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On a Network Creation GamePoA SeminarPresenting: Oren GilonBased on an article by Fabrikant et al

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Background

•Attempt to model the creation of a joint infrastructure▫Internet▫Cell towers

•What costs people money?▫Creating connections▫Slow communication

•Selfish-Routing inaccurate

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Outline

•Define a model for our problem•Prove upper and lower bounds on the PoA•Present a conjecture concerning structure

of NE•Prove tighter bounds on PoA if conjecture

is true•State later results that improve bounds on

PoA

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The Model

•N players, each represented by a vertex•Each player i chooses a set of “links”, or

edges (i,j) he wishes to add to the graph•The price of each edge is α•A graph G is created by the union of all

edges•Each player also pays the sum of all

distances to other players in G

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More Formally

•Player i’s strategy is some , the set of vertices he wishes to connect to.

•For some strategy profile , the resulting graph is

•Player i’s cost under s is

{ }is N i

1, , Ns s s

1

[ ] ,N

ii

G s i sN

1

  ,N

i i G sj

c s s d i j

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Example 1

•Assume that the price of a link is 3•NE?

1 2

6

4

3

5

4 6 7 13c s

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Example 1

•No!

1 2

6

4

3

5

4 3 8 11c s

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Example 2•Assume that the price of a link is 3•Star•NE?

1 2

6

4

3

5

4 0 9 9c s

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Example 3

•Assume that the price of a link is 4•NE?•Cost(6) = 8+15=23

1 3

45

2

7

8

90

6

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Example 3

•Cost(6) = 23•Cost(4) = 4+3+12=19

1 3

45

2

7

8

90

6

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Example 3

•Cost(4) = 4+3+12=19•Cost(7) = 8+2+10+6=26

1 3

45

2

7

8

90

6

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Example 3

•Cost(7) = 8+2+10+6=26•Cost(6) = 8+3+12=23

1 3

45

2

7

8

90

6

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Example 3

•Cost(6) = 4+2+12+3 = 21 < 23•“Transient”

1 3

45

2

7

8

90

6

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Best-Responses

•It is NP-hard to calculate a player’s best-response to a given strategy profile

•Reduction from dominating-set▫We take the price of a link to be 2▫A player’s best response to a given graph is

clearly a minimal dominating-set of the graph

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Social Cost

•Sum of all costs of all players, i.e. social welfare

•In NE

because no edge is purchased twice

,

,G si j

C s C G s E d i j

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Social Cost•Distance between unconnected vertices is

at least 2, so

•If diameter is at most 2, bound is achieved

2 2 1 2

2 1 2

C E E n n E

n n E

G

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Special Cases

•In a NE, the graph may not be missing any edge that would decrease the sum of a vertex’s inter-node distances by more than α

•α<1▫Only NE is complete graph▫Social optimum is complete graph▫PoA=1

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Special Cases cont.

•If 1<α<2▫Social optimum is still the complete graph▫Every NE is of diameter of at-most 2▫Worst when |E| is minimal, i.e. |E|=n-1▫Worst NE is the star

2 1 2C n n EG

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Special Cases cont.

C 1 2( 1)(n 2) 2(n 1)

1 12

2(n 2) 2 4 4 2α

2 n 2 α1

2

4 4

2 α 3

n

star n n

C Kn n

n

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Special Cases cont.

•2<α▫Social optimum’s diameter is 2▫Social optimum has minimal |E|▫Star is the social optimum!▫It is also a NE, but there may be others…

2 1 2C n n EG

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Upper Bound on PoA,•A strategy profile s is a NE iff G[s] is a

tree▫Otherwise, a player could remove one of

his links and profit•Again, social optimum is the star

•PoA is O(1)

2α n

3

,

, ( 1) ( )

C(star) (n 1)

G si j

C s E d i j n n n

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Upper Bound on PoA,•Theorem: PoA is •Proof: The social optimum is a star•For any G that is the result of a NE

2α < n

( )O

,

2

,Gi j

E d i jC G

C Star n n

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Upper Bound on PoA,•Clearly

▫Otherwise i would profit by connecting to j

▫Saves at least

2α < n

, 2Gd i j

1 52 3 64

1

0

(2 1 2 ) (2 1 1)2i

i

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Upper Bound on PoA,•Lemma: for any graph with diameter d,

the cost is at most O(d) times that of the optimum.

•Proof:

•We need to bound the cost of edges

2α < n

2

2

2C G E n dO

C Star n n

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Upper Bound on PoA,•Two types of edges:

▫Cut edges – removing them disconnects the graph

▫Non-cut edges – all others•Cut edges cost at most•What do non-cut edges cost?

2α < n

( 1)n

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Upper Bound on PoA,•Consider the edges out of a vertex v•For a non-cut edge out of v, e, let

•These groups must be disjoint

2α < n

: eT u V the shortest path from v to u goes through e

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Upper Bound on PoA,•Before adding e, the shortest path

between u and v is of length at-most 2d•If and v are connected after

removing e, then for any

•and the total improvement is

2α < n

eT

'( , ) ( , ) 2G Gd v u d v u d

'( ( , ) ( , )) 2 | |e

G G eu T

d v u d v u d T

| | ( )eT d

eu T

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Upper Bound on PoA,•Thus v has at most non-cut edges.

