vault backup: 2025-04-04 23:58:09

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Marco Realacci 2025-04-04 23:58:09 +02:00
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commit 1ce5aec8cd
13 changed files with 34 additions and 50 deletions

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@ -13,30 +13,15 @@
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@ -232,8 +220,6 @@
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@ -62,7 +62,7 @@ function withdraw() {
While `read()` and `write()` may be considered as atomic, their sequential composition **is not**.
![[/Concurrent Systems/notes/images/Pasted image 20250303090135.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250303090135.png)
#### Mutual Exclusion (MUTEX)
Ensure that some parts of the code are executed as *atomic*.
@ -100,7 +100,7 @@ Every solution to a problem should satisfy at least:
**Both inclusions are strict:**
$$\text{Deadlock freedom} \not \implies \text{Starvation freedom}$$
![[/Concurrent Systems/notes/images/Pasted image 20250303093116.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250303093116.png)
*p1 is starving!*
$$\text{Starvation freedom} \not \implies \text{Bounded bypass}$$
Assume a $f$ and consider the scheduling above, where p2 wins $f(3)$ times and so does p3, then p1 looses (at least) $2f(3)$ times before winning.

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@ -30,7 +30,7 @@ A configuration C obtained during the execution of all A is called:
If A wait-free implements binary consensus for n processes, then there exists a bivalent initial configuration.
*Proof:*
![[/Concurrent Systems/notes/images/Pasted image 20250401083747.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250401083747.png)
### CN(Atomic R/W registers) = 1
**Thm:** There exists no wait-free implementation of binary consensus for 2 processes that uses atomic R/W registers.
@ -135,7 +135,7 @@ propose(i, v) :=
Let us consider a verison of the compare&swap where, instead of returning a boolean, it always returns the previous value of the object, i.e.:
![[/Concurrent Systems/notes/images/Pasted image 20250401092557.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250401092557.png)
```
CS a compare&swap object init at ⊥

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@ -53,10 +53,10 @@ unlock(i) :=
**How has p0 entered its CS?**
a) `FLAG[1] = down`, this is possible only with the following interleaving:
![[/Concurrent Systems/notes/images/Pasted image 20250303100721.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250303100721.png)
b) `AFTER_YOU = 1`, this is possible only with the following interleaving:
![[/Concurrent Systems/notes/images/Pasted image 20250303100953.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250303100953.png)
##### Bounded Bypass proof (with bound = 1)
- If the wait condition is true, then it wins (and waits 0).
@ -140,7 +140,7 @@ Easy to generalize to k-MUTEX.
Peterson's algorithm cost $O(n^2)$
A first way to reduce this cost is by using a tournament of MUTEX between pairs of processes:
![[/Concurrent Systems/notes/images/Pasted image 20250304082459.png|350]]
![350](Concurrent%20Systems/notes/images/Pasted%20image%2020250304082459.png|)
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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@ -49,9 +49,9 @@ unlock(i) :=
##### MUTEX proof
How can pi enter its CS?
![[/Concurrent Systems/notes/images/Pasted image 20250304084537.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250304084537.png)
![[/Concurrent Systems/notes/images/Pasted image 20250304084901.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250304084901.png)
(*must finished before nel senso che $p_i$ deve aspettare $p_j$*)
##### Deadlock freedom
Let $p_i$ invoke lock
@ -72,6 +72,6 @@ Let $p_i$ invoke lock
- 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
![[/Concurrent Systems/notes/images/Pasted image 20250304090219.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250304090219.png)
esercizio: prova che NON soddisfa starvation freedom

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@ -50,5 +50,5 @@ By Deadlock freedom of RR, at least one process eventually unlocks
The worst case is when TURN = *i+1* mod n when FLAG[i] is set.
![[/Concurrent Systems/notes/images/Pasted image 20250304093223.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250304093223.png)

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@ -31,7 +31,7 @@ unlock(i) :=
*Proof:* by contradiction.
Let's consider the execution of $p_i$ leading to its CS:
![[/Concurrent Systems/notes/images/Pasted image 20250310172134.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250310172134.png)
**Corollary** (of the MUTEX proof)**:** DATE is never written concurrently.
@ -57,7 +57,7 @@ Let's consider the execution of $p_i$ leading to its CS:
**Theorem:** the algorithm satisfies bounded bypass with bound $2n-2$.
*Proof:*
![[/Concurrent Systems/notes/images/Pasted image 20250310103703.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250310103703.png)
