If you’re a psychiatrist (or psychiatric RNCS) in the US reading this, you are almost certainly aware that all of our billing codes changed on Jan 1, 2013. If you are like most of the psychiatrists I know—at least in private practice—you are at least somewhat freaked out by this. If so, keep reading. If not—for example, if you’re not a psychiatrist in the US—stop reading this

immediatelyand go do something more interesting, like… well, like just about anything other than memorizing a phone book. (There used to be things called phone books… never mind.)

Unlike 2013, the billing codes themselves have not changed, but as of January 1, 2021, the documentation requirements for these codes have gotten **significantly less complex and onerous**. Like then, this post should be of no interest to you if your are not a US psychiatrist or psychiatric nurse. But if you are—and especially if a significant portion of your practice is psychotherapy—the situation is very different from 2013 because these changes are:

- good news
- not widely publicized

The 2013 codes could be chosen based on either time or complexity. That is still the case, **but**:

- When choosing a code based on
**time**, you can now count**total**time, not just face-to-face time talking to the patient. That means you can include the time it takes to write your note, review previous notes or records, speaking with other providers, etc. The time periods are somewhat longer, but this definitely looks like a win. - When choosing a code based on
**complexity**, the choice is now based**only on medical decision-making (MDM)**; before it was based on MDM and also required documenting certain combinations of history and exam elements. This is huge.

The changes for complexity-based codes are particularly significant for physician or nurse psychotherapists, because if you are using a psychotherapy add-on code (90833, 90836, 90838), the E&M code that you are adding on to **must** be based on complexity.

The AMA has a set of slides with details about the changes.

Code | Total time |

99211 | < 10″ |

99212 | 10-19″ |

99213 | 20-29″ |

99214 | 30-39″ |

99215 | 40-54″ |

For follow-up visits of 55 minutes or longer, you can bill the code “99XXX” (seriously, that is what it seems to be called) once for each additional 15″ block. So 55-59″ is billed as 99215 + 99XXX, and 70-84″ is billed as 99215 + 2 units of 99XXX.

Code | Total time |

99202 | 15-29″ |

99203 | 30-44″ |

99204 | 45-59″ |

99215 | 60-74″ |

As above, for initial visits with total time 74″ (are your write-ups as time-consuming as mine?), you can bill one unit of 99XXX for each additional 15″.

Of course, **make sure that your note clearly documents the total time**.

Here is a link to a table which gives a clear overview of the criteria for MDM complexity.

The rows in the table address particular codes (such as 99213, 99214, 99215). Each row has three columns of “Elements of Medical Decision Making.” For a particular row to work, two of these three columns must apply. The first column is “Number and Complexity of Problems Addressed.” The second is “Amount and/or Complexity of Data to be Reviewed and Analyzed.” The third is “Risk of Complications and/or Morbidity or Mortality of Patient Management.” The second column has to do mostly with ordering and interpreting tests; as I read it, it is rarely relevant to our coding. So to use a particular code, we just need to make sure that columns 1 and 3 apply.

The 99213 applies to patients with, at a minimum, one stable chronic illness, and “low risk of morbidity from additional diagnostic testing or treatment.” If you are a psychiatrist doing psychotherapy, this applies to every visit with a patient. Presumably you need to document which illness you are treating.

For 99214, there must be “moderate risk of morbidity from additional diagnostic testing or treatment”; as an example, they cite “prescription drug management.” If someone is stable on their medications and you are merely prescribing a refill, I would not guess that this applies. But if you are actively considering a medication change with the patient, your action (or inaction) does have at least a moderate risk of morbidity, either by introducing new side-effects, or by worsening or failing to improve the patient’s condition (especially in the case of inaction or medication reduction). You can meet the criteria in the first column with either “one or more chronic illnesses with exacerbation, progression, or side-effects of treatment,” or “two or more stable or chronic illnesses.” Given the ubiquity of comorbidity in our diagnostic system, this should be pretty much automatic.

Any time you seriously consider hospitalizing a patient, you should be able to use 99215. Column one needs a “chronic illness with severe exacerbation, progression, or side-effects of treatment” or an “acute or chronic illness or injury that poses a threat to life or bodily function.” Column three refers specifically to “decision regarding hospitalization.”

Exactly the same criteria apply for the corresponding initial visit codes (99203, 99204, and 99205).

Given the simplicity of MDM system, I suspect that psychiatrists will be better off coding follow-up visits based on MDM instead of time, even when there is no psychotherapy add-on. For initial visits where hospitalization is not a consideration, time-based coding may allow you to bill using 99215 (with possible additional units of 99XXX).

]]>What does the Fong/Walters algorithm do differently? Let’s note what’s the same. Conway’s scheme of “Doomsdays” is still in place, and you still need to know the Doomsday for the start of the century. The difference is in computing the offset. There is no dividing by 12, remembering the quotient, dividing the remainder by 4, etc. (Apparently that part of the method was due to Lewis Carroll, as explained in Part 2 or in Fong’s Scientific American blog post.)

How does the alternate method , which they call “odd+11” work? If you love flowcharts, look here. But it’s pretty simple:

- Take the last two digits of the year.
- Is it odd? If so, add 11.
- Divide your number (which now must be even) by 2.
- Is your new number odd? If so, add 11 again.
- Reduce your number modulo 7.
- Subtract your number from 7 (“take the 7s complement”).

The number you come up with is your offset from the Doomsday at the start of the century. Let’s try it with 2020, and check the result against what we got in Part 1. Step 1: 20. Step 2: Not odd, so still 20. Step 3: 10. Step 4: Still not odd, so still 10. Step 5: 3. Step 6: 4. That is what we got in Part 1.

To be honest, I’m not wild about Steps 5 and 6. Really, it should be one step that says “Take the negative of your number, modulo 7.” If your number is 14, it is 0 modulo 7. The negative of 0 is still zero. That is better than subtracting it from 7, coming up with 7, and saying, “Well, 7 is 0 modulo 7, anyway, so let’s call it 0.”

Let’s also do 1929, because it illustrates another reason why I prefer the modified Step 5/6. Step 1: 29. Step 2: Odd, so 40. Step 3: 20. Step 4: Not odd, so still 20. Following Fong/Walters, you could say 20 is 6 modulo 7, and the 7s complement is 1. And that is the correct answer, as we saw in Part 1. But 21 is a multiple of 7, so you could just say “20 is -1 modulo 7,” and if we take the negative of that, we get 1 immediately.

How did Fong and Walters come up this? I have no idea, but my guess is that it involved a *lot* of time noodling around.

How do we know it works? Fong and Walters give a proof, but I think it’s possible to simplify the proof, a lot. [Edit: I would also have said I don’t really understand *why* it works, but see the addendum below.]

What happens to your answer if you add 4 to the year? How does that change your calculation? Well, the first thing to notice is that if the number was even before, it’s still even, so you will do the same thing you did before in step 2, and so the number you at the end of Step 2 is 4 more than what you had before. In Step 3, the number that you end up with will be 2 more than what you had before. So if your number was odd before, it still is, and at the end of Step 5, your number is still 2 more than what you had last time around. When you reduce modulo 7, your number is still 2 more (understood modulo 7). When you take the negative modulo 7, you wind up with something that is 2 **less**. But modulo 7, 2 less is the same as **5 more**! This is exactly what we want! Advancing the year by 4 should advance the day by 5—one for each year, and one for the leap year.

