vault backup: 2025-03-10 22:38:12

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Marco Realacci 2025-03-10 22:38:12 +01:00
parent 4287e12e31
commit 1b85c91be0

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@ -93,7 +93,12 @@ lock(i) :=
- $p_n$ invokes lock alone, completes its CS and its new DATE is n
**Lemma 1:** Suppose we have $n$ processes, then $\not \exists p_{j} : DATE[j]=DATE[i] \forall i \in [0, n]$
**Lemma 1:** Suppose we have $n$ processes, then $\not \exists p_{j} : DATE[j]=DATE[i] \forall i \in [0, n]$ (non esistono due processi con lo stesso valore per DATE)
*Proof:*
- Suppose $p_j$ is the last process to execute unlock, then `DATE[j] = n` and for each i, `DATE[i] < n`, as DATE is either set to $n$ or decreased.
- by iterating this reasoning, it is clear that until every DATE is > 0, we have no processes with the same value for DATE.
- What if we reach 0?
- Let's suppose that $p_i$ is the first process that reaches `DATE = 0`
**Lemma 2:** DATE's bounds are $[0, n]$
*Proof:*