Is Eve the cube or the light source

@@columbus8myhw Eve is holding the camera filming their math contest. For some reason Eve's camera makes everything look like a wire-frame rendering with hidden line removal. What I want to know is what Trent and Malory were up to while all this was going on.

ahha from the Bitcoin video I see

@@hoppingturtles Alice and Bob are just standard cryptography names, not specific to bitcoin

Ok

I FELT SAME ASÖALKSMAKAND

Me: Pauses the video and scribbles on a piece of paper for five minutes. " Yes that checks out".

@@JohnDoe33408whyd it take 5 minutes? had to think about all possable last digits of pi?

​@@TheAdhdGamingseems like lol

Try not😮t😮 r

A man of culture, I see.

Well you see... the outside and the inside of a sphere both have a turning number of one...

lmfaooo seems like it's haunting us all then

I can't believe I see the "outside in" community here.

​@@toxic-LI would rather not believe that there exists such a community that doesn't intersect with that particular video xD

It's great to see comments that are valuable and great additions to the video :)

does that mean that Alice is actually a genius?

@@aguyontheinternet8436 no exactly. as he said , it's incredibly hard to solve real life problems using top down method, the problem here was unrealistic where in real life example you would need do things down up because of sheer amount of variables. edit: spelling

Kind of like using the pythagoras(top down) vs creating it (bottom up)?

@@arlenboi7360 yeah so precisely because its hard but she still does it she's a genius? whether she can do it everywhere is irrelevant because even in the cases that can be done by this method others aren't able to

I agree! Most of my math teachers have made me feel that if i dont solve problems like alice does im bad at math

Stepping back we can see that that was the (actual) generalization this video meant to derive 😉

Yes I probably would be Bob for solving this for cube or tetrahedron and suspect the fact the answer is a quarter of its surface area then become Alice.

I would suppose that the brilliant, refined mathematical tools we have exist only because years of experimentation and work - sometimes even happy little accidents - were put into creating them. Do not feel bad that an obvious, to the instructor who has spent years reiterating the same lessons, proof doesn't come to you naturally. You only learned about it five minutes ago.

This. I have almost always taken this exact approach. Typically I will go from calculation to insight to verification of the insight with further calculations, at which point I'll either follow the insight through or perform more calculations to connect more deeply with the insight. Never do I ever utilize only one, because even if I it's simply too easy to miss something if following only one method.

Bob is the superior for school, Alice is the superior for the real world

@@tomepsilon As a quantum chemist, Bob is how you do things, Alice is how you report it.

@@tomepsilon I think it might be the other way around

Bob is a practitioner, Alice is a theoretician

Sounds a lot like the P = NP conjecture (which is almost certainly false). You use Bob (NP) to find the solution. You use Alice (P) to show that a proposed solution is a solution. Finding the solution in the first place is the hard part. Showing that it is a solution is much easier and fun.

this really is the essence of what my experience felt in olympiad mathematics, understanding the two methods is crucial

I’ve always been a Bob kinda guy and my inability to find Alice like patterns is why I didn’t pursue a PHD. I wish I could learn though

That's a really great way of putting it

Oh wow. This is a well put way of describing this. I'm gonna steal that thought pattern. The generalizability of the method vs the result.

dont try to be a genius why so serious 😡 you

well said! couldnt agree more

It is also useful to have Bob's approach on hand when looking for any logical holes in Alice's solution. At least that's how I usually discover & fix mistakes in my "slick" reasoning attempts.

That's true. Often you try something straightforward and after you finish all the work you get a nice answer. That's often indicative that there's a different way to think about the problem. Though sometimes the Bob approach actually doesn't work. Trying the most obvious or "just do it" evaluation of a problem or brute force sometimes gets you stuck in a world of computations that you can't actually compute well. Then you'd try to make more insights and think about how to do the problem a different way that requires more observations and understanding.

@@x0cx102 I have the same opinion

Ah yes, can the computer determine there is a solution without determining the solution.

@@x0cx102 It's fascinating to me that some general concepts can be argued for/against throughout the history as a lot of comments in this section (like yours) are actually talking about Occam's razor. I actually don't have anything to add to the discussion just found it interesting :)

That's a really nice way to put it. Begin by diving in with vigor to a few representative cases, and put on the Alice hat when you sit back to reflect.

