Comments on: Finding Primes http://cow.neondragon.net/index.php/1765-Finding-Primes Helping you to get the most out of modern technology and communications since 2004. Tue, 07 Feb 2012 14:38:07 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: Lewis http://cow.neondragon.net/index.php/1765-Finding-Primes/comment-page-1#comment-1165 Lewis Wed, 30 Nov -0001 00:00:00 +0000 #comment-1165 <p>Couldn't you just find it for 1001 and times it by 10? <br /></p><p> </p><p>hahaha I crack myself up! <img src="/ui/emoticons/pinch.png" alt="" /></p> Couldn’t you just find it for 1001 and times it by 10?

 

hahaha I crack myself up!

]]> By: Tim http://cow.neondragon.net/index.php/1765-Finding-Primes/comment-page-1#comment-1166 Tim Wed, 30 Nov -0001 00:00:00 +0000 #comment-1166 If you want to be a bit smarter, only use primes-up-to-sqrt(M) as denominators, too. Note that this leads to two mutually recursive functions, but runs hellish fast because of the way primes thin-out over increasing N. http://pig.sty.nu/temp/isprime-q.txt (Oh yeah. I'm using a hash in there of previously-computed primes; and I print out true or false for 10 numbers either side of N. Benchmark it! Let there be gnuplot graphs!) If you want to be a bit smarter, only use primes-up-to-sqrt(M) as denominators, too.

Note that this leads to two mutually recursive functions, but runs hellish fast because of the way primes thin-out over increasing N.

http://pig.sty.nu/temp/isprime-q.txt

(Oh yeah. I’m using a hash in there of previously-computed primes; and I print out true or false for 10 numbers either side of N. Benchmark it! Let there be gnuplot graphs!)

]]> By: Khlo http://cow.neondragon.net/index.php/1765-Finding-Primes/comment-page-1#comment-1167 Khlo Wed, 30 Nov -0001 00:00:00 +0000 #comment-1167 Interesting Tim :) Only checking existing primes sounds good; I'm gonna try implementing that. I also got told in IRC that all primes above 2 and 3 are either 1 greater or smaller than a multiple of 6.<br /> Interesting Tim :) Only checking existing primes sounds good; I’m gonna try implementing that. I also got told in IRC that all primes above 2 and 3 are either 1 greater or smaller than a multiple of 6.

]]> By: Michael http://cow.neondragon.net/index.php/1765-Finding-Primes/comment-page-1#comment-1168 Michael Wed, 30 Nov -0001 00:00:00 +0000 #comment-1168 <p>You can floor the square root, rather than ceil it, which will save (a very tiny amount of) time.</p><p>You can see the 6n±1 by doing a Sieve of Erastothenes six wide.  So that would winnow the search pretty well.<br /></p><p>The other speedup is that, as Tim said, every natural number greater than one has a unique prime factorisation, so you can do the modulus of the primes you've already found, 2, 3, 5, ... up to √n. You don't actually have to divide by either 2 or three if you've already selected only numbers matching 6n±1. This is part of the fundamental theorem of arithmetic.</p> You can floor the square root, rather than ceil it, which will save (a very tiny amount of) time.

You can see the 6n±1 by doing a Sieve of Erastothenes six wide.  So that would winnow the search pretty well.

The other speedup is that, as Tim said, every natural number greater than one has a unique prime factorisation, so you can do the modulus of the primes you’ve already found, 2, 3, 5, … up to √n. You don’t actually have to divide by either 2 or three if you’ve already selected only numbers matching 6n±1. This is part of the fundamental theorem of arithmetic.

]]> By: Xeen http://cow.neondragon.net/index.php/1765-Finding-Primes/comment-page-1#comment-1169 Xeen Wed, 30 Nov -0001 00:00:00 +0000 #comment-1169 <p>I'm not sure if already mentioned but wouldn't save it quite some time only checking every 2nd number (by changing n = n + 1 to n = n + 2)? Except of 2, every single prime is odd - of course, otherwise it's dividable by 2 hence it can't be a prime.</p><p>I don't have phyton installed, would be nice to know if it really saves that much time since it fails anyway after dividing by 2. Might only make a difference if you go for really big numbers...</p><p>Greetings<br />xeen <br /></p> I’m not sure if already mentioned but wouldn’t save it quite some time only checking every 2nd number (by changing n = n + 1 to n = n + 2)? Except of 2, every single prime is odd – of course, otherwise it’s dividable by 2 hence it can’t be a prime.

