cpsc 689: discrete algorithms for mobile and wireless systems
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CPSC 689: Discrete Algorithms for Mobile and Wireless Systems. Spring 2009 Prof. Jennifer Welch. Lecture 26. Topic: Maximal Independent Set Sources: Luby Schneider & Wattenhofer Linial MIT 6.885 Fall 2008 slides. Overview. - PowerPoint PPT PresentationTRANSCRIPT
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CPSC 689: Discrete Algorithms for Mobile and Wireless Systems
Spring 2009
Prof. Jennifer Welch
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Lecture 26 Topic:
Maximal Independent Set Sources:
Luby Schneider & Wattenhofer Linial MIT 6.885 Fall 2008 slides
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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)
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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)
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Unit Disk Graphs
R
R
Wireless networks often modeled as unit disk graphs
Wireless networks often modeled as unit disk graphs
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More Realistic Graphs
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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
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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.
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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).
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Maximal vs. Maximum IS
a maximum independent set a maximal independent set
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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)
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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|
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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?
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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
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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.
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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)
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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)
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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
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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
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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.
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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
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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)
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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…)
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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
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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.
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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.
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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.
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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.