Agreed. I think Bob would actually gain a lot of Alice's insights if the question were posed as "What is the average shadow of a sphere?" and forced themselves to do the integral calculus.

@@tezzeret2000 Isn't the average shadow of a sphere just a circle with the same radius? Since there's only one shadow for a sphere...

Exactly what I was thinking ... often it is not so easy to jump to the Alice mode, to get to that mode, it would need a Bob mode to have consumed and understood those special cases.

Lm

[Eve Lobachevsky has entered the chat]

love your vids! They really helped me on my calc 3 final.

Perhaps one uses the generality one is capable of. As one advances and the problems get harder one inevitably takes a more Bob approach.

I also think of it from a real world problem solving situation. This is a situation I encountered. Say you went to the doctor and your insurance company denied your claim. You call your insurance company and they say it looks like there was a mistake made. You will need to send a letter appealing the claim and explaining the mistake and requesting the claim be reviewed again. I was taking the call for the insurance company and I saw that there was a simple solution to the issue. Writing a letter explaining the mistake would almost guarantee that the claim would be paid. However the person who went to the doctor did not see it as a simple solution, because if it was guaranteed that the claim would be paid if he wrote the letter then why did he have to write the letter. In the end there were two issues. The first was that the claim was denied, that issue was easy to fix. The second was that he had to write a letter, but that was not as easily solved. It involved policies from the insurance company, laws passed by congress, and issues of ethics avoiding potential for fraud. The deeper you went the more you realized that it went even deeper. In the end the insurance company had applied a policy that worked most of the time, but was sometimes inconvenient. In the end I told the guy that we could spend weeks debating about why things are the way they are, and I was quite enjoying the conversation, but in the end the thing we are really wanting to solve is his denied claim and we already have a simple solution for that.

Not necessarily, Michael Micek. Olympiad questions (such as those from IMO) are good examples of brutally hard questions that primarily require Alice's mindset

@@ivarangquist9184 fair enough.

They really are, watching still in my last year of college

I'm in the exact same position

And therefore likely some of the best in the world :)

they are definitely very good (enjoying as a phd in cs)

I'm gonna go on a limb here and throw some "high praise" and say: some of the best in the world. If you've seen the one on Quaternions -- and specifically the webtool they developed to help teach about them -- I wish _really really wish_ down to my heart that schools would adopt that style of teaching. Within hours I got something that I've been struggling to even faintly grasp for years, and not for lack of trying. There's a lot of value in 3B1B's particular style of teaching, and I'm super happy to see he and whatever team may be lurking behind him in his productions are getting the eyeballs they deserve.

not just that, but in every video every animation is so well made that understanding the math behind it gets way easier

So pretty much exactly what you say at the end. I hadn't finished the video yet :D

Ditto. Physics guy here, too. Didn't get into much fancy math until a bit later and, when I did, was reminded of some of the tedious integrals from homework problems that ended up with nearly all of the terms canceling one another out, leaving something along the lines of a constant multiplied by an integral from zero to one/pi/2pi/etc with a constant integrand. Wish I had had the benefit of content like this back in those days... seems obvious (in hindsight and w/ Grant's awesome material) that calculations which eat themselves away into almost nothing are good signs of a more abstract method of reasoning about a problem. Grant and his team and supporters are a national treasure!

I think it only makes sense. Humans are pattern-matchers, not SAT solvers. It is much easier for us to come to conclusions by generalizing over discrete data, and only afterwards finding the justification.

@@NXTangl pattern is really true on note of vsauce face recognition

i’m not a physics major in the slightest, but what i did. was take the area of the shadow where it’s the smallest (1.00) and where it’s the largest (1.73) and took the mean of those two. i haven’t finished the video yet someone tell me if i’m either a genius or really really stupid

I am so unpretty 😭 When I go to the bank, they turn the cameras off. At least I am a big star on UKposts. So don't feel too bad for me, dear mat

As you said thats the way humans have evolved over time. The quicker more broad solution has prooven to be less energy consuming than having to calculate and check everything

That's true, most of my own videos aim to be more like Alice than Bob not because one is better, but because Alice-like videos somehow seem more natural of a fit for a UKposts audience.

I agree. Its hard to talk about "most people"... I particularly like to exercise my brain with those tedious calculations lol. But taking in cosideration the way math is taught at schools and more generally how we are evolving as a society, the rather creative ways to solve problems comes as candy for the minds that have had enough of systematic approaches.