Once we know that, it’s enough to show that it works for 00, 01, 02, and 03… and then you can get to any year up to 99 by repeatedly adding 4. (This is the wonder of “mathematical induction.”) I don’t have the stomach to drag either of us through a written blow-by-blow, but this table shows you the result of each step:

Year/Step 1 | Step 2 | Step 3 | Step 4 | Step 5 | Step 6 |

00 | 0 | 0 | 0 | 0 | 0 |

01 | 12 | 6 | 6 | 6 | 1 |

02 | 2 | 1 | 12 | 5 | 2 |

03 | 14 | 7 | 18 | 4 | 3 |

At the end of Step 6, we see that each year advances the day by 1, which is exactly what we need! And we saw above, that for 04, the day will advance by 5, again exactly what we need.

**Addendum,** later that same day. I sent an appreciative email to Fong and Walters. I heard back from Mike Walters almost immediately. He wrote:

I actually did come up with the idea by ‘noodling around’ for an hour or so, as described on my blog … I never liked Conway’s method, there was too much mental math involved for my tastes … One night I was bored and I just wrote out the grid of years (the one shown on my blog post) and stared at it until I saw the pattern with the leap years, then the rest came together.

He noticed two things:

- If you have a leap year, dividing by 2 and taking the negative modulo 7 gives you the correct offset. This is exactly what we saw above: every extra 4 years adds two to the quotient, taking the negative subtracts 2, which is the same modulo 7 as adding 5, which is exactly how much each subsequent leap year needs to add to the offset.
*If you have a non-leap year, the year 11 years later has the same Doomsday.**

So Walters’ original algorithm is “just add 11 until you get to a leap year [something divisible by 4], and then do the divide by 2 and negative-modulo-7 thing. You might have two add 11 up to 3 times. Chamberlain Fong picked this up and made it even easier to compute, resulting in the steps above.

Before taking apart Fong’s version, why is it that adding 11 years to a non-leap year gives a year with the same Doomsday? It’s just what we’ve seen. If you start with a non-leap year, the subsequent 11 years will include 3 leap years. How many days will we advance the Doomsday when we go 11 years forward? One for each year, with an additional one for each leap year, is 11+3=14. But 14 is 0 modulo 7, so the two years have the same Doomsday.

How does Fong’s version work? Start with the last two digits. If your number is odd, add 11. Let’s call what we end up with “y.” It will be even, and it will refer to a year with the correct Doomsday. This y might be divisible by 4, or it might not. If it’s divisible by 4, we just saw that the negative of y/2, modulo 7, is the correct offset for our Doomsday. But if y is divisible by 4, y/2 is even, so the Fong algorithm gives us the right answer. Now here’s the clever thing: if y/2 is odd, adding 11 to y/2 has the same effect as adding 22 to y, that is, it’s the same as adding 11 to y twice. And it gets you to a place where the original leap year calculation works. So essentially the algorithm has a way of adding 11 to your original year either: 0 times (for a leap year), once (in Step 2 but not in Step 4), twice (in Step 4 but not in step 2), or 3 times (in Step 2 and step 4). How many times it does it depends on what your year is **modulo 4**, but you don’t have to think about that, you just think about even and odd.

I subsequently heard back from Chamberlain Fong. He pointed me to a survey of calendar methods by Kamal Abdali, which contains an alternate verification of the Fong/Walters algorithm. The proof I’ve given here is more “walking you through the algorithm,” and it may work better as a conversation than it does written out. Abdali’s proof, and Fong’s original proof, involve equations, which may be more to your taste. But if you want to clean up the discussion of the previous paragraph, you would end up looking at your year modulo 4. In other words, you would look at the two “least significant bits” in its binary expansion, which are “b” and “c” in Abdali’s paper. And there is something to clean up here, because the “add 11 and get a year with the same Doomsday” trick does *not* work if you start from a leap year. So if you’re effectively adding 11 two or three times, you need to verify that those take you *to* a leap year, not *across* one.

[*Just to keep you from tearing your hair out, I say “leap year” above when I should say “year divisible by 4. If we start in 1897, the subsequent 11 years include 3 that are divisible by 4: 1900, 1904, and 1908. However, 1900 is not a leap year, so it includes only two bona fide leap years. But the math still gives the correct answer for 1897, even though 1908 does **not** have the same Doomsday as 1897.]

The first helpful thing Fong does is to give some history. Conway’s method has two parts: the identification of particular days of the year as “Doomsdays,” which all fall on the same day of the week. Once you know what day of the week any give year’s Doomsdays fall on, it is easy to figure out the day of the week of an arbitrary date. The other part is to figure out what day of the week the given year’s Doomsdays fall on.

The first part of the algorithm seems to be entirely due to Conway. The second part, interestingly, was apparently worked out by Lewis Carroll—one of the few people in the orbit of mathematicians who may have been more playful that Conway. Martin Gardner, the legendary author of Scientific American’s *Mathematical Games* column, was a serious student of Carroll, and when he came across Carroll’s work on perpetual calendars, challenged Conway to come up with something simpler.

The second helpful thing is some clarification of how the leap year issue arises:

Earth spins around its axis approximately 365.24219 times annually. This rate is known as the mean tropical year. The true number of days in the mean tropical year actually varies slightly over time.

He couples this with some helpful history:

Our modern calendar system of 12 months started with the Julian calendar circa 45 B.C. The Julian calendar approximated the mean tropical year as 365.25 days and proposed 366-day leap years for every one divisible by four. This approximation made the Julian calendar fairly accurate in tracking the annual seasons for several centuries. But eventually the approximation errors accumulated. By 1582 the Julian calendar was out of sync with the seasons by several days. Pope Gregory XIII mandated a reform to the calendar; his Gregorian calendar approximated the mean tropical year as 365.2425 days.

The Gregorian rule says that years ending in 00 are *not* leap years, unless they are divisible by 400. So every four centuries, the Julian calendar will advance by 3 days compared to the Gregorian.

Of course, not everyone adopted the Gregorian calendar immediately, and Church schisms seem to have a lot to do with it. Henry VIII separated the Church of England from Catholicism in 1534, about 50 years before the Papal calendar reform. Fong goes on:

The Gregorian calendar was not widely adopted across the world until much later. In fact, the British Empire and its colonies did not use the updated calendar system until 1752. By then the Julian calendar was out of sync with the seasons by 11 days. In order to synchronize with the Gregorian calendar, it was mandated that Wednesday, September 2, 1752, be followed by Thursday, September 14, 1752.

Russia did not adopt the “New Style” dates until their revolution in 1918; I have been told that their “October Revolution” (10/25/1917, Old Style) is celebrated annually on November 7! So the bottom line is: Conway’s Doomsday algorithm won’t help you with *any* date before 1582; in the anglophone world it won’t help you until 1752; and in Russia it won’t help you until 1918.

The final valuable thing that Fong does is to point out an even simpler Doomsday algorithm that he and Michael Walters developed to calculate the Doomsday for any given year, which I will discuss in Part 3.

]]>John Horton Conway, a great mathematician of unmatched playfulness, died recently at age 82 of Covid-19. His biographer, Siobhan Roberts, wrote an excellent New York Times obituary. Conway was something of a showman, in a good way, so I should give a spoiler alert up front, especially these days, that this has nothing to do with Doomsday as in the end of the world.

