Who cares about topology? (Inscribed rectangle problem)
Summary
TLDR视频介绍了一个有趣的数学问题——内嵌正方形问题,即在任意闭合曲线上是否总能找到四个点构成一个正方形。尽管这个问题尚未解决,但通过探讨内嵌矩形问题,视频展示了一个优雅的解决方案。通过定义一个函数,将曲线上的点对映射到三维空间中的点,利用莫比乌斯带的性质,证明了对于任何闭合曲线,总能找到两对点,它们共享中点且距离相等,从而构成一个矩形。这个过程不仅展示了拓扑学的美妙,也解释了数学家为何关注这些形状及其属性。
Takeaways
- 🔍 视频介绍了一个未解决的数学问题——内嵌正方形问题,即是否每个封闭环路都能找到一个内嵌的正方形。
- 🌟 视频中提到,对于内嵌长方形问题,虽然同样困难,但存在一个优雅且令人惊讶的解决方案。
- 🏷️ 拓扑学是数学的一个分支,它研究空间形状的性质和结构,这些性质在空间变形下保持不变。
- 📊 通过将问题的关注点从单个点转移到点对,视频中提供了一种新的视角来解决内嵌长方形问题。
- 📈 视频中定义了一个函数,将环路上的点对映射到3D空间中的一个点,以此来编码中点和距离信息。
- 🔧 通过将环路上的点对映射到三维空间,可以形成一个表面,这个表面有助于我们理解为什么这个函数的图像必须与自身相交。
- 🔄 视频中解释了如何将有序点对转换为无序点对,并展示了如何通过折叠和粘合来形成表示这些点对的几何形状。
- 🍩 通过将单位正方形对角线折叠,可以形成一个莫比乌斯带,它自然地表示环路上所有无序点对。
- 💡 莫比乌斯带的边缘代表重复的点对(xx),在将莫比乌斯带映射到3D空间的表面时,这些点必须恰好映射到环路上。
- 🎯 由于莫比乌斯带的形状特殊,无法避免它在映射过程中与自身相交,这证明了至少存在两个不同的点对,它们共享中点并且距离相等,从而形成长方形。
- 🧠 视频强调了拓扑学不仅仅是关于形状和变换的抽象概念,而是可以应用于解决具体问题的有力工具。
Q & A
拓扑学是什么,为什么人们关注它?
-拓扑学是数学的一个分支,研究空间中的对象在连续变形下的属性和关系。人们关注拓扑学,因为它能够帮助解决一些看似复杂的问题,例如视频中提到的内嵌正方形问题,以及通过直观和创新的方式理解形状和空间。
什么是内嵌正方形问题?
-内嵌正方形问题是一个未解决的数学问题,它询问是否每个封闭环路都可以找到四个点,这四个点构成一个正方形。这个问题揭示了即使在数学中,也存在一些看似简单但实际上难以解答的问题。
如何找到内嵌矩形的一个优雅解法?
-找到内嵌矩形的解法涉及将关注点从单独的点转移到点对。通过证明任何封闭环路上都存在两对点,它们的中点相同且距离相等,就可以确定这些点构成一个矩形。
视频中提到的函数是如何定义的?
-该函数定义为接受环路上的点对作为输入,并输出一个三维空间中的点。输出点位于由输入点对的中点确定的平面上的正上方,其高度等于点对之间的距离。
为什么说输出的三维图形会紧贴着环路?
-当点对在环路上越来越接近时,输出的点会降低高度,因为其高度等于点之间的距离。同时,中点也会更接近环路。当点对重合时,输出的点会精确地位于环路上。
如何从有序点对过渡到无序点对?
-通过将单位正方形对角线折叠,我们可以从有序点对过渡到无序点对。这个过程涉及到将表示相同点对的坐标对(例如(0.2, 0.3)和(0.3, 0.2))视为等同,并用一个连续的表面——即Möbius带——来表示它们。
Möbius带是什么?
-Möbius带是一个只有一个面和一个边界的二维表面。它通过将一个长条纸带的一端翻转180度后与另一端粘合而成。在拓扑学中,Möbius带是一个重要的非定向表面,常用于说明和理解复杂的拓扑概念。
为什么Möbius带能够用来解决内嵌矩形问题?
