the jsomers.net blog.

You could not say a word and still be thought a great conversationalist, so long as your interlocutor ends up doing the kind of talking they like to do. That's what it means to be good at conversation. It's not about saying interesting things or absorbing what someone else says. It's about extruding the right kind of talk.

The picture I have in mind is of a pasta maker. I imagine someone turning a little hand crank, working a glob of dough into clean hollow tubes of macaroni or waves of lasagna. That’s what conversation is like. Conversation is like making pasta the old-fashioned way, except that in conversation what you’re working into form is someone else’s ideas.

At the heart of the process, you, the listener, are just like a pasta machine—a machine so obnoxiously simple that you hesitate to call it one. Because all it is is a shape.

* * *

As a talker I have all kinds of configurable settings. Entering a conversation, there are many knobs I can turn:

Will I talk intensely, blood flushing my face, or calmly and casually?
Will I try to sound colloquial, professional, intelligent, ignorant?
Will I reference continental philosophers by name? Golfers? Poets?
Will I lean on analogies from math or physics, and microbiology?
Will I look for opportunities to deploy a sesquipedalian word?
Will I work the other guy in like my thoughts depend on his?
Will I let myself ooze with emotion, and romantical angst?
Will I consciously experiment with tones and inflection?
Will I be trying to find out what I genuinely believe?
Will I ask questions -- real and/or rhetorical ones?
Will I wait for that name on the tip of my tongue?
Will I be tendentious and polemical, out to win?
Will I use a long silence to plan what to say?
Will I speak delicate or with high abandon?
Will I be afraid to use the word "blog"?
Will I deploy pop culture references?
Will I be jokey? Punny? Sarcastic?
Will I play high status, or low?
Will I take it to a meta level?
Will I get fiercely political?
Will I play devil’s advocate?
Will I crank up my charisma?
Will I indulge digressions?
Will I talk fast or slow?
Will I try small talk?
Will I spin yarns?
Will I brood?
How much?
Etc.

I shine—I come into my own—under certain configurations of these knobs, and shrivel under others. For instance I enjoy riffs of sarcastic banter and quoting movies, but I tire quickly of pun-upmanship. I like to explain. I lean heavily on a stock of nerdy analogies and feel crippled when I can’t use them. I don’t know how to keep small talk going. I range from being very charismatic to having something like a stammer. I like it when the gossip knob is turned up high. I can’t riff about football or how many gigabytes a phone has. I prefer my talk to be salted with curse words. I don’t like talk that sounds like it’s coming out of an English classroom. I rarely argue. I don’t do well when I’m trying to impress.

Different speakers draw different kinds of talk out of me. Michael, fluent in most of my intellectual interests, is great for helping me feel out ideas. Rob gets me spilling insecurities. I have a friend, Carey, who leaves me thinking I’m inarticulate and wrong. With Nikhil I talk slow and philosophical. I get Seinfeldian with Matt. I spout bullshit with Sanders. There is a guy at work who encourages me to improv. An old roommate, Andy, always had me explaining things I didn’t understand well enough to explain. Drew gets my polemical side going. I’m made to feel young when I talk to my older brothers, and wise when I talk to my older friends. I’m at my most charming in the company of my good friends’ girlfriends.

Which is all to say that I configure myself in light of who I’m talking to—so much so that you could say they configure me.

* * *

Some talkers are no doubt more configurable than others, in the sense that they change themselves, chameleon-like, depending on whoever they’re talking to, while others are “just themselves” no matter what. But I’d bet most people are more pliable than they’d say. How easy it is to tell when a friend of yours picks up their phone that they’re talking to their girlfriend? Their employer? A mutual friend? Their parents?

What’s happening, I think, is some combination of “mirroring”—that phenomenon where I'll unconsciously mimic your posture, tone, level of intimacy, style of humor, and so on—and this thing where before each of my remarks I'll think about you and what I know about you and what I think I can say and then I'll triage your likely responses, and my responses to your responses, and so on, and make my conversational moves in light of this projected snap-analysis of where I think our talk might take us.

