Category Archives: offhand

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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I’m at Disney World with CJ, on a Pirates of the Caribbean-style ride, a car careening through a tunnel.  On the wall of the tunnel there are math posters, the kind you’d see in a high school classroom, about Pascal’s triangle, conic sections, etc.  And I feel sort of annoyed and depressed, because I know that Disney is going to make a big deal about how educational this ride is, but actually, nobody except me is looking at the posters, nobody who didn’t already know the math could get anything out of the posters, the way the car speeds down the track.

Please interpret and derive relevant policy prescriptions for math pedagogy in comments.

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Posthumous appreciation

Monday was the day of appreciating David Bowie on my Facebook feed.  That’s the way it is when people die; until that moment, you can, if you’re inclined to, say their best work is behind them, they were never that great anyway, etc etc.  Afterwards, for a while at least, it’s pure appreciation, love, honor.

Why do we do it that way?  It seems totally backwards.  For my own part, I want everyone to appreciate me and tell me I’m great right now, while I can enjoy it.  And if you think my work is overrated and anyway I’m kind of a jerk?  After I’m dead would be an awesome time to bring that up.  You have my permission, go for it.

Ranking mathematicians

I’m on the hiring committee, I chair the graduate admissions committee, and I’m doing an NSF panel, so basically I’ll be spending much of this month judging and ranking people’s mathematics.  There’s a lot I like about these jobs:  it’s a very efficient way to get a panorama of what’s going on in math and what people think about it.  The actual ranking part I don’t like that much — especially because the nature of hiring, admissions, and grant-making means you’re inevitably putting tons of very worthwhile stuff below the line.  I feel like a researcher when I read the proposals, like a bureaucrat when I put scores on them.

But of course the bureaucratic work needs to be done.  I’d go so far as to say — if mathematicians aren’t willing to rank each other, others will rank us, and that would be worse.

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My Erdos-Bacon-Sabbath number is 11

I am pleased to report that I have an Erdös-Bacon-Sabbath number.

My Erdös number is 3; has been for a while, probably always will be.  I wrote a paper with Mike Bennett and Nathan Ng about solutions to A^4 + B^2 = C^p; Mike wrote with Florian Luca; Luca wrote with Erdös.

A while back, I shot a scene for the movie Gifted.  I’m not on the IMDB page yet, but I play against type as “Professor.”  Also in this movie is Octavia Spencer, who was in Beauty Shop (2005) with Kevin Bacon.  So my Bacon number is 2.

That gives me an Erdös-Bacon number of 5; already pretty high on the leaderboard!

Of course it then fell to me to figure out whether I have a Sabbath number.  Here’s the best chain I could make.

I once played guitar on “What Goes On” with my friend Jay Michaelson‘s band, The Swains, at Brownies.

Jay performed with Ezra Lipp “sometime in 2000,” he reports.

Lipp has played with Chris Robinson of the Black Crowes.

From here we use the Six Degrees of Black Sabbath tool, written by Paul Lamere at EchoNest (now part of the Spotify empire.)

The Black Crowes backed up Jimmy Page at a concert in 1999.

Page played with David Coverdale in Coverdale.Page.

David Coverdale was in Deep Purple with Glenn Hughes of Black Sabbath.

So my Sabbath number is 6, and my Erdos-Bacon-Sabbath number is 11.







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