I wonder if the same effect applies to math cirricula. It would be interesting to chart the number of calculation drills done by primary school students over time.

Thanks so much!

@@3blue1brown I am sorry trying to get your attention like this, however would it be right if we would add the smallest surface of the shadow (1^2 =1) and the biggest surface shadow (~1,73 which is the surface when the shadow is a perfect hexagon) and devid them by two (end result ~1,366) to get the average surface shadow? By the way I like the amount of time spend into the animations👍.

@@renanokten6058 Why do you think that could be right?

@@renanokten6058 unfortunately not, the correct average* is 1.5 (see 31:40). What your approach misses is that a typical shadow is more likely to be closer to the higher end 1.73 than the lower end 1 (*using the rotation-invariant measure)

@@japanada11 you are absolutely right👍. Thank you!

Just use Haar measure.

Indeed, something even more subtle is that measuring on limits has its own dangers. E.g. the limit of stair cases (when stairs got smaller and smaller) is a triangle. But the length of the stair cases will always be total height + total length while the length of the triangle line will be sqrt((total height)^2 + (total width)^2). Of course, the limit of stair cases is topological something very different (e.g. nowhere differentiable, also never convex) vs a triangle (everywhere differentiable, of course convex etc). - I guess, Alice approach here works because she is assuming convexity (in the approximation of a sphere), but tbh, I couldn't really argument it here beside of a plausibilization and the world of math is full of paradox (like you can split a sphere into two spheres that have both the same volumina) if you miss a subtle point here when applying infinite limits. So, to be sure that Alice solution is right, you'll probably need already be a master in math (and even those have "failed" there). And you still need to do Bob's approach in most cases anywhere to detect "false" results (e.g. physical impossible solutions) or to make a plausibilitization of the results (these magic tricks can always be doubted by non maths, just as there are so many paradoxies, especially in combination with probabilities), while if you can calculate down in a numerical approximation, it's much more trustworthy for non mathematicians and also "easy" to double check via experiments (classic example: Buffons needle can be verified by by every 10year old).

Exactly. It would be easy for someone following Alice's reasoning to try to apply it to, say, a torus, without reasoning that it violates an assumption made along the way.

You have to be on your p’s and q’s to use Alice’s method. Your Einstein’s of the world can pull this off due to their vast understanding but your more introductory student certainly has to use a bit more caution for sure.

The issue is that many times the explanations you get from people are plainly bad and inaccurate and the math itself is simple. Many people who did not get to a deep level of understanding will talk with metaphors to explain it to you, while you understand it much better than the "internet teacher" or "colleague teacher" - all you need is to read the math. It's especially true for ML, what I do.

When he pointed out that Alice’s method could be generalized across convex shapes, my eyes actually went wide and I started grinning and mouthing the word “Sphere”.

@@bioboygamer same lol

i genuinely had to pause the video and i mouthed "oh my god is it gonna be the surface area of a sphere?" and i was almost gonna tear up

honestly so true

I agree. Great explanations & impressive animations

My thinking as well; I hope mathematicians have a greater respect for video editing in general as an art and skill.

I mean I think it's really impressive from his side but since he made Manim public.. anyone can do it pretty easily and it's not that hard anymore.

@@justafish5559 As a person quite familiar with how Manim works, I’m sure this was made with thousands of lines code. Not really easy.

The way it should be taught is that you come up with a recursive solution and then optimize it. A problem has a recursive solution when each nontrivial problem can be split into smaller problems.

@@MartinVillagra i sometimes prefer doing dp forwards though

Then you take a control theory course and the dynamic programming in that has nothing to do with your prior understanding and reconciling the two seems pretty hard since the CS version is so inherently discrete.

Very cool, thank

Ok

that's exactly what I said

@@simple3555 Hi. I have a question. Got a sec?

@@nenmaster5218 uuh sure?

@@simple3555 Sorry for being random but i love to always keep learning and never stop, so do you have some Recommendations for me? I do have some for you and i LOVE recommending sci-channel and edu-youtubers, but right now, i mainly wanted to ask for some myself. Got some?