I never met Conway, but in the mid-70s my math camp buddy Rob Indik had a summer in which he commuted to the UI Chicago campus “with Vera [Pless] driving, Conway in front, and me riding in the back and trying to ride along with Conway’s free flow of ideas.” I remember Rob regretting that he couldn’t remember a mental “perpetual calendar” that Conway had taught him. In the computer age we have no call for perpetual calendars, but back when dinosaurs roamed the earth they were novelty items: a small object with a few dials, which would let you figure out what day of the week any date falls on. A mental one sounded like fun, and I occasionally thought—not very hard—about how you might go about it. I never got very far. [I see that this is not the definition of “perpetual calendar” favored by Wikipedia, but that the American Heritage Dictionary has me covered.]

With Conway’s death, I decided to look it up, and found that it’s around the web in several places, but without a very clear explanation of why it works. So I thought I would explain both how it works and why it works.

The basis for any shortcut involving days of the week is to work “modulo 7.” If you move ahead (or back) by 7 days, any number of times, the day of the week will always be the same. Seven days from now (+7), seven days ago (-7), 14 days from now (+14) … they all fall on the same day of the week as today (which you can think of as “moving 0 days”). And all those numbers (7, -7, 14, and 0) are equal modulo 7. Similarly, tomorrow (+1) will fall on the same day as 8 days from now (+8), or 6 days ago (-6). You probably understand this intuitively: if I ask you, “What day of the week is 6 days from now,” chances are you won’t count up, you will realize that it is the same day as yesterday. That is, 6 is equal to -1 modulo 7.

As a first step, let’s notice that the perpetual calendar problem would not be very difficult if there were no leap years. There are 52 weeks in a year, and 52 x 7 is conveniently 364. So 365 is just 52 weeks plus one day. If today is Friday, 365 days from now will be Saturday. In terms of arithmetic modulo 7, 365 = 52×7 + 1, and because 7 is equal to 0 modulo 7, this is equal to 1 modulo 7. If every year has 365 days, then if January 1st is a Monday this year, it will be a Tuesday next year, Wednesday the year after that, and so on, until 7 years from now it is back to Monday. But leap years exist, and they make two problems: the 366-day year advances the next year by two days, **and** it advances everything after Feb 29 by one day.

There are two parts to Conway’s solution: the first part is to “anchor” a year to a particular day, and then figure out what day of the week some given date during that year falls on. The second part is to figure out what day any particular year is anchored to.

I will also mention that all the writeups I’ve seen of Conway’s method involve numbering the days modulo 7, and memorizing: 0 for Sunday, 1 for Monday, and so on until 6 for Saturday. You can do that, and it will probably help you do the calculations faster, but as you’ll see below, it isn’t necessary. We will come up with numbers that are “offsets” (like +4 or -1), and you will need to count 4 days forward from Tuesday, or one day back from Friday. If you have memorized that Tuesday is 2 and Saturday is 6, then you can tell me that Saturday is 4 days forward from Tuesday more quickly than someone who hasn’t, but you’ll both be able to tell me.

Conway anchors his year to what he calls “Doomsdays”: days on the calendar that always fall on the same day of the week. If you don’t like calling them Doomsdays, you could call them “anchor days.” Most people would probably anchor the year to January 1st, resulting in leap-year headaches for the 10 months starting in March, but Conway brilliantly chose to anchor it on *the last day of February*, confining headaches to the first two months. If the last day of February is the first “Doomsday,” then March 7, 14, 21, and 28th are all Doomsdays. The next one is… well you can either work it out by counting laboriously out loud, or you can say it’s March 35th. Subtract the actual 31 days of March and you will find that 4/4 is the next one. Proceeding in this way, you will find that the following are all Doomsdays:

- 4/4
- 5/9
- 6/6
- 7/11
- 8/8
- 9/5
- 10/10
- 11/7
- 12/12

Why did Conway list 10/10 rather than, say 10/3? This is another combination of brilliance and luck: for the even months, the Doomsdays **4/4, 6/6, 8/8, 10/10, and 12/12 **are all super easy to remember. The odd months on the list are covered by the mnemonic **“Working 9-5 at the 7-11”**: 5/9, 9/5, 7/11, and 11/7.

So let’s try something out. This year, 2020, is a leap year. I’m writing this on Friday, April 17. April 18 is a Doomsday, so all the Doomsdays this year are Saturdays. What day will Christmas be? 12/12 is a Doomsday, and then so are 12/19 and 12/26. So 12/26 is a Saturday, and 12/25 is, like today, a Friday. Let me check that… Yes, it’s true.

What about January and February? What day was Valentine’s day this year? Well, February is also pretty easy, because the last day of February is a Doomsday. This year is a leap year, so 2/29, 2/22, and 2/15 are Doomsdays, so, gee, 2/14 was also a Friday.

How about January 1? Well, in a leap year, 2/1 is a Doomsday, so 1/25 is, and 1/18, 1/11, and 1/4 are too. So the next mnemonic: **“Every 4th year 1/4 is a Doomsday.”** In non-leap years, the Doomsday is pushed one day back, to 1/3. So this year, January 4, being a Doomsday in this leap year, was a Saturday. Counting backwards, 1/3 was Friday, 1/2 was Thursday, and 1/1 was a Wednesday. (Checks again.) Yes! It works.

So if you have a date that you want to figure out, once you know what day of the week the Doomsday is for that year, you can find a Doomsday near to it and count up or down. That is how Conway solves that half of the problem. But how do you figure out the day of the week that a Doomsday falls on, given a particular year? Back to leap years.

As I said, it would be really easy if there were no leap years at all. Having leap years every fourth year makes it more complicated. And the fact that leap years *don’t* happen every fourth year, just *most* fourth years, makes it even more complicated! What is the actual rule? If a year is divisible by 4, it’s a leap year. Except… if it’s divisible by 100, it’s not. Unless… it is also divisible by 400, in which case it is. This is actually a fairly simple rule, when you consider that you are trying to match up two things that have essentially nothing to do with each other: the length of a year and the length of a day (more on that in Part 2).

Doomsday for 1700 was Sunday. Let’s take that as a given for now, but we’ll show below that it is correct. How can we figure out Doomsday for 1800? Each year will advance the Doomsday by one day, and each leap year will advance it by one additional day. And because Doomsday is the last day of February, moving from this year to next year we advance by 2 if *next* year is the leap year. So in a century, Doomsday will advance by 100 days, plus the number of leap years. 49 is a multiple of 7, so—working modulo 7—each 50 years advances by 1 day; 100 years advances by 2 days. In a century there would be 25 leap years (100 divided by 4), except that ’00 may or may not be a leap year, so there will either be 24 or 25. To go from one century to the next, the relevant question is whether the *following* ’00 is a leap year. 1800 is not, so there are 24 leap years to go from 1700 to 1800. Take away 7×3, and we advance 3 days due to leap years. That, plus the two days we advanced for the 100 years says we should advance by 5 days. Modulo 7, +5 equals -2, so it’s the same as going by 2 days.

Doomsday 1700 was Sunday, so Doomsday 1800 is Friday. 1900 is not divisible by 4, so again we go back 2, to Wednesday. 2000, being divisible by 400, **is** a leap year, so we advance by one additional day; instead of moving back by 2 days, we only move back 1 day. So Doomsday 2000 is Tuesday. Doomsday 2100: move back 2, and we have another piece of good luck. We are back to Sunday! Every four centuries, we have 3 centuries where we advance by 2, and one where we advance by 1, for a total of 7, so the pattern repeats.

