the other of the Type I gaps?

Let’s try to look at all the possibilities. For the Type I (rigid) gaps:

  1. There are no Type I gaps
  2. There is only one Type I gap
  3. There are two Type I gaps, and they’re the same size
  4. There are two Type I gaps, and they’re different sizes

And for the Type II (rigid) gap:

  1. There is no Type II gap
  2. There is a Type II gap, and it is also a Type I gap
  3. There is a Type II gap, and it is not a Type I gap

Type I gaps surround the last note. If the gaps adjacent to the last note aren’t rigid, that means there’s a note X above the last note, which would have to be the first note, and the only way that can happen is if nX = 0 \mod 1200. That is, X has to be rational, of the form 1200m/n (in lowest terms). Then the n note scale divides the octave into n equal intervals, and we have a 1-gap scale. In that case there aren’t any Type II rigid gaps, either. Every gap coincides with a gap when shifted.

If there is no note X above the last note, then the gap to the left of the last note is rigid and so is the gap to the right, and there are two Type I gaps unless those two gaps are one and the same. In other words, it’s a one-note scale with one (one octave) gap. Which is a degenerate sort of equal division of the octave, so in fact there aren’t any rigid gaps. In other words, if it’s not an equal division of the octave (or a one-note scale), then there must be two distinct Type I gaps. They can be the same size, or not.

Now suppose there is no Type II gap. No gap contains the first note in its shifted interior. The only way that can happen is if there is a note X to the left of the first note, and that would have to be the last note. Again, this means it’s an equal division of the octave, and there are no Type I gaps either.

So suppose there is a Type II gap, but it is also a Type I gap. That is, one of its end notes is the last note. Then when shifted, that end will not coincide with a note. If the two Type I gaps are different sizes, then all we can say is we have a 2-gap scale, and there is no particular relationship between the two sizes.

Example: the 7 note scale:

Here gap AB (the blue gap closest to the center) is a Type II rigid gap: if you shift it, it goes from E to where F♯ would be if there were an F♯, and it contains F. It’s also one of the Type I rigid gaps, since B is the last note; the other is BC. So it’s a 2-gap scale.

If the Type II gap is also Type I, and the two Type I gaps are the same size, well… then obviously we have a 1-gap scale, which means an equal division of the octave, but that has no rigid gaps at all, so by contradiction that case is impossible.

Finally, suppose there is a Type II gap, and it is not a Type I gap. That means there’s a note X to the right of each of its end notes, so when you shift the gap it’ll coincide with two gaps, the two surrounding the first note. The Type II gap is the sum of the gaps surrounding the first note. One of the two gaps surrounding the first note has its “older” end note (namely the first note) on the right end, the other has the older note on the left end. If you shift the first of these as many times as you can, the “newer” note on the left end reaches the last note first, so this gap matches up with the Type I rigid gap in which the last note is on the left end. But if you shift the second one, the “newer” note on the right end reaches the last note first, so that gap matches up with the Type I rigid gap in which the last note is on the right end. So if the two Type I gaps are different sizes, then so are the two gaps surrounding the first note, and we have a 3-gap scale in which the Type II rigid gap is the same size as the sum of the sizes of the two Type I gaps.

The only way the two gaps surrounding the first note can be the same size is if both Type I gaps are the same size, in which case the Type II gap is exactly twice that size and it’s a 2-gap scale. That would happen, for instance, if instead of ~702 cents, X were 700 cents, and you generated an 8-note scale. It would look like the 8-note scale above, except the gaps surrounding the first note both would be 100 cents in size, and so would the gaps surrounding the last note.

