The video you're about to watch was described to me as firing cannons at sparrows. The sparrow being the pretty simple mathematical problem we're about to tell you about, and the cannon being the heavy artillery, the mathematical machinery, that has been used to try to conquer it. Now in the middle of the video we will encounter some of that heavy artillery, but not for too long and stick with us because at the end you'll encounter a solution and some numbers you might be very surprised to see.
- I wanted to sort of show you a pretty little problem that I love a lot. The problem appears in a blog by some Indian engineer and amateur mathematician, Nandakumar, who said with his friend Ramana Rao: you're given any polygon and a number, let's say we could take I don't know a pentagon, doesn't have to be a regular anything, here's the pentagon. And we take a number, I don't know number six? And now we want to cut this pentagon into six pieces and the condition is that the six pieces have the same area, which is easy. So for example if you want to cut it into six pieces of equal area we could just slide over and could say, you know, this is one sixth, two sixths, three sixths, four,
five, six. Drawing isn't good but I think the idea should work out. But these pieces will have quite different perimeter. I think it's quite obvious that even if we really make them equal area they will have different perimeter. And the question is whether, you know, for any polygon and any number like 6 - I'll try it again - whether there's a different way to cut it up, perhaps something like like this; one, two, six pieces and we want to have them equal area and equal perimeter at the same time. It seems that this should be possible. It's not that easy to work out examples. Let's say even, not a pentagon let's take a triangle. And you want to take this triangle and cut it into three pieces, equal area equal perimeter - it's not so clear.
- (Brady: Is that not an easy one?) No, actually there's three mathematicians who then published a paper in advances in mathematics proving this problem but only for this special case of three pieces. And all kinds of machinery involved on a harmless-looking problem. So I think what we should do here is solve the problem but for n equals 2, just to see that given a polygon why can we cut it into two pieces of equal area and equal perimeter? So we take a polygon again and, I don't know, here is one. And we want to cut it into two pieces two pieces of equal area and equal parameter. So the pieces have to be convex so there's no indention or so. So if I cut something into two pieces, or two convex pieces, then this just means cutting by a line.
- (Brady: Professor I can't use) (curved lines and things like that?) No. Curved lines are not allowed because if you use a curved line then on one side of the curve it's not convex. So it has to be really straight lines. Let's just do a vertical line and I'll do the vertical line so that we have equal area, let's assume that's it all right? But now I think this left side will have larger perimeter than the right side, okay? And now the trick is what we do is we rotate this line. Which means we can rotate it in such a way that at every single moment there's the same area on the left side and on the right side, all right? And while we rotate it - so the next one will be here perhaps and next one will be here - always the same area but the perimeters will change. And the perimeter will change continuously. It may be that on the left side the perimeter gets larger, on the right side it gets smaller. So what we can do is look at the quantity which just says the perimeter on the left side minus the perimeter on the right side. And at the beginning, with the vertical line, perimeter on the left side was larger than on the right side so the difference is positive. And now we rotate. And while we rotate the perimeters change. And once we've rotated by 180 degrees we have the same vertical line again but left and right have interchanged; and which means this difference of perimeter to the left minus perimeter to the right has changed to its negative value. So if in the beginning this was three inches longer on the left side, then in the end it's three inches longer on the right side, and somewhere in between it must have been equal. End of proof. So what have we used? We've used that these quantities change continuously, and you have to think about it a bit that if you rotate the perimeter changes continuously but that's true. And we've used a theorem that people learn in high school and calculus classes call it the intermediate value theorem, which just says that if you have a quantity that changes from a positive value to a negative value, and if the change is continuous, then somewhere in between you get zero. And my point is that this is not a theorem of calculus it's a topology theorem; because it's not important that the quantities change smoothly, and probably they don't, it's only about continuously. That's the full story for n equals 2, and the problem is now that for any larger number than n these configuration spaces of all possible cuttings, let's say in two equal area, they are much more complicated. And by far not as well understood and there's all kinds of open questions; but it turns out that - well now comes the part where we throw in machinery. And the first part we throw in machinery is a theory called optimal transport. For us it tells the following: you take a polygon; let's assume I have a bunch of points in there, here's five points, and from optimal transport I get that I can give weights to these points so that if I look at the domains of influence, could look something like this - so that's what they call a weighted Voronoi
decomposition. They are always hard to draw. I'm not going to do this correctly but so I think it will be something like this. This is a partition associated with these points. You see this is not equal area, so I didn't
do it right, but this theorem by Kantorovich and others says that if I adjust the weights right, so like this point should get more
