master-degree-notes/Concurrent Systems/notes/6 - Atomicity.md

We have a set of n sequential processes p_{1},...,p_n , that access m concurrent objects X_1,...,X_m by invoking operations on the form Xi.op(args)(ret).

When invoked by p_j, the invocation Xi.op(args)(ret) is modeled by two events: inv[Xi.op(args) by pj] and res[Xi.op(ret) to pj].

A history (or trace) is a pair \hat{H}=(H, <_{H}) where H is a set of events and <_{H} is a total order on them.

The semantics (of systems and/or objects) will be given as the set of traces.

A history is sequential if it is of the form inv res inv res ... inv res inv inv inv ..., where every res is the result of the immediately preceding inv. (The last invocations do not have a return). A sequential history can be represented as a sequence of operations.

A history is complete if every inv is eventually followed by a corresponding res, it is partial otherwise.

Linearizability

A complete history \hat{H} is linearizable if there exists a sequential history \hat{S} s.t.

• \forall X :\hat{S}|_{X} \in semantics(X)
• \forall p:\hat{H}|_{p} = \hat{S}|_p
  • cannot swap actions performed by the same process
• If res[op] <_{H} inv[op'], then res[op] <_{S} inv[op']
  • can rearrange events only if they overlap

Given an history \hat{K}, we can define a binary relation on events ⟶_{K} s.t. (op, op’) ∈ ⟶K if and only if res[op] <K inv[op’]. We write op ⟶K op’ for denoting (op, op’) ∈ ⟶K. Hence, condition 3 of the previous Def. requires that ⟶H ⊆ ⟶S.

!

! there is another linearization possible! I can also push a before if I pull it before c! Of course I have to respect the semantics of a Queue (if I push "a" first, I have to pop "a" first because it's a fucking FIFO)

Compositionality theorem

\hat{H} is linearizable if \hat{H}|_{X} is linearizable, for all X in H (it is good to linearize complex traces)

For all X, let \hat{S}_{X} be a linearization of \hat{H}_{X} - \hat{S}_{X} defines a total order on the operations on X (call it \to_{X}) - a union of relations is a union of pairs!

Let \to denote \to_{H} \cup \bigcup_{X \in H} \to _{X}

We now show that \to is acyclic.

1. It cannot have cycles with 1 edge (i.e. self loops): indeed, if op \to op, this would mean that res(op) < inv(op), which of course does not make any sense.

2. It cannot have cycles with 2 edges:

   • let's assume that op \to op' \to op
   • both arrows cannot be \to_H nor \to_X (for some X), otw. it won't be a total order (and would be cyclic)
   • it cannot be that one is \to_X and the other \to_Y (for some X \neq Y), otherwise op/op' would be on 2 different objects.
   • So it must be op \to_X op' \to_H op (or vice versa)
     • then, op' \to op means that res(op') <_H inv(op)
     • Since \hat{S}_X is a linearization of \hat{H}|_X and op/op' are on X (literally because we have op ->x op'), this implies res(op') <_X inv(op), which means that op' \to_X op, and so it won't be a total order... So this is not possible either.

3. It cannot have cycles with more than 2 edges:

   • by contradiction, consider a shortest cycle
     • adjacent edges cannot belong to the same order (e.g. not both \to_X), otw. the cycle would be shortable, because of transitivity of the total order!
     • adjacent edges cannot belong to orders on different objects
       • this would mean that an operation is involved in both \to_X and \to_Y but it is not possible of course, so the cycle can only happen edges in \to_X and \to_H.
     • Hence, at least one \to_X exists and it must be between two \to_H i.e.: op1 \to_H op2 \to_X op3 \to_H op4, likely with op1 = op4
     • can this be a cycle?

       • op1 \to_H op2 means that res(op1) <_H inv(op2)

       • op2 \to_X op3 entails that inv(op2) <_H res(op3)

         • if not, as is a total order, we would have that res(op3) <_H inv(op2), but we then would have op3 \to_H op2, forming a cycle of lenght 2 because op2 \to_X op3, and we know cycles of lenght 2 are not possible...

       • op3 \to_H op4 means that res(op3) <_H inv(op4)

       • We would then have that res(op1) <_H inv(op2) <_H res(op3) <_H inv(op4)

       • So by transitivity res(op1) <_H inv(op4), i.e. op1 \to_H op4

         • IN CONTRADICTION WITH HAVING CHOSEN A SHORTEST CYCLE
         • as if op4 = op1, then this could not happen as \to is a total order.

This said, we can say that every DAG admits a topological order (a total order of its nodes that respects the edges), we will call \to' the topological order for \to

Let us define a linearization of \hat{H} as follows: \hat{S}=inv(op1)res(op1)inv(op2)res(op2)... we would have the topological order: op1\to'op2\to'... \hat{S} is clearly sequential.

Considero \to'|_X come \to' filtrato per gli eventi su X.

Osservazione 1: \to'|_X è come dire \to_{\hat{S}|X} perché l'ordinamento topologico è letteralmente la sequenza di eventi in S, e se filtro entrambi per gli eventi su X ottengo la stessa cosa...

Considero \to|_X come \to filtrato per gli eventi di X. Da non confondere con \to_X che invece è l'ordinamento su \hat{S}_X. Ricordiamoci che \to è l'unione di \to_X, \to_Y, \to_Z..., E DI \to_H.

Per come abbiamo appena definito \to|_X, è chiaro che sicuramente abbiamo che \to_X \subseteq \to|_X, in quanto \to contiene chiaramente \to_X, ma \to|_X può contenere anche elementi di \to_H.

Si ha poi che \to|_X \space \subseteq \space \to'|_X , perché \to' è un ordinamento topologico di \to, quindi deve rispettare tutti i vincoli di precedenza imposti da \to. Non sono necessariamente uguali perché l'ordinamento topologico può imporre ulteriori ordinamenti tra eventi che in $ non erano esplicitamente vincolati.

Posso quindi dire: $$<_{\hat{S}_X} \space = \space\to_X\space \subseteq\space \space\to|X\space \subseteq\space \to'|X\space = \space\to{\hat{S}|X}\space=\space<{\hat{S}|X}$$ ricordando: - \hat{S}_X lo storico ottenuto linearizzando \hat{H}|_X - definisce una relazione di ordinamento \to_X - \hat{S}|_X lo storico che ottengo filtrando per le operazioni su X, partendo da \hat{S}, che a sua volta viene ottenuto linearizzando \hat{H} - definisce una relazione di ordinamento \to|_{\hat{S}_X} - naturlamente, possiamo considerare \to_X \space = \space <_{\hat{S}_X} e \to|_{\hat{S}_X} \space = \space <_{\hat{S}|_X} (se l'ordinamento è uguale, allora anche le coppie in relazione tra loro sono le stesse)

E allora si ha come corollario che$$\forall X :\hat{S}|_{X} = \hat{S}_X (\in semantics(X))

Inoltre, dico che, For all process p, \hat{H}|_p=inv(op1_p)res(op1_p)inv(op2_p)res(op2_p)..., e questo è vero perché lo storico delle operazioni eseguite DA UN SOLO PROCESSO non può non essere sequenziale!

E siccome

• op1_p \to_H op2_p \to_H \dots
• \to_H \space \subseteq \space \to'

Allora \hat{H}|_p = \hat{S}_p ovverosia, se proiettiamo H e S solo per le operazioni eseguite da un solo processo p, allora gli storici saranno uguali. Infine, si ha che \to_H \space \subseteq \space \to \space \subseteq \space \to' \space = \space \to_S

[!PDF|red] class 6, p.6> we would have a cycle of length > we would contraddict op2 ->x op3