Re: combinatorics question, should be easy
- From: Ralf Goertz <r_goertz@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 25 Feb 2007 17:54:46 +0100
Jeremy Targett wrote:
Now I go to construct the set myself according to the CRC recipe. If
I'm reading correctly they say the set will be constructed of all i,
0<=i<v, such that SUM_[j=0;j=d]((a^i)^(q^j))==0 mod v, where a is a
generator of the group F_(q^(d+1)).
I'm taking a=2 to generate the group, so mod v I have the powers of a
are 2,4,8,3,6,12,11,9,5,10,7,1.
So for each b=a^i I add its 3^0-th power to its 3^1-th power to its
3^2-th power, i.e. b + b^3 + b^9 and I look for the sum to be 0 mod
13. Then the set of such i should be a difference set.
Problem is I can't find a single a^i that yields 0 mod 13 when added
to its 3rd and ninth powers.
Could someone tell me where I am going wrong please?
I'm not sure I understand everything you said, but I used to deal with
those sets and I wrote a program that generates them for the case y=1,
v=q^2+q+1 and k=q+1. One difference set for your example would be {0, 1,
4, 6}. This I construct by taking F_3, extending it using the
irreducible and primitive polynomial f=X^3+2*X^2+X+1 to F_3^3 which is
both a field an a three dimensional Vectorspace over F_3. Let alpha be a
root of f. Then the elements of the difference set are those exponents i
modulo 13 for which alpha^(i+1) is in the plane <alpha^2,alpha>.
HTH,
Ralf
.
- Follow-Ups:
- Re: combinatorics question, should be easy
- From: Jeremy Targett
- Re: combinatorics question, should be easy
- References:
- combinatorics question, should be easy
- From: Jeremy Targett
- combinatorics question, should be easy
- Prev by Date: A RSA Challenge:Finding two equal length prime factors when multiplied will be closest to rsa2048!
- Next by Date: A RSA Challenge:Finding two equal length prime factors when multiplied will be closest to rsa2048!
- Previous by thread: combinatorics question, should be easy
- Next by thread: Re: combinatorics question, should be easy
- Index(es):
Relevant Pages
|
