Computing with Concurrent Objects speaker sara tucci piergiovanni institution università di roma “la sapienza” dipartimento di informatica e sistemistica

Computing with Concurrent Objects

speaker

sara tucci piergiovanniinstitution

università di roma “la sapienza”dipartimento di informatica e sistemistica

midlab laboratory

June 8-9 2006 Seminars in Distributed Computing - Computing with Concurrent Objects

Concurrent Objects Definitionس

”The “outside” view-pointسLinearizability: what makes a concurrent object a “meaningful” programming

abstraction

The “inside” view-pointسWait-free implementation: what makes a concurrent object a “possible” programming

abstraction

A “global” look to the universe of concurrent objectsسObject hierarchy: what makes happy a “would-be theoretician” like me

Outline

Lecture1: the “outside” view-point

speaker



midlab laboratory


What is an object? .An object in languages such as Java and C++ is a container for dataس

أ Each object provides a set of methods that are the only way to manipulate that object’s state.

أ Each object has a class which describes how its methods behave

Object descriptionسأ The application programmer interface (API)

ر pre-condition (describing the object’s state before invoking the method) ر post-condition, describing the object’s state and return value after the method returns.

أ For example, if the FIFO queue object is non-empty (pre-condition), then the deq() method will remove and return the first element (post-condition), and otherwise it will throw an exception (another pre- and post-condition).


Pre and Post Conditions Defining objects in terms of pre-conditions and post-conditions makes perfect sense inس

a sequential model computation where a single thread manipulates a collection of objects.

س In this case methods are called once at a time, each method invocation is followed by the corresponding return and a sequence of method calls can be defined

enq(a;ok)

method invocation

method response

enq(b;ok) deq( ;a) deq( ;b) deq()

time

exception

p


Concurrent Modelس If an object’s methods can be invoked by concurrent threads, then the method

executions can overlap in time, and it no longer makes sense to talk about the order of method calls.

What does it mean, in a multithreaded program, if a and b are enqueued on a FIFOسqueue during overlapping intervals? Which will be dequeued first?

enq(a;ok) deq( ;?)

at the end of the invocation, the queue surely contains a, but

during the invocation what did it happen? God knows

p

q

a a a a a aa

enq(b;ok) deq( ;?)

enq(a;ok)

where is the trick?

method call

queue’s state


The Linearizability Manifesto.The Linearizability Manifestoس

Each method call should appear to “take effect” instantaneously at some moment between its invocation and response.

p

q

enq(a;ok)

enq(b;ok)

deq( ;b)

deq( ;a)

S time

aba b


Linearizability: scenario?Again, is this execution linearizableس

...try to put a point for each method call...

enq(a;ok) enq(b;ok)p

q deq( ;c)

enq(c;ok)r

deq( ;a)

deq( ;b)

S

aba

c ccba a


Linearizability: scenario?Again, is this execution linearizableس

...try to put a point for each method call...

enq(b;ok) enq(c;ok)p

q deq( ;a)

r

Sb

cb

deq( ;b)

acb

enq(a;ok)

deq( ;c)

no way...


Formalizing Linearizability Until now we had fun by putting points here and there...now the play is gettingس

harder...we should formalize what putting points would mean

-------------------------------- Definitions and Basic Notation ---------------------------------------- An execution of a concurrent system is modeled by a history H, which is a finiteس

sequence of method invocation and response events.

<A method invocation event is denoted as < inv (op(args), X) pس<A method response event is denoted as < res (op(res), X) pس

where op is the name of the method, args is a list of input arguments, res is a list of results (ok for void) , X is the name of the object, p the name of the process.

--------------------------------------------------------------------------------------------------------------------

enq(a;ok)p

tH: inv (enq(a), X) p res (enq(ok), X) p


Formalizing LinearizabilityConcurrent execution example and related Historyس

p

q

enq(a;ok)

enq(b;ok)

deq( ;a)

deq( ;b)

H: inv(enq(a),X)p, res(enq(ok)X)p, inv(enq(b)X)q, res(enq(ok)X)q,

inv(deq()X)p, res(deq(a)X)p, inv(deq()X)q, res (deq(b)X)q



p

q

enq(a;ok)

enq(b;ok)

deq( ;a)

deq( ;b)

H: inv(enq(a),X)p, inv(enq(b)X)q, res(enq(ok)X)p,

res(enq(ok)X)q, inv(deq()X)p, inv(deq()X)q, res(deq(a)X)p

res (deq(b)X)q


Formalizing Linearizability

-------------------------------------------- Definitions ---------------------------------------------

A response matches an invocation if their objects names agree and their processسnames agree.

