How many ways can circles overlap? - Numberphile 

From this video... 3-Circles T-Shirt: teespring.com/three-circles-numberphile 4-Circles Poster: teespring.com/four-circles-numberphile Other merch: teespring.com/stores/numberphile

Continuously transitioning between the 16951 combinations of 5 circles could make for a nice screensaver

4:38 *whispers* "Look at these. This is a very delicate...configuration..."

Today I'm sleeping early 3 am: _in how many ways can circles overlap_

4:34 I like how Neil whispers really gently here, as if to not disturb the very fragile circle arrangements

“Jonathan wild took this series out to 5” Oh that’s actually not that impress- “3: 14, 4: 173, 5: 16,951” Oh never mind

Did NOT at all expect hear that my classical sonata form professor is actually the goat of geometry!!! Very fascinating and I’m now even more a fan of Prof. Wild

The 4 Audi and 5 Olympic rings cannot be formed because of copyright violations.

"This cannot be realized by circles, it can be realized with ellipses" Time to recalculate all of these numbers but where ellipses are allowed

Whispers "this is very delicate" as if speaking loudly will blow the circles away. Lol

I like how this video went full circle.

4:30 sounding like a mathematical Attenborough

"I want to tell you about a really lovely problem..." Well I was doing something else, really, but now i'm going to stop and listen to this amazing man.

Well you have a _soft_ lower bound for 6. It's just 5's count.

Dr Sloane has one of those wonderfully relaxing-yet-compelling voices that are just ideal for podcasts.

"The answer for one circle is one. For two circles, three ways. What about three circles?" Me: Oh no, it's TREE(3), isn't it?

This is pure asmr. With the added bonus its really interesting

Neil Sloane. Always interesting stuff. The man always has a way of pulling me into something I would not otherwise find interesting. A lot of passion there.

I love Neil's passion for mathematics and his joy in showing us some of it. But for this video, the comments are just as gold.

This seems like it could work really well for some kind of fictional writing system. It almost reminds me of the one written language from Myst that used specific segments of circles to represent words. (Arayani, if I'm remembering right)

When a child paints supposedly random circles: "Oh look, it drew one of the countless configurations from n circles!"

1:58 Nice, Gray made a cameo!

How much paper do these guys go through

3:57 OEIS, On-Line Encyclopedia of Integer Sequences. A beatiful pleace on the internet, full of things all numberphile fans will love.

@8:19 I see a triangle and a hexagon wearing a bikini.

we do have a lowerbound for 6. it's atleast the total of 5 times 2, since we can take all configurations and put a circle next to it and a circle around it.

This is one of my favorite videos in a long time. Very beautiful.

Wouldn't a lower bound for the number of configurations be at least 2 times the previous number, because you would simply have all the previous configurations with another circle next to it, and all previous configurations with a circle around them?

Thanks for the animation efforts

Lower bound could be defined in terms of the previous value. but if you wanted to calculate it, you could get a rough estimate for a lower bound, let's say 2x the previous value, simply defined as the entire set inside of the new circle, and the entire set excluding the new circle. The value has to be greater than this, as there are other places that you can put the new configurations.

I haven't watched the video and I'm already excited, I love Neil's videos. Thanks for helping me end this crappy week with a smile.

9:17 - So is that the music of the spheres~?

I get hyped every time they upload a video

No one: Mathematician: "how many ways can I overlap circles"

I love your voice, & thank you for sharing this beauty. I'm going to have to look up affine plane again. I'm sure it made sense to me at some point in the past, but these days I only understand projective, as I've been obsessed with the projective plane for so long.

Thank you for publishing this video, I have been thinking about weary similar problem for two years. Now I know that I am not alone.

4:31 Neil Sloane turns into Bob Ross. 4:44 Neil Sloane is Neil Sloane again.

You're well on your way to becoming an osu! Rythm champion

It'd be really cool to watch this in three dimensions (with spheres). I'd love to see that in a collab with @3Blue1Brown... just sayin'

7:50 "Maybe you can draw it in more than one way." What's an example of two dissimilar arrangements with equivalent boolean truth tables with 5 circles? Does one exist for 4 circles?

A part of me asks, why would anyone wonder how many ways a circle can intersect, but a bigger part of me is impressed that people can think of this.

4:35 ASMR

Great video. One example (with only 3 circles) where there are two inequivalent arrangements with the same "truth table" situation (combinations of being inside or outside each circle that exist in the arrangement) is that the 14th (last) arrangement Slone drew is equivalent in that sense to the Venn Diagram, but in the 14th arrangement there are two non-contiguous regions where you are in one of the circles (the middle one) but neither of the other two.

I just love this guy’s quirky, quirky office

I loved this video, also realized how the configuration for 3 circle is 14, first 3 digits of pie. 🙌

4:50 ok I can't think of an upper bound but a lower bound is easy 16,951 you could take any arrangement of 5 an add a circle off to the side. If you want a bigger one then imagine the circle is arbitrarily small and could fit in any of the sections that the plain is cut into by the circles. Well for each configuration of 5 there are at least 6 sections (the reagon touching the inside of each circle and the outer reagon) so that makes a lower bound of 101,706 but you could also have a circle that surrounds the entire thing increasing the lower bound to 118,657.

