A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0

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Chapter1: Combinatorial Analysis
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A little twist in the problem of a configuration model for an exponential distribution of
degrees. Imagine that the probability pk of a node of degree k in this network is proportional
to kak (obviously with the constant a being a number smaller than one, otherwise that
expression would diverge for large k).
(1) Calculate the normalization of the degrees. In other words obtain an expression for
the Pk-
(2) Calculate the average degree of the network.
(3) Calculate the average number of second neighbors.
(4) Find out for which values of a does this neighbor have a giant component.
These three expressions might be useful
kak
k=0
a
(1-a)²
Σk² ak
k=0
=
a + a²
(1 - a)³
Σk²³ ak= ª +4a² + a²
(1-a)4
k=0
Transcribed Image Text:A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0
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