The iterated shared memory model of computation and an enrichment with safe-consensus tasks

The iterated shared memory model of computation and an enrichment

with safe-consensus tasks

Rodolfo CondeJoint work with Sergio Rajsbaum

Instituto de MatemáticasUniversidad Nacional Autónoma de México

GETCO 2010

The Model

15/01/2010 Rodolfo Conde and Sergio Rajsbaum

The Iterated Snapshot Shared Memory model

• We have n processes that communicate using a memory SM[i][0…n] (i ≥ 0) of Read/Write registers



• The computation proceeds in rounds



• In each round, a process P can atomically write to SM[i][P]


0 11


• each process can atomically read all of SM[i]


0 11


• In each round, the processes use a new memory array


0 11

0 10

Asynchronous

• The n processes are asynchronous– Arbitrary delays of any kind


Wait-Free

• The protocols are wait-free– All but one process can crash– A process cannot wait to hear from another

process


?

Generic Iterated Snapshot protocolinit r := 0; sm := input, dec := NULL;

loop foreverr := r + 1;SM[r].write(sm);sm := SM[r].snapshot();

/* Local computing */end loop





P writes sm to SM[r][P]





P reads all the array SM[r]


Two processes protocol

• One possible execution is the following: the two processes read and write concurrently


0 WRRD

1 WRRD


• We can represent this execution as a 1-simplex


0 WRRD

1 WRRD


• Each vertex represents the process view of the memory


0 WRRD

1 WRRD

01 01


• Another possible execution: One process is faster that the other


0 WRRD

1 WRRD

01 01

0 WRRD

1 WRRD


• The red process only sees itself, but the green can see both of them


0 WRRD

1 WRRD

01 01

0 WRRD

1 WRRD

0


• And the last possibility


0 WRRD

1 WRRD

01 01

0 WRRD

1 WRRD

0

0 WRRD

1 WRRD


• And the last possibility


0 WRRD

1 WRRD

01 01

0 WRRD

1 WRRD

0

0 WRRD

1 WRRD

1

Protocol complex (1 round)


01 01 01

The 2nd round

• The input for the 2nd round is any possible state after the first round


0 WRRD

1 WRRD

01 01

The 2nd round

• And the three possibilities repeat


0 WRRD

1 WRRD

01 01

0 WRRD

1 WRRD

The 2nd round



0 WRRD

1 WRRD

01 01

0 WRRD

1 WRRD

The 2nd round



0 WRRD

1 WRRD

01 01

0 WRRD

1 WRRD

Two processes protocols in the iterated model

• Given a possible input:– Each execution of a round is represented as a 1-

simplex– All possible executions are represented as a

simplicial complex (subdivision of a line)


Three processes protocols

• The state after an execution can be described by a triangle (2-simplex)


0 WRRD

0 WRRD

0 WRRD

Three processes protocols

• The state after an execution can be described by a triangle (2-simplex)


0 WRRD

0 WRRD

0 WRRD

000

00

00

Protocol complex (1st round)




0 WRRD

0 WRRD

0 WRRD



0 WRRD

0 WRRD

0 WRRD



0 WRRD

0 WRRD

0 WRRD



0 WRRD

0 WRRD

0 WRRD

Protocol complex (3 processes)

• For the 2nd round– Each triangle (state) of the 1st round subdivides in

the same way– Because we work in an iterated model– Recursive behaviour


Rercursive behaviour (2nd round)







Protocol complex (2nd round)

In general

• For n + 1 processes:– Each state of a protocol is represented as a n-

simplex– The executions of a protocol are represented as a

n-dimensional complex– A subdivision of the n-simplex !! [Gafni &

Borowsky]


The (n,k)-set agreement task[S. Chaudhuri, 90]


Set agreement

The (n,k)-set agreement task [S. Chaudhuri, 90]


Set agreement

2 7 9

Processes start with private input values from a domain I (|I| ≥ n)

The (n,k)-set agreement task [S. Chaudhuri, 90]


Set agreement

2 7 9

Their outputs must agree on at most k < n distinct values

7 7 2

Impossibility of (3,2)-set agreement in the Iterated model

• We can use the geometric view of distributed protocols to show this remarkable result.

• The basic idea is as follows:Assume a protocol exists.Find an execution of this protocol (using the protocol

complex) where processes decide 3 values !!!


Suppose a protocol exists

• Consider an input where processes have as input values their own ids

• Run the protocol until processes decide


Suppose a protocol exists

• Because we work in the iterated model– The protocol complex is a subdivision of the

triangle– We can colour the vertices with the decision each

process takes– This colouring satisfies the hypothesis of Sperner’s

Lemma


We apply Sperner’s lemma to the subdivided complex

• By Sperner’s Lemma, at least one simplex has all three colours

• This simplex corresponds to an execution where processes decide three distinct values !!!