It sounds effortful and conscious but of course it’s not. Having a sense of where you are in a conversation, of what’s apt and in play, is the bedrock social skill. It happens automatically. To be socially well-adjusted is to adjust well, to be highly responsive to the microdynamics of talk.

Of course this only underlines the leverage that you have as a listener. That is, it invites you to invert the picture, to consider what kinds of rejiggerings you cause others to make. What sort of conversational selves do you have a way of drawing out?

* * *

Like everybody else, I have an incalculable store of memories: remembered episodes, snatches of speech, images, trains of thought, and so on. The problem is, I can’t just enumerate this stuff. I need cues to call it up. That’s how my mind works: it all just sits there dead until I do something or see something or hear something or smell something—and then little memorical fragments effortlessly spring to life.

“Conversations take random walks through events and ideas in a manner determined by the associative networks of the participants,” writes the cognitive scientist Douglas Hofstadter. Conversations are “stimulating” in the concrete sense of arousing inert patches of thought. It’s by the random walks of conversation that I am made to access neglected memories: whole books I had forgotten I’d read, whole ways of thinking and talking, articles, relationships, moments, jokes, flavors, phrases, feelings, ambitions, tunes, stories, fears, ideas—all of which are stored so that they can be called into action in just the right context, and which otherwise might as well not exist.

And the point is this: different people activate different parts of that complex. What I get to be, the parts of me I get to see—it depends so much on the shape of a listener.

Of the 142 Project Euler problems I've battled so far, it may just be that #191 has put up the most interesting fight, not because the problem itself was all that difficult or unusual or enlightening, but because the road to solving it---seven to ten hours of work, seven sheets of paper, several trips to the online encyclopedia of integer sequences, three or four minor insights, and one major breakthrough---was for me almost a paradigm of puzzle-solving, or, closer to the point, a paradigm of how a mathematically crippled mistake-prone dabbler like myself solves puzzles.

Here's the problem statement that kicked it all off:

A particular school offers cash rewards to children with good attendance and punctuality. If they are absent for three consecutive days or late on more than one occasion then they forfeit their prize.

During an n-day period a trinary string is formed for each child consisting of L's (late), O's (on time), and A's (absent).

Although there are eighty-one trinary strings for a 4-day period that can be formed, exactly forty-three strings would lead to a prize:

  OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA
  OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO
  AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL
  AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA
  LAOO LAOA LAAO

How many "prize" strings exist over a 30-day period?

I encourage you to give it a shot. If you've solved lots of puzzles like this in the past, you'll probably be able to cook up a solution in under an hour. If not, expect to spend an afternoon or two scratching out all sorts of cases and conditions and combinatoric formulations. Either way it should be a great deal of fun.

When I first saw the problem I had only two approaches in mind:

1. If we let n be the number of days, we know that there are 3ⁿ possible strings of Os and As and Ls. To find the number of "prize" strings, then, all we have to do is subtract from 3ⁿ the number of strings that couldn't be winners.

2. Or, we could go the constructive route and try to only count prize strings.

Intuition told me that the first way would be a lot easier, so that's the way I went.

A first crack

The problem statement helps us out by giving the solution to a small case, n = 4, which we can use to validate our approach before ramping up to the much trickier case where n = 30.

What we're trying to do is enumerate all of the strings with either the substring AAA or two or more not-necessarily-adjacent Ls, because these are the strings that are ineligible for a prize. We know that we have 3⁴ = 81 possible strings to begin with, and from the problem statement we've been told that 43 of these are prize strings, so whatever rules or program we come up with to prune them should wipe out exactly 38 (=81 - 43).

At a first pass I mustered the following rough (and wildly incorrect) count:

• There are exactly two strings containing AAA.
• Six strings contain two Ls, a number I arrived at by taking "4 choose 2", or 4!/2!2!
• Four strings have three Ls (by the same argument as the two-L case).
• One string has four Ls (ditto).

Unfortunately that left me with only thirteen candidates for elimination, far short of required 43. Can you see what I was doing wrong?

