By Gerhard Larcher, Friedrich Pillichshammer, Arne Winterhof, Chaoping Xing

Harald Niederreiter's pioneering examine within the box of utilized algebra and quantity idea has resulted in very important and massive breakthroughs in lots of parts. This choice of survey articles has been authored via shut colleagues and best specialists to mark the party of his seventieth birthday. The ebook presents a contemporary evaluation of other study parts, masking uniform distribution and quasi-Monte Carlo equipment in addition to finite fields and their functions, specifically, cryptography and pseudorandom quantity new release. Many effects are released right here for the 1st time. The e-book serves as an invaluable place to begin for graduate scholars new to those parts or as a refresher for researchers desirous to persist with fresh traits.

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I) ⇔ (v) Let c be any (vector space) complement in Vn of the linear space of f . Then f (x+a)− f (x) p Da f (0) = f (y+z+a)− f (y+z) p = y∈ x∈Vn z∈ c f (y)+ f (z+a)− f (y)− f (z) p = y∈ z∈ c f (z+a)− f (z) . p =| | z∈ c Suppose that f is partially bent and a ∈ . Then Da f (0) = 0, hence f (z+a)− f (z) = 0. In particular, for all nonzero a ∈ c the derivative z∈ c p c Da f , restricted to is balanced. Consequently, f restricted to a complement c of is bent. Conversely suppose that f is bent on every complement of .

Since the argument of the proof is similar to that of the Boolean case, we only give a sketch. 3 Every function f ∈ F(Vn , F p ) satisfies |supp( f )||R( f )| ≥ p n . 4) Sketch of Proof. 2) we get p n |R( f )| ≥ Da f (0) = | f (0)|2 . 5) for all t ∈ Vn . By Parseval’s identity P1 we obtain ⎛ ⎞ |supp( f )| ≥ ⎝ | f (t)|2 ⎠ sup | f (t)|2 = p 2n sup | f (t)|2 . 6) completes the proof. The following theorem extends the properties given in [2, Theorem] to arbitrary primes and shows that the two definitions of partially bent functions, stated above are equivalent.

Clearly we have 2m possibilities for y = y ∈ c . If f (y ) = 0 ( f (y ) = 1) we require f (z ) = f (z ) = 1 ( f (z ) = f (z ) = 0) so that d , d ∈ T . 11 this implies f (z) = f (x) = 0, and then we have 2s−1 choices z , z for every choice of y . Consequently, we have 2m 2s−1 = 2m+s−1 solutions if f (x) = 0 and no solution if f (x) = 1. 12, all elements of Vn , except for those in , can be written as a difference of elements of T in precisely λ ways for an integer λ. The elements in the set {x ∈ : f (x) = 0} can be written as a difference of elements of T in precisely |T | ways, while for the elements in the set {x ∈ : f (x) = 1} there is no such representation.