.An invocation is pending in a history if no matching response follows the invocationس

س If H is a history, complete(H) is the maximal subsequence of H consisting only of invocations and matching responses.

-----------------------------------------------------------------------------------------------------------------------



p

q

enq(a;ok)

enq(b;ok)

deq( ;

deq( ;)

H: inv(enq(a),X)p, inv(enq(b)X)q, res(enq(ok)X)p, res(enq(ok)X)q,

inv(deq()X)p, inv(deq()X)q

complete(H): inv(enq(a),X)p, inv(enq(b)X)q, res(enq(ok)X)p, res(enq(ok)X)q


Formalizing Linearizability----------------------------------------- Definitions

----------------------------------------- A history H is sequential ifس

أ (1) The first event of H is an invocation.أ (2) Each invocation, except possibly the last, is immediately followed by a

matching response. A history that is not sequential is concurrent.

A process subhistory, H|p (H at p), of a history H is the subsequence of all events in Hسwhose process names are p. (An object subhistory H|X is similarly defined for an object X.)

.Two histories H and H’ are equivalent if for every process p, H|p = H’|pس

A history H is well-formed if each process subhistory H|p of H is sequential (in theسfollowing we will assume well-formed subhistories)

-----------------------------------------------------------------------------------------------------------------------------------------------------


Formalizing LinearizabilitySequential History Hس

p

q

enq(a;ok)

enq(b;ok)

deq( ;a)

deq( ;b)

H: inv(enq(a),X)p, res(enq(ok)X)p, inv(enq(b)X)q, res(enq(ok)X)q, inv(deq()X)p, res(deq(a)X)p, inv(deq()X)q, res (deq(b)X)q

H|p: inv(enq(a),X)p, res(enq(ok)X)p, inv(deq()X)p, res(deq(a)X)p

H|q: inv(enq(b)X)q, res(enq(ok)X)q, inv(deq()X)q, res (deq(b)X)q

well-formed subhistories


Formalizing Linearizability’Concurrent History Hس

p

q

enq(a;ok)

enq(b;ok)

deq( ;a)

deq( ;b)

H’: inv(enq(a),X)p, inv(enq(b)X)q, res(enq(ok)X)p, res(enq(ok)X)q, inv(deq()X)p, inv(deq()X)q, res(deq(a)X)p, res (deq(b)X)q

H’|p: inv(enq(a),X)p, res(enq(ok)X)p, inv(deq()X)p, res(deq(a)X)p

H’|q: inv(enq(b)X)q, res(enq(ok)X)q, inv(deq()X)q, res (deq(b)X)q

H and H’ are equivalent


Formalizing Linearizability---------------------------------- Definitions ---------------------------------------------------

A history H induces an irreflexive partial order H on methods: op1 H op2 ifس

res(op1) precedes inv(op2) in H. س If H is sequential, then H is a total order.

-----------------------------------------------------------------------------------------------------------------------

p

q

enq(a;ok)

enq(b;ok)

deq( ;a)

deq( ;b)

H’: inv(enq(a),X)p, inv(enq(b)X)q, res(enq(ok)X)p, res(enq(ok)X)q, inv(deq()X)p, inv(deq()X)q, res(deq(a)X)p, res (deq(b)X)q

enq(a) H’ deq(a)enq(a) H’ deq(b)enq(b) H’ deq(b)enq(b) H’ deq(a)


Formalizing Linearizability---------------------------------- Definitions -----------------------------------------------------.A set S of histories is prefix-closed if whenever H is in S, every prefix of H is also in Sس

A sequential specification for an object is a prefix-closed set of sequential histories forسthe object.

A sequential history H is legal if each object subhistory H|X belongs to the sequentialسspecification for X.

