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 arou
is is correct then I 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 2
Hi,
Is there a way to compute Groebner bases and varieties in parallel on
multiple processors or in a cluster?
Thanks.
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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
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
replac
does 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.
, x79 + x28, x79 + x29, x95 + x79 + x31 + 1, x36 + 1, 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 pos
I construct the ideal I generated from the three boolean polynomials
f1,f2 and f3:
sage: B. = BooleanPolynomialRing(3)
sage: f1 = x0*x1 + x2
sage: f2 = x1*x2
sage: f3 = x0*x1*x2 + x0*x2
sage: I = ideal(f1,f2,f3)
sage: I
Ideal (x0*x1 + x2, x1*x2, x0*x1*x2 + x0*x2) of Boolean PolynomialRing
in x0,
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. = 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: f.derivative(y)
x^
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]>
equations. This will be slightly more involving.
>
Anyway I was already thinking of developing a version of my code for
GF(2^n), so I would gladly help with this.
> Cheers,
> Martin
>
> --
> name: Martin Albrecht
> _pgp:http://pgp.mit.edu:11371/pks/lookup?op=get&search=0x8EF
thing 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
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). Ho
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 yo
ible 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, vpv <[EMAIL PROTECTED]> wrote:
>
> > I have two expressions - exp1 and exp2:
>
>
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
e
> 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
> >
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. = BooleanPolynomialRing(1)
sage: a
a
sage: A
Boolean PolynomialRing in a
sage: a+a
0
ement x:
>
> x = x0+x1*z^1+x2*z^2+x3*z^3+x4*z^4+x5*z^5+x6*z^6+x7*z^7
>
> # If all is well this should be true:
>
> assert x^256 == x
>
> #
>
> y = x^254
>
> # ideally, we should have x*y = x^255 = 1, but unfortunately not:
>
> assert not x*y==1
> assert
r*x1bar*x3bar*x4bar*x6bar*x7bar + (a^7 + a +
> 1)*x0bar*x2bar*x3bar*x4bar*x6bar*x7bar + (a^6 +
> a^3)*x1bar*x2bar*x3bar*x4bar*x6bar*x7bar +
>
> John Cremona
>
> 2008/5/28 David Joyner <[EMAIL PROTECTED]>:
>
>
>
> > It seems you should be able to represent multipli
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 of
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