Watch.

看。

And also pay attention to the age of the woman. The ages of the canon map are also adjacent. So they can be grouped as well.

另外，还要注意，这位女性的年龄。地图上的年龄也是相邻的。所以，它们也可以进行分组。

And groups can be overlapped. but they cannot be completely overlapped.

而且，这些组可以重叠。但不能完全重叠。

They can be partially overlapped. And there could be several correct solution. Depending on how you form a group.

它们可以部分重叠。而且，可能有多种正确的解法。这取决于你如何组建小组。

If you form a group or you draw the group in a different way, your final solution will also be different.

如果你以不同的方式组建小组，最终的解决方案也会有所不同。

So here you can see two different ways. Because the problem is slightly different. so the final solution is also slightly different. And we also proved that mathematically that these two expressions are actually identical. these two simple expressions are actually identical.

所以，这里你可以看到两种不同的方法。因为这个问题略有不同。所以最终的解决方案也会有些许差异。而且，我们还从数学上证明了，这两个表达式是实际上完全相同的。这两个简单的表达式其实是等价的。

And also, one last thing is, there's for the case of neutron, neutron, or we are amuon, amuon, there's some numerical precision. for the unitary combinations, you can either enter zero or one. depending on whether you can make it to be larger. if you can make it larger, you can just have a one.

另外还有一点是，对于中子或μ子的情况，存在一些数值精度的问题。对于幺正组合，结果只能是0或1。这取决于你是否能将其放大。如果能放大，那么结果就是1。

so here we use a to represent

所以在这里，我们用a来表示

the unitary combination. if you can make it larger, then you enter or have a one there. because the larger the qubit, the safer. the expression you are get.

幺正组合。如果能放大，那里就是1。因为量子比特越大，就越安全。这就是你得到的表达式。

so we have already finished the first three lectures. in the first lecture, it's mainly about the number system, right? we have learned what the binary number system is. what is the octal.

所以，我们已经完成了前三讲的课程。在第一讲中，主要是关于数制的，对吧？我们学习了什么是二进制数系统。还有八进制。

and in the second lecture, you mainly learned what a functional gate is. we have learned and gate, or gate. not gate, nor gate. these are all gates, right? and in the third lecture, it's mainly about the karnaugh map. it's just a very common paper.

而在第二讲中，大家主要学习了什么是逻辑门。我们学习了与门和或门。非门和或非门。这些都是逻辑门，对吧？第三讲主要是关于卡诺图的。那是一篇非常常见的论文。

and these three lectures are also the unit name. the main content in the first online case.

这三讲也是这门课的名称。这是第一次在线课程的主要内容。

the online case will be released soon. but we need to become us. our explaining later. online case will be one hour. and it will be opened for one day. so it's in the time window. log in your campus. and take the online case. the duration of the online test is one hour. and the online test will only from login hour for one day.

在线课程很快就会发布。但我们需要先完成这门课。稍后我们会详细讲解。在线课程时长为一小时。而且只开放一天。所以，它有一个时间限制。请登录你们的校园系统。然后参加在线课程。在线测试的时长是一小时。而且，从登录开始，只有一天的时间。

let's now to the tutorial of lecture 3.

现在，我们开始第三讲的教程。

we have just finished lecture 2.

我们刚刚讲完了第二讲。

next lecture of lecture 3, we have some exercise for practicing. and here are the exercises. it's not only focus on MATLAB, but also some knowledge. we have learned from lecture two and lecture one. so if you can finish the exercise in the exam, correct, you i think you should be have no problem. passing the online test. so this will be a excellent chance to enhance your understanding of the knowledge we have from lecture one, lecture two.

在第三讲中，我们会有一些练习题供大家实践。这些就是练习题。这些练习不仅针对MATLAB，还涉及一些其他知识。这些内容我们在第一讲和第二讲中已经学过了。所以，如果你能在考试中完成这些练习，那就说明你已经掌握了。我觉得你应该没问题。能够通过在线测试。所以，这将是一个很好的机会，来加深我们对所学知识的理解。也就是第一讲和第二讲的内容。

and okay, for now, I will give you any needs. okay, ten minutes. the first three are okay, first of all, let's first try to solve the first four questions. use your pens. use your notebook to solve the solution.

好的，现在，我来给大家布置作业。好的，十分钟。前三个是，好的，首先，我们先来尝试解决前四个问题。请用你们的笔。用笔记本来写出解题过程。

Hi.

你好。

So, question. First question is applying the moment theory to isolation equation. if you remember, if you don't remember, the moment is

那么，有一个问题。第一个问题是，将动量理论应用于隔离方程。如果你还记得的话，如果不记得的话，动量就是

a full form. First form is A or B and not.

一个完整的形式。第一种形式是A或B，但不是。

It's A or A not and B not.

它是A，或者非A且非B。

Second form is

第二种形式是

and we should use these two forms. to put each equation. First one is

我们应该使用这两种形式。来构建每个方程。第一个是

you use the first form. in your case, it is correct.

你使用了第一种形式。在你的例子中，这是正确的。

and second is correct as well.

第二个也是正确的。

third, yes.

第三个，是的。

it is also correct.

这也是正确的。

right?

对吧？

i is also correct.

I也是正确的。

j is also correct.

J也是正确的。

so any questions for any equations? question one.

那么，关于这些方程，有什么问题吗？第一个问题。

the solution for a is incorrect, what you have to check for a.

A的解是错误的，你需要检查一下A。

oh yeah.

哦，好的。

so for a, it's a small error here. a, it should be not n.

对于A来说，这里有一个小错误。应该是不是n。

so any questions?

那么，还有其他问题吗？

yes, so for question one.

是的，关于第一个问题。

do you remember the two forms of democracy? i think we should develop often. direct democracy or each expression. there is two forms in mind. democracy we are for sure.