-----------------------------------------------------------------------------------------------------------------------

Linearizability

A history H is linearizable if it can be extended to a history H’ (by appending zero or more response events to H) such that:

L1 : complete(H’) is equivalent to some legal sequential history S, and L2 : H’ S.


Linearizabilityس Informally, extending H to H’ captures the idea that some pending invocations may

have taken effect even though their responses have not yet been returned to the caller. This is visible when some successive method call returns a value set by a pending invocation.

Extending H to H’ while restricting attention to complete(H’) makes it possible toسcomplete pending methods, or just to ignore them.

L1 states that complete(H’) is equivalent to an apparent sequential interleaving ofسmethod calls that does not violate the specification of the object.

L2 states that this apparent sequential interleaving respects the precedence orderingسof methods.


Linearizability...Then, let’s try to find Sس

p

q

enq(a;

enq(b;ok)

deq( ;

deq( ; b)

H: inv(enq(a),X)p, inv(enq(b)X)q, res(enq(ok)X)q, inv(deq()X)p, inv(deq()X)q, res (deq(b)X)q, inv(deq()X)q, res (deq(a)X)q

step1: extending...H’: inv(enq(a),X)p, res(enq(ok)X)p, inv(enq(b)X)q,

res(enq(ok)X)p, inv(deq()X)p, inv(deq()X)q, res (deq(b)X)q,

inv(deq()X)q, res (deq(a)X)q

deq( ; a)


Linearizabilitystep2: completing...

complete(H’): inv(enq(a),X)p, res(enq(ok)X)p, inv(enq(b)X)q, res(enq(ok)X)p, inv(deq()X)p, inv(deq()X)q, res (deq(b)X)q,

inv(deq()X)q, res (deq(a)X)qp

q

enq(a;ok)

enq(b;ok) deq( ; b) deq( ; a)

step3: let me see the partial order...

enq(a) H’ deq(b) H’ deq(a)

enq(b) H’ deq(b) H’ deq(a)

step4: ordering what is not yet ordered... S: enq(b) S enq(a) S deq(b) S deq(a) we got it!


To do... :Linearizability has the following propertyس

Theorem : H is linearizable if and only if for each object x, H|x is linearizable.

To investigate if locality holds for the following alternative correctnessسcriteria:أ Sequential consistency: only L1أ Serizability: A history is serializable if it is equivalent to one in which transactions

appear to execute sequentially, that is, without interleaving. Transaction: finite sequence of methods to a set of objects

Lecture 2: the “inside” view-point

speaker



midlab laboratory


Objects Implementation .So, let us suppose now to have a concurrent object to implement, e.gس

stack, queue, etc.

how can we implement it? to get a linearizable execution, we can try toسuse some form of synchronization, e.g. to rule the access of the object by using of locks, mutex to define critical sections

The only one that can release the lock is the one who acquired the lockس

but we want also cope with failures...we want that a process gets aسresponse in a finite time, no matter the failures of others...

?So, if a process in the critical section failsس


Wait-free object implementation The meaning of wait-free computing is exaclty the following: eachس

process (that does not crash) calling a method must be able to get a response in a finite time no matter of how slow other proccesses are and failures of other processes

To introduce wait-free computing we will consider the wait-free سimplementation of two concurrent objectsأ A renaming object allows the processes to acquire new names from a smaller

name space despite possible process crashesأ A snapshot object provides the processes with an array-like data structure (with

one entry per process) offering two operations. The write operation allows a process to update its own entry. The snapshot operation allows a process to read all the entries in such a way that the reading of the whole array appears as it is was an atomic operation.


Setting We consider n processes, up to f are faulty (stop prematurely byس

crashing)

We will consider to use as building blocks for our implementation someسbasic concurrent objects, called atomic registers, that behave like registers accessed sequentially (so, we are implicitly assuming that the implementation of registers has been already done...later we will come back on this point)

Then processes can access these registers by invoking write() andسread() operations


M-Renaming Problem Let us assume that the n processes have arbitrarily large (and distinct)س

initial names id1, . . . , idn [0, . . . , N − 1], where n <<< N. In the M-renaming problem, each process pi knows only its initial name idi, and the processes are required to get new names in such a way that the new names belong to the set {0, . . . , M − 1}, M<< N, and no two processes obtain identical names.

