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Introduction to Complex Networks
V.A. Traag
KITLV, Leiden, the Netherlands
e-Humanities, KNAW, Amsterdam, the Netherlands
March 30, 2014
eRoyal Netherlands Academy of Arts and Sciences
Humanities
Overview
1 What are networks?
2 Classics: scale free & small worlds.
3 Percolation: giant components, failure & attack and epidemics.
Probability generating functions.
Examples
• Neural networks
• Power grids
• Gas networks
• Internet router network
• World Wide Web
• Road networks
• Airline networks
• Call networks
• Social networks
• Social media networks
Examples
• Neural networks
• Power grids
• Gas networks
• Internet router network
• World Wide Web
• Road networks
• Airline networks
• Call networks
• Social networks
• Social media networks
Examples
• Neural networks
• Power grids
• Gas networks
• Internet router network
• World Wide Web
• Road networks
• Airline networks
• Call networks
• Social networks
• Social media networks
Examples
• Neural networks
• Power grids
• Gas networks
• Internet router network
• World Wide Web
• Road networks
• Airline networks
• Call networks
• Social networks
• Social media networks
Examples
• Neural networks
• Power grids
• Gas networks
• Internet router network
• World Wide Web
• Road networks
• Airline networks
• Call networks
• Social networks
• Social media networks
Examples
• Neural networks
• Power grids
• Gas networks
• Internet router network
• World Wide Web
• Road networks
• Airline networks
• Call networks
• Social networks
• Social media networks
Basics
Network
• Graph or networks G = (V , E)
• Nodes V = 1, . . . , n (vertices)
Power station, webpage, intersection, person.
• Edges E ⊆ V × V (links, ties)
Power cables, hyperlinks, roads, friendships.
Can be directed, and possibly weighted
Essentials
• Degree ki is number of links at node i.
• If |E| = m number of edges, then i ki = 2m.
• Average degree k = 2m
n .
• Density p = m
(n
2)
= k
n−1 ≈ k
n .
• Most networks sparse: k , p low.
Picture worth . . . words
Visualisations essentially wrong, but sometimes insightful.
Need to assess statistics to understand networks.
Analysed properties
Analysis strategy
• Focus on some key (statistical) ingredients.
• Only overall general properties, no particulates.
• Compare to random graph: what can we expect?
• Modelling ⇒ replicate key properties.
Some key properties
• Degree distribution
• Degree correlations
• Path lengths
• Clustering
• Modularity
• Dynamics: inter event times
Small world
Milgram’s experiment (1960s)
• Ask people to reach specific person:
John Doe, Journalist, Kansas
• Send letter to acquaintance, who forwards, and so on
• Result: about 5 intermediaries to reach destination.
• Six degrees of separation.
Key question: is this different from a random graph?
Erd¨os-R´enyi (ER) graphs
Random graph
• Create empty graph G with n nodes.
• Every edge probability p of appearing.
• On average p n
2 = m edges.
• Average degree k = pn.
• Random graph essentially a (very simple) model.
• Was (and still is) used frequently.
Biology, epidemiology: well mixed population.
• Many interesting questions still.
Small world?
Path length
• Every node ki ≈ k , in steps, reach about k .
• When k = n reached whole network.
• Hence ≈ log n
log k : grows slowly!
Random edges create short paths.
Clustering
• Clustering, Ci =
ei
ki
2
.
• In ER graph Ci =
p3n(n − 1)
p2n(n − 1)
= p.
• Networks are sparse, low p, so low Ci .
Real world: both short paths & clustering. How to get that?
Small world?
Path length
• Every node ki ≈ k , in steps, reach about k .
• When k = n reached whole network.
• Hence ≈ log n
log k : grows slowly!
Random edges create short paths.
Clustering
• Clustering, Ci =
ei
ki
2
.
• In ER graph Ci =
p3n(n − 1)
p2n(n − 1)
= p.
• Networks are sparse, low p, so low Ci .
Real world: both short paths & clustering. How to get that?
Watts & Strogatz
Small world model
• Create lattice (connect to nearest neighbours).
• Rewire edge (or add) with probability p.
Watts & Strogatz
Watts & Strogatz
Small world model
• Create lattice (connect to nearest neighbours).
• Rewire edge (or add) with probability p.
Few shortcuts enough to create short paths
Degree distribution
0 20 40 60 80 100
ki ≈ k
Degree
Probability
• In real networks, power-law ki ∼ k−α, usually 2 < α < 3.
• In ER graphs, poisson ki ∼ k k
k! .
Degree distribution
100 101 102
Hubs
Degree
Probability
• In real networks, power-law ki ∼ k−α, usually 2 < α < 3.
• In ER graphs, poisson ki ∼ k k
k! .
Barab´asi & Albert
How to get power-law degree distribution?
Preferential attachment, cumulative advantage
Start with graph with q nodes
1 Add node
2 Add q links to previous nodes with probability pi ∼ ki
3 Repeat (1)-(2).
Results
• Analysis by master rate equation p(k) = k−1
2 p(k − 1) − k
2 p(k).