你还记得民主的两种形式吗？我认为我们应该经常发展它。直接民主或者代议制民主。这里指的是两种形式。民主，我们是可以确定的。

it included in the on my case. keep these two forms in mind.

它包含在我的案例中。请记住这两种形式。

the second one, convert the following expression to sum of products. so what is the sum of product form? it should be like this, right?

第二个，将下面的表达式转换为积之和形式。所以，什么是积之和形式？应该是这样的，对吧？

this is the product, this is sum, sum of product.

这是乘积，这是求和，这就是积之和。

so you should convert each expression to the following form. like this.

所以，你应该把每个表达式都转换成这种形式。就像这样。

for the a, you use distributing law.

对于a，你要使用分配律。

and here, c d and c d is c d. because you have repeat twice.

这里，c d和c d就是c d。因为你重复了两次。

and actually, you can simplify it. because there's c d here.

实际上，你可以把它简化一下。因为这里有一个c d。

so this should be one.

所以这应该等于1。

so this should be one.

所以这应该等于1。

for the second, you will get a b.

对于第二个，你会得到ab。

b and b not is zero. so the first function is zero actually.

b和b的补集等于0。所以第一个函数实际上等于0。

so you use distributing law for the second. second expression, b and b not is zero.

所以对于第二个，你要使用分配律。第二个表达式，b和b的补集等于0。

so b and b is b.

所以b和b就是b。

for the third one, if you use distributing law, you will get the formula to be this one. And then you can do the simplification.

对于第三个，如果你使用分配律，就会得到这个公式。然后你可以进行化简。

for the third one, if you use distributing law, you will get the formula to be this one. And then you can do the simplification.

对于第三个，如果你使用分配律，就会得到这个公式。然后你可以进行化简。

for example, here you have a

例如，这里有一个

form is e, right? here you have

形式是e，对吧？这里有一个

b b t is actually b.

b，b和t实际上就是b。

and there's also b t here.

这里也有一个b。

so any more questions?

那么，还有其他问题吗？

if there's no questions, let's continue.

如果没有问题的话，我们继续。

so this is the back of the podium. and expression is what i'm using to describe time. so x is one only if a is one and b is one.

这是讲台的背面。这个表达式就是我用来说明时间的。所以，只有当a和b都为1时，x才为1。

so x equals to the only time is when a is one and b is one.

也就是说，x等于1的唯一情况，就是a和b都为1的时候。

so from this, i think the most direct way, the most common is if you bought a, if you bought a no map or if you are in a newspaper.

所以，我认为最直接、最常见的方法，就是如果你买了一张地图，或者你是在报纸上看到的。

you have a, b, x, right? a, a, zero, zero, and zero. zero, one, zero, one, zero, zero. and only one must generate one.

这里有a、b、x，对吧？a是0，0，0。0、1、0、1、0、0。只有其中一种情况会生成1。

oh, okay.

哦，好的。

so if you remember the Karnaugh map,

如果你还记得卡诺图，

right? and you write each term.

对吧？然后你把每一项都写出来。

because for this one, all three is one, okay? you write a1, a0. or this one is ab. it's the way that you write the karnaugh map, right?

因为对于这个，所有三个都是1，明白吗？你就写a1，a0。或者这个是ab。这就是写卡诺图的方法，对吧？

so it's the opposite way. for the karnaugh map, first you have an expression table.

所以，这是相反的方法。对于卡诺图，首先你有一个表达式表。

but for this question, it seems they give us already. for the karnaugh map. and how can you use the karnaugh map to write the expression?

但对于这个问题，他们似乎已经直接给我们了。就是卡诺图。那你怎么用卡诺图来写出表达式呢？

is this solution clear?

这个解法清楚吗？

in our lecture, we have several examples. for some of the problem, we have several. and how can we simplify this expression? first, I draw karnaugh map like this. and how can we define where this term? this term is existing on karnaugh map.

在我们的课程中，有几个例子。对于其中一些问题，我们有几个。那我们该如何简化这个表达式呢？首先，我画出这样的卡诺图。那么我们如何确定这个项的位置呢？这个项存在于卡诺图上。

and in this side, it will give us the one.

在这一侧，它会给我们一个1。

and for this question, we just we first have this map. and we need to write down the expression. so that's the way. then this will be a term here. so it's exactly the opposite way. how work the karnaugh map? because we know the karnaugh map is actually rearrange. so i have identical.

对于这个问题，我们只需要先有这个图。然后我们需要写出表达式。就是这样。那么这里就会有一个项。所以，这恰恰是相反的方法。那么，卡诺图是如何工作的呢？因为我们知道，卡诺图实际上是对逻辑表达式的重新排列。所以，它们是相同的。

is this way logical clear?

这种方式逻辑清晰吗？

is there any problem?

有什么问题吗？

i do the same analogy. you have a b c and x, right?

我用同样的类比来说明。你有a、b、c和x，对吧？

x is zero if any of the variables is one. so 1 1 1 is one.

如果其中任何一个变量为1，x就是0。所以，1 1 1就是1。

0 1 0 is one. 0 0 1 is one. and actually it will be a, right? 1 1 0, 1. 1 0 1, 1. 0 1 1, 1.

0 1 0是1。0 0 1也是1。实际上，它就是a，对吧？1 1 0，1。1 0 1，1。0 1 1，1。

and 0 0 0, 0.

还有0 0 0，0。

x is zero if any of the variables is one.

如果其中任何一个变量为1，x就是0。

there's another, the last one.

还有一个，就是最后一个。

what is the last one?

最后一个是什么？

oh, 1 1 1.

哦，1 1 1。

it will be 0.

那它应该是0。

yes. so x 300.

是的。所以是x 300。

so this is the expression that you're gonna get. this is the expression we want to get. based on this graph.

这就是你得到的表达式。这就是我们想要得到的表达式。基于这张图。

is the solution clear? otherwise i will have to explain it again.

这个解清楚了吗？否则我得再解释一遍。

for example, it's actually we have those nature in the track. We use the different gate, different gate, different gate.

