评论
打印
Your interstellar police squad has tracked a group of dangerous rebels to a cluster of of seven small planets.
你的星际警队在一簇由七个小行星组成的星群上追踪到一帮危险的叛徒。
Now you must apprehend them quickly before their reinforcements arrive.
在他们的支援来到之前，你必须尽快地把他们一网打尽。
Of course, the rebels won't just stay put.
很自然的，这些叛徒不会呆着不动。
They'll try to dodge you by moving from planet to planet. But you have one major advantage.
为了逃避追捕，这些叛徒会从一个行星逃到另一个行星。虽然如此，你有一个巨大的优势。
Every hour, your state-of-the-art cruiser can warp between any two planets,
每个小时，你的超高科技警车可以扭曲空间，移动到任意一颗行星去，
while their beat-up smuggling ship can only jump to an adjacent planet in that same time.
而叛徒破旧的偷渡飞船每个小时只可以从一颗行星移动到相邻的行星。
These rebels don't like to stay put. Every time they can relocate, they will.
这些叛徒将不会坐以待毙。他们只要有机会，一定会移动。
Your scouts tell you that the approaching rebel fleet is 10 hours away.
侦察员通知你，叛徒援军距离你十小时之外。
You can't risk letting the rebels escape.
你绝对不能让叛徒逃走。
Can you devise a sequence for searching the planets that's guaranteed to catch them in 10 warps or less, no matter what moves they make?
在只移动十次或以下的条件下，你能想出不论叛徒怎么移动，保证可以把叛徒逮捕的搜索顺序吗？
Rounding up the rebels won't be easy.
要把叛徒抓拿归案并非容易。
For one, you have no way of knowing which planet they're on to begin with.
首先，你并不知道叛徒目前所在的行星。
And without that information, it's hard to determine where they'll move next. So where do you begin?
没这个信息，你是很难推断他们下一步会怎么走的。那么，你该从何着手？
When tackling problems of this kind it often helps to simplify things, so you can better understand their dynamics.
当处理这种问题时，简化问题非常有效，这样你就可以更好地了解事情的动态。
Let's imagine that this cluster has the same arrangement but no outermost planets, leaving only the four in the center.
假设那些行星的排列跟这里的一样，但是没有最外围的那些行星，所以剩下的只有中间那四颗。
We still don't know which planet the rebels start on.
我们仍旧不知道叛徒从哪颗行星开始。
But there's one key feature: the third planet is adjacent to all others,
不过，这儿有个关键的特点：第三颗行星毗邻所有的行星，
which means the rebels either start there and move somewhere else,
这表示叛徒要么从这颗开始然后移动到其他的行星，
or start on one of the other planets and have no choice but to move to planet three.
要么从其他的行星开始然后别无选择地移动到行星三号上。
Simply checking planet three twice in a row would do the trick.
我们只要搜索行星三号两次就能解决问题。
Adding the three outer planet adds a bit more complexity, but the same strategy remains.
把外围的那些行星加入问题中，也许会把情况变得稍微复杂，但是同样的方案仍然奏效。
We want to check the planets in an order that will eventually corner the rebels.
我们想出的搜索顺序必须最终使叛徒走投无路。
And there's another insight that can help us:
而这里有另一条信息可帮助我们：
each hour, the rebels move from an even-numbered planet to an odd-numbered planet, or vice versa.
每小时，叛徒会从偶数行星移动到奇数行星，或者反过来。
This gives us a way to simplify the problem by dividing the planets into two subsets, and tackling each one separately.
知道这点能帮助我们简化问题，也就是把行星分成两组，这样就能单独地处理每一组。
For starters, let's assume the rebels begin on an even-numbered planet: either two, four, or six.
作为开始，假设叛徒在偶数行星开始：二号、四号或六号。
So we'll search planet two first.
我们先从行星二号开始搜索。
If they're not there, they must have started on either four or six, which means they can move to three, five, or seven.
如果他们不在那儿，那么他们肯定在四号或六号开始，这表示他们会移动到三号、五号或七号。
Planet three at the center gives them the most options for their next move, so we'll want to check there next.
行星三号位于中央，会给他们最多移动的选择，所以我们接下来会搜索三号。
If we don't find them, they must have been at planet five or seven, meaning they'll next move to four or six.
如果他们不在那儿，他们肯定是在五号或七号，意味着下一步他们将去四号或六号。
Let's now search planet four.
现在我们搜索四号。
If they're not there, they must have gone to the sixth planet and can only flee to three or seven.
如果他们也不在那儿，他们肯定是去了六号，而且下一步只可去三号或七号。
If we next scour planet three and don't find them, we know they went to planet seven and are now cornered.
如果我们接下来搜寻三号但又找不到他们，我们知道他们去了七号，并且走投无路了。
They can only move to planet six, where we'll apprehend them on our fifth search.
他们只可以迁移到六号，这表示我们只需搜索五次就逮捕他们。
Of course, this plan only works assuming that the rebels were on an even-numbered planet in the first hour.
当然，要这个方案奏效我们必须假设叛徒第一个小时是在偶数行星开始的。
But what if that assumption was wrong? In that case, they must've started on an odd-numbered planet.
但如果这个假设是错的，怎么办呢？若是这样，他们肯定是在奇数行星开始的。
And because they move to an adjacent planet every hour,
由于叛徒每个小时都会移动到相邻的行星，
their location must alternate between odd and even-numbered planets.
他们的位置肯定是在奇数和偶数行星中交替。
This means that if they were on an odd-numbered planet to start, after five moves, they'd be on an even-numbered planet.
这表示如果他们在奇数行星开始，五次移动后，他们将会到偶数行星。
So if our first five searches missed them because our assumption that they started on an even-numbered planet was wrong,
如果我们前五次搜索不成功是因为假设他们从偶数的行星开始是错误的，
all we have to do now is repeat the sequence!
我们只需做的是重复之前的步骤！
Searching the planets in order two, three, four, three, six, two, three, four, three, six, leaves the rebels nowhere to run.
跟着这顺序搜索行星：二号、三号、四号、三号、六号、二号、三号、四号、三号、六号，就可以把叛徒逼得走投无路。
Thanks to your deductive reasoning, order is restored to the galaxy.
由于你的推理逻辑，宇宙秩序得以维持。
