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来自 University of California San Diego 的课程

Mathematical Thinking in Computer Science

228 个评分

University of California San Diego

Mathematical Thinking in Computer Science

228 个评分

课程 1（共 5 门，Specialization Introduction to Discrete Mathematics for Computer Science）

Mathematical thinking is crucial in all areas of computer science: algorithms, bioinformatics, computer graphics, data science, machine learning, etc. In this course, we will learn the most important tools used in discrete mathematics: induction, recursion, logic, invariants, examples, optimality. We will use these tools to answer typical programming questions like: How can we be certain a solution exists? Am I sure my program computes the optimal answer? Do each of these objects meet the given requirements? In the course, we use a try-this-before-we-explain-everything approach: you will be solving many interactive (and mobile friendly) puzzles that were carefully designed to allow you to invent many of the important ideas and concepts yourself. Prerequisites: 1. We assume only basic math (e.g., we expect you to know what is a square or how to add fractions), common sense and curiosity. 2. Basic programming knowledge is necessary as some quizzes require programming in Python.

从本节课中

Making Convincing Arguments

Why some arguments are convincing and some are not? What makes an argument convincing? How to establish your argument in such a way that there is no possible room for doubt left? How mathematical thinking can help with this? In this week we will start digging into these questions. We will see how a small remark or a simple observation can turn a seemingly non-trivial question into an obvious one. Through various examples we will observe a parallel between constructing a rigorous argument and mathematical reasoning.

One Example is Enough3:26

Splitting an Octagon1:24

Making Fun in Real Life: Tensegrities10:25

Know Your Rights5:49

Nobody Can Win All The Time: Nonexisting Examples8:27

与讲师见面

Alexander S. Kulikov
Visiting Professor
Department of Computer Science and Engineering
Michael Levin
Lecturer
Computer Science
Vladimir Podolskii
Associate Professor
Computer Science Department

Let's return to the existential statement and their proofs.

So what is important to understand here?

Let's start with some example, simple example.

Imagine you want to find the two-digit number that becomes

seven times smaller when the first digit is deleted.

Well, it's a very easy question because there is not so

many two-digit numbers, and also it should be divisible by seven.

So if you remember the multiplication table, the first two-digit number is 14,

and then they go, and then you get 63, and the next one is 70,

and then division by seven gives 10, which is not one digit.

So it's simple small list, and

you can check that 35 is the only thing, and if you're asked for

something more complicated, for example imagine we asked you for

a number that becomes 57 times smaller when the first digit is deleted.

It's then, it's not so easy, and what is important that if you come

next day and say, look here is the number, then you are done.

We have no right to ask you why you choose this number, and how you found it.

So it's just you give an example, we check the example,

and we should be completely satisfied, and this is from the student's view.

But of course the teachers is usually,

the teachers do not have the rights the student have.

So if you are teacher, then you should explain how you find that solution,

and now we'll try to explain how you can do this.

So, let's write again what do we want.

We, here is the number.

Here is the first digit, which should be deleted.

This is what remains, and

this is our equation shows what should be achieved.

Okay, but this is not very common for algebra because there is some dots here so

the number is written in a decimal notation,

it's not very good, so let's write something.

Note this part by X, and then on the right-hand side,

we have 57 X, it's easy, but on the left-hand side,

we have X + a0...0, and we need k zeros,

and what is important, that if we add one 0,

this means that we multiply a by 10.

So actually, this thing is just a x 10 to the number of zeros,

and we will write it now.

So we assume that x has k digits and we write this equation.

So this is a shifted to that k places to the left and

then we + X = 57 x X as we should.

Okay, now it's better equation because we can move X in the right-hand side and

get that a x 10 to the k = 56 x X, and 56 = 7 x 8,

and now, if you meditate a bit, you see how can you find a.

So from the equation you can find a.

Do you see how to do this?

Look here is, on the right-hand side, we have factor 7.

So in the left-hand side, also 7 should be somewhere, and here, 10 is 2 x 5,

so it's 2 to the K times 5 to the K, so here's 7 cannot hide.

So the 7 should be here, and a is just one digit.

So a should be 7, there is no other possibilities, and

this should be written in the next slide.

No, a should be divisible by 7, that's how first.

It should contain 7 as a factor, and only, a = 7, only one case.

So we know a now, and then if we know that this is 7,

so we can cancel the 7 and then we know that 10 to the k is 8 x X.

So the the question, what we don't know,

what we don't know the k, so 10 to the k should be divisible by 8.

This is the only thing we need, and you see, 10,

1000 are not multiples of eight, and 10 and

100 is not a multiple, but 1000 is a multiple.

So we know what X is for this case.

We know that X is 1000 divided by 8.

And we are done.

This is exactly the example we looked for,

and of course, we can take bigger powers of 10.

If we take not 1,000 but 10,000, we get just one zero, additional zero.

So there are other solutions, but

other solutions are just the same plus zero at the end.

So now we see how this problem can be solved, and

of course it's useful because it helps us to solve

similar problems with different, a different number.

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