Transcript snap.stanford.edu

CS224W: Social and Information Network Analysis
Jure Leskovec, Stanford University
http://cs224w.stanford.edu

Epidemic Model based on Random Trees
 (a variant of branching processes)
 A patient meets d other people
 With probability q>0 infects each
of them

Root node,
“patient 0”
Start of epidemic
d subtrees
Q: For which values of d and q
does the epidemic run forever?
𝑖𝑛𝑓𝑒𝑐𝑡𝑒𝑑 𝑛𝑜𝑑𝑒
 Run forever: lim 𝑃
>0
𝑎𝑡 𝑑𝑒𝑝𝑡ℎ 𝑛
𝑛→∞
 Die out:
-- || -=0
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
2



Too quick.
Nobody had a clue
what’s going on
pn = prob. there is an infected node at depth n
We need: lim 𝑝𝑛 =? (based on q and d)
𝑛→∞
Need recurrence for pn
𝑝𝑛 = 1 − 1 − 𝑞𝑝𝑛−1
𝑑
No infected node
at depth n by the root

lim 𝑝𝑛 = result of iterating
𝑛→∞
f x = 1 − 1 − 𝑞𝑥
𝑑
 Starting at x=1 (since p1=1)
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
3
f(x)
y=x
y=f x
Going to first
fixed point
When is this going to 0?
1
x
What do we know about f(x)?
f 0 =0
f 1 =1− 1−q d <1
f ′ x = qd 1 − qx d−1
f ′ 0 = qd ∶ f ′ (x) is monotone decreasing on [0,1]
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
4
f(x)
y=x
Reproductive
number R0=qd:
There is an epidemic
if R0  1
y=f x
1
x
We need f(x) to be bellow y=x!
f′ 0 < 1
lim 𝑝𝑛 = 0 ? to 𝑞𝑑 < 1
𝑛→∞
qd = expected # of people at we infect
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
5


Too quick.
Nobody had a clue
what’s going on
In this model nodes only go from
inactive  active
Can generalize to allow nodes to alternate
between active and inactive state by:
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
6
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
17

Blogs – Information epidemics
 Which are the influential/infectious blogs?
 Which blogs create big cascades?

Viral marketing
 Who are the influencers?
 Where should I advertise?

Disease spreading
vs.
 Where to place monitoring
stations to detect epidemics?
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
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
Independent Cascade Model
 Directed finite G=(V,E)
 Set S starts out with new behavior
 Say nodes with this behavior are “active”
 Each edge (v,w) has a probability pvw
 If node v is active, it gets one chance to
make w active, with probability pvw
 Each edge fires at most once

Does scheduling matter? No
 E.g., u,v both active, doesn’t matter which fires first
 But the time moves in discrete steps
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
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

Initially some nodes S are active
Each edge (v,w) has probability (weight) pvw
0.4
a
0.4
0.3
b
0.2
0.3
c

0.2
0.3
0.2
0.3
e
0.4
0.4

f
d
0.2
g
h
0.3
0.3
0.4
i
When node v becomes active:
 It activates each out-neighbor w with prob. pvw
Activations spread through the network
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
20


Have a little tree in
each circle so we
show which nodes get
infected
S: is initial active set
f(S): the expected size of final active set
graph G
b
a
d
c
… influence set
of a node

Set S is more influential if f(S) is larger
f({a,b} < f({a,c}) < f({a,d})
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
21
Problem:

a
Most influential set of
size k: set S of k nodes
producing largest
expected cascade size
f(S) if activated
[Domingos-Richardson ‘01]

0.4
0.4
d
0.2
0.3
0.3 0.2
0.3
b
f
0.2
e
h
0.4
0.4
0.2 0.3
0.3
0.3
g
i
0.4
c
Influence
set of a
Influence
set of b
f (S )
Optimization problem: max
S of size k
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
22


Most influential set of k nodes: set S on k
nodes producing largest expected cascade
size f(S) if activated
The optimization problem:
max f (S )
S of size k

How hard is this problem?
 NP-HARD!
 Show that finding most influential
set is at least as hard as a vertex cover
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
23

Vertex cover problem:
 Given universe of elements U={u1,…,un}
and sets S1,…, Sm  U
S3
S2
U
S1
S4
 Are there k sets among S1,…, Sm such that
their union is U?

