Advanced Network Security 2. Distributed Algorithms: Leader - - PowerPoint PPT Presentation

Jaap-Henk Hoepman

Digital Security (DS) Radboud University Nijmegen, the Netherlands

@xotoxot // * jhh@cs.ru.nl // 8 www.cs.ru.nl/~jhh

Advanced Network Security

• 2. Distributed Algorithms:

Leader Election

Jaap-Henk Hoepman // Radboud University Nijmegen //

Leader election: motivation

n IBM token ring (1970)

• For local area network
• Single token traversing ring
• Station with token was allowed

to send

n How to

• Start the network?

«0 tokens

• Recover from an error?

«>1 tokens

02-03-2020 // Distributed algorithms - Leader Election 2

Jaap-Henk Hoepman // Radboud University Nijmegen //

Leader election (1)

n Given a graph ! = ($, &) of nodes, design a protocol that will elect a single node as leader n Output stored in local variable ( ) . +,-.,/

• There is one node ) with ( ) . +,-.,/ = 0/1,
• For all 2 ∈ $, 2 ≠ ) we have ( ) . +,-.,/ = 5-+6,

n Assumptions

• ! is connected, i.e. nodes can reach each other; we assume a bidirectional

ring network here

• Nodes have unique identifiers ( ) . ).

«E.g. a MAC address «Note that nodes do NOT know V (i.e. the set of identities in the graph)

• All nodes simultaneously start the protocol

02-03-2020 // Distributed algorithms - Leader Election 3

Jaap-Henk Hoepman // Radboud University Nijmegen //

Leader election (2)

n Requirements

• Correctness: at most one leader is elected (and once elected stays

elected).

• Progress: eventually a leader is elected.

n Leader election used, for example to

• Recover from errors (the leader coordinates the repair)
• Initiate another higher-level distributed algorithm

02-03-2020 // Distributed algorithms - Leader Election 4

" Safety

" " livenest

"

Jaap-Henk Hoepman // Radboud University Nijmegen //

How would/could you solve this?

02-03-2020 // Distributed algorithms - Leader Election 5

pint

• to

message

• 4. ni
• )

passing

"

i

• n

t

→

s

(

FIFO :#

Hijs

k

^

KK

fijn

✓

←

n

km )

Jaap-Henk Hoepman // Radboud University Nijmegen //

Jaap-Henk Hoepman // Radboud University Nijmegen //

Some non-solutions

n Node ) with ( ) . ). = 0 becomes leader

• May not exist. E.g. if identifiers are based on MAC addresses

n Consider the following protocol for node )

• This protocol assumes that ( ) . ). = ), i.e. assigned increasing along

the ring, this is not necessarily the case

02-03-2020 // Distributed algorithms - Leader Election 6

Send clockwise (right) ( ) . ). Receive counterclockwise (left) ). ( ) . +,-.,/ = (( ) . ). < ).)

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann’s protocol: leader election on a ring

n Assumption

• FIFO message passing and unique identifiers
• Note: nodes do not know the size of the ring
• Unidirectional communication (clockwise only)

n Protocol for node )

02-03-2020 // Distributed algorithms - Leader Election 7

9: = ∅ ( ) . +,-.,/ = 5-+6, send right ( ) . ). while ( ) . ). ∉ 9: do receive left ). 9: = 9: ∪ {).} send right ). ( ) . +,-.,/ = (( ) . ). = min

C

2 ∈ 9:) Election point

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann

n Why does this work? n What is the message / round complexity? n What if message passing is not FIFO?

02-03-2020 // Distributed algorithms - Leader Election 8

9 = ∅ ( ) . +,-.,/ = 5-+6, send right ( ) . ). while ( ) . ). ∉ 9 do receive left ). 9 = 9 ∪ {).} send right ). ( ) . +,-.,/ = (( ) . ). = min

C

2 ∈ 9)

• single message

has constant

size

• 0 ( log N)
• # messages :

01N) for

a single node

⇒

total

message complexit

is

0 ( N')

Jaap-Henk Hoepman // Radboud University Nijmegen //

Le Lann

n Why does this work?

