Cannons and Sparrows - Numberphile

Featuring Professor Günter Ziegler, who was visiting at MSRI.
Extra footage at: • Cannons and Sparrows (...
More links & stuff in full description below ↓↓↓
This video features Professor Günter Ziegler from Institut für Mathematik, Freie Universität Berlin.
The blog post which kicked it off: nandacumar.blogspot.de/2006/09...
Convex Equipartitions via Equivariant Obstruction Theory: arxiv.org/abs/1202.5504
Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumberphile
We are also supported by Science Sandbox, a Simons Foundation initiative dedicated to engaging everyone with the process of science.
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* These slices (also known as "fair partitions") represented in drawings and animations are approximations for illustrative purposes - we've not calculated truly fair partitions, as you can probably see.

[@RobertTyrrellSlater]

"And 6, is 2 x 3" - My only moment of comfort during this video :P

[@GalexiDude]

It's not that hard to understand you take the intervals of separation and now I'm lost

[@kushla]

lol

[@joshyoung1440]

It really wasn't too hard... when he said this is a topology proof, not a calculus one, because the perimeter must be continuous but not necessarily smooth, I immediately knew what he meant. Because calculus can't deal with those corners. They're not part of a function. Cause functions don't have corners. But shapes do. And the intermediate value theorem applies to shapes.

[@AnonymousFreakYT]

Just the comparison "Cannons and Sparrows" makes me think of trying to help my kid to their math homework a few years ago. "Dad, how do I do this?" "Well, you take the integral of....... You know, you're only 8 years old, I doubt they want you to solve that with calculus... Let me think about that for a bit..."

[@corpsiecorpsie_the_original]

"Let's write a program to generate all possible real conceptually random homosedastic normally distributed data values. From there run the simulations with varying degrees of noise...."

[@antanis]

Got a real laugh out of me. Had that exact feeling trying to help my young brother with his algebra. Didn't remember how to solve it the they wanted him to and I was like hmm well just find the area under the curve with this neat trick involving infinity. The look reminded me that whoops.. This is algebra.

[@OlafDoschke]

"Firing cannons at sparrows" (In German: Mit Kanonen auf Spatzen schießen) could have been translated to "Using a sledgehammer to crack a nut." A related German proverb (also about sparrows) is to "prefer the sparrow in hand to the pigeon on the roof," which is kind of contradictory and dangerous if you know people fire cannons at sparrows. Welcome to the realm of German proverb wisdom.

[@pavphone2616]

Olaf Doschke "Mit Kanonen auf Spatzen feuern"...

[@MisterAppleEsq]

That seems more akin to the English proverb "a bird in the hand is worth two in the bush".

[@Kotoliszka]

This is completely unrelated to the video and possibly of interest to nobody but me, but the exact same proverb exists in Polish - as in the literal translation is identical. I found this interesting because it seems so rare to find a proverb that's the same in two or more languages; usually you'd find something with the same meaning, but phrased in a completely different way. Came here for the maths, learned a linguistic fun fact while the video was buffering. This is shaping up to be a fairly productive evening.

[@OlafDoschke]

The second proverb isn't related by similar meaning, just also being about sparrows...:)

[@MisterAppleEsq]

+Olaf Doschke Oh, I see.

[@johannespitz]

I am glad you're uploading more complex stuff from time to time. It's fascinating even if you can't understand everything! Thank you.

[@TokayGekko]

Well it depends. Is it an African or European sparrow?

[@sirjmo]

Swallow, not Sparrow.

[@blindleader42]

SirJMO, The Ministry of Truth has been notified of the historical error in the Monty Python document. Winston Smith has been assigned to review said document, and correct this and any other errors that can be discovered.

[@axjkalsok1058]

SirJMO whoosh

[@TrasherBiner]

super funny man, you should do stand up comedy. Nobody ever has made this Monty Python reference about sparrows.

[@caimacd]

TrasherBiner Screw you. that was funny

[@eideticex]

As much as I've worked with Pascal's Triangle. This really just blew my mind as it points out a property of mathematics I never even realized existed. Not just as the example stated but as a very basic truth about how numbers work at a very low level.

[@alphalil]

it's been 2 years and no one mentioned how firing cannons at sparrows seem just a bit similar to the concept of the angry birds game

[@hanssinaga747]

The first thing that comes in my head is Arsenal and Spurs.

