
### Hardware primitives
Atomic R/W registers provide quite a basic computational model.

We can strenghten the model by adding specialized HW primitives, that essentially perform in an atomic way the combination of some atomic instructions.
Usually, every operating system provides at least one specilized HW primitive.

##### Most common ones:
- **Test&set:** atomic read+write of a boolean register
- **Swap:** atomic read+write of a general register (generalization of the above)
- **Fetch&add:** atomic read+increase of an integer register
- **Compare&swap:** tests the value of a general register, returns a boolean (result of the comparison, true if it is the same).

#### Test&Set
```
Let X be a boolean register

X.test&set() :=
	tmp <- X
	X <- 1
	return tmp

(the function is implemented in an atomic way by the hardware by suspending the interruptions!)
```

###### How do we use it for MUTEX?
```
lock() :=
	wait X.test&set() = 0
	return


unlock() :=
	X <- 0
	return
```


#### Swap
```
X general register

X.swap(v) :=
	tmp <- X
	X <- v
	return tmp
```

###### How do we use it for MUTEX?
```
lock() :=
	wait X.swap(1) = 0
	return

unlock() :=
	X <- 0
	return
```

#### Compare&swap
```
X boolean register

X.compare&swap(old, new) :=
	if X = old then
		X <- new
		return true
	return false
```
###### How do we use it for MUTEX?
```
Initialize X at 0

lock() :=
	wait X.compare&swap(0, 1) = true
	return

unlock() :=
	X <- 0
	return
```

#### Fetch&add
Up to now, all solutions enjoy deadlock freedom, but allow for starvation. So let's use Round Robin to promote the liveness property!

Let X be an integer register; the Fetch&add primitive is implemented as follows:
```
X.fetch&add(v) :=
	tmp <- X
	X <- X+v
	return tmp
```

###### How do we use it for MUTEX?
```
Initialize TICKET and NEXT at 0

lock() :=
	my_ticket <- TICKET.fetch&add(1)
	wait my_ticket = NEXT
	return

unlock() :=
	NEXT <- NEXT + 1
	return
```

>[!note] It is bounded bypass with bound n-1>

### Safe registers
Atomic R/W and specialized HW primitives provide some form of atomicity. But is it possible to enforce MUTEX without atomicity?

A **MRSW Safe register** is a register that provides READ and WRITE such that:
1. every READ that does not overlap with a WRITE returns the value stored in the register
2. a READ that overlaps with a WRITE can return **any possible value** (of the register domain).

A **MRMW Safe register** behaves like a MRSW safe register, when WRITE operations do not overlap. Otherwise, in case of overlapping WRITEs, the register can contain any value (of the register domain).
*(Ci sta comunque lo stesso problema)*

> Si chiama safe nel senso che se ci sono letture overlapping siamo safe. Ma è letteralmente l'unica cosa safe che abbiamo!

This is the weakest type of register that is useful in concurrency.

### Bakery Algorithm
**The idea is that**
- every process gets a ticket
- because we don't have atomicity, tickets may be not unique
- tickets can be made unique by pairing them with the process ID
- the smallest ticket (seen as a pair) grants the access to the CS

```
Initialize FLAG[i] to down and MY_TURN[i] to 0, for all i

lock(i) :=
	FLAG[i] <- up
	MY_TURN[i] <- max{MY_TURN[1],...,MY_TURN[n]}+1
	FLAG[i] <- down
	forall j != i
		wait FLAG[j] = down
		wait (MY_TURN[j] = 0 OR ⟨MY_TURN[i],i⟩ < ⟨MY_TURN[j],j⟩)

unlock(i) :=
	MY_TURN[i] <- 0
```
Se il ticket number è minore si ottiene l'accesso, se il ticket number è uguale, allora si vede il process ID minore.

> It is possible that while reading, other processes are writing their turn! So it's possible that I read something unpredictable! We need to consider it in the proofs.

