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Transcript
Second case study:
Network Creation Games
(a.k.a. Local
Connection Games)
Introduction



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




n players: nodes V={1,…,n} 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=(s1,…, sn), the constructed
network will be G(S)
player u’s goal:


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




Each edge costs ≥0
distG(S)(u,v): length of a shortest path (in
terms of number of edges) in G(S) between
u and v
nu: number of edges bought by node u
Player u aims to minimize its cost:
costu(S) = nu +
v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
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,vV distG(S)(u,v)
Our goal



A network is optimal or socially efficient if it minimizes
the social cost
We aim to characterize the efficiency loss resulting
from selfishness, by using the Price of Stability (PoS)
and the Price of Anarchy (PoA)
We use Nash equilibrium (NE) as the solution concept: A
network (i.e., a formed graph) G(S)=(V,E) is stable (for
the given value ) if S is a NE, while a graph G=(V,E) is
stable if there exists a strategy vector S such that:



S is a NE
S forms G
Observe that any stable network must be connected,
since the distance between two nodes is infinite
whenever they are not connected
aspects of NCG



NCG are not potential games
Computing a best moves for a player is NPhard (as in the GCG)
The complexity of establishing the existence
of an improving moves 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 any 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 = |E| + u,vV distG(S)(u,v) ≥ 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
the SC of any star when m=n-1; indeed:
SC(Kn) =  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≥ LB(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
a 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