1. Viceroy: Scalable Emulation of Butterfly Networks For Distributed Hash Tables By: Dahlia Malkhi, Moni Naor & David Ratajzcak Nov. 11, 2003 Presented by Zhenlei Jia Nov. 11, 2004

2. Acknowledgments Some of the following slides are adapted from the slides created by the authors of the paper

3. Outline DHT Properties Viceroy Structure Routing Algorithm Join/Leave Bounding In-degree: Bucket Solution Fault Tolerance Summary

4. DHT What ’ s DHT Store (key, value) pairs Lookup Join/Leave Examples CAN, Pastry, Tapestry, Chord etc.

5. DHT Properties Dilation Efficient lookup, usually O(log(n)) Maintenance cost Support dynamic environment Control messages, affected servers Degree Number of opened connections Servers impacted by node join/leave Heartbeat, graceful leave

6. DHT Properties (cont.) Congestion: Peers should share the routing load evenly Load (of a node): the probability that it is on a route with random source and destination. If path length = O(log(n)) then on average, each node is on n 2 x O(log(n))/n = O(nlog(n)) routes. Average load = O(nlogn)/n 2 = O(log(n))/n

7. Previous Works

8. Intuition Route is a combination of links of appropriate size Chord: Each node has ALL log(n) links Viceroy Each node has ONE of the long-range links A link of length 1/2 k points to a node has link of length 1/2 k+1 Chrod

9. 000001010 011 100101 110 111 Level 1 Level 2 Level 3 Level 4 A Butterfly Network  Each node has ONE of the long-range links  A link of length 1/2 k points to a node has link of length 1/2 k+1  Nodes “ share ” each other ’ s long link  Routing 1.Route to root 2.Route to right group 3.Route to right level Path: O(log(n)) Degree: O(1)

10. A Viceroy network Level 1 Level 2 Level 3 Ideally, there should be log(n) levels There is not a global counter Later, we will see how a node can estimate log(n) locally 01 0001 0010 00110100 0101 01101000 100110111101 1110 1111

11. Structure: Nodes Node Id: 128 bits binary string, u Level: positive integer, u.level Order of ids b 1 b 2 … b k  ∑ i=1 … k b i /2 i Each node has a SUCCESSOR and a PREDECESSOR SUCC(u), PRED(u) Node u stores the keys k such that u≤k< SUCC (u)

12. Structure: Nodes 01 x SUCC (x) PRED (x) Keys stored on x  Lemma 2.1 Let n 0 = 1/d(x, SUCC (x)), then w.h.p. (i.e. p>1-1/n 1+e ) that log(n)-log(log(n))-O(1) <log(n 0 ) ≤ 3log(n)  Node x selects level from 1 … log(n 0 ) uniformly randomly

13. Structure: Links A node u in level k has six out links 2 x Short: SUCCESSOR,PREDECESSOR 2 x Medium: (left) closest level-(k+1) node whose id matches u.id[k] and is smaller than u.id. 1 x Long: the closest level-(k+1) node with prefix u 1 … u k-1 (1-u k )(?) u 1 … u k-1 (1-u k )u k+1 … u w * where w=log( n0)-log(log(n0)) 1 x Parent: closest level-(k-1) node Also keeps track of in-bound links

14. Structure: Links Level 1 Level 2 Level 3 01 0001 0010 00110100 0101 01101000 100110111101 1110 1111 Short link Parent link, to level k-1 Medium link Matches x[k]0* Long link, cross over about 1/2 k Matches u[w] except k th bit. (11*) Matches 1* Wrong!

15. Routing: Algorithm LOOKUP (x, y): Initialization: set cur to x Proceed to root: while cur.level > 1: cur = cur.parent Greedy search: if cur.id ≤ y < SUCC (cur).id, return cur. Otherwise, choose m from links of cur that minimize d(m, y), move to m and repeat. Demo: http://www.cs.huji.ac.il/labs/danss/anatt/viceroy.htmlhttp://www.cs.huji.ac.il/labs/danss/anatt/viceroy.html

16. Routing: Example Level 1 Level 2 Level 3 01 0001 0010 00110100 0101 01101000 100110111101 1110 1111 y x

17. One Observation Level 1 Level 2 Level 3 01 0001 0010 00110100 0101 01101000 100110111101 1110 1111

18. Routing: Analysis (1) Level 1 Level 2 Level 3 01 0001 0010 00110100 0101 01101000 100110111101 1110 1111 y x

19. Routing: Analysis (2) Expected path length = O(log(n)) log(n ) to `level-1’ node log(n ) for traveling among clusters log(n ) for final local search

20. Routing: Theorems Theorem 4.4 The path length from x to y is O(log(n)) w.h.p. Proof is based on several lemmas Lemma 4.1 For every node u with a level u.level < log(n)-log(log(n)), the number of nodes between u and u.Medium-left (Medium-right), if it exists, is at most 6log 2 (n) w.h.p.

21. Routing: Theorems (2) Lemma 4.2 In the greedy search phase of a lookup of value Y from node x, let the j ’ th greedy step v j, for 1 ≤ j ≤ m, be such that v j is more than O(log 2 (n)) nodes away from y. Then w.h.p. node v j is reached over a Medium or Long link, and hence satisfies v j.level = j and v j [j] = Y[j]. m = log(n)-2loglog(n)-log(3+e) W.h.p. within m steps, we are n/2 m = 6log 2 (n) nodes away from the destination

22. Routing: Theorems (3) Lemma 4.3 Let v be a node that is O(log 2 (n)) nodes away from the target y. Then w.h.p., within O(log(n)) greedy steps that target y is reached from v. Theorem 4.4 The total length of a route from x to y is O(log(n)) w.h.p. Theorem 4.6 Expected load on every node is O(log(n)/n). The load on every node is log 2 (n)/n w.h.p. Theorem 4.7 Every node u has in-degree O(log(n)) w.h.p.

