### Aravind’s algorithm
Problem with Lamport's "Bakery" 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 <- max_j{DATE[j]}+1
	if tmp >= 2n
		then foreach j, DATE[j] <- j
		else DATE[i] <- tmp
	STAGE[i] <- 0
	FLAG[i] <- down
```

#### MUTEX proof
**Theorem:** if $p_i$ is in the CS, then $p_j$ cannot simultaneously be in the CS.
*Proof:* by contradiction.

Let's consider the execution of $p_i$ leading to its CS:
![[Pasted image 20250310172134.png]]

**Corollary** (of the MUTEX proof)**:** DATE is never written concurrently.

#### Bounded bypass proof
**Lemma 1:** exactly after n CSs there is a reset of DATE.
*Proof:*
- the first CS leads $max_j{DATE[j]}$ to n+1
- the seconds CS leads ... to n+2
- ...
- the n-th read leads ... to n+n = 2n -> RESET

**Lemma 2:** there can be at most one reset of DATE during an invocation of a lock
*Proof:*
- let $p_i$ invoke lock, if no reset occurs, ok
- otherwise, let us consider the moment in which a reset occurs
	- if pi is the next process that enters the CS, ok
	- Otherwise let $p_j$ be the process that enters; its next date is $n+1 > DATE[i]$
		- $p_{j}$ cannot surpass $p_i$ again (before a RESET)
	- The worst case is then all processes perform lock together and $i = n$ (i am process n)
		- all $p_{1}\dots p_{n}$ surpass $p_{n}$
		- then $p_n$ enters and it resets the DATE in its unlock
			- only 1 reset and it is the worst case!

**Theorem:** the algorithm satisfies bounded bypass with bound $2n-2$.
*Proof:*
![[Pasted image 20250310103703.png]]
so by this, the very worst possible case is that my lock experiences that.

It looks like I can experience at most $2n-1$ other critical sections, but it is even better, let's see:
- $p_n$ invokes lock alone, completes its CS (the first after the reset) and its new DATE is n+1
- all processes invoke lock simultaneously
- $p_{n}$ has to wait all other processes to complete their CSs
- when $p_{n-1}$ completes its CS, its new DATE will be $n+(n-1)+1=2n$ -> RESET
- now all $p_{1}\dots p_{n-1}$ invoke lock again and complete their CSs (after that $p_i$ completes its CS, now it has `DATE[i] <- n+i`, because as everyone invoked lock after the RESET, max date was `n`)
- so $p_n$ has to wait n-1 CSs for the reset, and another n-1 CSs before entering again.  **Literally the worst case is when the process is the first of the first round, and the last of the last round.**

#### Improvement of Aravind’s algorithm
```
unlock(i) :=
	∀j≠i.if DATE[j] > DATE[i] then
		DATE[j] <- DATE[j]-1
		DATE[i] <- n
		STAGE[i] <- 0
		FLAG[i] <- down
```
Since the LOCK is like before, the revised protocol satisfies MUTEX. Furthermore, you can prove that it satisfies bounded bypass with bound n-1 -> EXERCISE!

>[!note] let's remember ourselves [[### Aravind’s algorithm|the locking function]]

(a causa di un bug, riavvia Obsidian se non vedi la dimostrazione qui sotto)
>[!question]- mostra mia soluzione
>-  $p_n$ invokes lock alone, completes its CS and so `DATE[n] = n`
>- then as DATE is either set to n or decreased, for each i, `DATE[i] < n`
>- *scenario 1*: every other $p_i$ keep invoking lock again immediately after the unlock
>	- every time some process exits the CS, `DATE[n]` is decreased
>	- after $n-1$ turns, `DATE[n]` will have `DATE[n] = 1`, with every other `DATE[i] > 1, i!=n`  -> ️✅
>- *scenario 2*: not every process invokes the lock
>	- eventually, more than one process will have its DATE set to 0
>	- but there will still be at most $n-1$ processes $p_i$ with `DATE[i] < DATE[n]`
>	- this ensures that after $n-1$ steps, every other process $p_i$ will have `DATE[i] > DATE[n]` -> ✅