比如说，实际上我们在电路中有这些元件。我们使用了不同的门，不同的门，不同的门。

use different logic gates.

使用不同的逻辑门。

the first one, if you have a b, and first you have to get the result.

第一个，如果你有a和b，首先你必须得到结果。

this is a not gate function. this is the not, right?

这是一个非门函数。这就是“非”，对吧？

a and the not.

a和它的非。

this is the and.

这是“与”。

a and b not, this is a. so the second part is a not.

a和b的非，这就是a。那么第二部分就是a的非。

this is the a not, right? a not and b.

这就是a的非，对吧？a的非和b。

and finally the or gate.

最后是与门。

okay, there are totally six expressions.

好的，总共有六个表达式。

i just want to repeat each one.

我想把每一个都重复一遍。

so each one, should i draw this?

那么，每一个，我应该画这个吗？

c and a. okay.

c和a。好的。

let's do c and a. maybe c and a are a little bit complex.

我们来处理c和a。也许c和a会稍微复杂一点。

for c, maybe you can first simplify the expression. that's a little bit.

对于c，也许你可以先简化一下这个表达式。这样会简单一些。

so this is the form for the expression c.

这就是表达式c的形式。

and here

然后这里，

and here

然后这里，

this is actually b.

这实际上是b。

and here we have b b over a.

这里我们有b b除以a。

and this is one.

而这是1。

we are finding this a or b or b.

我们要求的是a或b或b。

and now you can draw the boolean logic circuit.

现在，你可以画出这个布尔逻辑电路了。

you have a, you have b, b and b.

你有a，有b，还有b。

you have a, you have b, b and b.

你有a，有b，还有b。

first you have a or

首先，你有a或

b or b. okay, you have some more and here.

b或b。好的，这里还有一些与门。

this is and logic gate, you will basically see. and you have b, b.

这是与门，你基本上会看到它。然后你有b，还有b。

and then you or them together.

然后把它们一起或起来。

okay, i think we can now simplify it. because all the variables are different. from a to g.

好的，我想我们现在可以简化它了。因为所有的变量都是不同的。从a到g。

there's no overlap or same variables. it's the for each time. it's just a logical direction.

没有重叠或相同的变量。所以，对于每一次来说，都是不同的。这只是一个逻辑上的方向。

it helps you define.

这有助于你进行定义。

so you have a, b, c, d, e, f, g.

所以你有a、b、c、d、e、f、g。

the last one is over here. you have first a and b.

最后一个在这里。首先有a和b。

or a or b, sorry.

或者a，或者b，抱歉。

or a or b, sorry.

或者a，或者b，抱歉。

you have a or b and c or d.

你有a或b，以及c或d。

and then they are still and there.

然后它们仍然在那里。

and and and e. So it's not an e.

还有e。所以它不是e。

a and e or and e.

是a和e，还是and e？

i think they are all connected.

我认为它们都是连在一起的。

next one is a voltage circuit on person a.

下一个是A身上的电压电路。

Let's have a look at the question five here.

让我们来看一下这里的第五题。

So question five, we use a Turing architecture. Turing architecture simplify expressions. And question six, we use a kernel map. So there are two different ways. to simplify the expressions. The first one, question five a. you will use the distributive law, right?

那么，第五题，我们使用的是图灵架构。图灵架构可以简化表达式。第六题，我们使用了核映射。所以，有两种不同的方法。来简化表达式。首先是第五题的a部分。这里要用到分配律，对吧？

And the important thing here is that you can find common variable. You can do two terms. You can put the common variable outside. outside of the bracket.

这里很重要的一点是，你可以找到公因式。你可以对这两项进行操作。你可以把公因式提到括号外面。也就是括号外面。

One or anything is still one. So this way can be used to simplify. Finally, you will get the result like this.

1和任何东西相或，结果还是1。所以，这种方法可以用来简化。最后，你会得到这样的结果。

For the b, similarly. There's a common variable, a not and a b, right? in both terms. They all have a not and b.

对于b，也是同样的方法。这里有一个公因式，就是a非和b，对吧？在两项中都有。它们都包含a非和b。

So you can write the second expression like this. One or anything is one. So finally you will get a not and b.

所以，你可以把第二项表达式写成这样。1和任何东西相或，结果还是1。所以最后你会得到a非和b。

And for c, similarly, you have an ab. a and b or a is a, right?

对于c，同样地，你会得到ab。a和b相或a，结果就是a，对吧？

And if you find like this, you don't have to simplify anymore. It's already the simplest version.

如果你得到这样的结果，就不需要再化简了。这已经是最简形式了。

And for d, similarly. And firstly, you have a and b or a.

对于d，也是同样的道理。首先，你有a和b相或a。

And a is a.

而a就是a。

And for the first term, a and a, you will get a. And similarly for the second term, a and a, you will get a. and similarly for the same term. So you just get one.

对于第一项，a和a相或，结果还是a。同样地，对于第二项，a和a相或，结果也是a。对于第三项也是一样的。所以你最后得到的就是1。

Important is here. a not and a. is zero.

关键在于这里。a的非和a。结果是0。

So the third term and fourth term, they are always zero.

所以，第三项和第四项总是0。

And here you can find there are some variables. a and b.

在这里，你可以看到有一些变量。就是a和b。

because these two terms, you can get it form like this. And one more thing is always one. So you find it's a and b.

因为这两项，你可以得到这样的形式。还有一点，它总是1。所以你得到的是a和b。

and similarly, I think you must find.

同样地，我想你们也能找到。

you will get first term because a and a, you will get a.

第一项是a，因为a和a相或，结果还是a。

but a not c or b not c. or a b.

而a非c或b非c。或者ab。

we have two common terms, right? a b or a b is always a b.

我们有两项是相同的，对吧？ab或ab，结果总是ab。

so it's one.

所以就是1。

and here, i think we cannot simplify anymore, right? it's already the simplest version.