Summing up:

  1. If there are no Type I gaps, there are no Type II gaps, and vice versa; we have a 1-gap scale
  2. If there is a Type II gap, and it is also a Type I gap, then the two Type I gaps are different sizes, and we have a 2-gap scale, with no particular relationship between the two sizes
  3. If there is a Type II gap, and it is not a Type I gap, and the Type I gaps are the same size, we have a 2-gap scale with the large gap twice the small gap
  4. If there is a Type II gap, and it is not a Type I gap, and the Type I gaps are not the same size, we have a 3-gap scale with the large gap the sum of the small gaps

So there you go. I’ve framed this discussion in terms of musical scales, putting notes within an octave, at positions ranging from 0 to 1200 cents, but of course this all could be restated in terms of putting points on a unit line segment, at positions from 0 to 1, or on a circle, at angular positions from 0 to 2π. The Wikipedia article says in the latter form it has applications to phyllotaxis, although I’m not sure it has anything significant to say in that field. The theorem also has applications in the theory of Sturmian words, it says, and if I ever come to grips with what Sturmian words are and why one would care, maybe I’ll write about them here… don’t hold your breath, though…

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Three gaps, part 3

Let’s look at a proof of the three-gap theorem given by F. M. Liang and restated more clearly by Schiu [paywall].

Here’s the 6-note scale again:

We’re going to look for gaps that are “rigid”, by which we mean this: A rigid gap is one that, if shifted (right) by X, does not coincide with another gap.

Think about the gap from F to G (the leftmost blue gap). If you shift it (right) by X, you get the gap from C to D. Shift gap CD by X and you get the gap GA; shift that by X and you get the gap DE. But if you shift DE by X you don’t get a gap: you get a segment from A to where B would be if there were a B. The gap DE is rigid. EF (the purple gap) is rigid too, because shifting it by X doesn’t give you a gap (it goes from where B would be to C). And AC (orange) is rigid, because shifting it by X gives you the segment EG, which is two adjacent gaps (EF and FG), not one.

Generally, there are only two ways a gap can be rigid. One is if when you do the shift one or the other of its end notes doesn’t map to a note. That happens only when one or the other of the end notes is the last note added (A in this case), in which case there isn’t a note X to the right of it. We’ll call this a Type I rigid gap.

The other way is if when you do the shift there is a note, or more than one note, in the middle of the shifted segment. But that happens only if the note(s) in the middle have no notes X to the left of them. Otherwise there’d be one or more notes in the middle of the unshifted gap, and by the definition of a gap, there isn’t. And the only note with no note X to its left is the first note (F in this case), so there can be only one note in the middle of the shifted gap and there can be only one rigid gap of this sort. This is a Type II rigid gap.

And that means there are no more than three rigid gaps: two Type I gaps that start or end on the last point, and one Type II whose shifted version contains the first point. All the other gaps are nonrigid. By shifting any one of them one or more times, you can make it coincide with one of the rigid gaps, and so it must be the same size as one of the rigid gaps. Therefore there can be at most three gap sizes.

That’s the first part of the theorem. Second part, coming up soon.

 

 

Three gaps, part 2

Previously we saw if you build a musical scale by starting at F and adding notes, each a perfect fifth (702 cents, approximately) above the previous, modulo an octave, you start to see a pattern: Each scale has at most three kinds of intervals between consecutive notes, or what we call gaps. If there are three kinds, then the largest is equal to the sum of the two smaller ones. When you add a note to such a scale, it splits one of the large gaps into one of each of the two other kinds; eventually you split all the large gaps and are left with a scale having only two kinds of gaps.

At least that’s the pattern up through seven notes. Does it continue?

Adding an eighth note, F♯, splits the whole tone between F and G into two semitones — but, contrary to what you might expect, they’re not equal; one is a diatonic semitone and one is a slightly larger chromatic semitone, 104 cents. It’s yellow.So in the eight note scale there are three kinds of gaps again. (And the big one, the whole tone, is the sum of the two others.)

Add notes nine through twelve and you split the remaining whole tones, ending up with a twelve note scale with two kinds of gaps, diatonic and chromatic semitones.

And you can still go on. The thirteenth note splits a chromatic semitone into a diatonic semitone plus a Pythagorean comma (which is very small, about 24 cents, and grey), so we have three kinds of gaps again, with the chromatic semitone being the sum of the diatonic semitone and the Pythagorean comma. (In the diagram I’m running out of room, so I just show the sequence number of each note, not the letter name.)

