**Ten unresolved mathematical interesting puzzles**

Translator: July November February 26, 2011

References:**Mathematical puzzles**By Peter Winkler.

Book Douban address:Http://book.douban.com/subject/2483910/

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I have explained in my blog before,**Seven major mathematical difficulties in the world****23 problems with Hilbert**Many readers have strongly reflected that, on the one hand, it is difficult to solve or prove these problems. At the same time, they have to sigh, the beauty of mathematics and Its infinite charm.

Unlike classical mathematical problems such as Goldbach Conjecture and Riann hypothesis, some mathematical problems that are not resolved are difficult to solve, but at the same time, they are very interesting.

These problems are weak in mathematics. At first glance, they do not seem to have touched on profound mathematical theories. They seem to be a mathematical quandary that can be instantly killed, it makes mathematics enthusiasts "uncomfortable without a clever solution", but surprisingly, they are as difficult as the famous mathematical conjecture, this is perhaps more difficult than the difficult mathematical difficulties in various fields.

This article, from**Mathematical puzzles,**Book10 interesting mathematical puzzles are excerpted to help readers. At the same time, this article also referred to the blog of matrix67:Http://www.matrix67.com/blog/. Thank you. At the same time, I am not able to understand the latest progress in solving these problems or proving the situation. Therefore, if there are any mistakes, I hope you will criticize and correct them. Thank you.

OK. Let's start by thinking about it as soon as we are free.

**I. Questions about 3x + 1**

Problem description: repeat the following operations from any positive integer:

If the number is an even number, divide it by 2. If the number is an odd number, expand it to three times and then add 1.

Q: Will the sequence eventually become 4, 2, 1, 4, 2, 1 ,... ?

At first glance, the problem was very simple and there were a lot of breakthroughs, so mathematicians jumped into it. However, they did not know how easy it was to go out. Many mathematicians did not solve this problem when they died. Countless mathematicians have been recruited.

3X + 1 is also called Collatz conjecture, Syracuse, Kakutani, HasseAlgorithmAnd Ulam issues. Later, due to the name dispute, it was called the 3X + 1 issue.

An example shows how irregular the series convergence is. For example, starting from 26, the 10 steps fall into the "421 trap"

Trap ":

26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1 ,...

well, I understand this problem. OK. Let's take another example, starting from 27: at this time, the number will soar to several thousand or more, and you may think it is out of the "421 trap" at one time. However, after a hundred steps of operation, it will fall back:

27, 82, 41,124, 62, 31, 94, 47,142, 71,214,107,322,161,484,242,121,

364,182, 91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,

790,395,118 6, 593,178 0, 890,445,133 6, 668,334,167,502,251,754,377, 1132,

566,283,850,425,127 6, 638,319,958,479,143 8, 719,215 8, 1079,323 8, 1619,

4858,242 9, 7288,364 4, 1822,911,273 4, 1367,410 2, 2051,615 4, 3077,923 2, 4616,

2308,115 4, 577,173 2, 866,433,130 0, 650,325,976,488,244,122, 61,184, 92,

46, 23, 70, 35,106, 53,160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1 ,...

**July notes**: In this case, I think you have fully understood the description of the problem, how it is, what ideas, or comments, and leave a message to tell me. We look forward to your contribution. : D.

**Ii. Longest Common subsequence of random 01 string**

Problem description: if you delete some numbers from the number sequence A, you can get the number sequence B. We will say that B is the subsequence of. For example, 110 is a sub-sequence of 010010, but not a sub-sequence of 001011. There are many "common subsequences" of the two sequences, the longest of which is called "The Longest Common subsequence ".

Two 01 sequences with a length of N are randomly generated. The probability of occurrence of Number 1 is P, and the probability of occurrence of number 0 is 1-P. CP (n) is used to represent the length of their longest common subsequences, and CP is used to represent the limit value of CP (N)/n.

There is a clever proof of the existence of CP. However, this proof only shows the existence of CP, and it does not bring any useful tips to the calculation of CP.

No one can calculate the value of C1/2. Michael Steele guessed C1/2 = 2/(1 + √ 2) ≈ 0.828427. Later, V. chvátal and D. sankoff proved 0.773911 <C1/2 <0.837623. It seems that Michael Steele's conjecture is probably correct. In 2003, George lueker proved 0.7880 <C1/2 <0.8263, overthrowing Michael Steele's conjecture.

Even worse, "Reaching the minimum CP when P is 1/2" seems very reliable, but no one can prove this conclusion.

**July notes**: The longest public subsequence is described in this blog. But sinceArticle, Elaborated

It is not especially ideal and thorough. Here, we will not post it to show ugliness.

**Iii. Sorting question enhanced Edition**

Problem description:

There are n boxes numbered 1, 2,… from left to right ,..., N. Put two balls numbered N in the 1st boxes, two balls numbered n-1 in the 2nd boxes, and so on, put two balls numbered 1 in the nth box.

