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]]
(*must finished before nel senso che $p_i$ deve aspettare $p_j$*)
##### Deadlock freedom
Let $p_i$ invoke lock
- If it eventually wins -> √
- If it is blocked forever, where can it be blocked?
	1. In the second wait Y =  ⊥
		- in this case, it read a value in Y different from i
		- there is a $p_h$ that wrote Y after $p_i$
		- let us consider the last of such $p_h \to$ it will eventually win
	2. In the ∀j.wait FLAG[j] = down
		- this wait cannot block a process forever
		- if $pj$ doesn't lock, it flag is down
		- if $pj$ doesn't find Y at ⊥, it puts its flag down
		- if pj doesn't find X at j, it puts its flag down, otherwise pj enters its CS and eventually unlocks (flag down)
	- 3. In the first wait Y = ⊥
		- since pj read a value different from ⊥, there is at least one pk that wrote Y before (but has not yet unlocked)
		- if $p_k$ eventually enters its CS -> ok, otherwise it must be blocked forever as well. Where?
			- In the second wait Y = ⊥: but then there exists a $p_h$ that eventually enters its CS -> good
			- In the ∀j.wait FLAG[j]=down: this wait cannot block a process forever

![[Pasted image 20250304090219.png]]

esercizio: prova che NON soddisfa starvation freedom

#### From deadlock freedom to bounded bypass
-> Round Robin algorithm

Let DLF be any deadlock free protocol for MUTEX. Let's see how we can make it satisfy bounded bypass:

```
lock(i) :=
	FLAG[i] <- up
	wait (TURN = i OR FLAG[TURN] = down)
	DLF.lock(i)
	return

unlock(i) :=
	FLAG[i] <- down
	if FLAG[TURN] = down then
		TURN <- (TURN + 1) mod n
	DLF.unlock(i)
	return
```


###### Is it deadlock free?
Since DLF is deadlock free, it is sufficient to prove that at least one process invokes DLF.lock.
If TURN = k and $p_k$ invoked lock, then it finds TURN = k and exits its wait.
Otherwise, any other process will find `FLAG[TURN] = down` and exits from its wait.

**Lemma 1:** if TURN = i and `FLAG[i] = up` then $p_i$ enters the CS in at most n-1 iterations

Observation 1: TURN changes only when FLAG[i] is down (after pi has completed its CS)

Observation 2: FLAG[i] = up -> either pi is in its CS or pi is competing for its CS -> it eventually invokes (if not already) DLF.lock

Observation 3: if $p_j$ invokes lock after that FLAG[i] is set, $p_j$ blocks in its wait

Let Y be the set of processes competing for the CS (suspended on the DLF.lock)
- because of Observation 2, $i \in Y$
- because of Observation 3, once FLAG[i] is set, Y cannot grow anymore
- because DLF is deadlock free, eventually one $p_{y} \in Y$ wins if $y = i$.
	- if $y = i$ we are done
	- otherwise, Y shrinks by one. And because of Observation 1, TURN and FLAG[TURN] don't change, so $p_y$ cannot enter Y again.
- Iterating this reasoning we can see that $p_i$ will eventually win, and the worst case is when is the last winner.

**Lemma 2:** If FLAG[i] = up, then TURN is set to i in at most $(n-1)^2$ iterations.

If TURN=i when FLAG[i] is set, done
By Deadlock freedom of RR, at least one process eventually unlocks
- If FLAG[TURN] = down, then TURN is increased. Otherwise, by Lemam 1, $p_{TURN}$ wins in at most n-1 iterations and increases TURN.
- If now TURN = i then we are done. Otherwise, we repeat this reasoning.

The worst case is when TURN = *i+1* mod n when FLAG[i] is set.

![[Pasted image 20250304093223.png]]