And if you keep going, when you get to the 17th note it splits the last of the chromatic semitones and you have a scale with two kinds of gaps, diatonic semitones and Pythagorean commas. And so on. You can keep this up for weeks if you want to.

What’s intriguing is that the 5-note scale, called a pentatonic scale, is widely used especially in folk music; the 7-note diatonic scale is the basis of most mainstream Western music; the 12-note chromatic scale (in a slightly different tuning) is what you find on a piano keyboard; and while the 17-note scale has no significant role in western music, the 13th century Islamic music theorist Safi al-Din al-Urmawi developed scales based on division of the octave into 17 notes. Note those are all 2-gap scales. Meanwhile the 3-gap scales with 4, 6, 8, 9, 10…  notes don’t turn up much at all. Hm.

Now, all of this can be generalized. You can use tempered fifths (as opposed to pure), you can use other intervals like major or minor sixths, pure or not; heck, you can use any interval you want to generate scales. For that matter you can use a tempered octave as your circle, or a perfect twelfth or something else. And if you do you always find the same pattern. Every scale has one, two, or three kinds of gap, and if there are three kinds, the largest gap is the sum of the other two.

Yes, that’s what you observe, but is it always true? It is, and that’s the three-gap theorem.

Wikipedia states it as

if one places n points on a circle, at angles of θ, 2θ, 3θ … from the starting point, then there will be at most three distinct distances between pairs of points in adjacent positions around the circle. When there are three distances, the larger of the three always equals the sum of the other two.

An article by Peter Schiu [paywall] gives it as

Let α > 0 be an irrational number and n > 1. For 1 ⩽ m ⩽ n, order the fractional parts of mα to form an increasing sequence (bm):

Then there are at most three distinct values in the set of gaps gm defined by

Moreover, if there are three values, then the largest one is the sum of the other two.

Despite the musical roots going back to Pythagoras and Safi al-Din al-Urmawi, this theorem wasn’t proved until the late 1950s.

Shall we look at a proof? Sure. In the next part.

Three gaps, part 1

From David Eppstein’s blog I learned of a new Wikipedia article about the three-gap theorem. The who? I’d never heard of it myself, but was familiar with the general idea, having written about an application of it in a long article on scales for xenharmonic music.

In a more conventional music context, think about notes in the range from one note, say F, to the next F an octave higher. Scales, if you will. Here’s a representation of that. Ignore the black box for now and just focus on the white rectangle. That’s an octave, with our first note, F, at the left end.

(These diagrams may be rather small and hard to see; click on them to expand them.) Notes an octave apart are considered to be equivalent for our purposes, so you can think of the left end as being joined to the right.

One way to talk about the distances between musical pitches — intervals, as they’re called — is to use a unit called cents, where 1200 cents make an octave. We’re going to generate a scale based on an interval called a perfect fifth, which is X\equiv 1200\log_2 (3/2) cents — about 702 cents. So we’ll take our note F, shift it to the right by a distance X (see arrow below), and call that new note C.

We’ll call the interval between two consecutive notes a gap. Here there are two gaps, one from F to C which is a perfect fifth, 702 cents, and one from C to F which is a perfect fourth, 498 cents. These are shown in red and green respectively, and they add up to one octave, 1200 cents. This is a rather minimalist scale, two notes, with two kinds of gaps. (The number in the black box tells you how many notes are in the scale.)

Now go up a fifth again, starting from C. That puts you past the right end of the octave, but remember we identify the right end with the left end, so we can think of the arrow as going to the right end and then continuing from the left end to the new note, G.

G lands between F and C. It’s 2X above F, minus an octave, which works out to 204 cents, an interval called a whole tone. From G to C is 702 cents minus 204 cents or 498 cents, so G splits the perfect fifth from F to C into a whole tone (in blue) plus another perfect fourth (again in green). In this three note scale there again are two kinds of gaps.

