Codes from bent functions over finite fields
Sihem Mesnager University of Paris VIII, Department of mathematics and University of Paris XIII LAGA and Telecom Paris-Tech, France Seminar at Telecom Paris September 2016
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Codes from bent functions over finite fields Sihem Mesnager - - PowerPoint PPT Presentation
Codes from bent functions over finite fields Sihem Mesnager University of Paris VIII, Department of mathematics and University of Paris XIII LAGA and Telecom Paris-Tech, France Seminar at Telecom Paris September 2016 1 / 41 Outline
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p → Fp :
u∈Fn
p auxu, with xu = n
i=1 xui i and au ∈ Fp.
i=0 xpi.
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2 → F2 an n-variable Boolean function.
2 → F2 be a Boolean function. Then f can be expressed as :
2
n
i .
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1
1(x) of x over F2
1(x) := k−1
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1
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1
1(ax) + λ, a ∈ F2n, λ ∈ F2.
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2
2
2 defined by
i=1 xiyi, ∀x = (x1, . . . , xn) ∈ Fn 2,
2.
1(ax),
1" is the absolute trace function on F2n.
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n 2 ,
n 2
n−2 2
n 2
n−2 2
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a∈Fn
2
a∈Fn
2
2(a) = 22n
2 (
n 2 −1
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n 2 −1
n 2 −1
n 2 (−1)˜
f(ω), ∀ω ∈ Fn 2, defines the dual function ˜
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n 2 −1, 2n−2 ± 2 n 2 −1).
2 is bent if and only if
2 | f(x) = 1} is a Hadamard difference set in Fn 2.
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2 (n = 4) by
2.
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f(x)−Trpn/p(bx) p
2πi p is the complex primitive pth root of unity and elements of Fp
2 Sf (b) = ζf ⋆(b)
p
2 Sf (b) = ζf ⋆(b)
p
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2 Sf (b) =
p
p
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q with
a,¯ b∈C,¯ a=¯ bd(¯
2 → F2}
r
distance
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n 2 −1
n 2 −1
n 2 −1 and the associated n-variable
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1
2
3
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q ; α ∈ Fq, β ∈ Fq}.
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1 its dual function. Then the weight distribution of
1
1 (β)2 m 2 −1 (β ∈ F⋆
2m).
2
p and ψ∗ 1(¯
p )
m+1 2 p m−1 2
1 (¯
αβ) p
p and ψ∗ 1(¯
pm.
m 2 −1ǫ(p − 1) if α ∈ F⋆
p and ψ∗ 1(¯
m 2 −1ǫ if α ∈ F⋆
p and ψ∗ 1(¯
pm.
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p )
m+1 2 p m−1 2
m−1 2 ) (p−1)2
2
p )
m+1 2 p m−1 2
m−1 2 ) (p−1)2
2
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m 2 −1(p − 1)
m 2 −1(p − 1)2
m 2 −1
m 2 −1(p − 1)2
m 2 −1
m 2 −1
m 2 −1
m 2 −1
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p σω(σα(Sψ1(¯
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m 2 ξg∗(ω)
m 2 ξg(−ω)
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m 2 −1 + ǫp m 2 ;
m 2 −1, 1 ≤ j ≤ p − 1;
m−1 2 ( j
1 2 ;
1 2 ,
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2π√−1 p
p
p
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(m−2) 2
(m−2) 2
(m−2) 2
(m−2) 2
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nf 2 − 2
m−4 2
2m−1+nf 2− m−2
2
2 nf 2 + 2
m−4 2
2m−1+nf 2− m−2
2
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pm | Q(x) = 0}. Then if m is odd, we have #DQ = pm−1 − 1 and if m
m−2 2
m−2 2
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pm × Fpm.
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z
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r (αxG(yx2m−2)).
r (αγiG(ζiγi 2m−2)) | 1 ≤ i ≤ (2m − 1)2); α ∈ F2m}.
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r , 2m−r(2r − 1)(2m − 1)].
r (2r). Therefore, the hyperovals
r (2r) (where r is a divisor of m) whose duals are the
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