Post a reply
Username:
Note:If not registered, provide any username. For more comfort, register here.
Subject:
Message body:
Enter your message here, it may contain no more than 60000 characters. 

Smilies
:D :) :( :o :shock: :? 8) :lol: :x :P :oops: :cry: :evil: :twisted: :roll: :wink: :!: :?: :idea: :arrow: :| :mrgreen:
Font size:
Font colour
Options:
BBCode is ON
[img] is ON
[flash] is OFF
[url] is ON
Smilies are ON
Disable BBCode
Disable smilies
Do not automatically parse URLs
Confirmation of post
To prevent automated posts the board requires you to enter a confirmation code. The code is displayed in the image you should see below. If you are visually impaired or cannot otherwise read this code please contact the %sBoard Administrator%s.
Confirmation code:
Enter the code exactly as it appears. All letters are case insensitive, there is no zero.
   

Topic review - The largest "n" that GF(2^n) can support in Singular?
Author Message
  Post subject:  Re: The largest "n" that GF(2^n) can support in Singular?  Reply with quote
That depends on the way the ring is defined.
For the form
Code:
[ing r1=(2^n,a),....

the maximum for n is 15 (not 16: less than 2^16).
For the form
Code:
ring r2=(2,a),...; minpoly=.....

the maximal n is the maximal degree of the minpoly, which is
at least 65536.
Post Posted: Thu May 27, 2010 10:04 am
  Post subject:  The largest "n" that GF(2^n) can support in Singular?  Reply with quote
What is the largest n that GF(2^n) can support in Singular?

From 6.1 Limitations of Singular http://www.singular.uni-kl.de/Manual/la ... htm#SEC386 ,
"the number of elements in GF(p,n) must be less than 65536 " which means the "n" can at most be 16, is this correct?

But in my computation, I found "n" could be much larger. I am confused here. Could you please explain this?

Thanks
Gepo
Post Posted: Wed May 26, 2010 4:46 pm


It is currently Sun Nov 18, 2018 1:14 am
cron
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group