The ARCTIC CIRCLE THEOREM or Why do physicists play dominoes?

I only stumbled across the amazing arctic circle theorem a couple of months ago while preparing the video on Euler's pentagonal theorem. A perfect topic for a Christmas video.

Before I forget, the winner of the lucky draw announced in my last video is Zachary Kaplan. He wins a copy of my book Q.E.D. Beauty in mathematical proof.

00:00 Intro
00:35 Chapter 1: mutilated chessboards
07:23 Chapter 2: Monster formula
15:12 Chapter 3: Aztec gold
20:07 Chapter 4: Square dance
30:41 Chapter 5: Ice
34:35 Chapter 6: Hexagon
38:25 Credits
40:46 Mini masterclass

In response to my challenge here are some nice implementations of the dance:
Cannot fit any more links in this description because of the character limit. For lots of other amazing implementations check out the list in my comment pinned to the top of the comment section of the video.

For the most accessible exposition of iterated shuffling that I am aware of have a look at the relevant chapter in the book "Integer partitions" by Andrews and Eriksson. They also have a nice set of exercises that walk you through proofs for the properties of iterated shuffling that I mention in this video.

Around the same time that Kasteleyn published the paper I showed in the video, the physicists Temperley and Fisher published similar results, Dimer problem in statistical mechanics-an exact result, Philosophical Magazine, 6:68, (1961) 1061-1063. The way Kasteleyn as well as Temperley and Fisher calculated the numbers of tilings of boards with square tiles was a bit more complicated than the nice refinement that I show in the video which is due to Jerome K. Percus, One more technique for the dimer problem. J. Mathematical Phys., 10:1881–1888, 1969.

Some great articles and websites to check out:

An accessible article about tilings with rectangles by my colleague Norm Do at Monash Uni. In particular, it's got some more good stuff about the maths of fault lines in tilings that I only hinted at in the video:

An introduction to the dimer model by Richard Kenyon

As usual the music in the video is from the free YouTube audio library: Night Snow by Asher Fulero and Fresh fallen snow by Chris Haugen.

Today's t-shirts I got ages ago. Don't think they still sell those exact same ones. Having said that just google "HO cubed t-shirt" and "i squared keep it real t-shirt" ... :)

Jokes:
1. Aztec diamond = Crytek logo; 2. no. tilings of Arctic diamond: 2^(-1/12). 3. ℝeal mathematical magic, 4. (HO)³ : joke for mathematicians (HO)₃ : joke for chemists
Bug:
Nice insight:
If all holes in a mutilated board can be tiled with dominoes the determinant will work. Why is that?

Merry Christmas,

burkard

[WilliametcCook]

(HO)³ : A Christmas joke for mathematicians
(HO)₃ : A Christmas joke for chemists

[calebvuli]

When he said “what a crazy, crazy year right?” I’ve been conditioned to expect him to say “Wrong!” 😂

[victorquantum]

I find really wholesome this man's dedication to speak and explain so passionly for 50+ minutes straight. As a phisicist that I am, I love how matematicians like this one continiously inspire us all everytime they can. Keep on the good work, stay amazed and happy holidays!

[ChrisConnett]

20:00 Number of tilings of the Arctic Circle: 2^(-1/12). Got it. 😏

[pedrocrb]

Merry Christmas!

6:11 - You could pick the two blacks in the top left corner. It would isolate the corner green square, so not every combination of 4 squares removed is tileable.

10:13 - say m = 2p-1 and n=2q-1. The denominators in the cosines will be 2p and 2q. Carrying out the product, when j=p and m=q, we will have a term (4cos²(π/2)+4cos²(π/2)), which is 0, cancelling out everything else

13:50 - Lets say T(n) is the number of ways to tile a 2xn rectangle. First two are obviously T(1)=1 and T(2)=2. For the nth one, lets look at it from left to right. We can start by placing a tile vertically, which will isolate a 2x(n-1) rect. - so T(n-1) ways of doing it in this case. If we instead place a tile horizontally on the top, we will be forced to place another one directly below, so we don't isolate the bottom left square, this then isolates a 2x(n-2) - so T(n-2) ways of doing it in this case.
----
We have T(n) = T(n-1) + T(n-2). Since 1 and 2 are fibonacci numbers, the sequence will keep spitting out fibonacci numbers

14:33 - It's 666. I did it by considering all possible ways the center square can be filled and carrying out the possibilities. It was also helpful to see that the 2x3 rectangles at the edges are always tiled independently. I was determined to do all the homework in this video, but hell no I wont calculate that determinant, sorry

30:12 - I'll leave this one in the back of my mind, but for now I'm not a real math master. I'm also not a programmer, but this feels like somehting fun to program

37:28 - Just look at the cube stack straight from one of the sides, all you'll see is an nxn wall, either blue, yellow, or gray

[petemagnuson]

Ok, can we take a moment to appreciate the slide transition at 25:40? It's magnificent.