Goal:
Encode vertex cover as an instance of max f (S )
S of size k
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
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

Given a vertex cover instance with sets S1,…, Sm
Build a bipartite “S-to-U” graph:
S1
S2
1
S3
1
Sm

1
u1
u2
e.g.:
S1={u1, u2, u3}
u3
un
Construction:
• Create edge
(Si,u) Si  uSi
-- directed edge
from sets to their
elements
• Put weight 1 on
each edge
There exists a set S of size k with f(S)=k+n
iff there exists a size k set cover
Note: Optimal solution is always a set of Si
This is hard in general, could be special cases that are easier
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
25

Bad news:
 Influence maximization is NP-hard

Next, good news:
 There exists an approximation algorithm!

Consider the Hill Climbing algorithm to find S:
 Input: Influence set of each node u = {v1, v2, … }
 If we activate u, nodes {v1, v2, … } will eventually get active
 Algorithm: At each step take the node u that gives
best marginal gain: max f(Si-1{u})
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
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Algorithm:
 Start with S0={}
 For i=1…k
 Take node u that max f(Si-1{u})
 Let Si = Si-1{u}

a
d
Example:
b
 Eval f({a}),… f({e}), pick max
 Eval f({a,b}),… f({a,e}), pick max
 Eval f(a,b,c}),… f({a,b,e}), pick …
10/13/2009
f(Si-1{u})
b
a
c
c
e
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
d
e
27

Hill climbing produces a solution S
where: f(S) (1-1/e)*OPT (f(S)>0.63*OPT)
[Nemhauser, Fisher, Wolsey ’78, Kempe, Kleinberg, Tardos ‘03]

Claim holds for functions f() with 2 properties:
 f is monotone: (activating more nodes doesn’t hurt)
if S  T then f(S)  f(T) and f({})=0
 f is submodular: (activating each additional node helps less)
adding an element to a set gives less improvement
than adding it to one of its subsets: S  T
f(S  {u}) – f(S) ≥ f(T  {u}) – f(T)
Gain of adding a node to a small set
10/13/2009
Gain of adding a node to a large set
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
28
Diminishing returns:
f(X)

S  T
f(T {u})
f(T)
f(S {u})
f(S)
Adding u to T helps less!
Solution size, |X|
f(S  {u}) – f(S) ≥ f(T  {u}) – f(T)
Gain of adding a node to a small set
10/13/2009
Gain of adding a node to a large set
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
29


We must show our f() is submodular:
S  T
(trivially uT)
f(S  {u}) – f(S) ≥ f(T  {u}) – f(T)
Gain of adding a node to a small set

Gain of adding a node to a large set
Basic fact 1:
 If f1(x), …,fk(x) are submodular, and c1,…,ck  0
then F x = 𝑖 𝑐𝑖 ∙ 𝑓𝑖 𝑥 is also submodular
(Linear combination of submodular functions is a submodular function)
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
31

S  T: f(S  {u}) – f(S) ≥ f(T  {u}) – f(T)
Gain of adding u to a small set

Gain of adding u to a large set
Basic fact 2: A simple submodular function
 Sets A1, …, Am
 𝑓 𝑆 = 𝑖∈𝑆 𝐴𝑖
(size of the union of sets Ai, iS)
 Claim: f(S) is submodular!
T
S
u
ST
10/20/2010
The more sets
you already
have the less
new area a new
set will cover
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
32

