Fastest prime generating algorithm
$begingroup$
What is the fastest known algorithm that generates all distinct prime numbers less than n?
Is it faster than Sieve of Atkin?
algorithms prime-numbers
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
$endgroup$
add a comment |
$begingroup$
What is the fastest known algorithm that generates all distinct prime numbers less than n?
Is it faster than Sieve of Atkin?
algorithms prime-numbers
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
$endgroup$
$begingroup$
You can't generate all prime numbers nor an infinite subset of all prime numbers in finite time...
$endgroup$
– FUZxxl
Sep 29 '11 at 9:02
4
$begingroup$
Ok, time to reopen
$endgroup$
– grok_it
Sep 29 '11 at 13:09
$begingroup$
Sieve of Eratosthenes for reference takes O(n)+P(n)(n/k) time and P(n)log(n)+k space. Iteratively apply the sieve to the next k numbers; for each prime keep track of the smallest prime multiple you have seen.
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:27
add a comment |
$begingroup$
What is the fastest known algorithm that generates all distinct prime numbers less than n?
Is it faster than Sieve of Atkin?
algorithms prime-numbers
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
$endgroup$
What is the fastest known algorithm that generates all distinct prime numbers less than n?
Is it faster than Sieve of Atkin?
algorithms prime-numbers
algorithms prime-numbers
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
edited Nov 16 '12 at 17:32
Martin Argerami
128k1184184
128k1184184
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
asked Sep 29 '11 at 8:58
grok_itgrok_it
2621210
2621210
$begingroup$
You can't generate all prime numbers nor an infinite subset of all prime numbers in finite time...
$endgroup$
– FUZxxl
Sep 29 '11 at 9:02
4
$begingroup$
Ok, time to reopen
$endgroup$
– grok_it
Sep 29 '11 at 13:09
$begingroup$
Sieve of Eratosthenes for reference takes O(n)+P(n)(n/k) time and P(n)log(n)+k space. Iteratively apply the sieve to the next k numbers; for each prime keep track of the smallest prime multiple you have seen.
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:27
add a comment |
$begingroup$
You can't generate all prime numbers nor an infinite subset of all prime numbers in finite time...
$endgroup$
– FUZxxl
Sep 29 '11 at 9:02
4
$begingroup$
Ok, time to reopen
$endgroup$
– grok_it
Sep 29 '11 at 13:09
$begingroup$
Sieve of Eratosthenes for reference takes O(n)+P(n)(n/k) time and P(n)log(n)+k space. Iteratively apply the sieve to the next k numbers; for each prime keep track of the smallest prime multiple you have seen.
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:27
$begingroup$
You can't generate all prime numbers nor an infinite subset of all prime numbers in finite time...
$endgroup$
– FUZxxl
Sep 29 '11 at 9:02
$begingroup$
You can't generate all prime numbers nor an infinite subset of all prime numbers in finite time...
$endgroup$
– FUZxxl
Sep 29 '11 at 9:02
4
4
$begingroup$
Ok, time to reopen
$endgroup$
– grok_it
Sep 29 '11 at 13:09
$begingroup$
Ok, time to reopen
$endgroup$
– grok_it
Sep 29 '11 at 13:09
$begingroup$
Sieve of Eratosthenes for reference takes O(n)+P(n)(n/k) time and P(n)log(n)+k space. Iteratively apply the sieve to the next k numbers; for each prime keep track of the smallest prime multiple you have seen.
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:27
$begingroup$
Sieve of Eratosthenes for reference takes O(n)+P(n)(n/k) time and P(n)log(n)+k space. Iteratively apply the sieve to the next k numbers; for each prime keep track of the smallest prime multiple you have seen.
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:27
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
What is the fastest known algorithm that generates all prime numbers?
I assume you mean: Given $n$, what is the fastest known algorithm that generates all prime numbers $p le n$ ? Currently it is the Sieve of Atkin.
And what is the fastest known algorithm that generates any infinite subset of the prime numbers?
