Category Archives: offhand

The Omnivore's Hundred

A list of a hundred foods that went around the Internet a while back; the original source seems to be gone. Your score is how many you’ve eaten. Here’s how the family did:

  • Me: 68
  • Dr. Mrs. Q: 41
  • CJ: 38
  • AB: 36

I did pretty well except for the alcohol, which makes sense; I’m always inclined to try a food I haven’t eaten before, but the opportunity to try a new drink doesn’t move me at all. I wonder how many of those 68 I’ve only eaten once? Just glancing at the list again, I see: fugu, haggis, Jamaican Blue Mountain coffee, whole insects, horse. I have only eaten at a three Michelin-star restaurant once and the menu was selected for me so I counted that as “tasting menu at a three-star Michelin,” so that too.

The format reminds me of the “Purity Test” that was a mainstay of Usenet groups and, I’m pretty sure, FIDONet before that. Wikipedia suggests the version of the test I saw, like everything else weird on the early Internet, originated at MIT.

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A divergent sequence

Someone asked on Twitter:

Indeed this series diverges, just as the tweeter says: there’s a positive-density subset of n such that the summand exceeds n^{-1/2}.

More subtle: what about

\sum_{n=1}^{\infty} \frac{1}{n^{2 + \cos n}}?

This should still diverge. Argument: the probability that a real number x chosen uniformly from a large interval has \cos x < -1+\epsilon is on order \epsilon^{1/2}, not \epsilon; so there will be a subset of integers with density on order \epsilon^{1/2} where the summand exceeds n^{-1-\epsilon}, and summing over those integers along should give a sum on order \epsilon^{-1/2}, which can be made as large as you like by bringing \epsilon close to 0.

What’s not so clear to me is: how does

\sum_{n=1}^{N} \frac{1}{n^{2 + \cos n}}

grow with N?

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The American bar mitzvah, 1887

“And during the time when the Hungarian or Polish Jewish youngster was brought to a level where he could understand the Prophets, and listen to rigorous biblical and legal studies, the American youngster is merely brought to the magnificent level of being able to stammer a few words of English-style Hebrew, to pronounce the blessing over the Torah, and to chant half the maftir from a text with vowels and notes on the day he turns thirteen — a day that is celebrated here as the greatest of holidays among our Jewish brethren. From that day onward a youngster considers his teacher to be an unwanted article.”

Moses Weinberger, Jews and Judaism in New York, 1887 (Jonathan D. Sarna, trans.)
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Edith Wharton shipped Esther/Haman

That theory, now, that Odysseus never really forgot Circe; and that Esther was in love with Haman, and decoyed him to the banquet with Ahasuerus just for the sake of once having him near her and hearing him speak; and that Dante, perhaps, if he could have been brought to book, would have had to confess to caring a good deal more for the pietosa donna of the window than for a long-dead Beatrice — well, you know, it tallies wonderfully with the inconsequences and surprises that one is always discovering under the superficial fitnesses of life.

(Edith Wharton, “That Good May Come,” 1894.)

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The real diehard

Yesterday a guy saw my Orioles hat and said, “Wow, you’re a real diehard.”

He was wearing a Hartford Whalers hat.

The Americans can retire at 42

On the front page of my New York Times today, this capsule summary:

The French can retire at 62. Or 52. Sometimes 42. President Emmanuel Macron calls the tangle unsustainable. A million protesters disagree.

In the actual article, we learn that the retirement age of 42 applies to one group of workers; dancers in the national ballet. I find it very annoying when an article is teased with a number presented as normal when it’s actually extremely atypical. You could write the same teaser about the United States, having New York City firefighters in mind. But you would be misleading your audience even though the claim would be, I suppose, technically correct.

Khot,Minzer,Safra on approximate sections of sheaves

Subhash Khot is giving a talk at Current Developments in Math this year and his talk has the intriguing phrase “Grassmann graph” in it so I thought I’d look up what it is and what he and his collaborators prove about it, and indeed it’s interesting! I’m just going to jot down something I learned from “Pseudorandom Sets in Grassmann Graph have Near-Perfect Expansion,” by Khot, Dor Minzer, and Muli Safra, in a way that makes it sound like something algebraic geometers might be interested in, which, indeed, I think they might be!

Suppose you have a sheaf F on a space, and the space has a covering U_1, .. U_N. The sheaf axiom says that if we have a family of sections s_i of F(U_i) such that s_i and s_j agree on U_i \cap U_j for all i,j, then there is actually a global section s in F(X) which restricts to each s_i.

What if we only have an approximate section? That is: what if we have a family of s_i such that: if I select i and j uniformly at random, the probability that s_i and s_j agree on U_i \cap U_j is bounded below by some p > 0. Call such a family a “p-section.” (You should take the view that this is really a family of problems with X changing and N growing, so that when I say p > 0 the content is that p is bounded away from some 0 uniformly in X,N.)

