I really like the illusion that rhombuses can look like an isometric projection of a collection of cubes Edit: Congratulations on 100k subs!!! You deserve millions though, I can't wait for that

@I_Love_Learning

10 ай бұрын

Yeah. It can also look like you are looking at the cubes from the bottom, and tons of other weird configurations.

@maker0824

10 ай бұрын

I feel like it’s cheating a tad bit when talking about the toy there are “three different colors” then when switching to the simulation there are three different orientations, and the color is based on the orientation. This definitely is kinda explained in the video, so maybe I’m just the weird one, but it reminds confused me for a bit.

@isbestlizard

10 ай бұрын

There must be a mapping between the 2D tile state space and a 3D cube state space o.o

@puddlejumper3259

10 ай бұрын

It's the math of the universe conspiring against you

@Potato2017

10 ай бұрын

aops logo

A lot of the time when I watch videos like these I get very easily lost. This time I followed you the entire time. Great job

@alongal407

10 ай бұрын

Same, this was really easy to follow

@mihailmilev9909

9 ай бұрын

Nice

@mihailmilev9909

9 ай бұрын

​@@alongal407the formula threw me off for a sec after reading this lol but yeah

@mihailmilev9909

9 ай бұрын

271 1

@mihailmilev9909

9 ай бұрын

​@@alongal4071st like

Mathologer taught me about the arctic circle, but thank you for introducing me to other arctic curves and for discussing the generation strategy.

@danielthecake8617

10 ай бұрын

Mcdonalds is way better.

One thing I noticed is that “removing a cube” is the same as rotating that “vertex” by 180°! This is only allowable when all three colors are intersecting at the “vertex”!

@popodood

9 ай бұрын

Interesting. That actually showed me a lot ahaha

@mihailmilev9909

9 ай бұрын

Oooohhhhhh. Nice lol

@mihailmilev9909

9 ай бұрын

90 1

@mihailmilev9909

9 ай бұрын

90th like

@mihailmilev9909

9 ай бұрын

​@@popodood1st like

7:04 “a lot of the time, being aesthetic, or interesting, or beautiful, is the only thing you need to make learning about something worth it” holy crap, this is exactly why i like this channel without understanding most of the stuff in it.

One of the best new science channels on YT. Love it!

I may have not unterstood everything, but... I love your voice. It's so calm, so understanding, so... I just can't describe it

It reminds me a lot of a video from the french math youtuber Mickaël Launay called "La puissance organisatrice du hasard - Micmaths" ("The organizing power of randomness") It's about how seamingly organized randomness can sometimes be. And there is a specific part discussing the Aztec Diamond which reminds me a lot about how the liquid region of the rombuses is a circle but in the video in question the size of it was significant enough so the circle was blatantly obvious. It's really interesting if you don't mind french subtitles or auto translated english ones.

I have a few questions about the sampling method: if each of your moves is some local change of the state its not clear why this should generate a uniform distribution, especially when starting from such a special place in the state space. surely at early times this random walk will only fill some corner of the whole set of possible states. so, is it that you have to take many samples over a long time to generate a representative sample? even just starting at a random-enough initial state would make it more believable to me. otherwise, i would think you need to make a truly random permutation of all possible states to get a uniform distribution rather than only transitions which only differ in one position. this is probably well understood in mcmc but wasnt clear to me from the video.

@physicsforthebirds

10 ай бұрын

The most interesting result from Markov chains for us is that if a Markov chain is finite and any state can be reached by some number of transitions, then that chain has an "invariant distribution" that remains the same under transitions. The invariant distribution is the limit of applying transitions many times to any starting distribution. I wish I could give some intuition on why that is, but I just can't (that's why it's not in the video!) so I'll leave it up to a proof in a statistics textbook. Once we have this fact, we just need to prove that the invariant distribution of the Markov chain that I described in the video is the uniform distribution, and that proof is on page 231 of the PDF of Gorin's book in the description. It's not too hard, so you should give a look! If anybody has a more intuitive way of describing invariant distributions in Markov Chains (or maybe you've seen a video), please share it!