so by this, the very worst possible case is that my lock experiences that.
It looks like I can experience at most $2n-1$ other critical sections, but it is even better, let's see:

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@ -180,7 +180,7 @@ producer A:
- so producer A will write at `BUF[0]`
- but wait! Consumer B is still reading there
- **Producer A doesn't give a fuck.**
![[/Concurrent Systems/notes/images/Pasted image 20250312121828.png|200]]
![200](Concurrent%20Systems/notes/images/Pasted%20image%2020250312121828.png|)
*don't be like Producer A, be more like Bob, who always scans EMPTY before!*
So the issue here is that producers just assume that IN is the first available slot. But it its not necessarily the case.

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@ -3,7 +3,7 @@ The first real practical example of a concurrent system.
- one chopstick between each pair of philosophers
- a philosophers must pick up its two nearest chopsticks in order to eat
- a philosopher must pick up first one chopstick, then the second one, not both at once
![[/Concurrent Systems/notes/images/Pasted image 20250317100456.png|100]]
![100](Concurrent%20Systems/notes/images/Pasted%20image%2020250317100456.png|)
**PROBLEM:** *Devise a deadlock-free algorithm for allocating these limited resources (chopsticks) among several processes (philosophers).*

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@ -105,7 +105,7 @@ The **casual past** of a transaction T is the set of all T' and T'' such that
VWC allows more transactions to commit -> it is a more liberal property than opacity.
![[/Concurrent Systems/notes/images/Pasted image 20250317105355.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250317105355.png)
#### A Vector clock based STM system
We have m shared MRMW registers; register X is represented by a pair XX, with:

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@ -21,9 +21,9 @@ A complete history $\hat{H}$ is **linearizable** if there exists a sequential hi
Given an history $\hat{K}$, we can define a binary relation on events $⟶_{K}$ s.t. (op, op) ∈ ⟶K if and only if res[op] <K inv[op]. We write op K op for denoting (op, op) K. Hence, condition 3 of the previous Def. requires that H S.
![[/Concurrent Systems/notes/images/Pasted image 20250318090733.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250318090733.png)
![[/Concurrent Systems/notes/images/Pasted image 20250318090909.png]]But there is another linearization possible! I can also push a before if I pull it before c!
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250318090909.png)But there is another linearization possible! I can also push a before if I pull it before c!
Of course I have to respect the semantics of a Queue (if I push "a" first, I have to pop "a" first because it's a fucking FIFO)
#### Compositionality theorem

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@ -2,9 +2,9 @@ Let us define $op ->_{proc} op'$ to hold whenever there exists a process p that
### Sequential consistency
**Def:** a complete history is sequentially consistent if there exists a sequential history $𝑆$ s.t.
![[/Concurrent Systems/notes/images/Pasted image 20250324082534.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250324082534.png)
![[/Concurrent Systems/notes/images/Pasted image 20250324082545.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250324082545.png)
>[!warning]
>The problem with sequential consistency is that it is NOT compositional.

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@ -64,7 +64,7 @@ this implementation satisfies the three requirements for the splitter
- let us consider the last process that writes into LAST (this is an atomic register, so this is meaningful)
- if the door is closed, it receives R and √
3. let $p_i$ be the first process that receives $S \to LAST=i$ in its second if
![[/Concurrent Systems/notes/images/Pasted image 20250324091452.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250324091452.png)
### An Obstruction-free Timestamp Generator
A **timestamp generator** is a concurrent object that provides a single operation get_ts such that:
@ -98,7 +98,7 @@ this implementation satisfies the three properties of the timestamp generator
- every process that starts after its termination will find NEXT to a greater value (NEXT never decreases!)
3. Obstruction freedom is trivial
**REMARK:** this implementation doesnt satisfy the non-blocking property:![[/Concurrent Systems/notes/images/Pasted image 20250324092633.png]]
**REMARK:** this implementation doesnt satisfy the non-blocking property:![](Concurrent%20Systems/notes/images/Pasted%20image%2020250324092633.png)
### A Wait-free Stack
REG is an unbounded array of atomic registers (the stack)
@ -149,7 +149,7 @@ This is needed for the so called ABA problem with compare&set:
- with the compare&set you mainly test that the sequence_number has not changed
TOP : a register that can be read or compare&setted
![[/Concurrent Systems/notes/images/Pasted image 20250324100652.png]]
![](Concurrent%20Systems/notes/images/Pasted%20image%2020250324100652.png)
```
push(w) :=