You can work this out in your head, or you can memorize the repeat, and work either forward or backward, from this list of anchor days:

- 1700: Sunday
- 1800: Friday
- 1900: Wednesday
- 2000: Tuesday

Right. But the intervening years should be easy! Take the last two digits, figure out how many times 4 goes into it, and add those two numbers together modulo 7, because each year advances it by 1, and each leap year advances by an additional 1. Let’s look at October 28, 1929. We advance one for each of the 29 years. 29 is equal to 1 modulo 7, since 28 is a multiple of 7. So for the years, we advance by 1. For the leap years, 4 goes into 29 7 times, so we advance by 7, which is 0 modulo 7, in other words, no change. So Doomsday 1929 is advanced by 1 from 1900’s Wednesday: it is Thursday. 10/10 is Doomsday, so are 10/17 and 10/24 are, too. So to get to October 28, advance 4 from Thursday, and you get to Monday. Which checks out, because October 28, 1929 is known as “Black Monday,” due to the stock market crash.

That is actually pretty laborious, and not suitable for lightning-fast mental calculations for most of us, Conway included. So for the next part, Conway used a trick originated by Lewis Carroll (as described in Part 2.) Each 4 years will advance by 5 days (one for each year, and one for the leap year), so after 12 years, we will advance by 15 days, which is the same as advancing by 1.

So here’s what you do. Take the last two digits of the year. Let’s say 12 goes in q times, with a remainder of r (which will be between 0 and 11, inclusive); this is relatively easy to do in your head. For example, 29 = 12×2 + 5. For each 12 year block, we advance by 1, so we need to advance by q.

Then we can deal with r just as we did above: advance by r days, and one additional for each time 4 goes into r.

So for 1929: we advance 2 for q; another 5 for r; and another 1 because 4 goes into 5 once. Modulo 7 this is just 2+5+1, which is 1. So we get the same answer as the long way: advance 1900’s Wednesday by 1, to Thursday.

So let’s check 2020. 12 goes into 20 once (q=1) with a remainder of r=8. 4 goes into 8 twice. So for 2020, we need to advance from 2000 by 1+8+2 modulo 7, which is 4. We said above that Doomsday 2000 was Tuesday, so Doomsday 2020 is Saturday. Which is exactly what we said above, based on the actual calendar for 2020. Working backwards, that means that we were correct that Doomsday 1700 was Sunday!

And that’s all there is to it. Except that it doesn’t work for any year before 1582, and for any year in the Anglo-American world before 1752, as explained in Part 2. And as clever as Lewis Carroll was, there is actually a significantly simpler way to do this second part, as explained in Part 3.

]]>In the process, I revisited my stash of moribund computer programs, including a little web page with links to some of them. It was dusty and pathetic and pretty much nothing worked. So I rolled up my sleeves and prettied it up, and if you want to look any of it, you can find it here.

]]>Happy New Year! I just returned from a quick family trip to Seattle, and you would think that this post would be about my visit to the Living Computer Museum with my cousin Eric, and programming adventures now 50+ years ago in the RESISTORS. But it’s not, as worthy as those topics are.

Actually it all started when I read a post by Ben Orlin on his Math With Bad Drawings site about Evelyn Lamb‘s Page-a-Day Math Calendar, recently published by the AMS. One thing led to another, and my son Ben was kind enough to give me a copy for Christmas.

So I peeked ahead a bit, and on January 10 there is a “cryptarithmetic puzzle,” in which letters stand for single digits in an arithmetic problem, and you have to figure out which choice of digits will make the arithmetic problem correct. Her fun example, which she credits to Manan Shah, is this:

(Get it? 1/10 is a member of a small club of dates that can be written using only zeroes and ones.)

Well, this one sent me back into my programming archives. When Ben (my son, not Ben Orlin) was 10 or so, he came to me and asked if I would sign his math homework. He explained that in his class, if a student worked the homework for an hour and a half without being able to complete it, they could get their parents to sign it. This was the first and only time that this happened, and it turns out that it was a cryptarithmetic problem. I probably spent an hour on it myself without any progress.

I hadn’t programmed in a long time, and writing a program to solve this seemed like a potential way to get Ben interested in programming. At the suggestion of my friend David Coletta, I started learning Java, partly because you could embed code in web pages pretty easily. I eventually wrote a Java applet and posted it; as of today it has had almost 2600 hits according to the little hit-counter I found somewhere and embedded in the page. (These are undoubtedly not “unique hits.”)

This was around 2003, and it has gradually gotten increasingly decrepit. It worked fine in Internet Explorer, but when Chrome came along the text field for entering the formula became a slit. It still worked, if you didn’t care about actually *seeing* the formula you’d typed in. Then browsers stopped supporting the plugin that let you run Java applets, but you could still run it under Internet Explorer if you dusted that off. Today when I tried, even that didn’t work.

But I wondered if there was some way to resurrect it. I thought briefly about rewriting it in Javascript, but it’s actually a fairly large program, and not worth that much time. It turned out to be fairly easy to solve the text-slit problem, which was encouraging, fortunately or unfortunately. There is a not-yet-dead technology called Java Web Start, but apparently it will be dead soon and I had trouble cranking it up. It took me most of a day to accept the truth that “Running Java in the browser is dead.”

Just when it seemed that all was lost, I ran across a website (repl.it) that lets you write and post example code in many different languages. It supports not just Java, but Swing, the horrible graphics library that was in use at the time. So now it is up and running. If you try it, be patient, it takes quite a while after you hit the “Run” button to actually start up.

[Edit May 2021: repl.it is now replit.com, and they fixed quite a few of their Swing problems! So it works reasonably well, but you do need to wait 10+ seconds for it to start up.] ~~I almost said, “happily up and running,” but really does not quite make it to “happily.” The repl.it support for Swing still has some kinks to work out… like, the shift key doesn’t work. So you can’t type “zeroes+ones=binary” into the formula slot. Fortunately, it is a pretty good program, and you can type “zeroes=binary-ones” and get the answer. I slowed it up so that you can watch the digits flip, and that part looks better when it is running locally, but oh well. Also, the applet looked quite shrunken until I realized you could get it to be a reasonable size by dragging the dividers to resize the panes. ~~

To do this, I wound up putting the source code on GitHub. There is now a decent amount of explanation in the README file, including instructions for the simplest way to run it locally, which is in VS Code. [Edit Jan 22, 2020: this actually got me started on dusting out my old garage of programming projects; you can click to see the cleaned-up results.]

It turns out that Ben didn’t get interested in programming until a few years later, but since then it’s been an on-and-off hobby of mine. It turns out there were a few other projects whose technologies are obsolete, although I did migrate the Minesweeper game to Javascript, minus some of the more interesting features. When I complained about Swing, David Coletta suggested I learn Flex, and there were a couple small Flex projects in various states of incompletion. Alas, poor Flash player.

Although these technologies are not exactly shiny and new, I don’t think that there is any technical reason why they couldn’t still be made to work. These programs are obsolete as a result of commercial decisions by the colossi of Apple, Oracle, Google, etc. Not that it makes any difference in my case, but clearly many companies have had to devote huge amounts of time to finding alternate solutions when some of these options are no longer possible.

