谴责申明：世界十大数学难题的解答,圆球体层式的解答解读

1、标题：世界十大数学难题的解答.圆球体层式解答 .作者 : 百度里的昵称“蔡於竟道”Apology statement.Condemn stated:relatively speaking, this layer ball type theory.Is true, and I in baidus nickname, CAIin registration in May 2014, at the same time.Can other will not.B: yes.Other, opposite in other forms.0, 7 years, told the diameter of the

2、 sphere, and how the score, N, anything better to do one thing at the time, including the universe at the time, and so on.Principle of outer principle, principle of the inner principle.About is like a thing well done.The condemned man: baidus nickname, CAI YU JING DAO .道歉申明。谴责申明：相对来说，此圆球体层式理论。真正的，是和

3、本人在百度的昵称“蔡於竟道 ”，注册时 2014 年 5 月，同时发表的。其他的可以都将不算。是的。其他的，相对是以其他形式讲过。零七年时，讲过的0，圆球体的直径，怎样的参比数，N，任何当时不如做一件事，包括宇宙当时，等。原理外层有原理，原理里层有原理。讲了就是当时如同做好了一件事。谴责人：百度里的昵称，“蔡於竟道 ”.浙江省台州市。2015 年.10 月.3 日 . （机械翻译，不知道对吧？）摘要：此圆球体层式理论，能解答此世界十大数学难题。The Perfect Sphere Layer expression can solve the following ten difficult ma

4、thematic problems in the world.P=NP. 原理同源 .（也叫原里同源.）（P = NP.The principle of homology.The origin is the origin.）几何法论证中。 若，理论证回到这个原点， 实际证回到这个原点。 （是的。不能说理论和实际都各为一个平面。那只是几何法中时的一个演示的一道题时。当然也不能说成多大圆球体。）因为论证最后，理论和实际都证回到原点，则相等。那么，原理为等值原理（原理同源 .）。引言：零六年在网络上，以其它形式讲过。零九年曾经发表过。有的他们也认得。时在百度里，去年五六月份在老师微信群的帮助下，做好

5、的这十道题。其实它就像一个讲话的形式：逻辑。相对来说正确而正确。能通过任何考验而相对正确而正确。And in repeated argumentation, this sphere layer expression can demons trate any existing questions (including the location of the logic).The principle seems to begin with the origin and centers on the origin. If it is a plane in theory, and it could a

6、ctually be a plane as well,（ ） then an equation is achieved.Just like the argument in the fourth layer, it is the basic mathematical geometry method, and is very important. If the original strength at that moment coul d really keep up, that is the real original strength and it is called: This is the

7、 r eal original strength.As for the sphere theory, I once demonstrated it in Peking University sspace forum in 2006, also proved that “Zero”and many other relevant questions, and published it in 2009. It wasrecognized in some places. During May/Jun, 2014, under the assistance of wechat group of the

8、teachers office, I accomplished these ten questions on Baidu sspace.“ Cai Yu Jing Dao ”。This sphere expression is like a way of talking, the logic is relatively correct so it s correct. It can pass any tests and is relatively correct, so itscorrect.世界十大数学难题的题目：难题”之一：P （多项式算法）问题对NP （非多项式算法）问题 .难题”之二：

9、霍奇 (Hodge) 猜想 .难题”之三：庞加莱 (Poincare) 猜想 .难题”之四：黎曼 (Riemann) 假设 .难题”之五：杨米尔斯 (Yang-Mills)存在性和质量缺口 .难题”之六：纳维叶斯托克斯 (Navier-Stokes) 方程的存在性与光滑性 .难题”之七：贝赫 (Birch) 和斯维讷通戴尔 (Swinnerton-Dyer) 猜想 .难题 之八：几何尺规作图问题 .难题”之九：哥德巴赫猜想 .难题之十四色猜想 .The Perfect Sphere Layer expression can solve the following ten difficultma

10、thematic problems in the world.Problem 1: P (Polynomial Algorithm) Problem vs. NP (Non Polynomial Algorithm) ProblemProblem 2: Hodge ConjectureProblem 3: Poincare ConjectureProblem 4: Riemann HypothesisProblem 5: YangMills Existence and Mass gapProblem 6: The Existence and Smoothness of Navier - Sto