• Leads to p(k) = m(m+1)(m+2)
k(k+1)(k+2) ∼ k−3.
• Preferential attachment ⇒ scale free network.
Scale free
Scale free: so what?
Why does it matter?
• Scale free networks robust again random node failure.
• Vulnerable for targeted attacks (take out the hubs).
• No threshold for epidemic spreading.
Approach: percolation & generating functions.
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate mean k = k kpk
• Sum S = Si , pgf f (x) = E(xS )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate mean k = k kpk
• Sum S = Si , pgf f (x) = E(xS )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate mean k = k kpk
• Sum S = Si , pgf f (x) = E(xS )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate mean k = k kpk1k−1
• Sum S = Si , pgf f (x) = E(xS )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate mean k = g (1)
• Sum S = Si , pgf f (x) = E(xS )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate moment km = x ∂
∂x
m
g
x=1
• Sum S = Si , pgf f (x) = E(xS )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate moment km = x ∂
∂x
m
g
x=1
• Sum S = Si , pgf f (x) = E(xS )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate moment km = x ∂
∂x
m
g
x=1
• Sum S = Si , pgf f (x) = E(x Si )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate moment km = x ∂
∂x
m
g
x=1
• Sum S = Si , pgf f (x) = E(xSi )
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate moment km = x ∂
∂x
m
g
x=1
• Sum S = Si , pgf f (x) = g(x)
Generating functions
Definition (Generating function)
Let Pr(S = k) = pk. Then
g(x) = E(xS
) =
k
pkxk
is the probability generating function (pgf).
Properties
• Normalized g(1) = E(1S ) = k pk = 1
• Calculate moment km = x ∂
∂x
m
g
x=1
• Sum S = Si , pgf f (x) = g(x)m
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = k
n
k pk(1 − p)n−kxk
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = k
n
k pk(1 − p)n−kxk
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = k
n
k (xp)k(1 − p)n−k (binomial theorem)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = (px + (1 − p))n
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = (1 + p(x − 1))n (remember k = pn)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = (1 + k (x−1)
n )n (limn→∞, def. exp)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = e k (x−1)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = e k (x−1)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = e k (x−1)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (x) = k e k (x−1).
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = e k (x−1)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (1) = k .
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = e k (x−1)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (1) = k .
• Number of neighbours of m nodes g(x)m = em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = e k (x−1)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (1) = k .
• Number of neighbours of m nodes (g(x)m) = m k em k (x−1).
Degree generating function
Example, ER degree distribution
• Let pk be probability node has degree k.
• Take pk = n
k pk(1 − p)n−k (Erd¨os-R´enyi)
• Then pgf g(x) = e k (x−1)
• Normalized g(1) = e k (1−1) = 1.
• Mean g (1) = k .
• Number of neighbours of m nodes (g(1)m) = m k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k kpk
.
• Average neighbour degree k kpk
k .
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k kpk
.
• Average neighbour degree k kpk
k .
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k kpk
k .
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k kpk
k .
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2
k .
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2
k > k .
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2
k − k > 0.
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0.
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k qkxk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k(k + 1)pk+1xk
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k k kpkxk−1
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of second neighbours m k pmp2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of second neighbours m pm k p2(k|m)xk
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of second neighbours m pmg1(x)m
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of second neighbours g(g1(x))
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of third neighbours g(g1(g1(x)))
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of d neighbours g(g1(· · · g1(x) · · · ))
• Average number of second neighbours k2 − k .
Excess degree distribution
Earlier: random node
Now: follow random edge to node
• Probability that node has k links is kpk
k .
• Average neighbour degree k2 − k 2
k > 0 (friendship paradox).
• Probability that node has k other links is qk = (k+1)pk+1
k
• pgf g1(x) = 1
k g (x)
• Second neigbhours for m degree g1(x)m = k p2(k|m)xk
• Distribution of d neighbours g(g1(· · · g1(x) · · · ))
• Average number of second neighbours k2 − k .
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u).
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u).
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u).
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u).
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u).
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u).
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u).
Giant component
Giant component (GC)
• Always only one GC (and lots of small ones).
• Probability link does not connect node to GC u.
• Probability node of degree k not in GC uk
• Probability node not in giant component k pkuk = g(u)
• Size of giant component: S = 1 − g(u).
But what is u?
Self consistency
• Probability link not connects to GC is u.
• Connects to node with k other neighbours: excess degree.
• Average probability: k qkuk = g1(u) = u.
Giant component
How to solve g1(u) = u?
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
g1(u)
u = g1(u)
u
Giant component
If derivative g1(1) > 1 giant component appears.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
g (1)
u
Giant component
• GC appears when 1 < g1(1).
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = k kqk.
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = 1
k k k(k + 1)pk+1.
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = 1
k k(k − 1)kpk.
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = 1
k k k2pk − kpk.
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = k2 − k
k .
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = k2 − k
k .