weight so in the end its domain get's larger, and perhaps this one should get a smaller weight, and and so you know this would rather be the domain - and if I put the weights right then I get an equal area partition, that's the theorem. And if I move the points then these unique weights will move continuously. So it's again continuous and this means that the space of n distinct points in the polygon gives me basically a parameter space for nice partitions of the polygon into equal area. So which means I have the space. Doesn't give me all partitions but it gives me many of them and actually enough of them. This is just the equal area, now we have to get equal perimeters. And that basically means that for every configuration of points I know how to get equal area but and now I have to compare the perimeters. And that basically means I have- I get a map from this space of configurations of points to a space of values. We solve the problem if we can show that there is no good map which would never give me all perimeters equal. And not all perimeters equal means basically we map the space of points to the space of values and we never hit zero, which means we always go around that. And now the question is does that map exist? And now comes the next part, it's something that we take from algebraic topology. And somewhere in there is a theory called equivariant obstruction theory. Actually it's not that much big a theory, it's it's basically it's a theory which just consists of one theorem. But this one theorem says that this map doesn't exist if and only if a certain cohomology class is nonzero. So this sounds amazingly abstract, and it is, and the good place to learn equivariant obstruction theory is a guy- is a
book by a senior German topology professor,
Tammo tom Dieck. And it's one of these fat books without a single picture, but very precise. I always quote Alice in Wonderland who says, 'what's the value of a book without pictures or conversations?' So there's no pictures in there, there's no conversations - but the theory's in there. And I'm not really an algebraic
topologist; I admire these tools. If I want to use it I have to make it concrete. And I think I can show you on paper what really happens there, except I'll need a new sheet of paper for that. So we try and make it concrete, which means- our concrete now here means n equals three. Again we have a polygon and let's assume that this is a polygon that we cannot cut into three pieces of equal area and equal perimeter.
- (Brady: So you're assuming that) (the whole thing we're trying to do is) (impossible?)
- Yes, so proof by contradiction. And now if this is impossible this means that for any configuration of three points there's a unique equal area partition with the right weights, but then I get the values which is these perimeter 1, perimeter 2, perimeter 3, he perimeters of these three pieces. And these three parameters are not equal. What we really translate this in is that we say we have here, on the one hand we have the space of configurations of three distinct points, and over there we go to something that's the space of three numbers - numbers have to show up somewhere, right, because this is Numberphile. And these three numbers: not all equal. And actually we can assume that this is three numbers which sum to zero, because we just subtract the average. So if we work out what these spaces are then in the end it will turn out that here really we're talking about maps to a circle. You know, from any three numbers that are not equal you can make a point on the circle by subtracting the average, which gives you three numbers that sum to zero, and then you divide by the sum of squares, so you get three numbers where the sum of squares equals one and that's a circle. And now it turns out that one can understand the space of configurations of three distinct points in the plane; and basically this space can be glued together from hexagons. Where basically hexagons could look like this, I just draw one of these hexagons in the end it will be six of them. And the meaning of this hexagons is basically I can label this by 123, and 123 really just means I have three points in the plane: 1,2 and 3, and point 1 is to the left of point 2 and this is to the left of point 3. Whereas perhaps
this one here would be 132, which means again this is sort of symbolizes three points in the plane but 1 is to the left of 3 and left of 2. If I look at all configurations of three points in the plane then of course it could also happen that they're just not sorted left to right, but let's say point 2 is above point 3, and this basically means the space I get will more look like this; and we can work out the combinatorics of that we're not going to do the detail on that. This here basically, this red line corresponds to three points but 2 is above 3, and this is again the points 1, 2 and 3 in the plane but 1 and 2 is below point 3. So each of the elements of this space we can sort of interpret as certain types of point configurations in the
space - in the plane. (Brady: Alright, I'll take
your word for it.) Getting a bit too complicated?
- (We're starting to get the cannons
out I think) I think what we'll do is say
that now some magic happens. What comes out is that in the end this map doesn't exist. So this map
from space of configurations of three points to the space of three numbers that are not all equal. So we have the problem solved if this map doesn't exist. And it turns out that this is the case, if and only if the following equation does not have a solution. And the equation is 1 plus 3X plus 3Y equals zero; and where this X and this Y are supposed to be integers. And just to illustrate where the 3s come from, this 3 comes from- one 3 is this edge or these or these. It's sort of combinatorics of the hexagon. And the other thing is the other three edges of the hexagon. But that's sort of - that's where the magic happens. And now the questions is does this equation have a solution in the integers? And you stare at it for a while and you say, well obviously not. Because if I put in any integers there I get something that's divisible by three. And I add one and I get something that's not divisible by three. So we've learned that this equation does not have a solution, which means the counter-example does not exist, which means every polygon can be cut into three pieces of equal area and equal perimeter.
- (Brady: For n equals 3?)