Principle of deferred decision:
 Flip all the coins at the
b
beginning and record
which edges fire successfully.
 Now we have a
c
deterministic graph!
 Edges which succeed are live

a
f
For the i-th realization of coin flips
 fi(S) = size of the set reachable by
live-edge paths from nodes in S
d
e
g
h
i
Influence sets:
fi(a) = {a,f,c,g}
fi(b) = {b,c},
fi(d) = {d,e,h}
fi(i) = {i,h}, …
 fi(S={a,b}) = {a,f,c,g,b}
 fi(S={a,d}) = {a,f,c,g,d,e,h}
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
33




a
d
Fix outcome i of coin flips
fi(v) = set of nodes
b
f
e
reachable from v on
live-edge paths
g
i
fi(S) = size of cascades
c
from S given coin flips i
𝑓𝑖 𝑆 = 𝑣∈𝑆 𝑓𝑖 (𝑣)  fi(S) is submodular!
h
 fi(v) are sets and fi(S) is the size of the union

Expected influence set size:
𝑓 𝑆 = 𝑖 𝑓𝑖 (𝑆)  f(S) is submodular!
 f(S) is linear combination of submodular functions
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
34

One slide showing the whole process:




Network,
Then Coin flips,
Then influence sets
We do multiple coin flips and get multiple
influence sets
 Maximizing under coin flips i is submodular
 We “average” over coin flip scenarios i
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
35


END: 60 min, the Robyn Came
For the proof
7/17/2015
60 min
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
36
Claim:
If f(S) is monotone and submodular.
Hill climbing produces a solution S
where: f(S) (1-1/e)*OPT (f(S)>0.63*OPT)
 Setting






Keep adding nodes that give the largest gain
Start with S0={}, produce sets S1, S2,…,Sk
Add elements one by one
Marginal gain: i = f(Si) - f(Si-1)
Let T={t1…tk} be the optimal set of size k
We need to show: f(S)  (1-1/e) f(T)
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
38

𝑓(𝐴 ∪ 𝐵) − 𝑓(𝐴) ≤
𝑘
𝑗=1[𝑓(𝐴
∪ {𝑏𝑗 }) − 𝑓(𝐴)]
 where: B = {b1,…,bk} and f is submodular,

Proof:
 Let Bi = {b1,…bi}, so we have B1, B2, …, Bk=B
 𝑓 𝐴∪B −𝑓 𝐴 =
=
≤
𝑘
𝑖=1 𝑓
𝑘
𝑖=1 𝑓
By submodularity
since AX {b}  A{b}
10/13/2009
𝑘
𝑖=1 𝑓
𝐴 ∪ 𝐵𝑖−1 ∪ 𝑏𝑖
𝐴 ∪ {𝑏𝑖 } − 𝑓 𝐴
𝐴 ∪ 𝐵𝑖 − 𝑓 𝐴 ∪ 𝐵𝑖−1
− 𝑓 𝐴 ∪ 𝐵𝑖−1
Work out the sum.
Everything but 1st and
last term cancels out.
𝑓 𝐴 ∪ 𝐵1 − 𝑓 𝐴 ∪ 𝐵0
+ 𝑓 𝐴 ∪ 𝐵2 − 𝑓 𝐴 ∪ 𝐵1
+ 𝑓 𝐴 ∪ 𝐵3 − ⋯
+ 𝑓 𝐴 ∪ 𝐵𝑘 − 𝑓(𝐴 ∪ 𝐵𝑘−1 )
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
39

𝑓 𝑇 ≤ 𝑓 𝑆𝑖 ∪ 𝑇

= 𝑓 𝑆𝑖 ∪ 𝑇 − 𝑓 𝑆𝑖 + 𝑓 𝑆𝑖

≤
𝑘
𝑗=1
𝑓 𝑆𝑖 ∪ {𝑡𝑗 } − 𝑓 𝑆𝑖
𝑘
𝑗=1 𝛿𝑖+1

≤

Thus: 𝑓 𝑇 ≤ 𝑓 𝑆𝑖 + 𝑘 𝛿𝑖+1

10/13/2009
𝛿𝑖+1 ≥
(by monotonicity)
+ 𝑓(𝑆𝑖 )
+ 𝑓 𝑆𝑖 = 𝑓 𝑆𝑖 + 𝑘 𝛿𝑖+1
1
[𝑓
𝑘
𝑇 − 𝑓(𝑆𝑖 )]
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
(by prev. slide)
T = {t1, … tk}
tj is one choice
of a next
element, and
we greedily
choose the
best one, for a
gain of i+1
40