02-03-2020 // Distributed algorithms - Leader Election 9

9 = ∅ ( ) . +,-.,/ = 5-+6, send right ( ) . ). while ( ) . ). ∉ 9 do receive left ). 9 = 9 ∪ {).} send right ). ( ) . +,-.,/ = (( ) . ). = min

C

2 ∈ 9)

i

FIFO !

, →

40k

also

/

rijker

Identificeer

:

\

• s

Jaap-Henk Hoepman // Radboud University Nijmegen //

Le Lann

n What is the message / round complexity?

02-03-2020 // Distributed algorithms - Leader Election 10

9 = ∅ ( ) . +,-.,/ = 5-+6, send right ( ) . ). while ( ) . ). ∉ 9 do receive left ). 9 = 9 ∪ {).} send right ). ( ) . +,-.,/ = (( ) . ). = min

C

2 ∈ 9)

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann

n Why does this work?

• See proof further on

n What is the message / round complexity?

• Every node forwards messages until it receives it’s own id back.
• If the size of the ring is n, each node sends D + 1 messages
• Total number of messages sent it D D + 1
• Round complexity is D

n What if message passing is not FIFO?

• Homework ;-)

02-03-2020 // Distributed algorithms - Leader Election 11

9 = ∅ ( ) . +,-.,/ = 5-+6, send right ( ) . ). while ( ) . ). ∉ 9 do receive left ). 9 = 9 ∪ {).} send right ). ( ) . +,-.,/ = (( ) . ). = min

C

2 ∈ 9)

Jaap-Henk Hoepman // Radboud University Nijmegen //

Le Lann: proof

n Proof obligations

• Correctness
• Progress

02-03-2020 // Distributed algorithms - Leader Election 12

9 = ∅ ( ) . +,-.,/ = 5-+6, send right ( ) . ). while ( ) . ). ∉ 9 do receive left ). 9 = 9 ∪ {).} send right ). ( ) . +,-.,/ = (( ) . ). = min

C

2 ∈ 9)

NY

election ⇒

⑧

pour

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann proof of correctness (1)

n Correctness: at most one leader is elected (and once elected stays elected).

• We need to prove that for all nodes ), 2 that reach the election point,

we have 9: = 9

• C. Then the result follows as nodes have unique

identifiers.

• In fact we will show that we have 9: = 9

C = ( G . ). G ∈ 0, D − 1

(nodes are numbered clockwise around the ring, and D is the number of nodes – which is unknown to the number of nodes!)

• In what follows, let 9: be the list of values, in the order in which they

were received (instead of a set).

02-03-2020 // Distributed algorithms - Leader Election 13

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann proof of correctness (1)

n [correctness proof continued]

• We prove this using induction on the /-th message node ) receives; in

fact we show that when node ) receives the /-th message, it actually received 9: = (( ) − 1 mod D . )., … , ( ) − / mod D . ).) in that order.

• For round / = 0 (i.e. initially) the statement holds trivially: no

messages have been received so far and 9: = ()

02-03-2020 // Distributed algorithms - Leader Election 14

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann proof of correctness (2)

n [correctness proof continued]

• For round /L = / + 1 observe

«All message received by node ) by round /L = / + 1 must have been sent by its left hand neighbour 2 = ) − 1 mod D in or before round /. «At the end of round / node 2 has sent all values in 9

C to ), in the same order,

but first sent out ( 2 . ). «Because of the FIFO property node ) receives these in the same order «Using the induction hypothesis 9

C = ( 2 − 1 mod D . )., … , ([2 − / mod D] in that

• rder.

«Then ) receives ( 2 . )., … , ( 2 − r mod D = ( ) − 1 mod D . )., … , ( ) − /L . ). in that order as required.

