primality test speed
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Noticed using a lattice with line x+y=n for n prime, every grid point on the line is co-prime. For example:
Lattice with x+y=n
To test, I wrote a C program to specify a number (e.g., 104729) then travels along the line looking for a non co-prime point (x,y) and outputs prime or not prime:
#include "stdafx.h"
unsigned int gcd(unsigned int a, unsigned int b) {
unsigned int t;
while (b != 0) {
t = a;
a = b;
b = t % b;
}
return a;
}
int main(int argc, char *argv)
{
// x+y=n
unsigned int x, y, n;
// This number is prime:
n = 104729;
for (x = 1; x < n; x++)
{
y = n - x;
if (gcd(x, y) != 1)
{
printf("Not Prime: %un", n);
break;
}
}
if (x == n)
{
printf("Prime: %un", n);
}
return 0;
}
My maths background is very limited, so I'm wondering if it's possible to speed up the prime check using some fancy maths tricks.
Thanks.
primality-test
asked Nov 18 at 4:42
vengy
6
add a comment |
up vote
0
down vote
favorite
Noticed using a lattice with line x+y=n for n prime, every grid point on the line is co-prime. For example:
Lattice with x+y=n
To test, I wrote a C program to specify a number (e.g., 104729) then travels along the line looking for a non co-prime point (x,y) and outputs prime or not prime:
#include "stdafx.h"
unsigned int gcd(unsigned int a, unsigned int b) {
unsigned int t;
while (b != 0) {
t = a;
a = b;
b = t % b;
}
return a;
}
int main(int argc, char *argv)
{
// x+y=n
unsigned int x, y, n;
// This number is prime:
n = 104729;
for (x = 1; x < n; x++)
{
y = n - x;
if (gcd(x, y) != 1)
{
printf("Not Prime: %un", n);
break;
}
}
if (x == n)
{
printf("Prime: %un", n);
}
return 0;
}
My maths background is very limited, so I'm wondering if it's possible to speed up the prime check using some fancy maths tricks.
Thanks.
primality-test
asked Nov 18 at 4:42
vengy
6
2
I suspect this test is much slower than trial division.
– Lord Shark the Unknown
Nov 18 at 4:52
Your current algorithm will run in $O(n log n)$ time if $n$ is a prime (the $log $ comes from $mathsf{gcd}$), while you can decide primality in $O(n^{1/2})$ time in the worst case. Indeed, if $n$ is not prime, then its smallest divisor larger than $1$, must not exceed $n^{1/2}$, since if $d$ divides $n$ then $n/d$ also divides $n$, thus both cannot become larger than $n^{1/2}$. Thus, all you need is to check the numbers from $2...[n^{1/2}] + 1 $ to see if any of them divides $n$. If none of them does, then $n$ is prime, otherwise you have non prime.
– Hayk
Nov 18 at 5:32
Nice explanation...Thanks!
– vengy
Nov 20 at 0:22
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Noticed using a lattice with line x+y=n for n prime, every grid point on the line is co-prime. For example:
Lattice with x+y=n
To test, I wrote a C program to specify a number (e.g., 104729) then travels along the line looking for a non co-prime point (x,y) and outputs prime or not prime:
#include "stdafx.h"
unsigned int gcd(unsigned int a, unsigned int b) {
unsigned int t;
while (b != 0) {
t = a;
a = b;
b = t % b;
}
return a;
}
int main(int argc, char *argv)
{
// x+y=n
unsigned int x, y, n;
// This number is prime:
n = 104729;
for (x = 1; x < n; x++)
{
y = n - x;
if (gcd(x, y) != 1)
{
printf("Not Prime: %un", n);
break;
}
}
if (x == n)
{
printf("Prime: %un", n);
}
return 0;
}
My maths background is very limited, so I'm wondering if it's possible to speed up the prime check using some fancy maths tricks.
Thanks.
primality-test
asked Nov 18 at 4:42
vengy
6
Noticed using a lattice with line x+y=n for n prime, every grid point on the line is co-prime. For example:
Lattice with x+y=n
To test, I wrote a C program to specify a number (e.g., 104729) then travels along the line looking for a non co-prime point (x,y) and outputs prime or not prime:
#include "stdafx.h"
unsigned int gcd(unsigned int a, unsigned int b) {
unsigned int t;
while (b != 0) {
t = a;
a = b;
b = t % b;
}
return a;
}
int main(int argc, char *argv)
{
// x+y=n
unsigned int x, y, n;
// This number is prime:
n = 104729;
for (x = 1; x < n; x++)
{
y = n - x;
if (gcd(x, y) != 1)
{
printf("Not Prime: %un", n);
break;
}
}
if (x == n)
{
printf("Prime: %un", n);
}
return 0;
}
My maths background is very limited, so I'm wondering if it's possible to speed up the prime check using some fancy maths tricks.
