vault backup: 2025-03-10 10:10:13

.obsidian/workspace.json

@@ -34,9 +34,9 @@
               "type": "pdf",
               "state": {
                 "file": "Concurrent Systems/slides/class 3.pdf",
-                "page": 11,
+                "page": 12,
                 "left": -26,
-                "top": 49,
+                "top": 15,
                 "zoom": 0.57541567695962
               },
               "icon": "lucide-file-text",

Concurrent Systems/notes/3.md

@@ -194,4 +194,28 @@ 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)
+	- (indeed, since there is no deadlock, when pi is waiting somebody enters the CS)
+
+### Aravind’s algorithm
+Problem with Lamport's algorithm: registers must be unbounded (every invocation of lock potentially increases the counter by 1 -> domain of the registers is all natural numbers!)
+
+For all processes, we have a FLAG and a STAGE (both binary MRSW) and a DATE (MRMW) register that ranges from 1 to 2n.
+
+```
+For all i, initialize
+	FLAG[i] to down
+	STAGE[i] to 0
+	DATE[i] to i
+
+lock(i) :=
+	FLAG[i] <- up
+	repeat
+		STAGE[i] <- 0
+		wait (foreach j != i, FLAG[j] = down OR DATE[i] < DATE[j])
+		STAGE[i] <- 1
+	until foreach j != i, STAGE[J] = 0
+
+unlock(i) :=
+	tmp <-
+```
+