77 lines
No EOL
2 KiB
Markdown
77 lines
No EOL
2 KiB
Markdown
Initial idea:
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```
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Initialize Y at ⊥, X at any value (e.g., 0)
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lock(i) :=
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x <- i
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if Y != ⊥ then FAIL
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else Y <- i
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if X = i then return
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else FAIL
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unlock(i) :=
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Y <- ⊥
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return
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```
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Problems:
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- we don't want the FAIL
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- it is possible to have an execution where nobody accesses its CS, if repeated forever it entails a deadlock
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```
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Initialize Y at ⊥, X at any value (e.g., 0)
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lock(i) :=
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* FLAG[i] <- up
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X <- i
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if Y ≠ ⊥ then
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FLAG[i] <- down
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wait Y = ⊥
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goto *
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else Y <- i
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if X = i then return
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else FLAG[i] <- down
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∀j.wait FLAG[j] = down
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if Y = i then return
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else wait Y = ⊥
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goto *
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unlock(i) :=
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Y <- ⊥
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FLAG[i] <- down
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return
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```
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##### MUTEX proof
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How can pi enter its CS?
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![[Pasted image 20250304084537.png]]
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![[Pasted image 20250304084901.png]]
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(*must finished before nel senso che $p_i$ deve aspettare $p_j$*)
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##### Deadlock freedom
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Let $p_i$ invoke lock
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- If it eventually wins -> √
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- If it is blocked forever, where can it be blocked?
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1. In the second wait Y = ⊥
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- in this case, it read a value in Y different from i
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- there is a $p_h$ that wrote Y after $p_i$
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- let us consider the last of such $p_h \to$ it will eventually win
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2. In the ∀j.wait FLAG[j] = down
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- this wait cannot block a process forever
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- if $pj$ doesn't lock, it flag is down
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- if $pj$ doesn't find Y at ⊥, it puts its flag down
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- if pj doesn't find X at j, it puts its flag down, otherwise pj enters its CS and eventually unlocks (flag down)
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- 3. In the first wait Y = ⊥
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- since pj read a value different from ⊥, there is at least one pk that wrote Y before (but has not yet unlocked)
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- if $p_k$ eventually enters its CS -> ok, otherwise it must be blocked forever as well. Where?
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- In the second wait Y = ⊥: but then there exists a $p_h$ that eventually enters its CS -> good
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- In the ∀j.wait FLAG[j]=down: this wait cannot block a process forever
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![[Pasted image 20250304090219.png]]
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esercizio: prova che NON soddisfa starvation freedom |