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
```

Problems:
- we don't want the FAIL
- it is possible to have an execution where nobody accesses its CS, if repeated forever it entails a deadlock

```
Initialize Y at ⊥, X at any value (e.g., 0)


lock(i) :=
	* FLAG[i] <- up
	X <- i
	if Y ≠ ⊥ then FLAG[i] <- down
		wait Y = ⊥
		goto *
	else Y <- i
	if X = i then return
	else FLAG[i] <- down
	∀j.wait FLAG[j] = down
	if Y = i then return
	else wait Y = ⊥
	goto *

unlock(i) :=
	Y <- ⊥
	FLAG[i] <- down
	return
```

MUTEX proof

How can pi enter its CS?
![[Pasted image 20250304084537.png]]

![[Pasted image 20250304084901.png]]