Liu Juan said: "Why do you want you to take care of our affairs?"
Xiao Hong said: "If you treat our little water flower like this, I will take care of it. Little water flower is ours, our baby, and our princess. Over the years, I have never seen her treat anyone like this, and I don't know about you either."
Such a toad has cultivated such blessings in that life, I——."
Xiao Shuihua glanced at Liu Juan and said, "Xiaohong, let's go."
Liu Juan didn't know what to say. He looked at Xiao Shuihua and suddenly said: "Xiao Shuihua, we need to calm down. I——."
Xiao Shuihua said: "I know what you mean, okay, we are put together now, we -."
Liu Juan said: "Xiao Shuihua, I will definitely save you, but we should think about it carefully. Yes, you are beautiful, but beauty does not necessarily mean love. You understand what I say. I——."
Xiao Shuihua said: "Okay, don't say it. I know what you mean. Now we will get through the difficulties together, and I won't say anything else."
Liu Juan said: 'That's not what I meant, why don't you understand, I, I, I..."
Xiao Shuihua said: "I understand what you mean, do you hate our kind of life? You feel that living like that is a fake person, but this kind of life is real to us."
Liu Juan didn't say anything more. He had nothing more to say. He watched Xiao Shuihua's back slowly walk away, but he still didn't know what to do.
The poor have the life of the poor, and the rich have the life of the rich, but Liu Juan likes the life of the poor, because in his opinion, the life of the poor is real, while the life of the rich is so artificial and false.
In the afternoon, he focused on the arithmetic problems, but he could not solve any of them. It was not until 17:00 that Professor Zeng came over. When he saw that Liu Juan had answered a problem, he couldn't help but exclaimed happily: "Genius, what a genius."
genius."
Liu Juan looked at Professor Zeng strangely. Professor Zeng shed tears of joy and said, "Liu Juan, you don't know that this is a problem that many mathematicians in the world cannot solve."
It turns out that the first question of Liu Juan is: Start with any positive integer and repeat the following operations on it: if the number is an even number, divide it by 2; if the number is an odd number, expand it to the original 3
After doubling, add 1. Will the sequence eventually become a cycle of 4, 2, 1, 4, 2, 1, …?
This problem can be said to be a "pit" - at first glance, the problem is very simple and has many breakthroughs, so mathematicians jumped into it one after another. Little did they know that it is easy to get in and difficult to get out, and many mathematicians did not solve this problem until they died.
Come out. Countless mathematicians have been recruited, which can be seen from the various aliases of the 3x + 1 problem: the 3x + 1 problem is also called the tz conjecture, the syracuse problem, the kakutani problem, the hasse algorithm, the ulam problem, etc. Later.
, because the naming is too controversial, let’s just let no one take the credit and just call it 3x + 1. Forget it.
The 3x + 1 problem is not an ordinary difficulty. Here is an example to illustrate how irregular the convergence of the sequence is. Starting from 26, we fall into the "421 trap" in 10 steps:
….
However, starting from 27, the number will soar to more than a few thousand, and you may think that it has escaped the "421 trap" for a time; however, after hundreds of calculation steps, it still fell back:
….
Liu Juan’s second question is:
Longest common subsequence of random 01 strings
If you can get a number sequence b by deleting some numbers from the number sequence a, we say that b is a subsequence of a. For example, 110 is a subsequence of 010010, but not a subsequence of 001011. The "common subsequence" of the two sequences
"There are many, and the longest one is called the "longest common subsequence".
Randomly generate two 01 sequences of length n, where the probability of the number 1 appearing is p and the probability of the number 0 appearing is 1 - p. Use cp(n) to represent the length of their longest common subsequence, and use cp to
Represents the limit value of cp(n) / n.
There is a very clever proof about the existence of cp; however, this proof only illustrates the existence of cp, and it does not bring any useful hints for calculating cp at all.
Even the value of c1/2, no one can successfully calculate it. Michael Steele conjectured that c1/2 = 2/(1 + √2) ≈ 0.828427. Later, v. chvátal and d. sankoff proved..., it seems
Michael Steele's conjecture seems likely to be correct. In 2003, e Lueker proved 0.7880 < c1/2 < 0.8263, overturning Michael Steele's conjecture.
What's worse is that "cp reaches a minimum when p is 1/2" seems to be a very reliable thing, but no one can prove this conclusion.
Liu Juan’s third question is: the inscribed square of a curve
Prove or disprove that on any simple closed curve in the plane, four points can always be found, which can form a square.
