95
1 Leader Election in Rings
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Leader Election in Rings
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A Ring Network
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1 1
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1
11
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1
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Sense of direction
left
right
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1 1
1
1
11
1
1
2
2
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Links are bi-directional for messages
At most one message in each direction
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1 1
1
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11
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Anonymous Ring
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Eponymous (non-anonymous) Ring
1 1
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Leader Election
Initial state Final state
leader
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Leader election algorithms are affected by:
Anonymous RingEponymous Ring
The size of the network is knownThe size of the network is not known
n
Synchronous AlgorithmAsynchronous Algorithm
n
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Synchronous Anonymous Rings
Every processor runs the same algorithm
Every processor does exactly the sameexecution
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Initial state Final state
leader
If one node is elected a leader,then every node is elected a leader
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Final state
leader
Conclusion 1:
Leader electioncannot be solvedin synchronousanonymous rings
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Final state
leader
Conclusion 2:
Leader electioncannot be solvedin asynchronousanonymous ringseither
Why?
The asynchronous ring may behaveLike the synchronous ring
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Asynchronous Eponymous Rings
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5
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The maximum id node is elected leader
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12
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5
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Each node sends a message with its idto the left neighbor
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2
6
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3
5
15
12
34
5
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If: message received id current node id
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6
7
5
Then: forward message
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12
34
5
6
7
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If: message received id current node id
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7
Then: forward message
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12
34
5
6
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If: message received id current node id
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Then: forward message
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12
34
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6
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If: message received id current node id
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Then: forward message
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12
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5
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If: a node receives its own message
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Then: it elects itself a leader
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12
34
5
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If: a node receives its own message
Then: it elects itself a leader
leader
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8leader
The leader sends a message in the networkdeclaring itself as the “leader of the ring”
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12
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8leader
Time complexity: )(nO n nodes
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1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
worst case scenario:
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1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
n messages
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1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
1n messages
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1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
2n messages
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1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
1n
Total messages:
n
2n
1 )( 2n2
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Notes:
ndoes not need to be known to the algorithm
The algorithm can be converted to asynchronous
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An O(n log n) Mess. Algorithm
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Again, the maximum id node is elected leader
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12
34
5
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odneighborhok
nodesknodesk
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12
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Phase 1: send id to 1-neighborhood
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2
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3
5 8
1
5
3 74
6
2
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If: received id current idThen: send a reply
12
34
5
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8
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If: a node receives both repliesThen: it becomes a temporal leader
12
34
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12
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Phase 2: send id to 2-neighborhood
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If: received id current idThen: forward the message
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If: received id current idThen: send a reply
12
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At second step:
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12
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If: a node receives both repliesThen: it becomes a temporal leader
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12
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Phase 3: send id to -neighborhood22
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If: received id current idThen: forward the message
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12
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If: received id current idThen: send a reply
At the step:22
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If: a node receives both repliesThen: it becomes the leader
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12
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8leader
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12
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8leader
nodesn nlog phases
In general:
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12
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8leader
Phase i: send id to -neighborhood12 i
12 i
12 i
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Time complexity
The leader spends time in
Phase 1: 2Phase 2: 4…Phase i:…Phase log n:
i2
Total time: )()2(22 log1log nOO nn
nlog2
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Phase 1: 4Phase 2: 8…Phase i:…Phase log n:
Message complexity
Messages per leader
12 i
1log2 n
Max #leaders
n
2/n
12/ in
1log2/ nn
