“Ball Belt Conjecture”and Zhang's Proof

zhangyd2007@soh

                                                             “Ball Belt Conjecture”and Zhang's Proof
                                                                           “球带猜想”以及张式证法
                                                                                    张彧典         张利翀
                                                                             Zhang yudian      zhang lichong

摘要：作者把宽带覆盖球体表面的问题转化为覆盖过球体球心横截面-----圆的问题，并且运用映射理论把拉斯洛.菲杰斯.托特与姜子麟等人的宽带“交叉覆盖”转换成宽带“平行覆盖”，直观地证明了本文给出的两个不同含义的球带猜想的研究结果。
Abstract : The author transforms the problem of wide-band covering the sphere surface into the problem of covering the cross section of the sphere center-circle, and transforms the wide-band "cross-coverage" of Laszloffy Jestot and Jiang Zilin into wide-band "parallel coverage" by using mapping theory, which intuitively proves the research results of two different meanings of the spherical belt conjecture given in this        paper.
关键词：球带猜想，宽带，宽度，交叉覆盖，平行覆盖
Key words : spherical belt conjecture, broadband, width, cross coverage, parallel coverage

                                                                                一、球带猜想
                                                                                 一、 the ball belt conjecture
球带猜想：假如给你一些彩带和一个地球仪，你发现每条彩带可以围住地球仪的部分区域；如果用它们将整个地球仪包裹，这些彩带的宽度加起来至少是多少？
Ribbon conjecture: if you are given some ribbons and a globe, you will find that each ribbon can surround part of the globe; If you wrap the globe with them, what is the total width of these ribbons at least?
1973年，匈牙利数学家László Fejes Tóth（拉斯洛•菲杰斯•托特）指出：半径为1的单位球体被等宽的区域覆盖，所有区域的宽度总和的最小值是π。
In 1973, the Hungarian mathematician László Fejes Tóth (Jestot, Laszloffy) pointed out that a unit sphere with a radius of 1 is covered by areas of equal width, and the minimum value of the sum of the widths of all areas is π.
2017年的一个下午，姜子麟同俄罗斯同事亚历山大 • 波利扬斯基（Alexandr Polyanskii）在闲聊过后，用半天时间破解了这个长达44年的数学难题。
One afternoon in 2017, Jiang Zilin and his Russian colleague Alexandr Polyanskii solved this 44-year-old math problem in half a day after chatting.
至少是地球仪上赤道长度的一半——这是姜子麟给出的答案，也是他就球带猜想给出的通俗版解释。
At least half of the equator length on the globe-this is Jiang Zilin's answer, and it is also his popular explanation of the belt conjecture.


                          

                                                 二、        两个已知的研究成果
                                                 二、 two known research results

                                                                   1、        Tarski的研究成果
                                                                  1.        Tarski's research results
László Fejes Tóth 的区域猜想与离散几何中的一些其他问题也密切相关，这些问题已在20世纪就被解决，涉及到用条带覆盖表面。其中第一个就是所谓的木板问题（Plank Problem），涉及到用平行线组成的条带覆盖住圆盘。Alfred Tarski 和 Henryk Moese 用一个简洁的方式证明了用来覆盖圆面的条带（或木板）的宽度和至少等于圆的直径。也就是说，没有比用一个宽度与圆的直径相等的木板更好的方法用来覆盖圆盘。接着，Thøger Bang 解决了用长条覆盖任意凸体的问题。也就是说，他证明了覆盖凸体的条带的总宽度至少是凸体本身的宽度，即单个能覆盖凸体的单个条带的最小宽度。即  
László Fejes Tóth's regional conjecture is also closely related to some other problems in discrete geometry, which have been solved in the 20th century, involving covering the surface with strips. The first one is the so-called Plank Problem, which involves covering the disk with a strip composed of parallel lines. Alfred Tarski and Henryk Moese proved in a simple way that the width of the strip (or wooden board) used to cover the round surface is at least equal to the diameter of the circle. That is to say, there is no better way to cover the disc than to use a wooden board whose width is equal to the diameter of the circle. Then, th gerbang solved the problem of covering arbitrary convex body with long strips. That is, he proved that the total width of the strip covering the convex body is at least the width of the convex body itself, that is, the minimum width of a single strip that can cover the convex body. That is,
Tarski证明了：一个半径为1的单位圆不能被宽度和小于2（即圆的直径）的条带完全覆盖。
Tarski proved that a unit circle with a radius of 1 cannot be completely covered by a strip with a width less than 2 (that is, the diameter of the circle).