Great Video aus always! I just love the voice of this gentlemen! Please do more interviews with him he seems like a very interesting character.

This is exactly the content I subscribed for! I love the featured number videos, but when it boils down to modular/base arithmetic I find it so much less interesting than the elegant, complicated, and unknown curiosities such as this. Nonetheless, thank you for this amazing channel Brady!

I’ve got an elegant proof for this one, but this reply box is too small to contain it.

I love the UNIX book behind Mr. Sloane :)

This is really cool. I think the first step to figuring out how to take this further is to simplify what we know as far as possible. That chart of the configurations of four circles is painfully asymmetrical and makes random choices for the sake of visuals rather than uniform ones

We do have a lower bound for 6 circles; there have to be at least as many as the previous set, since you can just put another circle next to the previous configurations. You could also completely encompass the previous configurations, so really you can say that the next set has to have at least twice as many as the last set. TL:DR the lower bound for 6 circles is 33902

Neil Sloane (is that his name?) has the mind for math and the voice for story telling !

"We don't have a lower bound." Of course we do. Just make an arbitrary number of arrangements (e.g., 0) and you now have a lower bound. A slightly less useless lower bound is 16,951 (draw a circle around every arrangement of 5.

There are at least 118657 combinations for 6 circles. One for the 5 case surrounded by the 6th, one for the 6th being completely outside, and the others being the 6th combined with each of the other 5. I tried to find a way to represent each of the combinations uniquely, and while doing so found an upper limit of 2^(2^(n)-1) or 9223372036854775808 combinations for 6 circles. You can represent each of the combinations by describing the unique areas; as an example for the 3 case, 1: A,B,C 2: A, AB, B, BC, C and so on.

4:51 I think that there is a simple (bad) upper bound. If n is the number of circles, then there are at most k < 2^n regions in the final diagram. Consider the incidence graph of those regions. This graph has at most k < 2^n vertices, corresponding to the different regions. Now, we will think of our set of circles as [n] and we will label each vertex of our graph by the subset of [n] which consists of those circles that contain the corresponding region. I'm not going to prove this here, but I claim that this function from arrangements of circles to labeled graphs on at most 2^n vertices is injective. i.e. I'm saying that if two arrangements give the same labeled graphs, then they are really isotopic. Anyway, we now have an injective function from arrangements to the set of graphs on at most 2^n vertices labeled by 2^n labels. To get an upper bound on the number of arrangements, it is enough to upper bound this number of labeled graphs. The number of graphs on k vertices is trivially upper bounded by 2^{\binom{k}{2}}. There are 2^n labels, so given a graph on k vertices, the number of labelings is at most (2^n)^k. It remains to sum the product of these over k from 0 to 2^n. Each term can be upper bounded by the largest term, giving a bound of 2^{\binom{2^n}{2}}*(2^n)^{2^n}*2^n

Here some (bad) upper bound: For n circles, you can look into the 2^n ways to choose a subset of them. For each subset you look into the intersection of the circles and look whether it's empty or non empty (if you have a singleton, look on the part that so not intersect with all other circles). Each one of the 2^n subsets can be empty or non empty, so you have 2^(2^n) possiblities, and there can not be to different drawings with the same intersection tabke I described.

I wanna see that full circle animation already!

When I saw the title in my feed I thought, "Oh yeah, another Neil Sloane feature!" I love these.

Loving how it’s all animated, nice

3:06 neil you're not the only one laughing 😂

Round Like a circle in a spiral Like a wheel within a wheel Never ending nor beginning On an ever-spinning reel ... ... ... Like the circles that you find In the windmills of your mind

Thanks for teaching us how to draw unique venn diagrams

Something I've been thinking about is if I doodle a figure consisting of a number of loops which may intersect but are never co-linear, this divides the plane into regions bounded by the loops. I seem to be always able to colour these regions with only two colours such that no two adjacent regions are the same colour

Neil's videos always rock.

Hey Bradey, Can you ask Neil if the same is true for spheres? Great video, thanks!

Love this guy he makes learning so much fun

I think I will attempt that tomorrow. Thanks for the vid

4:20-4:45 when no one but you understands your doodling

Me: do you speak gallifreyan ? Neil: 😏

This is very similar to a problem or two I've been working on since January. I'm taking a multidimensional venn diagram of some number of "circles" x having all possible intersections, then filling the intersections with integers 0 to some number y that all sum to y. I'm creating a program to compute and create a growing tree of depth y representing all unique types or configurations such that there is no specified order of "circles". However the 1st problem I was working on is equivalent to this problem except that a circle's contents can be "inverted", meaning a slower growing tree. This will be used to show all unique types of Boolean functions given some number w inputs, which is an equivalent problem to solving the number of unique mutidimensional shapes that can be made by selecting vertices of a multidimensional cube of w dimensions all side lengths one. The selected vertices make a complete graph showing the squared distances between each vertice, and every equivalent graph represents an equivalent shape. Each unique Venn type, with inversions allowed, can easily be converted to a unique graph. I should make a RU-vid video of this in 1.5 months or so when I'm done; it'll be more clear.

