cpsc 689: discrete algorithms for mobile and wireless systems

CPSC 689: Discrete Algorithms for Mobile and Wireless Systems

Spring 2009

Prof. Jennifer Welch

Discrete Algs for Mobile Wireless Sys 2

Lecture 26 Topic:

Maximal Independent Set Sources:

Luby Schneider & Wattenhofer Linial MIT 6.885 Fall 2008 slides


Overview Recall that a minimum connected dominating set is a

useful substructure of a graph representing a network: routing medium access control coverage

Computing a MCDS in a general graph is NP-complete What about special-case graphs that still reflect the reality

of wireless networks? UDG too restrictive QUDG still too restrictive let's try growth-bounded graphs (GBG), a.k.a. bounded

independence graphs (BIG)


Overview In BIG model, a maximal independent set is a

constant approximation of a MCDS A MIS is an independent subset S of the nodes of a

graph (none of the nodes in S are neighbors), and no superset of S is independent

[SW] paper gives an O(log*n) time algorithm for MIS in BIG model log*n is number of times you can take the log of n until

reaching 1 algorithm is distributed, deterministic, and does not

require location information Running time is optimal (cf. paper by Linial)


Unit Disk Graphs

R

R

Wireless networks often modeled as unit disk graphs

Wireless networks often modeled as unit disk graphs


More Realistic Graphs


Bounded Independence Graph

Even more general than quasi unit disk graphs

No links between far-away nodes Close nodes tend to be connected In particular: Densely covered area

many connections bounded neighborhood bounded

independent set


Bounded Independence Graphs (BIGs)

Definition: Given a function f(r), a graph G=(V,E) is f(r)-independence bounded if for all nodes v in V and all r ≥ 0, the size of a maximum IS in the r-neighborhood of v is at most f(r).

Note that f is only a function of r and in particular independent of the number of nodes n.


Bounded Independence Graphs (BIGs)

Typically require that f(r) = poly(r). It can never be more than exponential.

UDGs and QUDGs are independence-bounded with f(r) = O(r2).


Maximal vs. Maximum IS

a maximum independent set a maximal independent set


MIS and DS

A MIS is a dominating set (DS) If S is an IS but does not dominate some

node, then the undominated node can be added to S while maintaining the independence property

But a DS is not necessarily independent two dominators are allowed to be

neighbors (not independent)


MIS and MDS

Theorem: On an f(r)-independence bounded graph G, a MIS is a f(1)-approximation of an MDS.

Proof: Consider any maximal IS S of G. Suppose T is a minimum DS of G. Every node in S is either in T or is a neighbor of

some dominator t in T Since G is a f(r)-BIG, t has at most f(1) elements of

S as its neighbors So |S| ≤ f(1) •|T|


Distributed MIS Algorithm

For general graphs [Luby]: A simple parallel algorithm for the MIS problem(similar algorithm in [Alon,Babai,Itai])

Randomized algorithm Runs in O(log n) rounds in expectation and

with high probability Can we do better in special-case graphs?


Log-Star MIS Algorithm for BIGs

Assumptions: Every node has a unique ID between 1 and n For simplicity, assume that all nodes know f(r)

and n (not necessary) For simplicity, synchronous model (not

necessary)

Main result of [SW]: O(log*n) time MIS algorithm for bounded independence graph


Algorithm: Basic Structure

During the algorithm, each node is always in one of 5 states: competitor: Node actively competes to be in MIS dominator: Node has joined the MIS dominated: Node has a neighbor in the MIS, will

definitely not join MIS ruler: Node not actively in competition, will compete

again actively if there are no neighboring competitors left

ruled: Neighbor of ruler, does not start competing again before all neighboring rulers become ruled themselves.