I left out a whole lot of permutations! Each of the cases above is actually under-specified, in the sense that there are multiple strings that could satisfy the given condition. For example, although I said there were only two strings containing AAA, there are actually a lot more, because each template---AAA_ or _AAA---can be completed with any one of O, L, or A. Similarly, in cases where we're counting Ls, we need to remember that the spots without Ls can have either an O or an A, so that each L-template really has 2^k possible fillings, where k is just the number of non-L spots.

Trouble looms

Are you beginning to smell the complexity?

Consider the rule for counting strings with an AAA in them. Even if you multiply 2 (for the two templates AAA_ and _AAA) by 3 (because in each template you can have three possible fillings: O, L, or A), you're still incorrectly counting the number of ineligible strings. In particular, you've double-counted the case where an A fills the blank, because regardless of which template you use, you end up with the same string AAAA.

At first this might not seem like a big deal, but that little AAAA snag and others like it will turn out to be major wrenches in the works, hitches enough to drive me into what an impartial observer might have described as hours of psychotic numerology.

The trick is to see what happens when you crank n up to 30. Remember what we're trying to do: ultimately, we want the number of eligible prize strings. To wit, we want to first count the non-prize strings---the number of strings with three or more As in a row (LOOAAAO..., ...AAAAO, L...AAAAAAOO...O, etc.) or two or more Ls (OAAL...AOLO, AOOLLALLL..., etc.)---and then subtract that number from 3³⁰, the number of total possible strings.

Suppose we start by trying to count the first type of string, the ones with a bunch of AAAs in them. At some point we're going to run into a string that also has a few Ls, and it's important that we somehow keep track of when this happens. If not, we'll end up counting such a string twice, first when we go through looking for AAA-style strings and second when we look for strings with multiple Ls.

This is exactly the kind of bookkeeping you don't want to run into when hacking on a combinatorics problem, because it almost always means you've taken a wrong turn. Project Euler is careful to choose problems that can be solved elegantly, usually with less than ten lines of code, so it's a major red flag anytime you've got pages of branching cases and ad hoc "rules."

Divide and conquer?

At this point I probably should have gone back to square one. Instead, though, I hatched what I thought was a workable game plan: I'd avoid the double-counting issue by breaking those non-prize strings into mutually exclusive sets.

I'd proceed in three steps:

1. First, count the strings of length 30 with two or more Ls in them. The non-L spots in these strings can filled any whichway. They can even have runs of three or more As in a row---it doesn't matter.
2. Second, count all of the strings with runs of As in them that also don't have any Ls. So we're just considering the length-30 strings with some mix of Os and As, and counting only the ones that include three or more As in a row.
3. Finally, count all of the strings with three or more As in a row that also have exactly one L in them.

Do you see how that works? The trick is that we're taking care of all the messy stuff in Step 1, and restricting ourselves in Steps 2 and 3 to small precise collections of ineligible strings---strings where only the AAAs matter. So our three sets are guaranteed, by construction, not too overlap with one another.

So much for the theory---at a certain point one has to actually count these things, either with a formula or a simple program. Will my sets be well enough behaved, or will they too have all sorts of quirks and caveats?

I tamed the first set rather quickly. It turned out that counting those strings, the ones with with two or more Ls, was simply a matter of taking a sum from i = 2 to i = 30 of "30 choose i" * 2^{(30 - i)}.

That first term, "30 choose i," just counts the number of ways of arranging i Ls in a string with 30 different spots. For i = 3 it would count the number of templates that look like __L...L_..._L or LL__...L..., i.e., templates with exactly three Ls in them. The second term just counts the number of possible fillings-in of each template: for each spot there are by assumption two choices (O or A), and each template has 30 - i open spots (spots without an L in them).

The second set is a lot more complicated, but once I thought about it I was immediately reminded of another Project Euler problem, one I'd worked on---without solving---just a few weeks before. It was problem #114.

Old friends and new

One way to see the correspondence is to imagine the red blocks in problem #114 as As and the black ones as Os. If you solve it for n = 30, then, you've implicitly solve my Step 2 above, because you've counted the number of strings with at least one run of three or more As and no Ls.

