The Gauntlet | Think Like A Coder, Ep 8

Translator: Reviewer: Thomas Tam

［TED-Ed 原創系列］

《像程式員思考噉》

［上回地點：198 樹林］

［今回地點：Huxenborg］

［第八集：挑戰］

Their fall from the tower sends Ethic and Hedge

Ethic、Hedge 從高塔墮下

spinning into the rapids of a river of pure energy.

陷入咗湍急嘅純能量河流

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

呢股洪流從 Bradbarrier 一路流向 Huxenborg

There an entire city’s worth of factories

呢度整個城市嘅工廠 制造機械人

build the robots and house the Node of Memory,

同時存放咗「記憶節點」：

係 Ethic 要收集 三大神器最後一件

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

After a long day and a longer night

佢哋渡過咗漫長嘅日 同夜

they find themselves in a canyon of brick and steel.

然後發現置身喺磚塊同鋼鐵嘅峽谷度

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

正喺急流盡頭 快要滑落峽谷

a rope catches them.

一根從天而降嘅繩 抓住咗佢哋

Their savior, Lemma, has been waiting for them.

Lemma 呢個救星 一直等緊佢哋

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

當 Ethic 從森林嘅塔度 奪取咗「創造節點」

radios all across the land came back to life.

通訊電波喺所有地方都恢復咗

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

反抗組織領袖 Adila 立即聯絡佢嘅盟友

none more important than Lemma,

至醒天才 Lemma ：

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

喺 Huxenborg 內部 做摧毀機械人嘅科學家

好不幸，無線電同樣驚動咗機械人

Unfortunately, the radios also tipped off the robots.

So they’ve taken defensive measures

所以機械人已採取咗防禦措施

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

保護喺城市中心基地 僅存嘅最後神器

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

其實只有一個方法嚟到呢度 係挑戰分叉路

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

喺 Luxembourg 下底 有一系列致命發光傳送帶

從而家呢個位置開始

Starting from the current position,

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

每段路前進一段距離 會分叉成兩條路

Every branch does the same thing, again and again.

每條分叉路，都不斷重覆之前嘅情況

There are thousands of branches.

分叉口會有成幾千個

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

只係進入咗正確嘅路徑 先搵到神器，否則都會被消滅

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

好彩「創造節點」 畀咗 Hedge 一種神奇力量：

he can produce slightly smaller versions of himself.

佢可以自我複製，不過細少少

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

複製品只可做兩件事： 將無線電信號傳回父級複製品

and produce slightly smaller versions of itself…

同時複製較少嘅自己……

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

全部複製品都可以做呢兩件事

for as many generations as needed.

按佢哋嘅需要製造複製品

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

一支巡邏隊正行近 Ethic 嘅時間有限

佢應該畀啲乜嘢嘅指示 Hedge 嚟搵到條安全嘅路徑呢？

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

Pause the video to figure it out yourself.

按停 video 等自己搞清楚

Hint in 3

[ 答案揭曉：3 ]

Hint in 2

[ 答案揭曉：2 ]

Hint in 1

[ 答案揭曉：1 ]

Programmers have an elegant tool in their arsenal called recursion.

喺程式員武器庫度 有個優雅嘅工具叫做遞歸

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

遞歸係一組會自我呼叫嘅程式

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

呢嗰就似係自己定義一個詞

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

除非遞歸遇到設定嘅離開條件 其實係勁有效率嘅

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

遞歸涉及重複，但同迴圈唔同

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

迴圈係就住一個動作，重複又重複

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

遞歸將開始一個動作 直至完成前

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

一再使用佢

It keeps doing this until some end state is reached.

佢一直噉做，直到遇見最後狀況

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

之後就一層層噉將資訊回報返父級

until it reaches the top and ends the cycle.

直到佢達到頂層同結束循環為止

Recursion is ideal for problems that involve self-similarity,

遞歸係理想嘅 處理涉及相似性循環嘅問題

where each part resembles the larger whole.

每部分都與更大整體邏輯相似

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

例如一個防禦系統嘅目的

係消滅任何膽敢闖進去嘅人

who dares tread upon it.

