Can you solve the seven planets riddle

你能解开七大行星之谜吗

来源视频

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.

你的星际警队 在一簇由七个小行星组成的星群上追踪到一帮危险的叛徒。

在他们的支援来到之前， 你必须尽快地把他们一网打尽。

很自然的，这些叛徒不会呆着不动。

为了逃避追捕，这些叛徒 会从一个行星逃到另一个行星。

虽然如此，你有一个巨大的优势。

每个小时，你的超高科技警车可以 扭曲空间，移动到任意一颗行星去。而叛徒破旧的偷渡飞船每个小时只可以 从一颗行星移动到相邻的行星。

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?

这些叛徒将不会坐以待毙。

他们只要有机会，一定会移动。

侦察员通知你，叛徒援军 距离你十小时之外。

你绝对不能让叛徒逃走。

在只移动十次或以下的条件下，你可想出不论叛徒怎么移动， 保证可以把叛徒逮捕的搜索顺序吗？

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.

要把叛徒抓拿归案并非容易。

首先，你并不知道叛徒 目前所在的行星。

没这个信息，你是很难推断 他们下一步会怎么走的。

那么，你该从何着手？

当处理这种问题时， 简化问题非常有效，这样你就可以更好地 了解事情的动态。假

设那些行星的排列跟这里的一样，但是没有最外围的那些行星， 所以剩下的只有中间那四颗。

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.

我们仍旧不知道叛徒 从哪颗行星开始。

不过，这儿有个关键的特点：第三颗行星毗邻所有的行星，这表示叛徒要么从这颗开始 然后移动到其他的行星，要么从其他的行星开始然后别无选择地移动到行星三号上。

我们只要搜索行星三号两次 就能解决问题。

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.

把外围的那些行星加入问题中， 也许会把情况变得稍微复杂，但是同样的方案仍然奏效。

我们想出的搜索顺序必须 最终使叛徒走投无路。

而这里有另一条信息可帮助我们：每小时，叛徒会从偶数行星移动到奇数行星，或者反过来。

知道这点能帮助我们简化问题，也就是把行星分成两组，这样就能单独地处理每一组。

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.

作为开始，假设叛徒在偶数行星开始：二号、四号或六号。

我们先从行星二号开始搜索。

如果他们不在那儿， 那么他们肯定在四号或六号开始，这表示他们会移动到 三号、五号或七号。

行星三号位于中央， 会给他们最多移动的选择，所以我们接下来会搜索三号。

如果他们不在那儿， 他们肯定是在五号或七号，意味着下一步他们将去四号或六号。

现在我们搜索四号。

如果他们也不在那儿， 他们肯定是去了六号，而且下一步只可去三号或七号。

如果我们接下来搜寻三号 但又找不到他们，我们知道他们去了七号， 并且走投无路了。

他们只可以迁移到六号，这表示 我们只需搜索五次就逮捕他们。

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.

当然，要这个方案奏效我们必须假设叛徒第一个小时 是在偶数行星开始的。

但如果这个假设是错的，怎么办呢？

若是这样， 他们肯定是在奇数行星开始的。

由于叛徒每个小时 都会移动到相邻的行星，他们的位置肯定是 在奇数和偶数行星中交替。

这表示如果他们在奇数行星开始，五次移动后，他们将会到偶数行星。

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.

如果我们前五次搜索不成功因为假设他们从 偶数的行星开始是错误的，我们只需做的是重复之前的步骤！

跟着这顺序搜索行星：二号、三号、四号、三号、六号、二号、三号、四号、三号、六号，就可以把叛徒逼得走投无路。

由于你的推理逻辑， 宇宙秩序得以维持。