Yeah, but there are 8 possible hexigon shadows (one for each corner)- and only 6 possible squares (one for each side)…so your average would skew in favor of being smaller than the average (of all possible orientations were equal). I think your idea would have better weight with 8h+6s/14 than h+s/2

Thank you, this was bothering me

I recently grappled with your example while driving, might as well take 2 long legs instead of many turns because the distance is the same BUT in driving there's a "cost" to turning lol!

yup. just pick faces that are planes tangent to the space and you’re okey dokey

Yeah, you have to be careful because there can be infinitely meny corners that add up to a non zero surface area, area can fit into zero volume

Also the pi=4 proof is a great example of this

Not sure that I understand what you are saying, but it’s lovely. Dive deeper 🙂

I literally don't understand this. But dude youre smart. Can U dumb it down for me 🤣

It's earilier than that. Around 17:45, he talks about sampling rotations and this is where probability distribution really matters. The hidden assumption is that it's a uniform distribution over the space SO(3) mentioned earlier.

@@DrAlexisOlson The OP is correct. Everything up to 20:07 has made no assumptions about the set of rotations being considered and would be true for any specified set of rotations. At 20:07 when c rather than c_j is introduced into the sum it requires that all the faces have the same average shadow coefficient under the chosen distribution over the set of rotations. That doesn't work for any arbitrary distribution over the rotations anymore, even for regular polyhedra since the faces start with different initial orientations. It only works for sets of rotations that respect the symmetries of the faces, for example the set of all rotations that exchange two equivalent faces, or (crucially) for SO(3). Oddly, any set of rotations would work again once you generalize to the sphere, but only a uniform distribution over SO(3) works for the infinite series of polyhedra that gets you to the sphere.

@@DrAlexisOlson It's even earlier than that. At 12:20, the notation f(R1)*A already assumes that the coefficient f only depends on the rotation R1 and not the original orientation of the shape. This is not true; luckily later Alice is only using the proposition that the average of the f(Ri)'s is not dependent on the original orientation of the shape. This is only true if all the rotations Ri are distributed in SO(3) uniformly.

Probably a volumetric fractal of some sort

a shape made of a perfectly clear material, so it has no shadow ;)

in the spirit of looking at things from different perspective, this is not the only possible definition of convexity. it takes a bit of working out to formally proof, but the following is also equivalent: "Any tangent line to the boundary of the set touches the set only at boundary points" and "The intersection of [any known convex shape (say a circle) tangent at any boundary point of the set], with [the set itself] is exactly 1 point". some even define a set convex if it is equal to its convex hull, defining the convex hull using the lines method, which I find really backwards but its still exactly as valid.

I've always focused on the edges and angles, personally- A convex shape is one in which all of the inside angles are less than (or equal to?) 180 degrees. There's some geometric truth somewhere that could explain why focusing on the angles and focusing on the points would give the same answer to the same question (something something... make triangles with disparate vertices of the shape...)

@@ferociousfeind8538 in some cases it is not clear what the relevant angles are. E.g. what if it is a smooth shape in many dimensions? The “are all line segments between points in the shape, contained entirely in the shape” can still be applied even to talk about an infinite dimensional thing, and I think it would be difficult to do that by talking above angles. So, I think the way OP/the video define it, is probably overall best?

This right here is why studying Real Analysis in my undergrad math curriculum was hell. It felt like so many of the proofs relied on having the right kinds of insight to generalize problems that you can't necessarily acquire outside of an organic process of exploring different kinds of problems of a type. And that kind of mathematical learning was really difficult to engage in while also trying to balance all the other commitments and responsibilities of approaching post-undergrad life

Its a shadow! 🤷

@@Dogzz13 But the sum itself isn't. You can, of course, relate it to integrals because of the 1/n and the limit you're taking, and you can very easily ask Bob to provide an easy upper bound you can use to show convergence, but that's not *completely* trivial.

But on the other hand, the problem being solved, although specific, reveals a very general perspective of what it means to think mathematically and how that can vary from person to person. Admittedly, the audience here is relatively niche (i.e. those who know linear algebra and calculus), but that’s his target audience anyway. There would be an extremely high chance of anyone with this demographic to click on the video. But even then, the issue at heart here, i.e. problem solving, is enticing to anyone who is mathematically inclined.

@@vlogsbyrow Good point - it's not like this video isn't doing well. Just nice to see a heavily invested product for a relatively niche audience.

Same, my blender cube is now an ultimate shadow-making cube

Ah, turns out there's a third perspective on the problem. Alice, Bob, and the artist.