[@nagoshi01]

Holy shit. So this means that (imagine the largest prime you know of, even the ones with millions of digits) you can split a polygon into that many slices with equal area and perimeter. That's crazy

[@hunterbelch2524]

Isn't it crazier knowing that it doesn't work for prime products? It's almost like encryption in reverse. We know that two primes equal a product that can't be deconstructed. But we don't know what the product looks like. As opposed to seeing a product and brute force decomposing into prime factors. I've been drinking. Pardon my french

[@MK-13337]

@@hunterbelch2524 This might still work for all numbers. We know for certain it can be done for prime powers, but we don't know for certain if it can be done for any other numbers. That is not the same as knowing for certain we _can't_ do it for other numbers.

[@EckiSchmecki]

Hold on a second - you _could_ do it, it's not like it's that easy

[@MyFilippo94]

I still get surprised by how you can in any given moment of your proof find a solution which is on a completely different place of this Math world, and how complex it can get to enstablish a link between two "simple" things like splitting a polygon in equal area and perimeter pieces and Pascal's triangle. My mind gets blown away every time

[@DaniPhii]

That Pascal triangle is taken from The Number Devil! I love that book!

[@neostrix]

That was the first thing I noticed in the thumbnail; this book made me the math lover I am now

[@rushikeshshinde2325]

This is one of my favourite numberphile videos. Günter Ziegler should feature in numberphile videos more often.

[@Aciek25]

i love this canon slowly moving into a frame.

[@PasseScience]

It's amazing how something so smoothly geometrical related to areas and perimeters (and the machinery context we use to partially solve it, with topological argument and voronoi partition) reveals itself to have a link with something so numerical than prime numbers.

[@lystic9392]

Wait I thought we were going to talk about cannons and birds... Aaah! You tricked me again!

[@jamesloving2357]

B Spits that is why I started watching.

[@ChilledfishStick]

Thanks for reassuring us of that there won't be too much math in a video about math.

[@numberphile]

The whole 22 minutes is math.... and besides, don’t I decide what my videos are about!?

[@nex]

Given that there is a solution for every prime n, can we not conclude that there also must be solutions for all composite n? Edit: The answer is no. Pif the Mestre's comment in another thread illustrates nicely what I'd been missing. Not sure how to link to that comment, so I'll quote them here: "Start with a polygon P, cut it in two pieces Q1 and Q2 with same area and same perimeter. If you cut Q1 in R1 and R2, and Q2 in R3 and R4 there is no reason we have the same perimeter for R1 and R3."

[@MrFrally]

nex I was looking for this comment. Thank you for looking it up!

[@bregonz]

The question is not "Can we be sure they have the same perimeter?". The question is "Can we assert that it doesn't exist a certain way of cutting Q1 in R1 and R2, and Q2 in R3 and R4, so that R1 has a the same perimeter of R3?" That is the assumption we want to prove or disproof.

[@bregonz]

And in the case of R1,R2,R3 and R4, it actually always exists a way of cutting them.

[@CBDroege]

I'm reminded of a previous episode with a similar solution for leveling a table at the biergarten by rotating it.

[@pegy6384]

I really enjoy these longer form videos.

[@mokopa]

Remember, the perimeters are parameters.

[@zockertwins]

And if the edges are cutvy they are even circumferences !

[@maulwurf9414]

mokopa not dyslexic but I can tell it could really annoy or confuse someone dyslexic

[@qawarful]

This reminds me the sandwich video.

[@numberphile]

It's different - wait until you see the result!

[@Czeckie]

only because the proof for two pieces uses the same idea. The problems are not related in any way, this is much harder.

[@benjaminwessel5290]

Of

[@LokiClock]

Benjamin Wessel Dative "me"

[@gameswoodmore5950]

+Czeckie They are related in some way, since they both use continuos changes between two numbers.

[@marsielo99]

Amazing video. He can explain such a difficult concept in a very simple way.

[@will1603]

This seems similar to the episode with Hannah fry

[@wierdalien1]

Will Turner which hannah fry?

[@drakoz254]

The ham sandwitch one? I have a feeling that the ham sandwitch theorem video was a prep for this one.

[@danielhamilton3496]

Thanks for your disclaimer Brady. Got very heavy in the middle the but a very elegant proof in the end.

[@sobanya_228]

I got math overdose, please call nine eleven

[@MrTridac]

I don't think nine would be happy to be called eleven. Also confusing.

[@kirkelicious]

Does it have to be 911 or will any other Sophie Germain prime do as well?

[@RWBHere]

Nice one, Kirk!

[@aDifferentJT]

It’s easier to think about it by keeping the perimeters equal on each side and then the difference in areas goes from negative to positive so must be 0 at some point.

[@myrec8883]

This is really good point. Because it's easy to see that end points of line move continuously.

[@nosmirck]

but, as soon as you modify the sub polygons to adjust the areas, the perimeters will change. with weighted points, you guarantee that the polygons will keep the same area no matter where you move them.