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = k2 − k
k .
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Giant component
• GC appears when 1 < g1(1) = k2 − k
k .
• For ER graphs k2 − k = k 2, so k > 1 the GC appears.
0 1 2 3 4 5 6
0
0.5
1
k
S
• For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node does not “fail”.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average k qk(1 − φ + φuk).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node “functions”.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average k qk(1 − φ + φuk).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average k qk(1 − φ + φuk).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average k qk(1 − φ + φuk).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average k qk(1 − φ + φuk).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average k qk(1 − φ + φuk).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average k qk(1 − φ + φuk).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average 1 − φ + φ k qkuk.
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average 1 − φ + φg1(u).
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average 1 − φ + φg1(u) = u.
• Solve for u gives solution.
Node failure
How fast is giant component destroyed if nodes are removed?
Same approach
• Probability φ node not removed from network.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φ).
• II: Neighbour is not removed (φ), but not in GC (uk).
• So, probability is 1 − φ + φuk.
• On average 1 − φ + φg1(u) = u.
• Solve for u gives solution.
Node failure
• Again, solving u = 1 − φ + φg1(u) not easy.
• But if ∂
∂u 1 − φ + φg1(u) > 1 GC exists.
• For ER φc = 1/ k , for scale free φc = 0.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ER
Scale Free
φ
S
Node failure
• Again, solving u = 1 − φ + φg1(u) not easy.
• But if ∂
∂u 1 − φ + φg1(u) > 1 GC exists.
• For ER φc = 1/ k , for scale free φc = 0.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ER
Scale Free
φ
S
Node failure
• Again, solving u = 1 − φ + φg1(u) not easy.
• But if φg1(u) > 1 GC exists.
• For ER φc = 1/ k , for scale free φc = 0.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ER
Scale Free
φ
S
Node failure
• Again, solving u = 1 − φ + φg1(u) not easy.
• But if φ > 1
g1(u)
GC exists.
• For ER φc = 1/ k , for scale free φc = 0.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ER
Scale Free
φ
S
Node failure
• Again, solving u = 1 − φ + φg1(u) not easy.
• But if φ > 1
g1(u)
= φc GC exists.
• For ER φc = 1/ k , for scale free φc = 0.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ER
Scale Free
φ
S
Node failure
• Again, solving u = 1 − φ + φg1(u) not easy.
• But if φ > 1
g1(u)
= φc GC exists.
• For ER φc = 1/ k , for scale free φc = 0.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ER
Scale Free
φ
S
Node failure
• Again, solving u = 1 − φ + φg1(u) not easy.
• But if φ > 1
g1(u)
= φc GC exists.
• For ER φc = 1/ k , for scale free φc = 0.
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ER
Scale Free
φ
S
0 0.2
Node attack
What if we attack specific nodes?
Same approach
• Probability φk node of degree k does not “fail”.
• On average φ = k φkpk.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φk).
• II: Neighbour is not removed (φk), but not in GC (uk−1).
• So on average u = k qk−1(1 − φk + φkuk−1).
• Define f (u) = k φkqk−1uk−1.
• Then u = 1 − f (1) + f (u), solve for u gives solution.
Node attack
What if we attack specific nodes?
Same approach
• Probability φk node of degree k does not “fail”.
• On average φ = k φkpk.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φk).
• II: Neighbour is not removed (φk), but not in GC (uk−1).
• So on average u = k qk−1(1 − φk + φkuk−1).
• Define f (u) = k φkqk−1uk−1.
• Then u = 1 − f (1) + f (u), solve for u gives solution.
Node attack
What if we attack specific nodes?
Same approach
• Probability φk node of degree k does not “fail”.
• On average φ = k φkpk.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φk).
• II: Neighbour is not removed (φk), but not in GC (uk−1).
• So on average u = k qk−1(1 − φk + φkuk−1).
• Define f (u) = k φkqk−1uk−1.
• Then u = 1 − f (1) + f (u), solve for u gives solution.
Node attack
What if we attack specific nodes?
Same approach
• Probability φk node of degree k does not “fail”.
• On average φ = k φkpk.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φk).
• II: Neighbour is not removed (φk), but not in GC (uk−1).
• So on average u = k qk−1(1 − φk + φkuk−1).
• Define f (u) = k φkqk−1uk−1.
• Then u = 1 − f (1) + f (u), solve for u gives solution.
Node attack
What if we attack specific nodes?
Same approach
• Probability φk node of degree k does not “fail”.
• On average φ = k φkpk.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φk).
• II: Neighbour is not removed (φk), but not in GC (uk−1).
• So on average u = k qk−1(1 − φk + φkuk−1).
• Define f (u) = k φkqk−1uk−1.
• Then u = 1 − f (1) + f (u), solve for u gives solution.
Node attack
What if we attack specific nodes?
Same approach
• Probability φk node of degree k does not “fail”.
• On average φ = k φkpk.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φk).
• II: Neighbour is not removed (φk), but not in GC (uk−1).
