Can you solve the rogue submarine riddle? - Alex Rosenthal

翻译人员: Ashley Wu 校对人员: Helen Chang

Smuggling yourself aboard the rogue submarine was the easy part.

偷溜上离群的潜水艇是最容易的一步。

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

黑进核导弹发射超控就有些难度了。

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

但是有个问题：你不知道 用来阻止发射的密码。

You know you need the same two numbers

你知道需要输入 混乱之敌特刚用来

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

开启发射的两个数字。

But one wrong answer will lock you out.

可是答错一次，超控就会上锁。

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

你从隐藏的地方获得了以下的信息：

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

老大不够信任他的走狗们，

to launch nuclear missiles on their own.

所以没把用来开启发射的全部信息 交给任何一个走狗。

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

于是他把开启密码的其中一个 交给走狗A，另外的交给走狗B，

and forbade them to share the numbers with each other.

并且禁止他们两互相分享密码。

When the order came,

老大命令下来后，

each entered their own number and activated the countdown.

他们各自输入了自己的密码， 启动了倒计时。

That was 50 minutes ago,

那已经是50分钟前的事了。

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

距离导弹发射只剩下10分钟。

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

老大突然道，“跟你们说件好笑的事情。 你们的开启密码其实是相关的。”

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

我选择了一组至少有两个元素的 不同正整数，各自小于 7，

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

把总和告诉了A，把乘积告诉了B。

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

尴尬的一瞬间后，A对B说：

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

“不知道你知不知道我的数字是多少。”

B thinks this over, then responds,

B想了想，回答道:

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

“我知道你的数字是多少， 现在也知道你也知道我的数字。”

That’s all you’ve got.

你获得的信息就这么多。

What numbers do you enter to override the launch?

输入什么数字才能阻止发射？

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

想自己解答就请暂停视频。 3秒后公布答案

Answer in 2

2秒后公布答案

Answer in 1

1秒后公布答案

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

像这样一个建立在无知的基础的谜题 是出了名的难解。

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

解题的窍门是通过两人的角度想，

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

然后根据他们知道和不知道的事 缩小可能性的范围。

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

先从A说的第一句话开始吧。

It means that B could conceivably have something with the potential

那代表B可能知道一件

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

有可能能透漏A的数字是多少的事，

但是不一定。

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

这听起来并不能确定， 却能使我们达到极重要的见解。

The only scenarios where B could know A’s number

仅当B的数字只有一种 因数分解方式的时候

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

B才会知道A的数字的大小。

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

你通过尝试整除几个数就会发现规律。

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

可能会是素数——也就是说， 它的因式必须是一和那个数字。

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

或者可能是一个有一和一个素数的二次方的因式，如四，的数字。

In both cases, there is exactly one sum.

在这两种情况下，正好有一个和数。

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

把一个像八一样的数字整除成二和四或一，二，和四的话，

creates too many options.

就有太多可能性了。

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

因为老大的数字都小于七，

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

A能列为有可能是B的数字 只包括这四个数字。

Here’s where we can conclude a major clue.

我们从此能判断出一个重要的线索。

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

A的数字必定是B数字的因式的和， B才能是这些数字之一。

so 3, 4, 5, or 6.

也就是说是三，四，五，或六。

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

可以排除三和四，因为和数 如果是其中的一个数的话，

the product could only be 2 or 3,

那乘积就只能是二或三。

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

在那种情况下，A就已经知道 B知道A的数字了，

contradicting A’s statement.

和A说的话恰恰相反。

5 and 6, however, are in play,

但是五和六都有可能，

because they can become sums in multiple ways.

因为这两个数字可以 以许多方式成为和。

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

考虑这个的需要就是 这个谜题最难的部分之一。

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

要记得的关键就是B的数字

that B’s number is on A’s list—

不一定在A列的数字里。

those are just the possibilities from A’s perspective

这些只是A觉得有可能的数字，

that would allow B to deduce A’s number.

B也就能借此推断出A的数字是多少。

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

这种模糊性迫使我们通过不直观、 多步骤的过程来考虑， 如：

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

考量一个乘积，看看乘积的因式 加起来是什么和数，

then break those apart and see what products can result.

然后把那些和数分解成因式， 再找出因式的乘积。

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

我们马上就要做一件相似的事，

and back to sums.

从和找到乘积，再回到和。

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

但我们现在知道A说出第一句话时，

he must have been holding either 5 or 6.

他的数字一定是五或六。

B has access to the same information we do,

B知道的信息和我们一样，

so he knows this too.

所以这他也知道。

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

我们回顾一下大家现在都知道些什么：

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

大家都知道很多关系和数的信息， 但只有B知道乘积是多少。

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

现在看看B说的话的第一部分。

What if A’s number was 5?

A的数字要是是五呢？

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

那可能是一加四或二加三的结果。

in which case B would have either 4 or 6.

在这种情况下， B的数字不是四就是六。

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

如果是四的话，B就知道 A的数字是什么了，

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

因为那样只有一种乘法，四乘以一。

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

六反而能以三种方法分解， 因式的和数如以上。

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

B觉得有可能的和数不包括七， 但包括五和六。

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

那代表B就不知道 A的数字是五还是六了，

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

所以我们就能排除这个可能性， 因为和他说的话自相矛盾。

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

那好，五和四可能就是密码，

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

但我们怎么才能确定是唯一一个？

Let’s consider if A’s number was 6—

假设A的数字是六，

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

也就是一加五，二加四， 或者一加二加三，

giving B 5, 8, or 6, respectively.

B的数字分别就是五，八，或六。

If B had 5, he’d know that A had 6.

B的数字如果是五， 他就会知道A的数字是六。

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

他的数字如果是八的话，

A的数字的可能性就是 二加四和一加二加四。

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

只有六是有可能的和数之一， 所以B就又知道A的数字是六了。

To summarize, if A had 6,

总结一下，A的数字如果是六，

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

他也不知道B的数字是五还是八。

That contradicts the second half of what B said,

这就与B说的话的后半段自相矛盾了，

and 5 and 4 must be the correct codes.

所以五和四一定就是正确的密码。

With seconds to spare you override the missile launch,

以几秒之差，你阻止了导弹发射，

shoot yourself out of the torpedo bay,

从鱼雷舱里射出去，

and send the sub to the bottom of the ocean.

还有让潜艇沉没到了海底。