Let v be a vertex in G(n, p). We are interested in the size of the...
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Let v be a vertex in G(n, p). We are interested in the size of the connected component that contains v, which we will call Cy. We explore C, as follows. Start at v and mark it as "live". While there are live vertices, choose a live vertex and mark it as "processed". Find it's neighbors among the vertices in G(n, p) that are not yet live or processed, and mark them as live. This ensures that we find a tree in G(n, p), and that each vertex is only added to the tree and processed at most once: no vertices are double counted. Continue until there are no more live vertices, and note that at this point, the processed vertices are exactly the set V(C,). We let Y" = 1, and Y9T + R – 1, if Y" > 0, if Y" = 0, gr t Ygr t+1 gr 0, where R represents the number of new neighbors discovered at time t. In th graph exploration process, there is a time dependence, since we have only n vertices total. We similarly define Yr = 1 and Yg" + Rt – 1. t+1 and T9" = min{t : Y" = 0} = T9" = min{t : Y" = 0}. Exercise 3. Suppose that we are at time t and that Y", = Yt–1· Show that Rt has a binomial distribution with n – Yt-1 – (t – 1) trials and success probability p. t-1 Let v be a vertex in G(n, p). We are interested in the size of the connected component that contains v, which we will call Cy. We explore C, as follows. Start at v and mark it as "live". While there are live vertices, choose a live vertex and mark it as "processed". Find it's neighbors among the vertices in G(n, p) that are not yet live or processed, and mark them as live. This ensures that we find a tree in G(n, p), and that each vertex is only added to the tree and processed at most once: no vertices are double counted. Continue until there are no more live vertices, and note that at this point, the processed vertices are exactly the set V(C,). We let Y" = 1, and Y9T + R – 1, if Y" > 0, if Y" = 0, gr t Ygr t+1 gr 0, where R represents the number of new neighbors discovered at time t. In th graph exploration process, there is a time dependence, since we have only n vertices total. We similarly define Yr = 1 and Yg" + Rt – 1. t+1 and T9" = min{t : Y" = 0} = T9" = min{t : Y" = 0}. Exercise 3. Suppose that we are at time t and that Y", = Yt–1· Show that Rt has a binomial distribution with n – Yt-1 – (t – 1) trials and success probability p. t-1
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