Going up another perfect fifth gives us D, between C and F and splitting that perfect fourth into a whole tone and a major third (294 cents). Now we have a four note scale with three kinds of gaps. The major third is orange.Notice the big gap (perfect fourth) is equal to the sum of the other two.

The next perfect fifth gives us A, which splits the remaining perfect fourth into a whole tone and a major third. This five note scale is back to having two kinds of gaps.

Add a sixth note, E,  and one of the major thirds gets split into a whole tone and a diatonic semitone (90 cents), shown here in purple. There are three kinds of gaps. The big one (major third this time) is again equal to the sum of the other two.

But adding a seventh note, B, splits the other major third into a whole tone and a diatonic semitone, and once again there are two kinds of gaps.

At this point we have a diatonic scale, or in less pedantic terms, a C major scale if you start on C. (Or A minor if you start on A.)

You start to see a pattern. Each scale has at most three kinds of gaps. If there are three kinds, then the largest is equal to the sum of the two smaller ones. When you add a note to such a scale, it splits one of the large gaps into one of each of the two other kinds; eventually you split all the large gaps and are left with a scale having only two kinds of gaps. But maybe we’re getting ahead of ourselves? That’s the pattern so far, but does it continue?

We’ll see in part 2.

If it’s the fourth this must be Thursday

What’s the day of the week of [insert date here]? Do it in your head.

There’s some similarity between calculating the phase of the moon for a given date and calculating the day of the week for a given date. Also some differences.

Similarly, how one does it depends on whether one’s doing it in one’s head or on a computer. (And how one does it on a computer depends on the software environment. Modern software libraries tend to make calculating the day of the week a complete non-problem, for instance. Whereas if you were programming in FORTRAN in the 1970s, it was on you.)

Differently, though, in terms of accuracy. The phase of the moon varies continually and slowly enough that getting it right to within a day or so is probably good enough, at least if you’re doing a mental calculation; if more precision matters, you probably want to do it in software anyway. Whereas if you want to know if June 17 is a Monday, finding out it’s probably within a day or so of Monday isn’t likely to be good enough. For the day of the week, either it’s right or it isn’t.

Anyway, if you were to do it, how would you do it?

We’ll specify a date with century, year, month, and day numbers. The day number D is just the number from 1 to 31 denoting which day of the month it is. Now, for month numbers, M, because leap years make things so complicated, let’s use years that start on March 1 and end on February 28 or 29. Within a March-to-February year we can figure out the number of days between any two dates without having to worry about whether it’s a leap year or not. So instead of M running from 1 (January) through 12 (December) we’ll use 3 (March) through 14 (February). January and February will be regarded as belonging to the preceding calendar year. For March through December the year number Y is the last two digits of the year (AD) and the century number C is the preceding digits. For January and February, use the same Y and C as for the preceding March.

What about BC years? Forget ’em. We’re dealing with Gregorian calendar only here because otherwise, ow. You can develop formulas for Julian calendar and BC and whatever else on your own, if you insist.

Okay. For 10 Apr 1937 we have C = 19, Y = 37, M = 4, D = 10. For 2 Feb 1600 we have C = 15, Y = 99, M = 14, D = 2. Got it?

If you know (x+1) mod 7, the weekday (encoded as an integer from 0, for Sunday, through 6, for Saturday) of the 1st of the month, then the formula for any other day in that month is simple: (D) mod 7.

How do you get x? If every month were 30 days then it’d be (y + 30(M–3)) mod 7 = (y + 2+ 1) mod 7, and therefore the weekday of month M day D would be (y + 2+ 1 + D) mod 7, where (y + 1) mod 7 is the weekday of 1 Mar. But unfortunately it’s more complicated. “Thirty days hath September…” (Or do you know this trick? Make a fist with your left hand. You have four knuckles, and three valleys between the knuckles. Label the knuckles and valleys with the month names starting from January; that and March and May and July are knuckles, February and April and June are valleys. Continue on the right fist: August, October, and December are knuckles and September and November are valleys, with a knuckle and a valley left over. The knuckle months have 31 days, the valley months don’t. You can get “JMMJ AONx” tattooed on your knuckles, if you want a good conversation starter.)

