Infinite Circles - Numberphile

Professor Holly Krieger discusses filling an infinite plane with circles - without them touching or overlapping. More links & stuff in full description below ↓↓↓
More videos with Holly (playlist): bit.ly/HollyKrieger
More videos about circles: • Circles on Numberphile
Dr Holly Krieger is the Corfield Lecturer at the University of Cambridge and a Fellow at Murray Edwards College: www.dpmms.cam.ac.uk/~hk439/
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[@numberphile]

Become a Patreon supporter by the end of this month to receive one of these: www.numberphile.com/a-prime-for-patrons

[@iwantosleepeatandruneveryday.]

wow

[@smamin6181]

I need your whatsaap numbar

[@mrosskne]

why can't a circle have zero radius?

[@charstringetje]

About that... It says "Numbers will be allocated randomly." What's the shape of the distribution those primes will be drawn from?

[@philh8829]

Whoa whoa whoa. I want a discussion on this immediately. Pump the breaks. I have so many issues with this video and it’s premise. Who can I talk to.

[@juliafloridausa]

Happy to see Dr Krieger back on the channel!

[@wtspman]

I came to say the same.

[@mihawk-yc7mx]

Krieger is german for warrior. What a last Name!

[@rollsreus2807]

Plus sounding a bit more British than before!

[@philstubblefield]

My favorite ginger mathematician, who hails from the US yet inexplicably says, "Zed"... 😁

[@jaybingham3711]

Affirmative

[@eliaslundheim1732]

Feedback: I think many of us were a bit confused about how a sphere of circles can have their poles (the two missing points) be on an arbitrary positions of the sphere. From what I understand from reading a comment, it is something to do with Möbius transformations on a Riemann sphere. I can understand that this would have been too much detail to go into, but it would have been nice to mention that this was a thing you skipped over, maybe even just some text or a footnote in the corner of the screen or something. Super interesting and good video though! You (Brady) are usually very good at asking questions about this sort of thing, and I don't blame you for missing this one (assuming you did). Keep up the good work! :)

[@rev6330]

Thank you, this is exactly the missing part of explanation I needed! I thought the poles had to be opposite (because that's what they are in the drawn example), but now I realize they can be anywhere. Let's say we take Earth, shift the North Pole to somewhere in Argentina, but leave the South Pole where it is. Now we can STILL draw latitudes between the poles, only now those latitudes are no longer parallel, but instead are further apart on one side of the Earth than on the other. But with infinitely many of them, we still hit every point.

[@deltalima6703]

Start drawing circles at one point, finish up at the other point. Thats how I would do it.

[@Nia-zq5jl]

Circles would have to have a slight (infinitesimal) tilt from its neighbouring circles if the empty points are not at the poles or something?

[@avz1865]

I think the easy way to think about it is this: A circle on a sphere can be obtained by the intersection of the sphere with any plane. Now, if you have two points on a sphere, we can take the two planes which are tangent to the sphere at each of the points. These two planes intersect at a line. Now we can sweep between the two planes by rotating around this line. This gives us a family of planes, which in turn gives us a family of circles on the sphere.

[@danielj.8876]

​@@avz1865That's indeed very clever and easy to visualize in imagination.

[@Fanny-Fanny]

I'm just impressed with how well she can free hand draw a circle. If I try it, it looks like I've dropped an elastic band on the page, had too much coffee and then tried to trace around it.

[@AdelaeR]

Stop drinking coffee and dropping elastic bands then.

[@digitalfootballer9032]

Next time, trace the coffee mug 😜

[@stevenmathews7621]

🤣

[@rosyidharyadi7871]

skill issue

[@icew0lf98]

the way to wrap the globe with circles when the removed two points A, B aren't polar opposites: construct a touching plane to the sphere at both of them, and consider the line p where the two of them will intersect (because they aren't parallel since A, B weren't polar opposites) and now for each plane going trough this line p that intersects the sphere, it gives one of the circles that cover the sphere

[@alephnull4044]

Great explanation!

[@finnwilde]

class, thank you sir

[@gabor6259]

But then on one side of the sphere the circles will be more tightly spaced than on the other side. Isn't that a problem?

[@terdragontra8900]

@gabor6259 Remember, we are talking about infinitely many circles that are already "infinitely tightly spaced", in just the same way all the points on a line segment are spaced. If you smoosh them together by a factor of 2, for example, they are actually still spaced in the same way, any two distinct circles still don't touch. Smooshing them or stretching by a factor of "infinity" would break it.

[@sm64guy28]

The fact that circles can cover 3D space was a shock, I would never have thought it was possible

[@colesweed]

The real shocking thing is that you can cover 3D space with circles that are all the same size

[@xenontesla122]

​@@colesweed Is there a construction out there with equal circles? Dr. Krieger used circles on spheres of different sizes.

