Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Maybe its the same thing. I am not sure.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

I could also write the routines to generate a problem in Mathematica with exact arithmetic. Leaving Geogebra out. The error would diminish greatly.

Did you have a fixed n you needed to be solved or were you just seeing what could be done?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

No, only I want to see...just being curious!

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

Hi;

Okay, I understand.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

Each set of polynomials of n degree can only intersect at n points. Each point will have 2 variables x and y. You will need 2n equations to solve for them.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,937

I thought we already got to this conclusion before.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

The evidence supports this but there are practical considerations.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

If you mean solved modulo 1...n, then Mathematica might be able to handle the job.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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...

*Last edited by Herc11 (2013-07-19 19:55:22)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

Hi;

Do you have a worked example?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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

*Last edited by Herc11 (2013-07-19 21:56:23)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

Hi;

Reading the pdf now.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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.Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

Isn't that covered in his "power extraction algorithm?" Of course, he has gone on and on with his existence proofs but I do not see an implementation so far.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Whats this?I ve never heard it. Can this solve the aforementioned problem?

You give me hope!

*Last edited by Herc11 (2013-07-20 00:32:35)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

He mentions it in the pdf you provided.

You should not have hope yet. Highbrows like him rarely explain anything. The chance that he will provide an example that a lowbrow like me can follow is 1 in 10000.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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...

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

Wouldn't a0, a1, a2,... all have to be members of the GF?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Yes all are members of the GF.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

Hi;

How can you be sure?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

About what?

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,095

That the a0, a1, a2... will all be integers let alone mebers of that set?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

Offline

**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

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.

Offline