“He’s as obstinate a young

gonophas I know.”

This is of course via the Hebrew *ganav* (thief) via the Yiddish gonif. Had no idea it had 19th century London demimonde currency. Dickens is generally said to have been the first writer to put it in print in English, though Judith Flanders found a somewhat obscure reference a decade older.

“Gonoph” is overtaken by “gonif” as preferred English spelling sometime in the 1940s:

2. During L’cha Dodi last week I was struck by the word “Hitna’ari” (“shake yourself off!”). That’s a word I know “Na’ar,” as in “Na’ar hayiti, gam zakanti,” (“I have been a youth and I have been old.”) But does it mean “young” or does it mean “shake yourself off?” Well, kind of both. It seems there are two words, the na’ar of youth and the na’ar of shaking off.

Or is there only one word? People have tried to connect the two senses:

I don’t know. Hebrew speakers should feel free to weigh in on either of these words!

]]>One theory of gerrymandering is that the practice isn’t much of a problem, because the power of a gerrymandered map “decays” with time — a map that suits a party in 2010 may, due to shifting demographics, be reasonably fair a few years later.

How’s the Wisconsin gerrymander doing in 2018? We just had a statewide election in which Rebecca Dallet, the more liberal candidate, beat her conservative rival by 12 points, an unusually large margin for a Wisconsin statewide race.

The invaluable J. Miles Coleman broke the race down by Assembly district:

Dallet won in 58% of seats while getting 56% of the vote. That sounds fair, but in fact a candidate who wins by 12 points is typically going to win in more seats than that. (That’s why the courts are right to say proportional representation isn’t a reasonable expectation!)

Here’s the breakdown by Assembly district, shown a little bigger:

Dallet won by 2 points or less in 8 of the Assembly districts. So, as a rough estimate, if she’d gotten 2% of the vote less, and won 54-46 instead of 56-44, you might guess she’d have won 49 out of 99 seats. That’s consistent with the analysis of Herschlag, Ravier, and Mattingly conducted last year, which estimates that under current maps Democrats would need an 8-12 point statewide lead in order to win half the Assembly seats. (Figure 5 in the linked paper.)

I don’t think the gerrymander is decaying very much. I think it’s robust enough to make GOP legislative control very likely through 2020, at which point it can be updated to last another ten years, and so on and so on. This isn’t the same kind of softcore gerrymandering the Supreme Court allowed to stand in 1986, and I hope the 2018 Supreme Court decides to do something about it.

]]>So I guess what I’m trying to say is, yes, subscribe to your local paper because local journalism badly needs financial support, and maybe actually take seriously the local events it alerts you to.

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There was some worry among liberal political types that voters who went for Burns, the vocally left candidate, would sit out the general rather than show up for the more conventionally liberal Dallet. Did that happen? Here’s something cool: Wisconsin offers full statewide ward-level election results, which helps us figure that out!

First of all, here’s a ward-by-ward picture of the primary:

Each circle is a ward and its position in the triangle shows the proportion of votes going to Screnock (top vertex), Burns (left vertex), and Dallet (right vertex.) The size of the circle is the total number of votes in that ward. You can see that there’s no visible clustering, and that Dallet did much better than Burns.

So what happened in the general?

Well, first of all, Dallet won, and won big: 56-44. But that doesn’t mean Burns voters showed up. We can’t really know! But the ward-by-ward data at least helps us make some guesses.

Quick and dirty: you can do a linear regression on Dallet’s share of the general in terms of Burns’s and Dallet’s share of the primary vote. I stripped out wards with fewer than 100 general-election votes, which still left 1827 wards. You get

Dallet general ~ 0.724*Burns primary + 0.892* Dallet primary + 0.112

with a pretty decent fit

The Burns coefficient is a little bit lower but I don’t see strong evidence that a lot of Burns voters skipped the general election.

Here’s a test I like a little bit more. There are 79 wards where Burns and Dallet together got between 54 and 56% of the vote in the primary. Among these wards, Burns’s voteshare ranged from 6.5% (Milwaukee ward 211) to 35% (Town of Moscow wards 1-2, a bit on the nose, don’t you think?) If Burns voters were skipping the general election, you might expect Dallet to do worse in April in those wards where Burns did better in February. Here’s the scatter. If there’s a downward trend here, it’s not very strong.

My conclusion: liberals gonna liberal.

**Update: **I got the last scatter wrong when I originally posted this; if you remember the post being a little different, you’re right!

You may know that I kind of hate driving so if I’m gonna drive all the way to Milwaukee it’s got to be for more than a Bucks game. When I thought about what the kids would *really* want to do it was pretty clear — see the Brewers, stay over, then see the Bucks. So that’s what we did!

