Introduction to complex networks. I discuss percolation and appearance of giant components and related issues such as epidemics and attack tolerance.

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.

17. Watts & Strogatz

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.

114. Failure and Attack

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?