1. Decomposing Networks and Polya Urns with the Power of Choice Joint work with Christos Amanatidis, Richard Karp, Christos Papadimitriou, Martha Sideri Presented By: Henry Lin

2. Overview Motivations Linked Decompositions Preferential Attachment Model Analysis of a Polya Urns Process Open Problems

3. Motivations The Internet is large and growing rapidly – difficult to manage – routers cannot store too much data Can we divide up the network into distinct regions that can be managed mostly independently and still route efficiently?

4. Linked Decomposition Decompose into c connected components Each node belongs to 1 or 2 components Each component has size ~ a Any two components intersect

5. Linked Decomposition Example Decompose into c = 3 connected (small diameter) components Each node belongs to 1 or 2 components Each component has size ~ a = 6 Any two components intersect

6. Routing with Linked Decompositions How to route to node v in your component(s)? Store each node in your component(s) uv

7. Routing with Linked Decompositions How to route to node v in your component(s)? Store each node in your component(s) How to route to node v outside of component? Store an intersection node in your component to reach v u v w

8. Routing with Linked Decompositions How to route to node v in your component(s)? Store each node in your component(s) How to route to node v outside of component? Store an intersection node in your component to reach v u v w Requires O(a) storage per node Requires O(n/a) storage per node

9. Linked Decompositions Desiderata Recall: we can route with O(a+n/a) tables Therefore, we want: size a ~  n and thus number of components c ~  n and diameter ~ log n Can we do this?

10. Note: Routing with  n tables can be achieved by other means For example, by compact routing [AGMNT ‘04] But a linked decomposition is much more than a routing scheme: – It breaks down a network into many largely independent components, with no harm to routing capabilities It is also very simple

11. What about the Internet? Can we decompose into c ~  n components Each node belongs to 1 or 2 components Each component has diameter ~ log n Each component has size a ~  n Any pair of components intersect

12. Surprise! Yes! Our experiments show that the actual Internet graph has a linked decomposition with these approximate parameters Main point of this paper: A theoretical justification for this phenomenon By analyzing a well-studied random model of the Internet

13. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

14. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

15. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

16. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

17. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

18. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

19. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

20. Preferential Attachment Model PA(m) : Nodes arrive one at a time, each new node: – selects m nodes iid at random, proportional to degree – adds one edge to each of the m selected nodes

21. Linked Decomposition in PA(m) Decompose graph into c connected components Each vertex belongs to 1 or 2 components Each component has diameter d Each component has size about a Any two components intersect Our Main Result: PA(m) graphs have linked decompositions whp with parameters above – (Note: m depends on ε) = Θ(n 1/2+ε ) = Θ(n 1/2-ε ) = log n

22. Decomposing PA(m) graphs For each node t є {1,...,n 1/2-ε }, we assign node t to its own component n 1/2-ε 1…

23. Decomposing PA(m) graphs 1. For each node t є {n 1/2-ε + 1,...,n/2}, we look at where the m edges of node t point, and assign node t to the component of lowest total degree n 1/2-ε 1…n 1/2-ε +1

24. Decomposing PA(m) graphs n 1/2-ε 1…n 1/2-ε +1 … 2. For each node t є {n 1/2-ε + 1,...,n/2}, we look at where the m edges of node t point, and assign node t to the component of lowest total degree

25. Decomposing PA(m) graphs n 1/2-ε 1…n 1/2-ε +1 … n/2 2. For each node t є {n 1/2-ε + 1,...,n/2}, we look at where the m edges of node t point, and assign node t to the component of lowest total degree

26. Decomposing PA(m) graphs 3. For each node t є {n/2 + 1,...,n}, we look at the first two edges of node t, and assign each node t to two components of the endpoints n 1/2-ε 1…n 1/2-ε +1 … n/2 n/2+1

27. Decomposing PA(m) graphs 3. For each node t є {n/2 + 1,...,n}, we look at the first two edges of node t, and assign each node t to two components of the endpoints n 1/2-ε 1…n 1/2-ε +1 … n/2 n/2+1

28. Decomposing PA(m) graphs 3. For each node t є {n/2 + 1,...,n}, we look at the first two edges of node t, and assign each node t to two components of the endpoints n 1/2-ε 1…n 1/2-ε +1 … n/2 n/2+1

29. Decomposing PA(m) graphs 3. For each node t є {n/2 + 1,...,n}, we look at the first two edges of node t, and assign each node t to two components of the endpoints n 1/2-ε 1…n 1/2-ε +1 … n/2 n/2+1

30. Decomposing PA(m) graphs 3. For each node t є {n/2 + 1,...,n}, we look at the first two edges of node t, and assign each node t to two components of the endpoints n 1/2-ε 1…n 1/2-ε +1 … n/2 n/2+1

31. Decomposing PA(m) graphs 3. For each node t є {n/2 + 1,...,n}, we look at the first two edges of node t, and assign each node t to two components of the endpoints n 1/2-ε 1…n 1/2-ε +1 … n/2 n/2+1n

32. Decomposing PA(m) graphs Decompose graph into c= Θ(n 1/2-ε ) connected components Each vertex belongs to 1 or 2 components Each component has diameter d= O(log n) Each component has size a= Θ(n 1/2+ε ) Any two components intersect n 1/2-ε 1…n 1/2-ε +1 … n/2 n/2+1n

33. Recall: Polya Urns Process Process starts with: – N bins/urns and1 ball in each bin At each step, pick a random bin with probability proportional to load – Add one ball to the selected bin Bin loads become unbalanced whp

34. Polya Urns with the Power of Choice Process P starts with: – N bins, O(N) balls – Each bin contains ≥ 1 ball Pick m bins iid at random, with replacement, probability proportional to load – Add one ball to the least loaded of the m bins Does this process balance bin loads?

35. Our Main Technical Result Theorem: For any ε > 0, there exists an m, s.t. if we run P for Θ(N 2+ε ) steps, the loads of all bins differ by at most by a factor of (1±ε). More complicated analysis shows the degree & size of components become balanced after Ω(c 2+ε ) = n nodes arrive in PA(m)  The main result follows

36. Polya Urns w/ Choice: Two Bins L t = fractional load of low bin at time t H t = fractional load of high bin at time t Key insight: Pr[add ball to high bin] = ( H t ) m < H t Pr[add ball to low bin] = (1 - (1- L t ) m ) > L t Thus both bins obtain the same load, and stay roughly balanced whp (Azuma’s Inequality) Can use Chernoff bounds to show fractional load of high bin decreases quickly

37. Very Rough Proof Sketch Define a marker bin M Bins with less load than M are low bins – Show fractional load of low bins increase Bins with more load than M are high bins – Show fractional load of high bins decrease  Eventually every bin obtains the same load as bin M, and subsequently stays roughly balanced with bin M

38. Open Problems Does Polya urns with m=2 choices balance after O(N 2+ε ) balls? Our proof for m=2 gives O(N 3 ). – New proof being checked… What are necessary and sufficient conditions for a linked decomposition to exist? Show linked decomposition routing has limited congestion and is robust against failures? Incentives for autonomous systems to form a linked decomposition?

39. Thanks! Questions?