1 Mar is day 1 of the year, or 0 days more than (3–3) x 30 + 1. 1 Apr is the 32nd day, or 1 day more than (4–3) x 30 + 1. 1 May is the 62nd day, or 1 day more than (5–3) x 30 + 1. And so on. The whole table for the number of days more than 30 x (M–3) + D days past the last day of the preceding year is:

March 0
April 1
May 1
June 2
July 2
August 3
September 4
October 4
November 5
December 5
January 6
February 7

So now if you know y, and you’ve memorized that table (call those numbers mM) then the formula becomes:

(y + 2M + 1 + mMD) mod 7

Memorizing that table’s a pain, but what else can you do? (Well, see below.) Note we can absorb additive constant values into y, so let’s just use

(y + 2M + mMD) mod 7

and just define y as whatever works.

Now, if you know the weekday of a date in some century, and if there were no leap years, then you could get the weekday for any date in that century just by using

(z + Y + 2MmMD) mod 7

because in any year the weekday for a given date is one later than it was the previous year. (365 mod 7 = 1.) Here z is, well, it’s what makes the formula come out right for the date you know.

But there are leap years, which add another 1 day shift from one year to the next. So we have to add 1 for each leap year from year 0 through Y–1, inclusive. Recall that Y = 0 is not a leap year, in any century, because by Y = 0 we mean 1 Mar of calendar year 0 through 28 Feb of calendar year 1. On the other hand Y = 3 is 1 Mar 03 through 29 Feb 04, and is a leap year! It won’t shift any days in that year but it’ll shift days starting in Y = 4. So — ignoring the century leap year rule for a moment — the number of leap years to take into account is just floor(Y/4). We have to add that number to our tally.

So for dates in that century we use:

(z + Y + floor(Y/4) + 2MmMD) mod 7

How about dates in the next century? Well, in most cases a century year (our Y = 99 year) is not a leap year so there are 24 leap years and 76 standard years in a century, or 24 x 2 + 76 x 1 mod 7 = 5 weekdays shift. If all centuries were like that it’d be

(z + 5CY + floor(Y/4) + 2MmM D) mod 7

I’m still calling the constant z, but it’s whatever makes things work.

But every 400 years we do have a leap year. So we have to add 1 day back in for every four centuries:

(z + 5C + floor(C/4) + Y + floor(Y/4) + 2MmMD) mod 7

Now we just need to determine z. Pick a day at random, oh let’s say Tuesday, 2 Jan 2018. Then it’s (remembering Y = 17 and M = 13 for January 2018)

(z + 5 x 20 + floor(20/4) + 17 + floor(17/4) + 2 x 13 + 7 + 2) mod 7

= (z + 2 + 5 + 3 + 4 + 5 + 6 + 2) mod 7

= (z + 6) mod 7

But this should equal 2 for Tuesday, so z = 3.

And now we can do any date. 17 Apr 2733 is:

(3 + 5 x 27 + floor(27/4) + 33  + floor(33/4) +  2 x 4 + 1 + 17) mod 7

= (3 + 2 + 6 + 5 + 1 + 1 + 1 + 3) mod 7 = 1 = Monday. Right, cal(1)?


$ cal 4 2733
     April 2733
Su Mo Tu We Th Fr Sa
                   1
 2  3  4  5  6  7  8
 9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30

Right!

Oh, one more thing:

The red dots are the values of mM plotted versus M. The blue dots are mM+1. The line is y = 0.6M – 1.3, and it passes between each pair of dots. So that means instead of memorizing mM and using

(3 + 5C + floor(C/4) + Y + floor(Y/4) + 2MmMD) mod 7

if you’d rather you could use

(3 + 5C + floor(C/4) + Y + floor(Y/4) + 2M + floor((6M – 13)/10) + D) mod 7

So just memorize the first formula and mM, or memorize the second formula, and keep in mind you have to use Y and for the preceding year and add 12 to M if the month is January or February, and know your Gregorian calendar limits for the particular geographical location in question (remember, February 1712 had 30 days in Sweden), and you can figure out the weekday in your head!