:More formally, the problem is defined by the three following propertiesس

Termination. Each correct process decides a new name.Validity.A decided name belongs to [0, . . . , M − 1].

Agreement.No two processes decide the same name.


Implementation Note that the renaming problem is a problem of allocation (one processس

for each name)We assume the presence of MultiWriterMultiReader registersس .We will see a simple and elegant algo by Moir–Andersonس It uses a mechanism called splitter that is particularly suited to wait-freeس

computing (the splitter has been used to implement wait-free mutual exclusion)

The renaming problem is trivial when no process cancommit a crash failure. Differently, it has been shownthat there is no wait-free solution to the M-renamingproblem when M < n+ f , where f is an upper bound

on the number of processes that can crash


Wait-free Splitter

X=undefined

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=true

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=true

stop= empty

right=emtpy

down=empty

yawn


Splitter

X=red

Y=true

stop= empty

right=emtpy

down=empty

zzzz


Splitter

X=red

Y=true

stop= empty

right=emtpy

down=empty

zzzz


Splitter

X=red

Y=true

stop= empty

right={red}

down=empty

zzzz


Splitter

X=red

Y=true

stop= empty

right={red}

down=empty

zzzz


Splitter

X=red

Y=true

stop= empty

right={red}

down=empty

awake!


Splitter

X=red

Y=true

stop= empty

right={red}

down=empty


Splitter

X=red

Y=true

stop= empty

right={red}

down={green}


Splitter

X=red

Y=true

stop= empty

right={red}

down={green}


Splitter

X=red

Y=true

stop= empty

right={red}

down={green}

Note that green was slow and red was in late, nobody got stop


Splitter

X=undefined

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=false

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=true

stop= empty

right=emtpy

down=empty


Splitter

X=green

Y=true

stop= {green}

right=emtpy

down=empty


Splitter

X=green

Y=true

stop= {green}

right=emtpy

down=empty


Splitter

X=green

Y=true

stop= {green}

right={red, orange}

down=empty

Note that green was on time, red and orange was in late, nobody was slow


Moir-Anderson: the grid of Splittersn(n+1)/2 renaming splitters


SnapshotThe object is made up of n SWMR atomic registers (one per process)س:Two operationsس

أ update(v) invoked by pi allows to update its register to the value vأ snapshot() returns the value of the n registers

All the operations appear as they were executed instantaneouslyس

The operation updates to inform others on its progress, the snapshotسallows a process to understand the what others are doing...

...collect vs snapshot, collect is not atomicس


Collect vs Snapshot

update(1, R1;ok)

update(2, R2;ok)

snapshot(?)possible values:

[000] [010] [012]

[000] [002] [012]

update(1, R1;ok)

update(2, R2;ok)

snapshot(?)

snapshot(?)

collect(?)

collect(?)

collect(?) possible values:

[000] [010] [002]

[000] [002] [010]


A simple non-wait free algo

key mechanism:

double collect

update(1, R1;ok)

update(2, R2;ok)

collect(?) collect(?)possible values:

[000] [010] [002]collect(?) collect(?) it does not return!

How many times is this condition false?


A wait free snapshot-Afek and et al.

key mechanism:

helping mechanism

double collect

If a process P sees two consecutive updates issued by the same process R, it knows that the second update began after its snapshot began..Then P borrows the snapshot that Q did here

Lecture 3: the “global” look

speaker



midlab laboratory


Wait-free computing issues The fundamental problem of wait-free computing is to characterizeس

circumstances under which synchronization problems have wait –free solutions and to derive efficient solutions when they exist.

To show that a wait-free implementation there exits, just draw an algoسand prove it.

?To show that such an implementation does not existس

Herlihy: tecnique to prove impossibility and an object hierarchyس


The Relative Power of Objects:Fundamental problemس

أ Given two concurrent objects X and Y, does there exists a wait-free implementation of X by Y?

أ Thanks to Herhily classification of objects based on their synchronization power, we can derive impossibility: if for example we have two objects X and Y and Y has less synchronization power than X, then there exists no implementation of X by Y.

أ How to define the synchronization power of an object?


The Consensus Number Each object is classified through its possibility of solving Consensus in aس

wait –free manner.

