Strategic Network Formation With Structural Holes

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By Jon Kleinberg, Siddharth Suri, Eva Tardos, Tom Wexler. Strategic Network Formation With Structural Holes. - PowerPoint PPT Presentation

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Strategic Network Formation With Structural Holes

By Jon Kleinberg, Siddharth Suri, Eva Tardos, Tom Wexler

Structural Holes

Structural holes theory suggests that node A is in a stronger position than the other nodes, because it can control the flow of information between the three otherwise independent groups of nodes

Structural Holes

The paper looks at what would happen to a social network graph if all the nodes were incentivized to become 'bridging' nodes

The Model

The payoff for a node u is

a |N(u)| + Σv,w N∈ (u) (β(rvw ) ) −Σv L(u)∈ (cuv ), where

a = the static benefit associated with having a link with another node

N(u) = the number of nodes connected to u

β = any decreasing function

rvw = the number of length 2 paths between v and w, if v and w are not connected, 0 otherwise

L(u) = the number of nodes u has bought a link to

cuv = the cost associated with the u,v edge

Computing a node's best move

Can be done in polynomial time

Proof in the paper, via a reduction to the largest weight ideal problem, which can be reduced to the minimum cut of a network

What kinds of graphs does this create?

Does equilibria exist for any number of nodes?

Can we always reach equilibria using best response updates?

Experiments: The possibility of cycling

a = .9β(r) = 2a/rcxy = 1

The Cost Matrix: Uniform

We first look at what would happen if the 'cost' of maintaining an edge was constant (in this case, cuv = 1 for every edge), and will try to answer the following questions:

Does there always exist some equilibrium, for a graph of n nodes?

If so, is it always reachable by round robin best response updates?

Does equilibrium exist: Uniform Metric

Let Gn,k be a multipartite graph of n nodes, where the nodes are split up into n/k roughly equal sized groups, and every node in the ith group buys connections to every node in the jth group, for all j<i

Does equilibrium exist: Uniform Metric

Can we chose k such that Gn,k is at equilibrium?

Yes - we do this by defining a benefit function B(n,k) = k(a-1) + Ck,2 β(n-k), and picking k' such that B(n, k')>0 and B(n, k'-1)<=0

Can we always reach equilibrium: Uniform Metric

We have shown that for any n, there is always a k, such that Gn,k is in equilibrium.

Will our algorithm for computing best response dynamics reach an equilibrium?

Do other Equilibria exist: Uniform

Yes! After running several

experiments, all equilibria were found to be dense, Ω(n^2) edges

The paper then proves that all equilibria are dense, assuming rβ(r) >0

The Cost Matrix: Hierarchical

Useful for situations like the dynamics of a large company's social network

Here, we let the cost cuv, be the unique simple path between nodes u,v in the tree

Does equilibrium exist: Hierarchical Metric

This is still an open question, for arbitrarily large n

However, running experiments suggest that when equilibrium does exist, it occurs with a small group of people with links to everyone, a few people with a significant number of links, and most with very few links

Average degree being O(√n)

Conclusions

In both hierarchical and uniform metrics, we end up with a network divided into social classes, where a small number of nodes maintain O(n) links, and most nodes have much less

Even starting from an empty graph, the bridging incentive causes a break in the symmetry, but what happens under different bridging conditions?

Other Research Sanjeev Goyal and Fernando Vega-Redondo's

'Structural holes in social networks' uses a model where a node u receives benefits from residing on arbitrarily long paths between two other nodes, w and v. Here, star networks turn out to be the most robust equilibrium, for a wide range of parameters

Vincent Buskens and Arnout van de Rijt's 'Dynamics of networks if everyone strives for structural holes' looks at only benefits from length 2 paths, but uses a stricter form of equilibrium, which they call unilateral stability

Questions?

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