谴责申明：世界十大数学难题的解答,圆球体层式的解答。

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 baidu's nickname, "CAI in" 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 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: baidu's nickname, "CAI YU JING DAO ".

Taizhou city, zhejiang province.China.In October 2015. 3. (machine translation, I do not know right?)

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 mathematic 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 actually be a plane as well,

（ Correction: can not be said to be a plane.Wrong.Should be, back to the sa me origin.） then an equation is achieved.

Just like the argument in the fourth layer, it is the basic mathematical geomet ry 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 r eal original strength".

As for the sphere theory, I once demonstrated it in Peking University’s space forum in 2006, also proved that “Zero”and many other relevant questions, and published it in 2009. It was recognized in some places. During May/Jun, 2014, under the assistance of wechat group of the teachers' office, I accomplished these ten questions on Baidu’s space.“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 it’s correct.

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：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 difficult mathematic problems in the world.

Problem 1: P (Polynomial Algorithm) Problem vs. NP (Non Polynomial Algorithm) Problem Problem 2: Hodge Conjecture

Problem 3: Poincare Conjecture

Problem 4: Riemann Hypothesis

Problem 5: Yang–Mills Existence and Mass gap

Problem 6: The Existence and Smoothness of Navier - Stokes Equations

Problem 7: Birch and Swinnerton-Dyer Conjecture

Problem 8:Geometric Ruler Gauge Construction Problem

Problem 9: Goldbach Conjecture

Problem 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，就能直接用上一个名词，“而物”。）。之，当时什么物。之，当时什么。

This sphere layer expression can solve the world’s ten difficult problems in mathematics. Firstly, we talk about the objects, which can be counted by the numbers (Perfect Sphere Layer Expression). The Perfect Sphere Layer Expression, is hereinafter referred to as this Sphere 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 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 have

something, 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 it’s what it is. It appears that it is an equation of anything to anything, any nature to nature, so it’s what it 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 speaking, 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 that time? When natural number N was used as a reference, it was in direct use. 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 number “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“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 Expression

From zero "0", the calculation starts. Just like from the perspective of any orientation, any angle orientation is a spatial orientation angle. While firstly on the plane, 0 starts the geometric equations 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 that time it was the first layer. Then what’s the next? The number 2 is the second layer, then followed by the third layer 3, 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 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 "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 Layer 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 operation

in due course. Similarly, this number, as an original point, 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 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 "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 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 condition.

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 the next, the geometric equations are applied to coming and going, so the amount of each layer is almost equal. Only the situation at 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 for 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, 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 sphere 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 the space by the geometry method, and one origin of the space can be proved to be the sphere expression by the geometric method. Then, the origin is equal to the origin, the original nature = the original nature: P = NP).（Answer this layer of ball type.）These ten questions are basically the same. This sphere expression could be similar to the P = NP in the geometric method. The original nature remains unchanged and the sphere expression can be explained by geometric equation operations, just like the origin "O" starts the mathematical operations in geometric method 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：“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,（Correction: 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: "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 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 that 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, the direction and its periodicity is "4". At that time it could be explained by similar reference objects. At that time, X, Y corresponded to the plane and it corresponded 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’t meet 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 appears 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 method of mathematics.

The followings are Goldbach and Riemann 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., 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.( This quantity expression should be mentioned here: a straight line segment from this point to that point. It’s like 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。

I n this sphere, 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’s like the area of each layer of the whole sphere. Expand it by a geometric method to become a plane, at that time it was 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 to 9, the three-dimensional volume of each whole layer of the sphere can be expressed by the multiplication of three corresponding odd prime numbers. At that time when they were added, the sum of three odd prime numbers is an odd number. It is also correct. (The expression of Goldbach Conjecture proves that his nature is still the same. They are the origin of the sphere expression and the sphere expression in the geometric method. Therefore, the original nature in essence remains unchanged).

(Diagram, in the final.)

接着，函数，可看出怎样的有限无限，对应着怎样(怎样的有限无限）。（如，对应变量，一平面图。如图们）。

之八.几何尺规作图问题.

解答：

没有刻度,角度数, 其实一样。就是等比等量的怎样等式来去：时一开始就用一定多少的长度，或一定多少的量。（相对来说一定多少的量。）只是到时候出现多少的等比等量等式的参用（翻译理解：参加应用，叫“参用”。），而参比着当时怎样情况（翻译理解：参照物一样，叫“参比”。而参比着当时怎样情况。）

之，尺规几何作图。（这样的等比等量等式的参用，在此运用中相当正确。）

Problem 8: Geometric Ruler Gauge Construction Problem.

There is no scale, the angle degrees, in fact, are the same. It is how Geometric equivalent does according to geometric equation. At that time, it started with a certain length, or a certain amount.(There must be a certain amount, relatively speaking). What matters is how many geometric equivalent equations will participate in the application and what situation they refer at that moment.