[@Smithers888]

​@@colesweed Not shocked, but still excited to hear that. One of my first mental exercises after the video was "what R^n does S^0 partition?" and concluded "R^1 or greater, because S^0 is just a pair and you can parition R into pairs". One obvious way: pair non-integers x with -x, pair integers 2n with 2n+1. Then I immediately noticed this was exactly how the x-axis looked in the video (excepting a scale factor of 2). When I also noticed that "pair x with x+1 if floor(x) is even, otherwise x+1" also partions R into S^0, I started wondering if that had an analogue in partitioning R^3 into S^1.

[@digitalfootballer9032]

It's all about using a lower dimension object in a higher dimension. The higher dimension gives you more flexibility to stack and manipulate your lower dimension objects. I would suspect that you can't cover a 3D space entirely with 3D objects like spheres or you would run into the same problem you had with 2D objects in 2D space. Just my take on it anyways.

[@thenoobalmighty8790]

​@@digitalfootballer9032i can fill your moms 3 dimensions with my 1 dimension

[@zilvarro5766]

I don't really get the 3D construction. I guess the initial circles are supposed to fill in the missing 2 points of each sphere. But so far, those points were on the oppsite ends of the sphere. Aren't we missing a step here to show that for ANY 2 points we can cover the rest of the sphere with circles?

[@skallos_]

I suppose it wasn't explained properly, but eyes, that is the case. The construction can be made by "sweeping" a circle across the sphere.

[@Adamreir]

@@skallos_what do you mean by sweeping? Thanx btw! Came here with the exact same question.

[@1CO1519]

I didn't get it either 😕

[@05degrees]

Here’s a general argument and a concrete construction. 1. There are transformations of a sphere that leave circles on it as circles but can map a pair of distinct points to any given pair of distinct points. They can be realized as Möbius transformations on a Riemann sphere (complex numbers plus an infinite point, stereographically projected onto a sphere). So we can take our “parallel circles” construction and map it into a construction where untouched “poles” are not antipodal. 2. Concretely, when two “poles” aren’t antipodal on a sphere, you just take a line outside the sphere and slice the sphere with planes that go through that line. Each intersection will give either a circle on the sphere or a point; the pair of points closer and closer together if the line is closer to the sphere. If you look at this construction in projective space, you can make the line infinitely distant. Then planes going through it will be parallel (in the usual non-projective space) and we retrieve the original construction with antipodal “poles”.

[@fowlerj111]

I guess it would have been too much detail for the video, but here's a slightly more formal way to describe it. Any plane intersecting a sphere either defines a point (if they're tangent) or a circle (otherwise). For the case where we're leaving out the poles, think of a plane tangent at the top, a plane tangent at the bottom, and all the intermediate planes between and parallel to the first two. Now for leaving out two points not opposite each other, think of the planes tangent at those points. They will intersect in a line outside the sphere. Think of that line as a hinge. Now the intermediate planes also go through that hinge line. They swing from one point to the other. Each intermediate plane defines a circle on the sphere, none of the circles intersect, and every point on the sphere is on one of the circles, except for the two original points on the tangent planes.

[@smoorej]

Thank you so much for bringing Dr. Krieger back to Numberphile.

[@WiseSquash]

insane circles drawing skills 😮

[@SlipperyTeeth]

Some viewers may not understand why the 2 points missing on the sphere can be moved around - why they don't have to be opposite each other. So I'll fill in that detail with a geometric argument that isn't too advanced: Circles on a sphere can be thought of as the intersection of the sphere and a plane that's not tangent to the sphere. For the diagram shown in the video, draw a straight line between the 2 points that don't get covered. Notice that the line is normal to those points on the sphere (perpendicular to the tangent planes of the sphere at both points). And notice that the planes we'll use to define the circles are all the planes that are normal to that line. So in the general case, for 2 arbitrary points on the sphere, draw a curve that's normal to those points on the sphere (make it a nice curve - not self-intersecting, smooth). And then the circles are those formed by the planes that are normal to that curve. There is one more detail we have to consider - if the curve isn't a straight line, then the planes will intersect with each other, since they're no longer parallel - but this is fine as long as they don't intersect ON the sphere, they can intersect without making the circles intersect or touch. If you're good at visualizing, you can imagine the circles being swept along the curve without intersecting - the curve acts like a hinge to move the circles through.

[@amirfeyzi566]

After a few attempts at drawing, I was able to get a visual that makes sense to me. My current best description is a paper fan that swats through the sphere.

[@gabor6259]

But then on one side of the sphere the circles will be more tightly spaced than on the other side. Isn't that a problem?

