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1 Maximal Independent Set
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Maximal Independent Set
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Independent Set (IS):
In a graph, any set of nodes that are not adjacent
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Maximal Independent Set (MIS):
An independent set that is nosubset of any other independent set
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Applications in Distributed Systems
•In a network graph consisting of nodes representing processors, a MIS defines a set of processors which can operate in parallel without interference
•For instance, in wireless ad hoc networks, to avoid interference, a conflict graph is built, and a MIS on that defines a clustering of the nodes enabling efficient routing
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A Sequential Greedy algorithm
Suppose that will hold the final MISI
Initially I
G
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Pick a node and add it to I1v
1v
Phase 1:
1GG
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Remove and neighbors )( 1vN1v
1G
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Remove and neighbors )( 1vN1v
2G
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Pick a node and add it to I2v
2v
Phase 2:
2G
10
2v
Remove and neighbors )( 2vN2v
2G
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Remove and neighbors )( 2vN2v
3G
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Repeat until all nodes are removed
Phases 3,4,5,…:
3G
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Repeat until all nodes are removed
No remaining nodes
Phases 3,4,5,…,x:
1xG
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At the end, set will be an MIS of I G
G
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Worst case graph (for number of phases):
n nodes
Running time of algorithm: m)O(n
Number of phases of the algorithm: )(nO
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A General Algorithm For Computing MIS
Same as the sequential greedy algorithm,but at each phase we may select any independent set (instead of a single node)
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Suppose that will hold the final MISI
Initially I
Example:
G
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Find any independent set 1I
Phase 1:
And insert to :1I I 1III
1GG
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1I )( 1INremove and neighbors
1G
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remove and neighbors 1I )( 1IN
1G
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remove and neighbors 1I )( 1IN
2G
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Phase 2:
Find any independent set 2I
And insert to :2I I 2III
On new graph
2G
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remove and neighbors 2I )( 2IN
2G
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remove and neighbors 2I )( 2IN
3G
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Phase 3:
Find any independent set 3I
And insert to :3I I 3III
On new graph
3G
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remove and neighbors 3I )( 3IN
3G
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remove and neighbors 3I )( 3IN
No nodes are left
4G
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Final MIS I
G
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The number of phases depends on the choice of independent set in each phase:
The larger the independent set at eachphase the faster the algorithm
Observation:
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Example:If is MIS, 1 phase is needed
1I
Example:If each contains one node, phases are needed
kI)(nO
(sequential greedy algorithm)
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A Randomized Sync. Distributed Algorithm
Same with the general MIS algorithm
At each phase the independent setis chosen randomly so that it includesmany nodes of the remaining graph
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Let be the maximum node degree in the whole graph
d
1 2 d
Suppose that is known to all the nodesd
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Elected nodes are candidates forindependent set
Each node elects itself with probability
At each phase :k
kI
dp
1
1 2 d
kGz
z
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However, it is possible that neighbor nodes may be elected simultaneously
Problematic nodeskG
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All the problematic nodes must be un-elected. The remaining elected nodes formindependent set kI
kGkI
kIkI
kI
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Success for a node in phase : disappears at end of phase(enters or )
Analysis:
kGz
kI
1 2 y
No neighbor elects itself
z
z
k)( kIN
k
A good scenariothat guaranteessuccess
elects itself
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Basics of Probability
E: finite universe of events; let A and B denote two events in E; then:
1. A B is the event that either A or B occurs;
2. A B is the event that both A and B occur.
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Probability of success in phase:
1 2 y
p
p1
p1p1
dy pppp )1(1
z
No neighbor should elect itself
At least
elects itself
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Fundamental inequalities
tn
t ent
nt
e
11
21n
nt ||
10 p
1k
k
kp
p
11
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Probability of success in phase:
ed
ded
dd
ppppd
dy
2
1
11
1
11
1
)1(1
At least
For 2d
First ineq. with t =-1
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Therefore, node will enterand disappear at the end of phasewith probability at least
1 2 y
z
z kI
ed21
k
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Expected number of phases until nodedisappears:
at most ed2phase in success of yprobabilit
1 phases
z
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after phases
Bad event for node :
ned ln4
node did not disappear
Probability (First ineq. with t =-1 and n=2ed):
2ln2
ln411
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1need n
ned
z
z
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after phases
Bad event for any node in :
ned ln4
at least one node did not disappear
Probability:
nnnx
Gx
11) for event bad of ty(probabili 2
G
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within phases
Good event for all nodes in :
ned ln4
all nodes disappear
Probability:
n1
-1event] bad of yprobabilit[1
(high probability)
G
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Total number of phases:
)log(ln4 ndOned
