Can you solve the egg drop riddle? - Yossi Elran

翻译人员: Nisi J 校对人员: Sylvia He

“许多人害怕高，我不怕，我怕宽。” ——Steven Wright

The city has just opened its one-of-a-kind Fabergé Egg Museum

这个城市刚刚开了一家 别具一格的俄罗斯彩蛋博物馆，

with a single egg displayed on each floor of a 100-story building.

一栋每层只展出一个蛋 的100层建筑。

And the world's most notorious jewel thief already has her eyes on the prize.

全世界最臭名昭著的珠宝大盗 已经锁定了她的目标。

Because security is tight and the eggs are so large,

因为安保很严，蛋又很大， 她只有一次得手的机会，

she'll only get the chance to steal one

by dropping it out the window into her waiting truck

把蛋丢到下面接应的卡车，

and repelling down before the police can arrive.

然后在警察到来之前撤退。

All eggs are identical in weight and construction,

每个蛋的重量和构造都是一样的，

but each floor's egg is more rare and valuable than the one below it.

但是每一层的蛋都比 下面一层的更值钱。

While the thief would naturally like to take the priceless egg at the top,

大盗自然想拿最顶层的无价之蛋，

she suspects it won't survive a 100-story drop.

她觉得这个蛋不会 在掉落一百层后还完好无损。

Being pragmatic, she decides to settle for the most expensive egg she can get.

务实的她决定去偷 她能成功偷到的最贵的蛋。

In the museum's gift shop, she finds two souvenir eggs,

在博物馆的礼品店， 她找到了两个纪念蛋，

perfect replicas that are perfectly worthless.

完全不值钱但复制得很完美。

The plan is to test drop them

她的计划是把它们从高处扔下来，

to find the highest floor at which an egg will survive the fall

看看它们能在最高的哪层掉落

without breaking.

还能完好无损。

Of course, the experiment can only be repeated

当然，两个纪念蛋都碎了的时候

until both replica eggs are smashed.

就不能继续试验了。

And throwing souvenirs out the window too many times

而且从窗户往外扔蛋太多次

is probably going to draw the guards' attention.

会引起保安的注意。

What's the least number of tries it would take

她最少实验几次

to guarantee that she find the right floor?

就能保证她找到对的层数？

Pause here if you want to figure it out for yourself!

【如果你愿意的话， 暂停一下，自己想一想。】

Answer in: 3

【答案在三秒后出现，

Answer in: 2

二秒，

Answer in: 1

一秒。】

If you're having trouble getting started on the solution,

如果你觉得无从下手的话，

it might help to start with a simpler scenario.

可以简化一下情景。

Imagine our thief only had one replica egg.

想像一下如果大盗只有一个蛋。

She'd have a single option:

她就只有一个选择：

To start by dropping it from the first floor

从第一层开始，

and go up one by one until it breaks.

一层层上去，直到蛋碎了。

Then she'd know that the floor below that

那她就知道下面那层

is the one she needs to target for the real heist.

就是她真正需要抢劫的目标。

But this could require as many as 100 tries.

但是要试一百次。

Having an additional replica egg gives the thief a better option.

有第二个复制蛋 给了她更好的选择。

She can drop the first egg from different floors at larger intervals

她可以先用大间隔测试，

in order to narrow down the range where the critical floor can be found.

以缩小实验范围。

And once the first breaks,

第一个蛋碎了以后，

she can use the second egg to explore that interval floor by floor.

她就可以用第二个蛋 在间隔内一层层试了。

Large floor intervals don't work great.

太大的间隔不太有效。

In the worst case scenario, they require many tests with the second egg.

在最糟的情况下， 第二只蛋要试好多层。

Smaller intervals work much better.

小间隔好用多了。

For example, if she starts by dropping the first egg from every 10th floor,

打个比方，如果她每十层丢一次，

once it breaks, she'll only have to test the nine floors below.

蛋碎了之后就只要试之下的九层。

That means it'll take at most 19 tries to find the right floor.

这就意味着最多十九次 就能找到正确的层数。

But can she do even better?

但是她还能做得更好吗？

After all, there's no reason every interval has to be the same size.

毕竟，每一间隔不需要一样大。

Let's say there were only ten floors.

如果只有十层，

The thief could test this whole building with just four total throws

大盗可以四下试完，

by dropping the first egg at floors four,

从四层开始，

seven,

七层，

and nine.

九层。

If it broke at floor four, it would take up to three throws of the second egg

如果在第四层碎了， 就只要扔三次第二个蛋，

to find the exact floor.

就能找到准确层数。

If it broke at seven,

如果在第七层碎了，

it would take up to two throws with the second egg.

就只要再扔两次。

And if it broke at floor nine,

如果在第九层碎了，

it would take just one more throw of the second egg.

就只要再试一次。

Intuitively, what we're trying to do here is divide the building into sections

实际上我们在试图将建筑物分区，

where no matter which floor is correct,

无论哪一层是对的，

it takes up to the same number of throws to find it.

都可以用同样的次数找到。

We want each interval to be one floor smaller than the last.

我们要每一个间隔 都比下面的少一层。

This equation can help us solve for the first floor we need to start with

这个公式能帮我们 找到一百层内

in the 100 floor building.

开始扔蛋的层数。

There are several ways to solve this equation,

这个式子有几种解法，

including trial and error.

包括反复试验。

If we plug in two for n, that equation would look like this.

如果我们代入 2，等式就是这样。

If we plug in three, we get this.

如果我们代入 3，等式就是这样。

So we can find the first n to pass 100

因此我们通过增加 n 的值， 就可以找到

by adding more terms until we get to our answer,

让左边总和超过 100 的 最小的 n，

which is 14.

也就是 14。

And so our thief starts on the 14th floor,

因此，我们的大盗 应该从第十四层开始，

moving up to the 27th,

再到二十七层，

the 39th,

三十九层，

and so on,

等等，

for a maximum of 14 drops.

最多十四下就可以找到正确的层数。

Like the old saying goes, you can't pull a heist without breaking a few eggs.

就像老话说的， 不打碎几个蛋就抢不了劫。