[@SlipperyTeeth]

@@gabor6259 Short answer: no. The problem was only to cover a sphere with 2 points missing by using circles that don't touch/intersect. The concept of "density" for the circles was never defined/needed. What you are describing would be a different problem. Here are some thoughts related to that: There is a sense in which the density is different for spheres with different missing pairs of points, but there is also a sense in which the density is always the same. On the sphere, around any circle, there is always a continuum of other circles. This might be easier to see on the height of a cylinder. Imagine the height of a cylinder being covered by circles in the most obvious way. Now imagine a line going through the cylinder in its center (parallel to the height). We can project the circles onto the line (imagine that each circle is equivalent to a point on the line at the same height). Then around each point, there is an entire continuum of points (a connected section of the real line). So in an analogous sense, around each circle, there is an entire continuum of circles. Consider that same setup, but this time make the cylinder have a height that ranges from 0 to 1. Now imagine a transformation of the cylinder that sends the height at value x to the height at value x^2. This transformation sends us back the exact same cylinder, but in a sense, it squished everything down. Although the circles would get packed down closer together, we can do the exact same projection onto the points on the same line as before and we see that around each point/circle, there is still an entire continuum of points/circles. This is related to Order Theory (and also Set Theory). In this sense, on the real line (and on stuff that can be made to look like the real line), density doesn't change. On the other hand, in Measure Theory, we can give a "size" to subsets of the continuum and say that although around each point is a continuum, they are different "sizes" of continuum. We can use this to compare the size before and after the transformation. Now to relate this to our sphere problem. We again project our circles onto the line (the one we used to act as a hinge for the planes we used to define the circles). So in the Order Theory sense, the density is the same everywhere. Now, if the density of this line varies (in the Measure Theory sense), we can always apply a transformation on the line (which is the same thing as applying a transformation on the sphere) that makes the density of the line constant. So even in that Measure Theory sense, there is a way to say that the density is the same. But, if I had to guess what you're really talking about: There are other projections you can do besides onto the hinge. If you consider the great circle that goes through the 2 missing points, each of the arcs (separated by the points) defines a separate projection of the circles on the sphere. While each of these projections can be made to have constant density, you're probably thinking about the more obvious fact that each of these arcs have different arclengths. So in this sense, one projection is more dense than the other - one side of the sphere is more dense than the other. Technically, this is still malleable, because measures are malleable - you could "artificially" say that the 2 arcs have the same arclength/measure and be done with it. But we really don't want to say that. There's something in us that screams out "One of the arcs is clearly larger than the other". It's our evolutionarily induced concept of distance/perception. Adhering to that, only the case where the 2 points are opposite yields equal density on either side, as you said.

[@johannschiel6734]

Thanks, was wondering about that detail, too. Came to some similar conclusion, but your description of the solution helped me a bit visualizing it.

[@TheCdizzle163]

Nice video. Can you explain why the point in the center does not meet the definition of a circle? As you say, "a circle is just the collection of points in a two-dimensional space which are some fixed distance away from a center". Is there an additional requirement that the distance is greater than 0 and/or does the word "distance" imply a non-zero value?

[@galoomba5559]

For some reason it wasn't explicitly mentioned, but yeah radius-0 circles don't count or else the problem would be trivial.

[@mrosskne]

@@galoomba5559 Then the problem is trivial.

[@TicoTimeCR]

I'm going to take a stab at this, not idea if it's correct. A point, by definition, is simply a specified coordinate with no dimensions. It fills no space, it's just a location. A circle (or square, or line, or any other shape) is an object that consists of a theoretically infinite number of connected points. A circle with a radius of 0 would consist of 0 points and would be indistinguishable from empty space. You cannot fill empty space with more empty space and consider it "filled", as the original problem required.

[@mrosskne]

@@TicoTimeCR Sure you can. That's just a space filling curve. The real number line is nothing but an uncountably infinite number of points of zero area, and they fill an infinite length.

[@galoomba5559]

@@TicoTimeCR Space is a set of points. You can definitely fill space with points, as long as you use uncountably many.

[@alexhawco2970]

When explaining why this doesn't work for the 2D example and how the centre of the circle is the sticking point Dr Krieger says 'at the end of the process we're going to end up with a single point here, just like in the original example, that cannot lie in any of the circles'. I don't understand why there has to be an end to the process of adding more and more circles here - after all there was clearly no end to the process of adding concentric circles outside the original circles to cover the exterior, and that wasn't an issue, and there are an infinite number of points inside the circle as well, so we're not going to run out of points to run circles through. Can anyone explain what I'm misunderstanding here?

[@wtspman]

A point is dimensionless. Any circle you can draw, however small will always be larger than a point. And by the definition of what makes a circle (as Dr Krieger says in the video) there will always be a point at the centre of the circle, equidistant from all points on the circle.

[@praxiquot]

I think the idea is that when you're expanding outward, you can always provide a valid circle that covers any given point. Moving inward, per the constraints of the problem, no valid circle can ever occupy the origin (0,0), even in the infinite limit. It's less about "reaching" the middle as it is that there is no solution in which the middle is covered with concentric circles.