Again, I assume you mean: Given $n$, how fast can I generate $n$ distinct primes? There might be a faster method than the Sieve of Atkin, but I don't know of any. A good question!
And is there a theoretical lowest possible O(n) of such programs?
Is $n$ the number of primes you want to generate? Then it would take $O(n)$ operations just to store them in memory. So yes. But if you want to generate all primes $p le n$ , the Sieve of Atkin has time complexity $O(n/log log n)$ . So no.
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
$endgroup$
7
$begingroup$
@Robert: So? The question was about: asymptotic time complexity. The OP specifically used the notation O(n).
$endgroup$
– TonyK
Sep 29 '11 at 11:26
1
$begingroup$
In practice, the Atkin-Bernstein sieve isn't the fastest at any range. Sieve of Eratosthenes variants outperform it at all sizes.
$endgroup$
– Charles
Sep 29 '11 at 13:19
1
$begingroup$
@Charles: the paper at cr.yp.to/papers/primesieves-19990826.pdf suggests otherwise - Atkin-Bernstein outperformed Eratosthenes by 15.0 seconds to 19.6 seconds in generating all primes up to $1000000000$ . The fact that the benchmark programs were written by one of the creators (Bernstein) of the Aktin Sieve means that we have to be cautious in interpreting them; but knowing Bernstein's work, and given that higher numbers are now reachable (the paper was written in 1999), I think you must be wrong.
$endgroup$
– TonyK
Sep 29 '11 at 13:31
1
$begingroup$
@TonyK: I have only the highest respect for Dr. Bernstein--he does great theoretical work that is very practically accessible. But if you download yafu and primegen right now I think you'll see that yafu outperforms slightly for 32-bit limits and greatly outperforms above that point.
$endgroup$
– Charles
Sep 29 '11 at 13:34
3
$begingroup$
In base ten we know anything with more than one digit ending in 0,2,4,5,6,8 is nonprime. For tractable $n$ why not just use base 2*3*5*7*11*13*17*19...k? You can then make a suffix list of known nonprimes in the base up to k which you avoid, and from this you throw it against a sieve that includes everything you have found. Every so often you rebase. Ideally you want to use only P(n)*log(n) space. Atkin takes O(n) memory too...
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:03
|
show 8 more comments
$begingroup$
I recently just chanced upon a particular logic. All prime numbers either fall in multiples of 6n+1 or 6n-1 except for 2 and 3.
Using this the search space could be considerably reduced for applying any sieve such as Atkins or Eratosthenes. Hence making it a slightly better form of finding primes.
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
$endgroup$
2
$begingroup$
This is en.wikipedia.org/wiki/Wheel_factorization for n=2*3.
$endgroup$
– Iiridayn
Nov 29 '14 at 19:12
$begingroup$
@sohaib, in essence it is enough to consider 2/6 = 1/3 of N to get all the primes below N (since we need to consider only the two progressions (6k+1) and (6k-1) and add 2 at the end to account for primes 2 and 3. One can even write pi(n)+c(n)=N/3. Here, c(n) is the number of composite within the two progressions. And there is an easy way to get those if you work with indices instead of numbers.
$endgroup$
– user25406
Dec 18 '16 at 12:18
add a comment |
$begingroup$
The sieve of Eratosthenes is one of the most efficient ways to find all the prime number less than n.
Implemenation of Algorithm
answered May 1 '16 at 11:52
BadalBadal
1
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add a comment |
$begingroup$
Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.
Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.
Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuting and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.
A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - n. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
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add a comment |
protected by dxiv Dec 22 '17 at 6:46
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What is the fastest known algorithm that generates all prime numbers?
I assume you mean: Given $n$, what is the fastest known algorithm that generates all prime numbers $p le n$ ? Currently it is the Sieve of Atkin.
And what is the fastest known algorithm that generates any infinite subset of the prime numbers?
Again, I assume you mean: Given $n$, how fast can I generate $n$ distinct primes? There might be a faster method than the Sieve of Atkin, but I don't know of any. A good question!
And is there a theoretical lowest possible O(n) of such programs?
