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Marco Realacci 2025-03-04 09:24:45 +01:00
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14
Concurrent Systems/notes/Lezione2.md
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@ -103,4 +103,16 @@ unlock(i) :=
###### Is it deadlock free? ###### Is it deadlock free?
Since DLF is deadlock free, it is sufficient to prove that at least one process invokes DLF.lock. Since DLF is deadlock free, it is sufficient to prove that at least one process invokes DLF.lock.
If TURN = k and $p_k$ invoked lock, then it finds TURN = k and exits its wait. If TURN = k and $p_k$ invoked lock, then it finds TURN = k and exits its wait.
Otherwise, any other process will find `FLAG[TURN] = down` and Otherwise, any other process will find `FLAG[TURN] = down` and exits from its wait.
**Lemma 1:** if TURN = i and `FLAG[i] = up` then $p_i$ enters the CS in at most n-1 iterations
Observation 1: TURN changes only when FLAG[i] is down (after pi has completed its CS)
Observation 2: FLAG[i] = up -> either pi is in its CS or pi is competing for its CS -> it eventually invokes (if not already) DLF.lock
Observation 3: if $p_j$ invokes lock after that FLAG[i] is set, $p_j$ blocks in its wait
Let Y be the set of processes competing for the CS
- because of Observation 2, $i \in Y$
- because of Observation 3, once FLAG[i] is set, Y cannot grow anymore