138 lines
4.4 KiB
Markdown
138 lines
4.4 KiB
Markdown
Peterson's algorithm cost $O(n^2)$
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A first way to reduce this cost is by using a tournament of MUTEX between pairs of processes:
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![[Pasted image 20250304082459.png|350]]
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Of course this is a binary tree, and the height of a binary tree is logaritmic to the number of leaves. A process then wins after $\lceil \log_{2}n \rceil$ competitions $\to O(\log n)$ cost.
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But we can do better. Let's see an idea of a constant-time algorithm.
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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
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#### From deadlock freedom to bounded bypass
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-> Round Robin algorithm
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Let DLF be any deadlock free protocol for MUTEX. Let's see how we can make it satisfy bounded bypass:
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```
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lock(i) :=
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FLAG[i] <- up
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wait (TURN = i OR FLAG[TURN] = down)
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DLF.lock(i)
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return
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unlock(i) :=
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FLAG[i] <- down
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if FLAG[TURN] = down then
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TURN <- (TURN + 1) mod n
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DLF.unlock(i)
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return
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```
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###### Is it deadlock free?
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Since DLF is deadlock free, it is sufficient to prove that at least one process invokes DLF.lock.
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If TURN = k and $p_k$ invoked lock, then it finds TURN = k and exits its wait.
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Otherwise, any other process will find `FLAG[TURN] = down` and exits from its wait.
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**Lemma 1:** if TURN = i and `FLAG[i] = up` then $p_i$ enters the CS in at most n-1 iterations
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Observation 1: TURN changes only when FLAG[i] is down (after pi has completed its CS)
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Observation 2: FLAG[i] = up -> either pi is in its CS or pi is competing for its CS -> it eventually invokes (if not already) DLF.lock
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Observation 3: if $p_j$ invokes lock after that FLAG[i] is set, $p_j$ blocks in its wait
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Let Y be the set of processes competing for the CS (suspended on the DLF.lock)
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- because of Observation 2, $i \in Y$
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- because of Observation 3, once FLAG[i] is set, Y cannot grow anymore
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- because DLF is deadlock free, eventually one $p_{y} \in Y$ wins if $y = i$.
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- if $y = i$ we are done
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- otherwise, Y shrinks by one. And because of Observation 1, TURN and FLAG[TURN] don't change, so $p_y$ cannot enter Y again.
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- Iterating this reasoning we can see that $p_i$ will eventually win, and the worst case is when is the last winner.
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**Lemma 2:** If FLAG[i] = up, then TURN is set to i in at most $(n-1)^2$ iterations.
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If TURN=i when FLAG[i] is set, done
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By Deadlock freedom of RR, at least one process eventually unlocks
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- If FLAG[TURN] = down, then TURN is increased. Otherwise, by Lemam 1, $p_{TURN}$ wins in at most n-1 iterations and increases TURN.
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- If now TURN = i then we are done. Otherwise, we repeat this reasoning.
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The worst case is when TURN = *i+1* mod n when FLAG[i] is set.
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![[Pasted image 20250304093223.png]]
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