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gackt yfc mp4 001 - By selecting one vertex from an n-vertex graph, we can embed ëlog2 nû bits. Therefore following theorems have been deduced, Lemma 2. gackt yfc mp4 001 - 2: Given a random graph Gn,p, Let. To create a convincible watermarking a large graph, we have to add w(n/log n) edges by the first technique. To embed long messages, we have to construct more MISes, which may result in huge overhead gackt yfc mp4 001. gackt yfc mp4 001 - ·The first vertex of any MIS have been selected randomly and the next vertex's choices are restricted to (1-p)n=qn as pn neighbors of the first vertex have been eliminated. 1: Given random graph Gn,p , almost all randomly selected MIS is of size logb n, where b = 1/(1-p). 1, at most log2 n log b n bits of information could be embedded into the MIS. be the event that in a random solution, all vertices in this MIS have the same color. Miodrag Potkonjak [7] discussed some technical aspects of "selecting MIS" approach, Whenever a MIS has been removed from the original graph, the graph became a random one again with the same edge probability gackt yfc mp4 001. gackt yfc mp4 001 - ·To generate a random graph Gn+1,p , add one new vertex into a random graph Gn,p and add an new edge in between the new vertex and the old vertex in Gn,p with probability P.