02-03-2020 // Distributed algorithms - Leader Election 15

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann proof of correctness (3)

n [correctness proof continued]

• If node ) reaches the election point then ( ) . ). ∈ 9:
• This happens when the /-th message node ) receives (so / > 0), i.e.the

message ( G . ). with G = ) − / mod D , equals ( ) . ). .

• As identities on the ring are unique, this implies G = ) and so / = D and

hence, 9: = (( ) − 1 mod D . )., … , ( ) − D mod D . ).)

• In other words 9: = ( G . ). G ∈ 0, D − 1] as required.

02-03-2020 // Distributed algorithms - Leader Election 16

Jaap-Henk Hoepman // Radboud University Nijmegen //

LeLann proof of correctness (3)

n Progress: eventually a leader is elected.

• Let node ) have the smallest ( ) . ).
• Initially node ) sends ( ) . ). to its right-hand neighbour
• This means a message ( ) . ). is either in transit on a link (meaning the next

node will eventually receive it) or received by the node (meaning it will be sent out to the right by that node)

• Whenever this message is sent, it moves one step closer back to node )
• Eventually node ) receives ( ) . ). (and sends it once more the right) and then

stops

• It determines that (( ) . ). = min

:

) ∈ 9) and hence becomes leader as required

02-03-2020 // Distributed algorithms - Leader Election 17

Jaap-Henk Hoepman // Radboud University Nijmegen //

What if nodes do not have unique identifiers?

02-03-2020 // Distributed algorithms - Leader Election 18

→ are

• id

µ

(

leider

¥

i

% !

÷ #

Jaap-Henk Hoepman // Radboud University Nijmegen //

What if nodes do not have unique identifiers?

n Then there exists a symmetric configuration (

• where all nodes have the same state, and all edges have the same state
• I.e. either all nodes are leaders, or no node is leader

n Starting in ( let all nodes take a step (the same) in turn, then

• all steps are local steps (changing the local state to a new state, the same for all

nodes)

• all steps are receive actions (receiving the same message), or
• all steps are send actions (sending the same message)

n Therefore the resulting configuration (L is again symmetric n We can repeat this forever, never reaching a state where there is exactly one leader n This is called a symmetry argument

02-03-2020 // Distributed algorithms - Leader Election 19

Jaap-Henk Hoepman // Radboud University Nijmegen //

Peterson’s protocol: leader election on a ring

n Bidirectional communication

• Nodes can send messages clockwise and anticlockwise

n Idea: algorithm proceeds in rounds

• First round all D nodes are active and participate
• If a round starts with G participants, at least G/2 and at most G − 1 will

be eliminated (and become passive)

• If a round start with 1 participant, it will declare itself leader at the

end of the round

02-03-2020 // Distributed algorithms - Leader Election 20

start with

N

modes :

Yom are garant eed

to

find

a

leader in

log ( N)

rounds

Jaap-Henk Hoepman // Radboud University Nijmegen //

Peterson’s protocol: leader election on a ring

n The protocol for node )

02-03-2020 // Distributed algorithms - Leader Election 21

( ) . -S0)T, = 0/1, ( ) . +,-.,/ = 5-+6, while 0/1, /* new round */ do if ( ) . -S0)T, == 0/1, ∧ ( ) . +,-.,/ == 5-+6, send left ( ) . ). send right ( ) . ). receive right /)Vℎ0). receive left +,50). if (( ) . ). == +,50).) ∧ (( ) . ). == /)Vℎ0).) ( ) . +,-.,/ = 0/1, else if (( ) . ). < +,50).) ∨ (( ) . ). < /)Vℎ0).) ( ) . -S0)T, = 5-+6, else /* passive or leader */ receive right ). ; send left ). receive left ). ; send right ).

Jaap-Henk Hoepman // Radboud University Nijmegen //

Peterson’s protocol

n Why does it work? n What is the message / round complexity? n What if message passing is not FIFO?