Notes on the Brewers:

- I got lost in the impossible off-ramp spaghetti surrounding Miller Park and we ended up not getting into the ballpark until the second inning. The Brewers were already down 4-0. 4-0! To the sad Miami Marlins, the team Derek Jeter is using as a tax dodge, the team so bad Marlins Man cancelled his season tickets!
- But as soon as we sat down, Travis Shaw muscled a huge home run to left center. Didn’t even look like he got all of it, he kind of sliced it. But Travis Shaw is a big strong man.
- Brewers just keep creeping back. Crowd stays in the game, at no point do you really feel the Brewers are out of it. Three straight Brewers hit what look like go-ahead home runs but each dies at the wall. (Ryan Braun at least gets a sacrifice fly out of it.) In the 8th, Derek Dietrich loses an Eric Sogard fly ball in the, I dunno, the lights? The roof? He plays for the Marlins and he just doesn’t care? Anyway the ball plunked down right next to him, Shaw hustles in from second to tie it, Eric Thames, who starts the play on first, tries to get in behind with the go-ahead run but is tagged out at or rather substantially before the plate because Eric Thames made a bad decision.
- Josh Hader looks like he should be playing bass in Styx.
- Then comes the bottom of the 9th and the play you might have read about. Still tied 5-5. Jesus Aguilar, who’s already warmed up twice in the on-deck circle, finally gets his chance to pinch-hit against Junichi Tazawa. Gets behind 0-2. And then just starts fouling, fouling, fouling. Takes a few pitches here and there. Full count. Foul, foul, foul. And on the 13th pitch, Aguilar launches it to center field. I thought it was gonna be one more death on the warning track. But nope; ball gets out, game over, fireworks. I felt like my kids got to see
*true baseball*.

On to Milwaukee. Bucks play the Celtics at noon, in what, if they lose, could be the last ever game played at Bradley Center. (This is a bit of a sore point for UW folks, who absorbed as a budget cut the $250m state contribution to the arena’s cost.) We have breakfast at the hotel and chat with a nice older couple in Packers/Celtics gear — what? — who turn out to be Boston forward Al Horford’s aunt and uncle from Green Bay.

This is only the third NBA game I’ve been to, CJ’s second, AB’s first. We wander around inside the arena for a bit. Two separate groups of Bucks cheerleaders come up to AB and applaud her curly hair. I think people are especially struck by it when they see us together, because *I* don’t have curly hair, except here’s a little-known fact: I do have curly hair! I just keep it short so it doesn’t curl. In 1995 or so it looked like this:

Anyway. The atmosphere, as I have promised AB, is more intense than baseball. Bucks build up a 19-point lead and seem poised to coast but the Celtics come back, and back, and back, and finally go ahead with 52 seconds left. Jaylen Brown plainly capable of taking over a game. Aron Baynes has a very dumb-looking haircut. Milwaukee’s Thon Maker is ridiculously skinny and has very long arms. He’s just 21, a former refugee from South Sudan. We saw his first game as a Buck, an exhibition against the Mavericks at Kohl Center. Those long skinny arms can block a shot.

Game tied at 102, 5 seconds left, Malcom Brogdon (called “The President” — why?) misses a layup, and there, rising like a Greek column above the scene, is the Greek arm of Giannis Antetokounmpo — the tip-in is good, Celtics miss the desperation last shot, Bucks win 104-102, crowd goes berserk.

I was going to blog about this last week but got busy so let’s throw in more sports. Bucks eventually lose this series in 7, home team winning every game a la Twins-Braves 1991. The next Friday, I’m giving a talk at Maryland, and the Orioles are playing that night. It’s been five years since I’ve seen OPACY. I brought CJ along this time, too. The Orioles are not in a good way; they’ve won 6 and lost 19, though 3 of those 6 were against New York at least. Attendance at the game, on a beautiful Friday night, was just over 14,000. The last baseball game I went to that felt this empty and mellow was the AAA Tucson Toros, several months before they moved to El Paso and became the Chihuahuas. Chris Tillman, tonight’s starter, was the Orioles’ ace five years ago. Now he’s coming off a 1-7 season and has an ERA over 9.

So who would have thought he’d toss seven shutout innings and take a no-hitter into the fifth? Never looked overpowering but kept missing bats. His first win in almost a year. Manny Machado, surely now in his last year as an Oriole, strokes a home run to dead center to get things started. It’s a beautiful thing. It doesn’t even look like he’s working hard. It’s like he’s just saying “Out there. Out there is where this ball should be.” Pedro Alvarez homers twice, in exactly the opposite manner, smashing the ball with eye-popping force. Jace Peterson, who the Orioles picked up off the Yankees’ scrap heap, steals third on the shift when the Tigers third baseman forgets to pay attention to him. He did the same thing against the Rays the night before. I am already starting to love him the way I love Carlos Gomez. Maybe now the Orioles are going to go back to being a bad team that makes good use of players nobody else wants, like Melvin Mora and Rodrigo López.