[@AhsimNreiziev]

+Luis Pulido Jonathan was (probably) talking about the case where you want to split the polygon into 2 parts using a single line. Not about splitting any into more than 2 parts.

[@aDifferentJT]

Ahsim Nreiziev indeed

[@nosmirck]

it's the same case I guess. Imagine you start with 2 perimeters equal, how can you balance the areas without breaking the perimeters? this basically means that if you start with 2 equal perimeters then you have a line that cuts the polygon creating 2 points (A and B). If you then want to modify the areas, you have to modify A or B or both! breaking entirely both perimeters (to actually try to maintain the perimeters, you HAVE to move both, moving one only will break both perimeters). So we end up in the same conclusion as having already a line dividing the shape in 2 equal areas and moving the points to find the equal perimeters by moving both points, which basically means rotating the line.

[@dreamteamsprediction2536]

For n=2 you have made a1=a2 and rotated it in such a way that p1-p2=0. But the problem is at what pivot point we should rotate it and secondly we can make p1=p2 easily because the straight line which divide the polygonal is in both shape and by moving the intersection equal distance in opposite side on polygon we can get a1-a2=0 and it is more obvious

[@bryceherdt2363]

Neat. Some other commenters thought that because the conjecture is proven for all primes, it must generalize directly to all composites. Well, it doesn't, but I have an incomplete workaround, and any counterexample to it would be fascinating in itself. Say we're constructing an answer for 2N, where N is some number already proven. Divide your starting convex shape into 2 equal-area halves as before with line L, but don't keep track of their perimeters this time. Instead, find all solutions for dividing each half into N pieces of the same area and perimeter. The smaller areas are guaranteed to match, which leaves only the smaller perimeters; these will match on either side of L, though maybe not across L. As you rotate L through 180 degrees, the perimeters may go from being A and B to being B and A. So if certain conditions hold, some orientation of L will have a solution where all shapes have not only the same area, but also the same perimeter. I can't just say I proved it, though. A given half might well have more than one solution with many values for the perimeter. I picture the perimeter values as multiple continuous segments and some points, not contiguous with each other--no clear reason those values can't completely dance around each other, but it would require some strange behavior. More details to come tomorrow.

[@bryceherdt2363]

I technically wrote that comment just after midnight, so technically this one is going up "tomorrow." Now, a little bookkeeping. Imbed the original shape in the Cartesian plane and parametrize the dividing line L as Xcos(theta) + Ysin(theta) = K, with K varying for different theta. This allows a couple of things. Less urgently, it shows that the appropriate dividing line varies smoothly with direction; moment to moment, it rotates around the center of the shared boundary to maintain the same area, and this midpoint too must vary continuously because the shape was assumed to be convex. More importantly, this allows those of us living in three dimensions to visualize the solution space. Admittedly, UVZ-space is not anyone's first choice, but it's better than reusing X and Y. In this graph, U=cos(theta) and V=sin(theta). (I'm using U and V instead of theta partly because this means graphing points on a cyclic function only once, on the wall of a bounded cylinder.) So L is the line UX+VY=K, and the UVZ graph will relate to the portion of the shape where UX+VY >= K. Specifically, for every UV pair, consider all possible dissections of the half-shape into P convex shapes of equal area and perimeter. The area is fixed; take Z to be every possible value of the equal perimeters on that side of the line. The problem reduces to finding, or proving there exists, some value of theta such that Z(theta) overlaps with Z(theta + pi). This... *ought* to work, since we already chose P to be solved, but it's not certain. The set of perimeters Z is nonempty for each theta, but is it continuous for each theta? One might imagine Z satisfies only a narrow band around Z=1+theta/3 where theta ranges from 0 to 3pi. The solutions would have to have some other quality, like continuity of the upper bound of Z, or no tapering, or somesuch. And that's still assuming we're looking for a solution for 2P regions. Even if it works for all P, that only proves some even numbers. To truly construct all solutions, you need to be able to multiply by *odd* prime powers too, and even 3P carries no guarantee. Two thirds switching, even with their perimeter values varying smoothly enough for the case of 2P, might not match the remaining third when they match each other.

[@TunaAlert]

The title of this video reminded me of a class mate of mine; in 11th grade in a maths major course (in Germany, from 11th grade to 13th grade [in my region] which is basically Highschool you choose 3 major and ~7 minor courses) we were talking about polynomials and polynomial division. We had a task similar to the following: Solve for x: 2x³ + 5x² - x He proceeded to solve for x using polynomial division, and my maths teacher used pretty much exactly this expression "Shooting with cannons at sparrows" to describe his solution.

[@daanwilmer]

It's mind-boggling, the connection between this topology problem and prime numbers. Even if it isn't inherent to the problem (for example, it might turn out that this partition is possible for any n), it's amazing how people can get a partial solution that somehow involves primes.

