1 maximal independent set. 2 independent set (is): in a graph, any set of nodes that are not...

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1

Maximal Independent Set

2

Independent Set (IS):

In a graph, any set of nodes that are not adjacent

3

Maximal Independent Set (MIS):

An independent set that is nosubset of any other independent set

4

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

5

A Sequential Greedy algorithm

Suppose that will hold the final MISI

Initially I

G

6

Pick a node and add it to I1v

1v

Phase 1:

1GG

7

Remove and neighbors )( 1vN1v

1G

8

Remove and neighbors )( 1vN1v

2G

9

Pick a node and add it to I2v

2v

Phase 2:

2G

10

2v

Remove and neighbors )( 2vN2v

2G

11

Remove and neighbors )( 2vN2v

3G

12

Repeat until all nodes are removed

Phases 3,4,5,…:

3G

13

Repeat until all nodes are removed

No remaining nodes

Phases 3,4,5,…,x:

1xG

14

At the end, set will be an MIS of I G

G

15

Worst case graph (for number of phases):

n nodes

Running time of algorithm: m)O(n

Number of phases of the algorithm: )(nO

16

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)

17

Suppose that will hold the final MISI

Initially I

Example:

G

18

Find any independent set 1I

Phase 1:

And insert to :1I I 1III

1GG

19

1I )( 1INremove and neighbors

1G

20

remove and neighbors 1I )( 1IN

1G

21

remove and neighbors 1I )( 1IN

2G

22

Phase 2:

Find any independent set 2I

And insert to :2I I 2III

On new graph

2G

23

remove and neighbors 2I )( 2IN

2G

24

remove and neighbors 2I )( 2IN

3G

25

Phase 3:

Find any independent set 3I

And insert to :3I I 3III

On new graph

3G

26

remove and neighbors 3I )( 3IN

3G

27

remove and neighbors 3I )( 3IN

No nodes are left

4G

28

Final MIS I

G

29

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:

30

Example:If is MIS, 1 phase is needed

1I

Example:If each contains one node, phases are needed

kI)(nO

(sequential greedy algorithm)

31

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

32

Let be the maximum node degree in the whole graph

d

1 2 d

Suppose that is known to all the nodesd

33

Elected nodes are candidates forindependent set

Each node elects itself with probability

At each phase :k

kI

dp

1

1 2 d

kGz

z

34

However, it is possible that neighbor nodes may be elected simultaneously

Problematic nodeskG

35

All the problematic nodes must be un-elected. The remaining elected nodes formindependent set kI

kGkI

kIkI

kI

36

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

37

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.

38

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

39

Fundamental inequalities

tn

t ent

nt

e

11

21n

nt ||

10 p

1k

k

kp

p

11

40

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

41

Therefore, node will enterand disappear at the end of phasewith probability at least

1 2 y

z

z kI

ed21

k

42

Expected number of phases until nodedisappears:

at most ed2phase in success of yprobabilit

1 phases

z

43

after phases

Bad event for node :

ned ln4

node did not disappear

Probability (First ineq. with t =-1 and n=2ed):

2ln2

ln411

21

1need n

ned

z

z

44

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

45

within phases

Good event for all nodes in :

ned ln4

all nodes disappear

Probability:

n1

-1event] bad of yprobabilit[1

(high probability)

G

46

Total number of phases:

)log(ln4 ndOned

Time duration of each phase: )1(O

Total time: )log( ndO

with high probability

47

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

48

Let be the degree of node)(vd

1 2 )(vd

v

v

49

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

50

)(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

51

)(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

52

Problematic nodeskG

Using previous rules, problematic nodesare removed

53

kG

The remaining elected nodes formindependent set kI

kI

kIkI

kI kI

kI

54

at least one neighbor enters

Analysis

2z

A good event for node

1 2)(vd

v

1z)(vdz

kI:vH

Consider phasek

v

55

1z)(vdz

At end of phasek

If is true, then andwill disappear at end of current phase

v)( kINv vH

56

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

57

2z

1 2)(vd

v

1z)(vdz

No neighbor of elects itself with probability

)(

))(1(vNz

zp

vPROOF:

(the elections are independent)

58

)(~

2

)()(

~2

)()(~

2)(

)(~

2

11

)(~

2

11 vd

vdvd

vdvdvd

evdvd

)()( )(21

1))(1(vNzvNz zd

zp

maximum neighbor degree)(max)(~

)(zdvd

vNz

59

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

60

If a node elects itself,then it enters with probabilityat least

z

1 2

)(zdz

1 2

)(zdzv v

LEMMA 2:

kI

21

61

1 2

)(zdz

)()( 1 zdud

1u

u)()( zdud

Node enters if no neighbor of same or higher degree elect itself

zkI

PROOF:

62

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

63

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

64

1 2

)(zdz

Thus, if elects itself, it enterswith probability at least

z kI

21

1 2

)(zdzv

v

END OF PROOF

65

2z

1 2)(vd

v

1z)(vdz

at least one neighbor of enters kI:vH v

)(

~2

)(

121

][ vd

vd

v eHPLEMMA 3:

66

and no node is elected

neighbor is iniz kI

121 ,,, izzz

2z

12

)(vdv

1z)(vdz

New event

iz1iz kI

1ii

:iY

PROOF:

67

)(21 ,,, vdYYY The events

are mutually exclusive

)(

1)(1][

vd

iii

vdiYPYP

68

i

vdiv YPHP

)(1][

It holds:

)(

1)(1][][

vd

iii

vdiv YPYPHP

Therefore:

69

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

70

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

71

)(

1

)(

1

)(

1

][21

][][][][vd

ii

vd

iii

vd

iiv APBPAPYPHP

21

][ iBP

72

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

73

)(~

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

74

)(

~2

)()(

1

121

][21

][ vd

vdvd

iiv eAPHP

Therefore node disappears in phasewith probability at least

v k

END OF PROOF

75

Let be the maximum node degree in the graph

kd

kG

Suppose that in : 2

)( kdvd

Then, )(2)(~

vdvd

ceeHP vd

vd

v

41

)(~

2

)(

121

121

][constant

kG

76

(thus, nodes with high degreewill disappear fast)

2)( kd

vd a node with degree

with probability at least c

Thus, in phasek

disappears

77

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

78

nc

1log3 1Take

Node does not disappear within phases with probability at most

v

3

1log3 1

)1()1( 1

ncc nc

79

Thus, within phases nc

1log3 1

v either disappears or its degree gets less than

with probability at least

2d

3

11

n

80

by the end of phases nc

1log3 1

there is no node of degree higher than2d

with probability at least (ineq. 2)

Therefore,

23

11

11

nn

n

81

In every phases,nc

1log3 1

the maximum degree of the graphreduces by at least half, with probability at least

2

11

n

82

)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

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