In summary

• The iterated model• Executions are represented as simplicial

complexes• Simple recursive structure


In summary

• The set agreement task is impossible to solve [Borowsky & Gafni, Saks & Zaharoglou, Herlihy & Shavit, 93]

• The iterated model is equivalent to the usual read/write model [Borowsky & Gafni, 97]

• Set agreement result is valid in the usual model (but easier to prove in the iterated model)


We can enrich the Iterated model with more powerful

objects


The safe-consensus task


The safe-consensus task[Afek, Gafni & Lieber, 09]


Safe-consensus



Safe-consensus

2 7 9

Processes start with private input values from a domain I



Safe-consensus

2 7 9

2 2 2

Their outputs values must be the same

The safe-consensus task [Afek, Gafni & Lieber, 09]


Safe-consensus

2 7 9

The safe-consensus has two special rules



Safe-consensus

2 7 9

7

(1) If a process starts executing the task and outputs before any other process starts executing the task



Safe-consensus

2 7 9

7 7 7

the task’s output is that process proposed input value.



Safe-consensus

2 7 9

(2) Otherwise, if two or more processes initially access the task concurrently



Safe-consensus

2 7 9

α α α

it can return any value.(even invalid values)

What happens if we enrich the iterated model with safe-

consensus tasks ?


The enriched Modelinit r := 0; sm, input, scret, dec := NULL;

loop foreverr := r + 1;SM[r].write(sm, scret);scret := safe-consensus[h(sm, scret)](id);sm := SM[r].snapshot();



The enriched Modelinit r := 0; sm, input, scret, dec := NULL;

loop foreverr := r + 1;SM[r].write(sm, scret);scret := safe-consensus[h(sm, scret)](id);sm := SM[r].snapshot();



Process access the object indicated by h(sm, scret)

What happens to the protocol complex ?


Protocol complex with safe-consensus

• 1 round• 3 processes• All processes invoke the

safe-consensus• Input values: Ids


A closer look


A closer look


A closer look

• Which executions are represented in this complex ?


A closer look


A closer look

• Why do we have only these executions ?


A closer look


WRSCRD

WRSCRD

WRSCRD

A closer look


WRSCRD

WRSCRD

WRSCRD

Safe-consensus =

A closer look

• Executions where the safe-consensus returns green


A closer look

• Why there cannot be more adjacent simplexes ?


A closer look



A closer look



WRSCRD

WRSCRD

WRSCRD

A closer look



WRSCRD

WRSCRD

WRSCRD

Safe-consensus =

A closer look



WRSCRD

WRSCRD

WRSCRD

Safe-consensus = Safe-consensus =

A closer look

• Because the safe-consensus does not allow it



A closer look

• Similar argument for other executions


A closer look

• Similary for other executions


WRSCRD

WRSCRD

WRSCRD

A closer look



WRSCRD

WRSCRD

WRSCRD

Safe-consensus =

A closer look



WRSCRD

WRSCRD

WRSCRD


A closer look




A closer look


Safe-consensus =

A closer look


Safe-consensus =

A closer look


Safe-consensus =

And the black complex

• It represents executions where the safe-consensus returns an invalid value



• At least two processes invoke the safe-consensus concurrently


WRSCRD

WRSCRD

WRSCRD


• At least two processes invoke the safe-consensus concurrently


WRSCRD

WRSCRD

WRSCRD

A closer look


Safe-consensus = a value different from valid ids

And again…

• Because we work in the iterated model• In the 2nd round• This behaviour is going to repeat


Remember, Iterated model






Some results for set agreement


(k+1,k)-set agreement is solvable in this model

proc (k+1,k)-set-agreement(val)SM.write(val);sc := safe-consensus(id);sm := SM.snapshot();

if (sc is in {1, …, k+1} Λ sm[sc] ≠ NULL) thendec := sm[sc];elsedec := sm[j] with j := min{ m | sm[m] ≠ NULL };end if

decide dec;end proc


Notice


• We omit here the correctness proof of the protocol

• Not difficult, but tedious

In particular

(3,2)-set agreement is solvable in the iterated model with safe-consensus.


In particular

(3,2)-set agreement is solvable in the iterated model with safe-consensus.

But we can prove that

(3,1)-set agreement (3-consensus) is not solvable in the Iterated model with safe-consensus.


Proof’s idea• Suppose a protocol exists.• Consider an input where

processes propose their ids• Take the gray subcomplex


In the protocol’s 1st round


WRSCRD

WRSCRD

WRSCRD



WRSCRD

WRSCRD

WRSCRD



There exists a path here

In the protocol’s 2nd round

• And because we work in an iterated model


2nd round


2nd round


2nd round


2nd round


WRSCRD

WRSCRD

WRSCRD

2nd round


WRSCRD

WRSCRD

WRSCRD

2nd round


There’s also a path

…And so on…

Because of the iterated model

• In any r-round partial execution:– a solo execution of – is “conected” to a execution of and

without


When the protocol finishes


must decide green

When the protocol finishes


must decide red or yellow

must decide green

Contradiction


This “connectivity” in all rounds lead us to a

contradiction, so no such protocol can exists

In general

There is no protocol in the Iterated Snapshot model with safe-consensus objects that can solve the (k, 1)-set agreement problem (k ≥ 3).


Summary

• There’s a deep conection between Distributed computing and Topology

• Impossibility results arise from this conection• We can derive algorithms by looking at the

geometric structure of protocol complexes• Shared objects can affect the topology of the

protocol complex (safe-consensus)


Thank you


Documents

The iterated shared memory model of computation and an enrichment with safe-consensus tasks