A faster way to compute the largest prime factor





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I am self-learning js and came across this problem(#3) from the Euler Project




The prime factors of 13195 are 5, 7, 13 and 29.



What is the largest prime factor of the number 600851475143 ?




Logic:




  • Have an array primes to store all the prime numbers less than number


  • Loop through the odd numbers only below number to check for primes using i



  • Check if i is divisible by any of the elements already in primes.




    • If yes, isPrime = false and break the for loop for j by j=primesLength

    • If not, isPrime = true




  • If isPrime == true then add i to the array primes and check if number%i == 0




    • If number%i == 0% update the value of factor as factor = i



  • Return factor after looping through all the numbers below number



My code:






function problem3(number){

	let factor = 1;

	let primes = [2];	//array to store prime numbers



	for(let i=3; i<number; i=i+2){		//Increment i by 2 to loop through only odd numbers

		let isPrime = true;

		let primesLength= primes.length;



		for(let j=0; j< primesLength; j++){

			if(i%primes[j]==0){

				isPrime = false;

				j=primesLength;	//to break the for loop

			}

		}



		if(isPrime == true){

			primes.push(i);

			if(number%i == 0){

				factor = i;

			}

		}

	}

	return  factor;

}



console.log(problem3(600851475143));





It is working perfectly for small numbers, but is quite very slow for 600851475143. What should I change in this code to make the computation faster?






javascript beginner programming-challenge time-limit-exceeded primes







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    1












    $begingroup$


    I am self-learning js and came across this problem(#3) from the Euler Project




    The prime factors of 13195 are 5, 7, 13 and 29.



    What is the largest prime factor of the number 600851475143 ?




    Logic:




    • Have an array primes to store all the prime numbers less than number


    • Loop through the odd numbers only below number to check for primes using i



    • Check if i is divisible by any of the elements already in primes.




      • If yes, isPrime = false and break the for loop for j by j=primesLength

      • If not, isPrime = true




    • If isPrime == true then add i to the array primes and check if number%i == 0




      • If number%i == 0% update the value of factor as factor = i



    • Return factor after looping through all the numbers below number



    My code:






    function problem3(number){
    
	let factor = 1;
    
	let primes = [2];	//array to store prime numbers
    

    
	for(let i=3; i<number; i=i+2){		//Increment i by 2 to loop through only odd numbers
    
		let isPrime = true;
    
		let primesLength= primes.length;
    

    
		for(let j=0; j< primesLength; j++){
    
			if(i%primes[j]==0){
    
				isPrime = false;
    
				j=primesLength;	//to break the for loop
    
			}
    
		}
    

    
		if(isPrime == true){
    
			primes.push(i);
    
			if(number%i == 0){
    
				factor = i;
    
			}
    
		}
    
	}
    
	return  factor;
    
}
    

    
console.log(problem3(600851475143));





    It is working perfectly for small numbers, but is quite very slow for 600851475143. What should I change in this code to make the computation faster?






    javascript beginner programming-challenge time-limit-exceeded primes







    share|improve this question









    New contributor




    Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$


      I am self-learning js and came across this problem(#3) from the Euler Project




      The prime factors of 13195 are 5, 7, 13 and 29.



      What is the largest prime factor of the number 600851475143 ?




      Logic:




      • Have an array primes to store all the prime numbers less than number


      • Loop through the odd numbers only below number to check for primes using i



      • Check if i is divisible by any of the elements already in primes.




        • If yes, isPrime = false and break the for loop for j by j=primesLength

        • If not, isPrime = true




      • If isPrime == true then add i to the array primes and check if number%i == 0




        • If number%i == 0% update the value of factor as factor = i



      • Return factor after looping through all the numbers below number



      My code:






      function problem3(number){
      
	let factor = 1;
      
	let primes = [2];	//array to store prime numbers
      

      
	for(let i=3; i<number; i=i+2){		//Increment i by 2 to loop through only odd numbers
      
		let isPrime = true;
      
		let primesLength= primes.length;
      

      
		for(let j=0; j< primesLength; j++){
      
			if(i%primes[j]==0){
      
				isPrime = false;
      
				j=primesLength;	//to break the for loop
      
			}
      
		}
      

      
		if(isPrime == true){
      
			primes.push(i);
      
			if(number%i == 0){
      
				factor = i;
      
			}
      
		}
      
	}
      
	return  factor;
      
}
      

      
console.log(problem3(600851475143));





      It is working perfectly for small numbers, but is quite very slow for 600851475143. What should I change in this code to make the computation faster?






      javascript beginner programming-challenge time-limit-exceeded primes







      share|improve this question









      New contributor




      Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I am self-learning js and came across this problem(#3) from the Euler Project




      The prime factors of 13195 are 5, 7, 13 and 29.