@@zbieramnakartonowyprzycisk8026 What if computers suddenly stopped existing. How will you draw your cube shadows then?

@@vigilantcosmicpenguin8721 When the light is closer? I guess using ray vectors would be easier if the light source was non-uniform.

“What if computer suddenly stopped working” Don’t worry that isn’t happening anywhere soon.

I need help finding a song that goes Put your poop on my shoulders (oh oh oh oh) and let your worries shut away I don’t remember what the second last words was, it was either shut or shit. Thanks!

Damn, I'm in the last year of my IMO days. I really want to pass the first stage of the exam on 9th January :(

@@adarshmohapatra5058 Good luck with it! My IMOs were a long time ago, 98 and 99. I later discovered academia wasn't for me, but no regrets.

@@sirgog Thanks!

Maybe grab yourself a beer.

My name Adam

@@constante7510 ok

@@constante7510 My name Jeff

@@eltyo340 ok

Lol. I immediately check out your comment after you mentioned. You could have just leave me your comment link :D. Anyway, I was able to find your comment after using browser search in this page. So glad that I find someone who think alike. I'm no physicist, and I don't have friends to talk about things like this. My approach is a little bit different but use the same electric flux principle (basically Gauss' Law). I'll just share this in the same thread. 1. The initial question only assumes parallel shadow projections, so I'll admit initially I didn't start as generic as yours. I immediately assumes there is a uniform charges very far away, causing the electric fields to be parallel. This is equivalent with your charges in sphere's surface with R infinity. 2. Electric fields will penetrate any kind of object (it doesn't have to be cube, but we can start with cube if we want to), but it will have net zero flux on the object (no charge inside in the object/cube). 3. The shadow is equivalent to flux penetrating this flat shadow casted by this cube. 4. Since flux on the object is net zero, that means the top half of the object and bottom half of the object have equal absolute flux value, which is the same as flux in the shadow. F_shadow = 1/2 * (sum of absolute value of the flux). Note that, single instance of shadow or average of shadows, this formula doesn't change. 5. For each small section of area dA in the cube surface, if you want to calculate the average flux over all the possible orientation, you iterate all the possible position and orientation of dA. It is just conveniently happens that when you rotate dA to all possible space, it will form a sphere. 6. Conveniently Electric field vector has the same value and orientation in this case. The total absolute flux is just twice the top half of the sphere, which is equivalent to twice the flux of the shadow. Shadow of a sphere is just a circle, with radius r which is the distance of dA to the center of the cube. total flux = 2 * pi * r^2. Since we are calculating the average over all orientation, we divide by our possible spaces, which is the sphere surface of dA. we have 4*pi*r^2. So the average total absolute flux is just 1/2. 7. Sum the average to all section of the surface dA. We got: average(F_shadow) = 1/2 * 1/2 * average(total absolute flux in all orientation) 8. What we want to have is the relationship between the shadow area and the the surface of the cube. Because the electric field value is the same, we can factor out E from both sides. We got average(shadow) = 1/4 * surface area of objects As you have said, the 1/4 factor is just stems out from the fact in step 6, and irrelevant on the object shape itself. It's just a ratio between the shadow and all possible spaces that makes the shadow (which is conveniently a sphere because our rotation is a sphere). The convex criteria is just so that the possible spaces is always a sphere (if it's not convex, some orientation doesn't have a flux). You can also generalize if the light source is a point near by the objects. Just treat it as charge source, then the electric field value and orientation will follow Gauss Law (spherical instead of parallel). That means the average shadow is only multiplied by the coordinate transform scale size of the sphere and the projection.

@@RizkyMaulanaNugraha very nice! Btw, I had no idea you can actually share a yt comment link, so thanks for the tip :)

Rings true especially considering the sphere case, where we take the limit of a sequence. Gives off the same vibe as swapping a limit and an integral

as soon as you combine three faces (areas) with indepentent normal vectors, and assume that the sum of their averages is the overall average, you already have to use this orientation distribution. this assumtion is the key step! the distribution which is used is the only one that is the same for every face, no matter how it is oriented relative to the other faces. (just what Pa Pa wrote, but stressing that it is not the infinities where the distribution gets fixed, but the finite steps)

I thought so too at first. But wouldn't both orders lead to the same result, no matter how the samples were picked? If I used a biased way to pick a thousand rotations, I could still average the shadow of each face and the sum of the averages of the faces will equal the average of the shadows of the entire shape.

