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!
**Lemma 1:** Suppose we have $n$ processes, then $\not \exists p_{j} : DATE[j]=DATE[i] \forall i \in [0, n]$ (non esistono due processi con lo stesso valore per DATE)
*Proof:*
- Suppose $p_j$ is the last process to execute unlock, then `DATE[j] = n` and for each i, `DATE[i] < n`, as DATE is either set to $n$ or decreased.
- by iterating this reasoning, it is clear that until every DATE is > 0, we have no processes with the same value for DATE.
- What if we reach 0?
- Let's suppose that $p_i$ is the first process that reaches `DATE = 0`