Besides me and CJ, this guy was at the game:

Never get tired of that flag.

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I actually spent much of today thinking about this so let me try to explain it in a down-to-earth way, because it involved me thinking about Bessel functions for the first time ever, surely a life event worthy of recording.

So here’s what we’re going to do. As I mentioned last week, you can express this problem as follows: suppose you have a map h: R^2 -> V, for some normed vector space V, which is a *unit-distance embedding*; that is, if |x-x’|_{R^2} = 1, then |h(x)-h(x’)|_V = 1. (We don’t ask that h is an isometry, only that it preserves the distance-1 set.)

Then let t be the radius of the smallest hypersphere in V containing h(R^2).

Then any graph embeddable in R^2 with all edges of length 1 is sent to a unit-distance graph in V contained in the hyperplane of radius t; this turns out to be equivalent to saying the Lovasz number of G (ok, really I mean the Lovasz number of the complement of G) is at most 1/(1-2t). So we want to show that t is bounded below 1, is the point. Or rather: we can *find *a V and a map from R^2 to V to make this the case.

So here’s one! Let V be the space of L^2 functions on R^2 with the usual inner product. Choose a square-integrable function F on R^2 — in fact let’s normalize to make F^2 integrate to 1 — and for each a in R^2 we let h(a) be the function F(x-a).

We want the distance between F(x-a) and F(x-b) to be the same for every pair of points at distance 1 from each other; the easiest way to arrange that is to insist that F(x) be a radially symmetric function F(x) = f(|x|); then it’s easy to see that the distance between F(x-a) and F(x-b) in V is a function G(a-b) which depends only on |a-b|. We write

so that the squared distance between F(x) and F(x-r) is

.

In particular, if two points in R^2 are at distance 1, the squared distance between their images in V is 2(1-g(1)). Note also that g(0) is the square integral of F, which is 1.

What kind of hypersphere encloses all the points F(x-a) in V? We can just go ahead and take the “center” of our hypersphere to be 0; since |F| = 1, every point in h(R^2) lies in (indeed, lies *on*) the sphere of radius 1 around the origin.

Hey but remember: we want to study a *unit-distance* embedding of R^2 in V. Right now, h sends unit distances to the distance 2(1-g(1)), whatever that is. We can fix that by scaling h by the square root of that number. So now h sends unit distances to unit distances, and its image is enclosed in a hypersphere of radius

2(1-g(1))^{-1}

The more negative g(1) is, the smaller this sphere is, which means the more we can “fold” R^2 into a small space. Remember, the relationship between hypersphere number and Lovasz theta is

and plugging in the above bound for the hypersphere number, we find that the Lovasz theta number of R^2, and thus the Lovasz theta number of any unit-distance graph in R^2, is at most

1-1/g(1).

So the only question is — what is g(1)?

Well, that depends on what g is.

Which depends on what F is.

Which depends on what f is.

And of course we get to *choose* what f is, in order to make g(1) as negative as possible.

How do we do this? Well, here’s the trick. The function G is not arbitrary; if it were, we could make g(1) whatever we wanted. It’s not hard to see that G is what’s called a *positive definite* function on R^2. And moreover, if G is positive definite, there exists some f giving rise to it. (Roughly speaking, this is the fact that a positive definite symmetric matrix has a square root.) So we ask: if G is a positive definite (radially symmetric) function on R^2, and g(0) = 1, how small can g(1) be?

And now there’s an old theorem of (Wisconsin’s own!) Isaac Schoenberg which helpfully classifies the positive definite functions on R^2; they are precisely the functions G(x) = g(|x|) where g is a mixture of scalings of the Bessel function $J_0$:

for some everywhere nonnegative A(u). (Actually it’s more correct to say that A is a distribution and we are integrating J_0(ur) against a non-decreasing measure.)

So g(1) can be no smaller than the minimum value of J_0 on [0,infty], and in fact can be exactly that small if you let A become narrowly supported around the minimum argument. This is basically just taking g to be a rescaled version of J_0 which achieves its minimum at 1. That minimum value is about -0.4, and so the Lovasz theta for any unit-distance subgraph on the plane is bounded above by a number that’s about 1 + 1/0.4 = 3.5.

**To sum up:** I give you a set of points in the plane, I connect every pair that’s at distance 1, and I ask how you can embed that graph in a small hypersphere keeping all the distances 1. And *you* say: “Oh, I know what to do, just assign to each point a the radially symmetrized Bessel function J_0(|x-a|) on R^2, the embedding of your graph in the finite-dimensional space of functions spanned by those Bessel translates will do the trick!”

That is cool!

*Remark: *Oliveira’s thesis does this for Euclidean space of every dimension (it gets more complicated.) And I think (using analysis I haven’t really tried to understand) he doesn’t just give an upper bound for the Lovasz number of the plane as I do in this post, he really computes that number on the nose.