@@CaesarsSalad not quite. at least not if you have the same constant by which the average scales the area. this constant would be different for different orientations of the faces.

@@CaesarsSalad But you can pick biased orientations which don' have the average for each face being equal. For example, if you pick a biased set so the top face must be at the top (i.e. a point on it must be the highest point on the cube) and can never be at the side, then it will contribute a larger amount. So a key part is going from adding up the faces to multiplying the average for a single face by the number of faces.

Entirely wrong way to look at it

Oh, but they have to use cryptography to make sure the other one doesn't see their problem solving.

More like, to share their solutions with each other, and nobody else.

Flashback to Khan Academy information theory XD

@@ekisacik And to detect when their solutions have been tampered with in transit.

Indeed. I do find it interesting that Alice never does assume a probability distribution. What she assumes is a _property_ of the distribution (isotropy, roughly speaking), and to me the fact that her property leads to the same answer is a noteworthy result itself.

i dont know if the content youll be making will be in russian, but im really genuinely interested in it. ig being curious about everyhting does have its perks haha (besides that, props to you for the channel! i hope you soon get time to make smth you like!! good luck)

Or to borrow from physics/the PhD approach: She imposes arbitrary rotational symmetry which locks her solution to SO(3).

Another way to explain (or possibly justify) her distribution here is to argue that for any given orientation of the cube, there are symmetrical orientations where each face can take the place of any other side with equal probability, therefore every term in one sum must appear in each of the other sums as well, or in other words each sum would be a permutation of the others and therefore be equal.

I don't think this is the complete answer. I'll use an intentionally terrible distribution to illustrate: if you sample random rotations by tossing a six-sided die and observing how it comes to rest on the table, the six faces of a cube are still symmetric under that distribution (with area of 0 or 1, 2/3 and 1/3 of the time respectively) and can be considered independently. It's only with the third constraint (asserting that it can be applied to other convex polyhedra) that the symmetry gets broken; tossing a six-sided die does not generate a face-symmetric distribution for an icosohedron, for example.

@@TC-cq7oc no, but rolling an icosohedron WOULD generate face-symmetric distribution for an icosohedron

@@phiefer3 - It would, but that'd be a different distribution of orientations than rolling a cube was using - it'd have 20 possible outcomes instead of 6. If you apply the icosahedron's distribution back to a cube, you'd get a constant factor on surface area much closer to 1/4 (matching the "correct" distribution) than my distribution which gives 1/6.

Prokhor Zakharov! That quote has remained with me for more than half my life since then.

A beautiful quote

God that game was so good

Nakamoto is the Alice of the 21st century

@@Corwin256 Hey, that is Alpha Centauri. I didn't remember the quote, but I faintly remembered the name Zakharov. Good game. I haven't played it in years.

Can you shere some more channels Like this??

Indeed. I was doing the exercise in parallel to the video (as an undergrad student this was delightful practice), and I did it in a very similar way to Alice. I defined the normal vectors of the different faces of the cube as "a series of linear transformations on the normal vector of 'face 1'." It quickly became clear that only with uniform probability distribution I could ignore those extra transformations and apply the formula for average shadow.

This was my first thought as well, but I think that's not it. I thought about a probability distribution similar to the one you suggested -- force two faces to be parallel to the ground, like rolling a die on the table. But the average shaded area of any given face is the same: 1/3, corresponding to the probability that the face is pointed up or down. I think the assumption is made when she generalizes her formula from the cube to other shapes -- around 21:16 in the video. Unlike a cube, an arbitrary shape can have faces pointing in any direction. So to apply her "1/2 * c * A" formula to each face, she must assume that turning the face around in space does not change the probability distribution. That is, the probability distribution must be invariant under rotation.

@@biggnate yes, you are perfectly right, thanks

@@Brainth1780 very clean way of doing it!

