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Re: [igraph] Question about finding power in a scale-free graph


From: Tamas Nepusz
Subject: Re: [igraph] Question about finding power in a scale-free graph
Date: Sun, 13 Feb 2011 14:45:18 +0100

AFAIK the network is not scale-free any more if the preferential attachment is not exactly linear. See, e.g.,: P.L. Krapivsky, S. Redner, and F. Leyvraz, Phys. Rev. Lett. 85, 4629 (2000).

T.

On 13 Feb 2011, at 14:23, Narges Zarabi <address@hidden> wrote:

Dear Tamas,

Thanks for your answer. Do you have any idea of how the value of alpha will change by changing the power argument? Is there a mathematical formula which defines the relationship?

Regards, Narsis

On Sun, Feb 13, 2011 at 1:21 PM, Tamás Nepusz <address@hidden> wrote:
> I get an alpha = 2.8 , which is different from the power I used in building the graph. Does anyone have any explanation for this?
The power argument does not specify the alpha exponent of the degree distribution; it specifies the exponent used when igraph determines the attachment probabilities after a new vertex is added. power=1 corresponds to linear preferential attachment, which should indeed correspond to alpha = 3 in the infinite limit, so the value of 2.8 you get seems to be correct.

--
Tamas


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