[@Jimbaloidatron]

But if a point is dimensionless, surely it is nonsense to consider filling it? What I think we are discussing here is that a circle must pass through every point, hence the effort to offset from the centre.

[@snarlbanarl1940]

As I understand it, the limit of centre points lies inside all the circles constructed through this process. Since the radii of the circles tend towards zero, that means that the limit of centres lies inside a circle with arbitrarily small radius. This means that if there was a circle that that point lies on, whatever radius it has, there is always a circle which contains that point which is smaller than that circle, so they must intersect.

[@MrBluelightzero]

It might be better to turn this around. Instead ask "Given a point, which circle does it lie on?"

[@katczar]

Welcome back Dr Krieger!

[@dave20874]

Wonderful! Bravo for bringing Professor Krieger back to Numberphile.

[@Mrfailstandstil]

drink every time you hear or see a circle in this vid and you'll get alcohol poisoning after around a 2-nd second

[@zaclaplant3001]

The way the dilemma is worded seems odd... The circles are not allowed to kiss, therefore cannot be tangent to each other, therefore, by definition, there is space between the two circles which can be infinitely divided... even if the space is infinitesimal. You have axiomatically made the game impossible. That's what happens when you try to establish a difference between tangential and infinitely near. They are different in concept in how you think of them, but they are the same in application. It's like saying you can never fill a glass with water because atoms and molecules are mostly empty space

[@therealax6]

I think the formal statement is defining a set of circles so that every point in the {plane|space} belongs to exactly one circle in the set.

[@JWentu]

I am so happy Dr. Krieger is back!

[@RJSRdg]

Great to have Doctor (now Professor!) Holly back - more please :-)

[@sebastianzander87]

I love the new "analogue" looking animation style with that nice brown paper texture and the hand drawn lines ❤ Very nice.

[@kenhaley4]

I think an important point was glossed over. My first instinct was to think there must be gaps between the concentric circles. But then I realized this: For any point, consider all possible circles of radius n centered at that point, where n is a positive real number. No two circles would touch each other, but every point in the plane would lie on one of these circles. This means not only is the number of circles infinite, it's uncountably infinite! With that in mind, the point of the video made more sense to me.

[@AthAthanasius]

Yeah, I had a "hang on a minute!" thought part way through the video. The 'line' of a circle has zero width, so it's not actually covering *anything*. You are, indeed, going to have to use an infinite number of them in order for their zero/infinitesimal line-width to collectively cover anything. Really all that's happening is defining a set of circles whose points leave no point in the plane unaccounted for.

[@SeanTBarrett]

Yes, I think the hand-wavey way the circles are added gives the impression of countably infinitely many circles, when uncountably infinitely many are required, and failing to explain this and all the other complexities of infinity makes the whole thing a lot harder to understand what's going on IMO.

[@EdLeeSB]

Hello @KenHaley4, thanks for your comment, because I also had the same “Hang on” moment as yours right after the start of the video. Your comment helped me understand why the original concentric circles in the 2D case could cover up every point (except the center). However, I’m not sure about: “for any point, consider all possible circles of radius n centered at that point”? The original “simple” concentric circles alone are sufficient to cover the entire 2D plane (except for the center); it seems these “new” circles only introduce “more new gaps” than without them. No? Thanks.

[@BujuArena]

This video makes no sense. They don't define what "covering" means. Infinitely many circles of radius epsilon (essentially the smallest non-zero number) "cover" the infinite plane, but even with any stroke width, you'd still need infinitely many circles of radius epsilon to "cover" the gap between any two non-infinitesimal circles, so what's the point of the video? The "solution" is dependent on the rules: that it's either impossible if circles of radius epsilon are not allowed, or possible if they are allowed.

[@olaf7441]

​@@BujuArena"Covering" the plane means: for every single point on the plane, there must be (exactly) one circle which contains that point. You can't use circles of radius epsilon because no matter what the circle's radius is, I can always pick a number x that's smaller than that radius and ask "is the point with coordinates (0,x) on this circle?", which it won't be.

[@jaminpeterson5171]

The part I feel like I'm missing is that the two holes in the sphere don't have to be at the poles. It seems like the fact that there are two and only two holes and insuring that the spheres keep hitting the circles exactly two times is important but I'm lost as to how those holes in the sphere get moved around.

[@madsoundcaddiehat]

This is where I realise that I'm not very smart because I thought it was easy, depending on one of two things; either you accept that it is infinite, or you define the resolution you're working to and at some point the smallest circle you can draw will fill the centre.

[@robshaw2639]

At first I thought there was a problem when the red sphere is tangent to a black circle, but now I see for those spheres, they are also tangent to a 2nd black circle, and so still have two points removed... partitioning R3 into circles is quite unexpected...

[@Calmerism]

To my simple mind even the basic idea of covering an area or a 3 dimensional space with something that has no width is mind boggling. Nice video!

