The Gauntlet | Think Like A Coder, Ep 8

譯者: Lilian Chiu 審譯者: H_L Au

《用程式設計師的方式思考》

地點：哈林堡 地點：198 森林

第八集：挑戰

Their fall from the tower sends Ethic and Hedge

艾希克和海吉從塔上落下之後，

spinning into the rapids of a river of pure energy.

便天旋地轉地落入 純能量的急流當中。

This torrent flows from the Bradbarrier all the way to Huxenborg.

這條洪流從八達圍 一路流到哈森堡。

There an entire city’s worth of factories

在那裡，有相當於 一整個城市的工廠，

build the robots and house the Node of Memory,

用來建造機器人和存放記憶之泉，

正是艾希克必須要收集的 三項強大寶物的最後一項。

the last of the three powerful artifacts Ethic needs to collect.

After a long day and a longer night

在經過漫長的一天 和更漫長的一夜之後，

they find themselves in a canyon of brick and steel.

他們到了一個由磚塊 和鋼鐵形成的峽谷。

Just when they’re about to reach the end of the line,

他們快要到達盡頭時，

a rope catches them.

一條繩索套住了他們。

Their savior, Lemma, has been waiting for them.

他們的救星雷瑪 一直在那裡等他們。

When Ethic claimed the Node of Creation from the forest tower,

當艾希克從森林中的高塔 取得創造之泉時，

radios all across the land came back to life.

各地的無線電都恢復通訊了。

Adila, the resistance leader, immediately started contacting her allies,

反抗軍領袖艾迪拉馬上 就開始聯絡她的盟友，

none more important than Lemma,

當中最重要的就是雷瑪了，

a brilliant scientist working from within Huxenborg to bring down the machines.

她是個聰明的科學家， 在哈森堡內部努力破壞機器。

Unfortunately, the radios also tipped off the robots.

不幸的是，無線電 也將消息走漏給機器人。

So they’ve taken defensive measures

於是，它們採取了防禦措施，

to protect the final artifact in its home in the very heart of the city.

保護身處在城市核心的 最後一項寶物。

There’s only one way to get there: the gauntlet of forking paths.

要到達只有一種方法，岔路挑戰。

It’s a deadly series of luminous conveyors that wind underneath Huxenborg.

這個挑戰是在哈森堡地下 一連串的致命迂迴發光輸送帶。

Starting from the current position,

從目前的位置開始，

each section runs for a distance, then splits in two.

每段輸送帶會輸送一段距離， 接著便一分為二。

Every branch does the same thing, again and again.

每個分支都會不斷重覆這個流程。

There are thousands of branches.

總共有數千條分支。

Only one path leads to the artifact; all the others to destruction.

只有一條路通往寶物，

其他則全部都通往死亡。

Fortunately, the Node of Creation has granted Hedge a strange power:

幸運的是，創造之泉 將一種奇特的力量賦予海吉：

he can produce slightly smaller versions of himself.

它能製造出它自己的 複製品，但比較小。

Each version can do only two things: radio information back to its parent,

每一個複製品都會做兩件事：

把資訊用無線電回傳給它的母體，

and produce slightly smaller versions of itself…

以及再複製出比較小的自己，

which can do the same two things, as can their children,

而這些複製品同樣也會 做這兩件事，以此類推，

for as many generations as needed.

有必要就可以一直產生新世代。

A patrol is closing in on their position, so Ethic’s time is limited.

巡邏兵正在靠近，所以時間有限。

What instructions should she give Hedge to find the one safe path?

她應該給海吉什麼指令， 才能找到那唯一一條安全的通路？

Pause the video to figure it out yourself.

如果想要自己解題，請在此暫停。

提示即將公佈：五

Hint in 3

Hint in 2

提示即將公佈：三

Hint in 1

提示即將公佈：一

Programmers have an elegant tool in their arsenal called recursion.

程式設計師有項利器，叫做遞迴。

Recursion is when you have a set of instructions that refers back to itself.

當你有一組指令，又會導至 它自己本身時，就叫做遞迴。

It’s like using a word in its own definition,

就像是在一個詞的定義中 用到這個詞一樣，

except where that’s frowned upon, this is incredibly effective.

差別在於不能使用它時， 這就是非常有效的方法。

Recursion involves repetition, but in a different way than loops.

遞迴會涉及重覆，但和迴圈不同。

Where a loop takes one action and repeats it again and again,

迴圈是不斷重覆同一個動作，

recursion will start an action, and before it’s finished, use it again,

遞迴會從一個動作開始， 在它結束之前，又再次使用它，

and before that’s finished, use it again, and so on.

在那結束之前， 又再使用它，以此類推。

It keeps doing this until some end state is reached.

遞迴會這樣不斷進行下去， 直到達成某種終止狀態。

It then passes the information back up, layer after layer,

接著，就會把資訊一層一層回傳，

until it reaches the top and ends the cycle.

直到資訊傳到頂層， 循環就此結束。

Recursion is ideal for problems that involve self-similarity,

遞迴很適合用來解決 和「自相似」有關的問題，

where each part resembles the larger whole.

也就是每個部分 都和更大的整體很相似。

Like, for example, a deadly defense system designed to end any person or thing

就像是這個致命的防禦系統，

設計的目的是要消滅 任何有膽踏上它的人或物。

who dares tread upon it.