… Either way, the lunar phase is easier.

 

Plausible phases

So why does that lunar phase method work anyway?

It’s mostly fairly obvious once you start thinking about it. For instance, suppose you know the age of the moon on a given day. Then its age a day later is, kind of obviously, one day more. Except when you hit new moon and it resets, every 29.53… days, and doing modulo 30 approximately accounts for that. So it makes sense for the formula to be something like (x+D) mod 30, where D is the day number and x doesn’t depend on the day (it’s approximately the age, minus 1, on the first of the month).

Now suppose you know the age of the moon on the first day of a given month. Then its age a month later is…? Well, in a 30 day month it’s about 0.47 days more, that is, 30–29.53; in a 31 day month it’s 1.47 days more. Or in an average month of about 365.25/12 = 30.52 days it’s very close to 1 day more. (Actually, since we’re working within the 2000 to 2018 Metonic cycle, the average year length that counts is 1/19 of five leap years and 14 normal years, which comes out to 365.263 days, not that the difference matters much.) If it weren’t for February it’d make sense for the formula to be something like (y+M+D) mod 30, where M is the month number and y doesn’t depend on the day or month (it’s the age at the start of the year, minus 2).

February’s short, though. How to deal with that? Hang on, we’ll get there.

Now suppose you know the age of the moon on the first day of a given year. Then its age a year later is… well, 365.25 days is 12.36 synodic months, so the age advance is 0.36 x 29.53 = 10.9 days. For a decent approximation, 11 days age advance per year, so a formula something like (z+(Y x 11)+M+D) mod 30, where Y is (last two digits of) the year number and z is a constant (the age on 1 Jan 2000, minus 2), would seem reasonable. If it weren’t for February.

It’s too bad February isn’t the last month of the year, then you wouldn’t have to deal with its shortness. Well, make it the last month of the year! Let’s pretend the year begins on March 1. Now within a month each day is a day greater age than the previous; within a (March-to-February) year each month is a day greater age than the previous; and each year is 11 days greater age than the previous. Looked at like this, we don’t have to worry especially about the age advance from February to March. So we could use a formula like the above, if we said March was month 1, April month 2, … January month 11, February month 12, and if we used the preceding year number in January and February. That is, for months March through December use

(z+(Y x 11)+(M–2)+D) mod 30

(here M is still 3 for March, et cetera) and for January and February use

(z+((Y–1) x 11)+(M+10)+D) mod 30 .

That gets a little complicated, though. But we haven’t defined z yet; we can absorb additive constants into it. That is, redefine z–2 as z so we have for March to December

(z+(Y x 11)+M+D) mod 30

and for January and February

(z+((Y–1) x 11)+M+12+D) mod 30 .

But now notice the latter is equal to

(z+(Y x 11)–11+M+12+D) mod 30

or

(z+(Y x 11)+M+1+D) mod 30

which means we don’t have to mess around with the year number or with month numbers like 13 and 14; just use the same formula for all months, except add 1 in January and February!

Nearly there. We just need to establish z. The age of the Moon at 00:00 UTC on 1 Jan 2000 was 23.882 days, so z is 20.882 or about 21. Or more conveniently, about 22, so we can replace z+(Y x 11) with ((Y+2)x 11), and that’s the formula. There are enough approximations here that one can justifiably worry about their accumulating to large errors, but in fact they tend to cancel out. (1 day per month is nearly exact, modulo 30 arithmetic gives you age increases that are too small, 11 days per year gives you age increases that are too large for non leap years and too small for leap years, z=22 means you start out the cycle with an age that’s too large. Sticking with z=21 would have made 1 Jan 2000 more accurate but would have led to too large errors later in the cycle.)

Depends on the phase of the moon

[Edited for corrections and updates]

What’s the phase of the moon on [insert date here]? Do it in your head.