-Möbius带能够表示环路上所有无序点对的连续一一对应关系。通过将Möbius带映射到三维空间中的特定图形上,我们可以证明存在至少两对点映射到同一个输出点上,这意味着它们构成一个矩形。
为什么说Möbius带不能不相交地映射到二维平面上?
-Möbius带的奇特形状和性质意味着,如果尝试将其边缘粘合到二维平面上,必然会在某处相交。这是因为Möbius带的边缘代表了环路上的重复点对,而这些点对在映射过程中必须映射到平面上的同一个位置。
视频中提到的数学问题和解决方案对我们有什么启示?
-视频中的问题和解决方案展示了数学的美和力量。它们启示我们,即使是非常抽象和复杂的数学概念,也能够用来解决具体的问题,并且通过创造性的思考和直观的理解,我们可以更深入地探索数学的世界。
拓扑学中的环面和Möbius带是如何帮助我们理解问题的?
-环面和Möbius带是拓扑学中的重要概念,它们提供了一种自然的方式来理解环路上的点对。通过将这些数学结构应用于实际问题,我们能够以一种直观和创新的方式来探索和解决问题。
为什么说这个视频的解决方案可能是作者最喜欢的数学片段?
-这个视频的解决方案结合了几何、拓扑和创新的思考方式,展示了数学的优雅和力量。作者可能因为它的直观性、创造性以及解决问题的简洁性而特别喜欢这个解决方案。
Outlines
🌟 引言:拓扑学的奇妙之旅
本段介绍了视频的主题,包括一个未解决的数学问题、一个优雅的解决方案,以及对拓扑学的基本介绍。作者分享了自己儿时对数学的热爱,以及如何通过拓扑学的一些直观例子(如莫比乌斯带和咖啡杯与甜甜圈的拓扑等价性)激发了对数学的兴趣。然而,这些例子并没有解释拓扑学如何实际解决问题,直到作者遇到了一个关于内切正方形的问题,才真正理解了数学家对这些形状和属性的关注。
📐 解决方法:内切矩形的探索
这一部分详细阐述了内切矩形问题,并提出了一个弱化版本的解决方案。作者解释了如何通过关注成对的点而不是单个点来解决这个问题,并介绍了一个关于矩形的有趣事实。通过定义一个函数,将成对的点映射到3D空间中的一个点,这个点编码了中点和距离信息。通过这种方式,可以构建一个3D图形,帮助理解为什么这个表面必须与自身相交,从而找到内切矩形。
🔍 深入探讨:有序与无序点对
本段深入探讨了有序和无序点对的概念,并解释了如何通过有序点对来理解无序点对。作者通过将循环展开成区间、构建一个单位正方形,并将其边缘粘合起来形成 torus(环面),来表示有序点对。然后,通过折叠正方形的对角线并粘合边缘,形成了 Möbius 带,这是表示无序点对的自然形状。这个带子的边缘代表重复的点对,而带子的扭曲性质意味着在将其映射到3D空间的表面时,必须自我相交。
🎉 结论:内切矩形问题的证明
最后一部分总结了如何利用 Möbius 带的性质来证明内切矩形问题。通过将 Möbius 带映射到3D空间的表面,可以证明如果带子在映射过程中自我相交,那么至少存在两对不同的点,它们共享中点并且距离相同,从而形成了一个矩形。这个证明直观上很清晰,但要使其严谨,需要发展拓扑学领域的知识。作者强调,探讨 torus 和 Möbius 带并非仅仅是为了好玩,而是为了解决具体的数学问题,即理解循环上的点对。
Mindmap
Keywords
💡拓扑学
💡内嵌正方形问题
💡莫比乌斯带
💡内嵌矩形
💡点对
💡中点
💡距离
💡连续性
💡相交
💡映射
💡数学工具
Highlights
视频介绍了一个未解决的数学问题,即内嵌正方形问题。
讲解了一个较弱版本的内嵌正方形问题的优雅解决方案。
探讨了拓扑学的基本概念以及为什么人们对此感兴趣。
通过拓扑学的例子,比如莫比乌斯带,激发了对数学的兴趣。
解释了如何通过拓扑学解决实际问题,而不仅仅是理论上的展示。
提出了内嵌矩形问题,并探讨了其解决方案。
介绍了如何通过关注点对而不是个别点来解决问题。
定义了一个函数,将环上的点对映射到3D空间中的一个点。
通过在3D空间中绘制点,形成了一个与环紧密相连的表面。
讨论了为什么随着点对接近,绘制的点会降低高度。
通过连续函数的概念,展示了如何找到两个不同的点对,它们映射到3D空间中的同一点。
通过思考点对的有序和无序,引出了如何表示环上点对的自然形状。
通过将单位正方形对角折叠,展示了如何得到表示无序点对的莫比乌斯带。
莫比乌斯带的边缘代表了重复的点对,如xx。
通过莫比乌斯带和3D空间中的表面之间的自然映射,解决了内嵌矩形问题。
如果莫比乌斯带在映射到二维平面时自我相交,那么至少有两个不同的点对映射到3D表面的同一点。
这个解决方案展示了拓扑学的实用性,不仅仅是理论上的思考。
通过这个视频,观众可以更好地理解拓扑学中的概念,如莫比乌斯带和环上的点对。
这个视频提供了一个直观的方式来理解复杂的数学概念,如内嵌图形问题。
Transcripts
I've got several fun things for you this video.
