Algebraic graphs and security of digital communications by Vasyl Ustimenko

By Vasyl Ustimenko

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Let as assume that affine transformation Af i , i = 1, 2 are known for adversary. Notice that finding them can be very difficult. Then the advern sary can compute di = Fd(x = (Af 1 )−1 C(xi )(Af 2 )−1 . The pass between i) vertices of the graph is unique. The Dijkstra algorithm is not suitable for finding the pass because the vertex space of the graph is the plainspace. But may be large group of automorphisms (see [118] and further references) will allow to find the pass. Then the adversary computes number bi = η(a(x))m modula known big prime.

Lemma 1. Let Γ be a k-regular graph of girth g. Then in the case of length n < g/2 of the decoding sequence, the probability of generating the correct message by applying the encoding sequence r times at random is r/(k(k − 1)n−1 ). Proof. Let us imagine that our message is the password to a certain account. We have the decoded message and r attempts to get into the account. So we are trying r different encoding sequences to recover the correct message. 2. Graphs with special walks, definitions and motivations that there are no C2n .

Then the adversary computes number bi = η(a(x))m modula known big prime. Still he is not able to find number m because of the complexity of discrete logarithm problem. So he has to take for the set {xi |i ∈ J} the totality of representatives from classes of watermarking equivalence (transversal). So |J| > O(|K|[1/4] ) because of the theorem 2. We use term hidden discrete logarithm to name modified algorithm because affine transformations do not allow the adversary to compute the class of watermarking equivalence containing the plaintext (base of the logarithm) and pass in the finite automaton corresponding to the value of the logarithm.

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