This is something I found on the Internet once, don’t know where; I learned it but eventually forgot it. Having put it back together again I’m posting it here for future reference. It’s more or less similar to the method, attributed to John Conway, described at much greater length at http://gmmentalgym.blogspot.com/2013/01/moon-phase-for-any-date.html. The two are not mathematically equivalent, though, giving slightly different results for some dates. See below.

To start with, assume the required date is in the range from 1 Jan 2000 to 31 Dec 2018. Take the last two digits of the year, add 2, and multiply by 11.

(Good thing multiplying by 11 is easy. You just add the two digits to the same two digits shifted one place left: e.g. 17 x 11 = 17 + 170 = 187. Or another way to think of it: the last digit of the result is the same as the last digit of the original number, the first digit of the result is the same as the first digit of the original number, and the middle digit of the result is the sum of the two digits of the original number. Of course if that sum exceeds 9, you have to add 1 to the first digit of the result.)

Now add the number of the month — 1 for January, 2 for February, and so on, 12 for December. If the month is January or February, add another 1.

Now add the day of the month.

Now divide that number by 30 and take the remainder. The result is, approximately, the age of the moon — the number of days since the new moon. 0 or 29 would be new moon, 14 or 15 would be full moon, about 7 would be first quarter, about 21 would be third quarter. It might be off a day or so (again, see below), but close enough for mental calculation, right?

What about dates outside that range? The moon returns to the same phase every 19 years exactly — or almost exactly; actually more like 19 years plus 2 hours. This is called a Metonic cycle. So add or subtract a multiple of 19 years to get into the 2000–2018 range and then proceed as above. Obviously for dates far in the past or future this breaks down, but for dates in the past or future few decades it’s fine.

Example: July 20, 1969. This is outside our 2000–2018 range but if you add 2×19=38 years you have July 20, 2007. So: for year 2007, (7+2)x11 = 99; for month 7 add 99+7=106; for day 20 add 106+20=126. To get the remainder when divided by 30, subtract the multiple of 30: 126-120=6. Age of the moon is 6 days, corresponding to just about first quarter. Of course July 20, 1969 is the date of the Apollo 11 landing. If you do the same for the other moon landing dates you get ages in the range 5 to 9. For reasons having to do with the angle of the Sun, NASA chose to do all the moon landings at about first quarter.

Example: January 1, 2018. (18+2)x11 + (1+1) + 1 = 223; 223-210=13. The moon is (about) 13 days old on New Year’s Day, so is close to full. In fact full moon occurs at about 2:00 UTC on January 2, or evening of January 1 in the United States. We get two full moons in January and March, none in February!

How well does this work? It depends on how you define “how well”. You’re getting an integer estimate of the age of the moon, which you can compare to the actual age of the moon… when? At 00:00 UTC? Noon?

Choosing to compare to the age at 00:00 UTC, here’s a plot of the error in this calculation for the years 2000 to 2018. There are just two dates where it’s more than 1.6 days off and most dates — 6269 out of 6940 —are less than 1 day off.

For years outside this range but within the nearest 4 Metonic cycles each way (i.e. from 1 Jan 1924 to 31 Dec 2094) I find there are only two dates where the error is (slightly) larger than 2 days, and average errors in each cycle are under 0.35 days. Outside that 171 year range it gets worse.

And here’s a comparison to the results of the method from gmmentalgym. The blue line is the same as above and the red line is gmmentalgym; they coincide for the first 10 years so you see only the red there. After that they differ by 1 day.

You can see gmmentalgym is systematically high in the second half of the cycle, so does worse. Note, though, that if we compared to the age at noon UTC, it would shift this whole graph down by half a day and the red line would be better than the blue.

On the other hand, here’s the 1962–1980 cycle (comparing at 00:00):

Here it looks like the gmmentalgym method is a little better, but would be much worse if comparing at noon UTC. On balance, I wouldn’t say either method has the clear upper hand — they’re both decent approximations, easily done in one’s head.