[@Aciek25]

That feeling when you have to skip the proof by contradiction because it's too hard.

[@MrStevenToast]

What the hell is going on here?

[@MrNicoJac]

Brain cells committing suicide

[@corpsiecorpsie_the_original]

Numberphile is flexing a bit.

[@CalculusPhysics]

This reminds me of that Ham Sandwich Theorem from a while ago! using the same idea that by rotating the “knife” you’re able to prove that there’s at least one point where there’s an equilibrium of sorts, like equal areas of both pieces of the sandwich (in this case equal area and perimeter)

[@MrMontugar]

a fun aspect of the rotating points eventually being equal is that the trick works with leveling tables as well. spin a wobbly table and eventually it will balance!

[@LanguagesWithAndrew]

He completely lost me when he started talking about space configurations because he did that without at all defining what a "space" was let alone a space configuration. This meant I couldn't understand the rest of it since the rest of it was predicated on the configured spaces. I've noticed this kind of thing happens a lot in these videos: experts will use a technical term or concept without defining it because they forget they're not speaking to fellow experts.

[@laurendoe168]

I agree... but I suspect the definition and ITS explanation would take well over an hour.

[@ignaciolarrea429]

When he says ‘space of configurations of 3 points’ he is just referring to all possible ways of choosing three distinct points on the figure. In mathematics ‘space’ can mean a lot of things. It’s true that in this case it’s not absolutely clear what kind of space he is talking about, but in general, space means a set of points with some additional structure (metric spaces, topological spaces, vector spaces...). In this case, the set of points would be the set of possible configurations. To each possible way of choosing three different points (a configuration) corresponds a point of the space. In this case it isn’t clear what kind of additional structure he has given to this set of points but is probably at least a topological space that is a way of giving the space a way of telling which functions are continuous by selecting some special subsets called open subsets.

[@EebstertheGreat]

A year after the fact, but this video used many cuts to almost completely skip the argument. Some of these cuts are obvious, but some might be lost in the way the video is put together. In any case, you are more than right that the video does not prove the theorem. Rather, the point seems to be to sort of demonstrate how complicated the mathematics behind the partial proof is. The idea is that a deep theorem is being used to prove results for something easily understood on the surface. Deep theorems practically by definition cannot be proved on this channel. But the idea of using a powerful result to prove a weak one is still sort of interesting, and I think that's where this video originally came from. As a small joke, I reposted a valid but silly proof from _American Mathematical Monthly_ in the comments to try to get the idea across. Maybe more to the point, imagine using the fundamental theorem of calculus to prove the area of a triangle. This is definitely not a great video on the channel. I feel like this one needs some help to be of any value at all.

[@user-kh5tv9rb6y]

It's not quite that they forget they're speaking to non-experts, it's more that they aren't sure how much non-experts know, so they sometimes assume incorrectly, either assuming too much or too little knowledge. That's something teachers in every field struggle with.

[@godfreytomlinson2282]

This was a really amazing video. Thank you so much.

[@mikejoecollins]

If it works for the primes, then why not for the composites? If ab is composite then first substitute n=a to give a partitions (each with equal area and perimeter), then solve for each partition independently with n=b

[@aljawad]

Brilliant, absolutely brilliant! Following the logic portrayed in this video makes me wish I had pursued a career in pure mathematics instead of choosing an engineering field of study, and eventually build a career working on environmental projects ...

[@wavedash-]

lmao I was expecting something boring because of the intro, but what I got was actual math! Numberphile should do more videos with actual math

[@deslomator]

What a journey!

[@abhijitborah]

A diploma of some kind must be awarded to whoever had completed the 22:32 minutes viewing.

[@yongmrchen]

Why hasn’t this video garnered the number of views it should? It is brilliant, even though I was lost most of the time.

[@bonecanoe86]

Numbers are pretty cool I guess

[@irzinfante]

Thank you Günter.

[@FutureNow]

I love that this channel has 2.3M subscribers.

[@nhatkiem]

Sounds like the engineer was trying to come up with a formula to break up a polygon into smaller areas for finite element analysis. While easy for rectangular and triangular shapes, this process becomes quite complex for higher geometries and smaller discretizations.

[@RevJR]

I'm just going out on a limb here but when you get above n=2, if n=4 for example, you could draw a triangle in the center, and then a line out directly from the center of that triangle which starts at its vertices and goes to the end of the original object. Then using the same math as n=2 you should be able to show that n=4 will have some point where each value is positive or negative compared with each other value to attain both the area and the perimeters of each new space, I think. You simply start with a proper shape of n-1 sides (where n is the number of areas you want to generate) and apply this. I think this also eliminates concave sides since any regular polygon will projected rays from its center point through its vertices at an angle such that the resultant shape on the outside will be proper. Neat video! I don't know how to say it any simpler than that though.

