How much can a few nodes affect the PageRank values

Tight bounds
on sparse perturbations of Markov Chains
Romain Hollanders
Giacomo Como
Jean-Charles Delvenne
Raphaël Jungers
MTNS’2014
UCLouvain
University of Lund
PageRank is the average portion of time spent in a node
During an infinite random walk
PageRank is the average portion of time spent in a node
During an infinite random walk
PageRank is the average portion of time spent in a node
During an infinite random walk
PageRank : πœ‹ = 𝑃𝑇 πœ‹, πŸπ‘‡ πœ‹ = 1
How much can a few nodes affect the PageRank values ?
PageRank : πœ‹ = 𝑃𝑇 πœ‹, πŸπ‘‡ πœ‹ = 1
How much can a few nodes affect the PageRank values ?
𝓦
PageRank : πœ‹ = 𝑃𝑇 πœ‹, πŸπ‘‡ πœ‹ = 1
How much can a few nodes affect the PageRank values ?
𝓦
PageRank : πœ‹ = 𝑃𝑇 πœ‹, πŸπ‘‡ πœ‹ = 1
How much can a few nodes affect the PageRank values ?
𝓦
~𝑇 ~
~
PageRank : πœ‹ = 𝑃 πœ‹, πŸπ‘‡~
πœ‹=1
How much can a few nodes affect a consensus ?
Consensus : π‘₯𝑑 = 𝑃π‘₯π‘‘βˆ’1
How much can a few nodes affect a consensus ?
the weight of each agent in the final decision
Consensus : πœ‹ = 𝑃𝑇 πœ‹, πŸπ‘‡ πœ‹ = 1
How much can a few nodes affect a consensus ?
𝓦
~𝑇 ~
~
Consensus : πœ‹ = 𝑃 πœ‹, πŸπ‘‡~
πœ‹=1
How large can πœ‹ βˆ’ πœ‹
be ?
Weak bounds already exist
They depend more on the size than the structure of the network / perturbation
πœ‹βˆ’πœ‹
𝑝
≀ πœ…π‘ƒ β‹… 𝑃 βˆ’ 𝑃
π‘ž
Condition number of 𝑃
Sensitive mainly to the magnitude of the perturbation
But what if perturbations only affect a few rows of 𝑃 ?
Typically blows up when the network size grows
Unless 𝑃 βˆ’ 𝑃
⟹
π‘ž
vanishes
We need better, tighter bounds, adapted to local perturbations !
Como & Fagnani proposed a bound for the 1-norm
a nice increasing function
escape time from 𝒲
πœ‹βˆ’πœ‹
1
πœπ’²β†’π’±
≀𝑓 πœβ‹…
πœπ’±β†’π’²
mixing time
hitting time from 𝒱 to 𝒲
Captures local perturbations
Provides physical insight
Difficult (impossible?) to extend to other norms
No reason to believe that it is tight
It is possible to compute the maximum of 𝝅 βˆ’ 𝝅
∞
Exactly and in polynomial time
1.
Compute πœ‹ from πœ‹ = 𝑃𝑇 πœ‹, πŸπ‘‡ πœ‹ = 1
2.
For all 𝑣, compute min πœ‹π‘£ βˆ’ πœ‹π‘£
𝑃
and max πœ‹π‘£ βˆ’ πœ‹π‘£
𝑃
Ξ”π‘šπ‘Žπ‘₯
𝑣
Ξ”π‘šπ‘–π‘›
𝑣
πœ‹π‘£
min πœ‹π‘£
𝑃
Ξ”π‘šπ‘–π‘›
𝑣
3.
max πœ‹π‘£
𝑃
Ξ”π‘šπ‘Žπ‘₯
𝑣
Return the largest Ξ” encountered over all 𝑣’s
Finding 𝐦𝐚𝐱 𝝅𝒗 is easy
𝑷
𝑣
probability 1
𝓦
1
πœ‹π‘£ =
expected time between two visits of 𝑣
Finding 𝐦𝐒𝐧 𝝅𝒗 is easy too but…
𝑷
𝑣
probability 1
𝓦
1
πœ‹π‘£ =
expected time between two visits of 𝑣
Finding 𝐦𝐒𝐧 𝝅𝒗 if we fix the escape time from 𝓦
𝑷
Let us add the constraint that πœπ’²β†’π’± = T
Furthest away from 𝑣
𝑣
𝑒
??
probability 1 βˆ’ 𝑝(𝑇)
probability 𝑝(𝑇)
𝓦
1
πœ‹π‘£ =
expected time between two visits of 𝑣
A counter example
𝑒′
𝑣
𝑀
𝑒
A counter example
distance 3.33
from 𝑣
𝑒′
𝑣
𝑀
𝑒
distance 4
from 𝑣
A counter example
distance 3.33
from 𝑣
𝑒′
𝑣
𝑒
probability 1 βˆ’ 𝑝(𝑇)
distance 4
from 𝑣
To minimize 𝝅𝒗
the optimal solution
is to go all-in to 𝒖′
and not to 𝒖
𝑀
probability 𝑝(𝑇)
⟹
We need to loop through every candidate β€œworst-node”…
The algorithm to compute the maximum of 𝝅 βˆ’ 𝝅
∞
over all perturbations of the nodes of 𝒲, under the escape time constraint πœπ’²β†’π’± = T
1.
Compute πœ‹ from πœ‹ = 𝑃𝑇 πœ‹, πŸπ‘‡ πœ‹ = 1
2.
For each 𝑣, compute:
Ξ”π‘šπ‘Žπ‘₯
= max πœ‹π‘£ βˆ’ πœ‹π‘£
𝑣
𝑃
1 computation of πœ‹π‘£
all nodes of 𝒲 go to 𝑣 with probability 1
Ξ”π‘šπ‘–π‘›
= min πœ‹π‘£ βˆ’ πœ‹π‘£
𝑣
𝑃
𝑛 computations of πœ‹π‘£
all nodes of 𝒲 go to some node 𝑒 with probability 1 βˆ’ 𝑝(𝑇)
and stay in 𝒲 with probability 𝑝(𝑇)
3.
Return the largest Ξ”π‘šπ‘–π‘›
or Ξ”π‘šπ‘Žπ‘₯
encountered
𝑣
𝑣
Perspectives
Improve the computation of Ξ”π‘šπ‘–π‘›
𝑣
by identifying the β€œworst-node” on the go, based on its distance to 𝑣
Extend the approach to other norms
especially the 1-norm
Compare the results with Como & Fagnani’s bound
to establish its quality
Thank you