*Edit Jan 8, 2020: no news from repl.it about bug fixes (not very surprising consider that they seem to support about 40 computer languages). I did some experimenting with Webswing, which seems to be very well done; at some point I may give it a real go. But then it occurred to me that since I have it running on my own computer, I could just make a video and post it. Duh. (It’s only 20 seconds long.)*

[Edit Feb 3, 2020: eerily enough, the same calendar, on my birthday, had the puzzle “TWO+TWO=FOUR,” to be solved in base 8, where it said there are 5 solutions. I fired up the program and, lo and behold, that is correct!]

]]>You are supposed to make as many words as you can. To count, a word must have at least 4 letters and must include the center letter. Four-letter words get one point, but longer words are worth one point per letter, and “pangrams,” which are words using all the letters, get an additional 7 points. Proper nouns and hyphenated words are out.

As you proceed, there is an ongoing, upbeat evaluation of how you are doing. At the start, with 0 points you are a, logically enough, a “beginner.” From there is moves through “good start,” “moving up,” “good,” “solid,” “nice,” “great,” “amazing,” and, in what sounds like a final destination, “genius.” When you hit genius, it gives you the option of stopping or continuing. If you choose to continue, and you manage to supply every *Times*-accepted word, you get the ultimate recognition:

Not that I would know, if I hadn’t cheated. The truth is: I suck at Spelling Bee. Claudia is really good and has gotten to “genius” on her own. I don’t think I’ve gotten beyond “nice.” Which is interesting to me, because I’m pretty good with words and I do very well on crossword puzzles. I think I could get better, with practice, if I wanted to (more on that below), but maybe not that much better. Even so, Claudia has not yet achieved “queen bee” status on her own. (I’m talking about in the game. Anyone who knows Claudia knows that she achieved queen bee status everywhere else long ago.)

So, to cut to the chase: I wrote a web app to cheat at Spelling Bee. You can find it at familykuhn.net/nat/spellingb.

Interestingly, there is quite a lot of empty space between “genius” and “queen bee.” I tracked this in obsessive detail only once, on May 9. I hit the 5th rung, “solid,” at 45 points. It took almost another hundred points to go three more steps to “amazing,” at 143. The next rung, “genius,” happened at 205, and I had to keep going to 280 to hit “queen bee.” The mileposts stretch further and further as the journey continues. I suspect they are awarded based on some percentage of the best possible score for a particular day, but that would be for another obsessive person to figure out, even I have my limits.

There are three things I find at least somewhat interesting about all this:

- What dictionary do they use to decide what’s a word and what isn’t? (Note: I wrote this question first, but my answer got somewhat tedious, even by own standards, so I moved it last.)
- How can you list all the possible solutions efficiently?
- Is it a good puzzle? What make a puzzle good, anyway?

And I’ll also add a local newspaper article I stumbled on a few years ago that suddenly has new relevance…

A big dataset, like a dictionary, needs some clever thinking. For example, how should you look up a word? The obvious answer is: go through all the words, one by one, and see if the word you want is in there. If you have a thousand words in your dictionary, the worst-case scenario is that you would need to check 999 words before matching on the thousandth. This known as a “brute-force” method. Is there a smarter way to do it?

The answer is yes: a process known as “binary search.” You don’t need to get freaked out by the word “binary”—there are no base-two calculations coming up. It just means you have a series of two-way choices. For example, let’s try to look something up in the “2of12inf” dictionary, described below. It has 81,883 total entries, so going through it one-by one is not an enticing process when done manually. Imagine you are walking down a road, and you hit a fork. In the middle of the fork is a sign, which says “liberates.” If that is the word you are looking for, you got extremely lucky and you got your answer on your first try: your word is in the dictionary. But that’s very unlikely. Suppose you want to look up “filially” (the adverbial form of “filial”). Because it comes before “liberates,” you take the left-hand fork, confident in your knowledge that the half of the words in the dictionary before “liberates” can be found there and only there. You walk a bit further and you come to a fork with sign saying “disguise.” Still not your word, but since your word comes after “disguise,” you take the right hand fork, which now leads to the half of the words on our current fork which are after “disguise,” which will be one-quarter of the words in the dictionary. You keep going until you either find your word or you come to the end of the road, that is, a place where there are no words left to compare. [Spoiler alert: “fillialy” is not in the dictionary. You would either get to “filial” and discover that there is no right turn, or to “filibuster” and discover that there is no left turn.] It occurred to me that there must be a good interactive example of this binary search algorithm on the web. They all involve a little more of a deep dive, but I particularly enjoyed VisuAlgo, with a runner up or two.

Each time you choose one fork over the other, the number of words ahead of you is cut in half (for us sticklers, very close to half). How many times can you cut 81,883 in half before you get to something less than 1? The answer turns out to be 17. This means that you will come to at most 17 forks in the road before you are guaranteed to get your answer. Doing 17 checks is a whole lot better than 81,883 (worst-case) or 40,000 or so on average.

If you have gotten this far, and this is new information to you, congratulations! You now understand an “algorithm.” And I hope it is clear that having a good algorithm (like binary search) makes certain problems do-able, which they may not have been if all you had was a bad algorithm (like brute-force search).

By the way, how do you know it takes at most 17 forks? Well, you could (a) take my word for it, (b) divide 81,883 by two, seventeen times (0.62….), or (c) compute the base-two logarithm of 81,883 (16.32…) and round up. OK, yes, I said no base-two calculations, but this is technically something different, plus it is optional.

In the case of the spelling bee game, I used a data structure that is a bit different from the “binary search tree” with half the words on either side of a “node.” I used a data structure called a “trie,” which I hadn’t heard of until I found this lovely post by Steve Hanov, while I was trying to figure out how to write my program.

The “root” node in the trie (apparently pronounced “tree” if you want to confuse people and “try,” if you don’t) has branches going to 26 nodes, depending on the first letter of your word. From there you have 26 further nodes, depending on the second letter. (Fewer than 26, actually: for example, if your word starts with “x,” there are only 3 possible choices for the second letter, “e,” “i,” and “y,” at least in the 2of12inf dictionary.)

The largest number of steps you need to go through in this process is whatever the length of your word is. And through the wonders of “recursion,” it is pretty easy to enumerate all the possible solutions. Take your list of allowable letters: “afilnty.” To search down the whole tree, start at your root node, take the “a” branch, and enumerate all the solutions from there. Then do the same for the “f” branch, and keep going one-by-one until you finish the “y” branch. Wait, what? Isn’t that circular reasoning? Well if it is circular reasoning, that is bad, and your program gets stuck in an infinite loop. But in this case it is not circular, it is “recursive.”

With our 2of12inf dictionary, we first go to the node that has all words starting with “a.” From there we follow to the node for all words starting with “aa.” From there we go to the node for all words starting with “aaa” except, wait! Our dictionary *doesn’t have any*. How about words starting “aaf.” Nope. “aai”? No. “aal”? “aan”? “aat”? “aay”? No, no, no, and no. OK, we are done with words starting “aa.” So no we go looking for words starting with “af.” Anything start with “afa”? And off we go. If you get to an actual word, you need to check that it includes the required center letter before you get too excited.

It takes a bit of thought to see that this will never get stuck in an infinite loop, because there is a limit to how long our dictionary words are. The beauty of recursion means that this whole process can be implemented in just a few lines of code.

Of course, to do this you need to put your dictionary into a “trie,” which means going through every single word. So if you are just doing one puzzle, you don’t save that much. But if you are doing a hundred, one after another, you have saved a huge amount of computing power! Well, some day the *Times* will publish a whole book of these, and we’ll be ready, right?

