vault backup: 2025-04-05 00:05:56

.obsidian/workspace.json

@@ -13,12 +13,12 @@
             "state": {
               "type": "markdown",
               "state": {
-                "file": "Concurrent Systems/notes/1b - Peterson algorithm.md",
+                "file": "Concurrent Systems/notes/2b - Round Robin algorithm.md",
                 "mode": "source",
                 "source": false
               },
               "icon": "lucide-file",
-              "title": "1b - Peterson algorithm"
+              "title": "2b - Round Robin algorithm"
             }
           }
         ]
@@ -191,8 +191,8 @@
   },
   "active": "7c5b0ca6f7687800",
   "lastOpenFiles": [
-    "Concurrent Systems/notes/images/Pasted image 20250405000438.png",
     "Concurrent Systems/notes/1b - Peterson algorithm.md",
+    "Concurrent Systems/notes/images/Pasted image 20250405000438.png",
     "Pasted image 20250405000428.png",
     "Concurrent Systems/notes/3b - Aravind's algorithm and improvements.md",
     "Concurrent Systems/notes/4 - Semaphores.md",

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

@@ -140,7 +140,7 @@ Easy to generalize to k-MUTEX.
 
 Peterson's algorithm cost $O(n^2)$
 A first way to reduce this cost is by using a tournament of MUTEX between pairs of processes:
-![350](Concurrent%20Systems/notes/images/Pasted%20image%2020250304082459.png|)
+![350](images/Pasted%20image%2020250304082459.png|)
 
 Of course this is a binary tree, and the height of a binary tree is logaritmic to the number of leaves. A process then wins after $\lceil \log_{2}n \rceil$ competitions $\to O(\log n)$ cost.
 

Concurrent Systems/notes/4 - Semaphores.md

@@ -180,7 +180,7 @@ producer A:
 - so producer A will write at `BUF[0]`
 - but wait! Consumer B is still reading there
 - **Producer A doesn't give a fuck.**
-	![200](Concurrent%20Systems/notes/images/Pasted%20image%2020250312121828.png|)
+	![
]()
 	*don't be like Producer A, be more like Bob, who always scans EMPTY before!*
 
 So the issue here is that producers just assume that IN is the first available slot. But it its not necessarily the case.

Concurrent Systems/notes/4c - Dining Philosophers.md

@@ -3,7 +3,7 @@ The first real practical example of a concurrent system.
 - one chopstick between each pair of philosophers
 - a philosophers must pick up its two nearest chopsticks in order to eat
 - a philosopher must pick up first one chopstick, then the second one, not both at once
-![100](Concurrent%20Systems/notes/images/Pasted%20image%2020250317100456.png|)
+![
]()
 
 **PROBLEM:** *Devise a deadlock-free algorithm for allocating these limited resources (chopsticks) among several processes (philosophers).*