Locality Sensitive Distributed Computing

Exercise Set 1David Peleg

Weizmann Institute

Exercises• Basic complexity considerations• Global function computation + pipeline• Termination detection for Dijkstra’s BFS• Fast DFS• Tgap in synchronizers and • 3-coloring bounded-degree graphs• MIS / coloring on unoriented rings

Basic complexity issues

1. Prove or disprove:In a graph G(V,E),if there are at least k edge-disjoint paths of length d between the nodes v and w,then it is possible to send m msgs from v to w in time O(d+m/k).

Exercises (cont)

2. Prove or disprove:In a graph G(V,E),if dist(v,w)=k and there are k2 edge-disjoint paths between the nodes v and w,then it is possible to send k2 msgs from v to w in time O(k).

Global function computationGoal: Compute global function f(Xv1,...,Xvn) where each node v holds input Xv

Semigroup function f:1. Well-defined for any input subset2. Associative and commutative

Efficiently computable on tree T by convergecast

Global function computation

During the process: The value sent upwards by each v in T= value of function on inputs of its subtree Tv

fv = f(v) where v = { Xw | w Tv }

Converge(f,X) process• Leaf v sends Xv to parent

• Intermediate v with k children w1,...,wk

- receives values fwi = f(wi) from all children

- applies fv (Xv,fw1,...,fwk)

Example: Addition

13

105

49

291329

50

Example: Maximum

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Global function computation

Claim:Assume f() is represented in O(p) bitsfor every input set

On tree T:

• Message(Converge(f)) = ?

• Time(Converge(f)) = ?

Pipelining

Separate broadcast / convergecast operations can be efficiently pipelined

Example: Pipelining 3 Converge(max) operations to get Mi = max {Xi(v) | v leaf } for i=1,2,3

Pipelining

Pipelining

Lemma:k global semi-group functions can be computed on tree in time ?

Level-synchronized BFS (Dijkstra)

Q. Prove the tightness of the message complexity analysis of Dijkstra's algorithm, by establishing the following:

Lower bound:For integers n and 1 D n-1, there exists n-node, D-diameter graph G=(V,E) on which the execution of Dijkstra's algorithm requires (nD+|E|) messages.

Level-synchronized BFS (Dijkstra)

Termination detection:Modify the Distributed Dijkstra algorithm so that the root can tell when the process is completed (and the entire graph is spanned by the constructed BFS tree)

Distributed Depth-First Search

DFS: Search process on G, traversing all vertices, progressing over edges, with preference to visiting new vertices

Distributed Depth-First Search

DFS algorithm• Search starts at origin v0 • Whenever search reaches vertex v:

- If v has neighbors not visited so far, then visit one of them next.- Else return to the vertex

from which visited first- If v = v0 then end

Distributed Depth First Search

Fact: DFS process visits every vertex in G.

Search defines DFS tree, with v0 as root:• v's parent = node from which v was visited first

Sequential time complexity = O(|E|)

Direct distributed implementation

- Completely sequential: one activity locus at any time. - Control carried via single message (“token”) traversing G in depth-first fashion

Note: For v to know whether neighbor w was visited or not,it must send message over edge (v,w)

Direct distributed implementation

every edge must be explored

both time and message complexities = (|E|)

Exercise

Q.• Modify the DFS algorithm to allow the traveler

to complete the tour (visiting all nodes) faster than O(|E|).

• Analyze the time complexity of the modified algorithm and prove your bound.

Synchronizers

Consider a 15-processor asynchronous G(V,E), V={0,…,14}, constantly running a synchronizer.v, v' = nodes in G. At time t, pulse counter Pv = 27.

What is the range of possible pulse numbers at Pv' in following cases:

Synchronizers (cont)

1. G = ring (with nodes arranged in order), v=11, v'=2, synchronizer used = .

2. G = full balanced binary tree (4 levels), v = root, v' = one of the leaves, synchronizer used =

3. The same as in (2), except both v and v' are leaves.

Synchronizers (cont)

4. Synchronizer used = . Clusters, spanning tree of each cluster, inter-cluster edges and locations of v, v' are as follows:

Synchronizer gaps

Gap of synchronizer Tgap() = maxv,p {t(v,p+1)-t(v,p)}

(max length of period (v,p) that some processor v stays in some pulse p)

Tgap()

Synchronizer gaps (cont)

1. How large could Tgap be for :a. synchronizerb. synchronizerc. synchronizer

2. For synchronizer , design a scenario realizing the worst-case waiting time, Tgap()

3-coloring bounded-degree graphs

Goal: Color arbitrary bounded degree G ((G)=O(1))

with +1 colors in time O(log*n)

Consistent orientation and MIS

Consistent orientation: the ring edges are oriented in a consistent manner(each node identifies its “left” and “right” neighbors, and each edge e = (v,w) is marked as going “left” by one of its endpoints and as going “right” by the other.

We have seen: Given an MIS on the ring, the nodes can be colored with 3 colors in a single round.

Consistent orientation and MIS(a) Show that algorithm might fail if the ring

does not enjoy consistent orientation(b) Prove that on an anonymous ring (no ID’s)

without a consistent orientation,it is impossible to deterministically 3-color the nodes, even given an MIS.

(c) Prove that on a non-anonymous ring without a consistent orientation, it is still possible to deterministically 3-color the vertices in a constant number of rounds given an MIS. (Try to use the smallest number of rounds.)

Consistent orientation and MIS(d) What lower bound can be proved for the

number of rounds required for computing MIS on a ring without consistent orientation? Specify precise constants (not only asymptotic lower bound).

(e) What is the smallest n for which the lower bound you got in the previous question is greater than 1? What is the smallest n for which the lower bound on 3-coloring is greater than 1?