135 lines
4.7 KiB
Markdown
135 lines
4.7 KiB
Markdown
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### Peterson algorithm (for two processes)
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Let's try to enforce MUTEX with just 2 processes.
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##### 1st attempt:
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```
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lock(i) :=
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AFTER_YOU <- i
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wait AFTER_YOU != i
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return
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unlock(i) :=
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return
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```
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This protocol satisfies MUTEX, but suffers from deadlock (if one process never locks).
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##### 2nd attempt:
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```
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Initialize FLAG[0] and FLAG[1] to down
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lock(i) :=
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FLAG[i] <- up
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wait FLAG[1-i] = down
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return
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unlock(i) :=
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FLAG[i] <- down
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return
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```
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Still suffers from deadlock if both processes simultaneously raise their flag.
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##### Correct solution:
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```
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Initialize FLAG[0] and FLAG[1] to down
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lock(i) :=
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FLAG[i] <- up
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AFTER_YOU <- i
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wait FLAG[1-i] = down OR AFTER_YOU != i
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return
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unlock(i) :=
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FLAG[i] <- down
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return
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```
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**Features:**
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- it satisfies MUTEX
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- it satisfies bounded bypass, with bound = 1
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- it requires 2 one-bit SRSW registers (the flags and 1 one-bit MRMW registers (AFTER_YOU)
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- Each lock-unlock requires 5 accesses to the registers (4 for lock and 1 for unlock)
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##### MUTEX proof
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Assume p0 and p1 are simultaneously in the CS.
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How has p0 entered its CS?
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a) `FLAG[1] = down`, this is possible only with the following interleaving:
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![[Pasted image 20250303100721.png]]
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b) `AFTER_YOU = 1`, this is possible only with the following interleaving:
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![[Pasted image 20250303100953.png]]
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##### Bounded Bypass proof (with bound = 1)
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- If the wait condition is true, then it wins (and waits 0).
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- Otherwise, it must be that `FLAG[1] = UP` **AND** `AFTER_YOU = 0` (quindi p1 ha invocato il lock e p0 dovrà aspettare).
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- If p1 never locks anymore then p0 will eventually read `F[1]` and win (waiting 1).
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- It is possible tho that p1 locks again:
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- if p0 reads `F[1]` before p1 locks, then p0 wins (waiting 1)
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- otherwise p1 sets A_Y to 1 and suspends in its wait (`F[0] = up AND A_Y = 1`), p0 will eventually read `F[1]` and win (waiting 1).
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### Peterson algorithm ($n$ processes)
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- FLAG now has $n$ levels (from 0 to n-1)
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- level 0 means down
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- level >0 means involved in the lock
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- Every level has its own AFTER_YOU
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```
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Initialize FLAG[i] to 0, for all i
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lock(i) :=
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for lev = 1 to n-1 do
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FLAG[i] <- lev
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AFTER_YOU[lev] <- i
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wait (∀k!=i, FLAG[k] < lev
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OR AFTER_YOU[lev] != i)
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return
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unlock(i) :=
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FLAG[i] <- 0
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return
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```
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We say that pi is at level h when it exits from the h-th wait -> a process at level h is at any level <= h
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##### MUTEX proof
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Lemma: for every $ℓ \in \{0,\dots,n-1\}$ , at most n-ℓ processes are at level ℓ, this implies MUTEX by taking ℓ = n-1
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Proof by induction on ℓ
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Base (ℓ=0): trivial
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Induction (true for ℓ, to be proved for ℓ+1):
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- p at level ℓ can increase its level by writing its FLAG at ℓ+1 and its index in $A_Y[ℓ+1]$
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- let $p_x$ be the last one that writes `A_Y[ℓ+1]`, so `A_Y[ℓ+1]=x`
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- for $p_x$ to pass at level ℓ+1, it must be that $∀k≠x. F[k] < ℓ+1$, then $p_x$ is the only proc at level ℓ+1 and the thesis holds, since 1<=n-ℓ-1
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- otherwise, $p_x$ is blocked in the wait and so we have at most n-ℓ-1 processes at level ℓ+1: those at level ℓ, that by induction are at most n-ℓ, except for px that is blocked in its wait.
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##### Starvation freedom proof
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**Lemma:** every process at level ℓ ($\leq n-1$) eventually wins $\to$ starvation freedom holds by taking $ℓ=0$.
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Reverse induction on ℓ
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Base ($ℓ=n-1$): trivial
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Induction (true for ℓ+1, to be proved for ℓ):
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- Assume a $p_x$ blocked at level ℓ (aka blocked in its ℓ+1-th wait) $\to \exists k\neq x, F[k]\geqℓ+1 \land A\_Y[ℓ+1]=x$
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- If some $p_{y}$ will eventually set $A\_Y[ℓ+1]$ to $y$, then $p_x$ will eventually exit from its wait and pass to level ℓ+1
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- Otherwise, let $G = \{p_{i}: F[i] \geq ℓ+1\}$ and $L=\{p_{i}:F[i]<ℓ+1\}$
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- if $p \in L$, it will never enter its ℓ+1-th loop (as it would write $A_Y[ℓ+1]$ and it will unblock $p_x$, but we are assuming that it is blocked)
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- all $p \in G$ will eventually win (by induction) and move to L
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- $\to$ eventually, $p_{x}$ will be the only one in its ℓ+1-th loop, with all the other processes at level <ℓ+1
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- $\to$ $p_{x}$ will eventually pass to level ℓ+1 and win (by induction)
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##### Peterson algorithm cost
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- $n$ MRSW registers of $\lceil \log_{2} n\rceil$ bits (FLAG)
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- $n-1$ MRMW registers of $\lceil \log_{2}n \rceil$ bits (AFTER_YOU)
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- $(n-1)\times(n+2)$ accesses for locking and 1 access for unlocking
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It satisfies MUTEX and starvation freedom. It does not satisfy bounded bypass:
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- consider 3 processes, one sleeping in its first wait, the others alternating in the CS
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- when the first process wakes up, it can pass to level 2 and eventually win
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- but the sleep can be arbitrary long and in the meanwhile the other two processes may have entered an unbounded number of CSs
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Easy to generalize to k-MUTEX.
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