Finally, the Steve Hanov post that I mentioned above is actually much more interesting than this. He talks not just about tries, but “succinct tries,” a way of encoding the trie that is space efficient, and can be accessed without re-expanding the whole thing. He posted his javascript source code, which I shamelessly plundered for its easiest part. He was kind enough to place it in the public domain, so I don’t need to feel bad about that. Thanks Steve!

I am sure that the fact that I don’t particularly like this puzzle is a classic example of sour grapes. So an honest list of why I don’t particularly like it would certain include #1.

- I’m not good at it.
- You need to be beyond-genius to actually complete a puzzle, which is where the little hit of dopamine for puzzle-solving comes from.
- “Completing” it depends on an unknown and sometimes arbitrary dictionary of allowable solutions.
- It gets harder as you go along, unlike, say a crossword puzzle which is easy at the beginning, gets harder, and then gets easier as you fill more stuff in.
- Doing well can take a long time.

Quick answer: I have no idea.

But even before that, how reasonable is their choice of what to include and exclude? I have my doubts.

For example, on May 9: how can they justify including “infantility” and “natality,” but exclude what seems to me to be the not-uncommon “flintily.” Why include “tinily” but exclude “tinnily?” I am having a hard time using “tinily” in a sentence, but maybe it’s because I’m distracted by my earbuds playing music tinnily. Why include “tali,” the plural of talus (ankle-bone), but exclude “ilia,” the plural of ilium (hip-bone) and “ilial.” I can understand including “lanai,” a darling of crossword enthusiasts, but excluding “latina”? Please.

Keep reading for a deep dive into what I discovered about dictionaries. Briefly: your mileage will vary a *lot* depending on you choice of dictionary. When you’ve had too much, skip to the cute ending. Here are the three dictionaries that I used:

- Official SCRABBLE Player’s Dictionary, 3rd ed (OSD3, below, and “s3” in the program output), which I also plundered from Steve Hanov.
- “2of12inf” from 12dicts. This is their dictionary that is supposed to be optimized for game playing. “12d” in the program output, followed by “!” if the word is flagged as a neologism, and “%” if it’s a word that should exist by the rules of word-formation but which no one actually uses.
- Collins Scrabble Words (2015) which fell off a truck somewhere; “c” in the program output.

For each solution, my program tells you which dictionaries it comes from. If it’s in all the dictionaries, the list shows up in green. (The word is in green if it’s a pangram.)

How closely do these agree with one another? And how closely do they agree with the *Times?* Read on, if you really want to know.

When I popped “mcehnta” (May 5) into the program, I found 133 “solutions.” Only 47 of them are in all three dictionaries.

You can find out the *Times’* opinion of acceptable solutions the day after the puzzle appears (and that seems to be it—no archive of past puzzles yet). Our puzzle has 55 accepted solutions:

The yellow check marks are the ones I entered. I missed six (enema, manhattan, matcha, meet, teammate, and teem), mostly through failing to type in things that appeared my list of 133. Four of the six were in all three dictionaries. One (manhattan) was only in the Collins dictionary. And how is that not a proper noun, which should be excluded? One of them, matcha, is not in *any* of my dictionaries.

Of the 47 solutions that were in all three of my dictionaries, six were not accepted by the *Times*: cementa, enemata, heme, mach, manana, and matt. OK, I can live without cementa and enemata (though honestly, if you include thema shouldn’t you include themata?). And maybe you have to be a doctor to think of heme as a widely accepted word. Doctors use both “heme” and “hamate” (which the *Times* accepted), but they probably use heme a hundred times for every hamate.

Of the 55 accepted by the *Times,* 12 were not in all three dictionaries. OK, the four pangrams (attachment, catchment, enchantment, and enhancement) are all too long to be in the OSD3, which is limited to seven (or is it eight?) letters. (There is another even more official Scrabble dictionary, the OCTWL, that includes words of up to 15 letters, the width of the board.) By the way it seems to be quite unusual to have *four* pangrams—most puzzles seem to have just one, but they are all supposed to have at least one.

The remaining eight *Times*-accepted non-consensus choices: emanant*, hamate, manhattan*, meme, mentee*, meta, meth, thema*. The four starred words are only in the Collins dictionary. So it seems that every *Times-*accepted word is in Collins, which is not surprising, because Collins is just much bigger, as noted below. There are, however, lots of words in Collins that the *Times* does *not* accept. Clearly I need a Venn diagram! Of the remaining four words, meme is in 12dicts, flagged as a neologism, but not in OSD3. Hamate, meta, and meth are in OSD3, but not 12dicts.

If you are still reading: the dictionary thing is complicated!

As a result, I decided to add my own dictionaries: one for additions, like “matcha,” that the *Times* accepts but aren’t in my dictionaries, and an anti-dictionary for exclusions, like the 79 words mentioned above. This will help make future solutions more accurate, since I have seen that the same excluded words, like “amah,” come up over and over. Anyway, “mcehnta” now yields the 55 *Times*-approved solutions.

While I was at it, I had the program count the words; originally all of them, and then only the words with four or more letters. So now I can tell you how many words in each dictionary have three or fewer letters:

- OSPD3: 1,075 out of 80,612 (1.33%)
- 2of12inf: 711 out of 81,882 (0.87%)
- Collins: 1,465 out of 276.643 (0.53%)

Of course, a Scrabble dictionary will have a great emphasis on two- and three-letter words, so it is not surprising that OSPD3’s percentage is high; although Collins is a Scrabble dictionary, it is so much larger that the added heft swamps the less-then-four-letter ones.

Whatever, dude.

In November 2015, our local paper, the Belmont Citizen-Herald, had a lead article, “Chenery Cheetahs win spelling bee,” talking about a team from the Chenery Middle School, our town middle school. I found it so amusing that anyone in the Boston area would name a team “the Cheetahs” that I scanned the article. But by a bizarre quirk of fate, it is less than four years later and I am a spelling bee “cheetah” in my own right.

One of those times was in medical school. For a quarter or two, I went to the pool regularly. After a vacation, I returned to the pool and ran into my friend Brad. When I asked how his vacation was, he said, “Great!” and proceeded to tell me about biking a century or maybe two, hiking, running. I realized that there two kinds of people in the world: the ones who exercise more when they’re on vacation, and the ones who exercise less. Both kinds were represented in that conversation, and you can see which I was.

I seem to have an abiding interest in getting something for, well, if not nothing, then “as little as possible” (hence: short-term therapy, no-knead bread, and no doubt a number of other hyphen-bearing shortcuts). So when I started to hear about 20-minute workouts using high-intensity interval training, I got interested. [Spoiler alert: 20 minutes is way too long.]

In traditional cardio exercise, you work out at a moderate level for 45 or more minutes. Research was showing, though, that by alternating, say, one-minute periods of high-intensity exercise with one-minute periods of easy exercise for a total of 20 minutes, you could get the same benefit. “Who doesn’t have time to exercise for 20 minutes?” I asked myself. So I borrowed my sister’s rowing machine, put it in my basement, and got started. I am happy to say I kept a log of more-or-less what I did. The rowing machine lets you read off the number of watts you are generating, which means you can gauge any progress you are making, which is the key to motivation and evaluation. If you decide to start exercising or to change your regimen, I really recommend that you log what you are doing, including not just what you are doing but also some measure of your performance. I’ve glanced back at this fairly frequently, although this is actually the first time I’ve looked it over in any detail. I started in September 2012.