An unsolved problem, a very elegant solution to a weaker version of the problem,
and a little bit about what topology is and why people care.
But before I jump into it, it's worth saying a
few words on why I'm excited to share this solution.
When I was a kid, since I loved math and sought out various mathy things,
I would occasionally find myself in some talk or a seminar where people
wanted to get the youth excited about things that mathematicians care about.
A very common go-to topic to excite our imaginations was topology.
We might be shown something like a mobius strip,
maybe building it out of construction paper by twisting a rectangle and gluing its ends.
Look, we'd be told, as we were asked to draw a line along the surface.
It's a surface with just one side.
Or we might be told that topologists view coffee mugs and donuts as the same thing,
since each has just one hole.
But these kinds of demos always left a lurking question.
How is this math?
How does any of this actually help to solve problems?
It wasn't until I saw the problem that I'm about to show you,
with its elegant and surprising solution, that I started to understand why
mathematicians actually care about some of these shapes and the properties they have.
So, there's this unsolved problem called the inscribed square problem.
If you have a closed loop, meaning you squiggle some line through space in a
potentially crazy way and you end up back where you started,
the question is whether or not you'll always be able to find four points on this
loop that make up a square.
If your closed loop was a circle, for example,
it's quite easy to find an inscribed square.
Infinitely many, in fact.
If your loop was instead an ellipse, it's still pretty easy to find an inscribed square.
The question is whether or not every possible closed loop,
no matter how crazy, has at least one inscribed square.
Pretty interesting, right?
I mean, just the fact that this is unsolved is interesting,
that the current tools of math can neither confirm nor deny that there's some loop with
no inscribed square in it.
Now, if we weaken the question a bit and ask about inscribed rectangles
instead of inscribed squares, it's still pretty hard, but there is a beautiful,
video-worthy solution that might actually be my favorite piece of math.
The idea is to shift the focus away from individual
points on the loop and instead onto pairs of points.
We'll use the following fact about rectangles.
Let's label the vertices of some rectangle ABCD.
Then the pair of points AC has a few things in common with the pair of points BD.
The distance between A and C equals the distance between B and D,
and the midpoint of A and C is the same as the midpoint of B and D.
In fact, any time you have two separate pairs of points in space, AC and BD,
if you can guarantee that they share a midpoint and that the distance between AC equals
the distance between B and D, it's enough to guarantee that those four points make up
a rectangle.
So what we're going to do is try to prove that for any closed loop,
it's always possible to find two distinct pairs of points on that
loop that share a midpoint and which are the same distance apart.
Take a moment to make sure that's clear.
We're finding two distinct pairs of points that share
a common midpoint and which are the same distance apart.
The way we'll go about this is to define a function that takes
in pairs of points on the loop and outputs a single point in 3D space,
which kind of encodes the midpoint and distance information.
It will be sort of like a graph.
Consider the closed loop to be sitting on the xy-plane in 3D space.
For a given pair of points, label their midpoint m,
which will be some point on the xy-plane, and label the distance between them d.
Plot the point, which is exactly d units above that midpoint m in the z-direction.
As you do this for many possible pairs of points,
you'll effectively be drawing through 3D space.
And if you do it for all possible pairs of points on the loop,
you'll draw out some kind of surface above the plane.
