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.

![[Pasted image 20250318090733.png]]

![[Pasted image 20250318090909.png]]But 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 b**e $op \to_X op' \to_H op$ *(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: $op1\to'op2\to'...$

$\to'_X$
$\to|_X$
$\to_X$

Si ha che $\to|_X \space \subseteq \space \to'_X$ perché l'ordinamento topologico è letteralmente la sequenza di eventi in $\hat{S}$

$\hat{S}$ is clearly sequential. Moreover:
1. $\forall X :\hat{S}|_{X} = \hat{S}_X (\in semantics(X))$, indeed:
	- $<_{\hat{S}_X} \space = \space\to_X\space \subseteq\space \space\to|_X\space \subseteq\space \to'_X\space = \space\to_{\hat{S}|X}\space=\space<_{\hat{S}|X}$
		- commento per non diventare scemi:
			- $\hat{S}_X$ 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 $\to|_X$ 
			- naturlamente, possiamo considerare $\to_X \space = \space <_{\hat{S}_X}$ e $\to|_X \space = \space <_{\hat{S}|_X}$ (se l'ordinamento è uguale, allora anche le coppie in relazione tra loro sono le stesse)
> [!PDF|red] class 6, p.6>  we would have a cycle of length 
> 
> we would contraddict op2 ->x op3

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