Can you solve the time travel riddle Dan Finkel

Your internship in Professor Ramsey’s
physics lab has been amazing.

Until, that is, the professor accidentally
stepped through a time portal.

You’ve got just a minute to jump through
the portal to save him before it closes

and leaves him stranded in history.

Once you’re through it,
the portal will close,

and your only way back will be
to create a new one

using the chrono-nodules from your lab.

Activated nodules connect to each other

via red or blue tachyon entanglement.

Activate more nodules and they’ll connect

to all other nodules in the area.

As soon as a red or blue triangle is
created with a nodule at each point,

it opens a doorway through time that
will take you back to the present.

But the color of each individual
connection manifests at random,

and there’s no way to choose
or change its color.

And there’s one more problem:

each individual nodule creates a
temporal instability

that raises the chances the portal
might collapse as you go through it.

So the fewer you bring, the better.

The portal’s about to close.

What’s the minimum number of nodules
you need to bring

to be certain you’ll create a red or
blue triangle and get back to the present?

Pause here if you want to figure it out for yourself!

Answer in: 3

Answer in: 2

Answer in: 1

This question is so rich that an entire
branch of mathematics

known as Ramsey Theory developed from it.

Ramsey Theory is home to some
famously difficult problems.

This one isn’t easy, but it can be handled

if you approach it systematically.

Imagine you brought just three nodules.

Would that be enough? No - for example,
you might have two blue

and one red connection,
and be stuck in the past forever.

Would four nodules be enough?
No - there are many arrangements here

that don’t give a blue or red triangle.

What about five?

It turns out there is an arrangement of
connections

that avoids creating
a blue or red triangle.

These smaller triangles don’t count because
they don’t have a nodule at each corner.

However, six nodules will always create a
blue triangle or a red triangle.

Here’s how we can prove that without
sorting through every possible case.

Imagine activating the sixth nodule,

and consider how it might connect
to the other five.

It could do so in one of six ways:

with five red connections, five blue
connections, or some mix of red and blue.

Notice that every possibility has at least
three connections of the same color

coming from this nodule.

Let’s look at just the nodules
on the other end

of those same three color connections.

If the connections were blue,

then any additional blue connection between
those three would give us a blue triangle.

So the only way we could get in trouble

is if all the connections
between them were red.

But those three red connections
would give us a red triangle.

No matter what happens,
we’ll get a red or a blue triangle,

and open our doorway.

On the other hand,

if the original three connections
were all red instead of blue,

the same argument still works,
with all the colors flipped.

In other words, no matter how the
connections are colored,

six nodules will always create a red or
blue triangle and a doorway leading home.

So you grab six nodules and jump through
the portal.

You were hoping your internship would
give you valuable life experience.

Turns out, that didn’t take much time.

你在 Ramsey 教授
物理实验室的实习非常棒。

直到,也就是说,教授不
小心穿过了一个时间传送门。

您只有一分钟的时间
跳过门户以在门户关闭

并让他陷入历史之前拯救他。

一旦你通过它
,门户就会关闭

,你唯一的回归方式
就是

使用你实验室的计时结节创建一个新的。

激活的结节

通过红色或蓝色的快子纠缠相互连接。

激活更多结节,它们将连接

到该区域的所有其他结节。

一旦
在每个点上创建一个带有结节的红色或蓝色三角形,

它就会打开一扇穿越时间的大门,
将您带回到现在。

但是每个单独连接的颜色是
随机显现的

,没有办法选择
或改变它的颜色。

还有一个问题:

每个单独的结节都会产生
时间上的不稳定性

,这会增加门户
在您通过它时可能坍塌的可能性。

所以带的越少越好。

门户即将关闭。

您需要携带多少个结节

才能确定您会创建一个红色或
蓝色三角形并回到现在?

如果您想自己弄清楚,请在此处暂停!

答案:3

答案:2

答案:1

这个问题是如此丰富,以至于被称为拉姆齐理论的整个
数学分支都

从它发展而来。

拉姆齐理论是一些
著名的难题的根源。

这并不容易,但如果你系统地处理它,它是可以处理的

想象一下,你只带了三个结节。

这样就够了吗? 不——例如,
您可能有两个蓝色

和一个红色连接,
并且永远停留在过去。

四个结节就够了吗?
不 - 这里有很多安排

没有给出蓝色或红色三角形。

五个呢?

事实证明,有一种连接排列

可以避免
创建蓝色或红色三角形。

这些较小的三角形不算数,因为
它们的每个角落都没有结节。

但是,六个结节总是会创建一个
蓝色三角形或一个红色三角形。

以下是我们如何证明这一点,而无需
对所有可能的情况进行分类。

想象一下激活第六个结节,

并考虑它如何连接
到其他五个结节。

它可以通过以下六种方式之一来实现:

使用五个红色连接、五个蓝色
连接或红色和蓝色的某种混合。

请注意,每种可能性至少有
三个来自该结节的相同颜色的连接

让我们看看

同样的三色连接另一端的结节。

如果连接是蓝色的,

那么这三个之间任何额外的蓝色连接
都会给我们一个蓝色三角形。

因此,我们遇到麻烦的唯一方法


它们之间的所有连接都是红色的。

但是这三个红色连接
会给我们一个红色三角形。

无论发生什么,
我们都会得到一个红色或蓝色的三角形,

然后打开我们的门。

另一方面,

如果最初的三个连接
都是红色而不是蓝色,

那么同样的论点仍然有效
,所有颜色都被翻转了。

换句话说,无论
连接的颜色如何,

六个结节总是会形成一个红色或
蓝色的三角形和一个通往家的门口。

因此,您抓住六个结节并
跳过门户。

你希望你的实习
能给你带来宝贵的生活经验。

事实证明,这并没有花费太多时间。