Or so I thought. What you've actually done is undercounted those strings by a huge margin, because you've left out strings where an A---a red block---fills one of the spots not already covered by a run of As. Problem #114 explicitly requires that the only red blocks you can have must appear in runs of three or more. You can't, in other words, have stragglers.

You can imagine what I did next. I tried not only to solve #114, a problem I couldn't crack weeks earlier (even with the help of a mathematically-inclined friend who wanted to chip in), but to do so knowing full well it'd only get me partway toward a solution to the problem at hand---thanks to those straggler As. It cost me about two hours of mostly unnecessary work, an expense I would have avoided if I had never seen #114 in the first place. It was an unlikely lesson: drawing on experience can sometimes do more harm than good.

After scrapping that angle I decided to do something a bit ignoble. I decided to write some code to exhaustively spit out trinary strings, not of length 30---such a program would take weeks to run---but for small ns. For each n I'd keep a count of the number of strings that would fall into each of the sets I named above, with a specific focus on the strings in Step 2: strings with zero Ls and at least one run of three or more As in a row. I'd then take that partial series and plug it into the Online Encyclopedia of Integer Sequences. If I hit a match I might be able to steal a nice clean formula.

(I've raved elsewhere about the OEIS, so I won't do that again here. Suffice it to say it's an invaluable resource and a constant temptation when working on Project Euler problems. One must use it sparingly and as a last resort.)

Much to my delight, the very sequence I had generated was chronicled in great detail on the OEIS as sequence #A050231, "the number of n-tosses having a run of 3 or more heads for a fair coin."

Yoink!

Descent into madness

#A050231 was a huge boon---rather than computing the count in Step 2 above, I could literally copy and paste the correct value from a list someone else had already generated and posted to the OEIS. And with that I was only one term away from a full solution. All that was left was to work through Step 3: counting strings just like the ones in Step 2 that also include a single L.

When you think about it, it actually seems like a trivial modification could get you from the strings that A050231 is counting to the strings that also include a single L (call them L-strings). To get the number of L-strings, all you have to do is to take each A050231-string---each string of As and Os---and multiply it by the number of O-spots (as opposed to A-spots), those being precisely the spots where you can put the L. As an example, for the A050231-string OAAAOAOAAAA I would have three corresponding L-strings, LAAAOAOAAAA, OAAALAOAAAA, and OAAAOALAAAA, one for each O-spot.

But since I wasn't man enough to generate A050231 myself, and so didn't understand its structure at anything but the shallowest level, I had no way of doing that simple multiplication, because for each n I only knew the number of A050231-strings, without knowing how many O-spots each of them contained.

So after fumbling around the OEIS for just over an hour trying to find some code or sequence or math that might get me those numbers, I was left seemingly without a way out. But instead of dropping the idea, I decided to work it out in painstaking detail. I decided to work out a sequence that tracked, for each string of length n, (a) the number of ways of arranging As, with the proviso that each "way" must include a run of three or more, and (b) the number of empty spots (O-spots) in each of these arrangements. I'd then look for patterns.

You can see the insane work that came out of this insane project by looking at pages 1 and 2 in this PDF of scratch paper. The "crazy-person triangles" that resulted include a list of comma-separated terms for each n, where the large numbers specify the number of O-spots we're working with (call it k) and the subscripts count the number possible arrangements of Type 2 strings---As and Os and at least one run of three or more---for each of those ks. (I'm pretty sure only the second triangle is correct, which is why I went on to larger ns with it. It's been long enough since I did this work, though, that I don't really know what I was thinking or what's actually going on in the first triangle.)

I used a helper script to compute those subscripts as n and k increased. I had to stop after n = 14 or so, both because I was running out of space on the page and because the subscripts were taking too long to generate. But even with eleven different rows of numbers, I wasn't able to find any meaningful patterns. I did find all sorts of unmeaningful patterns, mostly by subtracting sequences of numbers and then recursively subtracting the differences until I hit some sort of constant or steady state; but I'm pretty sure any sequence can degrade that way after enough "diffs of diffs." You can see on page 2 of my scrap work that I came upon a few promising generating formulas, both horizontally (as k varies) and vertically (as n varies), but nothing panned out.