The question is then: Is an approximate section approximately a section?

(This is meant to recall the principle from additive number theory that an approximate subgroup is approximately a subgroup, as in e.g. Freiman-Rusza.)

That is: if s_1, .. s_N from a p-section, is there some actual section s in F(X) such that, for i chosen uniformly at random,

\mathbf{Pr}(s | U_i) = s_i > p' > 0

for some p’ depending only on p?

The case which turns out to be relevant to complexity theory is the Grassmann graph, which we can describe as follows: X is a k-dimensional vector space over F_2 and the U_i are the l-dimensional vector spaces for some integer l. But we do something slightly weird (which is what makes it the Grassmann graph, not the Grassmann simplicial complex) and declare that the only nonempty intersections are those where U_i \cap U_j has dimension l-1. The sheaf is the one whose sections on U_i are the linear functions from U_i to F_2.

Speculation 1.7 in the linked paper is that an approximate section is approximately a section. This turns out not to be true! Because there are large sets of U_i whose intersection with the rest of X is smaller than you might expect. This makes sense: if X is a space which is connected but which is “almost a disjoint union of X_1 and X_2,” i.e. X_1 \cup X_2 = X and $\latex X_1 \cap X_2$ involves very few of the U_i, then by choosing a section of F(X_1) and a section of F(X_2) independently you can get an approximate section which is unlikely to be approximated by any actual global section.

But the good news is that, in the case at hand, that ends up being the only problem. Khot-Minzer-Safra classify the “approximately disconnected” chunks of X (they are collections of l-dimensional subspaces containing a fixed subspace of small dimension and contained in a fixed subspace of small codimension) and show that any approximate section of F is approximated by a section on some such chunk; this is all that is needed to prove the “2-to-2 games conjecture” in complexity theory, which is their ultimate goal.

So I found all this quite striking! Do questions about approximate global sections being approximated by global sections appear elsewhere? (The question as phrased here is already a bit weird from an algebraic geometry point of view, since it seems to require that you have or impose a probability measure on your set of open patches, but maybe that’s natural in some cases?)

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Pete Alonso was a Mallard

Pete Alonso of the New York Mets is the NL Rookie of the Year and the all-time home run king among both rookies and Mets. I’m proud to say I saw him hit a 407-foot three-run moonshot in June 2014, when he was a 19-year-old playing for the Madison Mallards in the summer-collegiate Northwoods League. Go Mallards!

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Morally problematic, even questionable

Today’s New York Times contains the phrase

is itself a morally problematic, even questionable, act.

Are there no editors anymore? What work is “even questionable” doing? Is it possible to imagine an act that was morally problematic but not morally questionable? And even if it is, is that thin distinction really what the writer of this piece about an HBO miniseries is going for? Or did they just think “is itself a morally problematic act” didn’t have enough heft, stuff another couple of non-nutritive words in there, admire the sentence’s new bulk, move on?

Boo, I say, boo.

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The quarter-circle game

Start at a lattice point inside the quarter-circle x^2 + y^2 < 10^6 in the positive quadrant. You and your opponent take turns: the allowable moves are to go up, right, or both at once (i.e. add (0,1), add (1,0), or add (1,1).) First person to leave the quarter-circle wins. What does it look like if you color a starting point black for “first-player loss” and yellow for “first-player win”? It looks like this:

I like the weird zones of apparent order here. Of course you can do this for any planar domain, any finite set of moves, etc. Are games like this analyzable at all?

I guess you could go a little further and compute the nimber or Grundy value associated to each starting position. You get:

What to make of this?

Here’s some hacky code, it’s simple.


M = 1000
def Crossed(a,b):
    return (a**2 + b**2 >= M*M)

def Mex(L):
    return min([i for i in range(5) if not (i in L)])


L = np.zeros((M+2,M+2))
for a in reversed(range(M+2)):
    for b in reversed(range(M+2)):
        if Crossed(a,b):
            L[a,b] = 0
        else:
            L[a,b] = Mex([L[a+1,b],L[a,b+1],L[a+1,b+1]])

plt.imshow(L,interpolation='none',origin='lower')
plt.show()

One natural question: what proportion of positions inside the quarter-circle are first-player wins? Heuristically: if you imagine the value of positions as Bernoulli variables with parameter p, the value at my current position is 0 if and only if all three of the moves available to me have value 1. So you might expect (1-p) = p^3. This has a root at about 0.68. It does look to me like the proportion of winning positions is converging, but it seems to be converging to something closer to 0.71. Why?

By the way, the game is still interesting (but I’ll bet more directly analyzable) even if the only moves are “go up one” and “go right one”! Here’s the plot of winning and losing values in that case:

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