11、kes EquationsProblem 7: Birch and Swinnerton-Dyer ConjectureProblem 8: Geometric Ruler Gauge Construction ProblemProblem 9: Goldbach ConjectureProblem 10: Four Color Conjecture附加：此球体层式，能解决此世界十道数学难题。先讲：物，物与数算起 （正圆球体层式），（正圆球体层式，在此以后就叫： 此圆球体或圆球体. ）原里，原来的里面。宇宙万物长河中的任何什么，而物（相对来说，也就是任何什么，而叫物。而在中文中，就能直接用上一

12、个名词，“而物”。）。之，当时什么物。之，当时什么。This sphere layer expression can solve the world sten difficult problems in mathematics. Firstly, we talk about the objects, which can be counted by the numbers (Perfect Sphere LayerExpression). The Perfect Sphere Layer Expression, is hereinafter referred to as this Sphere

13、or the Sphere. The original interior is the original interior part.Anythi ng in the universe is the object at a certain time and being under certain circumstances. Erwu, relatively speaking, is Any Thing and being called the object. In Chinese, we use Erwu, a noun to describe it.零“ 0”，相对来说没有动用任何，而任何

14、着。（此句的翻译，应是这样理解。就像力学，相对来说是平恒力学。似左边等于右边。为“ 0”时，则也构成了平恒时。啥都没有反应！）（）似任何没有动用，没有反应！一旦有什么了。（时，先参照物一样，参比一个数，或自然数 N 中。）的似，（翻译理解：似当时怎么样了。就似，就像。叫做“的似”。），动用了任何于任何 （此句， 翻译理解成： 任何为了任何。 “似电影片名中：勇敢的心。中， 说：为了自由！”），任何本质于任何本质，而怎么了着的。（翻译理解成：怎么样了着的。相对来说开始反应了一样。开始忙碌着。）。的似，动用了，任何于任何，任何本质于任何本质的怎式，而怎么了着的。 （翻译。这里时再阐明，也就是他还在

15、那讲那方式。又一次地阐明着。）。而似（“而似”两字。此时可翻译理解成：就像似.，什么什么 .。）任何于任何，任何本质于任何本质的等式，而怎么了着。（翻译此句理解。又，重新阐明了一下。相对来说讲完了。）。Zero 0, does not occupy anything relatively, and is being anything.（此处原来中文为不用翻译的红字，现在按绿字加上，其中“平恒”疑为“平衡”，按此翻译，如有其它意思，请说明）Just like mechanics, it is relatively a balance mechanics. It appears that the

16、left side equals the right side. When each side is 0, they also constitute a balance. There is not any response!It appears that it does not occupy anything, there s no response. At that time, we firstly referred to a number, or natural number N, just like a reference object. Once it does havesomethi

17、ng, just like using anything to anything, or any nature to any nature, it might begin to respond and do something. It appears that it uses anything to anything, any nature to any nature, so its what it is. It appears that it is an equation of anything to anything, any nature to nature, soit swhat it

18、 is.在这时讲一下平衡。都知道数算中不是等式，就像力学中不平衡一样，没法做事。之，（时“之”字，时总结了一下。翻译可理解成：则，怎么样了.。）相对来说，怎么样的等式被动用了，而怎么了着的。（翻译理解时。这里，先好比一个讲好了，总结一下。下面还有。）We will talk about the balance now. It is well known that nothing could be achieved without equations in calculations, just like the imbalance in mechanics. So, relatively spe

19、aking, when equations are used, then things are achieved.时，而怎么了着的。（先用自然数中（N ）参比用之，直接用上了！）自然数N 中，第一个参数为“ 1”（翻译理解：“参数”，参照物一样，而叫它“参数”。）。时又怎么了，时又似任何于任何，任何本质于任何本质的怎样的等式，而怎么了。时，第二个为 “ 2”。接着这样， 又怎么了， 为第三个， 为“ 3”。4，5，6，7，8，9，.。N 中。时，动用的都是怎样的等式。到那圆球体层式时，每一层的原来的量，依旧都是相等的。而每一层只是怎么的等式来回而已。What was it like at th