The clouds part

After all that---after all that---I decided to scrap the work I'd done so far and take another tack. Because it hit me like a ton of bricks: I should be solving this problem inductively.

Think about it this way. Suppose I know how exactly how many prize strings there are for a given n. Now increment n by one. What changes?

Well, there are only three letters that can be added to a given string: O, A, and L. If I add an L to a string that already has an L, all of a sudden that string becomes ineligible; same thing if I add an A to a prize string ending in AA. If I had an O to any string, it has no effect. If I add any letter to an ineligible string, it stays ineligible. Etc.

It's starting to sound like all we need to do, then, is keep counts of certain string "types"---strings ending with AA, strings with one L, ineligible strings, etc.---and formulate rules for how these counts change each time we increment n.

That's exactly what I did. After lots of trial and error, I determined that I needed to track six different types of special strings:

1. Eligible strings ending in AA.
2. Eligible strings with zero Ls.
3. Eligible strings ending in A (and only A).
4. Eligible strings ending in AA that also have exactly one L.
5. Eligible strings ending in A that also have exactly one L.
6. Eligible strings with one L that don't end in A.

It turns out that just this set of strings, call them a through f, are enough to keep an accurate accounting of prize strings as you increment n all the way up to 30.

An example should illustrate the method. Take c-strings, strings that end in a single A. Suppose that when n = 3 we have exactly five of them (we do: they're OOA, AOA, LOA, OLA, and ALA). What happens to that count (5) when we increment n? Well, we have to consider three cases, one for each of the possible new letters O, A, and L.

(Keep in mind that what we're doing in each case is essentially generating 3^{n + 1} new strings---that is, we're taking each of the 3ⁿ strings we have and appending one letter (O, for instance) to all of them. Do that three times and we have our full round of strings for the n + 1 case, because 3ⁿ * 3 is 3^{n + 1}. I say "essentially" above because what we're really doing is generating 3 * t new strings, where t is the number of eligible (prize) strings for a given n. We ignore ineligible strings.

Either way, the question is how many of these new -O strings and -L strings and -A strings are of type c.)

• Case O: If we add an O then all of the five eligible strings that used to end in a single A no longer end in a single A. And all of the strings that didn't end in a single A still don't. So the net effect of the Os is to decrease our count by 5.
• Case L: This works the same as the O, obviously. A net loss of 5 c-strings.
• Case A: If we add an A to one of the five c-strings, it'll no longer be a c-string. But if we add an A to any of the other strings it will become a c-string. So there's a net gain here of t - 5.

So we have a total number of ((t - 5) - 5) - 5 c-strings in our next round, which for n = 3 is ((27 - 5) - 5) - 5 = 12. If you were to enumerate all of the eligible trinary strings for n = 4, you'd see that there are in fact exactly 12 of them that end in a single A. Our method works!

What's nice is that we now have an exact formula that tells us how to get the number of c-strings in round n + 1 if we know the number of c-strings in round n. And what's more, that formula depends only on t, the total number of eligible prize strings in a given round; c; and a, where a tracks the number of strings ending in two As. (Where does that a come from? Well, when you add an A to a string that already ends in two As, you "kill" it (render it ineligible). So you need to know how many a-strings are floating around.)

Similarly, all the rest of a-f depend only on t and the values of a-f themselves. That's the sense in which the set a-f is complete.

The upshot of all this is that we can now write an obnoxiously simple program to find t when n = 30. We simply seed a list of values for n, t, a, b, c, d, e, and f---[1, 3, 0, 2, 1, 0, 0, 1]---and loop over it thirty times, applying our various formulas (like the one for c-strings above) with each iteration. Here are the simplified formulas for each element:

• n -> n + 1
• t -> 2 * t + b - a
• a -> c
• b -> 2 * b - a + d
• c -> t - (a + c)
• d -> e
• e -> f
• f -> t

I encourage you to try to derive these yourself. You can consult my scratch paper for hints.

