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
Acknowledgments Some of the following slides are adapted from the slides created by the authors of the paper
Outline • Outline • DHT Properties • Viceroy • Structure • Routing Algorithm • Join/Leave • Bounding In-degree: Bucket Solution • Fault Tolerance • Summary
DHT • What’s DHT • Store (key, value) pairs • Lookup • Join/Leave • Examples • CAN, Pastry, Tapestry, Chord etc.
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
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 n2 x O(log(n))/n = O(nlog(n)) routes. Average load = O(nlogn)/n2 = O(log(n))/n
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 1/2k points to a node has link of length 1/2k+1 Chrod
011 110 000 001 010 100 101 111 Level 1 Level 2 Level 3 Level 4 A Butterfly Network • Each node has ONE of the long-range links • A link of length 1/2k points to a node has link of length 1/2k+1 • Nodes “share” each other’s long link • Routing • Route to root • Route to right group • Route to right level • Path: O(log(n)) • Degree: O(1)
0101 1001 1011 1101 1111 0001 0010 0011 0100 0110 1000 1110 0 1 Level 1 Level 2 Level 3 A Viceroy network • 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
Structure: Nodes • Node • Id: 128 bits binary string, u • Level: positive integer, u.level • Order of ids • b1b2…bk ∑i=1…k bi/2i • Each node has a SUCCESSOR and a PREDECESSOR SUCC(u), PRED(u) • Node u stores the keys k such that u≤k<SUCC(u)
Keys stored on x 0 1 PRED(x) x SUCC(x) Structure: Nodes • Lemma 2.1 Let n0 = 1/d(x, SUCC(x)), then w.h.p. (i.e. p>1-1/n1+e) that log(n)-log(log(n))-O(1) <log(n0) ≤3log(n) • Node x selects level from 1…log(n0) uniformly randomly
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 u1…uk-1(1-uk)(?) u1…uk-1(1-uk)uk+1…uw* where w=log(n0)-log(log(n0)) • 1 x Parent: closest level-(k-1) node • Also keeps track of in-bound links
0101 1001 1011 1101 1111 0001 0010 0011 0100 0110 1000 1110 0 1 Short link Long link, cross over about 1/2k Matches u[w] except kth bit. (11*) Short link Level 1 Medium link Matches x[k]0* Matches 1* Wrong! Level 2 Parent link, to level k-1 Level 3 Structure: Links
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.html
y 0101 1001 1011 1101 1111 0001 0010 0011 0100 0110 1000 1110 x 0 1 Level 1 Level 2 Level 3 Routing: Example
0101 1001 1011 1101 1111 0001 0010 0011 0100 0110 1000 1110 0 1 Level 1 Level 2 Level 3 One Observation
y 0101 1001 1011 1101 1111 0001 0010 0011 0100 0110 1000 1110 x 0 1 Level 1 Level 2 Level 3 Routing: Analysis (1)
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
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 6log2(n) w.h.p.
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 vj, for 1 ≤ j ≤ m, be such that vj is more than O(log2(n)) nodes away from y. Then w.h.p. node vj is reached over a Medium or Long link, and hence satisfies vj.level = j and vj[j] = Y[j]. • m = log(n)-2loglog(n)-log(3+e) • W.h.p. within m steps, we are n/2m = 6log2(n) nodes away from the destination
Routing: Theorems (3) • Lemma 4.3 Let v be a node that is O(log2(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 log2(n)/n w.h.p. • Theorem 4.7 Every node u has in-degree O(log(n)) w.h.p.
Join: Algorithm • Choose identifier: select a random 128 bits x1x2…x128 • Setup short links: invoke LOOKUP(x), let x’ be the result node. Insert x between x’ and x’.SUUCESSOR. • 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. • Set parent link: If SUCC(x) has level x.level-1, set x.parent to it. Otherwise, move to SUCC(x) and repeat. • Set long link: p = x1…xk-1(1-xk)xk+1…xw Invoke LOOKUP(p), stop after a node at level x.level+1 and matches p is reached.
Join: Algorithm (cont.) 6. Set medium links: Denote p = x1x2…xx.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 = x1x2…xx.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.
Set Medium link: O(lg2n) w.h.p p = x1x2…xk (01) If y[k] != p: stop If y[k]=p and y.level=k+1: set Medium link Otherwise, move to succ(y) 0101 0111 1001 1011 1101 1111 0001 0010 0011 0100 0110 1000 1110 0 1 Lookup(x) Level 1 X Level 2 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. STOP Level 3 Set long link P = x1…xk-1(1-xk)…xw stop at level k+1? In this case, find 00* Join: Example Set Parent link: Following SUCC link, find a node has level k-1. 0111
Join: Analysis • LOOKUP takes O(log(n)) messages w.h.p. • Travels on short links during link setting phase is O(lg2n) w.h.p. • A Medium link is within 6log2(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(log2(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.
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
Level k-1 Level k Bucket Solution: Intuition ~log(n) • Node x has log(n) in-degree, assuming Medium Right x • Too many nodes at level k-1; Too few nodes at level k • Improve the level selection procedure
0101 1001 1011 0011 0110 1101 1110 1111 0001 0100 1000 0010 0 1 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)
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.
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
Max bucket size New bucket size clog(n) d1 d2 Log(n) min(c/2lgn, (c+1)/3lgn) max(c/2lgn, (2c+2)/3lgn) d1, d2 > (c-2)/3 Amortized analysis
Fault Tolerance • 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
Old New SMR SMR State Machine Replication Viceroy nodes Super node
Related Works • De Bruijn Graph Based Network • Distance halving • D2B • Koorde • Others • Symphony (Small world model) • Ulysses (ButterFly, log(n), log(n)/loglogn)
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(log2(n)/n) w.h.p. • Weakness/improvements: • Not Locality Aware • No Fault Tolerance Support • Due to the lack of flexibility of ButterFly network
Question Photo by Peter J. Bryant