Category Archives: offhand

91 white friends

Just ran across this hunk of data journalism from the Washington Post:

In a 100-friend scenario, the average white person has 91 white friends; one each of black, Latino, Asian, mixed race, and other races; and three friends of unknown race. The average black person, on the other hand, has 83 black friends, eight white friends, two Latino friends, zero Asian friends, three mixed race friends, one other race friend and four friends of unknown race.

Going back to Chris Rock’s point, the average black person’s friend network is eight percent white, but the average white person’s network is only one percent black. To put it another way: Blacks have ten times as many black friends as white friends. But white Americans have an astonishing 91 times as many white friends as black friends.

100 friends and only one black person!  That’s pretty white!

It’s worth taking a look at the actual study they’re writing about.  They didn’t ask people to list their top 100 friends.  They said to list at most seven people, using this prompt:

From time to time, most people discuss important matters with other people. Looking back over the last six months – who are the people with whom you discussed matters important to you?

The white respondents only named 3.3 people on average, of whom 1.9 were immediate family members.  So a better headline wouldn’t be “75% of white people have no black friends,” but “75% of whites are married to another white person, have two white parents, and have a white best friend, if they have a best friend”  As for the quoted paragraph, it should read

In a 100-friend scenario, the average white person has 57 immediate family members.

Who knew?

(Note:  I just noticed that Emily Swanson at Huffington Post made this point much earlier.)







Overfriendly balloon

A Mylar balloon I bought weeks ago for Dr. Mrs. Q’s birthday has been holding its helium astonishingly well, but has now lost enough that its density is almost exactly that of air.  It doesn’t hug the ceiling or the floor, but slowly traverses the room, carried by air currents too weak for me to feel, I guess.  And here’s the thing:  as I sit here working, it keeps — well — visiting me.  It’s kind of creepy, when it slowly floats into my peripheral vision and then taps me once or twice.  Hey.  Balloon here.  OK, moving on.  See you later.




Arithmetic progression joke

Why was 7.333… disgusted by 7.666….?

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I got a message for you

“I got a message for you, if I could only remember.  I got a message for you, but you’re gonna have to come and get it.”  Kardyhm Kelly gave me a tape of Zopilote Machine in 1995 and I played nothing but for a month.  “Sinaloan Milk Snake Song” especially.  Nobody but the Mountain Goats ever made do-it-yourself music like this, nobody else ever made it seem so believable that the things it occurred to you to say or sing while you were playing your guitar in your bedroom at home might actually be pop songs.   The breakdown at the end of this!

“I’ve got a heavy coat, it’s filled with rocks and sand, and if I lose it I’ll be coming back one day (I got a message for you).”  I spent a lot of 1993 thinking about the chord progression in the verse of this song.  How does it sound so straight-ahead but also so weird?  Also the “la la la”s (“Sinaloan Milk Snake Song” has these too.)

“Roll me in the greenery, point me at the scenery.  Exploit me in the deanery.  I got a message for you.”

The first of these I ever heard.  Douglas Wolk used to send mixtapes to Elizabeth Wilmer at Math Olympiad training.  This was on one of them.  1987 probably. I hadn’t even started listening to WHFS yet, I had no idea who Robyn Hitchcock was.  It was on those tapes I first heard the Ramones, Marshall Crenshaw, the Mentors (OK, we were in high school, cut us some slack.)

(Update:  Douglas denies ever putting the Mentors on a mixtape, and now that I really think about it, I believe Eric Wepsic was to blame for bringing the Mentors into my life.)

Why is this line so potent?  Why is the message never explicitly presented?  It’s enough — it’s better — that the message only be alluded to, never spoken, never delivered.

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Dissecting squares into equal-area triangles: idle questions

Love this post from Matt Baker in which he explains the tropical / 2-adic proof (in fact the only proof!) that you can’t dissect a square into an odd number of triangles of equal area.  In fact, his argument proves more, I think — you can’t dissect a square into triangles whose areas are all rational numbers with odd denominator!

  • The space of quadrilaterals in R^2, up to the action of affine linear transformations, is basically just R^2, right?  Because you can move three vertices to (0,0), (0,1), (1,0) and then you’re basically out of linear transformations.   And the property “can be decomposed into n triangles of equal area” is invariant under those transformations.  OK, so — for which choices of the “fourth vertex” do you get a quadrilateral that has a decomposition into an odd number of equal-area triangles? (I think once you’re not a parallelogram you lose the easy decomposition into 2 equal area triangles, so I suppose generically maybe there’s NO equal-area decomposition?)  When do you have a decomposition into triangles whose area has odd denominator?
  • What if you replace the square with the torus R^2 / Z^2; for which n can you decompose the torus into equal-area triangles?  What about a Riemann surface with constant negative curvature?  (Now a “triangle” is understood to be a geodesic triangle.)  If I have this right, there are plenty of examples of such surfaces with equal-area triangulations — for instance, Voight gives lots of examples of Shimura curves corresponding to cocompact arithmetic subgroups which are finite index in triangle groups; I think that lets you decompose the Riemann surface into a union of fundamental domains each of which are geodesic triangles of the same area.
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Dream:  I meet two 9-year-old boys with identical long curly hair.  They’re in a band, the band is called Bugbiter.  They explain to me that most of their songs are about products, and they share their songs via videos they post on Amazon.

I share this dream with you mostly because I think Bugbiter is actually a legitimately good band name.