We're left with this wonderful, short, extremely fast solution:

# [n, tot, -AA, 0Ls, -A, 1L-AA, 1L-A, 1LnotA]
lst = [1, 3, 0, 2, 1, 0, 0, 1]
while lst[0] < 30
  n, t, a, b, c, d, e, f = lst
  lst = [n + 1, 2 * t + b - a, c, 2 * b - a + d, t - (a + c), e, f, t]
end
p lst[1] # => 1918080160

When I was younger, I would occasionally sit down to my computer thinking I was about to compose a masterpiece, something that would all but earn my place in the canon. I would have a peculiar feeling brewing, that I was on the cusp of a radical insight, that I was about to see the soul of man. It was an epiphanic mood: everything I was seeing, I was seeing as crucial and human and connected to some secret the substance of which sat wonderfully unarticulated in the canny shadows of my mind. I had but to write it down and claim my prize.

I wouldn't need much. Ezra Pound said, "If a man writes six good lines he is immortal." A single stanza would do. Maybe a sonnet.

The mistake I was making was simple: looking at a blank page, thinking of myself as an above-average writer, a sensitive thinker, I felt like nothing material stood in the way of me and my six lines. It wasn't like I was trying to run a four-minute mile or prove the Collatz conjecture, where it would take years just to get the tools, the mechanics, the muscles, to level with the problem. All I had to do was type the right words.

But the page isn't blank, and my brain isn't the Library of Babel. There are only so many things I can think.

Insights---the kind that Shakespeare had; the kind that Joyce had---aren't just "in the air." They're the fruit of a lifelong process: seeds looked for and lucked upon, grown with steady care, seized when they ripen. I may never have one.

The first two weeks of college are exceptional for a lot of reasons. But there's one phenomenon that stands out. It's as peculiar and as powerful and as rare as magic. [1]

I'll call it "social annealing" because it so resembles the process by which metals are heated---their particles freed into a loose homogenous jumble---and then recrystallized.

It's the phenomenon whereby college freshmen, having just been ejected from the world they know and loosed upon a campus they don't, mutually decide, in the heat of their eager anxiety, to willingly engage every stranger as a friend.

It happens in the dorms, the cafeteria, coffee shops, classrooms, libraries, and just around town: students with no prior connection approach one another and strike up conversations. Everyone tries everyone out.

Groups do form, but they're not the kind of groups we're used to---they congeal and dissolve with remarkable ease. So a lone student can approach five others without feeling like he's intruding; two sets of roommates can combine; a pack can split with no friction.

Think of how bizarre that is. Think of what it would take, for instance, to introduce yourself to a group of four friends, none of whom you'd ever met. It's practically preposterous.

That's because in normal circumstances, confronting strangers without an overt excuse---the elevator breaks down, say, or a plane is canceled---is an act of aggression. If not outright threatening, it's intrusive, or at the very least distracting.

Even in settings that seem to encourage mixing, like parties or bars, it's not kosher for a person to engage just anybody---one must abide all kinds of cues and conventions and rules; contact must be made with a measure of finesse.

Which is to say that nothing you can find elsewhere in the workaday world even resembles the two-week college free-for-all, the strange fever in which everyone is basically pleased as hell to meet everyone else.

It almost sounds like a fantasy. But I assure you it happened. I'm not a spectacularly outgoing guy, but for the first two weeks of my freshman year at the University of Michigan, I introduced myself to just about everyone I saw. When I'd go down to the cafeteria, I could sit anywhere. At parties, on the way to class, in the dorms, etc., I---like everyone else---would flit from group to group in a crazy kind of convivial Brownian motion. Our social graph was effectively amorphous---fully connected. We were open to each other in a way that I imagine swingers must be open to sex, or hippies to psychedelics.

* * *

It's probably worth asking how this happens, or why. I don't think it's all that complicated:

1. Bizarre things are bound to happen when you throw a large number of eighteen year-olds into close quarters, especially if you don't give them any work to do.

2. For the most part, nobody knows anybody when they first arrive at college. And even if you did know some people, say, a few other kids from your high school, it's good sense to avoid them for a little while, if only to participate in the social madness I've been describing.

   Which means everyone is effectively looking to make a fresh start, to find replacements for their now-disbanded troupe of close dependable friends.