Now look at the surface, and notice how it seems to hug the loop itself.
This is actually going to be important later, so let's think about why it happens.
As the pair of points on the loop gets closer and closer, the plotted point gets lower,
since its height is by definition equal to the distance between the points.
Also, the midpoint gets closer and closer to the loop as the points approach each other.
Once the pair of points coincides, meaning the input of our
function looks like xx for some point x on the loop,
the plotted point of the surface will be exactly on the loop at the point x.
Okay, so remember that.
Another important fact is that this function is continuous,
and all that really means is that if you slightly adjust a given pair of points,
then the corresponding output in 3D is only slightly adjusted as well.
There's never a sudden discontinuous jump.
Our goal, then, is to show that this function has a collision,
that two distinct pairs of points each map to the same spot in 3D space.
Because the only way for that to happen is if they share a common midpoint,
and if their distance d apart from each other is the same.
So in some sense, finding an inscribed rectangle comes
down to showing that this surface has to intersect itself.
To move forward from here, we need to build up a
relationship with the idea of pairs of points on a loop.
Think about how we represent pairs of real numbers
using a two-dimensional coordinate plane.
Analogous to this, we're going to seek out a certain 2D surface
which naturally represents all pairs of points on the loop.
Understanding the properties of this surface will help to show
why the graph that we just defined has to intersect itself.
Now, when I say pair of points, there are two things that I could be talking about.
The first is ordered pairs of points, which would mean a
pair like AB would be considered distinct from the pair BA.
That is, there's some notion of which point is the first one.
The second idea is unordered points, where AB and BA would be considered the same thing,
where all that really matters is what the points are,
and there's no meaning to which one is first.
Ultimately, we want to understand unordered pairs of points,
but to get there, we need to take a path of thought through ordered pairs.
We'll start out by straightening out the loop,
cutting it at some point, and deforming it into an interval.
For the sake of having some labels, let's say that
this is the interval on the number line from 0 to 1.
By following where each point ends up, every point on the loop corresponds with a
unique number on this interval, except for the point where the cut happened,
which corresponds simultaneously to both endpoints of the interval,
meaning the numbers 0 and 1.
Now, the benefit of straightening out this loop like this is that we can start
thinking about pairs of points the same way we think about pairs of numbers.
Make a y-axis using a second interval, then associate each pair of values
on the interval with a single point in this 1x1 square that they span out.
Every individual point of this square naturally corresponds to a pair of
points on the loop, since its x and y coordinates are each numbers between 0 and 1,
which are in turn associated to some unique point on the loop.
Remember, we're trying to find a surface that naturally represents the set of
all pairs of points on the loop, and this square is the first step to doing that.
The problem is that there's some ambiguity when it comes to the edges of the square.
Remember, the endpoints 0 and 1 on the interval really correspond to
the same point of the loop, as if to say that those endpoints need to
be glued together if we're going to faithfully map back to the loop.
So all of the points on the left edge of the square, like 0, 0, 0.1, 0, 0.2,
on and on and on, really represent the same pair of points on the loop as the
corresponding coordinates on the right edge of the square, 1, 0.1, 1, 0.2,
on and on and on.
So for this square to represent the pairs of points on the loop in a unique way,
we need to glue this left edge to the right edge.
I'll mark each edge with some arrows to remember how the edges need to be lined up.
Likewise, the bottom edge needs to be glued to the top edge,
since y-coordinates of 0 and 1 really represent the same second point in a given pair
of points on the loop.
If you bend this square to perform the gluing,
first rolling it into a cylinder to glue the left and right edges,
then gluing the ends of that cylinder, which represent the top and bottom edges,
we get a torus, better known as the surface of a doughnut.
Every individual point on this torus corresponds to a unique pair of points on the loop,
and likewise, every pair of points on the loop corresponds to
some unique point on this torus.
The torus is to pair of points on the loop what the
xy-plane is to pairs of points on the real number line.
The key property of this association is that it's continuous both ways,
meaning if you nudge any point on the torus by just a tiny amount,
it corresponds to only a very slight nudge to the pair of points on the loop,
and vice versa.
So if the torus is the natural shape for ordered pairs of points on the loop,
what's the natural shape for unordered pairs?
After all, the whole reason we're doing this is to show that two distinct pairs
of pairs of points on the loop share a midpoint and are the same distance apart.