[@jessemckeown5628]

There seems to be an echo of the "Topological Tverberg" question, right down to the tool used in the affirmative case; the other cases of Top.Tverberg were given counter-examples just a couple years ago...

[@carnsoaks1]

same issue as shown on "infinite series" - PBS last week. FUN that its covered in many places. Proving Brouwer's Fixed Point Theorem | Infinite Series

[@n124lp]

There seems to me to be a piece missing from the argument for the n=2 case using the IVT. I agree that you can rotate the area bisecting line in such a way that the difference between the two areas varies continuously, but I do not see how you can guarantee that for any polygon, there exists such an continuous rotation that will get you to the line with the reversed endpoints. (The 180 deg. rotation.) I am not saying that I have a counterexample, I just don't see how to prove rigorously that it does work for all polygons.

[@matthewpurkis5079]

used an actual house sparrow chirp in the intro, well impressed top marks

[@juhaniu6371]

great video! enjoyed the whole 22 minutes

[@Hjerpower]

It hasn’t even been 22 minutes since upload

[@lambda9634]

Thank you for the great great great videos!

[@jonafladerer3074]

at about 15 min i had to remind myself what this whole Video was about :D love it

[@shakesmctremens178]

That is fecking mind blowing.

[@4jonah]

Correct me if I'm wrong. But you could 3D print pieces of equal area and perimeter, have them fit together, and in the end create any N polygon. So it would be possible to do the reverse and cut any N polygon into equal pieces. I use 3D printing as a way of reverse engineering the Problem

[@msmuffett1]

I cant help but be impressed by his curly brace drawing skills.

[@ashutoshchakravarty2669]

Mathematics is poetry

[@misterhat5823]

Just like poetry, it's really just a cryptic way to say something.

[@connorcriss]

There was a story about a guy who wanted to become a mathematician, but he wasn’t creative enough so he became a poet. Anyway this is one of the most complicated numberphile videos I’ve seen and I don’t understand it.

[@hybmnzz2658]

@@misterhat5823 true. And it shows that speaking poetically is easier.

[@hybmnzz2658]

@@connorcriss the story has to do with the legendary mathematician Hilbert. One of Hilberts students dropped out of math to pursue poetry. In response Hilbert said "He didn't have the imagination for maths".

[@rosiefay7283]

4:00 Neat. Alternatively, given that the shape is convex and has finite area, it has finite perimeter, so you could instead consider the behaviour of the partition by the straight line PQ, where P and Q move around the shape's perimeter, always with half the perimeter on one side and half on the other side. 5:17 So what do you mean by "smoothly", if not "continuously"?

[@anon8109]

@3:17 Why would rotating the line preserve areas? And how do you pick the point around which to rotate the line?

[@Daniel-fv1ff]

anon8109 you rotate it such that the areas remain the same. Not all rotations will do this but it is still possible. I don't think you would need to limit yourself to rotating it around a point.

[@steelwarrior105]

Exactly, it's just a ratio so you would need to put it say, in the middle of a regular polygon or solving for that point on other polygon. Of course it only works for one line in most cases

[@ZacharyMathematica]

You wouldn't pick a singular point around which to rotate the line. In a sense the line would "wander". But it can be said that for a polygon there is a like at angle x that bisects it into equal areas. This solution would require to prove that more rigorously and then rotate that line through multiple angles and comparing the resulting perimeters.

[@BobStein]

Right, for any given angle there must exist a line at that angle that divides the area evenly in two. The set of all such lines (for all angles) does not _necessarily_ contain a common point.

[@zenovanditzhuijzen]

Or another way to look at it: Take any point on a shape you're trying to divide, you can draw a line from that point to another point on the shape to divide it in half. Therefore, you can move the line or rather the two points in a continuous fashion around the shape you're dividing.

[@flymypg]

My initial intuition was to treat this as an optimization problem, specifically by using a duality such as Maxwell's Equations, perhaps in their integral form, using a path integral around the circumference of each subdivision and an area integral over the surface, then assume a common current (normalized to be 1) circulating through each perimeter, which will generate a given amount of magnetic flux. Given the initial polygon, then the subdivisions represent "equivalent" electromagnets, once a common rotational direction for the current is assumed. The solution would be a fairly messy ODE, though that's where my intuition ends. It's been far too long since I last tried to solve something like this.

