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

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? !

! (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

!

esercizio: prova che NON soddisfa starvation freedom