In particular, an object X has Consensus Number k, if withسobjects X and atomic registers (all initialized to appropriate values) it is possible to solve wait-free Consensus for k processes but not for k+1 processes.

For whom does not know Consensus: processes start with an inputسvalue and eventually agree on a common input value أ agreement: distinct processes never decide on distinct valuesأ wait-free: each process decides after a finite number of stepsأ validity: the common decision value d is the input of some process


Accessing Rules Defined by

the Consensus Protocol

Consensus26 6 75

6 6 6

ObjectObject


Wait- free Consensus26 6 75

26 26

ObjectObject


Objects and Consensus Numbers If there exits a wait-free implementation of an object X by Y, and Xس

has consensus number n, then Y has consesus number at least n

If an object X has consensus number n and an object Y has a consesusسnumber m<n, then there exists no wait-free implementation of X by Y in a system of k > m processes.

If two objects X and Y have consensus number n, then there exists aسwait-free implementation of X by Y (and viceversa) in a system with n processes.

If two objects X and Y have consensus number n , is it true that thereسexists no wait-free implementation of X by Y in a system of k>n processes? open issue...


Objects Classificationread-write register 1

list 2

stack 2

queue 2

test&set 2

fetch&add 2

swap 2

m-multipleassignment 2m-2

move and swap

augmentedqueue

compare&swap

fetch&cons

sticky byte

counters

1

atomic snapshot 1


Objects with Consensus number 1

Atomic registers, counters, other interfering objectsسthat don't return the old value (explained later)

-First observe that any type has consensus number at least 1, since 1سprocess consensus is trivial.

To prove that an object has consensus number 1, we should show thatسit is impossible to solve Consensus (by the object) among two processes (with initial values in the set {0,1})أ Then we show that an object has consensus number exactly 1, by running

FischerLynchPaterson with 2 processes.


Read-Write Registers & Consensus0 1

? ?

Read-Write Register

Read-Write Register


Proof: assume otherwise

Read-Write Register

Read-Write Register

0 0

Read-Write Register

Read-Write Register

0 0

Let us assume that there exits a Consensus Protocol that can use only read() and write() operations

In this case the outcome should be:

Initial State (0,0)



Read-Write Register

Read-Write Register

Initial State (1,1)

1 1

Read-Write Register

Read-Write Register

1 1




Read-Write Register

Read-Write Register

1 0

Read-Write Register

Read-Write Register

1

Initial State (1,0)




Read-Write Register

Read-Write Register

1 0

Read-Write Register

Read-Write Register

0

Initial State (1,0)




Read-Write Register

Read-Write Register

1 0

Read-Write Register

Read-Write Register

1 1

Read-Write Register

Read-Write Register

0 0

Initial State (1,0)

In this case the outcome could be:

or even:



Read-Write Register

Read-Write Register

0 1

Read-Write Register

Read-Write Register

0 0

Read-Write Register

Read-Write Register

1 1

Initial State (0,1)

In this case the outcome could be:

or even:


Proof: reasoning about any protocolBivalent&Univalent States: .A protocol state is bivalent if both decision values are still possibleس

أ the outcome is not fixed (like for the (0,1) and (1,0) initial states) أ the protocol execution can be extended to yield a decision value but can also be

extended to yield the other value

A protocol state is univalent if all protocol executions yields to the sameسvalue أ the outcome is fixed even if not yet known

A 0-valent state yields to decide 0سA 1-valent state yields to decide 1س


s’

Proof: reasoning about any protocol

()A “move” can be a read() or a write س

To reason about “any” protocol, we abstract the computation in this way:

we consider that Mr.Red and Mr.Green do “moves” against a protocol state. E.g. starting from the initial state either Mr. Red or Mr. Green do the first “move”, and we get in an other protocol state. At this point, again either Mr. Red or Mr. Green do the second “move” and so on...

s

s’’s

s’s

s’’’

s’v

sv

s’’s sv’

possible protocol executions



Initial state

Final states



decision:1 decision:0 decision:0 decision:1 decision:1 decision:0


Proof:reasoning about any protocol


bivalent states




1-valent states


Proof:reasoning about any protocol


0-valent states


Proof: reasoning about any protocolThe initial state is bivalent (by the fact that at least one failure may occur,

inputs are invisible and validity)

