Can you solve the computer virus riddle? - James Tanton

翻译人员: Su Wang 校对人员: Helen Chang

Your antivirus squad is up against a particularly sadistic bit

你的杀毒小队面对的是一个残暴

of malicious code that’s hijacked your mainframe.

且充满恶意的代码， 它劫持了你的主机。

What you’ve learned from other infected systems— right before they went dark—

你从其他受侵害的系统中学到 — 就在它完全摧毁主机之前 —

is that it likes to toy with antivirus agents in a very peculiar way.

是它喜欢和杀毒软件玩一种特殊的游戏。

It corrupts one of the 4 disks that run your mainframe,

它损坏了运行你主机的四个磁盘之一，

represented by lights showing which are on and which off.

以灯光表示磁盘的开关。

Then it selects one member of the antivirus squad— this’ll be you—

然后它会选择杀毒小队的一名成员 —— 就是你 ——

and brings them into the mainframe.

带入主机。

It tells them which disk it corrupted,

它告诉你损坏的磁盘是哪一个，

allows the agent to switch a single disk on or off,

允许你打开或关闭一个灯的开关，

then immediately de-rezzes the agent.

然后会立即消灭掉你。

Your squad can make an all-out attack to break into the mainframe

你的团队可以闯入主机

and destroy one disk before they’re wiped out.

并在被消灭之前摧毁一张磁盘。

If they destroy the corrupted one, the malware will be defeated.

如果摧毁了被损坏的磁盘， 那恶意软件就会被击败。

Any others, and the virus will erase the entire system.

如果摧毁任何其他一个， 病毒将清空整个系统。

The lights are only visible within the mainframe,

这些灯仅在主机内可见，

so you won’t know until you get there which, if any, are on.

因此在你到达那之前不会知道 亮灯的是哪个（如果有的话）。

How can you communicate, with your single action,

你将怎么用单一一个动作 将信息传达给你的团队，

which of the 4 disks has been corrupted?

四个磁盘中的哪一个是被损坏的呢？

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

请在这里暂停并独立思考。 倒计时 3

Answer in 2

2

Answer in 1

1

The setting is a big clue for one solution.

这个系统设置为解决方案 提供了明显线索。

Using binary code— the base two numbering system that only uses 1s and 0s—

运用二进制编码 —— 仅用 0 和 1 的编码系统 ——

we can represent each of the 4 disks with a 2-bit binary number

我们可以用一个二进制编码 来对磁盘进行编号，

ranging from 00 for zero to 11 for three.

从 00 代表 0，到 11 代表 3。

What we’re looking for now is some sort of mathematical operation

现在我们要做的就是一些

that can take the lit disks as input, and give the corrupted disk as an output.

能够通过亮灯作为输入， 损坏灯泡作为输出的数学运算。

Let’s consider one possibility.

我们来考虑一种可能性。

Say that the corrupted disk was this one,

假设这个是被损坏的磁盘，

and when you come in, no lights are on.

并且当你进入主机时， 没有磁盘是亮灯的。

You could turn 11 on to indicate that disk.

你可以将 11 的灯点亮 来通知你的团队。

Okay, what if you came in and 11 was already on?

好的，那么如果当你进入时 11 是亮灯的呢？

You have to switch one light.

你必须要操作一个开关。

Which seems like the most innocuous to change?

哪一个看起来像是最无关紧要的呢？

Probably 00, in that if you were to add 00 and 11,

可能是 00 ， 如果你用 00 加上 11，

you’d still get 11.

你仍然得到 11。

So maybe the key is to think of addition of binary numbers,

所以也许答案就是 将二进制数相加，

with the sum of the lit disks communicating the corrupted disk number.

通过亮灯的磁盘的总和 传递损坏磁盘的信息。

This works great, until we start with a different hypothetical.

这非常好用， 直到我们开始另一种假设。

What if 00 was the corrupted disk, and 01 and 10 were on?

如果 00 是被损坏的磁盘， 并且 01 和 10 是亮灯的呢?

Here, the sum of the lit disks is 11.

这里，亮灯磁盘的总和是 11 。

But we need to change this to a sum of 00 with the flip of one switch.

但是我们还需要操作开关一次 将总和变成 00 。

We have four options: turning switch 00 on gives us 11.

我们有四种选择： 点亮 00 得到 11 。

Turning 01 off takes us back to 10,

关掉 01 得到 10 。

and turning 10 off gives 01.

和关掉 10 得到 01。

None of those work.

全部没有用。

Turning switch 11 on gives us 110 by standard binary addition.

关掉 11 ， 通过二进制加法得到 110。

But we don’t really want three digit numbers.

但是我们并不想要三位数字。

So what if— to keep the result a two digit number—

那么如果 —— 只保留两位数字 ——

we break the rules a bit and let this sum equal 22.

我们稍稍打破规则让它得到 22 。

That’s not a binary number, but if we regard 2s as the same as 0s,

这不是一个二进制数字， 但是如果我们将 2 看作 0，

that does indicate the correct disk.

这将会指向正确的磁盘。

So this suggests a strategy:

所以这表明了一种策略：

look at the sum of all the lighted disks we see,

看亮灯磁盘的编号的总和，

regarding 2s as 0s.

将 2 看作 0，

If it’s already the correct result, flip 00,

如果它已经是正确的结果， 开关 00 号磁盘。

and if not, find the switch that will make the sum correct.

如果不是， 找到那个能够让总和正确的磁盘。

You can see for yourself that any starting configuration

你可以注意到任何初始情况

can sum to any number from 00 to 11 with a flip of a switch.

都可以通过改动一次开关， 用 00 和 11 加成任意总和。

The reason this works is related to a concept called parity.

这个策略跟一个 叫做奇偶校验的概念是相关的。

Parity tells you whether a given value is even or odd.

奇偶校验会告诉你一个已知量 是奇数还是偶数。

In this case, the values whose parity we’re considering

在我们的情况中， 我们要奇偶校验的数

are the number of 1s in each digit place of our binary sums.

是在每个位置上 所有二进制数中 1 的总和。

And that’s why we can say that 2 and 0, both even numbers,

这是为什么我们可以说 2 和 0，都是偶数，

can be treated as equivalents.

可以被看作是相同的。

By adding or subtracting 00, 01, 10, or 11,

通过相加或相减 00、 01、 10 或 11，

we can change the parity of either, both, or neither digit,

我们可以改变任一、 全部， 或没有位置的奇偶性，

and create the disk number we want.

得到我们想要的磁盘编码。

What’s incredible about this solution is that it works for any mainframe

令人惊叹的是这个方法 在任何主机中都适用，

whose disks are a power of two.

只要它的磁盘是 2 的幂。

With 64 you could turn each activated disk into a 6-bit binary number

如果 64 次幂，你可以将每一个磁盘 编码为一个 6 位二进制数

and sum the 1s in each column,

并且相加每个位置上的 1，

regarding any even sum as the same as 0 and any odd sum as 1.

然后将任意偶数看作 0 ， 任意奇数看作 1。

1,048,576 disks would be daunting, but entirely doable.

1,048,576 磁盘听起来令人生畏， 但完全可行。

Luckily, your mainframe is much smaller.

幸运的是， 你的主机小多了。

You make the valiant sacrifice and your team rushes in,

你英勇地牺牲了， 然后你的团队冲进来，

destroying the corruption and freeing the system.

摧毁了被损坏的磁盘 并挽救了系统。