[@hurktang]

For each point around the polygon, you can use it as a corner of a shape or area 1/6 of the polygon. I think it's possible to prove that some of these shapes would have a perimeter too big and some too small. So it's a continue thing. The remaining of the shape no matter what is proven to be able to be cut in 5 shape of the same area and even perimeter. So by logic, there is a solution.

[@mikeberezhkov7810]

so you can cut poligon into 3 pieces of equal area and perimeter, then you can cut that pieces into 2 pieces of equal area. May be any check for if the perimeters are equal? if it exists, you can cut poligon into 3*2=6 pieces.

[@EmilMacko]

For the n=2 solution, can you not just repeatedly cut up the polygon to get any power of 2 amount of polygons with the same area and perimeter?

[@PaulusFrank]

That should work also for all numbers n. For n=6, first divide the polygon into 2 parts and then divide each part into 3 smaller parts

[@pifdemestre7066]

It might not work so easily. Start with a polygon P, cut it in two pieces Q1 and Q2 with same area and same perimeter. If you cut Q1 in R1 and R2, and Q2 in R3 and R4 there is no reason we have the same perimeter for R1 and R3.

[@jairguilhermecertorio6362]

Yes, if the case n=2 works for any polygon you could just prove it recursively. Edit: Ops, Pif de Mestre is right, when dividing two different polygons the perimeter won't always be the same

[@codylee2818]

I don't think it's so easy because you have to ensure that each sub-polygon has equal perimeter to each other. Imagine cutting a very wide rectangle, say with x=8 and y=2 in half lengthwise so you have two rectangles with x=4 and y=2. You take the left one and cut it in half like that again so it's x=2, y=2, p=8, a=4 each. The one on the right you cut the otherwise, so they're x= 4, y=1, p=10, a=4 each.

[@MrDarkPage]

For this to work, we would need to prove that the perimeter of sub polygon in a solution depends only of the perimeter of the starting polygon

[@geizzlyandro]

I just want to say, Gunter makes some some really close to perfect Circles!!!

[@RWBHere]

I like your close-to-perfect sentence!

[@danielw.4876]

Please make a video about the new largest prime number discovered!

[@Mrsparky492]

Matt Parker has a similar video where he kills flies with nukes. His example took a simple mathematical statement and reduced it to Fermat's last theorem that has been proven and therefore is a valid proof.

[@ChenfengBao]

Fermat's last theorem itself is a classic example of killing flies with nukes. And the nukes are invented in the proof.

[@hetzz]

Interesting, then we have a classic teacher (boring as F) and a guy who makes a living making maths fun and lighthearted. Thanks for suggesting the better one of these two.

[@RaffaeleCanepa]

Big Congrats! This was really hardcore stuff: didn't fully understand it, but I surely grasped the general meaning of the topic. And please: tell me this has something to do with symmetry group SU3. If that is the case, I'd feel the smartest guy in the world for 30 secs....

[@Daniel-fv1ff]

You can't just skip the explanation, that's the whole point of the video.

[@mattwatson6259]

Daniel Macculloch fr fr though

[@shell_jump]

There is a zero percent chance of explaining cohomology off the cuff to a general audience. Even math undergraduates would find it very hard to understand.

[@scbunn]

And thus, because I say so these two sets are not equal and we have success! If you can't explain it then pick a different topic for a video. This taught me nothing.

[@satiethetutor3337]

If he explains high level Algebraic topology to you, you'll have a stroke! But you're right, he just asked us to believe a bunch of facts without giving a convincing argument.

[@texasray5237]

It's a very dull point.

[@agustingomez4516]

Sorry if im wrong, but when you prove the n=2 case you forget to mention that you rotate the line around the geometrical center of the poligon maybe?

[@awwkaw9996]

I would think the problem is solved for any system: If you take any number, which is not a power of a prime, then find the prime factorization for that number you then have some new cool numbers. Now you cut the polygon in the first prime of equal area and peremiter, and then you cut each piece in the number of pieces by the next prime number (which you can also do fairly). Then you should have cut your system in a fair way. example: (6) You cut your polygon in two pieces. This can be done fairly. then you take each of these two polygons and cut three times each. This can also be done fairly. Then you should have 6 pieces with equal perimeter and equal area.

[@cedros_]

You don't necessarily do, because the perimeter of your final three pieces in the first half is not necessarily equal to the perimeter of those you got from the second half. A quick example: Use a rectangle of side lengths 1 and 4. If the method you decribed were to work in all cases, you could cut that rectangle in 4 pieces of equal area and perimeter by cutting it in half and then again in half. So, let's try it: First, let's cut it into two rectangles A and B with a perimeter of 6 and an area of 2 each. Now, let's cut A into two sqaures S1 and S2 with a perimeter of 4 and an area of 1 each. Now, we cut B into two reactangles R1 and R2 (not squares this time!) with a perimeter of 5 and an area of 1 each. As you can see, the perimeters are not the same for all 4 of them. To illustrate the way you cut, here's a diagram: __________________________ |_____________| | | |_____________|______|______| You would have to prove first, that there is a way to have the length of each 'secondary cut' be the same for all pieces resulting from the 'primary cut(s)' for all possible numbers of pieces resulting from each step.

