Efficient way for generating coprime pairs
I need to print the number of coprime pairs (a,b), 0 < a <= b <= n and for n = 10^8 the program should run in less than 10 seconds. I have used this method : http://mathworld.wolfram.com/CarefreeCouple.html
But the program isn't as fast as I expected.
I have heard about an effient way of solving this problem by using something called 'Farey Sequence' but the code was written in PHP and I can only understand C.
So which method can help me solve the problem? thanks for the time.
primes greatest-common-divisor
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
add a comment |
I need to print the number of coprime pairs (a,b), 0 < a <= b <= n and for n = 10^8 the program should run in less than 10 seconds. I have used this method : http://mathworld.wolfram.com/CarefreeCouple.html
But the program isn't as fast as I expected.
I have heard about an effient way of solving this problem by using something called 'Farey Sequence' but the code was written in PHP and I can only understand C.
So which method can help me solve the problem? thanks for the time.
primes greatest-common-divisor
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
add a comment |
I need to print the number of coprime pairs (a,b), 0 < a <= b <= n and for n = 10^8 the program should run in less than 10 seconds. I have used this method : http://mathworld.wolfram.com/CarefreeCouple.html
But the program isn't as fast as I expected.
I have heard about an effient way of solving this problem by using something called 'Farey Sequence' but the code was written in PHP and I can only understand C.
So which method can help me solve the problem? thanks for the time.
primes greatest-common-divisor
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
I need to print the number of coprime pairs (a,b), 0 < a <= b <= n and for n = 10^8 the program should run in less than 10 seconds. I have used this method : http://mathworld.wolfram.com/CarefreeCouple.html
But the program isn't as fast as I expected.
I have heard about an effient way of solving this problem by using something called 'Farey Sequence' but the code was written in PHP and I can only understand C.
So which method can help me solve the problem? thanks for the time.
primes greatest-common-divisor
primes greatest-common-divisor
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
asked Nov 28 '18 at 21:25
SADBOYSSADBOYS
892
892
add a comment |
add a comment |
1 Answer
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Your stated interest is in co-prime pairs (a, b). The carefree couple adds an additional restriction that a is square-free. Therefore it is not the same problem, though some of the math is similar. As I understand your problem it is equivalent to summing the Euler totient function from 1 to n, the so-called Totient Summatory Function.
I do not know of any tricks that give one a simple closed form solution to come up with the answer. However, I think modifying a straightforward Sieve of Eratosthenes (SoT) should get you an answer in much less than 10 seconds in most programming languages.
Normally running the SoT simply yields a list of the primes <= n. Our goal will change, however, to computing the complete prime-power factorization of each integer between 1 and n inclusive. To do that we must store more than a single bit of information for each sieve entry, we must store a list. As we sieve a prime p through the array, we add (p, 1) to the list already stored at that entry. Then we sieve by p2 and change the (p,1) entries in each location we hit to (p,2), and so one for each power of p <= n and every p <= n. When it finishes, you
can compute the Totient function quickly for every value 0 <= x <= n and sum them up.
EDIT:
I see that there is already a much better set of answers to the question on math.stackexchange.com here. I'll leave this answer up for awhile until the disposition of the question is settled.
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
I have heard about a recursive way of solving my problem
– SADBOYS
Nov 30 '18 at 15:29
S(n)= (n*(n+1) )/2 -Σ{i=2..n}(S(n/i)) where S(x)=sum{i=1..x}(phi(x))
– SADBOYS
Nov 30 '18 at 15:30
add a comment |
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Your stated interest is in co-prime pairs (a, b). The carefree couple adds an additional restriction that a is square-free. Therefore it is not the same problem, though some of the math is similar. As I understand your problem it is equivalent to summing the Euler totient function from 1 to n, the so-called Totient Summatory Function.
I do not know of any tricks that give one a simple closed form solution to come up with the answer. However, I think modifying a straightforward Sieve of Eratosthenes (SoT) should get you an answer in much less than 10 seconds in most programming languages.
Normally running the SoT simply yields a list of the primes <= n. Our goal will change, however, to computing the complete prime-power factorization of each integer between 1 and n inclusive. To do that we must store more than a single bit of information for each sieve entry, we must store a list. As we sieve a prime p through the array, we add (p, 1) to the list already stored at that entry. Then we sieve by p2 and change the (p,1) entries in each location we hit to (p,2), and so one for each power of p <= n and every p <= n. When it finishes, you
can compute the Totient function quickly for every value 0 <= x <= n and sum them up.
EDIT:
I see that there is already a much better set of answers to the question on math.stackexchange.com here. I'll leave this answer up for awhile until the disposition of the question is settled.
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
I have heard about a recursive way of solving my problem
– SADBOYS
Nov 30 '18 at 15:29
S(n)= (n*(n+1) )/2 -Σ{i=2..n}(S(n/i)) where S(x)=sum{i=1..x}(phi(x))
– SADBOYS
Nov 30 '18 at 15:30
add a comment |
Your stated interest is in co-prime pairs (a, b). The carefree couple adds an additional restriction that a is square-free. Therefore it is not the same problem, though some of the math is similar. As I understand your problem it is equivalent to summing the Euler totient function from 1 to n, the so-called Totient Summatory Function.