02-03-2020 // Distributed algorithms - Leader Election 22

( ) . -S0)T, = 0/1, ( ) . +,-.,/ = 5-+6, while 0/1, /* new round */ do if ( ) . -S0)T, == 0/1, ∧ ( ) . +,-.,/ == 5-+6, send left ( ) . ). send right ( ) . ). receive right /)Vℎ0). receive left +,50). if (( ) . ). == +,50).) ∧ (( ) . ). == /)Vℎ0).) ( ) . +,-.,/ = 0/1, else if (( ) . ). < +,50).) ∨ (( ) . ). < /)Vℎ0).) ( ) . -S0)T, = 5-+6, else /* passive or leader */ receive right ). ; send left ). receive left ). ; send right ).

Jaap-Henk Hoepman // Radboud University Nijmegen //

Peterson

02-03-2020 // Distributed algorithms - Leader Election 23

( ) . -S0)T, = 0/1, ( ) . +,-.,/ = 5-+6, while 0/1, /* new round */ do if ( ) . -S0)T, == 0/1, ∧ ( ) . +,-.,/ == 5-+6, send left ( ) . ). send right ( ) . ). receive right /)Vℎ0). receive left +,50). if (( ) . ). == +,50).) ∧ (( ) . ). == /)Vℎ0).) ( ) . +,-.,/ = 0/1, else if (( ) . ). < +,50).) ∨ (( ) . ). < /)Vℎ0).) ( ) . -S0)T, = 5-+6, else /* passive or leader */ receive right ). ; send left ). receive left ). ; send right ).

etel

• "

Jaap-Henk Hoepman // Radboud University Nijmegen //

Peterson’s protocol: leader election on a ring

n There are at most log D rounds

• Node can only survive (remain active) if

both its left and right active neighbour are smaller

• Therefore at most half of the nodes can

survive

n In every round 2D messages are sent

• An active or passive node sends exactly 2

messages in each round

n So: message complexity is at most 2D log D

02-03-2020 // Distributed algorithms - Leader Election 24

( ) . -S0)T, = 0/1, ( ) . +,-.,/ = 5-+6, while 0/1, /* new round */ do if ( ) . -S0)T, == 0/1, ∧ ( ) . +,-.,/ == 5-+6, send left ( ) . ). send right ( ) . ). receive right /)Vℎ0). receive left +,50). if (( ) . ). == +,50).) ∧ (( ) . ). == /)Vℎ0).) ( ) . +,-.,/ = 0/1, else if (( ) . ). < +,50).) ∨ (( ) . ). < /)Vℎ0).) ( ) . -S0)T, = 5-+6, else /* passive or leader */ receive right ). ; send left ). receive left ). ; send right ).

Jaap-Henk Hoepman // Radboud University Nijmegen //

i

Öre

:

¥

Jaap-Henk Hoepman // Radboud University Nijmegen //

Peterson’s protocol: leader election on a ring

n There are at most log D rounds

• Node can only survive (remain active) if

both its left and right active neighbour are smaller

• Therefore at most half of the nodes can

survive

n In every round 2D messages are sent

• An active or passive node sends exactly 2

messages in each round

n So: message complexity is at most 2D log D

02-03-2020 // Distributed algorithms - Leader Election 24

( ) . -S0)T, = 0/1, ( ) . +,-.,/ = 5-+6, while 0/1, /* new round */ do if ( ) . -S0)T, == 0/1, ∧ ( ) . +,-.,/ == 5-+6, send left ( ) . ). send right ( ) . ). receive right /)Vℎ0). receive left +,50). if (( ) . ). == +,50).) ∧ (( ) . ). == /)Vℎ0).) ( ) . +,-.,/ = 0/1, else if (( ) . ). < +,50).) ∨ (( ) . ). < /)Vℎ0).) ( ) . -S0)T, = 5-+6, else /* passive or leader */ receive right ). ; send left ). receive left ). ; send right ).