Is $n$ the number of primes you want to generate? Then it would take $O(n)$ operations just to store them in memory. So yes. But if you want to generate all primes $p le n$ , the Sieve of Atkin has time complexity $O(n/log log n)$ . So no.
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
$endgroup$
7
$begingroup$
@Robert: So? The question was about: asymptotic time complexity. The OP specifically used the notation O(n).
$endgroup$
– TonyK
Sep 29 '11 at 11:26
1
$begingroup$
In practice, the Atkin-Bernstein sieve isn't the fastest at any range. Sieve of Eratosthenes variants outperform it at all sizes.
$endgroup$
– Charles
Sep 29 '11 at 13:19
1
$begingroup$
@Charles: the paper at cr.yp.to/papers/primesieves-19990826.pdf suggests otherwise - Atkin-Bernstein outperformed Eratosthenes by 15.0 seconds to 19.6 seconds in generating all primes up to $1000000000$ . The fact that the benchmark programs were written by one of the creators (Bernstein) of the Aktin Sieve means that we have to be cautious in interpreting them; but knowing Bernstein's work, and given that higher numbers are now reachable (the paper was written in 1999), I think you must be wrong.
$endgroup$
– TonyK
Sep 29 '11 at 13:31
1
$begingroup$
@TonyK: I have only the highest respect for Dr. Bernstein--he does great theoretical work that is very practically accessible. But if you download yafu and primegen right now I think you'll see that yafu outperforms slightly for 32-bit limits and greatly outperforms above that point.
$endgroup$
– Charles
Sep 29 '11 at 13:34
3
$begingroup$
In base ten we know anything with more than one digit ending in 0,2,4,5,6,8 is nonprime. For tractable $n$ why not just use base 2*3*5*7*11*13*17*19...k? You can then make a suffix list of known nonprimes in the base up to k which you avoid, and from this you throw it against a sieve that includes everything you have found. Every so often you rebase. Ideally you want to use only P(n)*log(n) space. Atkin takes O(n) memory too...
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:03
|
show 8 more comments
$begingroup$
What is the fastest known algorithm that generates all prime numbers?
I assume you mean: Given $n$, what is the fastest known algorithm that generates all prime numbers $p le n$ ? Currently it is the Sieve of Atkin.
And what is the fastest known algorithm that generates any infinite subset of the prime numbers?
Again, I assume you mean: Given $n$, how fast can I generate $n$ distinct primes? There might be a faster method than the Sieve of Atkin, but I don't know of any. A good question!
And is there a theoretical lowest possible O(n) of such programs?
Is $n$ the number of primes you want to generate? Then it would take $O(n)$ operations just to store them in memory. So yes. But if you want to generate all primes $p le n$ , the Sieve of Atkin has time complexity $O(n/log log n)$ . So no.
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
$endgroup$
7
$begingroup$
@Robert: So? The question was about: asymptotic time complexity. The OP specifically used the notation O(n).
$endgroup$
– TonyK
Sep 29 '11 at 11:26
1
$begingroup$
In practice, the Atkin-Bernstein sieve isn't the fastest at any range. Sieve of Eratosthenes variants outperform it at all sizes.
$endgroup$
– Charles
Sep 29 '11 at 13:19
1
$begingroup$
@Charles: the paper at cr.yp.to/papers/primesieves-19990826.pdf suggests otherwise - Atkin-Bernstein outperformed Eratosthenes by 15.0 seconds to 19.6 seconds in generating all primes up to $1000000000$ . The fact that the benchmark programs were written by one of the creators (Bernstein) of the Aktin Sieve means that we have to be cautious in interpreting them; but knowing Bernstein's work, and given that higher numbers are now reachable (the paper was written in 1999), I think you must be wrong.
$endgroup$
– TonyK
Sep 29 '11 at 13:31
1
$begingroup$
@TonyK: I have only the highest respect for Dr. Bernstein--he does great theoretical work that is very practically accessible. But if you download yafu and primegen right now I think you'll see that yafu outperforms slightly for 32-bit limits and greatly outperforms above that point.