- N equals 3. Now we get a bit overexcited and we say if we can do it for n equals 3; turns out with the same pattern if we take the equation 1 plus 2X equals zero, that's even easier to see that this doesn't have a solution, right? Because this is even plus 1 is odd, doesn't give zero. So this is n equals 2 - check
- (Brady: Which we already knew?) Which we already knew - even
without the missing details and the magic in the beginning. Then we did 1 plus 3X plus 3Y equals 0. And this doesn't have a solution because of divisibility by 3, and that solves the
problem for n equals 3. For n equals 4; and it turns out same pattern, the hexagon will be replaced by the permutohedron, which is a wonderful polyhedron to study. And we've solved the problem for n equals 4 if this equation doesn't have a solution; 1 plus 4X plus 6Y plus 4Z equals zero. And you stare at it for a while and you say, well 4 and 6 and 4 they're all even so if I plug in any integers there this is even. Plus 1 doesn't give zero. Which means we've solved the problem for n equals 4. We go on; 5x plus 10Y plus 10Z plus 5W equals zero, does this have a solution? No it doesn't, because all of this is divisible by five. Which means we've solved the problem for n equals 5. And so on, you might hope or guess and that's not true. Because the next step is 1 plus 6X plus 15Y plus 20Z plus 15W + 6U equals 0. So perhaps we should reveal the mystery, what are these numbers that show up here? And it turns out this is binomial coefficients, from Pascal's triangle. And if you don't want to memorise these numbers, actually every number here is just the sum of the two numbers above of them - that's the rule of how you do Pascal's triangle. So 1 plus 5 is 6, 5 plus 10 is 15, 10 plus 10 is 20, 10 plus 5 is 15, and 5 plus the missing 1 is 6. So that's the pattern. Which means this is the sixth row of Pascal's triangle except for a 1 in the end missing. And the question again is, now we are at n equals 6 does this equation have a solution in the integers? And indeed it does. If you, for example put 1 plus 6 times minus 1, so 1 minus 6, plus 15 times 1, plus 20 times minus 1; and the rest you take zeros- oh I think I got it wrong. I take plus 1 here, and I take minus 1 here. Then I find 1 plus 20 is 6 plus 15, right? This means we have a solution to this equation in the integers, which means we do not have a contradiction, which means this map might exist, which means the counter-example might exist. So we plainly have not solved the problem. There's not-
- (Brady: But only for n equals 6) That's n equals 6 and actually if
you continue n equal 7 again we can solve the problem easily by the same pattern. Now you asked- now so we've solved it for some numbers and we have not solved it for other numbers, what are the numbers? And now,
the magic is we started this problem- this problem came from India - Nandakumar and
Ramana Rao, 2006. Now in 1909 in the first issue of the first volume of the Journal of the math club in Madras, that's where Ramanujan came from. There's this paper which explains to us what we've done here, namely it solved the question for which values n does the nth row of Pascal's triangle have a common factor. That's exactly what we were looking at here. And it turns out that always if n is a power of a prime number. We look at this row and we disregard the 1s and we ask, do all these numbers have a common factor? 2 - yes. And here the common factor is 3. And here the common factor is 2. And here the common factor is 5. And these numbers don't have a common factor. It's not 2 because this is odd, and it's not 3 because this is not divisible by 3. Yeah so this one has a common factor again,
namely they are all divisible by 7. And if we do still the next one then this is 8, and this is 28, and this is 56, and this is 70, and 56, and 28 and 8 - and they are all even so it again has a common factor. And that's- if we have a common factor then this original problem is solved. And what are these numbers? So this is- 2 is a prime number. 3- (Brady: This is the row numbers
you're writing?) Yes that's the row numbers - so that's row number 2, there's row number 3, 3 is also a prime number. This is 4 and 4 is not a prime number but it's a power of a prime, it's 2 squared. And this is 5, which is again a prime number. And this here is 6, and 6 is 2 times 3. It's not a prime and it's not a prime power but it's a product of different primes. And that's where we don't solve the problem, that's [name]'s theorem. Now 7 again is a prime, 8 is 2 times 2 times 2 so it's a prime power, and 9 is 3 squared - it's a power of a prime, and 10 is 2 times 5 - which means again we have not solved our problem. Going back to a polygon, this means we do not know whether every polygon can be cut into ten pieces of equal area, equal perimeter. It may be possible but this method that we've used, and the whole machinery, and optimal transport and equivariant obstruction theory in the end; plainly does not answer this problem. Still an open problem for 6, for 10, for 12, for 14, for 15, for 21. But for
these other values it's solved, that's all the values where it's prime powers. I think the answer is also pretty magical. Of course someone at some point has to solve the problem for the other values, in the moment I don't know how to do that. If you'd like to see more from this interview I will put it on Numberphile2, our second channel. There'll be links on the screen and in the video description. This was filmed at the Mathematical Sciences Research Institute, in Berkeley California, where many of the world's best mathematicians come to visit and where we at Numberphile film a lot of these
interviews.