到这里，我想我们不能再化简了，对吧？这已经是最简形式了。

i would like to focus on the question because the question six it's exactly about panorma. how can we solve the panorma? first of all, if you are given an expression, how can you do it, for example, can you solve the panorma? you have three variables, right? you can use either c, a, b or a, b, c. it doesn't matter.

我想重点讨论一下第六题。它正是关于范德蒙矩阵的。我们该如何求解范德蒙矩阵？首先，如果给你一个表达式，你该如何处理呢？比如，你能解出范德蒙矩阵吗？这里有三个变量，对吧？你可以用c、a、b，或者a、b、c。这都无所谓。

if we put the number like this, this is 0, 1. and here will be 0, 0. here only one b can be changed. so who is this?

如果我们把数字这样写出来，这就是0和1。而这里是0和0。这里只有一个b可以改变。那么，这是谁呢？

binary number. only one bit can be changed. so it can be either 0, 1, 1, 1. or 1, 0. only one bit is changed, right?

是二进制数。只有一位可以改变。所以它可以是0、1、1、1。或者是1、0。只有一位发生了变化，对吧？

and then for the third time, they are not enough. so it's here, right? They are not enough in here. second time is a is 1. b is 0. a is 1, b is 0.

到了第三次，就不够了。所以是在这里，对吧？这里不够了。第二次，a是1。b是0。a是1，b是0。

c is 1. so it's here.

c是1。所以是在这里。

for the third time, a is 0. b is 1.

第三次，a是0。b是1。

c is 1.

c是1。

the fourth one, a is 1, b is 1, c is 0. so it's here.

第四个，a是1，b是1，c是0。所以是在这里。

okay, great. this is the karnaugh map for expression a. the second step is it's a group. the adjacent cells, this group of adjacent cells containing 1.

好的，很棒。这是表达式a的卡诺图。第二步是分组。相邻的单元格，这个包含1的相邻单元格组。

okay, there's no adjacent cells containing 1, right?

好的，没有包含1的相邻单元格，对吧？

that means the expression of expression a is already in the simplest version. there is nothing to be simplified anymore.

这意味着表达式a已经是最简形式了。不需要再进一步化简了。

and for the b, for the b, we should at first convert the expression of b. to make it in the form that means the sum of product.

至于b，我们首先要把b的表达式进行转换。把它变成“积之和”的形式。

we should at first convert the expression b to the sum of products, the form like this.

我们首先要把b的表达式转换成这种形式的积之和。

so if you simplify,

所以，如果你进行化简，

the expression b will be this.

b的表达式就会变成这样。

a b not a a b prime

就是a非b，a a非b。

a a c d

还有a a c d。

a a c not

以及a a c非。

so these are the terms. this is the expression we got it. let's simplify in order. if i have a c b not, and a and a is that's give one.

这些就是各项了。这就是我们得到的表达式。我们按顺序来化简。如果我有a c b非，而a和a，结果就是1。

a c d.

a c d。

c and c not is zero. so this term is gone already. we can simplify it further.

c和c非相与是0。所以这一项已经消失了。我们还可以继续化简。

a c b not or b.

a c b非或b。

this is one, right? finally what you get is

这等于1，对吧？最后得到的结果是：

so this is the question five. you use the boolean algebra. to simplify the expression. or you can also use the karnaugh map. but if you use karnaugh map, you should stop here. and not do the following steps. if you do the following steps, you already get the simplest solution.

这就是第五题。你使用了布尔代数。来化简这个表达式。或者，你也可以使用卡诺图。但是如果你用卡诺图，就应该停在这里。不要继续做下面的步骤了。如果你做了那些步骤，其实已经得到了最简解。

and if you have drawn the karnaugh map for expression b, i will stop. there is only a c d.

如果你已经为表达式B画出了卡诺图，我就停在这里。这里只有a、c、d。

so you have c, d, a.

所以是c、d、a。

that's only two points, right?

只有两个点，对吧？

zero, zero, zero, one, one, one.

0、0、0、1、1、1。

zero, zero, zero, one, one, one.

0、0、0、1、1、1。

a is one, d is three zeros. a is one, b is zero, c is one.

a是1，d是三个0。a是1，b是0，c是1。

yeah.

对。

and for the other, always zero.

至于其他的，总是0。

and then you do those little ones. it will be this one and zero.

然后你处理那些小的。那就是这个和0。

and you get the simple expression.

这样你就得到了一个简单的表达式。

and you get the simple expression.

这样你就得到了一个简单的表达式。

a is unchanged. so you are the middle.

a保持不变。所以你是中间的那个。

for efficiency, we have PEA, right?

为了提高效率，我们有PEA，对吧？

let's go like this.

我们就这样吧。

so you have 0, 0, 0, 1, 1, 1. for shy, it's 1, 1, 0.

所以你有0、0、0、1、1、1。对于shy来说，是1、1、0。

the first value is 1, 1, 0. Second value is 0, 1, 0.

第一个值是1、1、0。第二个值是0、1、0。

third value is 0, 0, 0. and for the rest,

第三个值是0、0、0。至于剩下的，

for the rest, we have 0.

都是0。

for the extra key, if you draw the canonical, it should look like this, right? we have three terms. one term here,

至于额外的键，如果你画出规范形式，它应该看起来像这样，对吧？我们有三项。这里有一项，

and for the rest of things, they are zero. and then you have to, you cannot put these three together because

至于其他部分，都是0。然后你必须，不能把这三个放在一起，因为

the number of things should be 2 to the power of n, right? so you should be at least two, four, eight. not just three. so we have to get two groups. This is the first group. and this is the second group.

项数应该是2的n次方，对吧？所以至少应该有2、4、8项。而不仅仅是三个。所以我们必须分成两组。这是第一组。这是第二组。

This is the first group. First one's here. This is the first group. and the two ones below. it is the second group.

