[sage-support] Complexity of Groebner Basis computation

2009-06-24 Thread vpv
I have a system of 300 quadratic boolean equations in 200 variables. I am able to find a single solution to the system using Groebner Bases (the PolyBori implementation) in time less than 2 minutes - 1 second for computing the Groebner Basis and 85 seconds for computing the variety and memory

[sage-support] Groebner bases and varieties computation on multiple CPUs

2008-11-24 Thread vpv
Hi, Is there a way to compute Groebner bases and varieties in parallel on multiple processors or in a cluster? Thanks. --~--~-~--~~~---~--~~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to [EMAIL

[sage-support] Re: Groebner bases and varieties computation on multiple CPUs

2008-11-24 Thread vpv
would appreciate if you can give me more information on how I can use @parallel/pyprocessing or DSage to parallerize the above code. Thanks for your help! Regards, vpv P.S. For the cluster I have access to, I think I do not have shared memory, but i can check this On Nov 24, 9:27 am, mabshoff

[sage-support] Groebner basis, Ideal dimension and Variety

2008-10-29 Thread vpv
Hello, I have a system of 11 quadratic equations over GF(2) in 8 variables. I compute it's groebner basis. The ideal generated by it has dimension 0 so I compute its variety. It results in 16 solutions (16 possible sets of values for the 8 variables). I take one of the solutions and then i

[sage-support] Re: Groebner basis, Ideal dimension and Variety

2008-10-29 Thread vpv
Thank you very much for the quick response, Simon! On Oct 29, 5:01 pm, Simon King [EMAIL PROTECTED] wrote: Hi! On Oct 29, 4:34 pm, vpv [EMAIL PROTECTED] wrote: The dimension of the ideal of the groebner basis of the new system is 4 and not 0? Why? Since you are still in a ring with 8

[sage-support] Re: solve() problem: raise ValueError, Unable to solve %s for %s%(f, args)

2008-10-14 Thread vpv
this necessarily mean that the system G (and hence e) is solvable for all variables composing it? Thanks a lot for your help. Regards, vpv P.S. My code for computing G is: N = 144 P = BooleanPolynomialRing(N, 'x',order='lex') x=[] for i in range(0,N): x.append(P.gen(i)) E = [x[20] + x[15

[sage-support] solve() problem: raise ValueError, Unable to solve %s for %s%(f, args)

2008-10-12 Thread vpv
, x37 + 1, x95 + x39, x44, x45, x95 + x46, x95 + x47 + x111] for ([x20, x21, x28, x29, x15, x63, x31, x36, x37, x39, x44, x45, x46, x47, x79, x95, x111],) Thanks for your help! Regards, vpv --~--~-~--~~~---~--~~ To post to this group, send email to sage-support

[sage-support] Re: Small Scale Variants of the AES (SR)

2008-09-24 Thread vpv
Thanks for the useful links, Martin. I downloaded and compiled the source for Sage (3.1.2). I'll have a look at crypto/mq/sr.py and i'll try to figure how to add my code there. If i have questions i'll ask. Cheers, vpv On Sep 19, 6:48 pm, Martin Albrecht [EMAIL PROTECTED] wrote:  * I'd

[sage-support] how to calculate the derivative of a boolean polynomial?

2008-09-24 Thread vpv
Is it possible to calculate the derivative of a boolean polynomial in Sage using PolyBoRi? I can do this for polynomials over GF(2): sage: R.x,y,x = PolynomialRing(GF(2),3) sage: f = 3*x^2*y + 2*x*y + y + 9*x^2 + 5*x - 3 sage: f x^2*y + x^2 + x + y + 1 sage: f.derivative(x) 1 sage:

[sage-support] Re: Small Scale Variants of the AES (SR)

2008-09-19 Thread vpv
-- name: Martin Albrecht _pgp:http://pgp.mit.edu:11371/pks/lookup?op=getsearch=0x8EF0DC99 _www:http://www.informatik.uni-bremen.de/~malb _jab: [EMAIL PROTECTED] Greetings, vpv --~--~-~--~~~---~--~~ To post to this group, send email to sage-support@googlegroups.com

[sage-support] Re: Small Scale Variants of the AES (SR)