• So on average u = k qk−1(1 − φk + φkuk−1).
• Define f (u) = k φkqk−1uk−1.
• Then u = 1 − f (1) + f (u), solve for u gives solution.
Node attack
What if we attack specific nodes?
Same approach
• Probability φk node of degree k does not “fail”.
• On average φ = k φkpk.
• Again u probability link does not connect to GC.
Self consistency
• I: Neighbour is removed (1 − φk).
• II: Neighbour is not removed (φk), but not in GC (uk−1).
• So on average u = k qk−1(1 − φk + φkuk−1).
• Define f (u) = k φkqk−1uk−1.
• Then u = 1 − f (1) + f (u), solve for u gives solution.
Failure and Attack
Epidemics
Disease spreading
• Standard models: Susceptable, Infected, Recovered.
• SIR: transmission rate β, recovery rate ν.
• Infect neighbour with probability φ = 1 − eβτ
• How far will it spread: giant component.
Percolation
• I: Disease not transmitted (1 − φ).
• II: Disease transmitted (φ), but not to GC (uk).
• Already solved: critical φc = 1
g1(u)
.
• Epidemiological threshold βτ = log k2 − k
k2 −2 k
Epidemics
Disease spreading
• Standard models: Susceptable, Infected, Recovered.
• SIR: transmission rate β, infectious time τ = 1/ν.
• Infect neighbour with probability φ = 1 − eβτ
• How far will it spread: giant component.
Percolation
• I: Disease not transmitted (1 − φ).
• II: Disease transmitted (φ), but not to GC (uk).
• Already solved: critical φc = 1
g1(u)
.
• Epidemiological threshold βτ = log k2 − k
k2 −2 k
Epidemics
Disease spreading
• Standard models: Susceptable, Infected, Recovered.
• SIR: transmission rate β, infectious time τ = 1/ν.
• Infect neighbour with probability φ = 1 − eβτ
• How far will it spread: giant component.
Percolation
• I: Disease not transmitted (1 − φ).
• II: Disease transmitted (φ), but not to GC (uk).
• Already solved: critical φc = 1
g1(u)
.
• Epidemiological threshold βτ = log k2 − k
k2 −2 k
Epidemics
Disease spreading
• Standard models: Susceptable, Infected, Recovered.
• SIR: transmission rate β, infectious time τ = 1/ν.
• Infect neighbour with probability φ = 1 − eβτ
• How far will it spread: giant component.
Percolation
• I: Disease not transmitted (1 − φ).
• II: Disease transmitted (φ), but not to GC (uk).
• Already solved: critical φc = 1
g1(u)
.
• Epidemiological threshold βτ = log k2 − k
k2 −2 k
Epidemics
Disease spreading
• Standard models: Susceptable, Infected, Recovered.
• SIR: transmission rate β, infectious time τ = 1/ν.
• Infect neighbour with probability φ = 1 − eβτ
• How far will it spread: giant component.
Percolation
• I: Disease not transmitted (1 − φ).
• II: Disease transmitted (φ), but not to GC (uk).
• Already solved: critical φc = k
k2 − k
.
• Epidemiological threshold βτ = log k2 − k
k2 −2 k
Epidemics
Disease spreading
• Standard models: Susceptable, Infected, Recovered.
• SIR: transmission rate β, infectious time τ = 1/ν.
• Infect neighbour with probability φ = 1 − eβτ
• How far will it spread: giant component.
Percolation
• I: Disease not transmitted (1 − φ).
• II: Disease transmitted (φ), but not to GC (uk).
• Already solved: critical φc = k
k2 − k
.
• Epidemiological threshold βτ = log k2 − k
k2 −2 k
Epidemics
Epidemic threshold
• For ER, threshold βτ = log k
k −1.
• For scale free, k2 diverges: always epidemic outbreak.
0 0.2 0.4 0.6 0.8 1
0
0.5
1
ER
Scale Free
φ
S
Conclusions
Models
• Short pats & clustering: small world model
• Scale free: preferential attachment
• Many other mechanisms: e.g. triadic closure, homophily, etc. . .
• Focus on stylistic features.
Analysis
• Scale-free networks robust, spread fast, but vulnerable for attack.
• Generating functions greatly help analysis.
• Compare observed network to random/model. How does it
deviate?
Questions?

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Introduction to complex networks

  • 1. Introduction to Complex Networks V.A. Traag KITLV, Leiden, the Netherlands e-Humanities, KNAW, Amsterdam, the Netherlands March 30, 2014 eRoyal Netherlands Academy of Arts and Sciences Humanities
  • 2. Overview 1 What are networks? 2 Classics: scale free & small worlds. 3 Percolation: giant components, failure & attack and epidemics. Probability generating functions.