Pause the video to figure it out yourself.

按停 video 搞清楚

Solution in 3

[ 答案揭曉：3 ]

Solution in 2

[ 答案揭曉：2 ]

Solution in 1

[ 答案揭曉：1 ]

Ethic’s conundrum seems sprawling on the surface,

Ethic 似乎撞到個辣手問題

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

但係使用遞歸係勁簡單嘅解決方法

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

為咗搵到神器 先睇下最簡單謎題嘅版本：

what if the entire maze were just two paths?

整個迷宮如果只係有兩條路？

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

如果 Hedge 嘅複製品走錯路 將會被消滅

So the other one, which will reach the artifact,

而另一個就會搵到神器

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

而且仲傳回佢行嘅正確路徑

Hedge 會得到呢個答案

that’s the answer Hedge will receive.

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

呢種稱為遞歸嘅「基本情況」

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

依家，假設迷宮從起點分叉兩次

and at every intersection, Hedge’s copies—

Hedge 嘅複製品喺每個路口

let’s call them Twig 1 and Twig 2—

姑且叫做 1 號、2 號——

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

製作更多複製品—— 佢哋為葉子 1 到 4

Three Leaves will be destroyed.

三個葉子會被消滅咗

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

當一個搵到神器 會傳回正確嘅答案

but only to its parent.

但僅限對上嘅一層

So if Twig 1 or 2 is waiting at an intersection

若果 1 同 2 號分枝喺分叉路口等候

and hears something over the radio,

其中一個收到回音

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

係從佢嘅位置 到搵到神器嘅正確方向

To tell Hedge the right answer from his perspective,

傳回畀 Hedge 嘅正確答案

the Twig should say which way it went,

分枝會報告佢先前嘅路徑

然後係佢啱啱通過無線電接收嘅路線

and then the route it just heard over the radio.

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

無論迷宮分支多少次 方法仍然有效

Any answer a copy hears on the radio

Hedge 會收到複製品嘅回音

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

必須係從佢嘅位置到控制室嘅路徑

and if it then adds the branch it took,

再添加佢所採用嘅分支路徑

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

佢告訴上一層點樣到達嗰度

We can sum up the instructions in an action called Pathfinder

我哋喺名為 Pathfinder 度總結指令

that every version of Hedge will follow:

Hedge 嘅每個複製品都將依照：

1. If you’ve reached the artifact,

1. 如果搵到神器嘅話

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

無論你係向左定向右 都可以通過無線電告訴上一層

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

2. 當佢到達十字路口時 要離開傳送帶

and send new copies down the left and right paths.

並沿左右路徑發送新副本

Have them each run Pathfinder.

等佢哋各自運行 Pathfinder

This is where recursion comes in,

呢個就係遞歸嘅用處

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

喺觸發最後指令之前會發生好多次

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

3. 當複製品收到傳回嘅無線電 就會報告上一層

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

無論你係向左定向右走 嚟到你現時位置

then repeat everything you just heard.

仲複述你啱啱聽到嘅一切

Pathfinder is an example of what programmers call functions,

Pathfinder 係程式員 呼叫函數、子程序

或者過程嘅一個例子

subroutines, or procedures.

No matter the terminology, the idea is the same—

唔理用乜嘢術語 諗法都係一樣——

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

佢係一組帶有說明嘅標籤 方便輕鬆噉重複使用——

perhaps even by itself.

甚至可能係佢自己噉簡單

And in our case that’ll work perfectly—

喺我哋嘅例子度完美噉工作——

an entire network of paths mapped using just three instructions.

僅係使用三個指令 就能夠探測到整個路徑嘅網絡

So here's what happens.

呢個就係會發生嘅嘢

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

當巡邏隊拐到拐角處時 Ethic 和 Lemma 已經即興偽裝

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

試圖蒙騙機械人 等 Hedge 有足夠時間

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

最後，Hedge 收到無線電 一系列嘅指示

三人跳上傳送帶逃命

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

一群執法嘅機器人跟得好貼

with a squadron of enforcer bots in hot pursuit.