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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，
she'll only get the chance to steal one by dropping it out the window into her waiting truck
她只有一次得手的机会，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, but each floor's egg is more rare and valuable than the one below it.
每个蛋的重量和构造都是一样的，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, she suspects it won't survive a 100-story drop.
大盗自然想拿最顶层的无价之蛋，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, perfect replicas that are perfectly worthless.
在博物馆的礼品店，她找到了两个纪念蛋，完全不值钱但复制得很完美。
The plan is to test drop them to find the highest floor at which an egg will survive the fall without breaking.
她的计划是把它们从高处扔下来，看看它们能在最高的哪层掉落还能完好无损。
Of course, the experiment can only be repeated until both replica eggs are smashed.
当然，两个纪念蛋都碎了的时候就不能继续试验了。
And throwing souvenirs out the window too many times is probably going to draw the guards' attention.
而且从窗户往外扔蛋太多次会引起保安的注意。
What's the least number of tries it would take to guarantee that she find the right floor?
她最少实验几次就能保证她找到对的层数？
If you're having trouble getting started on the solution, it might help to start with a simpler scenario.
如果你觉得无从下手的话，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: To start by dropping it from the first floor and go up one by one until it breaks.
她就只有一个选择：从第一层开始，一层层上去，直到蛋碎了。
Then she'd know that the floor below that is the one she needs to target for the real heist.
那她就知道下面那层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
她能以更大的间隔从不同的楼层扔下第一个鸡蛋
in order to narrow down the range where the critical floor can be found.
以缩小能找到临界值的范围。
And once the first breaks, she can use the second egg to explore that interval floor by floor.
第一个蛋碎了以后，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. But can she do even better?
这就意味着最多十九次就能找到正确的层数。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.
假设只有十层的话。
The thief could test this whole building with just four total throws by dropping the first egg at floors four, seven, and nine.
大盗只用四次投掷就可以试完整个建筑，从四层开始，七层，九层。
If it broke at floor four, it would take up to three throws of the second egg to find the exact floor.
如果在第四层碎了，就只要扔三次第二个蛋，就能找到准确层数。
If it broke at seven, it would take up to two throws with the second egg.
如果在第七层碎了，就只要再扔两次。
And if it broke at floor nine, it would take just one more throw of the second egg.
如果在第九层碎了，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, it takes up to the same number of throws to find it.
无论哪一层是对的，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 in the 100 floor building.
这个公式能帮我们找到一百层内开始扔蛋的层数。
There are several ways to solve this equation, including trial and error.
这个式子有几种解法，包括试错法。
If we plug in two for n, that equation would look like this. If we plug in three, we get this.
如果我们代入2，等式就是这样。如果我们代入3，等式就是这样。
So we can find the first n to pass 100 by adding more terms until we get to our answer, which is 14.
因此我们通过增加n的值，就可以找到让左边总和超过100的最小的n，也就是14。
And so our thief starts on the 14th floor, moving up to the 27th, the 39th, and so on, for a maximum of 14 drops.
因此，我们的大盗应该从第十四层开始，再到二十七层，三十九层，以此类推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。