Geometric Ruler Gauge Construction.(These geometric equivalent equations are fairly correct in this application).

（而一直都能用一定多少的等比等量参比着来去。）如图：（任意角的等分图角）三等分任意角。化圆为方。倍立方体。正十七边形.(正N边形)。时又有原点与有单位距离的点的全体.其实就是怎样的函数对应着怎样.(有限无限.)。

函数如图：（对应变量-平面图。如图们）。

(We can always use a certain amount of geometric equivalent to be the reference). As shown in figure: (The Equal Angle of Figure of Arbitrary Angle).The trisection of an arbitrary Angle. Turn a circle into a square. The duplication of the cube. The Regular Heptadecagon. (regular N polygon). At that time, it had the origin point and all the points with unit distance. In fact it is a case of how function corresponds to what it is. (Finite and Infinite.).

Function is as shown in figure: (Corresponding Variables - Planar Graphs as shown in figures).

先讲纳维叶-斯托克斯（Navier-Stokes)方程式存在性和光滑性,与霍奇（Hodge)猜想。

Then through the function, we can see what the finite and infinite is like, what they correspond. (what the finite and infinite is like).(e.g., the corresponding variables, a planar graph, as shown in figures).

Firstly we talk about Navier - Stokes equations existence and smoothness, and Hodge conjecture. 之六.（Navier-Stokes)方程式存在性和光滑性。

解答：

水面中行驶中的船，飞行中飞机，留下的后面跟着的小波浪和尾气流。

Problem 6: Navier - Stokes Equations Existence and Smoothness.

A moving vessel on the water surface and an aircraft in flight are followed by small waves and exhaust gas flow left behind.

时，似此圆球体中的上层到下层。（每一层厚度多一样，时直接压缩成一平面吧！）（每层的量都相等，先用自然数N参比之（翻译理解，就是参比之它，把它用N来参比之，用之。））。上层的量到下层，下层时层面依旧几何法展开表达，xy。而x=y 而一平面。时上层的量到下层xy上。而平均xy。那么，平均N。时有：N/xy=1/N。

At that time, it was like something between the upper layers and the under layers of this sphere. (The thickness of each layer is the same, and is directly compressed into a plane!)(The amount of each layer is equal, and we firstly use natural number N to compare it as a reference.

The amount of the upper layer goes to the under layer, the under layer is still expressed by geometric method expansion, xy. If x = y, then a plane. The amount of the upper layer goes to the under layer on the xy. Average xy, then average N. At that time : N/xy = 1 / N.

时若，xy一定了,（上层的量）量越大，则1/N而越小。时似真正的快刀划过蜡烛，火焰只晃了一下，而划过水面中。

At that time, with a certain xy, the greater the amount of the upper layer, the smaller is 1 / N. At that time it was like a real sharp knife across the candle, the flame only shook a little, and then it was across the water surface.

之二.霍奇（Hodge)猜想.

解答：

（力学，平恒似。）时似圆球体在一个空间中，有似力量从各个方位打来一样。就似这圆球体的面对着外面。

时若有上题，纳维叶-斯托克斯（Navier-Stokes)方程式存在性和光滑性中，N/xy=1/N。时1/N而无限小，若接近了“0”，就似此圆球体似。有理线性，时有筛子筛选一样，有了怎样。时又有似理论于实际，只求有当时情况。

Problem 2: Hodge Conjecture: (Mechanics, balance alike). At that time it was like a sphere in a

space, there was something like strength coming from each direction. It appears that the face of the sphere is facing the outside.

At that time, if the above-mentioned question was involved, then in Navier - Stokes equations existence and smoothness, N/xy=1/N.

At that time 1 / N was infinitesimal, if it was close to "0", it was just like the sphere body. Being rational and linear, at that time it was like something selected by a sieve screening and was what it was. At that time it was like a theory to a practice, the situation at that moment was the only concern.

而此圆球体式中函数。（如，对应变量，一平面图。如图们)。

The function in the sphere expression.(e.g., the corresponding variables, a planar graph. As shown in figures).

之三.庞加莱（Poincare）猜想.

之四.黎曼（Riemarr）假设.

之七.贝赫（Birch）和斯维纳讷渔-戴尔（Swinnerton-Dyer）猜想.

解答：

Problem 3: Poincare Conjecture.

Problem 4: Riemann hypothesis.

Problem 7: Birch and Swinnerton-Dyer Conjecture.

Answer:

时庞加莱（Poincare）猜想. 存有单位距离对应的点的全体。

时(对应变量.-平面图。如图们)，到时候是对应的变量。

若对应的都一定的量，就是一段距离。（时对回到起原点时，都按一定的量，参比之。这儿我就叫它“基比”。）那么，就是距离基比几何法与之相对应函数式。那么就是个圆球体。

Poincare Conjecture. There existed an entirety of points corresponding to the unit distance.