[@awwkaw9996]

Cedros thanks a lot, and nice example.

[@Torkye_G]

Well... this is really simple in my head, so it might be wrong... but, if I understood the problem right, the current situation is that if you have a power of a prime it is possible for any polygon, right? Well, what does multiplication means in this case? Means that you cut a polygon in p1 parts and then you take each cut polygon as a new problem dividing it in p2 parts (p1 and p2 are some primes). In the end, every part has the same area and perimeter. Ex. Imagine a square, we know it can be divided in 3 "equal" parts (equal perimeter and area). We also know that's possible for 2. Therefore we divide by 2 and take each of those polygons and treat it as a new problem and divide it by 3. In the end we get 6 "equal" parts. And thinking this way about 2,3, square, we can think for any polygon and for any number (because every number is a multiplication of primes) Any mathematician around who can check this?

[@JimBurke123123]

Well, I thought the simple end solution would be to use soap bubbles, and let them work it out. Wouldn't that solve the circumference. But would it also be the same area?

[@yongmrchen]

It’s brilliant!

[@sarahheikel]

For once not understanding a math problem doesn't give me crippling anxiety

[@corpsiecorpsie_the_original]

Technically you understood the problem but didn't understand the solution.

[@HunterJE]

Equal area and equal perimeter - in other words can you cut an arbitrary polygonal pizza so everyone gets the same amount of pizza AND the same amount of crust

[@OlafDoschke]

I just have one last question: Do we now call primes and prime powers cannon or sparrow numbers? And what about the numbers compound of at least two different primes?

[@babel_]

For attempts to partition by composite (non prime power) numbers, can we just partition by their prime factors recursively? So for six partition it twice, then subpartition each partition three times? Surely then it works?

[@ninjamaster224]

For n=2, if we move and pivot the cutting line while retaining equal area and perimeter on either side, we can get an infinite number of continuous solutions for the full 180 degrees. The perimeter value for each solution may vary. Then with each side for each solution, we can cut into two again. There's a continuous solution for this second part because the first part was continuous. Thus, if one side had too much perimeter, it will have too little perimeter on the other side, and due to the continuity, at some point in the middle, the perimeters will be equal. If we can prove that moving and cutting the lines by infinitely small amounts gives a continuous solution for all cases that n is prime, then all non-primes will have solutions.

[@octavzlatior7232]

Interestingly, the common factor is also the root of the prime power. When not a prime power, the factors are dividing exactly (each some, but not all all) the terms in the triangle, for example 2x3 -> 6, 15, 20. Does this happen only for the first rows shown in the video or is it a general rule of the triangle? Any thoughts on that?

[@TheEternalVortex42]

It must be so because the first value is equal to the row number. So it can only be divisible by p.

[@MooImABunny]

This is a really interesting video! I only understood some of it but somehow he makes it feel tangable even though it's super abstract. Nicely done!

[@MartiniComedian]

The most difficult Numberphile video yet! 😱

[@totaltotalmonkey]

You forgot to mention Transformation Groups by Tammo tom Dieck in the full description.

[@raulrodrigues9084]

Dude I smiled so hard when he put the equation for n=4, this shit is so cool. He even positioned the equations is such way to show what was happening, this is so cute. I never thought I would see the day when I call math cute, absolutely amazing.

[@5ucur]

EDIT: I've found that what I outlined below, easily does not hold true. I'm leaving it for posterity and for an exercise if someone else wants to prove where my though process went wrong. I'm not great at maths so please bear with me here and point out any flaws in my reasoning. Let's split a polygon into A, B, & C - three parts of equal area and perimeter (I also call them "equal parts" below, implying the same meaning). Prof. Ziegler in the video proves this to be possible, to cut a polygon into such thirds. Since we used straight lines, A, B, & C are all new polygons, ready to be split into parts as well. Let's take polygon A and use the method with the intermediate value theorem, spinning the line until the areas have equal area and perimeter. Now polygon A is split into halves in the way Prof. Ziegler proves to be possible. Repeat the process with B & C. Since A, B, & C are equal in area and perimeter, their halves themselves are equal in such ways as well. When we halve equal parts in this way, we get twice as many doubly smaller equal parts. This gives us 6 equal parts of the original polygon. If this holds (i.e. on the off-chance that there's not a flaw in my reasoning), then you can split a polygon into any number of equal parts that's a multiple of 2. I conjecture that by extension, this would hold for any multiple of a power of a prime - i.e. for any n that we are already sure can be used to split a polygon in that many equal parts. However, I suspect that I must have made some mistake somewhere, because greater minds would have solved this much earlier if it were as easy. As noted earlier, please point out any flaws in my reasoning - I'd like to learn from my mistakes!

