Can you solve the honeybee riddle Dan Finkel

You’re a biologist on a mission

to keep the rare honeybee
Apis Trifecta from going extinct.

The last 60 bees of the species
are in your terrarium.

You’ve already constructed wire frames
of the appropriate size and shape.

Now you need to turn
them into working beehives

by helping the bees fill
every hex with wax.

There are two ways to fill a given hex.

The first is to place a bee into it.

Once placed, a bee cannot
be removed without killing it.

The second:

if at any point an unfilled hex has three
or more neighboring wax-filled hexes,

the bees already in the hive
will move in and transform it.

Once the bees have transformed
every hex in a hive,

you can place an additional bee inside
and it’ll specialize into a queen.

The hive, if well cared for,
will eventually produce new bees

and continue the species.

If there are no hexes with three or more
transformed neighbors,

the bees will just sit and wait.

And once a bee transforms a hex,
it can never become a queen.

You could put 59 bees in one wire hive,
wait till they transform all the hexes,

and then create a queen.

But then just one collapse
would end the species.

The more viable hives
you can make now, the better.

So how many can you make with 60 bees?

Pause the video to figure
it out yourself

Answer in 3

Answer in 2

Answer in 1

Answer in 0

What you’re looking for here is some
kind of self-sustaining chain reaction,

where a small number of bees
will transform an entire hive.

The lower the number of bees needed,
the better.

So how low can we go,
and how can we engineer a chain reaction?

Let’s start with the first question.

There’s a really clever approach to this,

which involves counting the sides
of the filled-in hexes,

and examining their total perimeter.

Let’s suppose we put bees
in these three hexes.

The total transformed
perimeter has 18 sides.

But the middle hex
has three transformed neighbors,

so the bees will transform it too.

What happens to the perimeter?

It’s still 18!

And even after the bees transform the next
sets of hexes with three neighbors,

it still won’t change.

What’s going on here?

Each hex that has at least three sides
touching the bee-friendly space

will remove those sides from the perimeter
when it transforms.

Then it adds at most three new sides
to the perimeter.

So the perimeter of the transformed hexes
will either stay the same or shrink.

The final perimeter
of the entire hive is 54,

so the total perimeter of the hexes
we place bees in at the start

must be at least 54 as well.

Dividing that 54 by the six sides
on each non-adjacent hex

tells us it’ll take at least 9 bees
to transform the entire hive.

That’s a great start,

but we still have the tough question
of where the nine bees should go,

and if we’ll need more.

Let’s think smaller.

We already know that three bees could
completely transform a hive this big.

What about a slightly bigger one?

The perimeter of this hive is 30,

which means we’ll need
at least 5 bees to fill it in.

With 6 it’d be easy.

Placing them like this would fill out
the whole hive in just three steps.

But we can do better!

We don’t actually need to place
a bee on this hex,

since the other bees will transform
that spot on their own.

It looks like we have
the beginning of a pattern.

Can we extend it to our full hive?

That would mean placing
our 9 bees like so.

Once they get to work,
they’ll create a chain reaction

that fills in the center of the hive
and extend it to its edges.

Add a 10th bee to the completed hive
and it becomes a queen.

Repeat that process five more times

and you’ve helped the last 60 members
of Apis trifecta

create 6 producing hives.

All in all,
it’s a pretty good bee-ginning.

你是一名生物学家,肩负

着防止稀有蜜蜂
Apis Trifecta 灭绝的使命。

该物种的最后 60 只蜜蜂
在您的玻璃容器中。

您已经构建
了适当大小和形状的线框。

现在你需要帮助

蜜蜂用蜡填满每一个格子,把它们变成工作蜂箱。

有两种方法可以填充给定的十六进制。

首先是在里面放一只蜜蜂。

一旦放置,蜜蜂不能
在不杀死它的情况下被移除。

第二个:

如果在任何时候一个未填充的格子有三个
或更多相邻的蜡填充格子,

已经在蜂巢中的蜜蜂
会移动并改变它。

一旦蜜蜂改变
了蜂巢中的每一个格子,

你就可以在里面再放一只蜜蜂
,它就会专门化成蜂王。

蜂巢如果得到很好的照顾,
最终将产生新的蜜蜂

并延续该物种。

如果没有具有三个或更多
转化邻居的格子

,蜜蜂只会坐等。

而一旦蜜蜂转化了一个格子,
它就永远不可能成为蜂王。

你可以将 59 只蜜蜂放在一个蜂巢中,
等它们改变所有格子,

然后创建一个蜂王。

但只要一次崩溃
就会结束这个物种。

您现在可以制作的可行的荨麻疹越多越好。

那么,你可以用 60 只蜜蜂制作多少只蜜蜂?

暂停视频自己弄清楚

回答 3

回答 2

回答 1

回答 0

您在这里寻找的是
某种自我维持的连锁反应,

其中少量蜜蜂
会改变整个蜂巢。

需要的蜜蜂
数量越少越好。

那么我们可以走多低
,我们如何设计连锁反应呢?

让我们从第一个问题开始。

有一个非常聪明的方法,

它包括计算
填充的六边形的边,

并检查它们的总周长。

假设我们将蜜蜂
放在这三个格子中。

总变换
周长有 18 条边。

但是中间的格子
有三个变形的邻居,

所以蜜蜂也会变形它。

外围会发生什么?

还是18岁!

即使在蜜蜂
与三个邻居转换下一组格子之后,

它仍然不会改变。

这里发生了什么?

每个至少有三个边
与蜜蜂友好空间接触的六角格在变形时

将把这些边从周边移除

然后它最多在周边添加三个新
边。

所以转换后的六边形的
周长要么保持不变,要么缩小。 整个蜂巢

的最终周长
是 54,

因此我们在开始时放置蜜蜂的格子的总周长也

必须至少为 54。

将 54 除以
每个不相邻的六边形的六个边

告诉我们,至少需要 9 只蜜蜂
才能改造整个蜂巢。

这是一个很好的开始,

但我们仍然有一个棘手的
问题,九只蜜蜂应该去哪里,

以及我们是否需要更多。

让我们想想更小。

我们已经知道三只蜜蜂可以
完全改造这么大的蜂巢。

稍微大一点的呢?

这个蜂巢的周长是 30,

这意味着我们
至少需要 5 只蜜蜂来填满它。

如果有 6 只蜜蜂就很容易了。

像这样放置它们
只需三个步骤即可填满整个蜂巢。

但我们可以做得更好!

我们实际上不需要
在这个格子上放置一只蜜蜂,

因为其他蜜蜂会
自己改变那个地方。

看起来我们有
一个模式的开始。

我们可以将它扩展到我们的完整蜂巢吗?

那将意味着
像这样放置我们的 9 只蜜蜂。

一旦他们开始工作,
他们就会产生连锁反应

,填充蜂巢的中心
并将其延伸到蜂巢的边缘。

在完成的蜂巢中添加第 10 只蜜蜂
,它就会成为蜂王。

再重复该过程五次

,您已帮助 Apis trifecta 的最后 60 名成员

创建了 6 个生产蜂箱。

总而言之,
这是一个非常好的蜜蜂轧花。