The first thing I noticed is that I could not do all-out exercise for a minute, rest for a minute, and then do all-out exercise again. So I wound up setting a target for the exercise bursts that was about as high as I could do and still complete 20 minutes. On the last round, I would go all-out. Initially my target was to stay just above 160; by the end of the year, my target was to stay above 195.

I did this off and on for a number of years. Looking back, it seems that when I would stop I would lose capacity that I would regain over 3-4 months and then have more modest gains. Toward the end of 2016, I tried the “7-Minute Workout,” which is a collection of what used to be called calisthenics. It was great, I could see quick improvement in my capacity, but then I got all kinds of tendinitis and had to stop after a couple of months.

In August 2018 I read “The One-Minute Workout,” Martin Gibala’s summary of his and others’ research that was the basis for the NY Times coverage I’d read previously. It confirmed my reservations about the 20 minutes of one-minute-on-one-minute-off I had been doing. Apparently, what really matters is actually doing maximal exercise. Scaling back your efforts so that you can complete 20 minutes, as I had been doing, is probably counterproductive. It also seems that you don’t necessarily need one full minute of maximal exercise; 30 seconds works. Even 20 seconds may work, but even I am not going to go that minimal. Three repetitions is good. Even just doing one is good, but three is at least somewhat better.

So I came up with my own 7-minute workout. Three minutes of warm-up seems excessive, so I do two. I don’t do a cool-down, I just sit there panting and sometimes moaning or swearing until I can get up and go log my results. My wife, who is a devotee of Orange Theory, scoffs at what a—trying to find a paraphrase for “wimp,” which is not a word she uses—ah, yes, “lightweight” I am. She was happy when I told her that I did wind up lengthening the workout to 8 minutes, although she was less impressed when I told her that I had found the rest periods were not sufficiently long.

So here is what I came up with:

Time

Activity

Duration

0:00

Warm-up (light)

2 minutes

2:00

Max exercise

30 seconds

2:30

Light exercise

2 minutes

4:30

Max exercise

30 seconds

5:00

Light exercise

2″ 30 seconds

7:30

Max exercise

30 seconds

8:00

Stop, moan, etc.

up to you

When I started, my maximum effort was around 300 or below. Here it is 6 months later and my maximum is around 380 or a bit more. I’m pretty happy with the improvement, both in my numbers and in my body. I do this every other day; apparently it is not clear that doing it daily adds much benefit. I’m now alternating it with some even more modest weight-bearing exercise on the off days.

“Agony” is probably too strong a word for how I feel at the end, but it is definitely unpleasant. I do not feel motivated to increase my level of effort or time investment beyond what I am doing. “Ecstasy” would definitely be too strong for the up side, but for a guy who doesn’t like to exercise, “satisfaction” is a reasonable description.

]]>The cubic and quartic formulas were perhaps the greatest accomplishments of Renaissance mathematics, and they were a key spur to Evariste Galois’ beautiful study of the symmetry inherent in polynomial equations. Yet I got a PhD in mathematics without knowing much more than the bare fact that such formulas exist. It should certainly be possible to understand these formulas in light of Galois theory, but such explanations are not easy to find.

I wrote this paper to pass on what I learned about the cubic equation, and how it can be derived in the context of Galois theory.

It helped me understand the cubic formula, and it helped me understand Galois theory better by studying a semi-concrete example. The paper should be accessible to an undergraduate who has had a course in Galois theory; in any case, it contains a review of the necessary Galois theory, and an adventuresome reader who has not had such a course but is familiar with the basics of groups, fields, and linear algebra might get enough out of it to want to study Galois theory in more depth.

If you read it, I hope you enjoy it. Comments, questions, suggestions, etc. are welcome here.

Edit Jan 15, 2019: I edited the original version slightly to include a couple of diagrams.

]]>Since the book’s publication, Morris has been appearing here and there on the radio, and friends have been asking me about my take on the book. Rather than saying the same thing over and over, I have decided to put my thoughts together in one place. I haven’t read Morris’s book, for reasons that will become clearer to those who read further, but I was a careful reader of his five-part 2011 series on the New York Times Opinionator website, which the book is based on. I find Morris’s reading of my father’s work a vast oversimplification, to the point where the straw man is easy to knock down. Many people have taken that approach over the years, but Morris has a higher profile than most. There are many people more qualified to deal with his criticisms on whatever merits they may have, and I will mostly leave it to them. What impelled me to write this is my belief that the episode from which the book draws its title, in which my father supposedly threw an ashtray at Morris, never actually happened.

Is it important whether it happened or not? Everyone will have to come to their own conclusion on that. To be very clear up front: I would love to be wrong about this. I would love to hear from Morris’s classmates, family, or anyone else, that he told them that my father threw an ashtray at him at or close to the time that it allegedly happened. Because as misguided as I think his attacks on my father are intellectually, I believe that they come from a place of wanting to defend truthfulness. Defending truthfulness was important in 2011, and it is painfully obvious that it is more important than ever in 2018. I have great respect for Morris’s film work, and I think I share much of his political orientation. But just as the Buddha says hatred does not cease through hatred, falsehood does not cease through falsehood, and I believe that some of our current ills as a society stem from our delight in calling out falsehood among our opponents while ignoring it or even condoning it on “our team” and, more important, in ourselves.

Starting right from the book’s cover, Morris’s point of view is that my father denied that there was such a thing as truth or reality and that this is both wrong and dangerous. I agree that such a denial would be wrong and dangerous. The only difficulty is: that is not what my father believed or was trying to say; he would also have agreed that such a denial is wrong and dangerous. What he did say about truth and reality, among other things, is that they are more complicated than a naïve stance on them acknowledges. He didn’t always do a good job of reconciling the existence of an external reality with the limits of our ability to describe it. This was no doubt a result partly of his own limitations as a thinker and writer, but partly—maybe mostly—because it’s just a really hard problem. More concretely, while he may have believed that you could not say that a scientific theory was “true” in some absolute sense, my father certainly did not believe that any so-called scientific theory is just as good as any other. I would join Morris in finding such an opinion “revolting,” but that is not what my father believed; it is a caricature of what he believed. As I said, there are people more qualified than I am to focus on the merits of the discussion. Jacket blurbs aside, serious reviews do not seem to be coming down in Morris’s favor. In addition to the Kitcher piece cited above, which I highly recommend to anyone looking for a summary of the intellectual issues, there are also reviews by Steven Poole in *The Guardian*, and Laura Miller at *slate.com*.

The Times series starts with an argument between my father and Morris, then a graduate student studying with him at Princeton, which supposedly culminated in the throwing of the ashtray. I was disturbed by that story the moment I read it, because I simply didn’t believe it, and neither did anyone I spoke to who knew my father at all well. I wrote some comments on the series on the Times website and some blog posts (links below), as did my sister Sarah. My sister and I both respect Morris’s work, and we were reluctant to believe that he was peddling a false story, but that is where we ended up.

Of course, in the age of #metoo, we are all more aware than ever of the flimsiness of the “he just never seemed like that kind of guy” defense. But we are also more aware of how one might come to a more credible resolution of this kind of “he said-he said” disagreement. In particular, are there any accounts in the intervening time that shed light on it one way or the other? I do have one: when the 2011 series appeared, my stepmother, Jehane Kuhn, told us about an encounter with Morris that provides some reason to believe that his story is not true. Jehane was always quite a private person, and at the time her preference was to not make her account public. Sarah and I chose to respect her wishes, believing and hoping that the series would have a short half-life. Now that Morris’s claim has reappeared in more durable form, I am going to tell the story.