But if we consider a pair AB to be distinct from BA,
then that would trivially give us two separate pairs which have the same
midpoint and distance apart.
That's like saying you can always find a rectangle so
long as you consider any pair of points to be a rectangle.
Not helpful.
So let's think about this.
Let's think about how to represent unordered pairs
of points looking back at our unit square.
We need to say that the coordinates 0.2, 0.3 represent the same pair as 0.3, 0.2.
Or that 0.5, 0.7 really represents the same thing as 0.7, 0.5.
And in general, any coordinates x, y has to represent the same thing as y, x.
Once again, we capture this idea by gluing points together when they're supposed to
represent the same pair, which in this case requires folding the square over diagonally.
Now notice this diagonal line, the crease of the fold.
It represents all pairs of points that look like xx,
meaning the pairs which are really just a single point written twice.
Right now, it's marked with a red line.
And you should remember it.
It will become important to know where all of these pairs like xx live.
But we still have some arrows to glue together here.
We need to glue that bottom edge to the right edge.
And the orientation with which we do this is going to be important.
Points towards the left of the bottom edge have to be
glued to points towards the bottom of the right edge.
And points towards the right of the bottom edge have
to be glued to points towards the top of the right edge.
It's weird to think about, right?
Go ahead, pause and ponder this for a moment.
The trick, which is kind of clever, is to make a diagonal cut,
which we need to remember to glue back in just a moment.
After that, we can glue the bottom and the right like so.
But notice the orientation of the arrows here.
To glue back what we just cut, we don't simply
connect the edges of this rectangle to get a cylinder.
We have to make a twist.
Doing this in 3D space, the shape we get is a Möbius strip.
Isn't that awesome?
Evidently, the surface which represents all pairs
of unordered points on the loop is the Möbius strip.
And notice, the edge of this strip, shown here in red,
represents the pairs of points that look like xx,
those which are really just a single point listed twice.
The Möbius strip is to unordered pairs of points on
the loop what the xy-plane is to pairs of real numbers.
That totally blew my mind when I first saw it.
Now, with this fact that there is a continuous one-to-one association
between unordered pairs of points on the loop and individual points on this Möbius strip,
we can solve the inscribed rectangle problem.
Remember, we had defined this special kind of graph in 3D space,
where the loop was sitting in the xy-plane.
For each pair of points, you consider their midpoint m, which lives on the xy-plane,
and their distance d apart, and you plot a point which is exactly d units above m.
Because of the continuous one-to-one association between pairs
of points on the loop and the Möbius strip, this gives us a
natural map from the Möbius strip onto this surface in 3D space.
For every point on the Möbius strip, consider the pair of points on the loop
that it represents, then plug that pair of points into the special function.
And here's the key point.
When pairs of points on the loop are extremely close together,
the output of the function is right above the loop itself,
and in the extreme case of pairs of points like xx,
the output of the function is exactly on the loop.
Since points on this red edge of the Möbius strip correspond to pairs like xx,
when the Möbius strip is mapped onto this surface,
it must be done in such a way that the edge of the strip gets mapped right onto
that loop in the xy-plane.
But if you stand back and think about it for a moment,
considering the strange shape of the Möbius strip,
there is no way to glue its edge to something two-dimensional without
forcing the strip to intersect itself.
Since points of the Möbius strip represent pairs of points on the loop,
if the strip intersects itself during this mapping,
it means that there are at least two distinct pairs of points that correspond to the same
output on this surface, which means they share a midpoint and are the same distance
apart, which in turn means that they form a rectangle.
And that's the proof!
Or at least, if you're willing to trust me in saying that you can't glue the edge of
the Möbius strip to a plane without forcing it to intersect itself, then that's the proof.
This fact is intuitively clear looking at the Möbius strip,
but in order to make it rigorous, you basically need to start developing the
field of topology.
In fact, for any of you who have a topology class in your future,
going through the exercise of trying to justify this is a good way to
gain an appreciation for why topologists choose to make certain definitions.
And I want you to take note of something here.
The reason for talking about the torus and the Möbius strip
was not because we were playing around with construction paper,
or because we were daydreaming about deforming a coffee mug.
They came up as a natural way to understand pairs of points on a loop,
and that's something that we needed to solve a concrete problem.
Thank you.
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