Some method of determining primes summary (including C ++ code)

A prime number is defined

Prime (prime number), also known as a prime number , the number 1 only refers to its own nature and in number factor greater than 1. For example, a prime number is 2, 1 and 2 which is only a factor of two; 29 is a prime number, which is only a factor of two and 29; 51 is not a prime number, in addition to 1 and 51, 3 and 17, it has a factor of two, so that 51 is a composite number .

Second, the prime method for determining

Given a positive integer n (n≥2):

Method 1. Definitions

Divided by n is about [2, n-1] of all integers, if a remainder after which the operand is 0, that the number is a factor of n, so that n is not a prime number. code show as below:

bool isPrime(int n){
 bool yes=true;
 for(int i=2;i<n;i++){
  if(n%i==0){
   yes=false;
   break;
  }
 }
 return yes;
}

2. Define Method to Improve

Since the factor n always appear in pairs, and are distributed in [1, $\sqrt n$]with[ $\sqrt n$, The n] range (number Ruoyin $\sqrt n$We as two repeat factor), then we only need to determine a range. code show as below:

bool isPrime2(int n){
 bool yes=true;
 for(int i=2;i<=sqrt(n);i++){
  if(n%i==0){
   yes=false;
   break;
  }
 }
 return yes;
}

3. Method modulo

When n≥6, since n can be expressed as 6x + 1,6x + 2,6x + 3,6x + 4,6x + 5,6x (x≥1) a medium, then, if the expression for n 6x + 2,6x + 4 or 6x, it is clear that n is not a prime number; therefore, when n is 6x + 1 expression or 6x + 5, it may prime number, then the application method. code show as below:

bool isPrime3(int n){
 bool yes=false;
 if(n==2||n==3||n==5){ //当n<6时列举即可
  yes=true;
 }
 else if(n%6==1||n%6==5){ //通过判断余数的方式来判断n的表达式
  yes=true;
  for(int i=2;i<=sqrt(n);i++){
   if(n%i==0){
    yes=false;
    break;
   }
  }
 }
 return yes;
}