​@@biggnate You make a very interesting point, which prompted me to reevaluate a lot of what Alice did. Long wall of text incoming, the rabbit hole went deeper than I imagined. *TL;DR:* It's up to how you interpret the steps, both answers are right in their own way. What she claims for the "shadow of a cube" step is _not_ just that the shadow corresponds to "a constant times the surface area of the cube", she reaches the conclusion that that constant is 1/2 of *the same constant **_c_* that she got from calculating the average shadow of a square. As we know, for a Uniform probability distribution, c=1/2. Now, things get interesting. Using a different probability distribution would mean that a "side face" has a different average shadow than the top face, but you could get around that and say that "a face has a 1/3 chance to be one of the two that are parallel to the ground." This would still allow you to factorize in order to find a number for which the formula "1/2 * c * A" applies (meaning _c_ is independent of A), but _c_ would be particular to a cube and, arguably, not the same constant as the one you'd get for the shadow of a square. I say arguably because if you define the probability distribution p(θ) in a convenient way, the constant *can* be the same one. That is, using the same argument that a square has a 1/6 chance of being any given face of the cube, and modelling the distribution accordingly. Say that p_1(θ) is the base probability distribution of one of the faces, it's possible to define all p_i(θ) distributions, then average them for a "general" distribution. For the previous example it would look like this: > Let p_i(θ) be the probability distribution for "face i" of the cube. > p_1(θ) = { *1* if θ=0; *0* everywhere else} > p_2(θ) = { *1* if θ=π/2; *0* everywhere else} > ... > p(θ) = Sum(p_i(θ))/6 ; i∈{1,...,6} => p(θ) = { *1/6* if θ=0; *4/6* if θ=π/2; *1/6* if θ=π; *0* everywhere else} I personally don't like this too much, as I believe that the distribution should determine the position of the cube's faces, not the other way around. Still, the result is consistent, so there you go. What's cool is that from here it's very easy to point out that "uniform distribution => p_1(θ) = ... = p_6(θ) = p(θ)" which gets rid of the geometric dependency of _c_ and allows you to keep going from there. It's longer than assuming uniform distribution earlier on, but it's a valid alternative so that's cool.

I seem to recall this was a thing that Richard Feynman was infamous for: he would listen to a discussion of an abstract calculation and surreptitiously do it with some real numbers to see whether it made sense. The example which comes to mind was when a suggested formula for the radius of the observable universe would have come out to about half an inch or so… It's years since I read his book, so the details are fuzzy, maybe someone has their copy to hand to double-check me?

@@PhilBoswell that's fascinating.

15:00 UTC?

@@zbieramnakartonowyprzycisk8026 yes

@@ardaehi well, I can't disagree

Just btw, depending on how much drag variation between orientations you have in your cubesat (if it's a 3U, quite a bit), this problem is not a great analogue. The amount of time it spends in any given orientation depends on the aerodynamics of your satellite. Even in a 1U will likely have some preferred orientation depending on the center of mass placement. That said... it's a cubesat, if it lasts more than a year, it's probably done whatever you sent it up there to do.

@@JMurph2015 We were specifically targeting proximity operations with our 3U. The simulations I was running showed how long we could tumble before we were too far apart from our cohort to get back in formation (utilizing the 1U vs 3U faces) before the orbit decayed.

Schoolwork (and most achedemia) is geared around Bob. Alice's of the world get shoved to the side, much to our detriment.

It's fascinating and nothing Ive seen before. This cutting edge maths researwh is what makes your channel one of the best, if not the best math channel out there.

You can’t do it with convex polyhedra, though.

Isn’t the limit increasing the number of sides of one (regular) polyhedron? The complaint was that the previous sum over faces operator cannot be exchanged with the area operator for spheres because spheres don’t have faces. So the limit was just necessary to show that yes, the same area and sum operators can be exchanged even for spheres by showing spheres are infinite sided regular polyhedrons, not 0 sided shapes. It shouldn’t matter that dividing up cubes don’t become closer to a circle

I guess I don’t understand how dividing the cube into smaller cubes becomes a sphere in the limit

While you have a valid point, the tiny cubes stuck together to form a jagged surface, and they do not form a convex shape.