2、姜子麟和 Alexandr Polyanskii的研究成果
2. Research results of Jiang Zilin and Alexandr Polyanskii

姜子麟和 Alexandr Polyanskii 处理的问题有些不同，它涉及到用某类球带来覆盖一个单位球面。具体而言，每个球带都是球面与一个特定的三维板条的交集，其中板条是关于球心对称的夹在两个平行平面之间的空间区域。或者可以不用板条，而在球面测地线的度量空间里定义球带：一个在单位球表面的宽度为 ω 的球带，是距离大圆（球面上半径等于球体半径的圆弧）不超过 ±ω/2 的点的集合，测量点与点间距离的是连接它们的最短弧。数学家必须找到能覆盖单位球面的球带上的最小宽度和。因此，宽度测量方法不同于之前被解决的问题：它被定义为弧的长度，而不是平行线或面之间的欧几里德距离。
                             
Jiang Zilin and Alexandr Polyanskii deal with some different problems, which involve covering a unit sphere with a certain kind of ball belt. Specifically, each spherical belt is the intersection of a spherical surface and a specific three-dimensional slat, in which the slat is a spatial region sandwiched between two parallel planes symmetrically about the spherical center. Or you can define a spherical belt in the metric space of spherical geodesic instead of slats: a spherical belt with a width of ω on the surface of a unit sphere is a collection of points with a distance of no more than ω/2 from a great circle (an arc with a radius equal to the radius of the sphere), and the distance between points is measured by the shortest arc connecting them. Mathematicians must find the sum of the minimum widths on the belt that can cover the unit sphere. Therefore, the width measurement method is different from the previously solved problem: it is defined as the length of arc, not the Euclidean distance between parallel lines or faces.
姜子麟和 Polyanskii 所作出的证明是受到了 Bang 的启发，Bang 通过构造一个有限点集解决了用条带覆盖凸体的问题，该有限点集必有一个点不被任何条带覆盖。从某种意义上来说，无论是 Bang 还是姜子麟和 Polyanskii 都是通过反证法来证明的。在 Fejes Tóth 猜想的情况下，数学家假设完全覆盖球体的球带的宽度和小于 π，并试图得到矛盾点——即找到一个位于球体上的点，但又不在任何这些球带里。   
The proof made by Jiang Zilin and Polyanskii was inspired by Bang. Bang solved the problem of covering convex body with strips by constructing a finite point set, which must have a point not covered by any strips. In a sense, both Bang, Jiang Zilin and Polyanskii are proved by the method of disproof. In the case of Fejes Tóth conjecture, mathematicians assume that the sum of the widths of the spheres completely covering the sphere is less than π, and try to get the contradictory point, that is, to find a point on the sphere but not in any of these spheres.                             
姜子麟和 Polyanskii 在三维空间中构造了一组特别的点集，使得至少一个点不在木板覆盖的区域内。如果这整个点集都位于球内部，那么在球面上找另一个没被木板覆盖、也就是没被球带覆盖的点是相对容易的事。如果集合中的任何点碰巧位于球体之外，则可以用一个较大的球带代替几个较小的球带，其宽度和与较大球带的宽度相等。因此，我们可以做到在不影响宽度和的前提下，减少初始问题中球带的数量。最终，球体上的某个点会被确定为不在这些球带内。这与球带总宽度小于 π 的假设背道而驰，因此证明了 Fejes Tóth 的猜想。
Jiang Zilin and Polyanskii constructed a special set of points in three-dimensional space, so that at least one point was not in the area covered by wooden boards. If the whole point set is located inside the sphere, it is relatively easy to find another point on the sphere that is not covered by the board, that is, not covered by the ball belt. If any point in the set happens to be outside the sphere, several smaller bands can be replaced by one larger band, whose width sum is equal to that of the larger band. Therefore, we can reduce the number of balls in the initial problem without affecting the width sum. Eventually, a point on the sphere will be determined to be outside these spheres. This runs counter to the assumption that the total width of the spherical belt is less than π, thus proving Fejes Tóth's conjecture.
                                                                             三、        张式证明
                                                                             三、 the proof of Zhang
           
张彧典等人首先认为：在三维圆球的表面积公式S=4πR2中 ，包含通过球心截面----二维圆的面积πR2，所以，完全可以把三维的“球带猜想”演变成二维的“宽带覆盖圆猜想”。正如把立体几何问题演变成平面几何问题一样简单了   。                           
Zhang yudian and others first thought that the surface area formula of a three-dimensional sphere, S = 4π R, includes the cross section through the center of the sphere-the area π R of a two-dimensional circle, so it is possible to turn the three-dimensional "ball belt conjecture" into a two-dimensional "broadband covering circle conjecture". It is as simple as turning a solid geometry problem into a plane geometry problem.
证明：如图1所示，
张彧典等人接着，把姜子麟和 Polyanskii证明中的思路做了改变：
（1）、把5条等宽宽带的个数变成无限n条；
（2）、把n条宽带的宽度各不相同化；
（3）、把宽带的交叉覆盖垂直映射成平行覆盖。

Zhang yudian and others then changed the ideas in Jiang Zilin's and Polyanskii's proofs: (1) changing the number of five equal-width broadband into infinite n; (2) Differentiate the widths of N broadband; (3) Map the cross coverage of broadband vertically into parallel coverage. Proof: as shown in fig. 1,

                  