Such a seemingly basic problem has not yet been solved! It has been proved on this blog that there are always four points that can form a square on any convex polygon. By improving the proof method, the conclusion can be extended to concave polygons.
At present, there seems to be a certain conclusion about sufficiently smooth curves; but for arbitrary curves, this is still an unsolved problem. There are all kinds of curves on the plane, and it is uncertain whether we can really carefully construct one
Weird curves that don't meet the requirements.
Liu Juan’s fourth question is: the circular runway problem
There is a circular runway with a total length of 1 unit. n people start from the same position on the runway and run clockwise along the runway. The speed of each person is fixed, but the speed of different people is different. Prove or disprove,
For everyone, there will always be a moment when the distance between him and everyone else is greater than 1/n.
At first glance, this problem is no different from other very clever elementary combinatorial mathematics problems, but what is incredible is that this problem has not been solved until now. The best result so far is that when n ≤ 6, the conclusion is true
. Intuitively, for larger n , the conclusion should also hold, but no one has yet proven it.
Liu Juan’s fifth question is: Enhanced version of the sorting problem
There are n boxes, numbered 1, 2, ..., n from left to right. Put two small balls numbered n in the first box, and put two small balls numbered n - 1 in the second box.
Balls, and so on, put two small balls numbered 1 in the n-th box. Each time, you can take a small ball from each of the two adjacent boxes and exchange their positions. In order to put all the small balls
What is the minimum number of exchanges required to put it into the correct box?
In order to illustrate the trap behind this problem, we might as well take the case of n = 5 as an example. First, if there is only one ball in each box, the problem becomes a classic sorting problem: only adjacent elements can be exchanged,
How to turn 5, 4, 3, 2, 1 into 1, 2, 3, 4, 5 as quickly as possible? If a certain number in the front of a sequence is larger than a certain number in the back, we will say that these two
Number is a "reverse-order pair". Obviously, in the initial case, all pairs of numbers are reverse-order pairs. When n = 5, there are 10 reverse-order pairs. Our purpose is to reduce this number to 0. And exchange two adjacent pairs.
The number can only eliminate one reversed pair, so 10 swaps are necessary.
However, there are two balls in each box in the question, so do they have to be exchanged 20 times? Wrong! The following method can miraculously complete the sorting within 15 steps:
….
It may seem unbelievable at first, but if you think about it, you can still figure it out: two numbers can be placed in the same box, which indeed opens up many new possibilities. If a number in the box on the left is larger than a number in the box on the right
If the number in the box is large, we say that these two numbers form a reverse-order pair; but if two different numbers are in the same box, we regard them as half a reverse-order pair. Now let us take a look, a swap
The maximum number of reverse-order pairs that can be eliminated. Assume that a certain exchange step changes ab, cd into ac, bd. The best case is that the reverse-order pair bc is completely eliminated, and at the same time, half of the two reverse-order pairs ac and bd are eliminated, ab,
cd The two reverse pairs (half of which have been eliminated) are also eliminated in half, so a maximum of 3 reverse pairs can be eliminated in one exchange. Since the two identical numbers in each box at the beginning will be separated at some point in the middle
, will be merged together in the end, so we can treat the two identical numbers initially as a reverse-order pair. In this case, every two numbers are initially a reverse-order pair, and c(2n,
2) reverse order pairs. Naturally, we need at least c(2n, 2) / 3 steps to complete the sorting. When n = 5, c(2n, 2) / 3 = 15, which explains the n given above
The sorting scheme with = 5 is optimal.
This analysis is so clever that it is really amazing. Unfortunately, this lower bound is not always achievable. When n = 6, the lower bound obtained by the above analysis is 22 steps, but the computer exhaustively found that there are no 23 steps.
Exchange is not possible. Therefore, this problem has become a tempting pit that has not yet been filled.
Liu Juan’s sixth question is the thrackle conjecture:
If each edge in a graph intersects all other edges exactly once (intersections at vertices are also considered intersections), the graph is called a thrackle. Question, is there a thrackle graph in which the number of edges is greater than the number of vertices?
[The best known result so far is that the number of edges of a thrackle will not exceed twice the number of vertices minus 3.]
Professor Zeng wiped away his tears and said, "Okay, kid, go back and solve these math problems slowly."
Liu Juan didn't know why Professor Zeng was so excited. He packed up his things and said, "Professor Zeng, I'm going back then."
Professor Zeng looked at Liu Juan's retreating figure and couldn't help but said: "We will have another greatest mathematician in China."