[@vik24oct1991]

It has a width which is infinitesimally small, not zero, infinitesimals cover areas all the time in basic calculus.

[@joelbraun8584]

@@vik24oct1991I sympathise more with confusion or awe than this half-baked explanation. In what precise sense does a circle have an infinitesimal width, as opposed to zero width? Width could be crudely described along the lines of the length you can move orthogonal to the boundary, starting from a point on the boundary, until you exit the object (ie the width of a rectangle from a given point would be the length of one of its sides) By this definition (and any other definition that takes values in R), a circle would have 0 width. The reason we can cover the space is because we have an uncountable number of these circles - countable unions would preserve the measure 0, whereas these uncountable unions manage to allow area to be generated from area-free shapes. I think intuitively this should make sense as in general we can construct n+1 dimensions from n similarly, by stacking planes atop each other.

[@aachucko]

My Christmas present came early! Her laugh is like a warm blanket. 7:44 is when she lost me.

[@oncedidactic]

Dr Krieger and sphere packing, I’ve died and gone to heaven 😅

[@bectionary]

How do you get from the sphere with holes on opposite ends having a valid circle covering, to spheres with any arbitrary surface-placement for the two holes having the same?

[@nahblue]

Take the original sphere with holes on opposite ends, join the two holes with a straight line. The line is where the center points of the covering circles will be. Now imagine that you move the holes and that straight line smoothly deforms into a curve that still joins the holes. (And this curve guides you to where to place the circles.) Not sure if that visual feels intuitive, but maybe it helps.

[@chrslb]

Amazing video! Thank you and what a neat idea

[@BethShep1998]

Not related to this video, but I was talking to someone about deal or no deal and the monty hall problem. Should you switch boxes at the end of deal or no deal? Is the probability that you have the larger amount 1/2 or 1/26 (like the probability of having the door with the car if there were 26 doors then 24 with goats were revealed to you)? If the end conditions are basically the same, does it matter how the intermediate stages occurred (opening boxes at random vs someone who knows where the goats are revealing them to you)? Is there already a video like this?

[@TheReligiousAtheists]

This is a really interesting overlap between topology and elementary geometry!

[@phatst3r]

honestly that all went over my head :(

[@johnchessant3012]

That was an awesome solution in the 3-d case!

[@ladylhi360]

Clearly this video has everyone thinking about how to ask the most useful questions in math and what are the most useful explanations. It’s a great jumping off point that has peaked my curiosity! So which masterful, math KZbinr is going to make the hour-long video explaining this more thoroughly? ….. (I don’t have the skills, but I am curious and would definitely watch it!)

[@d4slaimless]

Now can someone explain how you partition the red sphere that is shown at 2:25 so there are to points on the side (where it intersects black circle) going to be left uncovered? On the example sphere we have 2 poles left uncovered - this is easy to understand. How do you get 2 points that are not on the opposing end of the diameter?

[@dewaard3301]

We haven't seen Holly Krieger in ages, right? Welcome back, I'd say.

[@DeathlyTired]

Feels like a perfect segue to cover circle packing, and thus do another video on Origami Maths.

[@nahblue]

For the 3D case, how do you handle the sphere kissing one circle exactly in one point? Is it also tangent to another circle in exactly one point in that case?

[@voinbobar]

I always knew Amy Adams was talented in all possible dimensions, even drawing perfect circles.

[@fierydino9402]

How is it used in other fields or in practice? It is very interesting so there may be a marvelous use of it!!

[@rossholst5315]

What if we invert the plane? Where we want to move the points that lie in the middle to the points that will now make the edge? So now all the points that were initially infinitely far away they are located in the single point in the middle of the plane, where all points that were in the middle are now on the outside of the plane. Apparently since going to infinity wasn’t a problem for making circles, we just need to invert until all of the points on the circle at infinity come together to create new origin point. Or you could rename all points 1 radius away from the circle 0. So a circle that previously had a radius of 1 will now represent a circle of radius 0. Any circle made within this circle would represent circles with negative radius values, and have negative surface areas.

[@kaushaltimilsina7727]

Nice informal intro to/demonstration of homotopy groups and weak equivalence.

[@Vannishn]

Nice video ! Are we talking about Hopf fibrations ?

[@johnferrara2207]

Welcome back Holly Krieger!!!

[@robertma6068]

This is perfect! I just bought the original spirograph kit today! I'm taking geometry in my math class right now. I'm shring this with my teacher. I'm interested to see how this might intersect with my spirographing :)

[@robertma6068]

I brought the spirograph kit to math class and my math teacher loved it! We had a blast learning about circular relationships!

[@adamcetinkent]

Was the answer "not at all"? 😂

[@Yzjoshuwave]

Interesting… But do we need to make another stipulation that we can’t have a circle with an infinitesimal radius? It seems like that could be more or less equivalent to an infinitesimal point.