4:35 my man sure was busting one with those circles 😳

0:58 **Super Smash Bros theme gets increasingly louder*

On the opposite side of the spectrum, my dog will turn around in circles trying to get comfy to sleep. But that’s his second choice. He chooses lap over circles every time.

this was so nice, just a guy sharing his fascination with the world

A video featuring the very celebrated Neil Sloane himself. Gold!

Very interesting. Two questions I am left with: Is this solved for triangles? I am genuinely unsure if it actually makes a difference, although I would imagine so. Seeing as "touches" is not acknowledged as a separate state, does that mean that this is approached as a geometrical problem and not a topological one?

the number of digits of each number of configurations up to 5 circles makes up the Fibonacci sequence (1,1,2,3,5), so maybe the configurations for 6 circles is 8 digits long?

I was actually just thinking about something like this on my own the other day!! so neat

I thoroughly enjoyed this

I actually thought of this 3 years ago when I was wondering how many ways I can overlap circles as I add them. I knew the first three, but when I tried to do the four circles, I gave up because there was so many ways to do it. I shouldn’t have gave up 😂. At least someone older than me did it 😂

An animation of these cycling through positions would make a very satisfying loading screen.

Upper bound: 3^(n*(n-1)/2) since there are only 3 ways for any pair of circles to overlap (either one is inside the other, they intersect, or they're entirely outside each other)

this was the golden age of numberphile

Ah it hurts to see a beautiful problem/question like that but not even knowing how to go about solving something like that~ (I wanna learn the problem solving mental gymnastics that's so integral to maths)

That circle intersection animation would make a great website spinner! Is it available anywhere?

This is so satisfying to contemplate. Thank you math friends!

The circle overlap configurations remind me of electron orbitals, of course the orbits of electrons aren't all strictly circular, but it got me thinking of elements

It seems like we should at least be able to put a lower bound on the problem. Let the number of ways to draw n circles be expressed as C(n). I assert, trivially, that for every arrangement C(n), if you remove one circle, you obtain one of the elements of C(n-1). Said another way, all elements of C(n) can be created by added one circle somewhere to an element of C(n-1). Therefore, a trivial lower bound is C(n) >= C(n-1). We can, of course, do better. All elements of C(n-1) with a circle disconnected from them are elements of C(n). All elements of C(n-2) with some element of C(2) disconnected from them are also elements of C(n). By extension, C(n) >= C(n-1)C(1) + C(n-2)C(2) + ... + C(n-k)C(k); k = floor((n-1)/2). We could improve this lower bound by also adding in all possible combinations of 3, 4, or more disconnected groups, for 'n' high enough to permit such. An upper bound might be constructed by finding a way to notate elements such that each notation has at most one element that corresponds to it - notations that correspond to impossible elements are allowed, since this is an upper bound, but notations that are ambiguous are not. I suggest the following - Order the circles. For each unordered pair of circles, provide a value - disjoint, intersecting, lower-contains-higher, higher-contains-lower. Ignoring a circle paired with itself, this gives 2^(n(n-1)) possibilities. Then, for each pair of (circle, (unordered pair of circles)) such that the pair does not include the initial circle, provide a value - circle does not contain the intersection of the pair, circle contains both intersections of the pair, circle contains only the intersection clockwise of the pair's overlap, circle contains the intersection anticlockwise of the pair's overlap. This gives 2^(n(n-1)(n-2)) possibilities. Combined, these give 2^(n(n-1)(n-1)) possibilities. Finally, since the initial ordering of the labels of the circles doesn't matter, divide by n!. This gives a first order upper bound of 2^(n(n-1)(n-1))/n!. We could improve this by taking more care to eliminate impossible notations from consideration - for example, if Circle A contains or is disjoint from Circle B, then no circle can contain their intersections. Taking this into account makes the number crunching more complex, but still doable, and could reduce the upper bound considerably, but this comment is already very long.

Not only the equation is hard, but also writing a program to brute force in efficient way is hard

Oh man, this is very cool stuff!

I love this man’s voice so much

“And for four, there are 173” *DON’T BLINK*

Im interested what would happen if you could also use ellipses? Also it will be interesting if you take the problem to the 3rd dimension and instead of circles you use spheres.

Would be interesting to check bigger knight moves as well. E.g, 1-5,1-7, 2-5 etc. also if prime moves behave differently (1-3, 1-5, 1-7 etc.) and what if we jumped up to bigger moves when stuck (1-3 to 1-5 etc.).

how did you do the 4 circles animations? could we some how calculate that by computer?

Those drawings of 4 circles look like "new" or "circular" Gallifreyan from Dr. Who

4:16 it looks like Futurama Alienese

9:15 mesmerising x) totally worth it to watch til the end.

Absolutely love this video