Can you solve the rogue submarine riddle? - Alex Rosenthal

譯者: Lilian Chiu 審譯者: Amanda Zhu

Smuggling yourself aboard the rogue submarine was the easy part.

偷搭上叛逃的潛水艇 還算是簡單的部分，

Hacking into the nuclear missile launch override— a little harder.

駭入核彈發射撤消系統 這就稍微難一點了。

But now you’ve got a problem: you don’t have the override code.

但，現在你遇到問題了： 你沒有撤消密碼。

You know you need the same two numbers

你知道你需要的兩個數字，

that the agents of chaos just used to authorize the launch.

就是混亂特務剛才 用來授權發射的數字。

But one wrong answer will lock you out.

但弄錯一個數字就會讓系統鎖住。

From your hiding spot, you’ve been able to learn the following:

從你躲藏的位置， 你得到了下列資訊：

The big boss didn’t trust any minion with the full information

大頭目不相信任何嘍囉， 不會把全部的資訊都給同一個人，

to launch nuclear missiles on their own.

以免樓囉自己發射核彈。

So he gave one launch code to Minion A, the other to minion B,

所以他把一個發射密碼給嘍囉 A，

另一個給嘍囉 B，

and forbade them to share the numbers with each other.

並禁止他們告訴對方自己的數字。

When the order came,

命令下來時，

each entered their own number and activated the countdown.

他們各自輸入他們的數字， 啟動發射倒數。

That was 50 minutes ago,

那是五十分鐘前的事，

and there's only 10 minutes left before the missiles launch.

只剩下十分鐘核彈就要發射了。

Suddenly, the boss says, “Funny story— your launch codes were actually related.

突然，頭目說：

「說件趣事——你們的 發射密碼其實是相關的。

I chose a set of distinct positive integers with at least two elements,

我選了一組正整數，

至少有兩個數字，

each less than 7, and told their sum to you, A, and their product to you, B.”

每個都小於 7，

把它們的加總告訴 A，

乘積告訴 B。」

After a moment of awkward silence, A says to B,

在短暫的尷尬沉默後，A 對 B 說：

“I don’t know whether you know my number.”

「我不知道你是否知道我的數字。」

B thinks this over, then responds,

B 想過之後，回應說：

“I know your number, and now I know you know my number too.”

「我知道你的數字，

且現在我知道你也知道我的數字。」

That’s all you’ve got.

你就只聽到這些。

What numbers do you enter to override the launch?

你要輸入什麼數字來撤銷飛彈發射？

Pause now to figure it out for yourself. Answer in 3

［若要自行解題，請在此暫停。 答案公佈倒數：三］

Answer in 2

［二］

Answer in 1

［一］

Ignorance-based puzzles like this are notoriously difficult to work through.

這種以「不知情」為基礎的謎題

是出名了難解。

The trick is to put yourself in the heads of both characters

技巧在於從兩位角色的角度來思考，

and narrow down the possibilities based on what they know or don’t know.

並根據他們知道什麼

或不知道什麼， 來找到更接近的答案。

So let's start with A's first statement.

咱們先從 A 的第一句陳述開始。

It means that B could conceivably have something with the potential

意即， B 可能有資訊 可以揭露出 A 的數字，

to reveal A’s number, but isn’t guaranteed to.

但不保證一定可以。

That doesn’t sound very definitive, but it can lead us to a major insight.

聽起來無法確定什麼，

但能讓我們窺見一件重要的事：

The only scenarios where B could know A’s number

只有一種情況下， B 會知道 A 的數字，

are when there’s exactly one valid way to factor B’s number.

就是 B 的數字只有 一種因數拆解的方式。

Try factoring a few and you’ll find the pattern—

試著拆解幾個數字， 你就會找到模式；

It could be prime— where the product must be of 1 and itself—

可能是質數

——乘積即 1 和質數本身——

or it could be the product of 1 and the square of a prime, such as 4.

或是一和質數的平方，比如 4。

In both cases, there is exactly one sum.

在這兩種情況下， 都剛好只有一個加總值。

For a number like 8, factoring it into 2 and 4, or 1, 2, and 4,

如果是八這類數字，

因數可能是 2 與 4，

或者 1、2、4，

creates too many options.

會有太多加總的可能性。

Because the boss’s numbers must be less than 7,

因為頭目說每個數字都小於七，

A’s list of B’s possibilities only has these 4 numbers.

A 把 B 可能的數字列成清單，

上面就只會有這四個數字。

Here’s where we can conclude a major clue.

這裡我們就能得到一條重要線索。

To think B could have these numbers, A’s number must be a sum of their factors—

A 認為 B 可能有這些數字，

所以 A 的數字一定是 其因數的總和——

so 3, 4, 5, or 6.