这是第一组。第一个在这里。这是第一组。下面还有两个。这是第二组。

okay, for the first group, you can see a is unchanged. b is unchanged. f is changing. so f will be removed. but b, b is always there, so b is not. f is always there.

好的，对于第一组，你可以看到a没有变化。b也没有变化。只有f在变化。所以f会被移除。但是b一直都在，所以b不是。f一直都在。

so this is the first group. the expression for the first group.

所以，这就是第一组。这是第一组的表达式。

and for the second group, the two ones below. b is changing. so b is eliminated.

至于第二组，就是下面的两个。b在变化。所以b被排除了。

e is one, f is zero.

e是1，f是0。

any questions?

有什么问题吗？

Pay attention to the edges of the Karnaugh map. The edges are also adjacent, so they can be grouped as well. Groups can be overlapped, but they cannot be completely overlapped—they can only be partially overlapped.

注意卡诺图的边缘。边缘也是相邻的，因此它们同样可以被分组。分组之间可以重叠，但不能完全重叠——只能部分重叠。

Here you can see two different ways to draw the group. Because the group is slightly different, the final solution is also slightly different. It has also been proved mathematically that these two simple expressions are actually identical.

这里你可以看到两种不同的方式来绘制群。由于群略有不同，最终解也略有不同。数学上已经证明，这两个简单表达式实际上是相同的。

Also, one of the strategies is to combine this beta mutation with an alpha mutation. There is only one such combination. In that combination, you can enter either zero or one, depending on whether you can make it larger. If you can make it larger, you simply enter one.

此外，其中一种策略是将这个β突变与一个α突变进行组合。这样的组合只有一种。在该组合中，你可以输入零或一，取决于你是否能将其增大。如果能增大，就直接输入一。

The language will be simpler; the expression will be clearer. We have already finished the first three lectures. In the first lecture, it was mainly about the number system. We have learned what the binary number system is and what the octal number system is. And in the second…

语言会更简洁，表达会更清晰。我们已经完成了前三次课。第一次课主要讲的是数制。我们学习了什么是二进制数制，什么是八进制数制。第二次……

One function of MIT is to have lectures—full-day, half-day, or no-day formats. The duration varies by day. Many of these lectures are about economics. When students want to attend lectures in person, note that these three lectures were also held at the main campus during the first online test period. The online test will be…

MIT的一项功能是开设讲座——有全天、半天或无讲座的安排，时长因日而异。其中许多讲座涉及经济学。当学生希望到场听课时，请注意这三场讲座也是在主校区于第一次在线测试期间举行的。在线测试将会……

The online test will be one hour in duration and will be open for one day. Within that time window, you should log in to your Canvas and finish the online test. The test will probably cover Lecture 1 to Lecture 3.

在线测试时长为一小时，开放时间为一天。在该时间窗口内，请登录你的Canvas并完成在线测试。测试内容大概涵盖第一讲到第三讲。

And the exercises do not only focus on the exam but also on some knowledge that we have learned from Lecture Two and Lecture One. If you can finish the exercises correctly, I think it should be no problem passing the online test. This is probably an excellent chance for you to enhance your understanding of the knowledge we have learned from Lecture One and Lecture Three.

这些练习不仅针对考试，还涉及我们在第二讲和第一讲中学到的一些知识。如果你能正确完成练习，我认为通过在线测试应该没有问题。这很可能是你加深对第一讲和第三讲所学知识理解的一个绝佳机会。

For now, I will give you ten minutes. Go to the first three problems. Use your pen and notebook to solve them.

现在给大家十分钟时间。请做前三道题，用笔和笔记本把答案解出来。

[No discernible lecture content detected in this segment. The audio consists primarily of background noise, shuffling, and coughing with no intelligible speech.]

[该片段未检测到可辨识的讲课内容。音频主要由背景噪音、翻动声和咳嗽声组成，无可理解的语言。]

[No speech content detected in this segment. The audio contains only non-verbal sounds such as coughing, throat clearing, and movement.]

[该片段未检测到语音内容。音频仅包含咳嗽、清嗓子和移动等非语言声音。]

No discernible speech content detected in this segment (1357.0–1417.0 s). The audio contains only ambient noise, breathing, and minor movement sounds.

该片段（1357.0–1417.0 秒）未检测到可辨识的语音内容。音频中仅包含环境噪音、呼吸声及轻微的动作声响。

No discernible speech content detected in this segment (1357.0–1417.0 s). The audio contains only ambient noise, breathing, and minor movement sounds.

该片段（1357.0–1417.0 秒）未检测到可辨识的语音内容。音频中仅包含环境噪音、呼吸声及轻微的动作声响。

No discernible speech content detected in this segment. The audio contains only ambient background noise such as paper rustling and chair movement.

该片段未检测到可辨识的语音内容。音频仅包含翻纸声、椅子移动等环境背景噪音。

Let us look at the first question: why is emotion related to expression? If you remember, there are two forms of emotion. The first form is A or B, and one is inherent to emotion.

我们来看第一个问题：为什么情感与表达相关？如果大家记得，情感有两种形式。第一种形式是A或B，其中一种是情感固有的。

We need to use these two forms for each information. The first one is…

我们需要对每条信息使用这两种形式。第一种是……

First of all, A is correct. Second, B is correct as well. Third, C is false. G is also correct.

首先，A是正确的。其次，B也是正确的。第三，C是错误的。G同样是正确的。

[No substantive teaching content in this segment. The speaker produced only incomplete, abandoned utterances that were never finished or confirmed.]

[本段无实质性教学内容。说话者仅产生了未完成的、被放弃的语句片段，从未说完或确认。]

This is correct. Eyes open, correct. It is open.

这是正确的。睁眼，正确。它是打开的。

Any questions for question one? If the information for A is incorrect, what does the system print?

关于第一题，大家有什么问题吗？如果A的信息是错误的，系统会打印什么？

[This segment consists primarily of keyboard typing sounds with no discernible lecture content.]