$endgroup$
– Charles
Sep 29 '11 at 13:34
3
$begingroup$
In base ten we know anything with more than one digit ending in 0,2,4,5,6,8 is nonprime. For tractable $n$ why not just use base 2*3*5*7*11*13*17*19...k? You can then make a suffix list of known nonprimes in the base up to k which you avoid, and from this you throw it against a sieve that includes everything you have found. Every so often you rebase. Ideally you want to use only P(n)*log(n) space. Atkin takes O(n) memory too...
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:03
|
show 8 more comments
$begingroup$
What is the fastest known algorithm that generates all prime numbers?
I assume you mean: Given $n$, what is the fastest known algorithm that generates all prime numbers $p le n$ ? Currently it is the Sieve of Atkin.
And what is the fastest known algorithm that generates any infinite subset of the prime numbers?
Again, I assume you mean: Given $n$, how fast can I generate $n$ distinct primes? There might be a faster method than the Sieve of Atkin, but I don't know of any. A good question!
And is there a theoretical lowest possible O(n) of such programs?
Is $n$ the number of primes you want to generate? Then it would take $O(n)$ operations just to store them in memory. So yes. But if you want to generate all primes $p le n$ , the Sieve of Atkin has time complexity $O(n/log log n)$ . So no.
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
$endgroup$
What is the fastest known algorithm that generates all prime numbers?
I assume you mean: Given $n$, what is the fastest known algorithm that generates all prime numbers $p le n$ ? Currently it is the Sieve of Atkin.
And what is the fastest known algorithm that generates any infinite subset of the prime numbers?
Again, I assume you mean: Given $n$, how fast can I generate $n$ distinct primes? There might be a faster method than the Sieve of Atkin, but I don't know of any. A good question!
And is there a theoretical lowest possible O(n) of such programs?
Is $n$ the number of primes you want to generate? Then it would take $O(n)$ operations just to store them in memory. So yes. But if you want to generate all primes $p le n$ , the Sieve of Atkin has time complexity $O(n/log log n)$ . So no.
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
answered Sep 29 '11 at 9:30
TonyKTonyK
43k356135
43k356135
7
$begingroup$
@Robert: So? The question was about: asymptotic time complexity. The OP specifically used the notation O(n).
$endgroup$
– TonyK
Sep 29 '11 at 11:26
1
$begingroup$
In practice, the Atkin-Bernstein sieve isn't the fastest at any range. Sieve of Eratosthenes variants outperform it at all sizes.
$endgroup$
– Charles
Sep 29 '11 at 13:19
1
$begingroup$
@Charles: the paper at cr.yp.to/papers/primesieves-19990826.pdf suggests otherwise - Atkin-Bernstein outperformed Eratosthenes by 15.0 seconds to 19.6 seconds in generating all primes up to $1000000000$ . The fact that the benchmark programs were written by one of the creators (Bernstein) of the Aktin Sieve means that we have to be cautious in interpreting them; but knowing Bernstein's work, and given that higher numbers are now reachable (the paper was written in 1999), I think you must be wrong.
$endgroup$
– TonyK
Sep 29 '11 at 13:31
1
$begingroup$
@TonyK: I have only the highest respect for Dr. Bernstein--he does great theoretical work that is very practically accessible. But if you download yafu and primegen right now I think you'll see that yafu outperforms slightly for 32-bit limits and greatly outperforms above that point.
$endgroup$
– Charles
Sep 29 '11 at 13:34
3
$begingroup$
In base ten we know anything with more than one digit ending in 0,2,4,5,6,8 is nonprime. For tractable $n$ why not just use base 2*3*5*7*11*13*17*19...k? You can then make a suffix list of known nonprimes in the base up to k which you avoid, and from this you throw it against a sieve that includes everything you have found. Every so often you rebase. Ideally you want to use only P(n)*log(n) space. Atkin takes O(n) memory too...
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:03
|
show 8 more comments
7
$begingroup$
@Robert: So? The question was about: asymptotic time complexity. The OP specifically used the notation O(n).