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Phase 1: 4Phase 2: 8…Phase i:…Phase log n:
Total messages:
n4
)log( nnO
12 i
1log2 n
n
2/n
12/ in
1log2/ nn
Messages per leader Max #leaders
n4
n4
n4
n4
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Notes:
The algorithm does not need to know n
It can be converted to anasynchronous algorithm
49
An Mess. Synchronous Algorithm
)(nO
The node with smallest id is elected leader
There are rounds:
If in round there is a node with id • this is the new leader• the algorithm terminates
i i
n is known
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Round 1 ( time steps): no message sentn
922
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24
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n nodes
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922
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n nodes
Round 2 ( time steps): no message sentn
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n nodes
Round 9
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new leader
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n nodes
Round 9 ( time steps): messages sentn n
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new leader
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22
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n nodes
Round 9 ( time steps): messages sentn n
9
new leader
Algorithm Terminates
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22
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24
15
16
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n nodes
Round 9 ( time steps): messages sentn n
9
new leader
Total number of messages:n
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Another Synchronous Algorithm)(nO
The node with smallest id is elected leader
n is not known
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The algorithm:
• Each node injects a message with its id
• Message with id is injected and transferred with rate
ii2
1
• Nodes which have seen smaller id absorb higher id messages
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42
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0
Time 1
0
Transfer rate120
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5
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0
Time 2
0
1
120
120 221
120 rate
60
42
31
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0
Time 3
0
1
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42
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0
Time 4
01
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0
Time 5
0
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Time 6
0
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0
Time 8
0
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Time 8
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00
New leader
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Message complexity
Assume leader has (smallest) idi
Total time of algorithm:in 2
Note that if then algorithmis exponentially slow
)(ni
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Take the node with immediately higher id
ik
Total number of messages:
222
edge each on delay messagealgorithm of time total nn
k
i
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Take the node with immediately higher id
ikl
Total number of messages:
422 nnl
i
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id
imessages
n
k 2/n
l 4/n
lower
higher
nnn
n 242
Total messages:
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An Lower Bound )log( nnAssume we have algorithms in which:
• the maximum identifier is elected leader
• all the nodes must know the leader
We will prove:at least messages are needed)log( nn
• the size of the network is not knownn
to elect a leader
• the network is asynchronous
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open edgeno messages cross it
There is a possible asynchronousexecution with an open edge
(i.e. a very slow edge)
Messages may bepending to crossopen edge
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open edge
If the edge remains open, then the executionwill eventually reach a quiescent statewhere not more messages cross the ring
Messages pendingto cross open edge
Quiescent State
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open edgeMessages pendingto cross open edge
Quiescent State
In the quiescent state, a node can send a message only after it receives a message(thus no messages cross the ring)
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This may cause a propagation of messagesuntil the algorithm terminates
Suppose thatthe edge closes
Time t
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This may cause a propagation of messagesuntil the algorithm terminates
Time t+1
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This may cause a propagation of messagesuntil the algorithm terminates
Time t+2
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Time t+k
k k
After k time steps the affected radius is k
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Time t+x
Eventually, the ring stabilizes with a leader
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We will show, that there is an executionwith nodes such that:n
• there is an open edge
• at least messages are received )(nM
1)2( Mwhere
422)(
nnMnM
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x ybasis case 2n
If then is the leader yx x
should know about thisy
Therefore a message is sent: 1)2( M
Proof by induction
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x y
open edge
The message can be sent on one edge
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The case 2n From induction hypothesis, we have an execution:
2n
2n
M
openedge
Messages sent
nodes
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The case 2n
openedges2
n2n
2n
M
2n
MMessages sent
nodes nodes
From induction hypothesis, we have an execution:
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Observation:
1R 2R
If no message is sent along
then ring cannotdistinguish the twoscenarios 1R
1R
(similarly for ring )2R
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2n
2n
2n
M
2n
MMessages sent
nodes nodes
Therefore, the same number of messageshave to be sent in sub-rings
True open edge
True open edge
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max id
1R 2R
All nodes in should learn about
x
2R x
(assume that max id is in )1R
open edge
open edge
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max id
1R 2R
All nodes in should learn about
x
2R x
Thus, after the open edges close,at least
2n
messages are sent
2n
(At least oneof these is sent)
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max idx
Suppose edges close at timet
At least one message is sent
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max idx
time 1t
At least one message is sent
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max idx
time 2t
At least messages have been sent since2 t
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max idx
time kt radius k
At least messages have been sent sincek t
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max idx
time4n
t
radius 4n
At least messages have been sent since4n t
Independent areas
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max idx
time4n
t
radius 4n
At least messages have been sent in one area
8n
Independent areas
8n
Messages
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max idx
time4n
t
radius 4n
open edge
Since areas are independent, we could haveclosed only one edge
8n
Messages
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max idx
time4n
t
radius 4n
open edge
8n
Messages
2n
M
2n
M
Induction Hypothesis
)log(82
2)( nnnn
MnM
Total Messages
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