Or another way of saying it is, we're going to use as the basic steps, those operations that run in constant time, so arithmetic operations.
我们用可以在恒定时间内完成的操作,算法,比较,内存读取。
And it's a topic I want to pick up on today, we're going to do for the next few lectures, and it's a topic I want to spend some time on because this idea of capturing data and methods, the term we're going to use for it, but data and functions that belong to that data, things that can be used to manipulate them, is a really powerful one.
我们会在今天讲这个话题,在后面几节课还会讲这个,这是个我愿意花点时间,在上面的话题因为这是个,捕获数据和方法的思想,我们用这个术语来形容它,数据中包括了数据和函数等,能用来操作它们的东西,真的是很强大的工具,我们真正要讲的东西是,我们要学会如何把信息。
So to be clear, we expect most students to very comfortably pursue the standard edition of the P set, but for this particular P set, and we have about 150 or so of these out on the hallway, and we'll have them at office hours throughout the coming week--this is a little logic board ; with lots of sensors; the sound sensor, a light sensor, a little joystick and the like.
我希望我们大多数同学,研究问题集的标准版本,可编程版有个特殊的问题集,我们会在接下来的几周里,在办公时间为大家做解答,这是个小的逻辑板,上面有很多传感器;,包括声音传感器,光传感器,小操作杆等。
So we're to assume we can get to any piece of data, any instruction in constant time, and the second assumption we're going to make is that the basic primitive steps take constant time, same amount of time to compute. Again, not completely true, but it's a good model, so arithmetic operations, comparisons, things of that sort, we're all going to assume are basically in that in that particular model.
因此如果我们假设在恒定的时间内,我们可以取得任何一块数据,任何一种数据结构的话,我们要做的第二个假设就是,基本的原始操作计算花费的时间是恒定的,这个假设也不是完全正确的,但这个模型其实挺不错的,因此算法操作,比较,这一类的事情,我们在这个特定的模型中都假设是基本的,操作,花费的时间是恒定相同的。
So I have n operations log n times, n log n there we go, n log n. Took us a long time to get there, but it's a nice algorithm to have.
所以我log,n遍的n次操作,就得到了,虽然花了不少时间得到了这个结论。
With this, if I can assume that accessing the i'th element of a list is constant, then you can't see that the rest of that analysis looks just like the log analysis I did before, and each step, no matter which branch I'm taking, I'm cutting the problem down in half.
读取数组中的第i个元素,是个常量时间的操作的话,我也就能像以前那样得到,这个算法是对数级复杂度的分析,并且每一步不管我选择哪个区间,我都可以把问题的规模缩小一半。
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