To solve consensus we have to reach a bivalent state C that has only univalent successors (otherwise we could stay bivalent forever and the protocol is not wait-free)

Now we assume that C has a 0-valent and a 1-valent successor produced by applying operations x and y of processes Mr.Red and Mr.Green:

Cx=0-valent and Cy=1-valent




Bivalent C


Proof: derive a contradiction Now, since we are looking at atomic registers, we have three casesس

consider cases:

أ (1) x and y are both readsأ (2) x is a read and y is a writeأ (3) x and y are both writes


Proof: derive a contradiction(1) x and y are both readsس If read() comes first then the protocol decides 1س If read() comes first then the protocol decides 0

The idea is the following: to get an univalent state the two processes should decide who is the winner, who came first.

Then we start from C: Mr.red does the first move, this means that both should decide for 0. Then they both decide for 0 if Mr.Red comes before Mr.Green. Now, think that Mr.Green reads before Mr.Red. Now this order should lead to a decision opposite to the other....


Proof: derive a contradictionBivalent CMr Green

reads first

1-valent state

Mr Red reads first

0-valent stateMr Red

reads 1-valent state

But for Mr Red these two states are indistinct (C=Cy): Mr Red cannot understand if Mr Green did it something or not before him...So, you can remove the green arrow and then get a contradiction!

We got that Cxy = 0-valent=Cyx=1-valent. Contradiction.


Proof: derive a contradiction(1) x is a read() and y is a writeس If write() comes first then the protocol decides 1س If read() comes first then the protocol decides 0

Let us suppose that Mr.Red runs before Mr. Green. Then, both will decide for 0. Let us suppose now that Mr. Green comes first. Now they both will decide 1. However, for Mr.Green the state C is indistinguishable from Cx. So running Mr. Green to completion gives the same decision value from both Cyx and Cxy, another contradiction.

We got that Cxy = 0-valent=Cyx=1-valent. Contradiction.


Proof: derive a contradiction(1) x is a write() and y is a writeس If write() comes first then the protocol decides 1س If write() comes first then the protocol decides 0

Let us suppose that Mr.Red runs before Mr. Green. Then, both will decide for 0. Let us suppose now that Mr. Green comes first. Now they both will decide 1. However, for Mr.Green the state C is indistinguishable from Cx because Mr.Green overwrites on the value written by Mr.Red. Then we got that Cxy=Cy for Mr.Green. Another contradiction.

We got that Cxy = 0-valent=Cy=1-valent. Contradiction.


Summarizing and Generalizing The consensus protocol should allow all processes discovering who was the processس

that access the object first, then all processes decides the value of the process that won.

Suppose that an object T has a read operation that returns its state and one or moreسmodify-write operations that don't return anything (uninformative).

We'll say that the T operations are interfering if for any two operations x and yسeither: أ x and y commute: Cxy = Cyx. أ one between x and y overwrites the other: Cxy = Cy or Cyx =

Cx.

Any T object with all operations uninformative and interfering has consensus number 1سsince for any two operations either they commute or the overwriter can't detect that the first operation happened


Objects with Consensus Number 2 Now we will see what are the characteristics of objects with consensusس

number 2

For all these objects there exists a wait-free consensus protocol amongسtwo processes

أ registers with interfering non trivial read-modify-write operations (RMW registers) أ objects with non interfering operations: queue, stacks, lists

We derive also the intuition of the impossibility to solve consensusسamong n>2 processes using these objects


Non-trivial RMW Registers What does it mean having non-trivial operations? It means that thereس

exists an operation such that it returns the current value and writes a value obtained through a certain function that is not the identity

أ The operation takes 2 arguments:

ر Register rر Function f

أ The operation has the following effects:ر Returns value x of rر Replaces x with f(x): x f(x)x

Then, it is not a simple read (obtained if f were the identity), but a readسthat leaves an evidence...


