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

I think that I can provide you with points of polynomials over GF(2^128). I can also choose the leading coefficient. But I think that in order the resuls to be right both of us have to employ the same irreducible polynomial.

I mean that if it is easy to "feed in" Mathematica with the irreducible polynomial, the required points the leading coefficient I can give you this info.

*Last edited by Herc11 (2013-07-21 22:36:12)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

I do not understand the part the irreducible polynomial plays in this.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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

In GF(p) all the operations are executed modulo p.

In a order to construct an extension field GF(2^128) an irreducible polynomial is employed in order the elements of the field to be generated. Then all the operations are executed modulo this ir. pol.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Do you have a general type of problem you want to try?

You originally wanted 2 points of intersection given one point on each quadratic and the leading, different coefficient. Is that type of problem we will be working on except over some GF?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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

Yes, its the same problem tha we were discussing in the previous post only that the elements now are selected by a GF(2^128).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

2^128 = 340282366920938463463374607431768211456

That is a huge number. Can we work with a GF that is smaller and still accomplish the same thing?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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

If all the parameteres of the problem are defined correctly I think that there will be no rpoblem. But I remind you that GF elements are not numbers the look like number. I used polynomial basis representation when I worked over GF @^128.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

What does the set look like?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Is this helpful? I ll try to find more...

http://mathworld.wolfram.com/FiniteField.html

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

This was very helpful to me

http://www.dragonwins.com/domains/getteched/crypto/polynomial_arithmetic_in_gf%28p%5En%29.htm

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Okay, I will look that over, see you later. I have some chores to do and need to go offline.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Hmm...I can create a polynomial and pre select its leading coefficient. Also I can pselect random points from it.

By I m not able to constrct polynomials which have intersection points...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

That is going to be a big problem. One of the reasons I am very skeptical that it will always be true.

Please provide what you have and I can work on that.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

I do not know if what I have is helpful. All I have is the abiltiy to select the coef. of the polynomial and to evaluate it to some x_i s.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

First are we solving for the two points of intersection of 4 quadratics with a given point and leading coefficient of each?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Yes

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

So the leading coefficients of each of the 4 given polynomials will be ∈ GF. The points given will have x's and y's that are ∈ GF. Now you are requiring the points of intersection of the 4 quadratics to also be ∈ GF?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

That right! All the problem is define over GF.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

I am thinking this GF looks like this

{1,2,3,4,5,..., p-1} in other words a set of elements.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

ok. but this is not the problem.the problem is to have the definition of addition multiplication division square root etc on GF.You cannot use integer arithmetic

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

You cannot use integer arithmetic

All the elements of the set are integers. Modular arithmetic deals with integers. We do not care how mathematica does the solving. We could not understand it anyway. Just a few more things and we can begin to deal with this

but this is not the problem.the problem is to have the definition of addition multiplication division square root etc on GF

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

The operations are not like those o f standard arithmetic. If mathematica does the solving may has their definitions. GF elements are not integers they seem like integers. Different properties

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

You mean that when I multiply two elements of this set, I do not first multiply then modulo p the result?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

You first multiply then tou do modulo p. The division, you first find the reverse of the number mod p and then multiply

I do not know how square root is computed.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Mathematica knows all that already.

For instance when solving an equation

Mathematica spits out x = 2 and x = 5 which corresponds to solving that equation over GF(7). True or false?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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