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Thread: Finding prime numbers code issue

06272013, 06:48 PM #16
Here's (nearly) the best I can do with the Sieve of Erastothenes:
Code:<script type="text/javascript"> var nums = [ ]; var primes = [2]; var MAXNUM = 500; // change this as desired // run the actual seive for ( var chk = 3; chk < MAXNUM; chk += 2 ) { if ( nums[chk] == null ) { primes.push(chk); for ( var mark = chk*3; mark < MAXNUM; mark += chk*2 ) { nums[mark] = 0; // any nonnull would do } } } alert(primes); </script>
(a) no need to initialize the array
(b) only checks odd numbers
(c) only marks odd increments of a found prime
(d) doesn't bother marking a found prime
The next step: Would only save memory, not speed. And probably not even that with JavaScript coding. Have each element of the array represent an ODD number. That is, the element numbering doesn't match the numbers represented. So the array can be half the size it is now.
Aside from changing algorithms, do you see a way to improve that?Be yourself. No one else is as qualified.

06272013, 06:50 PM #17
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Let's say that you want to check the first million primes, notice that I've conveniently chosen a limit that is the square of another number, and that number is composite ! Then you only need to check against primes that are less than 1,000. Make your life simpler by setting reasonable conditions.
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06272013, 07:00 PM #18
DrDos: Ummm...no.
(a) The first million *PRIMES* will not be found in the first million numbers. Not even close. I think you mean "all prime numbers that are less than one million".
(b) If that's a comment on my Sieve of Erastothenes implementation, I still don't see it. With the Sieve, any nonmarked number is prime. So if I don't continue the INNER loop all the way to MAXNUM, I won't be marking all nonprimes. And if I don't continue the outer loop all the way to MAXNUM, I won't *find* all the nonmarked numbers so won't find all the primes.Be yourself. No one else is as qualified.

06272013, 07:03 PM #19
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Welcome to http://www.myphotowizard.net
where you can edit images, make a photo calendar, add text to images, and do much more.
When you know what you're doing it's called Engineering, when you don't know, it's called Research and Development. And you can always charge more for Research and Development.

06272013, 07:06 PM #20
Okay, I do see that the marking and checking can be done separately, and then the marking only has to go to the square root of the max:
Code:<script type="text/javascript"> var nums = [ ]; var primes = [2]; var MAXNUM = 500; // change this as desired // run the actual seive for ( var chk = 3; chk < Math.sqrt(MAXNUM); chk += 2 ) { if ( nums[chk] == null ) { for ( var mark = chk*3; mark < MAXNUM; mark += chk*2 ) { nums[mark] = 0; // any nonnull would do } } } for ( var chk = 3; chk < MAXNUM; chk += 2 ) { if ( nums[chk] == null ) primes.push(chk); } alert(primes); </script>
Be yourself. No one else is as qualified.

06272013, 07:45 PM #21
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06282013, 05:52 PM #22
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A simple page for prime numbers which use the preceding primes to test a new number.
And a more subtle function (from rnd.me ? I can not find the post) that stores the previous calculations to go faster
Code:<script type="text/javascript"> (function(){ function isPrime(){ var n=this; if(isPrime[n]) return true; if( n<11 ! ((n%2)&&(n%3)&&(n%7)&&(n%9)) ) return false; for (var i=5, l=Math.sqrt(n)+1; i<l; i+=6){ if( !(n%i)  !(n%(i+2)) ) return false; } return isPrime.n++ < 1e6 ? (isPrime[n]=true) : true; }; var p=isPrime;p.n=0;p[2]=p[3]=p[5]=p[7]=true; Number.prototype.isPrime=p; }()); // Use var nmb=33353; alert(nmb.isPrime()) // true </script>
Strange prime numbers : 11113, 11117, 11119, 22229, 33331, 44449, 77773, 88883, 99991 ...Last edited by 007julien; 06282013 at 10:36 PM.

06282013, 05:59 PM #23
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Code:Is this number prime? <input type = "text" id = "pn" onblur = "return alert(Number.prototype.isPrime(this.value))"> <script type = "text/javascript"> Number.prototype.isPrime = function (which) { var num = Number(which), counter; if (num === 2) { return true; } if (num < 2  (num / 2) === Math.floor(num / 2)  num !== Math.floor(num)) { return false; } else { for (counter = 3; counter <= Math.sqrt(num); counter += 2) { if (num % counter === 0) { return false; } } return true; } } </script>
All the code given in this post has been tested and is intended to address the question asked.
Unless stated otherwise it is not just a demonstration.