[本段主要为键盘打字声，无可辨识的讲课内容。]

Regarding J.P. Morgan, I think we should not call him a prophet. J.P. Morgan made certain predictions. Keep this quote box in mind. J.P. Morgan's words were certainly included in the—

关于J.P.摩根，我认为我们不应称他为先知。J.P.摩根做出过某些预测。请记住这个引用框。J.P.摩根的话确实被包含在——

In the online case, there are these two forms defined. The second common polynomial expression is the sum of product form. What is the sum of product form? It should be an easy one.

在在线情况下，定义了这两种形式。第二种常见的多项式表达式是积之和形式。什么是积之和形式？这应该是一个简单的问题。

This is a product. You should convert each expression to the following form, the form like this. For part (a), you use substitution with Big-O.

这是一个乘积。你们应该把每个表达式转换成如下形式，就是这样的形式。对于第(a)小题，使用大O替换法。

Here we have CD and CE equals CD. You do not need to repeat it twice. Actually, you can simplify it because they are different—just CD here. So this should be one, and finally you get...

这里我们有 CD 和 CE 等于 CD。不需要重复写两次。实际上可以简化，因为它们不同——这里只写 CD。所以这应该等于一，最终你得到……

On the second condition, one gets p and p₀ being zero. So the first condition is satisfied. If you use this on the second equation on the black page, p and p₀ is zero.

在第二个条件下，得到 p 和 p₀ 为零。因此第一个条件得到满足。如果在黑板页上的第二个方程中使用这一点，则 p 和 p₀ 为零。

For the third one, if you spin the ball and want it to come back to you, you can do a simple calculation.

第三个问题，如果你旋转球并希望它回到你手中，你可以做一个简单的计算。

For example, here you have A, B, C, D. BGP is attached to C. There is also BGP here. So, are there more than all of these? Any questions?

例如，这里有 A、B、C、D。BGP 连接在 C 上。这里也有 BGP。那么，是否比这些全部还多？有什么问题吗？

If there are no questions, let's continue with the third one. The third one is the barycentric expression for the description. So x is 1 only if x is 1 and v is 1. So x equals to 1 if and only if x is 1 and v is 1.

如果没有问题，我们继续第三个。第三个是关于描述的重心坐标表达式。所以 x 等于 1 仅当 x 等于 1 且 v 等于 1。即 x 等于 1 当且仅当 x 等于 1 且 v 等于 1。

The most direct way to prove this is to use a truth table. You have A, B.

证明这一点最直接的方法是使用真值表。你有 A、B。

Eight zero zero one zero, zero one zero one zero zero, and one one zero zero one. Eight zero one zero one.

八零零一零，零一零一零零，以及一一零零一。八零一零一。

If you remember the Kano map, how can you read each part? For this one, all of this is one section. It reads from A1 to A2, or this one is labeled AB. This is the way you read the Kano map. Now, this is the opposite way for the Kano map. For the Kano map, the first quadrant and the next quadrant are different, and you have more quadrants. But for this question, it is like—

如果你还记得Kano地图，你该如何读取每个部分？对于这一个，所有这些属于一个区域。它从A1读到A2，或者这个标记为AB。这就是读取Kano地图的方式。现在，这是Kano地图的相反读法。对于Kano地图，第一象限和下一象限是不同的，并且你有更多的象限。但对于这个问题，它类似于——

They give us a range of the Carnot engine and how to use the Carnot engine to calculate efficiency. This is the projection for the year. In our lectures, we have several examples—some problems in general on how to identify the first law of thermodynamics and the Carnot engine, and how to define whether a given term is a heat engine or a Carnot engine or not.

他们给出了卡诺机的范围，以及如何利用卡诺机计算效率。这是本学年的教学规划。在我们的课程中，有多个例题——总体上涉及如何识别热力学第一定律和卡诺机，以及如何判断给定的术语是热机还是卡诺机。

One-to-one. For each equation, we first have this matrix, and then we write down an equation. This is the opposite way: if this one does not need to be a term here, this one needs to be a term here — it is exactly the opposite arrangement. However, we do not have the table; the expression — because we know we cannot rearrange the table. So we add them all, identical.

一一对应。对于每个方程，我们首先有这个矩阵，然后写出一个方程。这是相反的方式：如果这一项不需要出现在这里，那么这一项就需要出现在这里——这恰好是相反的排列。然而，我们没有这个表；这个表达式——因为我们知道无法重新排列这个表。所以我们将它们全部相加，作为相同项。

Wait, both must be here. Similarly, you have a, b, c, and x, right?

等一下，两者都必须在这里。类似地，你有 a、b、c 和 x，对吗？

X is zero if and only if Y equals one. So: one, one, one is one; zero, one, zero is one; zero, zero, one is one. Actually, for the A: one, one, zero, one; one, zero, one, one; zero, one, one, one; and zero, zero, zero, zero.

X 为零当且仅当 Y 等于一。所以：一、一、一为一；零、一、零为一；零、零、一为一。实际上，对于 A：一、一、零、一；一、零、一、一；零、一、一、一；以及零、零、零、零。

The exponential—the marble is... That's another thing, the last one.

指数——弹珠是……这是另一件事，最后一件。

Zero. One. The last one. One, one, one. Oh yes, one, one. One, one, one. Zero, zero, zero. Yes. So x, you have—that's actually—so this expression, the one case. This is the expression, the one case, based on this graph.

零。一。最后一个。一，一，一。对，一，一。一，一，一。零，零，零。是的。所以 x，你得到——实际上——这个表达式，对应的是 1 的情况。这就是该表达式，对应 1 的情况，基于这张图。

This is the solution to what we have seen. For example, next week we are going to use a different logic gate.

这就是我们所见问题的解决方案。例如，下周我们将使用一种不同的逻辑门。

First of all, if we have A prime — the first term here is the complement of B. This is a union. The second term is A complement.