      What is the largest prime factor of the number 600851475143 ?




      Logic:




      • Have an array primes to store all the prime numbers less than number


      • Loop through the odd numbers only below number to check for primes using i



      • Check if i is divisible by any of the elements already in primes.




        • If yes, isPrime = false and break the for loop for j by j=primesLength

        • If not, isPrime = true




      • If isPrime == true then add i to the array primes and check if number%i == 0




        • If number%i == 0% update the value of factor as factor = i



      • Return factor after looping through all the numbers below number



      My code:






      function problem3(number){
      
	let factor = 1;
      
	let primes = [2];	//array to store prime numbers
      

      
	for(let i=3; i<number; i=i+2){		//Increment i by 2 to loop through only odd numbers
      
		let isPrime = true;
      
		let primesLength= primes.length;
      

      
		for(let j=0; j< primesLength; j++){
      
			if(i%primes[j]==0){
      
				isPrime = false;
      
				j=primesLength;	//to break the for loop
      
			}
      
		}
      

      
		if(isPrime == true){
      
			primes.push(i);
      
			if(number%i == 0){
      
				factor = i;
      
			}
      
		}
      
	}
      
	return  factor;
      
}
      

      
console.log(problem3(600851475143));





      It is working perfectly for small numbers, but is quite very slow for 600851475143. What should I change in this code to make the computation faster?






      function problem3(number){
      
	let factor = 1;
      
	let primes = [2];	//array to store prime numbers
      

      
	for(let i=3; i<number; i=i+2){		//Increment i by 2 to loop through only odd numbers
      
		let isPrime = true;
      
		let primesLength= primes.length;
      

      
		for(let j=0; j< primesLength; j++){
      
			if(i%primes[j]==0){
      
				isPrime = false;
      
				j=primesLength;	//to break the for loop
      
			}
      
		}
      

      
		if(isPrime == true){
      
			primes.push(i);
      
			if(number%i == 0){
      
				factor = i;
      
			}
      
		}
      
	}
      
	return  factor;
      
}
      

      
console.log(problem3(600851475143));





      function problem3(number){
      
	let factor = 1;
      
	let primes = [2];	//array to store prime numbers
      

      
	for(let i=3; i<number; i=i+2){		//Increment i by 2 to loop through only odd numbers
      
		let isPrime = true;
      
		let primesLength= primes.length;
      

      
		for(let j=0; j< primesLength; j++){
      
			if(i%primes[j]==0){
      
				isPrime = false;
      
				j=primesLength;	//to break the for loop
      
			}
      
		}
      

      
		if(isPrime == true){
      
			primes.push(i);
      
			if(number%i == 0){
      
				factor = i;
      
			}
      
		}
      
	}
      
	return  factor;
      
}
      

      
console.log(problem3(600851475143));





      javascript beginner programming-challenge time-limit-exceeded primes




      javascript beginner programming-challenge time-limit-exceeded primes






      share|improve this question









      New contributor




      Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 1 hour ago









      200_success

      131k17157422




      131k17157422






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      asked 2 hours ago









      EagleEagle

      1085




      1085




      New contributor




      Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      Eagle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          There are many questions about Project Euler 3 on this site already. The trick is to pick an algorithm that…




          • Reduces n whenever you find a factor, so that you don't need to consider factors anywhere near as large as 600851475143

          • Only finds prime factors, and never composite factors, so that you never need to explicitly test for primality.


          Your algorithm suffers on both criteria: the outer for loop goes all the way up to 600851475143 (which is insane), and you're testing each of those numbers for primality (which is incredibly computationally expensive).






          share|improve this answer









          $endgroup$





















            1












            $begingroup$

            The first problem is that you are trying to find all prime numbers under number. The number of prime numbers under x is approximately x/ln(x) which is around 22153972243.4 for our specific value of x



            This is way too big ! So even if you where capable of obtaining each of these prime numbers in constant time it would take too much time.



