vault backup: 2025-03-10 11:51:08

Concurrent Systems/notes/3.md

@@ -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.
 
 #### 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
-- If pi has entered the CS, √
-- Otherwise, by Lemma1, $MY_TURN[i] < MY_TURN[j], then pj cannot bypass pi again!
+Then, $p_j$ enters its CS, completes it, unlocks and then invokes lock again
+- If $p_j$ has entered the CS, √
+- 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
 	- (indeed, since there is no deadlock, when pi is waiting somebody enters the CS)