Primes in P?

ruby-talk
1

There is an algorithm that tests primes with a polynomial running time:
http://fatphil.org/maths/AKS/

Has anyone coded it in Ruby?

···

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2

Well, I'm gonna guess that if anyone has, they would post it on that website.
I can't get to the original paper (server's offline, file removed?),
but a cursory overview of the c++ implementations makes me think a
decent ruby hacker could pound out an implementation in an afternoon.

Why don't you code it, post it here for everyone to enjoy, and to that website?

--Kyle

···

On Mon, Mar 31, 2008 at 9:42 AM, Charles Zheng <snarles2@gmail.com> wrote:

There is an algorithm that tests primes with a polynomial running time:
The AKS "PRIMES in P" Algorithm Resource

Has anyone coded it in Ruby?
--
Posted via http://www.ruby-forum.com/\.

3

Charles Zheng wrote:

There is an algorithm that tests primes with a polynomial running time:
The AKS "PRIMES in P" Algorithm Resource

Has anyone coded it in Ruby?

Hello Charles:

My head hurt trying to understand the wikipedia description for polynomial time so I stopped read it. That aside, a cool algorithm was pointed out by Tim Pease in March of 2007. It uses regular expressions to achieve, to a point, non-exponential solving times for prime numbers.

Here is an example of that algorithm demonstrated via a method which extends the Fixnum class.

class Fixnum
   def is_prime?
     ((("1" * self) =~ /^1$|^(11+?)\1+$/) == nil)
   end
end

irb(main):009:0> 2.is_prime?
=> true
irb(main):010:0> 113.is_prime?
=> true
irb(main):008:0> 123457.is_prime?
=> true

The turnaround time on solving is almost instantaneous for this algorithm until the numbers start gets really big (i.e. like the 123457 above). I don't know if this matches the criteria for "polynomial running time" but thought you might find this interesting if you didn't know about it.

Tim made reference to this web site for credit:

http://montreal.pm.org/tech/neil_kandalgaonkar.shtml

I used the above example to demonstrated Ruby's ability to modify base classes and support advanced regular expressions to a Python programming friend. He was both impressed and a little confused by the example :slight_smile: Prior to using this Ruby trick the equivalent Python program was about 2.25 times faster when solving into the low 100s on Windows. After using this trick the Ruby program was 1.1 times faster than Python which couldn't do the same trick according to my friend. FYI, both Ruby and Python were still 32 times slow than CodeGear's Delphi even though Delphi didn't use the regular expression trick.

Michael

4

Charles Zheng wrote:

There is an algorithm that tests primes with a polynomial running time:
The AKS "PRIMES in P" Algorithm Resource

Has anyone coded it in Ruby?

Yes. A few times. I included a tweak that greatly reduces runtime for
composites, at the cost of slightly increased runtime for primes.

The first time I used ruby-algebra for the polynomials. ruby-algebra
Z/nZ rings precalculate and cache all multiplicative inverses rendering
the runtime exponential. I changed a few lines, making precalculation
into calculate on demand and cache.

The second time I wrote my own Modular Polynomial class using an array
for the coefficients. This allowed slightly larger numbers before
running out of memory. Array became a Hash, and the maximum tolerable
number increased. Still run out of memory for many large primes. Time
is nearly instantaneous for all composites pq. And significantly slower
for some composites of the form pqr. And very slow for primes.

Under this version all the RSA challenge numbers return composite within
a few seconds.

Currently working on a libgmp version to speed up the polynomial math.

···

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5

For those of you interested in tackling this monster...

I'll try my hand at it eventually, but I don't think I could do it in
an afternoon.

Todd

···

On Mon, Mar 31, 2008 at 10:42 AM, Kyle Schmitt <kyleaschmitt@gmail.com> wrote:

Well, I'm gonna guess that if anyone has, they would post it on that website.
I can't get to the original paper (server's offline, file removed?),

6

Michael Brooks wrote:

Hello Charles:

Here is an example of that algorithm demonstrated via a method which
extends the Fixnum class.

class Fixnum
   def is_prime?
     ((("1" * self) =~ /^1$|^(11+?)\1+$/) == nil)
   end
end

...

The turnaround time on solving is almost instantaneous for this
algorithm until the numbers start gets really big (i.e. like the 123457
above). I don't know if this matches the criteria for "polynomial
running time" but thought you might find this interesting if you didn't
know about it.
Michael

That is pretty cool. It is not of polynomial running time since it tries
to factor the number brute-force, but that is a very nifty reg EXP
trick.

···

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7

Scratch that. I found a way to sort of "cheat" using a stdlib. But,
I think you should try it the hard way for fun.

Todd

···

On Mon, Mar 31, 2008 at 11:43 AM, Todd Benson <caduceass@gmail.com> wrote:

I'll try my hand at it eventually, but I don't think I could do it in
an afternoon.