            This tells us this approach is most likely unfixable.






            share|improve this answer








            New contributor




            Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$














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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              There are many questions about Project Euler 3 on this site already. The trick is to pick an algorithm that…




              • Reduces n whenever you find a factor, so that you don't need to consider factors anywhere near as large as 600851475143

              • Only finds prime factors, and never composite factors, so that you never need to explicitly test for primality.


              Your algorithm suffers on both criteria: the outer for loop goes all the way up to 600851475143 (which is insane), and you're testing each of those numbers for primality (which is incredibly computationally expensive).






              share|improve this answer









              $endgroup$


















                3












                $begingroup$

                There are many questions about Project Euler 3 on this site already. The trick is to pick an algorithm that…




                • Reduces n whenever you find a factor, so that you don't need to consider factors anywhere near as large as 600851475143

                • Only finds prime factors, and never composite factors, so that you never need to explicitly test for primality.


                Your algorithm suffers on both criteria: the outer for loop goes all the way up to 600851475143 (which is insane), and you're testing each of those numbers for primality (which is incredibly computationally expensive).






                share|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  There are many questions about Project Euler 3 on this site already. The trick is to pick an algorithm that…




                  • Reduces n whenever you find a factor, so that you don't need to consider factors anywhere near as large as 600851475143

                  • Only finds prime factors, and never composite factors, so that you never need to explicitly test for primality.


                  Your algorithm suffers on both criteria: the outer for loop goes all the way up to 600851475143 (which is insane), and you're testing each of those numbers for primality (which is incredibly computationally expensive).






                  share|improve this answer









                  $endgroup$



                  There are many questions about Project Euler 3 on this site already. The trick is to pick an algorithm that…




                  • Reduces n whenever you find a factor, so that you don't need to consider factors anywhere near as large as 600851475143

                  • Only finds prime factors, and never composite factors, so that you never need to explicitly test for primality.


                  Your algorithm suffers on both criteria: the outer for loop goes all the way up to 600851475143 (which is insane), and you're testing each of those numbers for primality (which is incredibly computationally expensive).







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  200_success200_success

                  131k17157422




                  131k17157422

























                      1












                      $begingroup$

                      The first problem is that you are trying to find all prime numbers under number. The number of prime numbers under x is approximately x/ln(x) which is around 22153972243.4 for our specific value of x



                      This is way too big ! So even if you where capable of obtaining each of these prime numbers in constant time it would take too much time.



                      This tells us this approach is most likely unfixable.






                      share|improve this answer








                      New contributor




                      Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$


















                        1












                        $begingroup$

                        The first problem is that you are trying to find all prime numbers under number. The number of prime numbers under x is approximately x/ln(x) which is around 22153972243.4 for our specific value of x



                        This is way too big ! So even if you where capable of obtaining each of these prime numbers in constant time it would take too much time.



                        This tells us this approach is most likely unfixable.






                        share|improve this answer








                        New contributor




                        Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The first problem is that you are trying to find all prime numbers under number. The number of prime numbers under x is approximately x/ln(x) which is around 22153972243.4 for our specific value of x



                          This is way too big ! So even if you where capable of obtaining each of these prime numbers in constant time it would take too much time.



                          This tells us this approach is most likely unfixable.






                          share|improve this answer








                          New contributor




                          Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$



                          The first problem is that you are trying to find all prime numbers under number. The number of prime numbers under x is approximately x/ln(x) which is around 22153972243.4 for our specific value of x



                          This is way too big ! So even if you where capable of obtaining each of these prime numbers in constant time it would take too much time.



                          This tells us this approach is most likely unfixable.







                          share|improve this answer








                          New contributor




                          Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          share|improve this answer



                          share|improve this answer






                          New contributor




                          Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          answered 52 mins ago









                          Jorge FernándezJorge Fernández

                          1645




                          1645




                          New contributor




                          Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          New contributor





                          Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          Jorge Fernández is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






















                              Eagle is a new contributor. Be nice, and check out our Code of Conduct.










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