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Concurrent Systems/notes/Lezione2.md

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+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:
+![[Pasted image 20250304082459.png|350]]
+
+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.
+
+But we can do better. Let's see an idea of a constant-time algorithm.
+
+```
+Initialize Y at ⊥, X at any value (e.g., 0)
+
+lock(i) :=
+	x <- i
+	if Y != ⊥ then FAIL
+	else Y <- i
+	if X = i then return
+	else fail
+
+ unlock(i) :=
+	 Y <- ⊥
+	 return