###### Why is the flag needed?
Without it, if two process $i$ and $j$ sets MY_TURN at the same time, then $i$ goes to sleep, $j$ enters the CS, but when $i$ wakes up, it will enter too!

#### MUTEX proof
**Lemma 1:** if $p_i$ enters the bakery before $p_j$, then `MY_TURN[i] < MY_TURN[j]`.

*Proof:*
- let t be the value of `MY_TURN[i]` after $p_{i}$ exits the doorway (after flag <- down)
- when $p_j$ computes its ticket, it reads $t$ from `MY_TURN[i]`, it reads t from `MY_TURN[i]` (no write overlapping with this read)
- Hence, `MY_TURN[j]` is at least $t+1$. (legge correttamente almeno t, gli altri boh ma non ci interessa ora)

**Lemma 2:** Let $p_i$ be in the CS and $p_j$ is in the doorway or in the bakery; then, $$⟨MY\_TURN[i] , i⟩ < ⟨MY\_TURN[j] , j⟩$$*Proof:*
If $p_i$ is in the CS, it has terminated its first wait for $j$
	let's consider the read of `FLAG[j]` done by $p_i$ that terminates such wait
	
W.r.t. the execution of $p_j$, it can be that
- this read overlaps with `FLAG[j] <- up` by Lemma 1, `MY_TURN[i] < MY_TURN[j]` and ok
- this read is contained within the computation of `MY_TURN[j]`, but it's not possible, since `MY_TURN` is computed with the `FLAG` up
- this read overlaps with `FLAG[j] <- down` or this read happens when $p_j$ is in the bakery:
	- `MY_TURN[j]` has been decided and no write will change it until $p_j$ is in the bakery, and `MY_TURN[j] > 0` of course
	- When $p_i$ evaluated the second wait for j, it found out that $⟨MY_TURN[i],i⟩ < ⟨MY_TURN[j],j⟩$, good!

**MUTEX:** $p_i$ and $p_j$ cannot simultaneously be in the CS:
*Proof:*
By contradiction, by Lemma 2 applied twice, we would have:
$$⟨MY\_TURN[i] , i⟩ < ⟨MY\_TURN[j] , j⟩ \land ⟨MY\_TURN[j] , j⟩ < ⟨MY\_TURN[i] , i⟩$$
Which is of course not possible, how in the world is it possible that x < y and y < x at the same time? √

#### Deadlock freedom proof
By contradiction, assume that there is a lock but nobody enters its CS
- All processes in the bakery (we will call this set Q) are blocked in their wait
- The first wait cannot block forever
	- All $p_i \in Q$ have their FLAG down
	- All $p_i \not \in Q$ have their FLAG down (if not in the doorway) or will eventually put their FLAG down (cannot remain in the doorway forever)
- The second wait cannot block all of them forever
	- Tickets can be totally ordered (lexicographically)
	- Let `<MY_TURN[i], j>` be the minimun
	- The second wait evaluated by $p_i$ eventually succeeds for all j
		- if $p_j$ is before the doorway, then `MY_TURN[j] = 0`
		- if $p_{j}$ is in the doorway, then `MY_TURN[i] < MY_TURN[j]` (bc of Lemma 1)
		- if $p_j$ is in the bakery, by assumption `⟨MY_TURN[i] , i⟩ < ⟨MY_TURN[j] , j⟩` since it is the minimum.

#### Bounded bypass proof (bound n-1)
Let $p_i$ and $p_j$ competing for the CS and $p_j$ wins

Then, $p_j$ enters its CS, completes it, unlocks and then invokes lock again
- If $p_i$ has entered the CS, √
- Otherwise, by Lemma1, $MY\_TURN[i] < MY\_TURN[j]$, then $p_j$ cannot bypass $p_i$ again!
- At worse, pi has to wait all other proceeses before entering its CS
	- (indeed, since there is no deadlock, when pi is waiting somebody enters the CS)