What we've done is taken this pitch and played it all the way down an octave below it so we're actually getting back to this configuration of the pitch right next to it, and we could--then of course we could go down one more step and we would get the octave, which is a duplication of two-to-one.

我们所做的是固定一个音高，向下一路弹奏一个八度,所以事实上我们又回到了与这个固定音高相邻的音上，我们当然也可以向下再弹一组，然后得到这个八度，这其实就是两个八度对一个八度的重复。

耶鲁公开课 - 聆听音乐课程节选

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。

哈佛公开课 - 计算机科学课程节选