We just showed: 𝛿𝑖+1 ≥

What is f(Si+1)?
1
[𝑓
𝑘
𝑇 − 𝑓(𝑆𝑖 )]
 𝑓 𝑆𝑖+1 = 𝑓 𝑆𝑖 + 𝛿𝑖+1
 ≥ 𝑓 𝑆𝑖 +
 = 1−

1
𝑘
1
𝑘
𝑓 𝑇 − 𝑓 𝑆𝑖
𝑓 𝑆𝑖 +
1
𝑓(𝑇)
𝑘
What is f(Sk)?
10/13/2009
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
41
  1 i 
 Claim: f ( Si )  1  1    f (T )
  k  
Proof by induction:
 𝑖 = 0:
 𝑓 𝑆0 = 𝑓({}) = 0
 1− 1
10/13/2009
1 0
−
𝑘
𝑓 𝑇 =0
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
42
  1 i 
 Claim: f ( Si )  1  1    f (T )
  k  
Proof by induction:
 At 𝑖 + 1:
 𝑓 𝑆𝑖+1 ≥ 1 −
 ≥ 1−
1
𝑘
𝑓 𝑆𝑖 +
1− 1−
 = 1− 1−
10/13/2009
1
𝑘
1 𝑖+1
𝑘
1 𝑖
𝑘
1
𝑓
𝑘
𝑓 𝑇
𝑇
1
+ 𝑓
𝑘
𝑇
𝑓(𝑇)
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
43

Thus:
𝑓 𝑆 = 𝑓 𝑆𝑘

≤
Then:
𝑓 𝑆𝑘
7/17/2015
1
≥ 1− 1−
𝑘
𝑘
𝑓 𝑇
𝟏
𝒆
1
≥ 1−
𝑓(𝑇)
𝑒
qed
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
44
We just proved:
 Hill climbing finds solution S which
f(S)  (1-1/e)*OPT
 this is a data independent bound
 This is a worst case bound
 No matter what is the input data (influence sets) we
know that Hill Climbing won’t do worse than 0.63*OPT
Data dependent bound:
 We want a bound whose value depends on
the input data
If the data is “easy”, we are likely doing better than 63% of OPT
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
45

Suppose S is some solution to
argmaxS f(S) s.t. |S|  k
f() is monotone & submodular
and let T = {t1,…,tk} be the OPT solution
 CLAIM:
For each u  S let u = f(S{u})-f(S)
Order u so that 1  2  …  n
Then: f(T)  f(S) + i=1k i
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
46
For each u  S let u = f(S{u})-f(S)
Order u so that 1  2  …  n
Then: f(T)  f(S) + i=1k i
 Proof:

 𝑓 𝑇 ≤𝑓 𝑇∪𝑆 =𝑓 𝑆 +
10/20/2010
𝑘
𝑖=1[𝑓
𝑆 ∪ 𝑡1 … 𝑡𝑖
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
−
47
7/17/2015
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
48
b
What do we know about
optimizing submodular
functions?
 A hill-climbing is near optimal
(1-1/e (~63%) of OPT)
c

Hill-climbing
reward
a
d
e
Add node with highest
marginal gain
10/20/2010
But
 Hill-climbing algorithm is slow
 At each iteration we need to reevaluate marginal gains
 It scales as O(n k)
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
49
[Leskovec et al., KDD ’07]

In round i+1: So far we picked Si = {s1,…,si}
 Now pick si+1 = argmaxu F(Si  {u}) - F(Si)
 maximize the “marginal benefit” u(Si) = F(Si  {u}) - F(Si)

By submodularity property:
𝑓 𝑆𝑖 ∪ 𝑢

− 𝑓 𝑆𝑖 ≥ 𝑓 𝑆𝑗 ∪ 𝑢
− 𝑓 𝑆𝑗 for i<j
Observation: Submodularity implies
i  j  x(Si)  x(Sj) since SiSj
Marginal benefits x only shrink!
u(Si)  u(Si+1)
u
Activating node u in step i helps
more than activating it at step j (j>i)
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
50
[Leskovec et al., KDD ’07]