Each time, you can take one ball in two adjacent boxes and exchange their positions. How many exchanges are required to put all the balls in the correct box?

To illustrate the traps behind this problem, let's take 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 can we set 5, 4, 3, 2 as quickly as possible, 1 to 1, 2, 3, 4, 5? If a number in the first part of a Series is larger than the number in the next part, we can say that these two numbers are a "reverse order pair ". Obviously, in the initial situation, all the number pairs are reverse order pairs. If n = 5, there are 10 Reverse Order pairs. Our goal is to reduce the number to 0. The exchange of two adjacent numbers can only eliminate one reverse order, so 10 exchanges are required.

However, if there are two balls in each box in the question, do you have to exchange them 20 times? Error! The following steps can be done in a miraculous manner within 15 steps:

55, 44, 33, 22, 11

54, 54, 33, 22, 11

54, 43, 53, 22, 11

54, 43, 32, 52, 11

54, 43, 32, 21, 51

54, 43, 21, 32, 51

54, 31, 42, 32, 51

41, 53, 42, 32, 51

41, 32, 54, 32, 51

41, 32, 42, 53, 51

41, 32, 42, 31, 55

41, 32, 21, 43, 55

41, 21, 32, 43, 55

11, 42, 32, 43, 55

11, 22, 43, 43, 55

11, 22, 33, 44, 55

It seems incredible for the first time, but I can still think about it:

There are many new possibilities for putting two numbers in the same box. If the number in the box on the left is larger than the number in the box on the right, we can say that these two numbers constitute a reverse order pair. But if two different numbers are in the same box, we will regard them as half Reverse Order pairs.

Now let's take a look at how many reverse pairs can be eliminated at most in one exchange. Assume that the AB, CD is changed to AC and BD in one step of exchange. The best case is that the reverse order of BC is completely eliminated. At the same time, both AC and BD are eliminated in half, both AB and Cd (half already) Reverse Order pairs are also eliminated, so a single exchange can eliminate up to three reverse order pairs. Since the two identical numbers in each box are separated at a certain time in the middle, and then merged, therefore, we can regard two identical numbers as a reverse order.

In this case, each of the two numbers is in reverse order, and C (2n, 2) Reverse Order pairs will be generated in N boxes. Naturally, we need at least C (2n, 2)/3 steps to complete the sorting. When n = 5, C (2n, 2)/3 = 15, this shows that the n = 5 sorting scheme is optimal.

This analysis is so clever that it is a great deal. It is a pity that this lower bound cannot always be reached. When n = 6, the lower bound obtained by the above analysis is 22 steps, but the computer does not find that there is no 23-step exchange. As a result, this problem has become an attractive trap and has not yet been filled in.

**July notes**: Sorting has always been one of the most widely used and difficult problems in the world. Currently,

This blog will explain the basic eight sorting algorithms step by step. read this article.

**Iv. Difficulties in annular Runway**

Problem description:

There is a circular runway with a total length of 1 unit. N people start from the same position on the runway and keep running clockwise along the runway. The speed of each person is fixed, but the speed of different people is different. Prove or overturn. For every person, there will always be a time when the distance between him and everyone else is greater than 1/N.

At first glance, it seems like a variety of other very clever elementary mathematics problems, but it is incredible that this problem has not been solved until now. At present, the best result is that when n ≤ 6, the conclusion is true. Intuitively, for a larger N, the conclusion should also be established, but no one has proved it.

**5. Angels and demons**

Problem description:

Angels and demons play games on an infinitely large chessboard. Each time, the devil can dig out any grid on the board, and the angel can fly 1000 steps on the board before landing; if the angel falls on a grid that is dug out, then the angel lost.

Q: Can a demon trap an angel (dig a circle around the angel and make a hole of 1000 thickness )?

This is a classic puzzle of John Conway. As a "normal" combination game, the problem of angels and demons has never been solved. The conclusion is that if an angel can only move one step at a time, the devil will surely win. However, as long as the angel can fly two steps each time, it seems that it is already invincible. Of course, the devil's advantage is not small-it does not have to worry about "going wrong". It is advantageous to dig more traps.

After that, Conway himself seems to still believe that the angel can win-He offered a reward of 1000 US dollars to seek proof that the devil will win, but only offered a reward of 100 US dollars to seek proof that the angel will win.

**July notes**: I heard that this problem has been solved. The solution is: n≥2, angel. For details, see here:Http://home.broadpark.no /~ Oddvark/angel/.

**6. thrackle Conjecture**

Problem description:

For example, each edge is exactly the same as all other edges (the vertex is also considered as the intersection). This graph is called a thrackle. Q: Is there a thrackle graph with the number of edges greater than the number of vertices?

John Conway proposed this conjecture. This is obviously a pitfall. If anyone wants to try this problem, they will crash. Conway offered a reward of 1000 US dollars to seek solutions, which shows how difficult this problem is. Currently, the best result is that the number of edges of a thrackle cannot exceed twice the number of vertices by 3.