23. Join: Algorithm 1. Choose identifier: select a random 128 bits x 1 x 2 … x 128 2. Setup short links: invoke LOOKUP (x), let x ’ be the result node. Insert x between x ’ and x ’. SUUCESSOR. 3. Choose level: let k be the maximal number of matching prefix bits between x and either SUCC (x) or PRED (x), choose level from 1 … k. 4. Set parent link: If SUCC (x) has level x.level-1, set x.parent to it. Otherwise, move to SUCC (x) and repeat. 5. Set long link: p = x 1 … x k-1 (1-x k )x k+1 … x w Invoke LOOKUP(p), stop after a node at level x.level+1 and matches p is reached.

24. Join: Algorithm (cont.) 6. Set medium links: Denote p = x 1 x 2 … x x.level. If SUCC (x) has prefix p and level x.level+1, set x.Medium-right link to it. Otherwise, move the SUCC(x) and repeat. 7. Set inbound links: Denote p = x 1 x 2 … x x.level. Set inbound Medium links: Following SUCC links, so long as successor y has a prefix p and a level different from x.level, if y.level = x.level-1, set y.Medium-left to x. Set inbound long links: Following SUCC links, find y that has a prefix matches p and has level x.level. Take any inbound links that is closer to x than y. Set inbound parent links: Following PRED link, find y such that y.level = x.level+1. Repeat until meet a node in same level as x.

25. Join: Example Level 1 Level 2 Level 3 01 0001 0010 00110100 0101 01101000 100110111101 1110 1111 0111 Lookup(x) 0111 Set Medium link: O(lg 2 n) w.h.p p = x 1 x 2 … x k (01) If y[k] != p: stop If y[k]=p and y.level=k+1: set Medium link Otherwise, move to succ(y) STOP Set Parent link: Following SUCC link, find a node has level k-1. Set long link P = x 1 … x k-1 (1-x k ) … x w stop at level k+1? In this case, find 00* Set inbound long links: Following short links, find y such that y[k]=x[k] and y.level = x.level, check y ’ s inbound links. X

26. Join: Analysis LOOKUP takes O(log(n)) messages w.h.p. Travels on short links during link setting phase is O(lg 2 n) w.h.p. A Medium link is within 6log 2 (n) nodes from x w.h.p. Similar for others Theorem 5.1: A JOIN operation by a new node x incurs expected O(log(n)) number of messages, and O(log 2 (n)) messages w.h.p. The expected number of nodes that change their state as a result of x ’ s join is constant, and w.h.p is O(log(n)). Because node x has O(log(n)) in-degrees w.h.p. Similar results holds for LEAVE.

27. Bounding In-degrees Theorem 4.7 Every node has expected constant in-degree, and has O(log(n)) in-degree w.h.p. In-degree=# of servers affected by join/leave How to guarantee constant in-degree? Bucket solution A background process to balance the assignment of levels

28. Bucket Solution: Intuition Level k-1 Level k Node x has log(n) in-degree, assuming Medium Right ~log(n) x Too many nodes at level k-1; Improve the level selection procedure Too few nodes at level k

29. Bucket Solution  The name space is divided into non- overlapped buckets.  A bucket contains m nodes, where log(n) ≤m ≤ clog(n), for c>2.  In a buckets, levels are NOT assigned randomly  For each 1≤j≤log(n), there are 1 … c nodes at level j in each bucket  In(x) < 7c (?? 2c) 01 0001 0010 00110100 0101 0110 1000 1001 1011 110111101111

30. Maintaining Bucket Size n can be accurately estimated When bucket size exceeds clog(n), the bucket is split into two equal size buckets. When bucket size drops below log(n), it is merged with a neighbor bucket. Further more, if the merged bucket is greater than log(n) x (2c+2)/3, the new bucket is split into two buckets. (c+1)/3 > 1 since c>2 Buckets are organized into a ring, which can be merged or split with O(1) message.

31. Maintain Level Property Node join/leave without merging or splitting O(1) Join: size < clog(n), choose a level that has less that c nodes Leave: If it is the only node in its level, find another level that has two nodes, reassign level j to one of them. Bucket merge or split may result in a reassignment of the levels to all nodes in the bucket(s) O(log(n)) Merging/splitting are expensive, but they do not happen very often After a merging or splitting of buckets, at least log(n) (c-2)/3 JOIN/LEAVE must happen in this bucket until another merging or splitting of this bucket is performed Amortized Overhead = c/((c-2)/3) = O(1) for c>2

32. Amortized analysis Log(n) clog(n) d1d1 d2d2 d 1, d 2 > (c-2)/3 New bucket size Max bucket size min(c/2lgn, (c+1)/3lgn)max(c/2lgn, (2c+2)/3lgn)

33. Viceroy has no built in support for fault tolerance Viceroy requires graceful leave Leaves are NOT the same as failures Performance is sensitive to failure External techniques: Thickening Edges State Machine Replication Fault Tolerance

34. State Machine Replication Old New SMR Super nodeViceroy nodes

35. Related Works De Bruijn Graph Based Network Distance halving D2B Koorde Others Symphony (Small world model) Ulysses (ButterFly, log(n), log(n)/loglogn)

36. Summary Constant out-degree Expected constant in-degree O(log(n )) w.h.p. O(1) with bucket solution O(log(n )) path length w.h.p Expected log(n )/n load: O(log 2 (n)/n) w.h.p. Weakness/improvements: Not Locality Aware No Fault Tolerance Support Due to the lack of flexibility of ButterFly network

37. Question Photo by Peter J. Bryant