[@turtlellamacow]

The radius can be arbitrarily small, but it has to be nonzero. Yes, it should really be stated explicitly, but it's taken for granted since otherwise every point would be its own circle and the problem would be trivial in any dimension.

[@dioptre]

The cutest mathematician on this channel

[@UnimatrixOne]

Sehr truely is❤️

[@rishigauswami1987]

what i don't get is what you mean by filling up? we are considering a covering that is dense (topologically) in the space (or plane), or we want it to be continuous?

[@MasterofNoobs69]

Continuous

[@sgrass111]

Yeah the problem’s definition is not clear

[@cassiemancassie]

They simply want to write R^2 as the a union of circles. You could formulate this as wanting a bijective map between R^2 (or R^3 in the second case) and a disjoint union of S^1 which is continuous (which under the right topology means continuous in every component), and such that the preimage of any of the S^1 is also a "circle", so of the form {y in R^2 : |x-y| = r}. Although actually, you could probably lose the last condition and still get the exact same results.

[@Matematikervildtsjov]

I think the definition is surely that every point in space lies on a circle. So a dense subset i not sufficient. And I don't know what you mean by continuous, it is a set of subsets covering the space, no notion of continuity is needed here.

[@xavierstanton8146]

I was also curious on the topological aspects.

[@languafranter3450]

It's more like a topic in point set topology, and I really enjoy this Intuitive fact about covering space with 1 dimentional spheres :D

[@iagocasabiellgonzalez7807]

1:11 if you go in the outside like in the inside, doesn't infinity relate to the problem at the center as being unreachable too?

[@user255]

I did not get this. Either the circles touch each other (forbidden) or there is empty space between them (therefore not filled).

[@ilyrm89]

Can we please appreciate her free hand circle drawing? Damn!

[@keescouprie5968]

Does this "covering the area with circles" even make sense, considering that the lines that make up the circles have zero width?

[@violetfactorial6806]

Instead of thinking of "covering", think of it as "defining" points on the 2d plane. A circle is just a group of points that satisfy a particular type of equation, and the 2d plane is just a larger group of points. So there's no "width" to worry about - a circle and a plane are both groups of points. The idea in the video is basically asking, "can we group all of the points on the plane into distinct, unique groups, such that each group can be fully described by a circle equation?" And the answer is not in 2d space, but we can do it in 3d space. Which is really weird and unintuitive.

[@keescouprie5968]

@@violetfactorial6806 that still doesn't make sense. There is no such thing as "all the points". We're not talking about pixels.

[@mayhemnc]

the goddess of math is back!

[@ghislainbugnicourt3709]

There appears to be a symmetry in the limits of this thought experiment if you allow : - circle radius=0 (then the central circle is a point no problem) - circle radius = infinity (then you cover the plane with straight lines)

[@JeffACornell]

Since the original problem statement didn't exclude circles of radius 0, the solution is actually trivial.

[@jakobthomsen1595]

In case a line counts as a circle (limit when radius goes to infinity) a simple construction to fill space with circles would be: (1) Use circles to cover a sphere without two points (at north pole and south pole). (2) Fill interior and exterior with scaled copies of this hole-y sphere, so that the holes form a (missing) line. (3) Fill the (missing) line with the infinite circle. Would be even simpler but more boring starting with the infinite-radius circle line and wrapping circles around it forming layered cylinders.

[@fowlerj111]

A lot of times problems are formulated precisely in the way that makes them interesting. "If lines are allowed it's simple, but is it possible if lines aren't allowed?"

[@dieterneumann7291]

Holly Krieger is back and my phone running low on battery😭

[@venkybabu8140]

Partition means numbers catagory. Some fall some don't. Depends on how you write a equation. Logic paste. Circle means number group. π x/y. Waves are considered number groups and they always show duality.

[@Firefoxav26]

I’m sure the intention was not to lose us when you went to three circles and a sphere, but the bus left me at the station 😅

[@notanotherpeterzhang7582]

That was a nice illustration

[@kikivoorburg]

I assume that radii of 0 (giving a point) and “infinity” (giving a line) aren’t included as valid “circles” - otherwise the problem is definitely doable and trivially so lol. If infinite radii are permitted (but 0 still isn’t), I have a potential solution for 3D. Tell me if you find some mistake in my logic: Take a circle lying in, say, the xy plane. Pick some point on this circle - say the point on the +x side. Make this the “limiting centre” of a bunch of circles in the xz plane that each have their centre offset a bit, with the radius constantly growing. (Basically the reverse of the proof in this video for why passing circles through consecutive centres doesn’t work - except this time the “limiting point” is covered by our original circle). Specifically make it so that the origin lies on a circle of “infinite” radius (a line). The image would look very similar to the equipotentials of an electric dipole, but only half the xz plane would be covered. Finally, just rotate the half-plane version around, making each point on the xy-circle a centre for a similar group of circles! I think the final solution would look like the magnetic field lines around a circular wire. Again though, it relies on allowing that central line being considered a “circle”. It’s cool that there’s a solution which doesn’t require that!