Algorithm: Basic Structure

Algorithm consists of f(f(2) + 3) stages Each stage consists of f(2) + 1 phases

Each phase consists of log*n + 2 competitions Each competition needs a constant number of rounds

So total number of rounds is

O(1)*(log*n + 2)*(f(2)+1)*f(f(2)+3) which is O(log*n) since f(c) = O(1) when c = O(1)


Competitions

Every competitor v starts a competition with a number rv and computes new rv’ initially rv = ID(v)

Computation of rv’: u: neighboring competitor with minimal ru

if ru > rv then rv’ = 0

else, rv’ is computed from the base-2 representations of rv and ru:rv’ is position of highest bit that is 1 in rv and 0 in ru

(position of least significant bit is 1)


Competitions

rv’ is position of highest bit that is 1 in rv and 0 in ru position of least significant bit is 1

Examples: rv = (10100010)2, ru = (10010110)2 rv’ = 6

rv = (00101000)2, ru = (00100101)2 rv’ = 4


Competition: New State

Compute new rv’ based on rv and min ru among neighboring competitors

Update state based on new values of v and neighbors: If rv’ < ru’ for all neighboring competitors v becomes

dominator Else if neighbor of v becomes dominator v becomes

dominated Else if rv’ · ru’ for all neighboring competitors v

becomes ruler Else if v has neighboring ruler node becomes ruled Else v stays competitor


Competition: New State

Lemma: Dominators always form an independent set. No 2 adjacent nodes can become dominator

together. Nodes that are dominated do not compete any

further. Only competing nodes can become dominator.


Reducing the Competitors

Lemma: After log*n + 2 competitions, no node is a competitor any more.

Proof: Initially, rv = ID(v), hence, rv uses at most log n bits

Hence, rv’ uses at most log log n bits

After log*n + 2 competitions, rv is in {0,1}

All nodes v with rv=0 become dominator or ruler Neighbors become dominated or ruled If rv=1 and all neighboring competitors u have ru=1, v

becomes ruler


Phase

log*n+2 competitions are called a phase For next phase:

All rulers become competitors again All rv are set back to ID(v)


Stage

Main technical lemma: No node becomes a ruler in the (f(2)+1)st phase. Thus, after f(2)+1 phases there are only nodes that are dominators, dominated, or ruled.

Proof: Read the paper. f(2)+1 phases are called a stage. In new stage, ruled nodes become

competitors again(note: there are no rulers any more…)


Proof of Progress

Lemma: Let v be a competitor at the beginning of a stage. During the stage, a node at distance at most f(2)+1 becomes dominator.

Proof: At the end of a stage, each node is ruled,

dominated, or a dominator Show that after i phases, there is a node at

distance at most i that is not ruled


Proof of Progress

Show that after i phases, there is a node at distance at most i that is not ruled

Induction on i: Clear for i=0 (v is not ruled) Let w be node that is not ruled at distance at most i

after i phases If w does not become ruled in (i+1)st phase, ok. If w becomes ruled in a competition of the (i+1)st phase,

some neighbor w’ becomes a ruler (w’ is at distance at most i+1).

w’ remains a ruler until the end of the phase and then becomes a competitor.


Proof of Progress After f(2) phases, there is a ruler at

distance at most f(2) or a dominator at distance at most f(2)+1.

If it is a ruler, itself or a neighbor of it becomes dominator in phase f(2)+1.

Thus if v is a competitor at the beginning of a stage, then during the stage, a node at distance at most f(2)+1 becomes a dominator.


Proof of Progress

Theorem: The algorithm terminates with a MIS after at most f(f(2)+1) stages.

Proof: The algorithm terminates as soon as there are no ruled nodes at

the end of a stage (i.e., all nodes are dominators or dominated) Suppose in contradiction there is still a ruled node v after stage

f(f(2)+1). v was a competitor in all f(f(2)+1) stages. In every stage, a node in (f(2)+1)-neighborhood of v joins the MIS At most f(f(2)+1) nodes in (f(2)+1)-neighborhood of v can join an

ind. set (because of BIG model) Hence, the ind. set is maximal and v cannot be a competitor any

more.


Comments

In the paper, the algorithm is described in a way that does not require knowledge of f(r) and n stages and phases need to be locally synchronized algorithm works for all graphs, time complexity depends

on graph Algorithm is asymptotically optimal:

Result in [Linial]: Any deterministic algorithm needs at least (log*n) rounds to color a ring with O(1) colors.

From a c-coloring, a MIS can be computed in c rounds. Since rings are bounded independence graphs,

algorithm is asymptotically tight.

cpsc 689: discrete algorithms for mobile and wireless systems

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