@Jojiho

10 ай бұрын

@@physicsforthebirds The "intuition" that I have for mcmc is the following: let's say you use some scheme where you jump from state to state to randomly sample from a probability distribution. Characterizing the scheme consists in giving the probability given the current state s_i to go to any ohter state s_j. Let's call this probability p(s_i -> s_j) = p_ij. If you succesfully managed to sample your target probability distribution P, then in the long run you should sample any state k with a certain probability P_k (which, because you did a good job, is equal to the target probability that you're trying to simulate). For this "empirical" (ie experimentally obtained) distribution to be stationnary, you want that for any states s_i and s_j, P_i*p_ij = P_j*p_ji. This corresponds to the fact that for this distribution to be stationnary, you should have "as many particles" (in the case of many independant experiments running in parallel) going from state s_i to s_j as the other way around (rigorously speaking, this is a sufficient but not necessary condition, also called "detailed balance condition", the true necessary condition, aka "global balance condition" is that the incoming flux of "particles"/probability entering s_i is equal to the outgoing flux, ie sum_k(P_k*p_ki) = sum_k(P_i*pik) for all i. The detailed balance condition is sufficient but not necessary but it is also often much easier to implement because it only involves balancing the fluxes between any pair of two states instead of having to sum over any accessible states). In the case of a uniform target distribution (all P_k's are equal), this "detailed balance condition" simplifies as p_ij = p_ji, which is to say as long as, when you choose a move with some probability, you have the exact same probability to return from the new state to the initial state, this satisfies the above condition and the experimental distribution converges towards the target one (but of course you should still wait long enough that you have completely forgotten your initial state, which might take a long time). The last thing that you have to be careful about is the case when some moves are not allowed, as described in the video. Starting from the state with no cubes at all (s_0), you only have one accessible state, the one with exactly one cube in the corner (s_1). Since this is the only available state from s_0, you might be tempted to move every time from s_0 to s_1 (with probability 1). But if you do the same reasoning with s_1, you now have 4 accessible states (either return to s_0, or add one cube in 3 possible locations), which means that if you only pick from these accessible states at random, the probability to return to s_0 is only 1/4. This breaks the detailed balance condition, as you would rather want to go back to s_0 with probability 1 (since you moved from s_0 to s_1 with probability 1). The solution here is to "play dumb" and try to add or remove a cube from anywhere in the hexagon without using the fact that most of these moves are impossible. This is very wasteful starting from s_0, as most moves will be rejected immediately, in which case you should sample s_0 again! This might make the simulation less efficient as you could repeat positions many times, but that is the easiest way to ensure that you sample from the correct (uniform) distribution. Of course for a "typical" state, which is very far from s_0 in general, many more moves are allowed so this is less of a problem.

@shadamethyst1258

10 ай бұрын

I think you can get an intuition on the MCMC algorithm by using a simpler game like snakes and ladders (with a board that loops back to the start). To find the standing distribution, you could play a billion games and watch where each end, or you could play one long game and watch where you're likely to end up in that game. The latter is a lot more efficient because you don't have to re-simulate the moves leading up to the final position

@TiagoTiagoT

10 ай бұрын

@@physicsforthebirds I don't remember who's video it was; but I have the memory of some weeks ago watching a video that used this markov chain distribution thing to calculate where it's more likely you would find the ghosts in the maze of the Pacman game when they get in their frightened state. edit: found it, it's by Retro Game Mechanics Explained

@physicsforthebirds

10 ай бұрын

@@shadamethyst1258 I'll be stealing that

Yesss yesss yessss more fun random math and physics to add to my repertoire of "Toast, How do you know this stuff????"

DUDE! The craziest thing happened with this video! I had an exam today. Yesterday night I decided that "screw it, I'll stop studying now and I'll relax a bit - let's see what's on KZread". I clicked on your video, and you talked about the Monte Carlo method I realized that is was a (niche) thing that the professor explained in class, but I didn't review: I opened my notes and reviewed it. Today at the exam, first question: "explain the Monte Carlo method, what it is and when it is used" my jaw dropped soooo thank you for helping me pass my exam! I wouldn't have answered that question if it wasn't for your video!

@physicsforthebirds

10 ай бұрын

That's awesome! Thank you so much for telling this story!

@oraziovescovi1922

10 ай бұрын

@@physicsforthebirds One could probably calculate the odds of this happening, but I wouldn't bother pretty low anyway😅

When you say lozenges I can't help but think of cough drops.

Losanges is what a rhombus is called in french. Also, this is something we should send to cgp grey. Also, this reminds me of those IQ tests where you're supposed to determine the number of cubes with a 2D view.

It seems kinda intuitive that the corners become frozen but I think you left out an actual explanation why that happens.

one of my favorite courses I have taken was a grad course in monte carlo so always excited to see it brought up in interesting applications like this one!