  • 3. Examples • Neural networks • Power grids • Gas networks • Internet router network • World Wide Web • Road networks • Airline networks • Call networks • Social networks • Social media networks
  • 4. Examples • Neural networks • Power grids • Gas networks • Internet router network • World Wide Web • Road networks • Airline networks • Call networks • Social networks • Social media networks
  • 5. Examples • Neural networks • Power grids • Gas networks • Internet router network • World Wide Web • Road networks • Airline networks • Call networks • Social networks • Social media networks
  • 6. Examples • Neural networks • Power grids • Gas networks • Internet router network • World Wide Web • Road networks • Airline networks • Call networks • Social networks • Social media networks
  • 7. Examples • Neural networks • Power grids • Gas networks • Internet router network • World Wide Web • Road networks • Airline networks • Call networks • Social networks • Social media networks
  • 8. Examples • Neural networks • Power grids • Gas networks • Internet router network • World Wide Web • Road networks • Airline networks • Call networks • Social networks • Social media networks
  • 9. Basics Network • Graph or networks G = (V , E) • Nodes V = 1, . . . , n (vertices) Power station, webpage, intersection, person. • Edges E ⊆ V × V (links, ties) Power cables, hyperlinks, roads, friendships. Can be directed, and possibly weighted Essentials • Degree ki is number of links at node i. • If |E| = m number of edges, then i ki = 2m. • Average degree k = 2m n . • Density p = m (n 2) = k n−1 ≈ k n . • Most networks sparse: k , p low.
  • 10. Picture worth . . . words Visualisations essentially wrong, but sometimes insightful. Need to assess statistics to understand networks.
  • 11. Analysed properties Analysis strategy • Focus on some key (statistical) ingredients. • Only overall general properties, no particulates. • Compare to random graph: what can we expect? • Modelling ⇒ replicate key properties. Some key properties • Degree distribution • Degree correlations • Path lengths • Clustering • Modularity • Dynamics: inter event times
  • 12. Small world Milgram’s experiment (1960s) • Ask people to reach specific person: John Doe, Journalist, Kansas • Send letter to acquaintance, who forwards, and so on • Result: about 5 intermediaries to reach destination. • Six degrees of separation. Key question: is this different from a random graph?
  • 13. Erd¨os-R´enyi (ER) graphs Random graph • Create empty graph G with n nodes. • Every edge probability p of appearing. • On average p n 2 = m edges. • Average degree k = pn. • Random graph essentially a (very simple) model. • Was (and still is) used frequently. Biology, epidemiology: well mixed population. • Many interesting questions still.
  • 14. Small world? Path length • Every node ki ≈ k , in steps, reach about k . • When k = n reached whole network. • Hence ≈ log n log k : grows slowly! Random edges create short paths. Clustering • Clustering, Ci = ei ki 2 . • In ER graph Ci = p3n(n − 1) p2n(n − 1) = p. • Networks are sparse, low p, so low Ci . Real world: both short paths & clustering. How to get that?
  • 15. Small world? Path length • Every node ki ≈ k , in steps, reach about k . • When k = n reached whole network. • Hence ≈ log n log k : grows slowly! Random edges create short paths. Clustering • Clustering, Ci = ei ki 2 . • In ER graph Ci = p3n(n − 1) p2n(n − 1) = p. • Networks are sparse, low p, so low Ci . Real world: both short paths & clustering. How to get that?
  • 16. Watts & Strogatz Small world model • Create lattice (connect to nearest neighbours). • Rewire edge (or add) with probability p.
  • 18. Watts & Strogatz Small world model • Create lattice (connect to nearest neighbours). • Rewire edge (or add) with probability p. Few shortcuts enough to create short paths
  • 19. Degree distribution 0 20 40 60 80 100 ki ≈ k Degree Probability • In real networks, power-law ki ∼ k−α, usually 2 < α < 3. • In ER graphs, poisson ki ∼ k k k! .
  • 20. Degree distribution 100 101 102 Hubs Degree Probability • In real networks, power-law ki ∼ k−α, usually 2 < α < 3. • In ER graphs, poisson ki ∼ k k k! .
  • 21. Barab´asi & Albert How to get power-law degree distribution? Preferential attachment, cumulative advantage Start with graph with q nodes 1 Add node 2 Add q links to previous nodes with probability pi ∼ ki 3 Repeat (1)-(2). Results • Analysis by master rate equation p(k) = k−1 2 p(k − 1) − k 2 p(k). • Leads to p(k) = m(m+1)(m+2) k(k+1)(k+2) ∼ k−3. • Preferential attachment ⇒ scale free network.
  • 22. Scale free Scale free: so what? Why does it matter? • Scale free networks robust again random node failure. • Vulnerable for targeted attacks (take out the hubs). • No threshold for epidemic spreading. Approach: percolation & generating functions.