$endgroup$
– TonyK
Sep 29 '11 at 11:26
1
$begingroup$
In practice, the Atkin-Bernstein sieve isn't the fastest at any range. Sieve of Eratosthenes variants outperform it at all sizes.
$endgroup$
– Charles
Sep 29 '11 at 13:19
1
$begingroup$
@Charles: the paper at cr.yp.to/papers/primesieves-19990826.pdf suggests otherwise - Atkin-Bernstein outperformed Eratosthenes by 15.0 seconds to 19.6 seconds in generating all primes up to $1000000000$ . The fact that the benchmark programs were written by one of the creators (Bernstein) of the Aktin Sieve means that we have to be cautious in interpreting them; but knowing Bernstein's work, and given that higher numbers are now reachable (the paper was written in 1999), I think you must be wrong.
$endgroup$
– TonyK
Sep 29 '11 at 13:31
1
$begingroup$
@TonyK: I have only the highest respect for Dr. Bernstein--he does great theoretical work that is very practically accessible. But if you download yafu and primegen right now I think you'll see that yafu outperforms slightly for 32-bit limits and greatly outperforms above that point.
$endgroup$
– Charles
Sep 29 '11 at 13:34
3
$begingroup$
In base ten we know anything with more than one digit ending in 0,2,4,5,6,8 is nonprime. For tractable $n$ why not just use base 2*3*5*7*11*13*17*19...k? You can then make a suffix list of known nonprimes in the base up to k which you avoid, and from this you throw it against a sieve that includes everything you have found. Every so often you rebase. Ideally you want to use only P(n)*log(n) space. Atkin takes O(n) memory too...
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:03
7
7
$begingroup$
@Robert: So? The question was about: asymptotic time complexity. The OP specifically used the notation O(n).
$endgroup$
– TonyK
Sep 29 '11 at 11:26
$begingroup$
@Robert: So? The question was about: asymptotic time complexity. The OP specifically used the notation O(n).
$endgroup$
– TonyK
Sep 29 '11 at 11:26
1
1
$begingroup$
In practice, the Atkin-Bernstein sieve isn't the fastest at any range. Sieve of Eratosthenes variants outperform it at all sizes.
$endgroup$
– Charles
Sep 29 '11 at 13:19
$begingroup$
In practice, the Atkin-Bernstein sieve isn't the fastest at any range. Sieve of Eratosthenes variants outperform it at all sizes.
$endgroup$
– Charles
Sep 29 '11 at 13:19
1
1
$begingroup$
@Charles: the paper at cr.yp.to/papers/primesieves-19990826.pdf suggests otherwise - Atkin-Bernstein outperformed Eratosthenes by 15.0 seconds to 19.6 seconds in generating all primes up to $1000000000$ . The fact that the benchmark programs were written by one of the creators (Bernstein) of the Aktin Sieve means that we have to be cautious in interpreting them; but knowing Bernstein's work, and given that higher numbers are now reachable (the paper was written in 1999), I think you must be wrong.
$endgroup$
– TonyK
Sep 29 '11 at 13:31
$begingroup$
@Charles: the paper at cr.yp.to/papers/primesieves-19990826.pdf suggests otherwise - Atkin-Bernstein outperformed Eratosthenes by 15.0 seconds to 19.6 seconds in generating all primes up to $1000000000$ . The fact that the benchmark programs were written by one of the creators (Bernstein) of the Aktin Sieve means that we have to be cautious in interpreting them; but knowing Bernstein's work, and given that higher numbers are now reachable (the paper was written in 1999), I think you must be wrong.
$endgroup$
– TonyK
Sep 29 '11 at 13:31
1
1
$begingroup$
@TonyK: I have only the highest respect for Dr. Bernstein--he does great theoretical work that is very practically accessible. But if you download yafu and primegen right now I think you'll see that yafu outperforms slightly for 32-bit limits and greatly outperforms above that point.
$endgroup$
– Charles
Sep 29 '11 at 13:34
$begingroup$
@TonyK: I have only the highest respect for Dr. Bernstein--he does great theoretical work that is very practically accessible. But if you download yafu and primegen right now I think you'll see that yafu outperforms slightly for 32-bit limits and greatly outperforms above that point.