You are one of my favorite KZread channels right now

congrats on 100k. Well deserved :)

So ... This is kinda funny. KZread's algorithm just randomly proposed this video out of nowhere. And while I don't work with this stuff in a *long* time, my thesis started on planar partitions - the thingies that @vye9431 pointed to - and whenever they could be mapped to quantum mechanics / condensed matter problems. Mainly, we used the planar partition (and rhombus tiling) equivalence to dimers on an hexagonal lattice. The latter problem has an equivalent system in quantum physics / condensed matter - double carbon bonds on a graphene sheet - and we wanted to see if there was a way to exploit this equivalence to study the quantum problem. In the end, the "quantum" version of these tilings has some interesting properties - like how you can recover something similar to the Arctic circle, but with some "ripple" effects. But the hexagonal shape of the full tiling add some *really* strong boundary conditions, so AFAIK you can't really use them to model a graphene sheet. And, well, maybe (surely) I'm not good enough on math to build the whole equivalence 😅. We ended up pivoting to doing numerical simulations on more "boring", periodic graphene sheets / tilings. But those have some mathematically interesting phase transition properties! Anyways, that was a fun walk down memory lane 🙂

I love your videos so so much, thank you for taking the time to make them!!

you make incredible videos man. i seriously appreciate it

Man, I love your videos, are really fascinating and educative at the same time. Keep it up!! Greetings from Argentina 😊

I love the conclusion to this video, appreciating something nice is enough reason to do anything

awesome video! keep making this stuff, it really makes my day

Good science communicator, ive been binging all now!

I have no clue what I was supposed to learn from this, but the visuals were fun.

First I've seen from you, fantastic work and remarkably gripping, thank you

the most underrated youtuber right now, i always expect your videos or channel to have millions of views or subscribers

Classic MCMC can only approximate the stationary distribution, and it can be very difficult to predict how many steps we have to run to ensure convergence to a certain degree. But in some problems including this lozenge tiling one, there is an elegant technique called coupling from the past that allows relative efficient *exact* sampling from the stationary distribution of a Markov chain! Basically the idea is you run the Markov chain from past infinity. This is possible when the state space is finite using the following trick: You randomly sample single step transition function of the MC, going backwards in time (sample transition t=-1 -> 0, then t=-2 -> -1, etc.), and you compute the composition of these step transitions. When you see the transition function from t=-N -> 0 becomes a constant function (which will happen almost surely), you know all the transition steps before t=-N will not matter, since precomposing a constant function with any function gives the same constant function. So your MC has "converged from past infinity", and the sampling will be exact. Of course, the space demand for storing a complete step transition function over the entire state space will be intractable if the state space is huge. But here comes another trick: In some problems, like the lozenge tiling one, the state space is actually partially ordered (a tiling is above another one if its corresponding cube filling contains the latter's), and we can sample the step transitions in such a way that preserves the ordering (Pick a vertex and flip a coin. If it's head then fill in a cube at that vertex if possible, if tail then remove a cube if possible. All these step decisions will be stored). The partial order has a top element (all cubes filled in) and a bottom element (no cube filled in), and we only need to track their destinations after all the sampled single step transitions. When they rendezvous, we know the order preserving single step transitions must have squeezed all other states into that state as well, so the whole chain has converged. This allows relatively efficient sampling for a certain class of MCs with this order structure. For details, check out the original paper Exact Sampling with Coupled Markov Chains and Applications to Statistical Mechanics.

Realy cool video, comenting for algorithm. More people should see this 😊

This is an excellent topic for a video and there is definitely a million view youtube video within this concept. Good work.

This is amazing!! Thank you for sharing

Love these videos, keep making 'em!

Great content as always!

This is so beautiful!!

congrats on 100k! :D

this is amazingly interesting, thank you for sharing this!

Really liked the video. It kinda feels like it could have been longer though. Feels like there's so much that has been touched but not really expanded upon. Also I didn't quite understand the connections between the tilings and physics. I mean I'm really interested in learning more so that part you did get brilliantly. Overall love your channel. Please keep making videos!

I may not understand things 100%, but seeing your animations, the bird and hearing your voice is calming

Those colors hurt my eyes 😄 "Look at these solutions" I can't!

I could smell and feel the foam instantly. What a throwback

I'm about to start a PhD on occupation processes, which are in bijective correspondence with lozenge tilings. Isn't it strange when you just learn about something and then you start to notice it everywhere? Great video!

whu....what....dude.....Arctic curve vocabulary is so perfect. it just make so much sense you are so very good at educating. I hope you teach at a university or something.

Watching your videos really wish I had kept it together long enough in uni to get my math minor.

YEAH NEW VIDEO YEAH!!!!!!!!!!

Great video!

