By Vasyl Ustimenko
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Whereas cryptography can nonetheless be a arguable subject within the programming neighborhood, Java has weathered that typhoon and gives a wealthy set of APIs that permit you, the developer, to successfully comprise cryptography in applications-if you recognize how.
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Extra info for Algebraic graphs and security of digital communications
As it follows from the results of  functions Pt and Rt deﬁne the arithmetical dynamical system. It means that we can use general private and public keys algorithms which were considered in Chapter 1. Let us consider the description of such algorithms in the particular case of the dynamical system corresponding to the family of directed graphs RDE(n, K). We assume that the ﬁnite ring K contains at least 3 regular elements (non zero divisors). We start from the public key encryption. The set of vertices of the graph RDE(n, K) is a union of two copies free module K n+1 .
At (x)) which allow us to determine the class of watermarking equivalence containing x. Let as consider the symbolic key which is chosen sequence of functions ki ∈ K[y1 , y2 , . . , yt−1 ], i = 1, 2, . . , s. We say that the symbolic key is regular where s ≤ [(n + 5)/2]. ,at (x)) , i = 1, 2, . . , s. ,ks ) × Af2 . It is clear that the map is invertible and ”hidden discrete logarithm” method is a particular case of the above general encryption with special regular symbolic key. Example 1. Let K = Fp for the prime p.
Let DF (I) be the directed graph (double directed ﬂag graph) on the disjoint union of F1 with F2 deﬁned by the following rules (i) (l1 , p1 ) → [l2 , p2 ] if and only if p1 = p2 and l1 = l2 , (ii) [l2 , p2 ] → (l1 , p1 ) if and only if l1 = l2 and p1 = p2 . Let DE(n, K) (DE(K)) be the double directed graph of the bipartite graph D(n, K) (D(K), respectively). Remember, that we have the arc e of kind (l1 , p1 ) → [l2 , p2 ] if and only if p1 = p2 and l1 = l2 . Let us assume that 1 − l2 . the colour ρ(e) of arc e is l1,0 1,0 Recall, that we have the arc e′ of kind [l2 , p2 ] → (l1 , p1 ) if and only if l1 = l2 and p1 = p2 .