20、at time?When natural number N was used as a reference, it was in directuse. Within the natural number N, the first parameter is 1.What was it like again at that time? It was like an equation of anything to anything, any nature to any nature, and it is what it is.At that time, the second was the numb

21、er “ 2” , then what, the third one is the number 3, and then 4,5,6,7,8,9 respectively among N.At that time, what kinds of equations were used? When it came to the sphere layers, the original amount of each layer was still equal. It is just an issue of certain equations for each layer.(此正圆球体层式似的构成：)零

22、“ 0”，起数算起。（翻译理解为：开始数学中的演算了。）的似（翻译理解：就似，就像 .。）从任何的角度方位来一样。任何角度方位，那就是一空间方位角度。而先在平面上，“ 0”开始起动了任何于任何，任何本质于任何本质的等比等式，而怎么了的。N 中，时第一个参比之，为“1”。时为第一层似。又怎么了。接着“2”，为第二层。接着第三层为“3”。 4， 5， 6， 7 ,8 , 9， .，N 层中。而每一层中N 都平均平铺开。The Structure of the Perisphere Layer ExpressionFrom zero 0, the calculation starts. Just l

23、ike from the perspective of any orientation, any angleorientationis a spatial orientationangle. While firstly on the plane, 0 starts the geometricequations of anything to anything, and any nature to any nature, so that is the case.Among N, the one that came first to compare at that time was“1”.At th

24、at time it was the firstlayer. Then what sthe next? The number 2 is the second layer, then followed by the third layer3, and 4,5,6,7,8,9.in layer N.In each layer, N tiles averagely.时每一层都动用怎样的等比等式到下层。 相对来说， 每一层的量都还相等的。 高度多一样，为一个参数 “1”。似时又一个平面，与他( 疑为“它” )一样。并同样的，原点与原点“0”，垂直相交。又有几何法可构成圆球体层式。为正圆球体层式。At

25、that time, each layer commands a certain geometric equation to the lower layer. Relatively speaking, the amount of each layer is still equal. It has the same height of a parameter 1.It is like that at that time there is another plane as well, the same as it, the original point and the original point

26、 0 also intersect vertically. Sphere layers may be constituted by geometry methods.（这里讲一下，此圆球体层的厚度应该这样证明：）此圆球体层式中，到时候哪一层中的一个数碰到对应的运算。则同样， 这个数作为一个原点运作成此圆球体层式，（但，时这个数原来的本质和量不变。）（这时也看出P=NP。）。就这样证明出： 一定的条件下， 每次怎样的碰到一个距离的数，就用一个此圆球体式参比之。时可看出，一定条件下为一个同等的参比值（参照物一样，叫它“参比值”。下面的还有，叫它“参数”。理解一样。），This Sphere Lay

27、er Expression is a perfect sphere layer Expression.We will talk about it here. The thickness of the sphere layer should be proved like this: in this sphere layer expression, a number in a certain layer will encounter the corresponding operationin due course. Similarly, this number, as an original po

28、int, could be operated to form this sphere layer expression. (However, at that time, the original nature and amount of this number remained unchanged). (At that time we could also see P = NP). So it is proved that under certain conditions, each time we encounter a distance number, we use this sphere

29、 expression as a reference. At that time we could see that, under certain conditions, it was the same reference value.只是在当时怎么样了。说回来，时可为一个参数“1”。就是说，在一定条件下，每隔一个距离。的似（就似）隔着一个“1”。那么这样，在一定条件下，每一层圆球体层的高度先可以一样。）The only concern is what it was like at that time.Again, at that time it could be a parameter of

30、1. That is to say, under certain conditions, it appears to be separated by a 1 for every other distance. Therefore, under certain conditions, the height of each sphere layer can firstly be the same.也再讲一下每一层此圆球体层的厚度。也就是直接证明了一下： 时因为在一个一定条件下，时都是等比等式来回。时自然数中直接一个参数，可为“1”。时一个一定条件下的厚度多一样。We also talk about

31、 the thickness of each sphere layer and it directly proves that: under a certain condition, at that time it came back and forth according to some geometric equations. At that time, a direct parameter among the natural numbers could be 1. At that time, the thickness was all the same under a certain c