*Update:* DeCorte, Oliveira, and Vallentin just posted a relevant paper on the arXiv this morning!

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Here’s an equivalent formulation: If G is a graph and V(G) its vertex set, I try to find a function f: V(G) -> R^d, for some d, such that

|f(x) – f(y)| = 1 whenever x and y are adjacent.

This is called a *unit distance embedding*, for obvious reasons.

The *hypersphere* number t(G) of the graph is the radius of the smallest sphere containing a unit distance embedding of G. Computing t(G) is equivalent to computing the Lovász number, but let’s not worry about that now. I want to generalize it a bit. We say a finite sequence (t_1, t_2, t_3, … ,t_d) is *big enough* for G if there’s a unit-distance embedding of G contained in an ellipsoid with major radii t_1^{1/2}, t_2^{1/2}, .. t_d^{1/2}. (We could also just consider infinite sequences with all but finitely many terms nonzero, that would be a little cleaner.)

Physically I think of it like this: the graph is trying to fold itself into Euclidean space and fit into a small region, with the constraint that the edges are rigid and have to stay length 1.

Sometimes it can fold a lot! Like if it’s bipartite. Then the graph can totally fold itself down to a line segment of length 1, with all the black vertices going to one end and the white vertices going to the other. And the big enough sequences are just those with some entry bigger than 1.

On the other hand, if G is a complete graph on k vertices, a unit-distance embedding has to be a simplex, so certainly anything with k of the t_i of size at least 1-1/k is big enough. (Is that an if and only if? To know this I’d have to know whether an ellipse containing an equilateral triangle can have a radius shorter than that of the circumcircle.)

Let’s face it, it’s confusing to think about ellipsoids circumscribing embedded graphs, so what about instead we define t(p,G) to be the minimum value of the L^p norm of (t_1, t_2, …) over ellipsoids enclosing a unit-distance embedding of G.

Then a graph has a unit-distance embedding in the plane iff t(0,G) <= 2. And t(oo,G) is just the hypersphere number again, right? If G has a k-clique then t(p,G) >= t(p,K_k) for any p, while if G has a k-coloring (i.e. a map to K_k) then t(p,G) <= t(p,K_k) for any n. In particular, a regular k-simplex with unit edges fits into a sphere of squared radius 1-1/k, so t(oo,G) < 1-1/k.

So… what’s the relation between these invariants? Is there a graph with t(0,G) = 2 and t(oo,G) > 4/5? If so, there would be a non-5-colorable unit distance graph in the plane. But I guess the relationship between these various “norms” feels interesting to me irrespective of any relation to plane-coloring. What is the max of t(oo,G) with t(0,G)=2?

The intermediate t(p,G) all give functions which upper-bound clique number and lower-bound chromatic number; are any of them interesting? Are any of them easily calculable, like the Lovász number?

**Remarks:**

- I called this post “What is the Lovász number of the plane?” but the question of “how big can t(oo,G) be if t(0,G)=2”? is more a question about finite subgraphs of the plane and their Lovász numbers. Another way to ask “What is the Lovász number of the plane” would be to adopt the point of view that the Lovász number of a graph has to do with extremizers on the set of positive semidefinite matrices whose (i,j) entry is nonzero only when i and j are adjacent vertices or i=j. So there must be some question one could ask about the space of positive semidefinite symmetric kernels K(x,y) on R^2 x R^2 which are supported on the locus ||x-y||=1 and the diagonal, which question would rightly be called “What is the Lovász number of the plane?” But I’m not sure what it is.
- Having written this, I wonder whether it might be better, rather than thinking about enclosing ellipsoids of a set of points in R^d, just to think of the n points as an nxd matrix X and compute the singular values of X^T X, which would be kind of an “approximating ellipsoid” to the points. Maybe later I’ll think about what that would measure. Or you can!

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The idea: given a set S of points in the plane, its unit distance graph G_S is the graph whose vertices are S and where two points are adjacent if they’re at distance 1 in the plane. If you can find S such that G_S has chromatic number k, then the chromatic number of the plane is at least k. And de Grey finds a set of 1,567 points whose unit distance graph can’t be 4-colored.

It’s known that the chromatic number of the plane is at most 7. Idle question: is there any chance of a “polynomial method”-style proof that there is no subset S of the plane whose unit distance graph has chromatic number 7? Such a graph would have a lot of unit distances, and ruling out lots of repetitions of the same distance is something the polynomial method can in principle do.

Though be warned: as far as I know the polynomial method has generated no improvement so far on older bounds on the unit distance problem (“how many unit distances can there be among pairs drawn from S?”) while it has essentially solved the distinct distance problem (“how few distinct distances can there be among pairs drawn from S?”)

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