1 Consensus Hierarchy Part 1. 2 Consensus in Shared Memory Consider processors in shared memory:...

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Consensus HierarchyPart 1

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Consensus in Shared Memory

Consider processors in shared memory:n

10,..., npp

which try to solve the consensus problem

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0

1

0

0p

1p

2p

0

1

0

3p

4p

5p

Every process starts with an initial valuestored in local memory (0 or 1)

Shared memory

Local memory

Local memory

4

0

1

0

0p

1p

2p

0

1

0

3p

4p

5p

communication through shared memory

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1

1

1

0p

1p

2p

1

1

1

3p

4p

5p

At the end of execution, every processhas decided the same value (0 or 1)

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Validity condition:

If every process starts with the same value, the every process should decide that value

Additional condition:The decided value is one of the initial values

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Wait-freedom in asynchronous systems:

A process should be able to finishexecution of an algorithmeven if all other processes fail

Wait-freedom captures:•Asynchronous executions•Crash failures

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Object Types

Read/Write

FIFO

Compare&Swap

Test&Set

n-register assignment

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Consensus Number

Consensus Number of an object type:

The maximum number of processes for which the object can be used tosolve the wait-free consensus problem(together with read/write objects)

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Object Type Consensus Number

Read/Write 1

FIFO, Test&Set 2

Compare&Swap (infinity)

n-register assignment 2n-2

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int compare_and_swap ( int* register, int oldval, int newval) { int old_reg_val = *register; if (old_reg_val == oldval) *register = newval; return old_reg_val;}

int Test-and-Set(boolean lock) { boolean initial = lock; lock = true; return initial;}

(Shared ) boolean lock = false;

function Critical_Section() { while TestAndSet(lock) skip //spin until lock is acquired //Critical-section code - only one process can be in this section at a time begin { … } //end-of critical section – release lock lock = false //release lock when finished with the critical section }

Mutual Exclusion

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Simulation:Object Type B

Object Type A

Object Type A

Read/Write Object

Object of type A simulates object of Type B(using auxiliary read/write objects)

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Theorem:Objects of Type A with consensus numbercannot simulate in wait-free manner another object of Type B withconsensus number

n

nm

Proof: Since otherwise, object A wouldhave consensus numberm

End of Proof

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can simulate in a wait-free mannerany other arbitrary object

Universal object:

Compare&SwapExample:

(infinity consensus number)

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Objects with consensus number ncan simulate in a wait-free mannerany other arbitrary object of up toprocessors

n

We can show:

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Read/Write

Shared Memory

Suppose that the sharedmemory can only be accessedthrough Read or Write operations

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Theorem:

Proof of Theorem:

The consensus number of the Read/Write object is 1

Trivially, any consensus algorithm with 1 process using read/write variables is wait-free.

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Wait-free consensus cannot besolved using only read/write objectsfor processors 2n

It remains to show:

We will show that any algorithmthat solves wait-free consensus for has an execution that never terminates

Approach:

2n

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System configuration:

Is the set of all variables in the system, including local and shared

0

1

0

0p

1p

2p

0

1

0

3p

4p

5p

C

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A distributed system execution can be always be viewed as a: sequence of configurations

0C fC1C 2C

Initialconfiguration

Finalconfiguration

0ip

1ip

2ip

Processor action: Read or Write

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1

C

D

C D

1 0 1 0 0 0

bivalent

univalentunivalent

bivalent

Valence of system configurations

1-valent0-valent

Consensus value at possible execution paths

consensus reached consensusnot reached

consensus reached

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A terminating execution:

0C 1C 2C


0ip

1ip

2ip

Bivalent Bivalent Univalent

fC

Univalent

UnivalentBivalent

Finalconfiguration

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To prove the theorem, we will show that there is always an executionwhere every configuration is bivalent

0C 1C 2C


0ip

1ip

2ip

Bivalent Bivalent Bivalent

Never-ending execution

Bivalent

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1

1

1

0p

1p

2p

1

0

0

0p

1p

2p

Similar configurations for processor 0p

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198

76

23

198

76

Same shared variablesLocal variables of others may differ

1C 2C21

0

CCp

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21 CCip

Lemma:If there exist univalent configurations

1C and such that2C

then if is -valentthen is -valent too

1C

2Cv

v

Proof of Lemma:

