Election in the Complete Graph
3
6
8
2
1
5
7
4
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Trivial Algorithm.
Ask neighbours one at a time
3
2
I am 1
You win
1
5
7
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6
Trivial Algorithm.
Ask neighbours one at a time
3
2
1
5
7
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6
Message Complexity
O(n2)
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Better Algorithm
Ideas:
- In stages
- Territory acquisition (capture neighbours)
ensuring that a node is captured by at most
one candidate in the same stage
- Disjoint territories
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A node attacks another node, if successful
it captures the node increasing the size of its territory
(= stage number)
Defeated nodes become captured (belonging
to a owner) and stop attacking
CANDIDATE: still playing trying to increase the territory
PASSIVE: transitional phase, will not attack anymore,
will eventually become captured
CAPTURED: belong to a territory, owned by a candidate
An
attack could reach candidate passive captured
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4
territories
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8
3
The Attack
Bigger territories win over smaller ones (i.e. higher stages)
In case of tie, smaller Ids win
8
8
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4
3
The Attack
Bigger territories win over smaller ones (i.e. higher stages)
In case of tie, smaller Ids win
8
8
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4
3
Attacking a candidate
Send capture message to one neighbour
(stage,Id:3)
captured
8
3
8
3
Increase territory and stage (+1)
(stage,Id:3)
reject
8
3
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3
If still candidate, 3 becomes passive
8
Attacking a passive
3
8
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Attacking a passive
captured
3
8
Increase territory and stage (+1)
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Attacking a captured
8
3
3
If I know I belong
to a bigger territory
8
3 becomes passive, if still candidate
Not sure, ask my owner
8
3
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3
8
Attacking a captured
8
3
3
If I know I belong
to a bigger territory
8
Not sure, ask my owner
8
3
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3
8
Increase territory and stage (+1)
Attacking a captured
8
3
3
If I know I belong
to a bigger territory
8
Not sure, ask my owner
8
3
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3
8
3 becomes passive, if still candidate
passive
candidate
captured
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When to terminate ?
When does a candidate become a leader ?
When it captures more than n/2 nodes
If a candidate has captured more than n/2 nodes
nobody else can become leader
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Important
The territories of any two candidates are disjoint
Because at any time, any node has only ONE owner.
3
Each territory is
rooted in its owner
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8
2
1
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7
4
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We need: number of stages and messages per stage
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Messages per attack
candidate --- candidate
2 msgs
candidate --- passive
2 msgs
candidate --- captured
4 msgs
At most 4 messages per attack
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Number of stages
A candidate with n/2 +1 captured nodes becomes
leader and notify
n/2 +1 stages
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How many candidates in each stage ?
Stage i ---> territory of size i
With disjoint territories
There cannot be more than n/i
candidates in stage i
ni  n/i
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i i
i
i
i
Message Complexity
At most 4 messages per attack
ni  n/i
Messages in stage i  4 n/i
Harmonic number Hn/2
n/2
O( 
1
4 n/i) =
O(
n/2
4n
 1/i )
1
M(completeElect) = O(n log n)
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= O(log n)
Election in a Complete Graph with
Chordal Sense of Direction
1
5
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4
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5
Any ring algorithm
IDEA: Put information in messages. At the next step, use a smaller
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ring.
(Id:3, originLink: 1)
6
1
8
7
6
1
2
1
7
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5
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2
6
4
3
5
8-1=7.
the message
was originated from
my neighbour
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1
4
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3
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6
1
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7
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1
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3
7
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4
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2
1
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7
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7
2
7
2
1
(Id:3, originLink: 1)
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7
6
1
8
1
7
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7
6
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5
(Id:3, originLink: 2)
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4 3
1
2
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3
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6
1
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5
3
7
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1
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3
6
2
2
5 4
7
1
1
3
4
4
5
4
2
1
6
7
5
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7
7
2
7
2
1
Modulo 8
3
7
6
1
8
1
7
2
7
6
5
5
(Id:3, originLink: 2)
6
4 3
1
2
4
3
6
5
4
4
3
5
6
1
2
3
4
5
6
5
3
7
2
1
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3
6
2
2
5 4
7
1
1
3
4
4
5
4
2
1
6
7
5
6
7
7
2
Which means
my link 8-2=6
7
2
1
Similarly for the other candidates
Modulo 8
3
7
6
1
8
1
7
2
7
6
5
5
6
4 3
1
2
4
(Id:2, originLink: 6)
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5
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4
3
5
6
1
2
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4
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6
5
3
7
2
1
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3
6
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2
5 4
7
1
1
3
4
4
5
4
2
1
6
7
5
6
7
7
2
7
2
1
Similarly for the other candidates
Modulo 8
(Id:2, originLink: 6+6=4)
3
7
6
1
8
1
7
2
7
6
5
5
6
4 3
1
2
4
(Id:2, originLink: 6)
3
6
5
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4
3
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6
1
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6
5
3
7
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1
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3
6
2
2
5 4
7
1
1
3
4
4
5
4
2
1
6
7
5
6
7
7
2
7
2
1
Similarly for the other candidates
Modulo 8
3
7
6
1
8
Which means
my 8-4=4
1
7
2
7
6
5
5
6
4 3
1
2
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3
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6
1
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3
7
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1
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3
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2
5 4
7
1
1
3
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4
5
4
2
1
6
7
5
6
7
7
2
7
2
1
Similarly for the other candidates
Modulo 8
3
7
6
1
8
1
7
2
7
6
5
5
6
4 3
1
2
4
3
6
5
4
4
3
5
6
1
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3
7
2
1
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3
6
2
2
5 4
7
1
1
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4
2
1
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2
7
2
1
Message Complexity
Planar graph - 2 messages per edge
O(n)
[because a planar graph has O(n) edges]
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