vault backup: 2025-03-10 16:55:21

.obsidian/workspace.json

@@ -34,9 +34,9 @@
               "type": "pdf",
               "state": {
                 "file": "Concurrent Systems/slides/class 3.pdf",
-                "page": 9,
+                "page": 10,
                 "left": -26,
-                "top": 381,
+                "top": 499,
                 "zoom": 0.57541567695962
               },
               "icon": "lucide-file-text",

Concurrent Systems/notes/3.md

@@ -167,12 +167,12 @@ W.r.t. the execution of $p_j$, it can be that
 - this read overlaps with `FLAG[j] <- down` or this read happens when $p_j$ is in the bakery:
 	- `MY_TURN[j]` has been decided and no write will change it until $p_j$ is in the bakery, and `MY_TURN[j] > 0` of course
 	- When $p_i$ evaluated the second wait for j, it found out that $⟨MY_TURN[i],i⟩ < ⟨MY_TURN[j],j⟩$, good!
-continuare
 
 **MUTEX:** $p_i$ and $p_j$ cannot simultaneously be in the CS:
 *Proof:*
 By contradiction, by Lemma 2 applied twice, we would have:
 $$⟨MY\_TURN[i] , i⟩ < ⟨MY\_TURN[j] , j⟩ \land ⟨MY\_TURN[j] , j⟩ < ⟨MY\_TURN[i] , i⟩$$
+Which is of course not possible, how in the world is it possible that x < y and y < x at the same time? √
 
 #### Deadlock freedom proof
 By contradiction, assume that there is a lock but nobody enters its CS