首先，如果我们有A的补集——这里的第一个项是B的补集。这是一个并集运算。第二个项是A的补集。

Handoff under tree, or bin. There are six expressions, and each one has four leaves.

在树结构下进行交接，或者用二叉结构。共有六个表达式，每个表达式有四个叶节点。

CNA. What is CNA of the complex? For C and E, maybe you can first simplify the expression with it.

CNA。复数的CNA是什么？对于C和E，也许你可以先化简含它的表达式。

This is a form for expression C. And here—

这是表达式C的一种形式。而这里——

This is actually B. And here we have B again, or sigma 3. And this is 1. We are writing it as A, or BC, or B. Now, if I go to the original circuit, we have A.

这实际上是 B。这里我们又得到了 B，或者说是 σ₃。而这里是 1。我们把它写成 A，或者 BC，或者 B。现在，回到原始电路，我们有 A。

You have b, c, d, e. First, you have a or, and b, c, d, e. You have more than one here. And what is a? Your a is c. And you have e. You have e, d. And then e or a is better. e or a is better.

你有 b、c、d、e。首先，你有 a 或，以及 b、c、d、e。这里不止一个元素。那么 a 是什么？你的 a 就是 c。你还有 e。你有 e、d。然后 e 或 a 更好。e 或 a 更好。

I am saying that you cannot simply find A. You cannot simply find A because all the variables are different—all the variables are different from A to G. There is no…

我要说的是，你不能简单地求出A。你不能简单地求出A，因为所有变量都不同——从A到G的所有变量都不同。没有……

A line or same matrix is the point that's defined. So you have a, b, c, d. The box is not a circle. You have first a and b.

一条线或同一矩阵即为所定义的点。因此你有 a、b、c、d。方框不是圆。你首先有 a 和 b。

This is A and B. Or A or B. Not A or B. And C or D. And not E. So this is not E.

这是 A 与 B。或者 A 或 B。非（A 或 B）。并且 C 或 D。并且非 E。所以这里是非 E。

This is A and B. Or A or B. Not A or B. And C or D. And not E. So this is not E.

这是 A 与 B。或者 A 或 B。非（A 或 B）。并且 C 或 D。并且非 E。所以这里是非 E。

Consider a point in 3D space, R³. The position vector goes from the origin to that point. Now, the dot product a · b equals the magnitude of a times the magnitude of b times the cosine of the angle between the two vectors.

考虑三维空间 R³ 中的一个点。位置向量从原点指向该点。现在，点积 a · b 等于 a 的模乘以 b 的模再乘以两个向量之间夹角的余弦值。

No discernible lecture content detected in this segment. The audio consists primarily of ambient room noise, movement sounds, and indistinct background activity with no intelligible instructional speech.

该时间段内未检测到可辨识的课堂教学内容。音频主要由环境噪音、移动声及模糊的背景活动声组成，无可理解的授课语音。

[No speech content detected in this segment. The audio contains only non-verbal sounds such as coughing, breathing, and object handling.]

[该片段未检测到语音内容。音频中仅包含咳嗽、呼吸及物品翻动等非语言声音。]

[No discernible speech content detected in this segment. The audio contains only ambient room noise, movement sounds, and occasional coughing.]

[该片段未检测到可辨识的讲课内容。音频中仅包含环境噪音、移动声和偶尔的咳嗽声。]

Let's have a look at question five.

我们来看第五题。

We use algebra to simplify expressions. For question six, we use Kano map. There are two different ways to simplify an expression. For question 5.8, you use the distributive law. The important thing here is that you can find common variables between two terms, and you can put the common variable outside the bracket.

我们使用代数来简化表达式。第六题，我们使用卡诺图。简化表达式有两种不同的方法。第五点八题，你使用分配律。这里的关键在于，你可以在两项之间找到公共变量，然后将公共变量提到括号外面。

One more, and you see here one. This is the way we use to simplify. Finally, when we get the result like this. For b, similarly, x is a common variable, a naught and b, right? For both terms, they all have a naught and b.

再看一项，这里出现了一个一。这就是我们用来化简的方法。最终，当我们得到这样的结果时。对于 b，类似地，x 是一个公共变量，a零 和 b，对吧？对于这两项，它们都含有 a零 和 b。

You set in the expression that not B is 1 for any case when not B is 1. For C, similarly, you have A and B or A. A and B or A is just A.

在表达式中，当 not B 为 1 时，not B 的值就是 1。对于 C，类似地，你有 A and B or A。A and B or A 就等于 A。

If you are not interested in giving the same kind of story in a different version, I will skip it. And for B, similarly. First you have—which parts of that were BA and AB, right?

如果你们不想听同一类论证换一种说法再讲一遍，我就跳过。对于B，类似地。首先你有——其中哪些部分是BA和AB，对吧？

For the first department, you will get both A and B. Similarly, for the second department, they are all common modules, so you just take one.

对于第一个部门，你将同时获得A和B。类似地，对于第二个部门，它们都是公共模块，所以你只需选其中一个。

A and B are not always zero. The third term is not always zero. From variables a and b, you can find that there are always two terms. One coordinate is always one when you translate this end.

A 和 B 并不总是零。第三项也不总是零。从变量 a 和 b 可以看出，始终有两项。当你平移这一端时，其中一个坐标始终为一。

First, we have C. Second, we have A or C, or A and B. There are two common elements. A and B, or A and B, is always A and B. That gives one.

首先，我们有C。其次，我们有A或C，或A与B。其中有两个公共元素。A与B，或A与B，始终等于A与B。结果为一。

You cannot simplify anymore. It is already the simplest version. This form is already the simplest version; you cannot simplify anymore. I would like to focus on question six.

你无法再简化了。它已经是最简形式。这个表达式已经是最简版本，不能再简化了。我想把重点放在第六题上。

This is an exercise about automata. The question is: how can we find an automaton? Here is the first thing. If you are given an expression, how can we draw a corresponding automaton? For A, we have three variables. You can use either C or ABC; it does not matter.

