Section 3.4 Let us denote with u=1-N/N the fraction of nodes that are not in the giant component(GC),whose size we take to be N.If node i is part of the GC,it must link to another node j,which must also be part of the GC.Hence if i is not part of the GC,that could happen for two reasons: There is no link between i and j(probability for this is 1-p). There is a link between i and j,but j is not part of the GC(probability for this is pu). Therefore the total probability that i is not part of the GC via node j is 1-p+pu.The probability that i is not linked to the GC via any other node is therefore(1-p+pu)s,as there are N-1nodes that could serve as potential S=1-ek)s links to the GC for node i.As u is the fraction of nodes that do not belong to the GC,for any p and N the solution of the equation u=(l-p+pu)N- 3.30) provides the size of the giant component via NN(1-u).Using p=<k>/ (N-1)and taking the log of both sides,for<>N we obtain tu-(N-Dh (3.31) Taking an exponential of both sides leads to u=exp[-<k>(1-u)].If we denote with S the fraction of nodes in the giant component,S=N/N,then S=1-u and (3.31)results inSection 3.4