  • 23. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate mean k = k kpk • Sum S = Si , pgf f (x) = E(xS )
  • 24. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate mean k = k kpk • Sum S = Si , pgf f (x) = E(xS )
  • 25. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate mean k = k kpk • Sum S = Si , pgf f (x) = E(xS )
  • 26. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate mean k = k kpk1k−1 • Sum S = Si , pgf f (x) = E(xS )
  • 27. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate mean k = g (1) • Sum S = Si , pgf f (x) = E(xS )
  • 28. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate moment km = x ∂ ∂x m g x=1 • Sum S = Si , pgf f (x) = E(xS )
  • 29. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate moment km = x ∂ ∂x m g x=1 • Sum S = Si , pgf f (x) = E(xS )
  • 30. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate moment km = x ∂ ∂x m g x=1 • Sum S = Si , pgf f (x) = E(x Si )
  • 31. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate moment km = x ∂ ∂x m g x=1 • Sum S = Si , pgf f (x) = E(xSi )
  • 32. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate moment km = x ∂ ∂x m g x=1 • Sum S = Si , pgf f (x) = g(x)
  • 33. Generating functions Definition (Generating function) Let Pr(S = k) = pk. Then g(x) = E(xS ) = k pkxk is the probability generating function (pgf). Properties • Normalized g(1) = E(1S ) = k pk = 1 • Calculate moment km = x ∂ ∂x m g x=1 • Sum S = Si , pgf f (x) = g(x)m
  • 34. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = k n k pk(1 − p)n−kxk • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 35. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = k n k pk(1 − p)n−kxk • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 36. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = k n k (xp)k(1 − p)n−k (binomial theorem) • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 37. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = (px + (1 − p))n • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 38. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = (1 + p(x − 1))n (remember k = pn) • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 39. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = (1 + k (x−1) n )n (limn→∞, def. exp) • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 40. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = e k (x−1) • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 41. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = e k (x−1) • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 42. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = e k (x−1) • Normalized g(1) = e k (1−1) = 1. • Mean g (x) = k e k (x−1). • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 43. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = e k (x−1) • Normalized g(1) = e k (1−1) = 1. • Mean g (1) = k . • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 44. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = e k (x−1) • Normalized g(1) = e k (1−1) = 1. • Mean g (1) = k . • Number of neighbours of m nodes g(x)m = em k (x−1).
  • 45. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = e k (x−1) • Normalized g(1) = e k (1−1) = 1. • Mean g (1) = k . • Number of neighbours of m nodes (g(x)m) = m k em k (x−1).
  • 46. Degree generating function Example, ER degree distribution • Let pk be probability node has degree k. • Take pk = n k pk(1 − p)n−k (Erd¨os-R´enyi) • Then pgf g(x) = e k (x−1) • Normalized g(1) = e k (1−1) = 1. • Mean g (1) = k . • Number of neighbours of m nodes (g(1)m) = m k .
  • 47. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k kpk . • Average neighbour degree k kpk k . • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 48. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k kpk . • Average neighbour degree k kpk k . • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 49. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k kpk k . • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 50. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k kpk k . • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 51. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 k . • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 52. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 k > k . • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 53. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 k − k > 0. • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 54. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0. • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 55. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 56. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 57. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k qkxk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 58. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k(k + 1)pk+1xk • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 59. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k k kpkxk−1 • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 60. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 61. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 62. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 63. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of second neighbours m k pmp2(k|m)xk • Average number of second neighbours k2 − k .
  • 64. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of second neighbours m pm k p2(k|m)xk • Average number of second neighbours k2 − k .
  • 65. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of second neighbours m pmg1(x)m • Average number of second neighbours k2 − k .
  • 66. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of second neighbours g(g1(x)) • Average number of second neighbours k2 − k .
  • 67. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of third neighbours g(g1(g1(x))) • Average number of second neighbours k2 − k .
  • 68. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of d neighbours g(g1(· · · g1(x) · · · )) • Average number of second neighbours k2 − k .
  • 69. Excess degree distribution Earlier: random node Now: follow random edge to node • Probability that node has k links is kpk k . • Average neighbour degree k2 − k 2 k > 0 (friendship paradox). • Probability that node has k other links is qk = (k+1)pk+1 k • pgf g1(x) = 1 k g (x) • Second neigbhours for m degree g1(x)m = k p2(k|m)xk • Distribution of d neighbours g(g1(· · · g1(x) · · · )) • Average number of second neighbours k2 − k .
  • 70. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u).
  • 71. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u).
  • 72. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u).
  • 73. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u).
  • 74. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u).
  • 75. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u).
  • 76. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u).
  • 77. Giant component Giant component (GC) • Always only one GC (and lots of small ones). • Probability link does not connect node to GC u. • Probability node of degree k not in GC uk • Probability node not in giant component k pkuk = g(u) • Size of giant component: S = 1 − g(u). But what is u? Self consistency • Probability link not connects to GC is u. • Connects to node with k other neighbours: excess degree. • Average probability: k qkuk = g1(u) = u.