•Using this, the total cost of non-cut edges is

•And we get

2α < n

( )nd

O

2( n ) O(n )nd

O d

2 2

2

2( )

C G n n d n dO O d

C Star n n

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Upper Bound on PoA,•As we have seen, the diameter of a NE is

•Using the lemma, this means that the PoA is also as desired

2α < n

( )O

( )O

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Tighter Upper Bound on PoA

•The Lemma we proved gives us a method of bounding the PoA

•We simply need to bound the diameter of NE graphs

•Lin proved a tighter bound on the diameter, giving a tighter bound on the PoA

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Tighter Upper Bound on PoA

•Theorem: the PoA is

•Proof: suffices to show the bound on the graph’s diameter

•Let G be a NE graph with diameter d•Let u, v be two nodes at distance d•Let

•Let B the set of nodes at distance at most d’ from u

( )On

1'

4

dd

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Tighter Upper Bound on PoA

v

u

w

G

B

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Tighter Upper Bound on PoA

•For some node w in B, we look at how d(v,w) changes after adding the edge (v,u)

•Before adding the edge:•After adding the edge:•And in total, v saves at least

( , ) d d'd v w ( , ) d' 1d v w

1( 2 ' 1) | B | | |

2

dd d B

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Tighter Upper Bound on PoA

•For some node w in B, let be all nodes t such that the u-t shortest path leaves B after w

•If is nonempty then d(u,w)=d’•Therefore, if u would connect to w it

would save

•There must exist a node w such that

wA

wA

| | ( ' 1)wA d

( | |)| |

| |w

n BA

B

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Tighter Upper Bound on PoA

•Combining these we get

•Which implies

( | |)( ' 1)

| |

n Bd

B

n | B | ( 1)' 1d

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Tighter Upper Bound on PoA

•And since

•Recalling

•We get

'd d ' 1

| B | n( )2

d

1| |

2

dB

2 ' 1( )

1

dn

d

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Tighter Upper Bound on PoA

•Which means

•And finally

as promised

2 2( ' 1)(d 1)( ) n(d' 1)

2

dn

4( ' 1) 1 4 9 ( )d d On n

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Tree Conjecture

•Experimentally, the only non-tree equilibrium found is the Petersen graph.

•Conjecture – for some constant A, if the price of a link is greater than A then all NE are trees.

•Could give a very strong bound on the PoA

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Tree NE

•Theorem: for any tree NE, the approximation ratio to the social optimum is at most 5

•Proof: Let s be the tree NE, and T be G[s]•Let L(i) be the largest connected

component after removing node i from the graph

•Let z be T’s central node•Let d be the tree’s depth when rooted at z

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Tree NE Proof

•If d is 1, T is a star•Otherwise, there is some leaf l at depth d

that decided not to link to z•Connecting to z would yield a profit of at

least

•And so we get

1 12

nd

42 2

2diam T d

n

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Tree NE Proof

•Since the distance between two nodes is at most the diameter of T,

and the approximation ratio

2

( ) ( 1) 2( 1) ( )( 2)( 1)

5 ( 1) 2( 1)

C T n n diam T n n

n n

25 n 1 2 n 1C T 5 n 1

5C Star 1 2 1 n 1n n n

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Tree Conjecture

•Sadly the conjecture was disproven in an article by Albers et al

•Proven: if then all NE are trees•Proven: the PoA is also bounded by 1.5 in

this case! •Tighter bound was shown in an article by

Mihalak et al, showing NE are trees for all

α 12 log logn n

α 273n

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A Lower Bound

•Theorem: For any there is an instance of the network connection game where the PoA is at least

•Proof: For any we look at the family of complete k-ary trees of depth d, . We call the number of nodes in this tree n, and set

0

3 4, 2k d

,k dT

( 1)d n

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k-ary trees of depth d

1

1 k…

d1 k…

1 k…

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A Lower Bound

•Lemma: If all links are purchased by the parent node, this is a NE

•Proof: The graph is a tree, every node must connect at least once to its subtree (otherwise an infinite penalty would be incurred)

•If only one link is sent to the subtree, clearly linking to the direct child is optimal

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A Lower Bound

•The node will not connect more than one link to any of its subtrees▫Additional links would shorten paths to at

most n nodes by at most d-1, but would cost

•A node will not add more than 1 link•A node will not send a link to the root•A node will not send a link to a sibling

( 1)d n

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A Lower Bound

•For any other node i and another node j not in its subtree and not i’s parent (trivially wasteful) we will see i does not link to j

•We look at the node that is an ancestor of j and a sibling of i, or the root if no such node exists

• i does not want to link to that node

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A Lower Bound

i…

d…

j …

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A Lower Bound

•In conclusion a node’s best response is:▫Connect exactly once to every subtree, at

the root.▫Don’t connect to any vertex outside your

subtrees

•The strategy is a NE!

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A Lower Bound

•The cost of the social optimum (the star) is

•By counting only distances between leaves, the cost of is at least

( 1) 2 ( 1) ( 1) n(n 1)n n n d

,k dT2

2

( 1) ( 1)( 1) 2 ( 1)( 1)

n k n kn d

k k

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A Lower Bound

•Thus, we get that the approximation ratio

2

,

,

2

( 1)( 1) 2 ( 1)( )

lim( ) ( 1) ( 1)

( 1) n(d 1) 2 (n 1)lim

( 1) ( 1)

3

k d

k d

d

n kn dC T k

C star d n n

n d

d n n

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Interesting Expansions

•Players buy fractions of links.▫More room for cooperation between the

players▫More room for inefficiency

•Edges are directed▫Less earned by being a “central node”▫Building edges is less worthwhile

•Weighted data▫Each player suffers differently from QoS▫Players with more data build more links

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Summary

Known Bound Proven Bound α

1 1 α<1

4/3 4/3 1<α<2

α<273n

5 273n<α<12nlogn

1.5 12nlogn<α<

O(1) O(1) <α

(1 )On

2n

2n

(1 )On

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Summary

•We presented a new game•We saw it is hard to compute best

response sequences in this game•We proved an upper bound on the PoA for

different parameters of the game•We showed a conjecture, and explained

why it would give us a good upper bound on the PoA

•We proved a lower bound on the PoA

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Questions?