[@markphc99]

so a circle cant have a radius of zero, or a sphere , etc

[@dwall2]

well it would exist if it did lol

[@PhilBoswell]

No, because then it's a point.

[@russellbeaubien7430]

I have a question. Which can plain the 5th dimension best (or easiest), circles or spheres?

[@notnek12]

1:09 probably a stupid question but don't the circles drawn outside the circle overlap with the first circle? I thought overlapping was part of the restriction?

[@batlrar]

Not a stupid question at all! You're thinking of discs, where the inside of a circle is a surface, but she's referring to circles where only the drawn part is considered the shape. The reason that you're able to (theoretically) cover the whole paper with them other than a center point (and the edges of the paper, but I think she means an infinite plane) is because you can draw infinitely many of them between any two concentric circles. Similar to her example with the globe and the different parallel lines of latitude she shows, you can draw one large circle and one small circle that share the same center point. Then draw the circle that would be exactly halfway between them, and keep drawing the circles that are halfway between those until you've done so infinitely.

[@colonelburak2906]

Since the circles just are the 1D lines, there's no problem. Concentric circles do not overlap. You are probably thinking about the *discs* (interior of the circles) overlapping, but that's allowed by these constraints.

[@notnek12]

@@colonelburak2906 thanks that makes sense.

[@Nethershaw]

I kind of fell in love with the view out the window in the background.

[@charstringetje]

This makes me think of two things: kissing numbers and the optimal packing of conical glasses in the cupboard when emptying the dishwasher in the office, to fill the time I wait on my coffee.

[@michael169chapman]

Amazing video

[@limitlessLtd]

im not sure i understand this one, doesnt the line that denote the edge of the circle have a diameter? And why cant you just reduce the radius of the circle by the smallest possible amount and use it as the same center point?

[@olaf7441]

Mathematical lines, circles, etc. are deemed to have no width, unlike the physical drawings of them we can make. So for example the circle centred at the origin with radius 1 contains the point (0,1) but it doesn't contain (0,1.0000001) even if you add a billion 0s in there.

[@algolin]

"Do not disturb my circles!" - Professor Krieger, possibly.

[@karthisscdm]

Great work

[@ShakalDraconis]

There seem to be a couple unclear definitions with this exercise. How are we defining a circle? Specifically, wouldn't a circle of radius 0 cover a single point? Are we defining circles as having to have a radius greater then 0? If the circles "cannot touch", does that not inherently require that there must be some empty space between them, and thus even the nesting circles idea for covering the space "outside the circle" fails to cover the plane?

[@killerbee.13]

Degenerate circles (with zero radius) are usually implicitly disallowed by mathematical constructions because they behave in ways that are both boring and inconsistent, so yes, r>0 is a constraint. If r=0 is allowed the problem is trivial (place a circle of radius 0 at every single point) and nobody enjoys solving a trivial problem. The mathematical definition of "covering the space" (or at least something logically equivalent to it) is that, for any point you can identify, it is on some circle. Covering without overlap means that the construction must assign exactly one circle to each point, not two or more. So any individual point you identify in the gap between two arbitrary circles has some other circle passing through it. Just like how there's a distance between any two nonequal real numbers, but there are still real numbers 'covering' the entire 'number line' without gaps.

[@enricosilvestri224]

I didn't know Amy adams retired from acting and started teaching maths

[@todddembsky8321]

Good to see Holly again on the channel. Now, what if you used a circle that was twisted into a figure 8?? On the option of removing two points of a sphere -- the sphere is no longer a contiguous surface. If I can remove points on a sphere, why is the option to remove points from a circle invalid?

[@fowlerj111]

The problem statement doesn't require that the function from points to circles be continuous, only that each point is on one and only one circle

[@killerbee.13]

The problem is not "can you cover a sphere in circles" answered by "yes, if you remove two points". The sphere without poles is just an intermediate step in the filling of all 3D space with circles. So the removed points are still filled in, they're just filled in by *different* circles.

[@Ikkarson]

Uh… isn’t a point a circle with radius zero? I don’t get why a point is forbidden for the center.

[@LifelineSoF]

If you allow a circle with zero radius, you can just pick infinitely many of these and put them on every point of space in any number of dimensions and completely trivialize the problem.

[@nicksamek12]

A point is that which has no part.

[@robertolson7304]

it dawned on me. Every digget in PI is slope formula. You got 1 dimension (line math) equally is absorbed by 2D math (squared). Like 2x3+5. 2x3 gives me two lines of 3 or a told value of 6 squared. +4 (line) can be added evenly on both lines. Like 3+2. What about odd numbet? It would be like 1 split into two (.5 squared). How do you get 1 dimension to slip into 2. That in it self should cause some issues. Like anti resonance in couple oscillator. Like two spirals over laping what we have is an average. That is the only way i can think of that would produce a infinate slope that gets smaller yet resonates.