$endgroup$
– Charles
Sep 29 '11 at 13:34
3
3
$begingroup$
In base ten we know anything with more than one digit ending in 0,2,4,5,6,8 is nonprime. For tractable $n$ why not just use base 2*3*5*7*11*13*17*19...k? You can then make a suffix list of known nonprimes in the base up to k which you avoid, and from this you throw it against a sieve that includes everything you have found. Every so often you rebase. Ideally you want to use only P(n)*log(n) space. Atkin takes O(n) memory too...
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:03
$begingroup$
In base ten we know anything with more than one digit ending in 0,2,4,5,6,8 is nonprime. For tractable $n$ why not just use base 2*3*5*7*11*13*17*19...k? You can then make a suffix list of known nonprimes in the base up to k which you avoid, and from this you throw it against a sieve that includes everything you have found. Every so often you rebase. Ideally you want to use only P(n)*log(n) space. Atkin takes O(n) memory too...
$endgroup$
– Chad Brewbaker
Nov 16 '12 at 16:03
|
show 8 more comments
$begingroup$
I recently just chanced upon a particular logic. All prime numbers either fall in multiples of 6n+1 or 6n-1 except for 2 and 3.
Using this the search space could be considerably reduced for applying any sieve such as Atkins or Eratosthenes. Hence making it a slightly better form of finding primes.
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
$endgroup$
2
$begingroup$
This is en.wikipedia.org/wiki/Wheel_factorization for n=2*3.
$endgroup$
– Iiridayn
Nov 29 '14 at 19:12
$begingroup$
@sohaib, in essence it is enough to consider 2/6 = 1/3 of N to get all the primes below N (since we need to consider only the two progressions (6k+1) and (6k-1) and add 2 at the end to account for primes 2 and 3. One can even write pi(n)+c(n)=N/3. Here, c(n) is the number of composite within the two progressions. And there is an easy way to get those if you work with indices instead of numbers.
$endgroup$
– user25406
Dec 18 '16 at 12:18
add a comment |
$begingroup$
I recently just chanced upon a particular logic. All prime numbers either fall in multiples of 6n+1 or 6n-1 except for 2 and 3.
Using this the search space could be considerably reduced for applying any sieve such as Atkins or Eratosthenes. Hence making it a slightly better form of finding primes.
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
$endgroup$
2
$begingroup$
This is en.wikipedia.org/wiki/Wheel_factorization for n=2*3.
$endgroup$
– Iiridayn
Nov 29 '14 at 19:12
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@sohaib, in essence it is enough to consider 2/6 = 1/3 of N to get all the primes below N (since we need to consider only the two progressions (6k+1) and (6k-1) and add 2 at the end to account for primes 2 and 3. One can even write pi(n)+c(n)=N/3. Here, c(n) is the number of composite within the two progressions. And there is an easy way to get those if you work with indices instead of numbers.
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– user25406
Dec 18 '16 at 12:18
add a comment |
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I recently just chanced upon a particular logic. All prime numbers either fall in multiples of 6n+1 or 6n-1 except for 2 and 3.
Using this the search space could be considerably reduced for applying any sieve such as Atkins or Eratosthenes. Hence making it a slightly better form of finding primes.
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
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I recently just chanced upon a particular logic. All prime numbers either fall in multiples of 6n+1 or 6n-1 except for 2 and 3.
Using this the search space could be considerably reduced for applying any sieve such as Atkins or Eratosthenes. Hence making it a slightly better form of finding primes.
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
answered Jan 23 '14 at 16:07
Sohaib ISohaib I
2611211
2611211
2
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This is en.wikipedia.org/wiki/Wheel_factorization for n=2*3.
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– Iiridayn
Nov 29 '14 at 19:12
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@sohaib, in essence it is enough to consider 2/6 = 1/3 of N to get all the primes below N (since we need to consider only the two progressions (6k+1) and (6k-1) and add 2 at the end to account for primes 2 and 3. One can even write pi(n)+c(n)=N/3. Here, c(n) is the number of composite within the two progressions. And there is an easy way to get those if you work with indices instead of numbers.
