vault backup: 2025-03-08 18:11:57

Concurrent Systems/notes/1b - Peterson algorithm.md

@@ -48,9 +48,10 @@ unlock(i) :=
 - Each lock-unlock requires 5 accesses to the registers (4 for lock and 1 for unlock)
 
 ##### MUTEX proof
-Assume p0 and p1 are simultaneously in the CS.
+>[!info] 
+>Remember that, in the [[#Correct solution|correct implementation]], there is this line: 	`wait (FLAG[1-i] = down OR AFTER_YOU != i)`. Thus, to access the critical section (the `return` instruction of the `lock` function), the possibilities are just the two discussed above.
 
-How has p0 entered its CS?
+**How has p0 entered its CS?**
 a) `FLAG[1] = down`, this is possible only with the following interleaving:
 ![[Pasted image 20250303100721.png]]
 
@@ -89,7 +90,8 @@ unlock(i) :=
 	FLAG[i] <- 0
 	return
 ```
-We say that pi is at level h when it exits from the h-th wait -> a process at level h is at any level <= h
+We say that:
+if $p_i$ is at level $h$ when it exits from the $h$-th wait $\to$ a process at level $h$ is at any level $<= h$
 
 ##### MUTEX proof
 Lemma: for every $ℓ \in \{0,\dots,n-1\}$ , at most n-ℓ processes are at level ℓ, this implies MUTEX by taking ℓ = n-1