Pause the video to figure it out yourself.

如果想要自己解題，請在此暫停。

Solution in 3

Solution in 2

解答即將公佈

Solution in 1

Ethic’s conundrum seems sprawling on the surface,

艾希克的難題在表面上 看似雜亂無規則，

but there’s a remarkably simple solution to it using recursion.

但其實若會用遞迴， 就有一個非常簡單的解法。

In order to find it, let’s first look at the simplest version of this puzzle:

為了要解題，我們先來看看 這個謎題最簡單的版本：

what if the entire maze were just two paths?

如果整個迷宮只有兩條路呢？

If Hedge copies himself, the copy that goes the wrong way will be destroyed.

如果海吉複製它自己， 走錯路的複製品會被摧毀。

So the other one, which will reach the artifact,

所以，另一個複製 就會抵達寶物所在，

can radio back the path it took, and then no matter which way is correct,

用無線電回傳它走過的路徑資訊，

不論哪條路是對的路， 那就是海吉會收到的答案。

that’s the answer Hedge will receive.

This is called the "base case" of the recursion.

這就叫做遞迴的「基本情況」。

Now, suppose the maze branches twice from the starting point,

假設迷宮從起始點開始 就有兩倍的岔路，

and at every intersection, Hedge’s copies—

在每個路口，海吉的複製品——

let’s call them Twig 1 and Twig 2—

咱們就稱它們為 小枝一號和小枝二號——

make more copies— let’s call them Leaves 1 through 4.

會產生更多複製品—— 稱為小葉一號到小葉四號。

Three Leaves will be destroyed.

會有三個小葉複製品被摧毀。

The one that reaches the artifact will radio back the right answer,

抵達寶物所在的那個小葉 會將正確答案用無線電回傳，

but only to its parent.

但只會傳給它的母體。

So if Twig 1 or 2 is waiting at an intersection

如果小枝一號 或小枝二號在路口等候，

and hears something over the radio,

從無線電聽到消息，

that’s the right way to go to the artifact from where it is.

那就是從它的所在 通往寶物的正確路線。

To tell Hedge the right answer from his perspective,

小枝要從它的角度 把正確答案告訴海吉，

the Twig should say which way it went,

就應該要說出它去過哪裡，

and then the route it just heard over the radio.

以及它從無線電所聽到的路線。

This same process will work no matter how many times the maze branches.

不論迷宮的分岔有多少， 同樣的流程都行得通。

Any answer a copy hears on the radio

任何複製品從無線電聽到的答案

must be the way to the control room from its location,

都是從它的所在 到達控制室的路徑，

and if it then adds the branch it took,

如果再加上它自己走的岔路，

it can tell its parent how to get there as well.

它就可以告訴它的母體 如何到達那裡。

We can sum up the instructions in an action called Pathfinder

我們可以把指令總整成 一個動作，叫做「找路程式」，

that every version of Hedge will follow:

每個版本的海吉複製品 都要遵循這個程式：

1. If you’ve reached the artifact,

一、如果你到了寶物的所在，

radio to your parent whether you got there by going left or right.

用無線電告訴你的母體你怎麼走到 那裡（用左轉、右轉來表述）。

2. When you reach an intersection, move off the conveyor

二、當你到達路口時， 離開輸送帶，

and send new copies down the left and right paths.

派遣新的複製品 去走左、右兩條路。

Have them each run Pathfinder.

要求它們都要執行「找路程式」。

This is where recursion comes in,

這就是要用到遞迴的地方了，

and this may happen many times before the last instruction triggers, which is:

這個指令可能要執行許多次， 才會啟動最後一個指令，即：

3. If you hear anything over the radio, you should radio to your parent

三、如果你從無線電收到資訊， 你就要用無線電聯絡你的母體，

whether you got to your spot by going left or right,

說明你是怎麼到你的所在 （用左轉、右轉來表述），

then repeat everything you just heard.

接著再重述你從無線電 聽到的所有資訊。

Pathfinder is an example of what programmers call functions,

「找路程式」就是程式設計師

所謂的函數、子程式，或程序。

subroutines, or procedures.

No matter the terminology, the idea is the same—

不論用哪個術語稱呼， 背後的想法都相同——

it’s a set of instructions given a label so that it can be easily reused—

把一組指令包裝起來， 貼上標籤，方便重覆使用——

perhaps even by itself.

甚至可以被它自己使用。

And in our case that’ll work perfectly—

在我們的例子中， 採用它就非常完美——

an entire network of paths mapped using just three instructions.

只用三條指令 就能繪製出整個網絡。

So here's what happens.

接著發生的狀況是：

By the time the patrol rounds the corner, Ethic and Lemma have improvised disguises.

當巡邏兵到轉角時，

艾希克和雷瑪臨時湊合出偽裝。

They try to confuse the bots long enough to buy Hedge time.

她們試圖混淆機器人， 為海吉拖延時間。

Finally, Hedge’s radio crackles to life with a series of directions.

最後，海吉的無線電傳來 一連串指示，帶來了生機。

The three dive onto the conveyor and flee for their lives,

三人跳上傳輸帶，開始逃命，

with a squadron of enforcer bots in hot pursuit.

而一整隊執法機器人正緊追在後。