Math bracket 2016

It’s that time again — March Math Madness, where we fill out an NCAA men’s tournament bracket with the best math department winning every game.  As always, this bracket was filled out by a highly trained team consisting of myself and a group of procrastinating grad students, making decisions by voice vote, and if you disapprove of one of our choices, I’m sure it’s somebody else’s fault.  This is Berkeley’s first championship after falling to Harvard in 2012; meanwhile, Michigan sees its second final in three years but falls short again…

Screen Shot 2016-03-16 at 16 Mar 10.51.PM

Update: In the 34th percentile at ESPN after one day of play — thanks, Yale!

Update:  Down to the 5th percentile and only Duke and UVa are left out of my final 8 picks.  Not gonna be the math bracket’s finest year.

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Metric chromatic numbers

Idle thought.  Let G be a (loopless) graph, let M be a metric space, and let b be a nonnegative real number.  Then let Conf(G,M,b) be the space of maps from the vertices of the graph to M such that no two adjacent vertices are within b of each other.

When b=0 and G is the complete graph K_n, this is the usual ordered configuration space of n points on M.  When G is the empty graph on n vertices, it’s just M^n.  When M is a set of N points with the discrete metric, Conf(G,M,b) is the same thing for every b;  a set of points whose cardinality is χ_G(N), the chromatic polynomial of G evaluated at N.  When M is a box, Conf(G,M,b) is the “discs in a box” space I blogged about here.  If M is (Z/2Z)^k with Hamming distance, you are asking about how many ways you can supply G with k 2-colorings so that no edge is monochromatic in more than k-b-1 of them.

Update:  Ian Agol links in the comments to this paper about Conf(G,M,0) by Eastwood and Huggett; the paper points out that the Euler characteristic of Conf(G,M,0) computes χ_G(N) whenever M has Euler characteristic N; so M being N points works, but so does M = CP^{N-1}, and that’s the case they study.

When b=0 and G is the complex plane, Conf(G,C,0) is the complement of the graphic arrangement of G; its Poincare polynomial is  t^-{|G|} χ_G(-1/t).  Lots of graphs have the same chromatic polynomial (e.g. all trees do) but do they have homeomorphic Conf(G,C,0)?

This is fun to think about!  Is Conf(G,M,0), for various manifolds other than discrete sets of points, an interesting invariant of a graph?

You can think of

vol(Conf(G,M,b)) vol(M)^{-n}

as a sort of analogue of the chromatic polynomial, especially when b is small; when M = C, for instance, Conf(G,M,b) is just the complement of tubular neighborhoods around the hyperplanes in the graphical arrangement, and its volume should be roughly b^|G|χ_G(1/b) I think.  When b gets big, this function deviates from the chromatic polynomial, and in particular you can ask when it hits 0.

In other words:  you could define an M-chromatic number χ(G,M) to be the smallest B such that Conf(G,M,1/B) is nonempty.  When M is a circle S^1 with circumference 1, you can check that χ(G,M) is at least the clique number of G and at most the chromatic number.  If G is a (2n+1)-cycle, the clique number is 2, the chromatic number is 3, and the S^1-chromatic number is 2+1/n, if I did this right.  Does it have anything to do with the Lovasz number, which is also wedged between clique number and chromatic number?  Relevant here:  the vector chromatic number, which is determined by χ(G,S^{v(G)-1}), and which is in fact a lower bound for the Lovasz number.





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The furniture sentiment

Today’s Memorial Library find:  the magazine Advertising and Selling.  The September 1912 edition features “How Furniture Could Be Better Advertised,” by Arnold Joerns, of E.J. Thiele and Co.

Joerns complains that in 1911, the average American spend $81.22 on food, $26.02 on clothes, $19.23 on intoxicants, $9.08 on tobacco, and only $6.19 on furniture.  “Do you think furniture should be on the bottom of this list?” he asks, implicitly shaking his head.  “Wouldn’t you — dealer or manufacturer — rather see it nearer the top, — say at least ahead of tobacco and intoxicants?”

Good news for furniture lovers:  by 2012, US spending on “household furnishings and equipment” was  at $1,506 per household, almost a quarter as much as we spent on food.  (To be fair, it looks like this includes computers, lawnmowers, and many other non-furniture items.)  Meanwhile, spending on alcohol is only $438.  That’s pretty interesting:  in 1911, liquor expenditures were a quarter of food expenditures; now it’s less than a tenth.  Looks like a 1911 dollar is roughly 2012$25, so the real dollars spent on alcohol aren’t that different, but we spend a lot more now on food and on furniture.

Anyway, this piece takes a spendidly nuts turn at the end, as Joerns works up a head of steam about the moral peril of discount furniture:

I do not doubt but that fewer domestic troubles would exist if people were educated to a greater understanding of the furniture sentiment.  Our young people would find more pleasure in an evening at home — if we made that home more worth while and a source of personal pride; then, perhaps, they would cease joy-riding, card-playing, or drinking and smoking in environments unhealthful to their minds and bodies.

It would even seem reasonable to assume, that if the public mind were educated to appreciate more the sentiment in furniture and its relation to the Ideal Home, we would have fewer divorces.  Home would mean more to the boys and girls of today and the men and women of tomorrow.  Obviously, if the public is permitted to lose more and more its appreciation of home sentiment, the divorce evil will grow, year by year.

Joerns proposes that the higher sort of furniture manufacturers boost their brand by advertising it, not as furniture, but as “meuble.” This seems never to have caught on.

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AI challop

When I was a kid people thought it would be a long time before computers could adequately translate natural language text, or play Go against a human being, because you’d need some kind of AI to do those things, and AI seemed really hard.

Now we know that you can get pretty decent translation and Go without anything like AI.  But AI still seems really hard.

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