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
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
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
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
> 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
> >
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
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:
>
>
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
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
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
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
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]>
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^
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,
, 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
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.
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
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
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 PRO
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
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
21 matches
Mail list logo