I do not know of any tricks that give one a simple closed form solution to come up with the answer. However, I think modifying a straightforward Sieve of Eratosthenes (SoT) should get you an answer in much less than 10 seconds in most programming languages.
Normally running the SoT simply yields a list of the primes <= n. Our goal will change, however, to computing the complete prime-power factorization of each integer between 1 and n inclusive. To do that we must store more than a single bit of information for each sieve entry, we must store a list. As we sieve a prime p through the array, we add (p, 1) to the list already stored at that entry. Then we sieve by p2 and change the (p,1) entries in each location we hit to (p,2), and so one for each power of p <= n and every p <= n. When it finishes, you
can compute the Totient function quickly for every value 0 <= x <= n and sum them up.
EDIT:
I see that there is already a much better set of answers to the question on math.stackexchange.com here. I'll leave this answer up for awhile until the disposition of the question is settled.
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
I have heard about a recursive way of solving my problem
– SADBOYS
Nov 30 '18 at 15:29
S(n)= (n*(n+1) )/2 -Σ{i=2..n}(S(n/i)) where S(x)=sum{i=1..x}(phi(x))
– SADBOYS
Nov 30 '18 at 15:30
add a comment |
Your stated interest is in co-prime pairs (a, b). The carefree couple adds an additional restriction that a is square-free. Therefore it is not the same problem, though some of the math is similar. As I understand your problem it is equivalent to summing the Euler totient function from 1 to n, the so-called Totient Summatory Function.
I do not know of any tricks that give one a simple closed form solution to come up with the answer. However, I think modifying a straightforward Sieve of Eratosthenes (SoT) should get you an answer in much less than 10 seconds in most programming languages.
Normally running the SoT simply yields a list of the primes <= n. Our goal will change, however, to computing the complete prime-power factorization of each integer between 1 and n inclusive. To do that we must store more than a single bit of information for each sieve entry, we must store a list. As we sieve a prime p through the array, we add (p, 1) to the list already stored at that entry. Then we sieve by p2 and change the (p,1) entries in each location we hit to (p,2), and so one for each power of p <= n and every p <= n. When it finishes, you
can compute the Totient function quickly for every value 0 <= x <= n and sum them up.
EDIT:
I see that there is already a much better set of answers to the question on math.stackexchange.com here. I'll leave this answer up for awhile until the disposition of the question is settled.
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
Your stated interest is in co-prime pairs (a, b). The carefree couple adds an additional restriction that a is square-free. Therefore it is not the same problem, though some of the math is similar. As I understand your problem it is equivalent to summing the Euler totient function from 1 to n, the so-called Totient Summatory Function.
I do not know of any tricks that give one a simple closed form solution to come up with the answer. However, I think modifying a straightforward Sieve of Eratosthenes (SoT) should get you an answer in much less than 10 seconds in most programming languages.
Normally running the SoT simply yields a list of the primes <= n. Our goal will change, however, to computing the complete prime-power factorization of each integer between 1 and n inclusive. To do that we must store more than a single bit of information for each sieve entry, we must store a list. As we sieve a prime p through the array, we add (p, 1) to the list already stored at that entry. Then we sieve by p2 and change the (p,1) entries in each location we hit to (p,2), and so one for each power of p <= n and every p <= n. When it finishes, you
can compute the Totient function quickly for every value 0 <= x <= n and sum them up.
EDIT:
I see that there is already a much better set of answers to the question on math.stackexchange.com here. I'll leave this answer up for awhile until the disposition of the question is settled.
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
edited Nov 29 '18 at 0:06
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
answered Nov 28 '18 at 23:43
James K PolkJames K Polk
30.5k116997
30.5k116997
I have heard about a recursive way of solving my problem
– SADBOYS
Nov 30 '18 at 15:29
S(n)= (n*(n+1) )/2 -Σ{i=2..n}(S(n/i)) where S(x)=sum{i=1..x}(phi(x))
– SADBOYS
Nov 30 '18 at 15:30
add a comment |
I have heard about a recursive way of solving my problem
– SADBOYS
Nov 30 '18 at 15:29
S(n)= (n*(n+1) )/2 -Σ{i=2..n}(S(n/i)) where S(x)=sum{i=1..x}(phi(x))
– SADBOYS
Nov 30 '18 at 15:30
I have heard about a recursive way of solving my problem
– SADBOYS
Nov 30 '18 at 15:29
I have heard about a recursive way of solving my problem
– SADBOYS
Nov 30 '18 at 15:29
S(n)= (n*(n+1) )/2 -Σ{i=2..n}(S(n/i)) where S(x)=sum{i=1..x}(phi(x))
– SADBOYS
Nov 30 '18 at 15:30
S(n)= (n*(n+1) )/2 -Σ{i=2..n}(S(n/i)) where S(x)=sum{i=1..x}(phi(x))
– SADBOYS
Nov 30 '18 at 15:30
add a comment |
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