Download Algorithmic Mechanism Design

Second case study:
Network Creation Games
(a.k.a. Local
Connection Games)
Introduction
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

Introduced in [FLMPS,PODC’03]
A LCG is a game that models the exnovo creation of a network
Players are nodes that:


pay for the links they personally activate
benefit from short paths on the created
network
[FLMPS,PODC’03]:
A. Fabrikant, A. Luthra, E. Maneva, C.H. Papadimitriou, S. Shenker,
On a network creation game, PODC’03
The model
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n players: nodes in a graph to be built
Strategy for player u: a set of incident edges (intuitively,
a player buys these edges, that will be then used
bidirectionally by everybody; however, only the owner of
an edge can remove it, in case he decides to change its
strategy)
Given a strategy vector S, the constructed network will
be G(S)
player u’s goal:
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
to spend as little as possible for buying edges (building cost)
to make the distance to other nodes as small as possible (usage
cost)
The model
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Each edge costs ≥0
distG(S)(u,v): length of a shortest path (in
terms of number of edges) between u and v
nu: number of edges bought by node u
Player u aims to minimize its cost:
costu(S) = nu + v distG(S)(u,v)
Cost of a player: an example
2
-1 3
-3
4
1
+
2
1
u

cu=+13
cu=2+9
(notice that if <4 this is an improving move for u)
Convention: arrow from the node buying the link
The social-choice function

To evaluate the overall quality of a
network, we consider the utilitarian social
cost, i.e., the sum of all players’ costs.
Observe that:
1.
2.
In G(S) each term distG(S)(u,v) contributes to
the overall quality twice
Each edge (u,v) is bough at most by one player
Social cost of a network G(S)=(V,E):
SC(S)=|E| + u,vdistG(S)(u,v)
Our goal
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We use Nash equilibrium (NE) as the solution concept
A network is optimal or socially efficient if it
minimizes the social cost
A graph G=(V,E) is stable (for a value ) if there
exists a strategy vector S such that:
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S is a NE
S forms G
Observe that any stable network must be connected,
since the distance between to nodes is infinite
whenever they are not connected
We aim to bound the efficiency loss resulting from
stability, by using the Price of Stability (PoS) and the
Price of Anarchy (PoA)
Some (bad) computational
aspects of NCG
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NCG are not potential games
Computing a best response for a player is NPhard
The complexity of establishing the existence
of an improving strategy for a player is open
Stable networks: an example

Set =5, and consider:
+2
-2
-1
-5
-1
+2
-1
+5
+5
+5
-5
+4
+1
-1
-5
-5
+1
That’s a stable network!
How does an optimal
network look like?
Some notation
Kn: complete graph
with n nodes
A star is a tree
with height at most 1
(when rooted at its
center)
Lemma
Il ≤2 then the complete graph is an optimal solution,
while if ≥2 then any star is an optimal solution.
proof
Let G=(V,E) be an optimal solution;
|E|=m and SC(G)=OPT
OPT≥ m + 2m + 2(n(n-1) -2m) =(-2)m + 2n(n-1)
LB(m)
Notice: LB(m) is equal to SC(Kn) when m=n(n-1)/2, and to
SC of any star when m=n-1; indeed:
SC(Kn) = |E| + u,vdistG(S)(u,v) =  n(n-1)/2 + n(n-1)
SC(star)= (n-1) + 2(n-1) + 2(n-1)(n-2) =  (n-1) + 2(n-1)2
and it is easy to see that they correspond to LB(n(n1)/2) and to LB(n-1), respectively,
Proof (continued)
G=(V,E): optimal solution;
|E|=m and SC(G)=OPT
LB(m)=(-2)m + 2n(n-1)
≥ 2
min m
LB(n-1) = SC of any star
OPT≥ min LB(m) ≥
m
≤ 2
max m
LB(n(n-1)/2) = SC(Kn)
Are the complete graph
and stars stable?
Lemma
Il ≤1 the complete graph is stable, while if ≥1 then
any star is stable.
proof
≤1
If a node removes any k owned
edges, it saves k in the building
cost, but it pays k≥k more in the
usage cost
Proof (continued)
≥1
c cannot change its strategy
v
c
u
If v buys any k edges it pays k more
in the building cost, but it saves only
k≤k in the usage cost
u’s cost is +1+2(n-2); if it removes (u,c) and buys any k edges,
it pays k in the building cost, and its usage cost becomes
k+2+3(n-k-2), and so its total cost increases to:
+ (k-1)+k +2+3n-3k-6 = +1+2(n-2)+[(k-1)-2k+n] ≥
+1+2(n-2)+[k-1-2k+n] = +1+2(n-2)+[n-k-1]
since the quantity in square brackets is not negative, being
1≤k≤n-1.
Theorem
For ≤1 and ≥2 the PoS is 1. For 1<<2 the PoS is at
most 4/3
proof
≤1 and ≥2 …trivial!
1<<2
…Kn is an optimal solution, any star T is stable…
maximized when   1
PoS ≤
SC(T)
SC(Kn)
-1(n-1) + 2n(n-1)
(-2)(n-1) + 2n(n-1)
=  n(n-1)/2 + n(n-1) ≤ n(n-1)/2 + n(n-1)
2n - 1
4n -2
= 3/2n = 3n
< 4/3
What about the
Price of Anarchy?
…for <1 the complete graph is the
only stable network,
(try to prove that formally)
hence PoA=1…
…for larger value of ?
Some more notation
The diameter of a graph G
is the maximum distance
between any two nodes
diam=2
diam=1
diam=4
Some more notation
An edge e is a cut edge of a
graph G=(V,E) if
G-e is disconnected
G-e=(V,E\{e})
A simple property:
Any graph has at most n-1 cut
edges (otherwise they would induce
a cycle, which is impossible…think
about it)
Theorem
The PoA is at most O().
proof
It follows from the following lemmas:
Lemma 1
The diameter of any stable network is at most 2 +1 .
Lemma 2
The SC of any stable network with diameter d is at most
3d times the optimum SC.
proof of Lemma 1
G: stable network
Consider a shortest path in G between two nodes u and v
u
v
2k≤ distG(u,v) ≤ 2k+1
for some k
…since G is stable:
≥k2
k ≤ 
distG(u,v) ≤ 2  + 1
k vertices reduce
their distance
from u
from ≥2k to 1
 ≥ 2k-1
from ≥2k-1 to 2
 ≥ 2k-3
●
●
●
from ≥ k+1 to k
 ≥1
k-1

(2i+1)=k2
i=0
Proposition 1
Let G be a network with diameter d, and let e=(u,v) be a
non-cut edge. Then in G-e, every node w increases its
distance from u by at most 2d
Proposition 2
Let G be a stable network, and let F be the set of
non-cut edges bought by a node u. Then |F|≤(n-1)2d/
Lemma 2
The SC of any stable network G=(V,E) with diameter d is
at most O(d) times the optimum SC.
proof
OPT ≥  (n-1) + n(n-1)
SC(G)= u,vdG(u,v) +  |E| ≤ d OPT+2d OPT = 3d OPT
≤dn(n-1) ≤d OPT
|E|=|Ecut| + |Enon-cut| ≤(n-1)+n(n-1)2d ≤ 2d OPT
≤(n-1)
≤n(n-1)2d/
Prop. 2