Thanks.
primality-test
primality-test
asked Nov 18 at 4:42
vengy
6
asked Nov 18 at 4:42
vengy
6
asked Nov 18 at 4:42
vengy
6
asked Nov 18 at 4:42
vengy
6
asked Nov 18 at 4:42
vengy
6
6
2
I suspect this test is much slower than trial division.
– Lord Shark the Unknown
Nov 18 at 4:52
Your current algorithm will run in $O(n log n)$ time if $n$ is a prime (the $log $ comes from $mathsf{gcd}$), while you can decide primality in $O(n^{1/2})$ time in the worst case. Indeed, if $n$ is not prime, then its smallest divisor larger than $1$, must not exceed $n^{1/2}$, since if $d$ divides $n$ then $n/d$ also divides $n$, thus both cannot become larger than $n^{1/2}$. Thus, all you need is to check the numbers from $2...[n^{1/2}] + 1 $ to see if any of them divides $n$. If none of them does, then $n$ is prime, otherwise you have non prime.
– Hayk
Nov 18 at 5:32
Nice explanation...Thanks!
– vengy
Nov 20 at 0:22
add a comment |
2
I suspect this test is much slower than trial division.
– Lord Shark the Unknown
Nov 18 at 4:52
Your current algorithm will run in $O(n log n)$ time if $n$ is a prime (the $log $ comes from $mathsf{gcd}$), while you can decide primality in $O(n^{1/2})$ time in the worst case. Indeed, if $n$ is not prime, then its smallest divisor larger than $1$, must not exceed $n^{1/2}$, since if $d$ divides $n$ then $n/d$ also divides $n$, thus both cannot become larger than $n^{1/2}$. Thus, all you need is to check the numbers from $2...[n^{1/2}] + 1 $ to see if any of them divides $n$. If none of them does, then $n$ is prime, otherwise you have non prime.
– Hayk
Nov 18 at 5:32
Nice explanation...Thanks!
– vengy
Nov 20 at 0:22
2
2
I suspect this test is much slower than trial division.
– Lord Shark the Unknown
Nov 18 at 4:52
I suspect this test is much slower than trial division.
– Lord Shark the Unknown
Nov 18 at 4:52
Your current algorithm will run in $O(n log n)$ time if $n$ is a prime (the $log $ comes from $mathsf{gcd}$), while you can decide primality in $O(n^{1/2})$ time in the worst case. Indeed, if $n$ is not prime, then its smallest divisor larger than $1$, must not exceed $n^{1/2}$, since if $d$ divides $n$ then $n/d$ also divides $n$, thus both cannot become larger than $n^{1/2}$. Thus, all you need is to check the numbers from $2...[n^{1/2}] + 1 $ to see if any of them divides $n$. If none of them does, then $n$ is prime, otherwise you have non prime.
– Hayk
Nov 18 at 5:32
Your current algorithm will run in $O(n log n)$ time if $n$ is a prime (the $log $ comes from $mathsf{gcd}$), while you can decide primality in $O(n^{1/2})$ time in the worst case. Indeed, if $n$ is not prime, then its smallest divisor larger than $1$, must not exceed $n^{1/2}$, since if $d$ divides $n$ then $n/d$ also divides $n$, thus both cannot become larger than $n^{1/2}$. Thus, all you need is to check the numbers from $2...[n^{1/2}] + 1 $ to see if any of them divides $n$. If none of them does, then $n$ is prime, otherwise you have non prime.
– Hayk
Nov 18 at 5:32
Nice explanation...Thanks!
– vengy
Nov 20 at 0:22
Nice explanation...Thanks!
– vengy
Nov 20 at 0:22
add a comment |
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2
I suspect this test is much slower than trial division.
– Lord Shark the Unknown
Nov 18 at 4:52
Your current algorithm will run in $O(n log n)$ time if $n$ is a prime (the $log $ comes from $mathsf{gcd}$), while you can decide primality in $O(n^{1/2})$ time in the worst case. Indeed, if $n$ is not prime, then its smallest divisor larger than $1$, must not exceed $n^{1/2}$, since if $d$ divides $n$ then $n/d$ also divides $n$, thus both cannot become larger than $n^{1/2}$. Thus, all you need is to check the numbers from $2...[n^{1/2}] + 1 $ to see if any of them divides $n$. If none of them does, then $n$ is prime, otherwise you have non prime.
– Hayk
Nov 18 at 5:32
Nice explanation...Thanks!
– vengy
Nov 20 at 0:22