)1 or 0( v

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1C

Univalent

Executionwith onlytaking actions

ip

ip

v v v

All possible executionsfrom 1C

final decision for each Possible execution

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1C

Univalent


ip

ip

v v v

2C

Univalent


ip

ip

x x x

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1C

Univalent


ip

ip

v v v

2C

Univalent


ip

ip

x vx x

21 CCip

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1C

Univalent


ip

ip

v v v

2C

Univalent


ip

ip

21 CCip

v v v

End of Lemma Proof

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Lemma:There exists a bivalentinitial configuration

Proof of Lemma:

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Possible Initial Configurations

Shared Memory

Empty

Local Memory

10p

Initial Configuration1I

11p

1np 1

0

0I

0

0

0

01I

1

1

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Shared Memory

Empty

Local Memory

10p

11p

1np 1

0

0I

0

0

0

01I

1

1

0-valent 1-valent?


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Shared Memory

Empty

Local Memory

10p

11p

1np 1

0

0I

0

0

0

01I

1

1

0-valent 1-valent1-valent?

No, because 010

0

IIp


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Shared Memory

Empty

Local Memory

10p

11p

1np 1

0

0I

0

0

0

01I

1

1

0-valent 1-valent0-valent?

No, because 101

1

IIp


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Shared Memory

Empty

Local Memory

10p

11p

1np 1

0

0I

0

0

0

01I

1

1

0-valent 1-valentbivalent

End of Lemma Proof


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Critical processor for a configuration:

the configuration is bivalent,and after the processor takes stepthe configuration becomes univalent

C C ipBivalent Univalent

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Lemma:If is a bivalent configurationthen, there is at least one processorwhich is not critical

C

Proof of Lemma:

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Assume for contradiction that all processors are critical

C

0C

1C

1nC

bivalent0p

1p

1np

univalent

univalent

univalent

Possibleexecutions

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C

bivalent0p

1p

1np

valent

It cannot be that all have the same valence

v

)1 or 0( v

valentv

valentv

0C

1C

1nC

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C

bivalent0p

1p

1np

valent

It cannot be that all have the same valence

v

)1 or 0( v

valentv

valentv

valentv

Contradiction 0C

1C

1nC

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C

bivalent0p

jp

1np

There must exist two processors with different valences

iC

jC

ip

0C

1nC

valent-0

valent-1

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C

bivalent

jp

iC

jC

ip

valent-0

valent-1

Case 1: suppose that they access different shared variables

Read x

Read y

x

y

ip

jp

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C

bivalent

jp

iC

jC

ip

valent-0

valent-1

Read x

Read y

two possible executions

C

C

jp

ip

Read y

Read x

valent-0

valent-1

impossible since CC different valence

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same result holds for any kind of operation (Read or Write) that the processors apply to x and y

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C

bivalent

jp

iC

jC

ip

valent-0

valent-1

Case 2: suppose that they access the same shared variable

Read x

Read x

xip

jp

subcase: read/read

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C

bivalent

jp

iC

jC

ip

valent-0

valent-1

Read x

Read x

C

C

jp

ip

Read x

Read x

valent-0

valent-1

impossible since CC different valence

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C

bivalent

jp

iC

jC

ip

valent-0

valent-1

Read x

Write x

subcase: read/write

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C

bivalent

jp

iC

jC

ip

valent-0

valent-1

Read x

Write x

C

C

jp

ip

Write x

Read x

valent-0

valent-1

impossible since j

p

CCj

different valence

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subcase: write/write

C

bivalent

jp

iC

jC

ip

valent-0

valent-1

Write x

Write x

C

C

jp

ip

Write x

Write x

valent-0

valent-1

impossible since j

p

CCj

different valence

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In all cases we obtained contradictionTherefore, there exists a processorwhich is not critical

C

0C

1C

1nC

bivalent0p

1p

1np

univalent

univalent

univalent

kC bivalent(not critical)

End of Lemma Proof

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Therefore, we can construct an execution

0C 1C 2C


0ip

1ip

2ip Never

ends

bivalent bivalent bivalent

Consensus can never be reached

End of Theorem Proof

1 Consensus Hierarchy Part 1. 2 Consensus in Shared Memory Consider processors in shared memory:...

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