Thanks for the lovely and informative video❤❤❤ I know the maths displayed at 0:34 is probably quite easy but it'd be amazing if you explained them a bit😅

@physicsforthebirds

10 ай бұрын

This notation isn't always taught in school, so sorry for not explaining it! Capital Pi at 0:34 means "the product of all of the terms on the right from a=1 to a=A", so it works the same way as capital Sigma does for sums. That means you do the fraction with a=1, b=1, and c=1, then you multiply it by the fraction again with a=1, b=1, c=2, and you keep on going until you've gone through every combination of a, b, and c. That's a lot of terms to multiply, so it gets big really fast.

@ozen.m8161

10 ай бұрын

@@physicsforthebirds thank you sooooo much🙏🌹🌹🌹

This is a great video. Have you ever spotted that the unbounded rhombus tiling maps onto the triangular ising antiferromagnet. This is why the middle of the model is 'liquid' (it maps onto the spin liquids).

congratulations on 100k

I use MCMC for work and simulations! Cool to see it come up here

Which tool do you use for your math/ drawing sections?

Thanks for knowledge, will be great to pass it down to grandkids in the future

@hypenheimer

10 ай бұрын

Hello again heisenberg! For those who don't know, Heisenberg is the fresh account of the "NMRIH is a great source mod" which was banned for botting/violating KZread TOS -Same mannerisms, Over 800+ subs to only popular/viewed channels, popped up right when the previous account was banned about four months ago, this account is a bot that spams and like baits channel's comment sections for subs.

could you do a an optimization for the big ones where you start with a low res one then subdivide and the markov chain then subdivide and repeat until you have high detail, but still random even at both low and high frequencies?

Is there an easy explanation for why the total number of rhombus tilings is the product of all permutations of (a+b+c-1)/(a+b+c-2)?

6:55 this is true even in other fields. engineers sometimes have Legos and other toy sets as quick stand ins for things. not often, but it's done all the same. it gives a person another perspective to a problem they had been considering.

The first thought that leapt into my mind to get a uniformly arbitrary tiling would be to 1: work up an algorithm that uniquely maps a tiling number to a tiling output (EG in the side length 3 case maps 0..979 -> a tiling output such that each number gives a different output, no other restriction to ordering required) and 2: select uniformly arbitrary number in the range of possible numbers, which I already know how to do with time and space complexity of roughly O(log(N)). 3: use algo to transform random number into random layout 4: sip margaritas on the beach at sunset. Optionally start with point 4 and do the rest on a laptop while fending off the gulls, but if I make my rhombuses out of tortillas and make extra I can toss those to keep 'em busy. Wait what were we talking about again? 😎

I didn't realise this was a physics video until you mentioned 'frozen' and I got statphys flashbacks, lol.

Markov chain. A nice call back to a stat mech coding homework.

You see an empty cube in the initial state. I see an upside down "filled" cube. We are not the same.

New physics for the bird video released!!!!!

@Asterism_Desmos

10 ай бұрын

After watching, I would like to add, the cubes can be inverted I my mind to where it’s not an empty cube, but the outside of one being chipped away. I hope others can see this because I thought it was odd.

Is there some way of knowing you've done enough iterations to be suitably random? Because if I was running this on my own without knowing about the frozen region, I would just assume I cut it off before changes could propagate out to the edges

@physicsforthebirds

10 ай бұрын

That's a very good question, and there is. If you keep track of the chain that you generate and then you apply that sequence of steps starting at both the "empty cube" tiling and the "full cube" tiling, and if that sequence leads to the exact same sample once it's been applied starting at both of those tilings, then you are sampling from the uniform distribution exactly. You can read about it in the last chapter of Gorin's book in the description. For the sake of runtime though, I just plugged in a big number of steps and ran it (because I already know what a random sample should look like. Not a good way of doing science, but a good way of making pictures!)

Why, even in the limit, would your random add/remove procedure approach uniform across all tilings?

Would you please give an intuition of why doing the random steps you mentioned is (approximately) equivalent to generating a uniformly sampled arrangement?

Does markov chain monte carlo not encourage the behaviour of the deadspots? Since all new cubes start from the center and not the edges. And the reason behind the deadspaces is also not very clear to me. if every space has an equal chance of being swapped, why do we see deadspaces around the circle?

so cool

4:31 can you explain why this shows the same behaviour?

