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When there's a contradiction in our heads we're not happy and will take steps to make the contradiction go away.
当我们的心理活动出现矛盾时,我们会不安，会采取办法,让这种矛盾消失。
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Be disciplined in order to continue to take the small steps that you're taking in order to really work towards that.
做好计划使你能够不断地朝这个方向慢慢进步。
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Our goal is to take problems and break them down into these computational steps, these sequence of instructions that'll allow us to capture that process.
我们的目标是得到问题,然后将问题分解为这些计算步骤，这些指令集,可以让我们看到这个过程。
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now let's go back to each of our individual steps and look, based on what we know about how to evaluate the thermodynamic changes that take place here, let's look at each one of the steps and see what happens.
让我们回到循环中的每一步,然后看看，基于我们已有的知识,来估计一下,发生的热力学变化，让我们看,循环中的每一步发生了什么。
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Well how many more steps does this algorithm take?
用同样的算法需要多几个步骤？
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And also be well-informed as to what you need to do and the steps you need to take
对于你要做什么，要采取什么步骤，你都要保持消息灵通，
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It's where I come to relax and to read on the steps or to take the train.
我来这儿休息放松，坐在台阶上看书或从这儿搭乘火车。
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and what steps they need to take in order to graduate high school,
以及他们下一步要做什么来完成高中毕业，
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Take small steps.
慢慢地，一步一步来。
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So I take 2 plus 3 b steps to go through this loop OK. So if b is 300, it takes 902 steps.
好，如果b是300的话就是902步,如果b是3000那就有9002步。
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One way to think about this intuitively if the derivative is very large the function is changing quickly, and therefore we want to take small steps.
关于这个方法很直观的一点想法是,如果导数非常大,函数也就变化的非常快，因此我们想一小步一小步的来。
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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。
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On the next step though, this, we get substituted by that. Right, on the next step, I'm back in the even case, it's going to take six more steps, plus t of b minus 1. Oops, sorry about that, over 2.
这一步就是偶数了，这一步会让我们得到，6+t这样的结果，因为b-1现在是偶数了,别忽略这里的细节。
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I know it's gonna take me a lot of work on the order of N squared steps to sort these things.
要对它们排序就要进行大量的工作,大概需要N平方步。
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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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