无穷多的素数会成对的出现

2024-03-01 18:38
文章标签 素数 无穷 会成

本文主要是介绍无穷多的素数会成对的出现,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

上午公司培训,请了一个腾讯的大牛做数据挖掘的讲座。我本来的目的是想了解一下hadoop与hbase相关的运维与部署,或多或少能了解到点什么,没想到讲的全是数据挖掘建模方面的知识。虽然现在做的是SA&DBA的事情,还好之前今天的职位是数据分析工程师,接触过一点数据分析挖掘的知识,能听懂一点。大牛讲了很多,基本的概念包括数据分类、聚类、拟合。算法也讲了很多分段建模,logit(logistic)变换,Slope One,推荐ranking=scoring+sorting+filtering,SVD对分类准确性的提高,SVM,Deep Learning,还有关于Netflix的百万美金与神经网络科学家的Hinton的故事(restricted boltzmann machines for collaborative filtering)。系统方面讲到hadoop、hbase、hive、pig与google的三架(不知道用什么单位!!!)火车:GFS、Big Table、Map/Reduce。数据、算法、系统在数据挖掘过程中的重要程度依次递减的,好的数据也不需要复杂的算法。

听完讲座之后,决定在办公室待到下午3点多再回去,叫了份外卖。准备等小伍哥一个过来,晚上和老陆,总管一起吃个饭。四个人都是从一个实验室毕业的研究生,老陆是我的师兄(或者说是师叔,汗!),小伍哥、总管和我是同一届毕业的。其实很久没见面了,也难得都在上海工作,大部分的研究生熟人都在杭州那边。

吃完饭,无聊打开facebook上看到一篇关于素数的理论(http://www.scientificamerican.com/article.cfm?id=first-proof-that-infinite-many-prime-numbers-come-in-pairs),中国人在学术上确实厉害。好久没面对数学问题,会不会脑子生锈,虽让写程序也要动脑子,耐心的看了下来。素数在笔试或者面试的时候是经常碰到的一个问题,觉得有点意思,整理了一下自己想法和网上的一些关于的素数的东西。

素数:只能被1和自己本身整除的正整数。(a positive whole number that is divisible only by 1 and by itself

合数:能唯一表示成素数的乘积,如15=3*5。

1:一个比较特殊的数字,不能说是素数也不能说是合数。

理论:素数以2*n-1,2*+1孪生的形式出现,如(3,5),(5,7),(11,13),(17,19),(29,31) 这样成对出现,而且无穷多个

不存在最大的素数

很简单,反证法,假设存在最大的数据P

那么存在一个合数M,M=2*3*5*7*......*P,即M为所有的不重复的素数的乘积。

M+1 = 2*3*5*7*......*P+1按照假设“M是最大的素数”推断,M+1是合数。

实际上(M+1)不能被2、3、5,......P中任何一个素数整除,余数均为1。

很明显M+1是一个素数,M+1>P,存在比P的素数,因为假设不成立。

因而不存在最大的素数。

下面是我个人无聊时一个简单分析,不能算作证明。如果我证明了,我也不是屌丝程序员了^.^。

假设M+1=2*3*5*......P + 1是一个素数,如果存在M+3或者M-1是一个素数,那么素数孪生出现的的结果成立。

M+3 =  2*3*5......P + 3 = 3*(2*5*7......P+1)很明显是一个合数。

M-1=2*3*5*7......P-1 可能为一个素数。

假设M-1不可能为素数:

1)p=3,M-1=5,假设不成立

M-1有可能是一个素数。

下面是英语原文

First ProofThat Infinitely Many Prime Numbers Come in Pairs

A U.S.mathematician claims a breakthrough toward solving a centuries-old problem

By Maggie McKee

Mathematician Yitang Zhang has outlined a proof of a"weak" version of the twin prime conjecture.Image: Maggie McKee

From Nature magazine

Cambridge,Massachusetts

It’s a resultonly a mathematician could love. Researchers hoping to get ‘2’ as the answerfor a long-sought proof involving pairs of prime numbers are celebrating thefact that a mathematician has wrestled the value down from infinity to 70million.

“That’s only [afactor of] 35 million away” from the target, quips Dan Goldston, an analyticnumber theorist at San Jose State University in California who was not involvedin the work. “Every step down is a step towards the ultimate answer.”

That goal is theproof to a conjecture concerning prime numbers. Those are the whole numbersthat are divisible only by one and themselves. Primes abound among smallernumbers, but they become less and less frequent as one goes towards largernumbers. In fact, the gap between each prime and the next becomes larger andlarger — on average. But exceptions exist: the ‘twin primes’, which are pairsof prime numbers that differ in value by 2. Examples of known twin primes are 3and 5, or 17 and 19, or 2,003,663,613 × 2195,000 − 1 and2,003,663,613 × 2195,000 + 1.

The twin primeconjecture says that there is an infinite number of such twin pairs. Someattribute the conjecture to the Greek mathematician Euclid of Alexandria, whichwould make it one of the oldest open problems in mathematics.

The problem haseluded all attempts to find a solution so far. A major milestone was reached in2005 when Goldston and two colleagues showed that there is an infinite numberof prime pairs that differ by no more than 16. But there was a catch. “Theywere assuming a conjecture that no one knows how to prove,” says DorianGoldfeld, a number theorist at Columbia University in New York.