Non-trivial RMW Registerstest&setس

أ The operation has the following effects:ر Returns value x of rر Replaces x with 1

fetch&incسأ The operation has the following effects:

ر Returns value x of rر Replaces x with x+1


Non-trivial RMW Registersswapس

أ The operation takes an additional argument yأ The operation has the following effects:

ر Returns value x of rر Replaces x with y

fetch&addسأ The operation takes an additional argument yأ The operation has the following effects:

ر Returns value x of rر Replaces x with x+y


Non-trivial RMW Registers?How a consesus protocol among two process can be designedس

....try to thinkسأ remember, both processes should discover who was the winner...أ first step: each one writes the proposed value in a registerأ second step: accessing the register object initialized to some value v with the

non-trivial operation أ who is the winner?

ر the one who reads vأ who is the loser?

ر the one who does not read vأ Note that both processes can deduce who is the loser and the winner, then they

go to the register to pick up the value proposed by the winner


Consensus among two processes

Two read-write

Registers

Two read-write

Registers

1 0

Initial State (1,0)

Non-trivial RMW register

Non-trivial RMW register

1 0v f(v)

Two read-write

Registers

Two read-write

Registers

1 1

winner loser

write write

read(1) read(1)


What about Queues? By their consensus number, there exists a consensus protocol usingس

read/write registers and queues

Wait-free queue object with enqueue and dequeue operations, whereسdequeue returns empty if the queue is empty. To solve 2-process consensus with a wait-free queue:

أ Initialize the queue with a single value (it doesn't matter what the value is).أ A process wins if it successfully dequeues the initial value and loses if it gets

empty.


Non-trivial RMW Registers: no Consensus for n>2

?why a third process cannot understand who was the winnerسأ Let F be a set of functions such that for all fi and fj, either

ر They commute: fi(fj(x))=fj(fi(x))

ر They overwrite: fi(fj(x))=fi(x)

test&setسأ f(x)=1 f(f(x))=1 overwrite

swapسأ f(v,x)=x f(y’(f(y,x))=f(y’,x) overwrite

fetch&incسأ f(x)=x+1 f(f(x))=x+1+1 commutative


Non-trivial RMW Registers: no Consensus for n>2

The Impossibility Intuition for Non-trivial interefering RMWسأ Let us suppose that Mr.green accesses the object with an operation x and

Mr. Red with an operation y (e.g. both fethc&inc). The third process Mr. Orange that accesses (reading) the object after Mr.Green and Mr.Red cannot understand who was the winner between these two (in both cases, i.e. Mr. Red the winner, or Mr. Green the winner, the state is the same).

أ So if we run Mr. Orange to completion we get the same decision value after both Cx and Cy, which means that Cx and Cy can't be 0-valent and 1-valent. Contradiction.


What about queues? Queues, stacks,lists are not interfering objects: no owerwriting andس

commutative operationsأ es enq(g)enq(r) enq(r)enq(g) enq(g)

they seems more powerful...if the Mr. Orange sees by dequeingسg,r or r,g it can understand who arrives first (Mr.Green in the first case, Mr.Red in the second)

But even for queues no wait-free Consensus implementation thereسexists for n>2, i.e. Mr. Orange is not able to understand who was the winner...why?


Non-interfering Queues: no consensus for n>2:Here we give the intuition behind the impossibilityس

Start from C enq(g) enq(r) (1-valent) and C enq(r) enq(g)س(0-valent) أ Run Mr.Red until its first deq() أ Run Mr.Green until it does its first deq() . أ Mr. Orange cannot distinguish between C, C enq(g) enq(r) and C

enq(r) enq(g).

Start from C deq() enq(r) and C enq(r) deq() on a non-emptyسqueue. أ To lose the trace of this order (we reach indistinguishable states), it suffices that

ony 2 witnessess (two dequeuers) fail. Then, the queue has number ≤ 2.


Objects with consensus number Augmented Queueس

أ Has operations enq(x) and peek(), which returns the first value enqueued but does not remove it. Protocol is to enq my input and then peek and return the first value into the queue.

Fetch-and-consسأ Returns old cdr and adds new car on to the head of a list. Use preceding protocol

where peek() = tail(car::cdr)

:Sticky bitsسأ Has write operation that fails unless register is in the initial state. Protocol is to

write my input and then return result of a read.

Compare-and-swapسأ has CAS(old, new) operation that writes new only if previous value = old. Use it to

build a sticky bit.


