Can you solve the seven planets riddle Edwin F. Meyer

Your interstellar police squad
has tracked a group of dangerous rebels

to a cluster of of seven small planets.

Now you must apprehend them quickly
before their reinforcements arrive.

Of course, the rebels won’t just stay put.

They’ll try to dodge you
by moving from planet to planet.

But you have one major advantage.

Every hour, your state-of-the-art cruiser
can warp between any two planets,

while their beat-up smuggling ship

can only jump to an adjacent planet
in that same time.

These rebels don’t like to stay put.

Every time they can relocate, they will.

Your scouts tell you that the approaching
rebel fleet is 10 hours away.

You can’t risk letting the rebels escape.

Can you devise a sequence
for searching the planets

that’s guaranteed to catch them
in 10 warps or less,

no matter what moves they make?

Rounding up the rebels won’t be easy.

For one, you have no way of knowing which
planet they’re on to begin with.

And without that information, it’s hard
to determine where they’ll move next.

So where do you begin?

When tackling problems of this kind
it often helps to simplify things,

so you can better understand
their dynamics.

Let’s imagine that this cluster
has the same arrangement

but no outermost planets,
leaving only the four in the center.

We still don’t know which planet
the rebels start on.

But there’s one key feature:

the third planet
is adjacent to all others,

which means the rebels either start there
and move somewhere else,

or start on one of the other planets

and have no choice
but to move to planet three.

Simply checking planet three
twice in a row would do the trick.

Adding the three outer planet
adds a bit more complexity,

but the same strategy remains.

We want to check the planets in an order
that will eventually corner the rebels.

And there’s another insight
that can help us:

each hour, the rebels move from
an even-numbered planet

to an odd-numbered planet,

or vice versa.

This gives us a way
to simplify the problem

by dividing the planets into two subsets,

and tackling each one separately.

For starters, let’s assume the rebels
begin on an even-numbered planet:

either two,

four,

or six.

So we’ll search planet two first.

If they’re not there, they must have
started on either four

or six.

which means they can move to three,

five,

or seven.

Planet three at the center gives them
the most options for their next move,

so we’ll want to check there next.

If we don’t find them, they must have been
at planet five

or seven,

meaning they’ll next move to four

or six.

Let’s now search planet four.

If they’re not there, they must have gone
to the sixth planet

and can only flee to three

or seven.

If we next scour planet three
and don’t find them,

we know they went to planet seven
and are now cornered.

They can only move to planet six, where
we’ll apprehend them on our fifth search.

Of course, this plan only works

assuming that the rebels were on
an even-numbered planet in the first hour.

But what if that assumption was wrong?

In that case, they must’ve started
on an odd-numbered planet.

And because they move to an adjacent
planet every hour,

their location must alternate between
odd and even-numbered planets.

This means that if they were on
an odd-numbered planet to start,

after five moves, they’d be
on an even-numbered planet.

So if our first five searches missed them

because our assumption that they started
on an even-numbered planet was wrong,

all we have to do now
is repeat the sequence!

Searching the planets in order

two,

three,

four,

three,

six,

two,

three,

four,

three,

six,

leaves the rebels nowhere to run.

Thanks to your deductive reasoning,
order is restored to the galaxy.

你的星际警察小队
已经追踪了一群危险的叛军

到一个由七个小行星组成的星团。

现在你必须
在他们的援军到达之前迅速逮捕他们。

当然,叛军不会只是原地踏步。

他们会试图
通过从一个星球移动到另一个星球来躲避你。

但你有一个主要优势。

每小时,您最先进的巡洋舰
可以在任意两个行星之间穿梭,

而他们破旧的走私船

只能在同一时间跳到相邻的行星

这些叛军不喜欢呆在原地。

每次他们可以搬迁时,他们都会搬家。

你的侦察兵告诉你,即将到来的
叛军舰队距离我们还有 10 小时的路程。

你不能冒险让叛军逃跑。

你能否设计一个
搜索

行星的序列,保证
在 10 次或更少的经线内捕捉到它们,

无论它们采取什么行动?

围捕叛军并不容易。

一方面,你无法知道
他们从哪个星球开始。

如果没有这些信息,就
很难确定他们下一步会搬到哪里。

那么你从哪里开始呢?

在处理此类问题时,
它通常有助于简化事情,

因此您可以更好地了解
它们的动态。

让我们想象一下,这个星团
有相同的排列,

但没有最外层的行星,
只剩下四个在中心。

我们仍然不知道
叛军从哪个星球开始。

但是有一个关键特征

:第三颗行星
与所有其他行星相邻,

这意味着叛军要么从那里开始,
然后搬到其他地方,

要么从其他行星之一开始,

别无选择,只能搬到第三颗行星。

只需连续两次检查行星三
就可以了。

添加三个外行星
会增加一点复杂性,

但相同的策略仍然存在。

我们希望以
最终将反叛者逼入绝境的顺序检查行星。

还有另一个
可以帮助我们的见解:

每个小时,叛军都会
从偶数行星移动

到奇数行星,

反之亦然。

这为我们提供了一种

通过将行星分成两个子集

并分别处理每个子集来简化问题的方法。

首先,让我们假设叛军
从一个偶数行星开始:

两个、

四个

或六个。

所以我们将首先搜索行星二。

如果他们不在那里,他们一定是
从四个

或六个开始的。

这意味着他们可以移动到三个、

五个

或七个。

位于中心的第三行星
为他们的下一步行动提供了最多的选择,

所以我们接下来要检查那里。

如果我们没有找到它们,它们一定
在第五

或第七行星,

这意味着它们接下来将移动到第四

或第六。

现在让我们搜索第四行星。

如果他们不在那里,他们肯定
去了第六颗星球

,只能逃到

三七颗。

如果我们接下来搜寻第三颗行星
但没有找到它们,

我们就知道它们去了第七颗行星
,现在已经走投无路了。

他们只能移动到第六行星,
我们将在第五次搜索中逮捕他们。

当然,该计划仅在

假设叛军
在第一个小时内位于偶数行星上的情况下才有效。

但如果这个假设是错误的呢?

在那种情况下,他们一定是
从一个奇数星球开始的。

而且由于它们每小时都会移动到相邻的
行星,因此

它们的位置必须在
奇数和偶数行星之间交替。

这意味着,如果他们
在奇数行星上开始,

五次移动后,他们将
在偶数行星上。

因此,如果我们的前五次搜索错过了它们,

因为我们假设它们开始
于偶数行星是错误的,

那么我们现在要做的
就是重复这个序列!

二、

三、

四、

三、

六、

二、

三、

四、

三、

六的顺序搜索行星,

使叛军无处可逃。

由于您的演绎推理,
银河系恢复了秩序。