   The trick is that everyone knows this. They know that everyone is in the same boat. And that pervasive common knowledge---where for any two people, A knows that B knows that A knows... that they're both looking to make friends---is enough to fuel cold approaches.

3. Freshmen are called "freshmen" for a reason: they're fresh; they don't yet have a reputation. Everyone understands that, and they understand that first impressions can stick. So it makes sense that they're unusually warm and friendly. It's the best way to keep their social options open.

4. There is what I'll call the "New Year's Resolution Effect," where kids just entering college decide, in light of their radical dislocation and the discontinuity from their life at home, to change themselves in fairly significant ways. In particular, they tend to commit to being more extroverted than they were in high school. It's a common enough ambition to accelerate the pandemonium.

* * *

I mostly wanted to articulate this phenomenon because of something that happened last weekend. After seventeen months away I was back in Ann Arbor, that great college town and the site of my alma mater, ostensibly for a big football game, but really to reunite with lots of old friends, many of whom I hadn't seen since the day we walked the Big House in our robes.

It was quite nostalgic. I really loved that place---still love it---and a lot of what I did that weekend was to reminisce, to reconnect with a mass of pleasant memories and in some cases to relive them.

But I also thought of all the things I didn't do, of all the people I never met. I thought of how little I took advantage of Ann Arbor's unbelievable density of young, curious kids with lots of free time and energy, all part of that same proud collective: students of the University of Michigan.

Walking around campus and the town, then, I had this remarkable urge, much like the one I had as an incoming freshman---but here I was older and more confident and more capable---to engage with everyone I saw.

But not much came of it. I wasn't quite as bold as I could have been, for sure, but nor was the place as ripe as I'd imagined. I didn't understand why until a friend explained it: the kids I'd seen that day had done the same thing we'd done, what we would later come to vaguely regret---they had annealed, and settled, and made themselves a home among a certain set of friendly faces. In the bargain they'd retired from the frenzy of their freshman year, the thrill of radical openness. And I had become a stranger once again.

Notes

[1] Why only two weeks? There are a few reasons: things in general have a way of lasting two weeks; school starts to get serious after about two weeks; and two weeks is just enough time for solid social bonds to form, for kids to get comfortable in their surroundings, and for everyone to pretty much sample everyone else in their little pocket of the campus. After two weeks, the magic's over and the metal cools.

What do you do when you want the last piece of bread? Do you just take it? Or do you offer it to the table?

You offer it. Everybody knows the score: they know that you want the last piece, and that you're only offering it to everyone else to be polite. So they're going to decline, and you're going to end up with the bread.

Why not be more direct? Because then you'd be skipping a ritual that gives everyone a chance to demonstrate how cooperative they are. Rituals like that are important.

Of course, making the offer opens you up to someone accepting it. That's the price you pay for coming off as polite and cooperative; what you've done, effectively, is to wager the bread to earn a bit of social credit.

Nine times out of ten you'll win---you'll look good and keep your baguette---but occasionally you'll lose the bread. You should be prepared for that.

The odds of pulling off a successful perfunctory offer are usually worse than in this bread situation. The reason is that the bread situation happens so often, and is so well-understood, that people rarely deviate from the script: offer, refuse, eat. Whereas in other cases---like when you offer a friend a ride but don't want him to accept---the game isn't so clear; your friend might suspect that your offer is just for show, but he can't be sure. And free rides are attractive. So he's more likely to accept.

The good news is that you get paid for taking on this extra risk. Offering your friend a ride is a bigger deal than offering your last piece of bread to the table, which means you stand to earn more social credit.

The stakes are higher for the other guy, too. He also earns points by turning down your empty offers; we saw that in the bread situation. It's because turning down an empty offer (a) lets you off the hook, which is a nice and cooperative move for him to make, and (b) demonstrates his ability to detect empty offers in the first place, which feeds his reputation as a skilled coordinator.

That, then, is the arithmetic of perfunctory offers: you balance the cost of giving up some X against the points you'd earn for seeming generous, while he weighs the value of receiving X against the points he'd earn for skillfully detecting---and then abiding---your true intentions.