[Mathologer]

Implementations of the crazy dance:
In response to my challenge here are some nice implementations of the dance:

Before I forget, the winner of the lucky draw announced in my last video is Zachary Kaplan. He wins a copy of my book Q.E.D. Beauty in mathematical proof. Congratulations! Zachary please get in touch with me via a comment in this video or otherwise.

I really did not think I could finish this video in time for Christmas. Just so much work at uni until the very last minute and I only got to shoot, edit, and upload the video on the 23rd, a real marathon. But it’s done :)

The arctic circle theorem, something extra special today. I’d never heard of this amazing result until fairly recently although it’s been around for more than 20 years and I did know quite a bit about the prehistory. Hope you enjoy it. As usual please let me know what you liked best. Also please attempt some of the challenges. If you only want to do one, definitely try the what’s next challenge. What’s the number of tilings of the 2x1, 2x2, 2x3, etc. boards? Fairly doable and a really nice AHA moment awaits you.

And let’s do another lucky draw for a chance to win another one of my books among those of you who come up with animations/simulators of the magical crazy dance that I talk about in this video.

Apart from that, I hope you enjoy the video. Merry Christmas, Fröhliche Weihnachten.

[mronewheeler]

For the m x n board with m and n being odd numbers:
Since m and n are odd, the denominators (m + 1, n + 1) in the fractions inside cos will always be even. And, since we round up in the expression above the PIs, j and k will in one factor both be exactly half of m + 1 and n + 1 respectively. When this happens we get cos(Pi/2) for both terms. Squaring the cos of course changes nothing. And when the product has one zero factor the entire thing will equal zero. Much fun this one!

[jackgartner]

I just realized: for the hexagon, if you turned it into a 3D broken cube, then looked at it from any of the open faces (assuming you're looking directly at the face from a single point), you would see a complete square of one color. This would be the same from all three open sides, and there would be no hidden faces. Thus, the sides being equal, there will always be an equal number of each color domino

[BenSpitz]

First challenge: The chessboard cannot always be tiled after removing 2 black & 2 green. Suppose we remove the two black squares adjacent to the upper-left corner and the two green squares adjacent to the lower-left corner. Then the left corner squares are no longer adjacent to any other squares, so the board cannot be tiled.
EDIT:
Second challenge: if m and n are odd, then ⌈m/2⌉ = (m+1)/2 and ⌈n/2⌉ = (n+1)/2. Now the j=(m+1)/2, k=(n+1)/2 term of the product is 4cos²(π/2) + 4cos²(π/2) = 0, so the product is 0. Moreover, if m is not odd, then 0 ≤ j < (m+1)/2 in all terms of the product, hence 0 ≤ jπ/(m+1) < π/2 in all terms, so cos²(jπ/(m+1)) > 0 in all terms, so each term of the product is nonzero. This means the formula gives a nonzero answer whenever m is even -- symmetrically, the answer is nonzero whenever n is even. Thus, the formula returns 0 if and only if m and n are both odd.

[omrizemer]

For the hexagon puzzle: looking at the picture as a 3D stack of blocks, it is obvious that each tiling can be gotten by adding one block at a time. This corresponds to rotating a hexagon with side length 1 by 180 degrees. Thus the number of tiles in each color doesn't change. But the numbers are equal when there are no stacked blocks.

[arigiancaterino]

I feel like I watch this guy 20% for his amazing math demonstrations and 80% for him laughing at his own jokes <3 thanks for the incredible content! Math is an amazing thing!

[Zephei]

Never thought I'd see such a detailed video on this topic. I've heard a little bit about all of these concepts (Aztec squares, Kesteleyn's formula, rhombic tilings, etc.) while watching Federico Ardila's great lecture series on combinatorics on YouTube, and I really think the accessibility of this subject benefits from visual-oriented, thorough, and intuition-driven videos like these. As always, great video.

[yufanzheng]

The determinant or closely relevant tricks are still intriguing topics nowadays. Leslie Valiant even introduced the name and opened a new subarea, “Holographic algorithms”, for these types of reductions (from seemingly irrelevant problems to linear algebraic ones).

[reecec]

I've just discovered there's no greater way to start a Christmas day than with a Mathologer video x

[charlottedarroch]

I was wondering what could possibly be interesting about the number of ways to tile the glasses. I couldn't have guessed it would be the number of the beast! I also didn't realise that the number of the beast could be written in a neat little expression involving only the first 4 primes: 666 = 3^2(5^2+7^2)

[andreiandrei]

Second challenge: it is obvious that in this hexagon there is the same number of dominoes of each color because transposing this 3D volume(from isometric axonometry) into projects on the XOYZ axes we obtain identical squares on OX, OY and OZ planes so an identical number at any scale of dominoes ;(and whenever we place the dominoes, the projections will always be squere).

[prabkiratsingh]

It's 12 : 00 AM in India
Looking forward to the following 50 minutes
And
Merry Christmas!!

[baoboumusic]

9:45 12 million "and change"?? I'll take your change then, thank you!

[livedandletdie]

Fun Fact: If that Hexagon tiling were blocks in Minecraft, then if that represented a sloped hill/mountain in Minecraft, it would be scalable as there is a path from the bottom to the top. Although it's not really obvious that that would be the case.