“this is a baby toy” _proceeds to explain math functions and probability distribution_

Does it look like it, or you can actually represent it as cubes stacked in the corner? Because that's a 3d Young diagram, for number of variations of which you can provide a recursive formula (or call it "dynamic programming" if you prefer CS terminology). For every such counting formula it's possible to provide a reverse formula, that, given some number from 0 to totalCount-1, traces back a path from the furthermost corner of the diagram back to the origin, and by uniformly producting a number from 0 to count-1 you get a uniformly taken rhombus tiling. Edit. Okay, that's a "plane partition in a box", and the only thing missing in a formula from wikipedia article is a recursive formula that relates larger box to a set of smaller boxes in some interpretable way.

Now I'm just waiting for Qbert to hop into that cube shape and start bouncing around.

Now I wonder how do you get that fomular to count possible arrangements. I should try it out later...

when you say that a particular solution "took a while" to generate, do you mean that you had to run MCMC for a long enough period of time to diffuse away from some low-entropy solution and reach a representative sample of the uniformly distributed state space?

The thumbnail ❤🥰

It's all well and good when you see it as a room iwth boxes pushed into the corner, but sometimes it looks like a cube with smaller cubes carved out, and I get dizzy.

If I recall, this sort of behaviour is very much analogous to Random Matrices, possibly even related For example, if you take a particular random matrix, known as a GUE (Gaussian Unitary Distribution) The matrix can have many different eigenvalue distributions. However, the typical behaviour of the eigenvalue distribution is in the form of a semicircle distribution. Its very spooky

this video was on my watch later tsice release

YES!!!!

Happy at 3:42

Funny how you said at 5:35 you add cubes to an empty cube while to me it looked like removing cube from a full cube the whole time

The human mind is so weird sometimes, I can’t believe that the rhombuses look like cubes so easily.

Cool👍

MORE PHYSICS FOR THE BIRBS

I'm curious to see how this concept could be linked to entropy

Can you elaborate on the combinatorial calculation in the beginning please

@physicsforthebirds

6 ай бұрын

Proving that equation for the number of valid tilings is not at all an easy problem, so I left it out of the video. If you want to see the proof, go to page 11 of the "great textbook on lozenge tilings" PDF I linked in the description. It takes the whole second chapter to derive it!

@goldstarlazerbeam8562

6 ай бұрын

@@physicsforthebirds Can't believe you replied, thank you 👑 sounds like a valid question in my combinatorics final a couple years back

What is the music at 6:39

@physicsforthebirds

10 ай бұрын

The music I use is mine, made for these videos. I keep telling people I'll release it... maybe someday

@Kaleisbord

10 ай бұрын

@@physicsforthebirds It sounds cool, you definitely should

is this just a popular topic now suddenly or were you in the same public lecture as me recently... the coincedence is coincedencing... anyway a good video to recap since the last time i heard about this was quite late at night :D

Something about this feels off to me. If you are making a random tiling of the hexagon, you should also be able to "rotate" and the colors of the rhombi which could break the cube illusion. Or, with sufficiently large enough hexagon allow points where the minor angles meet which also breaks the illusion of cubes.

Ive always intuitively known that there'd be the same number of rhombuses cause i image them as the number of "ceiling units" and number of "wall units". Painting any one side will always only require that number of losanges

Waiting the MattParker comment, Hey Matt!

Thank you mister bird

why is the arctic curve a circle, and not, say, the hexagonal cut that bisects the cube?

@michaeldamolsen

10 ай бұрын

Mathologer made a great video on that. I don't remember the exact details, so I can't offer a good summary.

i was confused at the part with the 3 rhombuses being cubes because i thought the cube was a removed part of a bigger cube and it looked soo weird when another cube apeared

Now tile the hexagon using lozenges of 3 different colors in such a way that no same colors touch.

tetriiiiiis yeeee YEEET

The visual here reminds me a lot of Minecraft terrain when it's generating.

I want a room made like that

-(Yeah same. And)- I learned the real definition of isometric projection now! That's actually so cool I might consider more for game design or something like that. What do you guys think? (Was my reply to one of the top comments btw,)

I watched this video before!

Where would you even find lozenge tilings in physics?

@physicsforthebirds

10 ай бұрын

Although you might not find literal lozenges in very many places, random tilings are useful toy models for other systems in statistical mechanics. For example, simple models of water molecules in ice or electron spins in magnets share many of the same "freezing" and "melting" features as lozenges.

@TheTablet314

10 ай бұрын

@@physicsforthebirds Thanks!

"EY! WHO MADE A UPSIDE DOWN Y WITH MY CHILDREN RHOMBUS TOY 😡" 💀

My brain had a hard time seeing the intended outcome. That cube was rotating, flipping, translating, and flipping inside out often during this whole thing

+