Test, time consuming

### 4.筛选法（Eratosthenes筛选）
我们可以很容易的判断出2、3、5等较简单的数是素数，而且，当一个数是素数后，它的倍数一定为合数。因此，我们可以将一定范围的整数筛选掉合数，剩下的即是素数。代码如下：

```cpp
bool isPrime4(int n){
 bool yes=false;
 //先生成一定范围的整数表，再应用筛选法，得到素数表，最后与素数表比对
 //生成0~100000内的素数表
 int num[100000]={0}; //0表示素数，1表示合数
 for(int i=2;i<100000;i++){
  if(!num[i]){
   for(int j=i+i;j<100000;j+=i){
    num[j]=1;
   }
  }
 }
 if(!num[n]){
  yes=true;
 }
 return yes;
}

5. Improved screening

First of all, we can be sure that all primes except 2 are odd, it is possible to establish a means greater than the odd array 2, so that both expanded the judgment, but also reduce the number of judges; and then, we get the odd array of 3, 5, 7,9 ...... to 3 as an example, according to the method we should be screened out 4 3 × 2,3 × 3,3 × 4 ...... but where the 3 × 2,3 × 4 ...... is an even number, has been screened out, and Thus only starts from 3 × 3, screen out 3 × 3,3 × 5,3 × 7 ...... when screening a multiple of 5, the same should start with 5 × 2, but 5 × 2,5 × 4 ...... is even, then you should consider starting from 5 × 3, 5 × 3 but already sieve at screening out a multiple of 3, so screening should be a multiple of 5 from 5 × 5 began, screened out 5 × 7,5 × 9 ...... when we observe the number of filter groups, there is such a rule: if a prime number, then the filter starts from a × a, screen out a × a + 2 × a × i (i = 0,1, 2……).
A × a reason from the start screen: 4 by methods we should be screened out a × 2, a × 3, a × 4 ...... wherein a × 2, a × 4 ...... is an even number, has been screened out, and when from a when the start × 3 filter, if a> 3, to be sure, when screening multiple of 3 has been screened out in this 3 × a number, then the filter starts a × 5, the same token, if a> 5, then the filter 5 when multiples have been screened out this number of 5 × a; so until the beginning screening a × a, and it did not process before being screened out.
Why screen out a × a + 2 × a × i (i = 0,1,2 ......)? We can look at the conversion equation: a × a + 2 × a × i = a × (a + 2 × i). Screening can be seen that when starting from a × a, screen out a × (a + 0), a × (a + 2), a × (a + 4) ...... line with the law in the previous procedure.
The above process is summarized as Code:

bool isPrime5(int n){
 bool yes=false;
 int num[100000]={0}; //生成3~2*99999+3范围内的奇数数组
 //先判断当n=2时的情况
 if(n==2){
  yes=true;
 }
 else{
  for(int i=0;i<100000;i++){
   if(!num[i]){
    for(int j=(2*i+3)*(2*i+3);j<(2*100000+3);j+=2*(2*i+3)){
     num[(j-3)/2]=1;
    }
   }
  }
 }
 if((n-3)%2==0){
  if(!num[(n-3)/2]){
   yes=true;
  }
 }
 return yes;
}

Third, the summary

The above-mentioned methods are put together, as follows:

#include <iostream>
#include <math.h>
#include <time.h>
#define SIZE 10000
using namespace std;
bool isPrime(int n) {
 bool yes = true;
 for (int i = 2; i < n; i++) {
  if (n % i == 0) {
   yes = false;
   break;
  }
 }
 return yes;
}
bool isPrime2(int n) {
 bool yes = true;
 for (int i = 2; i <= sqrt(n); i++) {
  if (n % i == 0) {
   yes = false;
   break;
  }
 }
 return yes;
}
bool isPrime3(int n) {
 bool yes = false;
 if (n == 2 || n == 3 || n == 5) {
  yes = true;
 }
 else if(n%6==1||n%6==5) {
  yes = true;
  for (int i = 2; i <= sqrt(n); i++) {
   if (n % i == 0) {
    yes = false;
    break;
   }
  }
 }
 return yes;
}
bool isPrime4(int n) {
 bool yes = false;
 int num[SIZE] = { 0 };
 for (int i = 2; i < SIZE; i++) {
  if (!num[i]) {
   for (int j = i + i; j < SIZE; j += i) {
    num[j] = 1;
   }
  }
 }
 if (!num[n]) {
  yes = true;
 }
 return yes;
}
bool isPrime5(int n) {
 bool yes = false;
 int num[SIZE] = { 0 };
 if (n == 2) {
  yes = true;
 }
 else {
  for (int i = 0; i < SIZE; i++) {
   if (!num[i]) {
    for (int j = (2 * i + 3) * (2 * i + 3); j < (2 * SIZE + 3); j += 2 * (2 * i + 3)) {
     num[(j - 3) / 2] = 1;
    }
   }
  }
 }
 if ((n - 3) % 2 == 0) {
  if (!num[(n - 3) / 2]) {
   yes = true;
  }
 }
 return yes;
}
int main() {
 cout << "function isPrime():" << endl;
 time_t before = clock();
 for (int i = 2, j = 0; i < 10000; i++) {
  if (isPrime(i)) {
   cout << i << " ";
   j++;
  }
  if (j == 9) {
   cout << endl;
   j = 0;
  }
 }
 time_t after = clock();
 cout << endl;
 cout << static_cast<double>((after - before) * 1000 / CLOCKS_PER_SEC) << "(ms)" << endl;
 cout << "function isPrime2():" << endl;
 before = clock();
 for (int i = 2, j = 0; i < 10000; i++) {
  if (isPrime2(i)) {
   cout << i << " ";
   j++;
  }
  if (j == 9) {
   cout << endl;
   j = 0;
  }
 }
 after = clock();
 cout << endl;
 cout << static_cast<double>((after - before) * 1000 / CLOCKS_PER_SEC) << "(ms)" << endl;
 cout << "function isPrime3():" << endl;
 before = clock();
 for (int i = 2, j = 0; i < 10000; i++) {
  if (isPrime3(i)) {
   cout << i << " ";
   j++;
  }
  if (j == 9) {
   cout << endl;
   j = 0;
  }
 }
 after = clock();
 cout << endl;
 cout << static_cast<double>((after - before) * 1000 / CLOCKS_PER_SEC) << "(ms)" << endl;
 cout << "function isPrime4():" << endl;
 before = clock();
 for (int i = 2, j = 0; i < 10000; i++) {
  if (isPrime4(i)) {
   cout << i << " ";
   j++;
  }
  if (j == 9) {
   cout << endl;
   j = 0;
  }
 }
 after = clock();
 cout << endl;
 cout << static_cast<double>((after - before) * 1000 / CLOCKS_PER_SEC) << "(ms)" << endl;
 cout << "function isPrime5():" << endl;
 before = clock();
 for (int i = 2, j = 0; i < 10000; i++) {
  if (isPrime5(i)) {
   cout << i << " ";
   j++;
  }
  if (j == 9) {
   cout << endl;
   j = 0;
  }
 }
 after = clock();
 cout << endl;
 cout << static_cast<double>((after - before) * 1000 / CLOCKS_PER_SEC) << "(ms)" << endl;
 return 0;
}

We Run the code (operating environment: vs2019, debug, x64), time-consuming results are as follows:
the isPrime (): 379ms
isPrime2 (): 370ms
isPrime3 (): 400ms
isPrime4 (): 1008ms
isPrime5 (): 1028ms
discovery efficiency of the process 2 highest. I may be the code to be optimized, the efficiency of the method 3 is not high. Method 4 and 5 and the method of Mr. table needs to be a prime number, and then to compare, the time consumed will increase. If the table is to generate a prime number, an integer is determined whether or not a prime number, the efficiency of the two methods may be relatively high. Wherein the method of generating a larger prime number table range.