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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 thannumber
Loop through the odd numbers only below
number
to check for primes usingi
Check if
i
is divisible by any of the elements already inprimes
.
- If yes,
isPrime = false
and break the for loop forj
byj=primesLength
- If not,
isPrime = true
- If yes,
If
isPrime == true
then addi
to the arrayprimes
and check ifnumber%i == 0
- If
number%i == 0%
update the value offactor
asfactor = i
- If
Return
factor
after looping through all the numbers belownumber
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
New contributor
$endgroup$
add a comment |
$begingroup$
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 thannumber
Loop through the odd numbers only below
number
to check for primes usingi
Check if
i
is divisible by any of the elements already inprimes
.
- If yes,
isPrime = false
and break the for loop forj
byj=primesLength
- If not,
isPrime = true
- If yes,
If
isPrime == true
then addi
to the arrayprimes
and check ifnumber%i == 0
- If
number%i == 0%
update the value offactor
asfactor = i
- If
Return
factor
after looping through all the numbers belownumber
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
New contributor
$endgroup$
add a comment |
$begingroup$
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 thannumber
Loop through the odd numbers only below
number
to check for primes usingi
Check if
i
is divisible by any of the elements already inprimes
.
- If yes,
isPrime = false
and break the for loop forj
byj=primesLength
- If not,
isPrime = true
- If yes,
If
isPrime == true
then addi
to the arrayprimes
and check ifnumber%i == 0
- If
number%i == 0%
update the value offactor
asfactor = i
- If
Return
factor
after looping through all the numbers belownumber
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
New contributor
$endgroup$
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 thannumber
Loop through the odd numbers only below
number
to check for primes usingi
Check if
i
is divisible by any of the elements already inprimes
.
- If yes,
isPrime = false
and break the for loop forj
byj=primesLength
- If not,
isPrime = true
- If yes,
If
isPrime == true
then addi
to the arrayprimes
and check ifnumber%i == 0
- If
number%i == 0%
update the value offactor
asfactor = i
- If
Return
factor
after looping through all the numbers belownumber
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
New contributor
New contributor
edited 1 hour ago
200_success
131k17157422
131k17157422
New contributor
asked 1 hour ago
EagleEagle
1085
1085
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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.
New contributor
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
add a comment |
$begingroup$
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).
$endgroup$
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).
answered 1 hour ago
200_success200_success
131k17157422
131k17157422
add a comment |
add a comment |
$begingroup$
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.
New contributor
$endgroup$
add a comment |
$begingroup$
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.
New contributor
$endgroup$
add a comment |
$begingroup$
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.
New contributor
$endgroup$
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.
New contributor
New contributor
answered 40 mins ago
Jorge FernándezJorge Fernández
1645
1645
New contributor
New contributor
add a comment |
add a comment |
Eagle is a new contributor. Be nice, and check out our Code of Conduct.
Eagle is a new contributor. Be nice, and check out our Code of Conduct.
Eagle is a new contributor. Be nice, and check out our Code of Conduct.
Eagle is a new contributor. Be nice, and check out our Code of Conduct.
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