I don’t have phyton installed, would be nice to know if it really saves that much time since it fails anyway after dividing by 2. Might only make a difference if you go for really big numbers…

Greetings
xeen

]]> By: Tim http://cow.neondragon.net/index.php/1765-Finding-Primes/comment-page-1#comment-1170 Tim Wed, 30 Nov -0001 00:00:00 +0000 #comment-1170 <p>Well, yes they are: in order for a number not to be divisible by either 2 or 3, it must not be divisible by 2*3; if you consider n%6, then the n for which n%6 ==2 or ==4 are obviously divisible by 2, and n%6 ==3 => it's divisible by 3, so you're left with the n for which n%6 is 1 or 5.</p><p>This is tantamount to saying that 2 & 3 are the two most-frequent denominators that can "catch out" a composite the quickest when you start counting from 2. Not exactly surprising and so I don't expect that peeling-off the "is n%6 == 1 or 5?" test is going to save you anything over and above simply starting from 2 and 3 and 5... :) </p><p>However,  you can scale this further. By the time you've examined the number 2, you know every second number is out of the running (duh). By the time you've examined the number 3, you know every 3rd number is out of the running. By the time you've examined the number 5, .... You're now looking at http://en.wikipedia.org/wiki/Sieve_of_eratosthenes . </p><p>Now, if you were to implement the Sieve inefficiently, you'd set aside a huge array to store [1..N] in, and mark boxes prime, composite, or unknown as you count up from 2. This is where the hash (dict) in my script comes in: as it goes along computing isprime 4, isprime 5, isprime 6, ... etc, it memoizes the results for later. So you can loop your n from 2..N happily knowing that, while each primality test will involve a whole bunch more tests-for-primes-in-the-range-2..sqrt(n), as time goes on those will become simple instantaneous lookups rather than calculations.  </p><p>If you want it to look pretty, plot http://en.wikipedia.org/wiki/Ulam's_spiral as you go along :) </p> Well, yes they are: in order for a number not to be divisible by either 2 or 3, it must not be divisible by 2*3; if you consider n%6, then the n for which n%6 ==2 or ==4 are obviously divisible by 2, and n%6 ==3 => it’s divisible by 3, so you’re left with the n for which n%6 is 1 or 5.

This is tantamount to saying that 2 & 3 are the two most-frequent denominators that can "catch out" a composite the quickest when you start counting from 2. Not exactly surprising and so I don’t expect that peeling-off the "is n%6 == 1 or 5?" test is going to save you anything over and above simply starting from 2 and 3 and 5… :)  

However,  you can scale this further. By the time you’ve examined the number 2, you know every second number is out of the running (duh). By the time you’ve examined the number 3, you know every 3rd number is out of the running. By the time you’ve examined the number 5, …. You’re now looking at http://en.wikipedia.org/wiki/Sieve_of_eratosthenes . 

Now, if you were to implement the Sieve inefficiently, you’d set aside a huge array to store [1..N] in, and mark boxes prime, composite, or unknown as you count up from 2. This is where the hash (dict) in my script comes in: as it goes along computing isprime 4, isprime 5, isprime 6, … etc, it memoizes the results for later. So you can loop your n from 2..N happily knowing that, while each primality test will involve a whole bunch more tests-for-primes-in-the-range-2..sqrt(n), as time goes on those will become simple instantaneous lookups rather than calculations. 

If you want it to look pretty, plot http://en.wikipedia.org/wiki/Ulam's_spiral as you go along :)  

]]> By: Marc http://cow.neondragon.net/index.php/1765-Finding-Primes/comment-page-1#comment-1870 Marc Wed, 30 Nov -0001 00:00:00 +0000 #comment-1870 <p>Hi,</p><p>The following statement caught my attention: "To my knowledge, there isn't anyway of telling if a number is prime except from dividing it by all the numbers between 1 and the number and seeing if they are factors."</p><p>There is actually a way, described at site <a href="http://www.primestructure.com/">www.primestructure.com</a>. This does not work for all integers, but it does work for infinitely many ranges of integers.</p><p>Marc</p> Hi,

The following statement caught my attention: "To my knowledge, there isn’t anyway of telling if a number is prime except from dividing it by all the numbers between 1 and the number and seeing if they are factors."

There is actually a way, described at site http://www.primestructure.com. This does not work for all integers, but it does work for infinitely many ranges of integers.

Marc

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