这是一道关于自动机的练习题。问题是：我们如何找到一个自动机？首先，如果给定一个表达式，我们如何画出对应的自动机？对于A，我们有三个变量。你可以使用C或ABC，这没有关系。

If we put a number like this—0, 1—and here it can be 0, 0. The important point here is that only one bit can be changed. For two adjacent binary numbers, only one bit is changed. It can be either 0 1, 1 1, or 1 0—only one bit will change.

如果我们把数写成这样——0、1——这里可以是0、0。关键在于，只能改变一位。对于两个相邻的二进制数，只有一位会发生变化。它可以是0 1、1 1或1 0——只有一位会改变。

For the first time, A equals zero, B equals zero, C equals zero. For the second time, A is one, B is zero, C is one. For the third time, A is zero, B is one, C is one. For the fourth time, A is one, B is one, C is zero.

第一次，A等于零，B等于零，C等于零。第二次，A为一，B为零，C为一。第三次，A为零，B为一，C为一。第四次，A为一，B为一，C为零。

This is the Karnaugh map for expression A. The second step is to group adjacent cells containing 1.

这是表达式A的卡诺图。第二步是将包含1的相邻单元格进行分组。

That means Expression A is already in a state where we do not need to do anything anymore. As for Expression B, you should first convert it into the sum of products form.

也就是说，表达式A已经处于无需再做任何处理的状态。对于表达式B，你应当首先将其转换为积之和的形式。

The sum of products, the product leaves. So if you define that this condition is limited: A B C not, A A C, A C D, A A C C not.

积之和，积项。如果我们将此条件定义为有限制：A B C非，A A C，A C D，A A C C非。

This expression can be simplified further. We have a times t naught, and a and x—that is just one. Since c naught is zero, the expression simplifies to a times t naught over [the remaining denominator]. This equals one.

这个表达式可以进一步简化。我们有 a 乘以 t₀，以及 a 和 x——那就是 1。由于 c₀ 为零，该表达式简化为 a 乘以 t₀ 除以[剩余分母]。这等于 1。

This is exactly the question: use the Fourier transform to simplify an equation. Alternatively, you can use the normal method. However, if you use the normal method, you should stop here and not proceed to the following steps, because once you go through those steps, you already obtain the simplest solution.

这正是本题的关键：利用傅里叶变换来简化方程。当然，你也可以使用常规方法。但如果采用常规方法，应当在此处停止，不要再继续后面的步骤，因为一旦走完那些步骤，你就已经得到了最简解。

If you have a control system, the output R(s) and C(s)—there is only A(s), B(s), C(s).

如果你有一个控制系统，输出 R(s) 和 C(s)——只有 A(s)、B(s)、C(s)。

So let's see. That's only two points, right? 0, 0, 0, 1, 1, 1. A equals 1, B equals 0, C equals 0. A equals 1, B equals 0, C equals 1. A equals 1, B equals 1, C equals 1.

我们来看一下。只有两个点了，对吧？0，0，0，1，1，1。A等于1，B等于0，C等于0。A等于1，B等于0，C等于1。A等于1，B等于1，C等于1。

So let's see. That's only two points, right? 0, 0, 0, 1, 1, 1. A equals 1, B equals 0, C equals 0. A equals 1, B equals 0, C equals 1. A equals 1, B equals 1, C equals 1.

我们来看一下。只有两个点了，对吧？0，0，0，1，1，1。A等于1，B等于0，C等于0。A等于1，B等于0，C等于1。A等于1，B等于1，C等于1。

A normal subgroup of G. For each coset, there is a single representative. Every element in the coset can be represented by one representative. If a is an element, then the coset aH equals bH if and only if a⁻¹b is in H.

G 的正规子群。对于每个陪集，存在唯一的代表元。陪集中的每个元素都可以由一个代表元来表示。若 a 是其中一个元素，则陪集 aH 等于 bH 当且仅当 a⁻¹b 属于 H。

The sequence is 01101101101101. The first three is 110, the second three is 010, the third three is 010, giving 000.

该序列为 01101101101101。第一组三位是 110，第二组三位是 010，第三组三位是 010，结果为 000。

Consider the input 010. Under normal operation, the output should return to 010. However, due to a race condition, the output may momentarily become 000. There are three terms in this circuit; one of them is 11. The race occurs because of the timing mismatch among these terms.

考虑输入 010。在正常工作情况下，输出应返回 010。然而，由于竞争条件，输出可能会短暂变为 000。该电路中有三个项，其中一项为 11。竞争的产生是因为这些项之间存在时序不匹配。

For the remaining sets, they are available. However, you cannot put these three together because the number of sets must equal the power of n. So it should be at least 4n minus 3. Therefore, we need two groups: this is the first group, and this is the second group.

对于剩余的集合，它们是可用的。然而，你不能将这三个放在一起，因为集合的数量必须等于 n 的幂。所以它至少应该是 4n 减 3。因此，我们需要两个组：这是第一组，这是第二组。

This is the first group. Here it is the first group; below it is the second group. For the first group, we can see that E is unchanged and F is changing. What is F? F is always zero. E is not zero. F is always zero.

这是第一组。这里是第一组；下面是第二组。对于第一组，我们可以看到E不变，F在变化。F是什么？F始终为零。E不为零。F始终为零。

For the first group, and for the second element—the first one is oxygen—P is given by n = 1, l = 0. Any questions?

对于第一组，以及第二个元素——第一个是氧——P 由 n = 1, l = 0 给出。有问题吗？

[No lecture speech content detected in this segment. The audio contains only non-verbal sounds such as coughing, ambient room noise, and incidental sounds.]

[该片段未检测到讲课语音内容。音频中仅包含咳嗽声、环境噪音及偶发声响等非语言声音。]

Break time. Any questions?

休息时间。有什么问题吗？