How to implement Emrips anti-prime enumeration in javascript

The editor below will share with you a sample code for implementing Emrips anti-prime enumeration in JavaScript. It has a good reference value and I hope it will be helpful to everyone. Let’s follow the editor to take a look.

I saw a kata today and proposed the concept of “emirps”: after a prime number is inverted, a different prime number is obtained. This number is called “emirps”.

For example: 13,17 are prime numbers, 31,71 are also prime numbers, 13 and 17 are "emirps". But the prime numbers 757,787,797 are palindromic primes, which means the inverted number is the same as the original number, so they are not considered "emirps".

The question requires writing a function that inputs a positive integer n and returns the number of "emirps" less than n, the largest "emirps", and the sum of all "emirps" less than n.

The idea of ​​​​solving the problem is to first enumerate all prime numbers less than n, and then eliminate palindrome prime numbers and numbers that are composite numbers after being reversed.

First write the function to determine prime numbers

Mainly based on three mathematical conclusions:

All composite numbers are the product of several prime numbers

If a number can be factored, then the two factors must be one less than or equal to sqrt(n), and one greater than or equal to sqrt(n) ).

All prime numbers greater than 3 are in the form 6X 1 or 6X-1, which are adjacent numbers that are multiples of 6, but not all 6X 1 or 6X-1 are prime numbers.

The first conclusion can be proved by proof by contradiction

The third conclusion is proved:

We express all the numbers It is in the following form 6X-1, 6X, 6X 1, 6X 2, 6X 3, 6X 4 (X is a positive integer) 6X => 2*3x 6X 2 => 2(3x 1) 6X 3 => 3( 2x 1) 6X 4 => 2(3x 2) It can be proved that these are definitely not prime numbers, that is, the prime numbers can only be 6X-1 or 6X-1

Code:

function isPrimeNumber(num){
  
 if(num == 2 || num == 3){
  return true;
 }／*2、3特殊处理*／
  
 if(num % 6 != 1 && num % 6 != 5){
  return false;
 }／*根据结论三排除*／
  
 for(var i=5;i<=Math.sqrt(num);i+=6){
  if(num % i == 0 || num % (i+2) == 0){
   return false;
  }
 }／*根据结论二、结论三排除*／
  
 return true;
}

Then eliminate palindromic prime numbers and numbers that are composite after inversion

Code:

function emirpNumber(num){
 
 var reverseNumber = Number(String(num).split('').reverse().join(''))
  
 if(reverseNumber != num && isPrimeNumber(reverseNumber)){
  return true;
 }
 else{
  return false;
 }
}

Finally output the desired result

Code:

function findEmirp(n){
 
 var emirpGroup = [];
 
 for(var i=1;i<n;i++){
  if(isPrimeNumber(i) && emirpNumber(i)){
   emirpGroup.push(i);   
  }
 }
  
 return [
  &#39;n为：&#39; + n,
  &#39;数量为：&#39; + emirpGroup.length,
  &#39;最大数：&#39; + emirpGroup[emirpGroup.length - 1],
  &#39;求和：&#39; + emirpGroup.reduce(function(total,current){
   return total + current;
  })
 ]
}

Look at the output results and time

n=1000000:

n=10000000：

##The above is what I compiled Everyone, I hope it will be helpful to everyone in the future.

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