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– user25406
Dec 18 '16 at 12:18
add a comment |
2
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This is en.wikipedia.org/wiki/Wheel_factorization for n=2*3.
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– Iiridayn
Nov 29 '14 at 19:12
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@sohaib, in essence it is enough to consider 2/6 = 1/3 of N to get all the primes below N (since we need to consider only the two progressions (6k+1) and (6k-1) and add 2 at the end to account for primes 2 and 3. One can even write pi(n)+c(n)=N/3. Here, c(n) is the number of composite within the two progressions. And there is an easy way to get those if you work with indices instead of numbers.
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– user25406
Dec 18 '16 at 12:18
2
2
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This is en.wikipedia.org/wiki/Wheel_factorization for n=2*3.
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– Iiridayn
Nov 29 '14 at 19:12
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This is en.wikipedia.org/wiki/Wheel_factorization for n=2*3.
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– Iiridayn
Nov 29 '14 at 19:12
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@sohaib, in essence it is enough to consider 2/6 = 1/3 of N to get all the primes below N (since we need to consider only the two progressions (6k+1) and (6k-1) and add 2 at the end to account for primes 2 and 3. One can even write pi(n)+c(n)=N/3. Here, c(n) is the number of composite within the two progressions. And there is an easy way to get those if you work with indices instead of numbers.
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– user25406
Dec 18 '16 at 12:18
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@sohaib, in essence it is enough to consider 2/6 = 1/3 of N to get all the primes below N (since we need to consider only the two progressions (6k+1) and (6k-1) and add 2 at the end to account for primes 2 and 3. One can even write pi(n)+c(n)=N/3. Here, c(n) is the number of composite within the two progressions. And there is an easy way to get those if you work with indices instead of numbers.
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– user25406
Dec 18 '16 at 12:18
add a comment |
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The sieve of Eratosthenes is one of the most efficient ways to find all the prime number less than n.
Implemenation of Algorithm
answered May 1 '16 at 11:52
BadalBadal
1
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add a comment |
$begingroup$
The sieve of Eratosthenes is one of the most efficient ways to find all the prime number less than n.
Implemenation of Algorithm
answered May 1 '16 at 11:52
BadalBadal
1
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add a comment |
$begingroup$
The sieve of Eratosthenes is one of the most efficient ways to find all the prime number less than n.
Implemenation of Algorithm
answered May 1 '16 at 11:52
BadalBadal
1
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The sieve of Eratosthenes is one of the most efficient ways to find all the prime number less than n.
Implemenation of Algorithm
answered May 1 '16 at 11:52
BadalBadal
1
answered May 1 '16 at 11:52
BadalBadal
1
answered May 1 '16 at 11:52
BadalBadal
1
answered May 1 '16 at 11:52
BadalBadal
1
1
add a comment |
add a comment |
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Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.
Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.
Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuting and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.
A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - n. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
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add a comment |
$begingroup$
Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.
Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.
Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuting and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.
A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - n. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
$endgroup$
add a comment |
$begingroup$
Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.
Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.
Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuting and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.
A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - n. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
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Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.
Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.
Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuting and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.
A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - n. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
edited Dec 25 '18 at 21:14
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
answered Dec 25 '18 at 20:46
S SpringS Spring
1723
1723
add a comment |
add a comment |
protected by dxiv Dec 22 '17 at 6:46
Thank you for your interest in this question.
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You can't generate all prime numbers nor an infinite subset of all prime numbers in finite time...
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– FUZxxl
Sep 29 '11 at 9:02
4
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Ok, time to reopen
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– grok_it
Sep 29 '11 at 13:09
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Sieve of Eratosthenes for reference takes O(n)+P(n)(n/k) time and P(n)log(n)+k space. Iteratively apply the sieve to the next k numbers; for each prime keep track of the smallest prime multiple you have seen.
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– Chad Brewbaker
Nov 16 '12 at 16:27