Idea:
 Use i as upper-bound on j (j>i)

Lazy hill-climbing:
 Keep an ordered list of marginal
benefits i from previous
iteration
 Re-evaluate i only for top node
 Re-sort and prune
Marginal gain
a
b
c
d
e
f(S  {u}) – f(S) ≥ f(T  {u}) – f(T)
10/20/2010
S1={a}
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
ST
51
[Leskovec et al., KDD ’07]

Idea:
 Use i as upper-bound on j (j>i)

Lazy hill-climbing:
 Keep an ordered list of marginal
benefits i from previous
iteration
 Re-evaluate i only for top node
 Re-sort and prune
Marginal gain
a
b
c
d
e
f(S  {u}) – f(S) ≥ f(T  {u}) – f(T)
10/20/2010
S1={a}
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
ST
52
[Leskovec et al., KDD ’07]

Idea:
 Use i as upper-bound on j (j>i)

Lazy hill-climbing:
 Keep an ordered list of marginal
benefits i from previous
iteration
 Re-evaluate i only for top node
 Re-sort and prune
Marginal gain
a
S1={a}
d
S2={a,b}
b
e
c
f(S  {u}) – f(S) ≥ f(T  {u}) – f(T)
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
ST
53
[Leskovec et al., KDD ’07]

Given a real city water
distribution network

And data on how
contaminants spread
in the network

Detect the
contaminant as quickly
as possible

S
Problem posed by the
US Environmental
Protection Agency
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
55
[Leskovec et al., KDD ’07]


Given a graph G(V,E)
Data on how outbreaks spread over the
network:
 for each outbreak i we know the
time T(i,u) when outbreak i contaminated node u

Select a subset of nodes A that maximize
the expected reward:
Reward for detecting
outbreak i

Reward: Save the most people
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
56
[Leskovec et al., KDD ’07]

Observation: Diminishing returns
New sensor:
S1
S’
s’
S1
S3
S2
S4
Placement A={s1, s2}
Adding s’ helps a lot
10/20/2010
S2
Placement A’={s1, s2, s3, s4}
Adding s’ helps
very little
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
57
[Leskovec et al., KDD ’07]

Claim:
 The reward function is submodular

Consider outbreak i:
 Ri(uk) = set of nodes saved from uk
 Ri(A) = size of union Ri(uk), ukA
outbreak i
Ri is submodular

fi(u2)
u2
u1
fi(u1)
Global optimization:
 R(A) = i Prob(i) Ri(A)
 R(A) is submodular
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
58
[Leskovec et al., KDD ’07]

Real metropolitan area
network
water
 V = 21,000 nodes
 E = 25,000 pipes



Use a cluster of 50 machines for a month
Simulate 3.6 million epidemic scenarios
(152 GB of epidemic data)
By exploiting sparsity we fit it into main
memory (16GB)
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
59
1.4
“Offline”
Solution quality F(A)
Higher is better
1.2
the (1-1/e) bound
Data-dependent
bound
1
0.8
0.6
Hill Climbing
0.4
0.2
0
0
5
10
15
20
Number of sensors placed
Data-dependent bound is much tighter
(gives more accurate estimate of alg. performance)
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
60
[Leskovec et al., KDD ’07]

Placement heuristics perform much worse
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
61
= I have 10 minutes. Which
blogs should I read to be
most up to date?
?
= Who are the most
influential bloggers?
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
62
Want to read things
before others do.
Detect blue & yellow
soon but miss red.
Detect all
stories but late.
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
63
[Leskovec et al., KDD ’07]

Online bound is much tighter:
 87% instead of 63%
(1-1/e) bound
Data dependent
bound
Hill Climbing
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
64
[Leskovec et al., KDD ’07]

Heuristics perform much worse
10/20/2010
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
65
[Leskovec et al., KDD ’07]

Naïve
hill climbing
Lazy
hill climbing
10/20/2010
Lazy evaluation
runs 700 times
faster than naïve
Hill Climbing
algorithm
Jure Leskovec, Stanford CS224W: Social and Information Network Analysis, http://cs224w.stanford.edu
66