在单位圆O中，设A1， A2 ，A3 ，… ，An ，An+1 为半圆弧上的n+1个分割点，
A1-An+1 为单位圆的直径，
弦A1 A2 ,A2 A3 ，… , An An+1表示n条宽带的宽度，Wi(i=1,2,3,…,n,n+i)表示n条宽带覆盖了的单位圆的面积 。
  In the unit circle o, let A1, A2, A3, ..., An, An+1 be n+1 dividing points on the semi-circular arc. A1-An+1 is the diameter of the unit circle, Chords A1 A2 ,A2 A3, ..., an an+1 represent the width of n broadband, and wi (I = 1, 2, 3, ..., n, n+I) represents the area of the unit circle covered by n broadband.   
                                                           
再设A1 A2’ ，  A2’A3’  ，… ，An’An+1为n条宽带宽度在直径上的垂直射影，那么，有
A1 A2 ≥ A1 A2’ ，…，   A2 A3  ≥ A2’A3’ ，An An+1 ≥An’An+1 .
Let A1 A2', A2'A3', …, An'An+1 be the vertical projection of n broadband widths in diameter, then, there are
A1 A2 ≥ A1 A2’ ，…， A2 A3 ≥ A2’A3’ ，An An+1 ≥An’An+1 .
这样一来，以A1 A2为宽度的宽带（平行虚线）W1 与 以A2 A3为宽度的宽带（平行点）W2的交叉覆盖圆，就会变成以A1A2’为宽度的宽带W1’ 与 以A2’A3’为宽度的宽带W2’的平行覆盖圆，…，直到以An An+1为宽度的宽带（平行小横线）Wn与前n-1条宽带的交叉覆盖圆，就变成以An’An+1为宽度的宽带Wn’ 的平行覆盖圆。
In this way, the cross coverage circle of broadband (parallel dashed line) W1 with width A1 A2 and broadband (parallel point) W2 with width A2 A3 will become the parallel coverage circle of broadband W1' with width A1A2' and broadband W2' with width A2'A3', …, until the cross coverage of broadband (parallel small horizontal line) Wn with width AN+1 and the first n-1 broadband.
这时，n条宽度不尽相同却平行的宽带完成了单位圆O面积的全覆盖。因此有：
At this time, n parallel broadband with different widths complete the full coverage of unit circle O area. Therefore, there are:                             
如果计算n条宽带覆盖单位圆的总长度的话，那就是Tarski证明了的：
“一个半径为1的单位圆不能被宽度和小于2（即圆的直径）的条带完全覆盖”的定理成立：即
A1 A2+A2 A3 +… + An An+1  ≥ A1 A2’+A2’A3’ + ,.. + An’An+1=2R
If you calculate the total length of n broadband covering unit circles, it is proved by Tarski: The theorem that "a unit circle with a radius of 1 cannot be completely covered by a strip with a width less than 2 (that is, the diameter of the circle)" holds: that is,
A1 A2+A2 A3 +… + An An+1 ≥ A1 A2’+A2’A3’ + ,.. + An’An+1=2R
如果计算n条宽带覆盖单位圆的面积的话，那就是姜子麟两人证明了的：
W1+W2+…+Wn +Wn+1≥ W1’ + W2’  + ,.. + Wn’ =πR2 =πx12=π。  
If you calculate the area of the unit circle covered by N broadband, it is proved by Jiang Zilin and his wife:
W1+W2+…+Wn +Wn+1≥ W1’ + W2’ + ,.. + Wn’ =πR2 =πx12=π。                        
以上两个结论中，当且仅当：整个圆周只被直径上的两个端点A1与An+1分割，即宽度恰好为直径A1An+1的一条宽带覆盖时，取等号。
                           
In the above two conclusions, if and only if the whole circumference is only divided by two endpoints A1 and An+1 in diameter, that is, a wide band whose width is just the diameter A1An+1 is covered, take the equal sign.
如果有人怀疑以上降级证明的话，那么，可以把图1中的直径A1An+1看作地球赤道的半圆弧，把直径A1An+1对应的半圆弧看作垂直于赤道的最大截面圆的半圆弧，分割点A1，A2…， An分布其中，由此产生的n条平面宽带就可以看作圆弧状宽带了。这样的变换，就可以把降级证明还原为原来的三维“球带猜想”证明了。
If someone doubts the above degradation proof, then the diameter A1An+1 in Figure 1 can be regarded as the semi-circular arc of the earth equator, and the semi-circular arc corresponding to the diameter A1An+1 can be regarded as the semi-circular arc of the maximum cross-section circle perpendicular to the equator, in which the dividing points A1, A2, ..., An are distributed, and the resulting N planar broadband can be regarded as circular-arc broadband. With this transformation, the degradation proof can be turned into a three-dimensional "ball-belt conjecture" proof.
                        

    作者简介：
张彧典，男，1943年9月生，山西省盂县人，大专学历，汉族，退休前在盂县县委党校从事教学工作（县城秀水桥南，邮编045100），数学高级讲师，一生从事数学教学与研究，研究方向：1980年开始研究世界数学名题“四色猜想”的人工证明。   
    联系方法：   手机18335385319      email:1633409368@qq.com
【参考文献】
44年前的一个数学猜想终被破解　

图1