[@sky173]

Well, this video made my brain explode...

[@007bistromath]

Unless I'm misunderstanding his accent very badly, he mentions the "permutagon" at one point. I can find very few places that mention this where it's not actually saying "permutation" with some kind of character encoding error. Places that do mention it seem to assume you already know what it is. What's a permutagon?

[@caimacd]

Love the show. I really appreciate the style and tone of the format. It makes it easier for dumbasses like myself 😛

[@wjb40]

Intermediate value can be used to prove that, at any instant, there exists along any great circle on Earth (take equator for instance) a set of antipodes with the exact (to infinite precision) temperature (or air pressure, or any other continuously variable metric that can be measured at the surface). I love it. After Intermediate Value Theorem... I got lost.

[@eduardogomes4865]

The common factor thing is pretty trivial in the "if" direction. One have to derive it to prove some elementary results. "Only if" sounds harder, though.

[@tisajokt7676]

Cut the polygon into N shapes of equal area, and then create small zig-zags on the dividing lines to add perimeter to the two shapes sharing that edge (but in a way that maintains the same area) until they're all the same? Would that work?

[@lostname605]

James Tisajokt it must be a convex polygon, so I would say no.

[@tisajokt7676]

+LostName Ah! I forgot that bit.

[@piguy314159]

That illustration of Pascal’s triangle in the thumbnail... is that from “The Number Devil”?

[@joshuahuntington1888]

Can't you always find 6 equal areas and perimeters by first finding 2 equal areas and perimeters and then for each of those, find 3 equal areas and perimeters? Same for 10 (First find 2 equal, then 5), 12 (first find two, then two from those then 3 of those 4) - etc.

[@45pierro]

imposing the perimeter was not useful when dividing the polygon into two pieces. You can thus do the trick of rotating the cut line until p=q, which seems to solve the 6-case. More generally, this technique works for p^j•2^i

[@shruggzdastr8-facedclown]

Can the same concept be expanded up in dimension to include equal volume as well as equal area (surface area, that is)?

[@peterchan6082]

This video is all about cutting polygons into equal areas and perimeters, and the Pascal Triangle. But what's that with "Cannons and Sparrows" at all ?

[@RWBHere]

Having watched and listened to this, I've no idea how he managed to progress to the triangle, or why he is using the triangle. Maybe a simple explanation in layman's terms might help? Proving by trying to disprove something sounds suspiciously like a mathematical fudge, since it often doesn't work in everyday life; you have to try to disprove every single case. Any exceptions only mean that another method of proving the theorem has to be found.

[@atrumluminarium]

Didn't Cedric Villani also work on Optimal Transport theory? Maybe you can get him to do a video about that Brady :D

[@md-sl1io]

so could you do similar things with 3d shapes splitting them up to have equal surface areas and equal volumes

[@QuasiELVIS]

This is one of those ones where I don't need extra footage.

[@retepaskab]

How do we get from a^2+b^2+c^2 = 1 to a cricle?

[@moustaffanasaj1584]

@retepaskab Suppose you want to plot a semi circle around the origin in an (x,y) graph. The way to do this is by using the function y = √(1-x²) Go ahead and plot it in Wolfram alpha to see what it looks like. Note that the plot only has positive y-values in its reach, due to the square root in the function. Now square both sides of this equation to get y² = (1-x²) This equation has the first equation as a solution, as well as the negative of that first equation. The latter plots the negative y-values, thus making another semi-circle. Rearranging the second equation yields x²+y²=1 Note that by adding variables we go into higher dimensions, so with 3 variables we're plotting on a sphere. Or a 3-dimensional circle, but the basic premise that sums of squares add up to a "circle" holds. Hope this helps!

[@8bit_pineapple]

@Moustaffa Nasaj ... That's half arsed backwards. y = ±√(1-x²) is the equation you get from rearranging x²+y²=1 in the first place. If you wanted to explain why x²+y²=1 is an equation of a circle just explain that - After applying the Pythagorean theorem to the coordinates (x,y) you find x²+y² = r² Where r is the distance to the origin (0,0) So, the equation x²+y²=1 is just the previous with r = 1 So the solutions to x²+y²=1 are all the points with a distance of 1 from the origin (0,0) This is, by definition, a circle with radius 1 centered at (0,0).