**7. traverse all "intermediate subsets"**

Problem description:

Proof or overturn. You can add or delete an element each time to traverse the {1, 2 ,..., All 2n + 1} values are subsets of N or N + 1.

After reading the line above, I can imagine that you already have a kind of restrained impulse, pick up a pencil, draft paper and computer, and start to look for the law of N when it is not big. This can be said to be the biggest pitfall of all the questions in this article-this is very tempting. When everyone sees this question for the first time, they will think that there is a constructor suitable for all N, as a result, everyone jumped one by one into the pit, and they couldn't stop it.

No one thinks this conjecture is wrong. The simple computer enumeration shows that with the increase of N, the number of schemes for Traversing these subsets not only increases, but also increases very quickly. To a certain N, the number of solutions suddenly fell to 0, which is obviously very impossible. However, after decades, no one can prove it!

**8. More than half of the elements appear**

Problem description:

Make u a finite set, S1, S2 ,... And Sn are all non-empty subsets of U. They satisfy any number of sets and are still in these sets. It is proved that an element can be found in at least half of the set.

Incredible, even the most basic and discrete mathematical research object, the finite set, has a broken-down unsolved problem.

In 1999, Piotr Wojcik proved with a clever method that an element exists in a set of at least N/log2n. However, this is still a long way from the target.

**9. Curve inner Square**

Problem description:

To prove or overturn any simple closed curve in the plane, we can always find four points, which can exactly Form

Square.

There are always four points on any convex polygon that can form a square. By improving the proof method, we can extend the conclusion to the concave polygon. At present, there seems to be a positive conclusion for a fully smooth curve; but for any curve, this is still an unsolved problem. The curves on the plane are no surprise. We cannot say that we can really carefully construct a strange curve that does not meet the requirements.

**10. Two interesting geometric problems:**

1. Venn Diagram

Problem description:

As shown in, the Venn diagram of the Three sets is used. Many people think that the four circles are the Venn diagram of the four sets in the same shape as a flower. In fact, this is not correct-four circles can only generate 14 regions, and four sets will generate 16 cases. If four circles are arranged like the middle graph, two areas are missing: the circle in the lower left corner and the circle in the upper right corner, and only the circle in the upper left corner and the circle in the lower right corner.

So, isn't the Venn diagram of the four sets impossible to draw? No. If you are not a perfectionist, you can extend the Venn diagram of the three sets to four sets like the picture on the right. Although it looks pretty bad, it looks like a topology, as long as the logic is correct, who cares about the circle.

Naturally, you will come up with a question: can the diagram on the right be expanded into a Venn diagram of five sets? More generally, can any Venn map of N sets be extended to n + 1 sets?

It is incredible that this problem has not been solved yet. In fact, the research on the Venn graph that meets various conditions is a persistent topic. The conjecture related to the Venn graph is more than this.

**2. polygon formed by a plane mirror**

Problem description:

Prove or overturn. If any of the interior walls is a mirror polygon, you can always find a point in it, so that the light source at this point can illuminate the interior of the polygon.

This is a very creative issue, but it is a pity that the earliest source of the problem is no longer necessary. Interesting question: the polygon condition is required: If curves are allowed, we can create a plane image composed of mirrors (seeLeft.

For polygon, tokarsky gave a 26-side room (Right), Place the light source on one of the points, it will not be able to shine to another point (assuming that the vertex does not reflect the light ). Therefore, there is only one problem: there are no polygon, and the light source at any position cannot illuminate the entire image.?**(Ten unresolved mathematical interesting puzzles have been completed ).**

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**Finally, I will share a simple, but equally interesting geometric proof question.**:

For example, to create an arbitrary Triangle ABC's incircle ⊙ I, it is tangent to AC at point N. Over N for the vertical line of AC, and another point of intersection with ⊙ I for M (So Mn is a diameter of this circle ). Connect and extend BM, and submit it to point L. Proof: Cn = Al.

The following proves that the method is very good: the M point is used as the tangent of ⊙ I, and the AB and BC are handed over to vertex E and F respectively. Therefore, EF is parallel to AC. Take B as the center and enlarge △bef to △bac, then M points will overlap with L points, and ⊙ I (as the subcircle of △bef) then it will become the subcircle of △bac ⊙ I '.

The following two theorems related to the tangent length are used:

(1) equal length of two outer public tangent of two circles

(2) The two tangent lengths from a point outside the circle to the circle

From (1) We can see that xy = ZW, that is, ax + AY = cz + CW. (2) It can be seen that ax, ay, CZ, and CW are equal to an, Al, CN, and Cl respectively. So there is an + Al = CN + Cl. If there is 2 * Al = 2 * CN, the conclusion is displayed.

OK. How is it interesting? Welcome and provide comments. This article is complete.

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