master-degree-notes/Concurrent Systems/notes/2 - Fast mutex by Lamport.md

77 lines
No EOL
2 KiB
Markdown

Initial idea:
```
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