8

The author is Perl hacker Abigail, it first appeared in comp.lang.perl.misc:

-- fxn

···

On Apr 1, 2008, at 4:23 , Charles Zheng wrote:

    ((("1" * self) =~ /^1$|^(11+?)\1+$/) == nil)
  end
end

...

The turnaround time on solving is almost instantaneous for this
algorithm until the numbers start gets really big (i.e. like the 123457
above). I don't know if this matches the criteria for "polynomial
running time" but thought you might find this interesting if you didn't
know about it.
Michael

That is pretty cool. It is not of polynomial running time since it tries
to factor the number brute-force, but that is a very nifty reg EXP
trick.

9

Charles, take a peak at the mathn library for your greatest prime factor...

require 'mathn'
n = 12345654321
p n.prime_division.last[0]

hth a little,
Todd

···

On Mon, Mar 31, 2008 at 9:23 PM, Charles Zheng <snarles2@gmail.com> wrote:

Michael Brooks wrote:
> Hello Charles:

>
> Here is an example of that algorithm demonstrated via a method which
> extends the Fixnum class.
>
> class Fixnum
> def is_prime?
> ((("1" * self) =~ /^1$|^(11+?)\1+$/) == nil)
> end
> end
...

> The turnaround time on solving is almost instantaneous for this
> algorithm until the numbers start gets really big (i.e. like the 123457
> above). I don't know if this matches the criteria for "polynomial
> running time" but thought you might find this interesting if you didn't
> know about it.
> Michael

That is pretty cool. It is not of polynomial running time since it tries
to factor the number brute-force, but that is a very nifty reg EXP
trick.

10

Funny, I tried this with the same number with a 1 attached to the end
(123456543211). It took about 20 minutes, but determined it was prime
(i.e. returned [123456543211, 1]) which means 123456543211 to the
power of 1. Coincidence :slight_smile:

Todd

···

On Tue, Apr 1, 2008 at 5:19 AM, Todd Benson <caduceass@gmail.com> wrote:

Charles, take a peak at the mathn library for your greatest prime factor...

require 'mathn'
n = 12345654321
p n.prime_division.last[0]

11

Oh, btw, "n" was a bad choice of variable name here if you go by the
wikipedia article. In their notation, you want to find the greatest
prime factor of (r-1).

Just pointing out what probably is obvious.

cheers,
Todd

···

On Tue, Apr 1, 2008 at 5:19 AM, Todd Benson <caduceass@gmail.com> wrote:

On Mon, Mar 31, 2008 at 9:23 PM, Charles Zheng <snarles2@gmail.com> wrote:
> Michael Brooks wrote:
> > Hello Charles:
>
> >
> > Here is an example of that algorithm demonstrated via a method which
> > extends the Fixnum class.
> >
> > class Fixnum
> > def is_prime?
> > ((("1" * self) =~ /^1$|^(11+?)\1+$/) == nil)
> > end
> > end
> ...
>
> > The turnaround time on solving is almost instantaneous for this
> > algorithm until the numbers start gets really big (i.e. like the 123457
> > above). I don't know if this matches the criteria for "polynomial
> > running time" but thought you might find this interesting if you didn't
> > know about it.
> > Michael
>
> That is pretty cool. It is not of polynomial running time since it tries
> to factor the number brute-force, but that is a very nifty reg EXP
> trick.

Charles, take a peak at the mathn library for your greatest prime factor...

require 'mathn'
n = 12345654321
p n.prime_division.last[0]

12

Posted by Todd Benson (Guest) on 01.04.2008 14:08

Charles, take a peak at the mathn library for your greatest prime factor...

require 'mathn'
n = 12345654321
p n.prime_division.last[0]

Funny, I tried this with the same number with a 1 attached to the end
(123456543211). It took about 20 minutes, but determined it was prime
(i.e. returned [123456543211, 1]) which means 123456543211 to the
power of 1. Coincidence :slight_smile:

Todd

Well, you could use the Miller-Rabin prime test for a speed up! See:

http://snippets.dzone.com/posts/show/4636

You may check the outcome with primegen, http://cr.yp.to/primegen.html

time -p primes 123456543211 123456543211 # done in well under a second

Cheers,
j. k.

···

On Tue, Apr 1, 2008 at 5:19 AM, Todd Benson <caduceass@gmail.com> wrote:

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Posted via http://www.ruby-forum.com/\.

13

Well, you could use the Miller-Rabin prime test for a speed up! See:

http://snippets.dzone.com/posts/show/4636

You may check the outcome with primegen, http://cr.yp.to/primegen.html

time -p primes 123456543211 123456543211 # done in well under a second

Cheers,
j. k.

Yes, try Miller-Rabin or some other test like it. They're not
absolutely perfect but very fast and usually good enough.

The original AKS is like O(d^12) , where d is the number of digits of
n. It is more of a theoretical result than a practical algorithm.
Simply testing all numbers up to sqrt(n) is 2^(d/2), which is faster
for numbers up to about 200 digits (at which they both already take
far too long, of course).