Consensus for n processes with CASn registers RiسFirst step: pi writes on its register Ri the value proposedس Second step: see if nobody have already access the object, if so, writeس

the identifier of i: CAS(-1,i) and set j to the previous value of the object j=CAS(-1,i).

If j==-1 (i is the first)س decide the value in Ri else

decide the value in Rj


m-multiple assignment objectSnapshot meansس

أ Write any array elementأ Read multiple array elements atomically

:What about the dual problemسأ Write multiple array elements atomicallyأ Scan any array elements

This problem is called multiple assignmentس

It has been proved that m-multiple assignment has consensus numberس2m-2


2-multiple assignment has at least number 2 Here we have a (large) collection of atomic registers augmented by an m-register writeس

operation that performs all the writes simultaneously.

The intuition for why this is helpful is that if Mr.Red writes atomically R1 and R whileسMr.Green writes atomically R2 and R, then any process can look at the state of R1, R2 and R and tell which write happened first: أ If Mr. Orange reads R1 = R2 =empty, then we don't care which went first, because the

Mr.Orange (or somebody else) already won. أ If Mr.Orange reads R1 = 1 and then R2 = empty, then Mr.Green went first. أ If Mr.Orange reads R2 = 2 and then R = empty, then Mr. Red went first. (This requires at

least one more read after checking the first case.) أ Otherwise if Mr.Orange see R1 = 1 and R2 = 2. Now it reads R: if it's 1, Mr. Green went

first, and vice versa.


Universality of Consensus ,Universality: any type that can implement n-process consensus canس

together with atomic registers, give a wait-free implementation of any object in a system with n processes.

the processes repeatedly use consensus to decide between candidateسhistories of the simulated object, and a process successfully completes an operation when its operation (tagged to distinguish it from other similar operations) appears in a winning history


Universality of Consensusس Have a n-process consensus protocol instance for each of a series of

phases 0, 1, ... . س The algorithm works as follows for a process p:

1. Post a list of the operation it wants to its register.2. Reads all the last-phase values and takes their max. 3. Runs the consensus protocol for the max phase to get the history decided on up

to that phase. 4. If the max phase history includes the process's pending operation, returns the

result that operation would have had in the winning history. 5. Otherwise, constructs a new history by appending all announced operations to

the previous history, and tries to win with that history in phase max+1. 6. Returns to step 2 if its operation doesn't make it into the winning history.

س This terminates because even if process i doesn't get its value into the winning history, eventually some other process will pick up the announced value and include it.


Summary?What have we learnedس

Per fare un albero ci vuole un seme, per fare un seme ci vuole un albero...questo loسsapevamo...ora sappiamo anche che

.Per fare un coda, una pila o una lista, non bastano uno o piu’ registri atomiciس

Per fare una coda in un sistema di 2 processi ci vogliono una o più pile. E’ vero cheسprendendo tante pile quante voglio non posso comunque fare una coda in un sistema con piu’ di due processi???? Nessuno lo ha mai dimostrato...

Per fare un compare&swap non solo non ce ne facciamo niente dei registri atomici maسneanche delle code...

Per fare un compare&swap ci vuole un oggetto con numero di consenso infinito...vaسbene consenso implementato tra un numero non noto di processi.


Sources M. Herlihy and J. Wing, "Linearizability: A Correctness Condition forس

Concurrent Objects", ACM Transactions on Programming Languages and Systems (TOPLAS), Volume 12 , Issue 3 (July 1990), pp. 463-492

M. Herlihy, “Wait-free synchronization” ACM Transactions onسProgramming Languages and Systems (TOPLAS) Volume 13 , Issue 1 (January 1991) pp: 124 – 149.

M. Raynal, “Wait-free computing: an introductory lecture” FutureسGeneration Computer Systems, Volume 21 , Issue 5 (May 2005) Special issue: Parallel computing technologies, pp: 655 – 663.

.M. Herlihy and N. Shavit. Concurrent Objects and Linearizabilityسwww.cs.tau.ac.il/~shanir/multiprocessor-synch-2003

Grazie per l’attenzione!

Documents

Computing with Concurrent Objects speaker sara tucci piergiovanni institution università di roma “la sapienza” dipartimento di informatica e sistemistica