  • 78. Giant component How to solve g1(u) = u? 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 g1(u) u = g1(u) u
  • 79. Giant component If derivative g1(1) > 1 giant component appears. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 g (1) u
  • 80. Giant component • GC appears when 1 < g1(1). • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 81. Giant component • GC appears when 1 < g1(1) = k kqk. • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 82. Giant component • GC appears when 1 < g1(1) = 1 k k k(k + 1)pk+1. • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 83. Giant component • GC appears when 1 < g1(1) = 1 k k(k − 1)kpk. • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 84. Giant component • GC appears when 1 < g1(1) = 1 k k k2pk − kpk. • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 85. Giant component • GC appears when 1 < g1(1) = k2 − k k . • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 86. Giant component • GC appears when 1 < g1(1) = k2 − k k . • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 87. Giant component • GC appears when 1 < g1(1) = k2 − k k . • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 88. Giant component • GC appears when 1 < g1(1) = k2 − k k . • For ER graphs k2 − k = k 2, so k > 1 the GC appears. 0 1 2 3 4 5 6 0 0.5 1 k S • For scale free graphs k2 → ∞ , so always GC (if 2 < α < 3).
  • 89. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node does not “fail”. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average k qk(1 − φ + φuk). • Solve for u gives solution.
  • 90. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node “functions”. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average k qk(1 − φ + φuk). • Solve for u gives solution.
  • 91. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average k qk(1 − φ + φuk). • Solve for u gives solution.
  • 92. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average k qk(1 − φ + φuk). • Solve for u gives solution.
  • 93. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average k qk(1 − φ + φuk). • Solve for u gives solution.
  • 94. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average k qk(1 − φ + φuk). • Solve for u gives solution.
  • 95. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average k qk(1 − φ + φuk). • Solve for u gives solution.
  • 96. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average 1 − φ + φ k qkuk. • Solve for u gives solution.
  • 97. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average 1 − φ + φg1(u). • Solve for u gives solution.
  • 98. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average 1 − φ + φg1(u) = u. • Solve for u gives solution.
  • 99. Node failure How fast is giant component destroyed if nodes are removed? Same approach • Probability φ node not removed from network. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φ). • II: Neighbour is not removed (φ), but not in GC (uk). • So, probability is 1 − φ + φuk. • On average 1 − φ + φg1(u) = u. • Solve for u gives solution.
  • 100. Node failure • Again, solving u = 1 − φ + φg1(u) not easy. • But if ∂ ∂u 1 − φ + φg1(u) > 1 GC exists. • For ER φc = 1/ k , for scale free φc = 0. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ER Scale Free φ S
  • 101. Node failure • Again, solving u = 1 − φ + φg1(u) not easy. • But if ∂ ∂u 1 − φ + φg1(u) > 1 GC exists. • For ER φc = 1/ k , for scale free φc = 0. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ER Scale Free φ S
  • 102. Node failure • Again, solving u = 1 − φ + φg1(u) not easy. • But if φg1(u) > 1 GC exists. • For ER φc = 1/ k , for scale free φc = 0. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ER Scale Free φ S
  • 103. Node failure • Again, solving u = 1 − φ + φg1(u) not easy. • But if φ > 1 g1(u) GC exists. • For ER φc = 1/ k , for scale free φc = 0. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ER Scale Free φ S
  • 104. Node failure • Again, solving u = 1 − φ + φg1(u) not easy. • But if φ > 1 g1(u) = φc GC exists. • For ER φc = 1/ k , for scale free φc = 0. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ER Scale Free φ S
  • 105. Node failure • Again, solving u = 1 − φ + φg1(u) not easy. • But if φ > 1 g1(u) = φc GC exists. • For ER φc = 1/ k , for scale free φc = 0. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ER Scale Free φ S
  • 106. Node failure • Again, solving u = 1 − φ + φg1(u) not easy. • But if φ > 1 g1(u) = φc GC exists. • For ER φc = 1/ k , for scale free φc = 0. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ER Scale Free φ S 0 0.2
  • 107. Node attack What if we attack specific nodes? Same approach • Probability φk node of degree k does not “fail”. • On average φ = k φkpk. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φk). • II: Neighbour is not removed (φk), but not in GC (uk−1). • So on average u = k qk−1(1 − φk + φkuk−1). • Define f (u) = k φkqk−1uk−1. • Then u = 1 − f (1) + f (u), solve for u gives solution.
  • 108. Node attack What if we attack specific nodes? Same approach • Probability φk node of degree k does not “fail”. • On average φ = k φkpk. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φk). • II: Neighbour is not removed (φk), but not in GC (uk−1). • So on average u = k qk−1(1 − φk + φkuk−1). • Define f (u) = k φkqk−1uk−1. • Then u = 1 − f (1) + f (u), solve for u gives solution.
  • 109. Node attack What if we attack specific nodes? Same approach • Probability φk node of degree k does not “fail”. • On average φ = k φkpk. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φk). • II: Neighbour is not removed (φk), but not in GC (uk−1). • So on average u = k qk−1(1 − φk + φkuk−1). • Define f (u) = k φkqk−1uk−1. • Then u = 1 − f (1) + f (u), solve for u gives solution.