The new result,from Yitang Zhang of the University of New Hampshire in Durham, finds thatthere are infinitely many pairs of primes that are less than 70 million unitsapart without relying on unproven conjectures. Although 70 million seems like avery large number, the existence of any finite bound, no matter how large,means that that the gaps between consecutive numbers don’t keep growingforever. The jump from 2 to 70 million is nothing compared with the jump from70 million to infinity. “If this is right, I’m absolutely astounded,” saysGoldfeld.

Zhang presentedhis research on 13 May to an audience of a few dozen at Harvard University inCambridge, Massachusetts, and the fact that the work seems to use standardmathematical techniques led some to question whether Zhang could really havesucceeded where others failed.

But a refereereport from the Annals of Mathematics, to whichZhang submitted his paper, suggests he has. “The main results are of the firstrank,” states the report, a copy of which Zhang provided to Nature. “The author has succeeded to prove a landmarktheorem in the distribution of prime numbers. … We are very happy to stronglyrecommend acceptance of the paper for publication in the Annals.”

Goldston, whowas sent a copy of the paper, says that he and the other researchers who haveseen it “are feeling pretty good” about it. “Nothing is obviously wrong,” hesays.

For his part,Zhang, who has been working on the paper since a key insight came to him duringa visit to a friend’s house last July, says he expects that the paper’smathematical machinery will allow for the value of 70 million to be pusheddownwards. “We may reduce it,” he says.

Goldston doesnot think the value can be reduced all the way to 2 to prove the twin primeconjecture. But he says the very fact that there is a number at all is a hugebreakthrough. “I was doubtful I would ever live to see this result,” he says.

Zhang willresubmit the paper, with a few minor tweaks, this week.

This article isreproduced with permission from the magazine Nature.The article wasfirst published on May 14, 2013.

这篇关于无穷多的素数会成对的出现的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/763427

相关文章

js算法判断是否为素数

/*判断一个数字是否是质数: 质数(prime number)又称素数,有无限个。除了1和它本身以外不再被其他的除数整除。*/ function isPrime(number){ //判断输入是否为number类型,是否为整数       if (typeof number!=='number'||!Number.isInteger(number))      {

【时时三省】c语言例题----华为机试题< 查找组成一个偶数最接近的两个素数>

山不在高,有仙则名。水不在深,有龙则灵。                                                                         ----CSDN 时时三省 1,题目 HJ60 查找组成一个偶数最接近的两个素数 描述 任意一个偶数(大于2)都可以由2个素数组成,组成偶数的2个素数有很多种情况,本题目要求输出组成指定偶数的两个

题目:求100以内的素数,全部打印出来。

题目:求100以内的素数,全部打印出来。 public class ZhaoSuShu {public static void isPrime1(){int i,j,count = 0;//System.out.println("2");for(i = 1; i <= 100; i++){for(j = 2; j <= i; j++){if(i % j == 0){break;}}if(j ==

素数判定和分解质素数

1.素数判定   public static boolean isPrime(int n) {if (n <= 1) return false;if (n == 2) return true;if (n % 2 == 0) return false;int limit = (int)Math.sqrt(n) + 1;for (int i = 3; i <= limit; i += 2) {i

Java-计算素数

判断输入的数字是不是素数: public class SuShu{public static void main(String[] args){java.util.Scanner s=new java.util.Scanner(System.in);int i=s.nextInt();boolean isSuShu=true; //标记;for(int j=2;j<i;j++){if(i%j=

常见素数筛法

列出几种常用的素数筛选法,附上计时器。。。 #include<cstdio>#include<cstdlib>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<algorithm>#include<cstring>#include<string>#inclu

【编程基础C++】素数判定、最小公倍数与最大公因数的实现方法

文章目录 素数法一法二 最大公因数辗转相除法另一写法 最小公倍数直接枚举法根据GCD算LCM 素数 素数 是指大于1的自然数,且只能被1和自身整除。例如,2、3、5和7都是素数。它们在数学中非常重要,因为任何大于1的自然数都可以唯一地表示为素数的乘积,这被称为素数分解。 法一 #include <iostream>using namespace std;bool IsPr

求素数的几个方法(最朴素版、n*sqrt(n)版、埃氏筛、欧拉筛)

最朴素版O(n^2) #include <bits/stdc++.h>using namespace std;int n, cnt, prim[6000000];bool flag; //true 表示质数int main(){scanf("%d", &n);for(int i=2; i<=n; ++i){flag=true; //默认为质数for(int j=2; j<=i-

【素数】-HDU-2521-反素数

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2521 题目描述:求区间内因数最多的数是哪个? 解题思路: 九野大神划到素数这类题里的,我看过的人是 4 / 8 就去做了一下。。晕。看起来好水,没想到真的很水。。1A 了。 AC代码: #include <iostream>#include <cstdio>#include <alg

统计素数并求和 / 求奇数和

练习4-11 统计素数并求和   (20分) 本题要求统计给定整数MM和NN区间内素数的个数并对它们求和。 输入格式: 输入在一行中给出两个正整数MM和NN(1\le M\le N\le 5001≤M≤N≤500)。 输出格式: 在一行中顺序输出MM和NN区间内素数的个数以及它们的和,数字间以空格分隔。 输入样例: 10 31 输出样例: 7 143 #inclu