2008-09-18 Thread vpv
interesting to be added to SAGE, please let me know. I'd be happy to share my code. Greetings, vpv On Sep 17, 7:36 pm, Martin Albrecht [EMAIL PROTECTED] wrote: On Wednesday 17 September 2008,vpvwrote: I've successfully created a 4x4 4-bit variant of SR sage: sr = mq.SR(1,4,4,4

[sage-support] Small Scale Variants of the AES (SR)

2008-09-17 Thread vpv
I've successfully created a 4x4 4-bit variant of SR sage: sr = mq.SR(1,4,4,4,allow_zero_inversions=True) sage: sr SR(1,4,4,4) Next I would like to create a plaintext/key pair, which is composed of variables (say x0,x1,...,x15,k0,k1,...,k15) rather than actual values (eg. 0,1,...,1,0,0,0,1,1).

[sage-support] Re: symbolic substitution

2008-06-20 Thread vpv
Thanks for your help! Both solutions worked for me. On Jun 16, 9:56 pm, William Stein [EMAIL PROTECTED] wrote: On Mon, Jun 16, 2008 at 9:43 AM, vpv [EMAIL PROTECTED] wrote: Hello William, Thanks for your reply! I think I can use your idea to solve my problem. I tried to run your example

[sage-support] Re: symbolic substitution

2008-06-16 Thread vpv
(1') must be alpha- numeric and begin with a letter sage: I wonder if I have an error in the syntax or if maybe it is not possible to do such declaration in Sage? Thanks for your help! Greetings, vpv On Jun 15, 8:41 pm, William Stein [EMAIL PROTECTED] wrote: On Sun, Jun 15, 2008 at 5:45 AM

[sage-support] symbolic substitution

2008-06-15 Thread vpv
I have two expressions - exp1 and exp2: sage: exp1 = (x1*x2+x3) sage: exp2 = (x1+x3*x5*x7) sage: exp1 x1*x2 + x3 sage: exp2 x1 + x3*x5*x7 I try to substitute exp1 and exp2 for the symbolic variables e1 and e2 respectively: sage: exp1 = var('e1') sage: exp2 = var('e2') sage: exp1 e1 sage: exp2

[sage-support] boolean arithmetic in BooleanPolynomialRing

2008-06-04 Thread vpv
How can I define a polynomial boolean ring in 'a' in which it holds that a+a=0 both cases: (1) when 'a' is an unknown parameter and (2) when 'a=1'. Please see below: sage: A.a = BooleanPolynomialRing(1) sage: a a sage: A Boolean PolynomialRing in a sage: a+a 0

[sage-support] Re: boolean arithmetic in BooleanPolynomialRing

2008-06-04 Thread vpv
you just overwrote the previous definition of a which was in the boolean polynomial ring Ok. Now I understand. Try A(1) for the constant 1 in A for which A(1) + A(1) holds. Yes, this works for me. Thanks! On Jun 4, 1:06 pm, Martin Albrecht [EMAIL PROTECTED] wrote: sage: a=1 sage: a+a

[sage-support] Re: multiplicative inverse of a polynomial in a ring

2008-05-30 Thread vpv
not: assert not x*y==1 assert not x^255==1 # However one = x*y assert one^2==one 2008/5/28 vpv [EMAIL PROTECTED]: Hello, I am trying to solve the following equation for y in SAGE: x*y = 1 (mod z^8+z^4+z^3+z+1) where x = x0+x1*z^1+x2*z^2+x3*z^3+x4*z^4+x5*z^5+x6*z^6+x7*z^7 y

[sage-support] Re: multiplicative inverse of a polynomial in a ring

2008-05-29 Thread vpv
]: It seems you should be able to represent multiplication by y as a matrix equation, which you might have luck inverting. On Wed, May 28, 2008 at 5:27 PM, vpv [EMAIL PROTECTED] wrote: Hello, I am trying to solve the following equation for y in SAGE: x*y = 1 (mod z^8+z^4+z^3+z+1) where

[sage-support] multiplicative inverse of a polynomial in a ring

2008-05-28 Thread vpv
Hello, I am trying to solve the following equation for y in SAGE: x*y = 1 (mod z^8+z^4+z^3+z+1) where x = x0+x1*z^1+x2*z^2+x3*z^3+x4*z^4+x5*z^5+x6*z^6+x7*z^7 y = ? x0,...,x7 are elements of GF(2). I do not know their values. I am searching for y in parametric form i.e. as a polynomial of z