Online game I am Sun Wukong

Liu Juan said, "What do you want to do with our affairs?"

Xiaohong said: "I will take care of you treating our little water flower like this. Little water flower is ours, our baby, and our princess. I have never seen her treat anyone like this for so many years, and I don't know you 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 Shui Hua, and suddenly said, "Little Shui Hua, let's calm down, I—"

Xiao Shuihua said: "I know what you mean, well, we are put together now, we——."

Liu Juan said: "Xiao Shuihua, I will definitely save you, but we should think about it. Yes, you are beautiful, but beauty does not necessarily mean love. You understand what I mean. I——."

Xiao Shuihua said: "Okay, don't talk about it, I know what you mean, now we are going through the difficulties together, so I won't talk about other things."

Liu Juan said: "That's not what I mean, why don't you understand, I, I, I... . . . "

Xiao Shuihua said: "I understand what you mean, do you hate our life like that, and 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, and he had nothing to say. He watched Xiao Shuihua's back slowly walk away, but he still didn't know what to do.

The poor have their lives, and the rich have their lives, but Liu Juan likes the lives of the poor, because in his opinion, the lives of the poor are real, while the lives of the rich are so contrived and false.

In the afternoon, he put all his energy into those math problems, but he didn't solve any of them. When Professor Zeng came over at 17:00, he saw that Liu Juan had actually accepted a problem, and he couldn't help shouting happily: "Genius, what a genius!" genius."

Liu Juan looked at Professor Zeng strangely, and 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 can't solve."

It turns out that Liu Juan's first question is: Starting from any positive integer, repeat the following operations: if the number is even, divide it by 2; if the number is odd, expand it to the original 3 Add 1 after doubling.Does the sequence always end up being 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 there are many breakthroughs, so mathematicians jumped into it one after another; as everyone knows, it is easy to get in and difficult to get out, and many mathematicians have not solved this problem until their death. come out.Countless mathematicians have been recruited, which can be seen from the various aliases of the 3x + 1 problem: 3x + 1 problem is also called collatz conjecture, syracuse problem, kakutani problem, hasse algorithm, ulam problem and so on.Later, because the naming controversy was too great, it was simply let no one get involved, and it was directly called the 3x + 1 problem.

3x + 1 problems are not of ordinary difficulty.Here is an example to illustrate how irregular the sequence of numbers converges.Counting from 26, 10 steps fell into the "421 trap":

......

However, starting from 27, the number will soar all the way to more than a few thousand, and you may think that it has escaped the "421 trap" for a while; however, after hundreds of calculations, it still falls back:

......

Liu Juan's second question is:

Longest common subsequence of random 01 string

A sequence of numbers b is said to be a subsequence of a if some numbers are removed from a sequence of numbers a.For example, 110 is a subsequence of 010010, but not 001011.There are many "common subsequences" of the two sequences, 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 represent the limit value of cp(n)/n.

There is a very neat proof of the existence of cp; however, this proof only shows the existence of cp, it does not bring any useful hints for computing cp at all.

Even the value of c1/2 has not been successfully calculated by anyone. Michael Steele guessed that c1/2 = 2/(1 + √2) ≈ 0.Later, v. chvátal and d. sankoff proved... and it seems likely that michael steele's conjecture was right. In 828427, george lueker proved 2003. 0 lt; c7880/1 lt; 2. 0, which overturned michael steele's conjecture.

To make matters worse, "cp is minimized 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 the curve

Prove or disprove that on any simple closed curve in the plane, four points can always be found, and they can just form a square.

Such a seemingly basic problem has not been solved!It has been proved on this blog that there are always four points that can form a square on any convex polygon; the proof method can be improved to extend the conclusion to concave polygons.At present, for sufficiently smooth curves, there seems to be a firm conclusion; but for arbitrary curves, this is still an open problem.There are all kinds of curves on the plane, maybe we can really carefully construct a weird curve that does not meet the requirements.

Liu Juan's fourth problem 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.Everyone's speed is fixed, but different people's speed is different.Prove or disprove that for each person there is a moment when he is more than 1/n away from everyone else.

At first glance, this problem is no different from other very clever elementary combinatorial mathematics problems, but it is incredible that this problem has not been solved until now.The best result so far is that the conclusion holds when n ≤ 6.Intuitively, the conclusion should also hold for larger n, but no one has proved it yet.

Liu Juan's fifth question is: the enhanced version of the sorting problem

There are n boxes, numbered 1, 2, ..., n from left to right.Put two balls numbered n in the first box, put two balls numbered n - 1 in the second box, and so on, put two balls numbered 2 in the nth box .Each time, you can take a small ball in each of the two adjacent boxes and exchange their positions.What is the minimum number of swaps required to get all the balls into the correct boxes?

In order to illustrate the trap behind this problem, we might as well take the case of n = 5 as an example.First of all, if there is only one ball in each box, the problem becomes a classic sorting problem: only adjacent elements can be exchanged, how to change 5, 4, 3, 2, 1 into 1, 2, 3 as quickly as possible , 4, 5 ?If a number in the front of a sequence is greater than a number in the back, we say that the two numbers are a "reversed pair".Obviously, in the initial case, all pairs are reversed pairs, and when n = 5, there are 10 reversed pairs.Our goal is to reduce this number to zero.And exchanging two adjacent numbers can only eliminate one reversed pair, so 0 exchanges are necessary.

However, there are two balls in each box in the question, so does it have to be exchanged 20 times?wrong!The following approach miraculously completes the sort in 15 steps:

......

It may seem unbelievable for the first time, but you can still figure it out after thinking about it: being able to put two numbers in the same box does have a lot of new possibilities.If a number in a box on the left is greater than a number in a box on the right, we say that the two numbers form a reverse pair; but if two different numbers are in the same box, we treat them as half pairs in reverse order.Now let's see how many reversed pairs can be eliminated by one exchange.Assume that ab and cd are changed into ac and bd in a certain step. In the best case, the reverse pair of bc is completely eliminated, and half of the two reverse pairs of ac and bd are eliminated, and the two reverse pairs of ab and cd (half eliminated ) reversed pairs are also eliminated in half, so a swap can eliminate up to 3 reversed pairs.Since the two identical numbers in each box at the beginning will be separated at some point in the middle, and finally merged together, we can treat the initial two identical numbers as a reverse pair.In this case, every two numbers are reversed pairs at the beginning, and c(2n, 2) reversed pairs will be generated in n boxes.Naturally, we need at least c(2n, 2) / 3 steps to complete the sort.When n = 5, c(2n, 2) / 3 = 15, which shows that the sorting scheme for n = 5 given above is optimal.

This analysis is so ingenious, it is really amazing.It's just a pity that this lower bound is not always achievable.When n = 6, the lower bound obtained from the above analysis is 22 steps, but the computer exhaustively finds that the exchange without 23 steps is not feasible.As a result, this problem has become a tempting pit, which has not been filled up to now.

Liu Juan's sixth question is the thrackle conjecture:

A graph is said to be a throttle if every edge intersects every other edge exactly once (connecting vertices counts as intersection).Ask, is there a thrackle graph with more edges than vertices?

[The best result known 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, you can 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, then I'm going back."

Professor Zeng looked at Liu Juan's receding back, and couldn't help saying: "We will have one of the greatest mathematicians in China."