vault backup: 2025-03-10 10:05:14

.obsidian/workspace.json

@@ -34,9 +34,9 @@
               "type": "pdf",
               "state": {
                 "file": "Concurrent Systems/slides/class 3.pdf",
-                "page": 10,
+                "page": 11,
                 "left": -26,
-                "top": 603,
+                "top": 49,
                 "zoom": 0.57541567695962
               },
               "icon": "lucide-file-text",

Concurrent Systems/notes/3.md

@@ -178,4 +178,20 @@ By contradiction, assume that there is a lock but nobody enters its CS
 - All processes in the bakery (we will call this set Q) are blocked in their wait
 - The first wait cannot block forever
 	- All $p_i \in Q$ have their FLAG down
-	- All $p_i \not \in Q$ have their FLAG down
+	- All $p_i \not \in Q$ have their FLAG down (if not in the doorway) or will eventually put their FLAG down (cannot remain in the doorway forever)
+- The second wait cannot block all of them forever
+	- Tickets can be totally ordered (lexicographically)
+	- Let `<MY_TURN[i], j>` be the minimun
+	- The second wait evaluated by $p_i$ eventually succeeds for all j
+		- if $p_j$ is before the doorway, then `MY_TURN[j] = 0`
+		- if $p_{j}$ is in the doorway, then `MY_TURN[i] < MY_TURN[j]` (bc of Lemma 1)
+		- 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
+
+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!
+- 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)