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As the problem was set the a0,a1 a2 where knowns as the x3 y3 all the otheres wew unknowns.Alll the variables should be Gf(17) elements.
Yes!if the system is sovlable the intersection points will be defined and there will be x0,y0...x2y2={0,...,16}. Remember that the operations are not the same as on real...
Hmmm ok. I used an extension field GF 2^128 I used a polynomial basis representation and a pentanomial irreducible polynomial for the generation of the field/. But I think that if the system can be solved on GF(p) will be solved at GF(2^128) .
So we can assume p=17.Why not?
p is a prime number. The GF is constituted by the elements 0,1...p-1.
All the variables known and unknown will be elements of a Galois Field. Multiplication Division, addition(=substraction) will be defined over the GF. The result of the operations will be GFs too. They are not integers but elements of the GF. I mean that alla the problem will be defined and solved over the GF.
About what?
Yes all are members of the GF.
If we try a different approach and instead replacing the
to the formula we writethen the formula will be e.g for 3 degree polynomials
So If again we know
we need a ste of 2x3=6 such polynomials to solve the system and the system remains linear i.e. to get the afterwards that we have definedwe can solve
and recover the missing intersection points.
So it seems that it can be solved over GF. I think...
Whats this?I ve never heard it. Can this solve the aforementioned problem?
You give me hope!
I think that when the set of equations is constituted by polynomials of degree-n, an unknown
will come up.I dont know how
can be computed over GFs.No..It is not easy to test it. I have written in C a galois field multiplier and divider for an irreducilbe polynomial of degree 128.
The addition is simply an xor.
The equations are the same that I was posted in previous posts.
I google it and the results that wre obtained confused me. e.g. I tried to read the following thesis but with no result.
If i undersstood correctly the problem is not easily solvable but I am not sure that this thesis copes the same problem as mine
https://openaccess.leidenuniv.nl/bitstream/handle/1887/4392/Thesis.pdf?sequence=1
Yes I meant a Finite feild generated by a prime n or an irreducible polynomial.
I am not sure that is solvable if the equations of the set are nonlinear i.e. te polynomials are of 2-degree 3- etc...
I have the following set of equations:
Does the previous set of equations is solvable?
And more...
If I have the following set of equations:
where
are the unknown GF elements.Is that set of equations solvable over the GF?
Hey bobby m,
Do you know if the set of equations that were talking about can be solved over Galois Fields?Or where to search?
To be honest, I want to know if by having 2n equations
(of the previous form==leading coef.+ 1 point of the polynomial)
I can define the intersection point of the n degree polynomials.
You and anomnimistefy prove that for the case of n=2, n=3 the previous case stands.
I think it stands for every n.
No, only I want to see...just being curious!
Maybe its the same thing. I am not sure.
I am familiar with Galois Fields and the operations executed over GFs but it is not ot easy to explain...
Something like that...
The problem is that when you work over Finite fields,
addition multiplication and division are not equal to standard operations. operations are executed modulo the prime number etc..etc..
Ok, if the operations were over a galois field will have any difference?
What do you mean "difficult"?
Thats nice!
So I think one can say that If you have n intesection points, and someone provides you with the lead coeficient and a point of 2n polynomials you can define the intersection point.
Scool student?
Hi anonimnystefy,
See the posts #133,135 and 146,147.
The aim is to solve a set of 6 equations of a_3 with the information provided in post 133 and 135.
I thinh that you can with mathematica.
You have 6 equations, with 6 unknown variables. I cant see why this is not sovlable.
By the way are you students or college students?
Thans both of you!