32、ondition.相对来说到时候的量似在每一层圆球层里一样，只是怎样的平均平铺着。也就是平均平铺着。因为每一层到下一层，都动用的等比等式来去。那么，每一层量多相等。来去只求有当时情况。之 .当时情况怎样。之当时可怎样。 之，当时怎样。 (而 N 似，此圆球体还在运作似。 )Relatively the amount in due course is like that in every sphere layer, the only concern is just how they tile, how they tile averagely. Because from each layer to

33、 the next, the geometric equations are applied to coming and going, so the amount of each layer is almost equal. Only the situationat that moment concerns us regarding coming and going. So what was the situation at the moment? How could they be that possibly? What was the case at that moment? (As fo

34、r N alike, the sphere appears to be still operating).相对来说： 此圆球体内，时又有哪个点要怎么样的。时又可圆球体式起，而起怎样对应的数学运算。： 又有能在一平面上的任意点的对应情况。也就一平面上的怎样对应的情况。： 圆层表面 .可平面几何。每一层的圆球层体，可立体几何。： P 与 NP 。此圆球体层式又可平面几何来解释。只是到时候又有怎样当时的情况而已。Relatively speaking, inside the sphere, at that time which point was the concern?At that time,

35、starting from the sphere expression, we could also begin a certain corresponding mathematical operation.There is a corresponding situation of an arbitrary point in the plane, that is to say, how they correspond in one plane.The sphere surface can be explained by plane geometry. Each layer of the sph

36、ere body can be explained by solid geometry.P and NP. This sphere layer expression can be explained by plane geometry, the only concern is what the situation is in due course.( 时可看出 P 与 NP ：此圆球体用几何法证明回到空间中的一原点。空间中的一原点，又几何法证成此圆球体式。则，原点等于原点，原里本质=原里本质： P=NP 。）解答这圆球体层型式，这十道题基本上一样。而几何法中的此圆球体式能与P=NP相似。原里的

37、本质不变与此圆球体式的等比等式的运算，时似那数学几何法中的原点“O起”动了数学运算。(At that time we could see P and NP: when this sphere can be proved back to one origin of thespace by the geometry method, and one origin of the space can be proved to be the sphereexpression by the geometric method. Then, the origin is equal to the origin,

38、the original nature =the original nature: P = NP). （ Answer this layer of ball type.）These ten questions are basically thesame. This sphere expression could be similar to the P = NP in the geometric method.Theoriginal nature remains unchanged and the sphere expression can be explained by geometriceq

39、uation operations, just like the origin O starts the mathematicaloperations in geometricmethod of mathematics.小言一下：而且在反复的论证中，此圆球体层式 , 能演示的任何出来的题。（包括“那个逻辑”也给定位出来。）。原理好像就围绕这个原点开始的。有，理论为一个平面,实际也可为一个平面。（纠正：不能为一个平面。错误了的。只是当时的一个题目。）则 ,相等。似在第四层论证时 ,即为数学几何法的基本 ,而且又很重要。若 ,原里有当时的那个力量真的跟上了。 那么才是真正的原里力量。 叫它为： “

40、这才是真正的原里力量。”。And in repeated argumentation, this sphere layer expression can demonstrate any existing questions (including the location of the logic).The principle seems to begin with the origin and centers on the origin. If it is a plane in theory,and it could actually be a plane as well, （ Correct

41、ion: can not be said to be a plane.Wrong. ） then an equation is achieved. Just like the argument in the fourth layer, it is the basic mathematical geometry method, and is very important. If the original strength at that moment could really keep up, that is the real original strength and it is called

42、: This is the real original strength.解答题目：（ 这十道题，能用此圆球体层式， 能解释任何一道题。 的从下说起。）之十 . 四色猜想 .解答：此圆球体内 .一层二层三层 . 多知道不行 . 平行时就碰到 .相同的边的着色了。（不能碰到有相同边的，而怎样排列着整个平面 . ）而第四层 . 好像刚刚好：几何法展开 ,似平面上。似幅度.频率方向和自己的周期性，并组成怎样的正方形格子平面。Theses ten problems can be explained by this sphere layer expression, each problem can be