Time duration of each phase: )1(O
Total time: )log( ndO
with high probability
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Luby’s MIS Distributed Algorithm
Runs in time in expected case)(lognO
)log(log ndO with high probability
this algorithm is asymptoticallybetter than the previous
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Let be the degree of node)(vd
1 2 )(vd
v
v
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Each node elects itselfwith probability
At each phase :k
kI
)(21
)(vd
vp
kGv
1 2 )(vd
v
degree ofin
Elected nodes are candidates for theindependent set
v
kG
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)(vd
v
z
)(zd
If two neighbors are elected simultaneously,then the higher degree node wins
Example:
)(vd
v
z
)(zd
12
1
2
12
1
2)()( vdzd
if
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)(vd
v
z
)(zd
If both have the same degree, ties are broken arbitrarily
Example:
)(vd
v
z
)(zd
12
1
2
12
1
2)()( vdzd
if
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Problematic nodeskG
Using previous rules, problematic nodesare removed
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kG
The remaining elected nodes formindependent set kI
kI
kIkI
kI kI
kI
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at least one neighbor enters
Analysis
2z
A good event for node
1 2)(vd
v
1z)(vdz
kI:vH
Consider phasek
v
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1z)(vdz
At end of phasek
If is true, then andwill disappear at end of current phase
v)( kINv vH
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At least one neighbor ofelects itself with probability at least
2z
1 2)(vd
v
1z)(vdz
LEMMA 1: v
)(~
2
)(
1 vd
vd
e
)(max)(~
)(zdvd
vNz maximum neighbor degree
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2z
1 2)(vd
v
1z)(vdz
No neighbor of elects itself with probability
)(
))(1(vNz
zp
vPROOF:
(the elections are independent)
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)(~
2
)()(
~2
)()(~
2)(
)(~
2
11
)(~
2
11 vd
vdvd
vdvdvd
evdvd
)()( )(21
1))(1(vNzvNz zd
zp
maximum neighbor degree)(max)(~
)(zdvd
vNz
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2z
1 2)(vd
v
1z)(vdz
Therefore, at least one neighbor of Elects itself with probability at least
v
)(~
2
)(
1 vd
vd
e
END OF PROOF
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If a node elects itself,then it enters with probabilityat least
z
1 2
)(zdz
1 2
)(zdzv v
LEMMA 2:
kI
21
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1 2
)(zdz
)()( 1 zdud
1u
u)()( zdud
Node enters if no neighbor of same or higher degree elect itself
zkI
PROOF:
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1 2
)(zdz
Probability that some neighbor of with same or higher degree elects itself
)()( 1 zdud
1u
u
neighbors of same or higher degree
2
1
)(2
)(
)(2)(
itself)] elects (node[
1i
k
zd
zd
zdup
uP
i
k
)()( zdud
z
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Probability that that no neighbor ofwith same or higher degree elects itself
21
21
-1itself)] elects (nodeP[-1
itself)] elects node (no[
k
k
kk
u
uP
neighbors of same or higher degree
z
1 2
)(zdz
)()( 1 zdud
1u
u
)()( zdud
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1 2
)(zdz
Thus, if elects itself, it enterswith probability at least
z kI
21
1 2
)(zdzv
v
END OF PROOF
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2z
1 2)(vd
v
1z)(vdz
at least one neighbor of enters kI:vH v
)(
~2
)(
121
][ vd
vd
v eHPLEMMA 3:
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and no node is elected
neighbor is iniz kI
121 ,,, izzz
2z
12
)(vdv
1z)(vdz
New event
iz1iz kI
1ii
:iY
PROOF:
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)(21 ,,, vdYYY The events
are mutually exclusive
)(
1)(1][
vd
iii
vdiYPYP
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i
vdiv YPHP
)(1][
It holds:
)(
1)(1][][
vd
iii
vdiv YPYPHP
Therefore:
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and no node elects itself
elects itselfiz
kI
121 ,,, izzz
2z
12
)(vdv
1z)(vdz
iz1iz
1ii
:iA
][][][ iii BPAPYP
:iB izafter elects itself, it enters
1u 2u ku
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21
][ iBP (from Lemma 2)
2z
12
)(vdv
1z)(vdz
iz1iz
1ii
2z
12
)(vdv
1z)(vdz
iz1iz kI
1ii
1u 2u ku1u 2u ku
kI:iB izafter elects itself, it enters
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)(
1
)(
1
)(
1
][21
][][][][vd
ii
vd
iii
vd
iiv APBPAPYPHP
21
][ iBP
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and no node elects itself
elects itselfiz
121 ,,, izzz :iA
)(
1d(v)i1][]P[
elected] is of neighbor one least at[vd
iii APA
vP
The events are mutually exclusive)(21 ,,, vdAAA
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)(~
2
)(
1elected] is of neighbor one least at[ vd
vd
evP
We showed earlier (Lemma 1) that:
Therefore:
)(~
2
)()(
1
1][ vd
vdvd
ii eAP
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)(
~2
)()(
1
121
][21
][ vd
vdvd
iiv eAPHP
Therefore node disappears in phasewith probability at least
v k
END OF PROOF
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Let be the maximum node degree in the graph
kd
kG
Suppose that in : 2
)( kdvd
Then, )(2)(~
vdvd
ceeHP vd
vd
v
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)(~
2
)(
121
121
][constant
kG
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(thus, nodes with high degreewill disappear fast)
2)( kd
vd a node with degree
with probability at least c
Thus, in phasek
disappears
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Suppose that the degree of remains at least for the next phases
Consider a node which in initial graphhas degree
2)(
dvd
Gv
v
2d
Node does not disappear within phases with probability at most
v
)1( c
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nc
1log3 1Take
Node does not disappear within phases with probability at most
v
3
1log3 1
)1()1( 1
ncc nc
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Thus, within phases nc
1log3 1
v either disappears or its degree gets less than
with probability at least
2d
3
11
n
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by the end of phases nc
1log3 1
there is no node of degree higher than2d
with probability at least (ineq. 2)
Therefore,
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11
11
nn
n
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In every phases,nc
1log3 1
the maximum degree of the graphreduces by at least half, with probability at least
2
11
n
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)log(log1
log3log 1 ndOn
d c
Maximum number of phases until degreedrops to 0 (MIS has formed)
with probability at least (ineq. 2)
nnn
nd 11
11
11
22
log