​@@coolcax99 Imagine the 2D case of a circle in a square. The circle's perimeter is 2πr, by definition, and the squares is 8r. If I take off the corners of the square, up to the circle's circumference, I'm left with a dodecagon that looks like a "+" with a bit of thickness, with a circle at the centre of the plus. The perimeter of the new + shape is still 8r. If we keep cutting off corners, we get the same perimeter 8r each time. In the limit, it **appears** as though this jagged shape approaches a circle, and its perimeter is therefore 8r. Therefore, 8r=2πr, so π=4. Yoav is extending this to the 3D case (which would get the bogus result π=6) and using it to poke holes in the proof (which is good practice), but as Vaibhav points out, it would cease to be convex. The reason why the 'proof' of π=4 is false would most likely have to do with cantor sets and the fact that the resulting curve would not be smooth (I've seen the staircase paradox, but this was the first time I've heard of the π=4 paradox, an interesting thought)

@GlenXoseph Nah man I totally get you

I think I understand what you’re saying but the proof relies on much more strict uniformity than what you’re proposing. The idea that the average shadow of a face is a constant times the area of the face depends on the distribution of shadows being invariant under arbitrary orientations. That is equivalent to specifying the uniform distribution on orientation’s. The bit is at 12:30 where this assumption is used.

This is a fun challenge and an excellent video. I agree that you've nailed where the assumption is being used *first* in Alice's proof. Spinning the cube (and other shapes) around and claiming this preserves the constant is fixing the probability distribution. However I think @3Blue1Brown could've been even more devious. Imagine the cube didn't spin but was kept in a fixed orientation. Then you could still build polygons with more and more sides that still "point" in the same direction and get the approximation to a sphere used later in the video. This would give c=1/2 no matter what the initial orientation of the cube is. Now where is the probability distribution being assumed?!

No, I think it's very early in the video - when she's considering the linearity. Imagine if the probability distribution of rotations was bonkers and valued certain angle much more than all the others. It's not at "limit of this shapes is a sphere", where things break down. They break down way in the beginning, when she says that the average shadow of the single face is linearly proportional to its area. You can remember that she has a rotation matrix there. Chosing a random orientation in her case boils down to chosing a random matrix. And the probability distribution being "uniform" plays into preserving the linearity there.

@@DukeBG But how exactly is the uniform probability tied to the linearity? I don't see any contradiction in choosing one particular rotation matrix and simultaneously not assuming that this one is as likely to occur as the others. Moreover, Bob also proves the linearity right in the beginning but he needed the sphere of normal vectors for the probability to be uniform.

@@krcprc Ok, I thought about it more and I was wrong. The uniform-ness (uniformity?) of the distribution plays a role later - more when the "rows into columns" trick is happening - we take get a "random choice" of N rotations and take a limit as N goes to infinity. And while we have a sum of 6 things N times, we turn it into 6 sums of N things and argue that the limit is applied to "both sides" and we get the resulting new formulae. Well, the limit of both sides needs to _exist_. I mean that the "random choice" of N rotations shouldn't change how the sequence of an average over them converges to the limit. That's what Grant is talking about in the numberphile video - how the rotations should be invariant under symmetries. Let's consider again what exactly breaks if we make the distribution favor some rotations over the others. Like, assigning some rotations a bigger probability than others we con construct a sequence that won't converge to a limit. Sorry if this still feels handwavey.

I think this is a great way of looking at it because the distribution of normal vectors Bob integrates over is exactly that of the normal vectors on the surface of the sphere. I think this goes to show that you would actually gain more insight as Bob if the question were posed as "what is the average shadow of the sphere", but you forced yourself to go through the integral calculus. I think you would reach a lot of the insights gained by Alice along the way.

@@DukeBG I don’t see why you wouldn’t get a limit if you chose a lopsided distribution? Taking the expectation of a random variable is linear. If we let R be the rotation to be applied, which is a random variable with some distribution, and S(R(F_i)) is the projected area of the rotation R applied to face F_i , and S(R(F_i)) is also a random variable, (but this time is real-valued, rather than matrix-valued) and E is used for the expectation, then, for any finite number of faces, E[sum_i S(R(F_i))] = sum_i E[S(R(F_i))] I think the issue is in concluding that the average for each surface depends only on the area of the surface, and not on the orientation of the surface. For example, suppose our shape is a very short box, like, a 1 by 1 by (1/100) box, with the (1/100) dimension being initially the up/down dimension. Then, suppose that in our distribution over the rotations, we essentially only include the orientations which are very close to the original orientation of the box (perhaps someone has written “this side up!” On the top of the box? Haha). It is still perfectly valid to distribute the operation of taking the expectation across the finite sum over the 6 sides, but we can see now that when the distribution isn’t uniform, the average for each side will depend on the initial orientation for that side. When all rotations are equally likely, then we can disregard the original orientations of the different parts.

good catch. was going to comment the same thing. A wonderful video nonetheless.