43、 explained. We will begin with the last problem:Problem10. Four color conjecture: There are layer one, layer two, and layer three inside the sphere body. They encounter when they parallel and the same edges are colored. (Do not touch the same edges, how would you array the whole plane?). It seems th

44、at the fourth layer is just good enough.Expand it by a geometric method, just like the amplitude, frequency direction and periodicity in a plane, and see what square grid plane can be formed.时从左到右，和从上到下，似坐标线X,Y ，方向上，再和自己的周期性，都是“4 ”。时参比之。（翻译理解：参照物一样，叫做“参比”。时参比之。）时，X,Y 对应着这一个平面。 而且无论从哪个 X,Y 方向都能以相同周期性

45、的对应参比着整个平面。那么，“ 4”个着色时，刚好且故意能排列成不碰到相同边的四色平面着色。From the left to the right and from the top to the bottom, it is like the coordinate lines X, Y direction and its periodicity is 4. At that time it could be explained by similar reference objects., theAt that time, X, Y corresponded to the plane and it c

46、orresponded to the entire plane according to the same periodicity no matter from which X, Y direction. So, when 4 is colored, it is just fine to intentionally array to form a four colors plane that won tmeet the same edges.第 5 层 .对自己的周期个数 . 到时候出现似有角度的怎样平行而不平行。就是多少数出来似的。后面的都似有怎样角度的参比之。四色着色 , 似看清了有角度在

47、平行参比中有怎样。 到时候平面上时的怎样等比等式, 而参比了多少。（这一层时的推论在数学几何法中很基本，而且，很重要。）接着 ,哥德巴赫 ,黎曼等函数 ,都等比等式怎样。先讲几何尺规作图问题。Layer 5. Cycle Numbers. There appears to be angles that parallel and not parallel. They are just like many numbers. Please refer to this for the following similar angles.Through four color shading, It app

48、ears to see clearly what the angles are like in the parallel reference, and what the geometric equation is while on the plane in due courses, and to what extend they correspond. This layer of inference is basic and very important in geometric methodof mathematics.The followings are Goldbach and Riem

49、ann functions, all geometric equations. Firstly we will talk about geometric ruler gauge construction problem之九 .哥德巴赫猜想 .解答：（如，对应变量，一平面图。如图们）右小图们.素数，（先不说 A 到 B 了的线段。）一点到另一点，而且只被这两端点整除。的似看作一个量表达。 （在这里， 时要提一下这样的一个量的表达：就是从这一点到那一点的一条直线段。看成一个量的表达。若，在哥德巴赫猜想中，就表达了一个素数。）Problem 9: Goldbach Conjecture: (e.g.

50、, the Corresponding Variables, a Planar Graph as shown in figures). Small figures to the right.A prime number, (regardless of the line segment from A to B), from one point to another, could only be divided exactly by the two end points. It could be regarded as a quantity expression.( Thisquantity ex

51、pression should be mentioned here: a straight line segment from this point to that point. It slike that a prime number was expressed in the Goldbach conjecture.在此圆球体中。圆球层面上：时 6 时。（如图），（对应变量，一平面图。如图们）。似每个整个圆球体层面的面积。几何法展开，成一平面，时 XY 。时有对应量的 X 与 Y 的对应互换。时又有相加时， X 与 Y 相对应的两个奇素数的和，为偶性。是对的。In this sphere,

52、on the layers of the sphere:At that time, when it is greater than or equal to 6. (as shown in the figure), (the corresponding variables, a planar graph as shown in the figures). I t slike the area of each layer of the whole sphere. Expand it by a geometric method to become a plane, at that time it w

53、as XY. At that time it had an amount corresponding the X and Y exchange. At that time when they were added, the sum of two odd prime numbers corresponding the X and Y was an even. It is correct.而9时，每层圆球体层整层的立体的体积，也可以用怎样对应的三个奇素数相乘表达。时也又有相加时， 三个奇素数的和，为奇性。 也对的。（哥德巴赫猜想的表达，证明他的本质，依旧一样。是几何法中，此圆球体式的原点，和此圆球体式。之，原里的本质不变。)（这里的“原里”两字，翻译理解是：原来的里面。）（图解的在下面。）When it is greater than or equal