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Concurrent Systems/notes/3.md
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@ -188,11 +188,11 @@ By contradiction, assume that there is a lock but nobody enters its CS
- if $p_j$ is in the bakery, by assumption `⟨MY_TURN[i] , i⟩ < ⟨MY_TURN[j] , j⟩` since it is the minimum. - if $p_j$ is in the bakery, by assumption `⟨MY_TURN[i] , i⟩ < ⟨MY_TURN[j] , j⟩` since it is the minimum.
#### Bounded bypass proof (bound n-1) #### Bounded bypass proof (bound n-1)
Let pi and pj competing for the CS and pj wins Let $p_i$ and $p_i$ competing for the CS and $p_j$ wins
Then, pj enters its CS, completes it, unlocks and then invokes lock again Then, $p_j$ enters its CS, completes it, unlocks and then invokes lock again
- If pi has entered the CS, √ - If $p_j$ has entered the CS, √
- Otherwise, by Lemma1, $MY_TURN[i] < MY_TURN[j], then pj cannot bypass pi again! - Otherwise, by Lemma1, $MY\_TURN[i] < MY\_TURN[j]$, then $p_j$ cannot bypass $p_i$ again!
- At worse, pi has to wait all other proceeses before entering its CS - At worse, pi has to wait all other proceeses before entering its CS
- (indeed, since there is no deadlock, when pi is waiting somebody enters the CS) - (indeed, since there is no deadlock, when pi is waiting somebody enters the CS)