  • 110. Node attack What if we attack specific nodes? Same approach • Probability φk node of degree k does not “fail”. • On average φ = k φkpk. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φk). • II: Neighbour is not removed (φk), but not in GC (uk−1). • So on average u = k qk−1(1 − φk + φkuk−1). • Define f (u) = k φkqk−1uk−1. • Then u = 1 − f (1) + f (u), solve for u gives solution.
  • 111. Node attack What if we attack specific nodes? Same approach • Probability φk node of degree k does not “fail”. • On average φ = k φkpk. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φk). • II: Neighbour is not removed (φk), but not in GC (uk−1). • So on average u = k qk−1(1 − φk + φkuk−1). • Define f (u) = k φkqk−1uk−1. • Then u = 1 − f (1) + f (u), solve for u gives solution.
  • 112. Node attack What if we attack specific nodes? Same approach • Probability φk node of degree k does not “fail”. • On average φ = k φkpk. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φk). • II: Neighbour is not removed (φk), but not in GC (uk−1). • So on average u = k qk−1(1 − φk + φkuk−1). • Define f (u) = k φkqk−1uk−1. • Then u = 1 − f (1) + f (u), solve for u gives solution.
  • 113. Node attack What if we attack specific nodes? Same approach • Probability φk node of degree k does not “fail”. • On average φ = k φkpk. • Again u probability link does not connect to GC. Self consistency • I: Neighbour is removed (1 − φk). • II: Neighbour is not removed (φk), but not in GC (uk−1). • So on average u = k qk−1(1 − φk + φkuk−1). • Define f (u) = k φkqk−1uk−1. • Then u = 1 − f (1) + f (u), solve for u gives solution.
  • 115. Epidemics Disease spreading • Standard models: Susceptable, Infected, Recovered. • SIR: transmission rate β, recovery rate ν. • Infect neighbour with probability φ = 1 − eβτ • How far will it spread: giant component. Percolation • I: Disease not transmitted (1 − φ). • II: Disease transmitted (φ), but not to GC (uk). • Already solved: critical φc = 1 g1(u) . • Epidemiological threshold βτ = log k2 − k k2 −2 k
  • 116. Epidemics Disease spreading • Standard models: Susceptable, Infected, Recovered. • SIR: transmission rate β, infectious time τ = 1/ν. • Infect neighbour with probability φ = 1 − eβτ • How far will it spread: giant component. Percolation • I: Disease not transmitted (1 − φ). • II: Disease transmitted (φ), but not to GC (uk). • Already solved: critical φc = 1 g1(u) . • Epidemiological threshold βτ = log k2 − k k2 −2 k
  • 117. Epidemics Disease spreading • Standard models: Susceptable, Infected, Recovered. • SIR: transmission rate β, infectious time τ = 1/ν. • Infect neighbour with probability φ = 1 − eβτ • How far will it spread: giant component. Percolation • I: Disease not transmitted (1 − φ). • II: Disease transmitted (φ), but not to GC (uk). • Already solved: critical φc = 1 g1(u) . • Epidemiological threshold βτ = log k2 − k k2 −2 k
  • 118. Epidemics Disease spreading • Standard models: Susceptable, Infected, Recovered. • SIR: transmission rate β, infectious time τ = 1/ν. • Infect neighbour with probability φ = 1 − eβτ • How far will it spread: giant component. Percolation • I: Disease not transmitted (1 − φ). • II: Disease transmitted (φ), but not to GC (uk). • Already solved: critical φc = 1 g1(u) . • Epidemiological threshold βτ = log k2 − k k2 −2 k
  • 119. Epidemics Disease spreading • Standard models: Susceptable, Infected, Recovered. • SIR: transmission rate β, infectious time τ = 1/ν. • Infect neighbour with probability φ = 1 − eβτ • How far will it spread: giant component. Percolation • I: Disease not transmitted (1 − φ). • II: Disease transmitted (φ), but not to GC (uk). • Already solved: critical φc = k k2 − k . • Epidemiological threshold βτ = log k2 − k k2 −2 k
  • 120. Epidemics Disease spreading • Standard models: Susceptable, Infected, Recovered. • SIR: transmission rate β, infectious time τ = 1/ν. • Infect neighbour with probability φ = 1 − eβτ • How far will it spread: giant component. Percolation • I: Disease not transmitted (1 − φ). • II: Disease transmitted (φ), but not to GC (uk). • Already solved: critical φc = k k2 − k . • Epidemiological threshold βτ = log k2 − k k2 −2 k
  • 121. Epidemics Epidemic threshold • For ER, threshold βτ = log k k −1. • For scale free, k2 diverges: always epidemic outbreak. 0 0.2 0.4 0.6 0.8 1 0 0.5 1 ER Scale Free φ S
  • 122. Conclusions Models • Short pats & clustering: small world model • Scale free: preferential attachment • Many other mechanisms: e.g. triadic closure, homophily, etc. . . • Focus on stylistic features. Analysis • Scale-free networks robust, spread fast, but vulnerable for attack. • Generating functions greatly help analysis. • Compare observed network to random/model. How does it deviate? Questions?