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The other obvious one to do would be worst case. Again, over all possible inputs to this function, what's the most number of steps it takes to do the computation?
很明显另一种就是针对最坏的情况了,也就是在所有可能的输入,里面选出用的,步骤最多的那个输入？
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You rotate the letters by a deterministic amount, by a fixed number of steps so that the end result is actually something that can be reversed.
你用确定的数量旋转那些字母，通过一个固定数量的步骤,这样，最终结果是可以把它,反加密出来。
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In fact, it took the same number of steps as it did in the other case, because each time I'm cutting it down by a half.
因为每次我都把问题的规模,缩小一半，这很棒，好，接下来让我们这么做。
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And the Bacon number is the number of steps it takes for them to get to Kevin Bacon.
这个Bacon数就是通过几度,他们就能与Kavin，Bacon连接上。
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I'm going to let t of b be the number of steps it takes to solve the problem of size b.
我会设立一个t作为,计算指数为b的时候解决问题需要的步骤数。
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You can actually say big O of 1, big O of 1 being constant time, the same number of steps.
其复杂度为O，表示时间是一个常量，所用的步数是相同的。
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t In the b even case, again I'm going to let t of b be the number of steps I want to go through.
如果b为偶数，那么我还是要用,来代表解决这个问题需要的步骤数。
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I've got one test, I've got a subtraction, I've got a multiplication, that's three steps, plus whatever number of steps it takes to solve a problem of size b minus 1.
我进行了一次比较，一次减法，一次乘法，一共是三个步骤,再加上t的步骤数。
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We're going to be counting the number of basic steps it takes to solve the problem.
我们还要去数着,解决问题的的基本步骤有多少。
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As I said, what we want to do is, we want to count the number of basic steps it takes to compute a computation as a function of input size.
我刚才提到了，我们要建立，一个根据输入大小的方法来计算,一个计算过程需要的时间,建立这个方法需要几步呢？
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So this line or these lines of code up here are arguably constant time steps to say if N is less than 2 in return, that it will always take maybe one step, maybe two steps, some number of fixed CPU cycles.
如果N小于2并返回,那么这些行所对应的代码,通常只需要执行一步，或者两步，具体数字与CPU周期有关。
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