At that time(corresponding variables- planar graph, as shown in figures), they are the corresponding variables in due course.

If it corresponds to a certain amount, it is a distance.(At that time, it returned to the original point according to a certain amount as a reference. Here I will call it " Kibbee "). So it is the functional expression corresponding to the distance Kibbee geometric method. It is a sphere body.

另外，又有距离几何中，到时候又有怎样的距离几何与对应的变量函数式。构成当时情况。（其中如，直角三角形的勾股定理，等。）。

本来就是等比等式的怎样来去，时又对应着怎样的有限无限：之四. 黎曼（Riemann）假设。和之七贝赫（Birch）和斯维纳讷渔-戴尔（Swinnerton-Dyer）猜想。

In addition, there is distance geometry, then the distance geometry in due course and the corresponding variables functional expression will constitute the situation at that time.(e.g., The Pythagorean theorem of a right triangle, etc.).

Originally it comes and goes according to the geometric equation and corresponds to certain finitude and infinity.

Problem 4. Riemann Hypothesis and Problem 7. Birch and Swinnerton-Dyer Conjecture

引用：

(而著名的黎曼假设.断言. 方程Z(S)=0的所有有意义的解都在一条直线上。这点已经对开始的1,500,000.000.个解验证过，证明它对于每一个有意义的解都成立。）

Reference:

The famous Riemann hypothesis asserted that the meaningful solutions of the equation Z (S) = 0 all lie in a straight line. This has been validated by the starting 1,500,000.000 solutions, proving that it works well for every meaningful solution).

之五杨—米尔斯（Yang-Mills）存在性和质量缺口.

解答：

似此圆球体层式中，似原点能同时碰到的几个有距离的点（而这些点，又能和此圆球体一样。里面又有哪距离的点，又能比作小圆球体。接着，小圆球体里面又有那更小的小圆球体。就都有这样下去。）。那么，时有距离几何法和相对应函数式与之相对应。

如果质子，夸克，它们里面都还能继续解剖一样分解下去。那么，显微镜一样看它。一样，依旧有怎样的距离几何法和它相对应的函数式。

Problem 5: Yang–Mills Existence and Mass gap

Just like in this sphere layer expression, it appears that the origin point can also meet several points of the distance simultaneously (and these points are the same as this sphere. There are points of distance inside it, which can be regarded as small sphere. Then there are smaller spheres inside the small spheres. This continues.). So, there was a distance geometry method and the corresponding functional expression that could correspond at that time.

If the protons and quarks could continue the anatomical decomposition inside them, then it is the same under the Microscope, it still has that distance geometry method and its corresponding functional expression.

难题之一. P与NP.(多项式算法问题.与非多项式算法问题.) 解答：

P=NP，时为同一个原点，就是那个圆球体层式的原点。和此圆球体式。

证明时只能这样说：空间的原点，用几何法证明成此圆球体式。时此圆球体式用几何法证明回到原点。这样来回而说：P=NP。

Problem 1. P and NP. (Polynomial Algorithm and Non Polynomial Algorithm)

Answer:

P = NP, at that time shared the same original point, it is the original point of that sphere expression and this sphere expression.

We can only say this when we are proving: the original point of the space is proved to be this sphere expression by the geometric method. At that time it was proved to be back to the original point in this sphere expression by the geometry method. So it goes back and forth and we can say: P = NP.

另则有，原点等于原点，又是本质等于本质。之：本质不变。

而且反复的论证中，原里（翻译理解：原来的里面，叫它“原里”。）的原理，（这里的“原理”两字，翻译理解：原来里面的道理。。）就跟这原点起的此圆球体式的运作，相似（翻译理解：基本上一样。）。P=NP.。之，原里同源.（原理同源.）（这里的“原里”两字。翻译理解：原来的里面。）。比如那物理的十道难题。时也五六月份做过了。

In addition, the original point is equal to the origin point, and the nature is equal to the nature. Therefore, the nature remains unchanged.

After repeated argumentation, the principle of the original interior is basically the same as the operation of this sphere originating from the original point. P = NP. So the original interior is from the same origin, for example, the ten difficult physical problems. It was discussed and done in May and June.

小言：P与NP.（P=NP.) 有这样讲话一样的形式：

一个区域上（区域上的事情）.等于,一个区域上所发生的事情。一条路上（路上的事情）.等于.一条路上所要发生的事情。说说说就说成这样，自己的事情自己做。

P and NP. There is such way of talking: a region (things in the region) equals to what

happens in the region. A road (things on the road) equals what is going to happen on that road. You say something repeatedly, they will be as you say, do your own things by yourselves.

此圆球体方式还可解释物理十道难题。相对来说，当时情况怎么样！。基本上差不多。

This sphere mode could also explain ten physical problems. Relatively speaking, how was it at that time! It is basically the same.

有图对照的解答题：

Figure compared with answer the questions:

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