[@SmileyMPV]

Is there a set of circles such that every point is covered exactly twice?

[@SmileyMPV]

In fact i just realized all you need is a set of unit circles whose centers form parallel lines that are 2 units apart. Next challenge: cover every point exactly thrice :)

[@dougdimmedome5552]

I mean the question feels obvious if you think about equivalence relation. Just define an equivalence relation on the plane with points (x_1, y_1) and (x_2, y_2) being considered equivalent if and only if x_1^2 + y_1^2 = x_2^2 + y_2^2. As equivalence relations correspond to a partition of the underlying set, we know that this partitions the plane and particularly partitions the plane in terms of equations of circles, the only problem is whether or not you can consider x^2 + y^2 = 0 a circle and so I’m guessing obviously the question becomes more interesting if the radius is positive definite, also it feels bad to call a one point set a circle.

[@jonathanlange5769]

The defintion of a circle I know is: The set of all points with the distance x from the origin. Doesn`t this allow for x to be zero and the point being a circle?

[@2eanimation]

That's a degenerate circle(seriously, look it up), the limiting case as r -> 0. It doesn't behave like a circle but, as you mentioned, like a point. Hence, when you look up the definition of a circle, there's an "r must be a positive real number"-statement somewhere. 0 is neither positive nor negative. You can think about it another way: if you allow degeneracy, e.g. a line can be anything, say a rectangular cuboid with two side-lengths being 0, a circle with infinite radius, a cone with an angle of 0°/base circle with radius 0. Even objects in higher dimensions could be lines, or better say, a line could be an object of infinite dimensions once it collapses to a line in the limiting case.

[@TheSuperiorQuickscoper]

4:06 No "paper change" slide? Unfortunate.

[@AM-sg4zg]

I see Dr. Krieger, I press like. Life's that simple.

[@Bluhbear]

My thought about three dimensions before the video finishes: Make 2D planes as normal, each having just that one center point that isn't covered, then line all those planes up so the missing points form a circle in 3D, then draw that circle through them. Then just repeat that forever.

[@JustinFriello]

My favorite is back!

[@curtiswfranks]

That was an interesting construction!

[@LuckyLuke79a]

It's been a while!

[@TheOriginalJohnDoe]

Only mathematicians can make an easy to solve problem impossible to solve

[@Cernoise]

Something about Holly Krieger’s voice or accent or laugh or mannerisms (or maybe hair?) always reminds me of Marian Call.

[@ManSubhu]

I don't get it, for expanding the ciricles outwards; for the circles not to be touching there has to be a point between them, if there is a point between them then the space between them is not filled.

[@dalekerr5091]

More Holly please

[@nigh7swimming]

My fav mathematician!

[@jamesroseii]

Ok, so remove 2 points on 3d space to cover the circle, but we don't remove any points in 2d? Also, isn't the center of the 2d circle analogous to the 2 points that we removed in 3d space? Am I missing something? Why the different rules?

[@semitangent]

You don't remove the points from 3d space, they still have to be covered, and the are - by the flat circles. You remove the points from the sphere, but because of the flat circles, that isn't a problem. Removing points in 2d actually changes the space and that's not happening in 3d space, there the whole of 3D space can be covered.

[@transientaardvark6231]

Given that you're quite happy to have an infinite number of infinitely thin circles infinitely close to one another without counting them as touching, I have to say it feels a bit arbitrary to say you can't have a zero diameter circle at the middle of the first 2D circle.

[@Jaggerbush]

That first circle was damn near a perfect circle!

[@justb4116]

To everyone explaining how two taken out points on the sphere can move around and stuff... I'm still stuck at the "how can we take out TWO inconvenient points from the sphere but not ONE on the flat circle" 😭 any tips what am I missing?

[@TheRealSeus]

i like the Map of the Mandelbrot Set in the background 😃

[@Nopillcaddy]

Yay Holly Krieger!

[@unvergebeneid]

So the circle must have r>0? Was that ever explicitly stated?

[@charlieangkor8649]

The center point is simple a circle radius 0. Circle is points with the same distance from the centre point, if you set the distance to 0 you get just the centre point.

[@susanwoodcarver]

Absolutely fascinating, but I found myself wondering why I would want to cover a surface with circles. If anyone knows the practical application, besides this being fascinating, could you let me know please. Thank you.

[@x022zdd]

There doesn't have to be a practical application!

[@COLATO_com_br]

well done !

[@benjaminpedersen9548]

{{x-1, x+1} | x \in \R} covers the real line with spheres from 1d, so the sequence begins 1 3 5.