3、4、5，或 6。

We can eliminate 3 and 4, because if the sum was either,

我們可以拿掉 3 和 4， 因為如果總和是 3 或 4，

the product could only be 2 or 3,

那乘積就只有可能是 2 或 3，

in which case A would know that B already knows A’s number,

不論是 2 或 3，A 都能知道 B 已經知道 A 的數字。

contradicting A’s statement.

這和 A 的說法矛盾。

5 and 6, however, are in play,

不過，5 和 6 都有可能，

because they can become sums in multiple ways.

因為有多種方式加總出這兩個數字。

The need to consider this is one of the most difficult parts of this puzzle.

知道要考量這一點

是這類謎題最困難的地方。

The crucial thing to remember is that there’s no guarantee

重點是要記住，

B 的數字並不保證 一定在 A 的清單，

that B’s number is on A’s list—

those are just the possibilities from A’s perspective

清單上的只是從 A 的角度 認為可能的數字，

that would allow B to deduce A’s number.

A 認為這樣 B 就能 推理出 A 的數字。

That ambiguity forces us to go through unintuitive multi-step processes like:

這種模糊性，迫使我們要進行

非直覺式的多步驟過程，

比如：先想一個乘積，

consider a product, see what sums can result from its factors,

看看它的因數能加出哪些總和，

then break those apart and see what products can result.

接著再把這些總和拆開，

看能乘出哪些乘積。

We’ll soon have to do something similar going from sums to products

我們很快就得要採用類似的 過程，從總和到乘積，

and back to sums.

再回到總和。

But now we know— when A made his first statement,

但現在我們知道， 當 A 說出第一句陳述時，

he must have been holding either 5 or 6.

他的數字一定是五或六。

B has access to the same information we do,

我們能得到的資訊 B 也都能得到，

so he knows this too.

所以他也知道這一點。

Let’s review what’s in each brain at this point:

咱們來回顧一下 每個人的腦中有什麼：

everyone knows a lot about the sum, but only B knows the product.

大家都對總和知道很多，

但只有 B 知道乘積。

Now let’s look at the first part of B’s statement.

咱們來看看 B 陳述的第一部分。

What if A’s number was 5?

如果 A 的數字是 5 呢？

That could be from 1+4 or 2+3,

5 有可能是 1+4 或 2+3，

in which case B would have either 4 or 6.

這兩種情況下， B 的數字分別會是 4 或 6。

4 would tell B what A had, like he said,

若是 4，B 就會知道 A 的數字，如他所言，

because there’s only one option to make the product: 4 times 1.

因為要讓乘積是 4 ，只有一種可能：

四乘以一。

6, on the other hand, could be broken down three ways, which sum like so.

另一方面，6 則有 3 種拆解方式，

其總和如圖所示。

7 isn’t on B’s list of possible sums, but 5 and 6 both are.

B 的可能總和清單中沒有 7，

但有 5 和 6。

Meaning that B wouldn’t know whether A’s number was 5 or 6,

意思就是 B 不會知道

A 的數字是 5 或 6，

and we can eliminate this option because it contradicts his statement.

我們可以剔除這個選項， 因為和 B 的陳述矛盾。

So this is great— 5 and 4 could be the override code,

這樣很棒，5 和 4 有可能是撤銷密碼，

but how do we know it's the only one?

但我們怎麼知道只有這個可能性？

Let’s consider if A’s number was 6—

咱們來看看 A 的數字 若是 6 會如何——

which would be 1+5, 2+4, or 1+2+3,

6 可能是 1+5、2+4、1+2+3

giving B 5, 8, or 6, respectively.

則 B 分別會是 5、8、6。

If B had 5, he’d know that A had 6.

如果 B 的數字是 5， 他就會知道 A 的是 6。

And if he had 8, the possibilities for A would be 2+4 and 1+2+4.

如果 B 是 8，

A 可能會是 2+4 或1+2+4。

Only 6 is on the list of possible sums, so B would again know that A had 6.

只有 6 出現在可能的總和清單上，

所以同樣的， B 會知道 A 的數字是 6。

To summarize, if A had 6,

總結一下，如果 A 是 6，

he still wouldn’t know whether B had 5 or 8.

他不會知道 B 是 5 還是 8。

That contradicts the second half of what B said,

不符合 B 陳述的後半段，

and 5 and 4 must be the correct codes.

所以 5 和 4 就一定是正確的密碼。

With seconds to spare you override the missile launch,

在只剩幾秒的時候， 你撤銷了核彈發射，

shoot yourself out of the torpedo bay,

從魚雷艙把你自己發射出去，

and send the sub to the bottom of the ocean.

送潛水艇到海底長眠。