'Ff'lo (fflo) wrote,

I already sent it around at work, but you might like it too.

NPR piece on Donald Knuth

Mostly I think of Knuth in association with TEX, the mark-up language for mathematical notation, but I see the occasional piece by or about him. Often his stuff has a context we lay folk understand, so he stands out for that; I can sometimes relate not only to his desire for aesthetically pleasing typesetting, but also to the curiosity about everyday stuff behind some of the problems he looks into. Here's a funny one that came from his musings in the restrooms at Stanford:

MR0761401 (86a:05006)
The toilet paper problem
Amer. Math. Monthly 91 (1984), no. 8, 465--470.

The toilet paper dispensers are designed to hold two rolls of tissues, and a person can use either roll. There are two kinds of users. A big chooser always takes a piece from the roll that is currently larger, while a little chooser does the opposite. When the two rolls are the same size, or when only one is nonempty, everybody chooses the nearest nonempty roll. Assume that people enter the toilet stalls independently at random, with probability $p$ that they are big choosers and probability $q=1-p$ that they are little choosers. If two fresh rolls of toilet paper, both of length $n$
[copy eds---and ALG---missed the comma that should have gone here!] are installed, let $M\sb n(p)$ be the average number of portions left on one roll when the other one first empties. The purpose of this paper is to study the asymptotic value of $M\sb n(p)$ for fixed $p$ as $n\to\infty$. Let $M(z)=\sum\sb {n\ge1}M\sb n(p)z\sp n$, and $C(z)=\sum\sb {n\ge1}c\sb nz\sp n$, $c\sb n={2n-2\choose n-1}/n$ (Catalan numbers) be the generating functions. It is proved that $M(z)=(z/(1-z)\sp 2)((q-C(pqz))/q)$. Let $r$ be any value greater than $4pq$; then $M\sb n(p)=p/(p-q)+O(r\sp n)$ if $q[...html freaked out here, where a "<" appeared...]p$, $M\sb n(p)=((q-p)/q)n+p/(q-p)+O(r\sp n)$ if $q[...html freaked out here, where a ">" appeared...]p$, and $M\sb n({1\over 2})=2\sqrt{n/\pi}-\frac 14\sqrt{1/\pi n}+O(n\sp {-3/2})$.

S'okay, when it turns into what is apparently a version of Banach's matchbook problem, most of us just toss our hands up in the air & let it go, but it's fun anyway. Some part of me still likes a puzzle, especially when it has that for-its-own-sake purity that seems in part to embody a childlike innocence in its pursuit of joyous cognitive play.

From the story this morning: "He wears his bike helmet indoors because, well, he's going to have to put it back on a little later anyway." Not entirely unlike my personal resistance to, say, closing kitchen cabinets I'm just going to be opening again soon enough. (For the record, I've been much better at closing them lately than I was in the first months after H's departure. I really do like the feel of the kitchen better when it's orderly, yet it's an ongoing struggle---though prob'bly not so much against the mistaken instinct to save energy at all costs as against that old deleteriously rebellious reaction against housekeeping as oppression.)

Knuth: "There's ways to amuse yourself [chuckle] while you're doing things, and that's the way I look at efficiency: it's an amusing thing to think about, but not that I, that I'm obsessed that it's got to be efficient, you know, or else I, uh, go crazy."

He makes it sound easy to strike a balance there, doesn't he.

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