As Jehane related it, she and my father went to a restaurant in the Boston area, where they then lived. There they ran into Morris, whom she had not previously met. Morris recounted a meeting with my father during his student days, culminating with his saying, rather gleefully in her recollection, something like, “He pushed me out of his office! He pushed me!” This is quite a different story from the much more inflammatory “He threw an ashtray at me.” I can’t think of a credible reason why, if my father had in fact thrown an ashtray at him, he would have focused on being pushed. Jehane is no longer able to speak for herself on this matter, and I don’t know that she told the story to anyone other than my sister and me. But there are people who can tell you that she would not fabricate a story of this sort to defend my father’s “reputation.”

Are there other accounts closer to the event in which Morris tells the ashtray story? I have seen Morris quoted from the 1980s as saying “my adviser actually assaulted me,” but a person nursing a sense of grievance could well describe being pushed out of an office in those terms. Despite requests from my sister through his publishers at the Times and the University of Chicago Press, Morris has not produced any contemporaneous or subsequent interlocutors who could vouch for his story. He did say something to her along the lines of “I’m sorry if I offended you,” entirely missing—or evading—the point. According to the *slate* review, Morris “has a track record of misrepresenting the statements of others in order to gin up a rhetorical antagonist to battle.” Perhaps he has misrepresented not just statements, but actions.

Despite our hopes that the story would sink out of view, Morris resurfaced in May of 2017 in a piece by John Horgan on the *Scientific American* website, “Did Thomas Kuhn Help Elect Donald Trump?” Horgan does a reasonably good job demonstrating that even “anti-Kuhnians” think that this claim gives an absurd amount of credit to the influence of intellectuals. He also published a follow-up piece in which he gives others an opportunity to present a more nuanced version of Kuhn.

Now that Morris’s book is here, I have taken the occasion to dust off the pieces I wrote back in 2011. The main one is My comment on Errol Morris’s 4th installment of “The Ashtray” ran over 5,000 characters. It still seems apt, except that my bemoaning of the state of our public discourse may feel rather quaint in light of more recent events. I had remembered talking about why I found the story impossible to believe in the first place; I had forgotten that I had highlighted a passage in the midst of a long digression in Part 3: “One of the oddities of history is that legends often supersede facts. Historical evidence accumulates, monographs are written; but the number of popular accounts retelling the apocryphal story of that non-crisis proliferate. Why? Because we love to read about crisis and conflict. It’s drama. It makes a better story.” There is no doubt that having an ashtray thrown at you makes a better story than being pushed out of an office. He goes on to say, “A legend that is not true can never become fact, but it can get printed as fact, anyway.”

While that blog post of mine was gestating, I posted some extemporaneous remarks that I had made after my father’s death at a symposium in honor of his intellectual legacy. Initially, I was tempted to say that I posted it in order to assert my bona fides as a non-apologist for Kuhn. Re-reading it now, I am struck by its additional relevance, on two points. The original remarks were in response to a presentation by a former student of my father’s who, like, Morris, seems to have been deeply wounded by their interactions. I don’t doubt for a minute that they may have been wounded in this way, and if so I don’t doubt that my father bore significant responsibility for it. That may well be a story worth telling, but I believe that if you are telling it in public on a broad stage it is incumbent on you to do so in a way that is as honest as possible. The remarks I made also speak to the difficulty of holding what may seem like irreconcilable truths in tension, rather than collapsing into oversimplification. This was a difficulty my father was quite explicit about in titling a collection of essays “The Essential Tension.” (For die-hards, there is a third post in which I respond to some of the other comments that were made, but unfortunately the links to the original comments no longer work.)

In the process of writing this piece, I recalled an anecdote which is germane to my father’s views on the “one scientific theory is just as good as another” caricature. In the 1970s, he was approached, as a renowned “expert on science,” to be a witness for the evolution side of an evolution-vs-creationism-in-the-classroom lawsuit. As a teenager, I thought this was pretty cool, and when my father declined, I asked him why. It was not (as I already knew) because he thought that evolution and creationism were “equally valid” and should be presented side-by-side in high school classes. He told the people who approached him that he agreed with them, and hoped they would prevail, but that in the face of an antagonist who is determined to oversimplify what he is saying and discredit it for the sake of discrediting it, his testimony might well do more harm to their cause than good. Of course, in our adversarial legal system, that is exactly what the lawyers on the other side are paid to do. It doesn’t mean that’s what the rest of us should be doing.

So I am left with the uncomfortable belief that Morris’s story, which he’s made the title piece of his book, is false. As I said above, I would love to be convinced that I’m wrong, apologize to Morris for casting aspersions on his honesty, and just forget the whole thing. If on the other hand I’m right, perhaps he sincerely believes that the story is true; perhaps someone else threw an ashtray at him, and it has become conflated in his memory with the painful interaction with my father. Or perhaps it is a deliberate fabrication. For that to be a plausible conclusion, there would need to be a motive. What motive could there be? To sell more books? Well, it *is* a good “hook,” as he has demonstrated. But he is perfectly capable of selling lots of books without it. Why would you undermine your defense of truth by marketing it around a falsehood which is in no way essential either to your argument or to your goal of selling books? Especially when your legacy as a documentary maker is largely dependent on your credibility. It makes no sense at all. Perhaps the goal is not to sell books, but to achieve the sort of immortality that goes along with being part of a legend that is printed and reprinted as fact because “we love to read about crisis and conflict.” One needn’t go to ancient history for examples—Morris’s anecdote has already been compared to Wittgenstein’s Poker (which despite being a celebrated dust-up took 55 years to appear in book form). And perhaps the sly satisfaction of using untruth to settle a vendetta against someone who you think is denying the very existence of truth (and therefore also of untruth) is irresistible. These are all speculations; there may be truth to all, some, or none of them. All I can say is that, as a psychiatrist and psychotherapist, I am a frequent witness to the fact that when people act out of their wounds they often do so in ways that don’t make sense and may in fact be highly self-defeating.

So if I believe that Morris is relating a false episode and titled his book on that basis, what should I do about it? It doesn’t speak to Morris’s intellectual argument one way or another. The allegation has no bearing on my father’s intellectual legacy; it may do some damage to his reputation as a human being, but that was already not pristine. What does not sit well with me is that, in the process of supposedly defending the truth from debasement, Morris seems to be taking his own shot at debasing it. I just don’t believe you can release yourself from your wounds unless you do it with integrity. Hatred does not cease through hatred. That goes for people, and it goes for countries. And silence in the face of truth’s debasement does not seem like the thing to do.

[I am grateful for editorial feedback from Sarah Kuhn, Claudia Matzko, and Anna Walker.]

[March 29, 2019. A friend alerted me to John Horgan’s review of The Ashtray; I was interested to note that sitting with Morris’s bile for an extended period seems to have caused Horgan to come closer to defending my father’s work. Horgan was kind enough to add a link to this piece, and also alerted me to David Kordahl’s review, which I think is wonderful.]

[April 15, 2019: John Horgan has posted a thoughtful letter from a colleague and former student of my father’s under the title Thomas Kuhn Wasn’t So Bad.]

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