View
220
Download
0
Category
Tags:
Preview:
Citation preview
1
Leader Election in Rings
2
A Ring Network
3
1 1
1
1
11
1
1
2
2
2
2
2
2
2
2
Sense of direction
left
right
4
1 1
1
1
11
1
1
2
2
2
2
2
2
2
2
Links are bi-directional for messages
At most one message in each direction
5
1 1
1
1
11
1
1
2
2
2
2
2
2
2
2
Anonymous Ring
6
Eponymous (non-anonymous) Ring
1 1
1
1
11
1
1
2
2
2
2
2
2
2
2
12
34
5
6
7
8
7
Leader Election
Initial state Final state
leader
8
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
9
Synchronous Anonymous Rings
Every processor runs the same algorithm
Every processor does exactly the sameexecution
10
Initial state Final state
leader
If one node is elected a leader,then every node is elected a leader
11
Final state
leader
Conclusion 1:
Leader electioncannot be solvedin synchronousanonymous rings
12
Final state
leader
Conclusion 2:
Leader electioncannot be solvedin asynchronousanonymous ringseither
Why?
The asynchronous ring may behaveLike the synchronous ring
13
Asynchronous Eponymous Rings
12
34
5
6
7
8
The maximum id node is elected leader
14
12
34
5
6
7
8
Each node sends a message with its idto the left neighbor
18
2
6
47
3
5
15
12
34
5
6
7
8
If: message received id current node id
8
6
7
5
Then: forward message
16
12
34
5
6
7
8
If: message received id current node id
8
7
Then: forward message
17
12
34
5
6
7
8
If: message received id current node id
8
7
Then: forward message
18
12
34
5
6
7
8
If: message received id current node id
8
Then: forward message
19
12
34
5
6
7
8
If: a node receives its own message
8
Then: it elects itself a leader
20
12
34
5
6
7
8
If: a node receives its own message
Then: it elects itself a leader
leader
21
12
34
5
6
7
8leader
The leader sends a message in the networkdeclaring itself as the “leader of the ring”
22
12
34
5
6
7
8leader
Time complexity: )(nO n nodes
23
1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
worst case scenario:
24
1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
n messages
25
1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
1n messages
26
1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
2n messages
27
1n-1
n-3
2
n-2
n
Message complexity: )( 2nO n nodes
1n
Total messages:
n
2n
1 )( 2n2
28
Notes:
ndoes not need to be known to the algorithm
The algorithm can be converted to asynchronous
29
An O(n log n) Mess. Algorithm
12
34
5
6
7
8
Again, the maximum id node is elected leader
30
12
34
5
6
7
8
odneighborhok
nodesknodesk
31
12
34
5
6
7
8
Phase 1: send id to 1-neighborhood
18
2
6
47
3
5 8
1
5
3 74
6
2
32
If: received id current idThen: send a reply
12
34
5
6
7
8
33
If: a node receives both repliesThen: it becomes a temporal leader
12
34
5
6
7
8
34
12
34
5
6
7
8
Phase 2: send id to 2-neighborhood
8
85
5
7
7
6
6
35
12
34
5
6
7
8
8
8
5
5
7
76
6
If: received id current idThen: forward the message
36
If: received id current idThen: send a reply
12
34
5
6
7
8
At second step:
37
12
34
5
6
7
8
If: a node receives both repliesThen: it becomes a temporal leader
38
12
34
5
6
7
8
Phase 3: send id to -neighborhood22
77
8
8
39
12
34
5
6
7
8
77
8
8
If: received id current idThen: forward the message
40
12
34
5
6
7
8
If: received id current idThen: send a reply
At the step:22
41
12
34
5
6
7
8
If: a node receives both repliesThen: it becomes the leader
42
12
34
5
6
7
8leader
43
12
5
6
8leader
nodesn nlog phases
In general:
44
12
5
6
8leader
Phase i: send id to -neighborhood12 i
12 i
12 i
45
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
46
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
47
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
48
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
50
Round 1 ( time steps): no message sentn
922
33
24
15
16
57
48
n nodes
51
922
33
24
15
16
57
48
n nodes
Round 2 ( time steps): no message sentn
52
22
33
24
15
16
57
48
n nodes
Round 9
9
new leader
53
22
33
24
15
16
57
48
n nodes
Round 9 ( time steps): messages sentn n
9
new leader
54
22
33
24
15
16
57
48
n nodes
Round 9 ( time steps): messages sentn n
9
new leader
Algorithm Terminates
55
22
33
24
15
16
57
48
n nodes
Round 9 ( time steps): messages sentn n
9
new leader
Total number of messages:n
56
Another Synchronous Algorithm)(nO
The node with smallest id is elected leader
n is not known
57
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
58
42
31
5
6
7
0
Time 1
0
Transfer rate120
59
42
31
5
6
7
0
Time 2
0
1
120
120 221
120 rate
60
42
31
5
6
7
0
Time 3
0
1
61
42
31
5
6
7
0
Time 4
01
62
42
31
5
6
7
0
Time 5
0
63
42
31
5
6
7
0
Time 6
0
64
42
31
5
6
7
0
Time 8
0
65
Time 8
42
31
5
6
7
00
New leader
66
Message complexity
Assume leader has (smallest) idi
Total time of algorithm:in 2
Note that if then algorithmis exponentially slow
)(ni
67
Take the node with immediately higher id
ik
Total number of messages:
222
edge each on delay messagealgorithm of time total nn
k
i
68
Take the node with immediately higher id
ikl
Total number of messages:
422 nnl
i
69
id
imessages
n
k 2/n
l 4/n
lower
higher
nnn
n 242
Total messages:
70
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
71
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
72
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
73
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)
74
This may cause a propagation of messagesuntil the algorithm terminates
Suppose thatthe edge closes
Time t
75
This may cause a propagation of messagesuntil the algorithm terminates
Time t+1
76
This may cause a propagation of messagesuntil the algorithm terminates
Time t+2
77
Time t+k
k k
After k time steps the affected radius is k
78
Time t+x
Eventually, the ring stabilizes with a leader
79
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
80
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
81
x y
open edge
The message can be sent on one edge
82
The case 2n From induction hypothesis, we have an execution:
2n
2n
M
openedge
Messages sent
nodes
83
The case 2n
openedges2
n2n
2n
M
2n
MMessages sent
nodes nodes
From induction hypothesis, we have an execution:
84
Observation:
1R 2R
If no message is sent along
then ring cannotdistinguish the twoscenarios 1R
1R
(similarly for ring )2R
85
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
86
max id
1R 2R
All nodes in should learn about
x
2R x
(assume that max id is in )1R
open edge
open edge
87
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)
88
max idx
Suppose edges close at timet
At least one message is sent
89
max idx
time 1t
At least one message is sent
90
max idx
time 2t
At least messages have been sent since2 t
91
max idx
time kt radius k
At least messages have been sent sincek t
92
max idx
time4n
t
radius 4n
At least messages have been sent since4n t
Independent areas
93
max idx
time4n
t
radius 4n
At least messages have been sent in one area
8n
Independent areas
8n
Messages
94
max idx
time4n
t
radius 4n
open edge
Since areas are independent, we could haveclosed only one edge
8n
Messages
95
max idx
time4n
t
radius 4n
open edge
8n
Messages
2n
M
2n
M
Induction Hypothesis
)log(82
2)( nnnn
MnM
Total Messages
Recommended