Can you solve the seven planets riddle

你能解開(kāi)七大行星之謎嗎

來(lái)源視頻

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.

你的星際警隊(duì)?在一簇由七個(gè)小行星組成的星群上追蹤到一幫危險(xiǎn)的叛徒。

在他們的支援來(lái)到之前，?你必須盡快地把他們一網(wǎng)打盡。

很自然的，這些叛徒不會(huì)呆著不動(dòng)。

為了逃避追捕，這些叛徒?會(huì)從一個(gè)行星逃到另一個(gè)行星。

雖然如此，你有一個(gè)巨大的優(yōu)勢(shì)。

每個(gè)小時(shí)，你的超高科技警車可以?扭曲空間，移動(dòng)到任意一顆行星去。而叛徒破舊的偷渡飛船每個(gè)小時(shí)只可以?從一顆行星移動(dòng)到相鄰的行星。

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?

這些叛徒將不會(huì)坐以待斃。

他們只要有機(jī)會(huì)，一定會(huì)移動(dòng)。

偵察員通知你，叛徒援軍?距離你十小時(shí)之外。

你絕對(duì)不能讓叛徒逃走。

在只移動(dòng)十次或以下的條件下，你可想出不論叛徒怎么移動(dòng)，?保證可以把叛徒逮捕的搜索順序嗎？

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.

要把叛徒抓拿歸案并非容易。

首先，你并不知道叛徒 目前所在的行星。

沒(méi)這個(gè)信息，你是很難推斷 他們下一步會(huì)怎么走的。

那么，你該從何著手？

當(dāng)處理這種問(wèn)題時(shí)， 簡(jiǎn)化問(wèn)題非常有效，這樣你就可以更好地 了解事情的動(dòng)態(tài)。假

設(shè)那些行星的排列跟這里的一樣，但是沒(méi)有最外圍的那些行星， 所以剩下的只有中間那四顆。

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.

我們?nèi)耘f不知道叛徒 從哪顆行星開(kāi)始。

不過(guò)，這兒有個(gè)關(guān)鍵的特點(diǎn)：第三顆行星毗鄰所有的行星，這表示叛徒要么從這顆開(kāi)始 然后移動(dòng)到其他的行星，要么從其他的行星開(kāi)始然后別無(wú)選擇地移動(dòng)到行星三號(hào)上。

我們只要搜索行星三號(hào)兩次 就能解決問(wèn)題。

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.

把外圍的那些行星加入問(wèn)題中， 也許會(huì)把情況變得稍微復(fù)雜，但是同樣的方案仍然奏效。

我們想出的搜索順序必須 最終使叛徒走投無(wú)路。

而這里有另一條信息可幫助我們：每小時(shí)，叛徒會(huì)從偶數(shù)行星移動(dòng)到奇數(shù)行星，或者反過(guò)來(lái)。

知道這點(diǎn)能幫助我們簡(jiǎn)化問(wèn)題，也就是把行星分成兩組，這樣就能單獨(dú)地處理每一組。

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.

作為開(kāi)始，假設(shè)叛徒在偶數(shù)行星開(kāi)始：二號(hào)、四號(hào)或六號(hào)。

我們先從行星二號(hào)開(kāi)始搜索。

如果他們不在那兒， 那么他們肯定在四號(hào)或六號(hào)開(kāi)始，這表示他們會(huì)移動(dòng)到 三號(hào)、五號(hào)或七號(hào)。

行星三號(hào)位于中央， 會(huì)給他們最多移動(dòng)的選擇，所以我們接下來(lái)會(huì)搜索三號(hào)。

如果他們不在那兒， 他們肯定是在五號(hào)或七號(hào)，意味著下一步他們將去四號(hào)或六號(hào)。

現(xiàn)在我們搜索四號(hào)。

如果他們也不在那兒， 他們肯定是去了六號(hào)，而且下一步只可去三號(hào)或七號(hào)。

如果我們接下來(lái)搜尋三號(hào) 但又找不到他們，我們知道他們?nèi)チ似咛?hào)， 并且走投無(wú)路了。

他們只可以遷移到六號(hào)，這表示 我們只需搜索五次就逮捕他們。

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.

當(dāng)然，要這個(gè)方案奏效我們必須假設(shè)叛徒第一個(gè)小時(shí) 是在偶數(shù)行星開(kāi)始的。

但如果這個(gè)假設(shè)是錯(cuò)的，怎么辦呢？

若是這樣， 他們肯定是在奇數(shù)行星開(kāi)始的。

由于叛徒每個(gè)小時(shí) 都會(huì)移動(dòng)到相鄰的行星，他們的位置肯定是 在奇數(shù)和偶數(shù)行星中交替。

這表示如果他們?cè)谄鏀?shù)行星開(kāi)始，五次移動(dòng)后，他們將會(huì)到偶數(shù)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.

如果我們前五次搜索不成功因?yàn)榧僭O(shè)他們從 偶數(shù)的行星開(kāi)始是錯(cuò)誤的，我們只需做的是重復(fù)之前的步驟！

跟著這順序搜索行星：二號(hào)、三號(hào)、四號(hào)、三號(hào)、六號(hào)、二號(hào)、三號(hào)、四號(hào)、三號(hào)、六號(hào)，就可以把叛徒逼得走投無(wú)路。

由于你的推理邏輯， 宇宙秩序得以維持。