master-degree-notes/Concurrent Systems/notes/10 - Implementing Consensus.md

Which objects allow for a wait free implementation of (binary) consensus? The answer depends on the number of participants

The consensus number of an object of type T is the greatest number n such that it is possible to wait free implement a consensus object in a system of n processes by only using objects of type T and atomic R/W registers.

For all T, CN(T) > 0; if there is no sup, we let CN(T) := +∞

Thm: let CN(T1) < CN(T2), then there exists no wait free implementation of T2 that only uses objects of type T1 and atomic R/W registers, for all n s.t. CN(T1) < n <= CN(T2).

Proof:

• Fix such an n; by contr., there exists a wait free implementation of objects of type T2 in a system of n processes that only uses objects of type T1 and atomic RW reg.s.
• Since n ≤ CN(T2), by def. of CN, there exists a wait free implementation of consensus in a system of n processes that only uses objects of type T2 and atomic RW reg.s.
• Hence, there exists a wait free implementation of consensus in a system of n processes that only uses objects of type T1 and atomic RW reg.s.
  • contradiction with CN(T1) < n

Schedules and Configurations

Schedule: sequence of operation invocations issued by processes.

Configuration: the global state of a system at a given execution time (values of the shared memory + local state of every process).

Given a configuration C and a schedule S, we denote with S(C) the configuration obtained starting from C and applying S.

Let's consider binary consensus implemented by an algorithm A by using base objects and atomic R/W registers; let us call S_A a schedule induced by A.

A configuration C obtained during the execution of all A is called:

• v-valent if S_A(C) decides v, for every S_A
• monovalent, if there exists v \in \{0,1\} s.t. C is v-valent
• bivalent, otherwise.

Fundamental theorem

If A wait-free implements binary consensus for n processes, then there exists a bivalent initial configuration.

Proof:

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.

Proof: Assume by contradiction A wait-free, with processes p and q.

By the previous result, it has an initial bivalent configuration C

• let S be a sequence of operations s.t. C’ = S(C) is maximally bivalent (i.e., p(C') is 0-valent and q(C') is 1-valent, or viceversa)
  • partendo da C' posso ancora avere due possibili computazioni dove una decide 0 e una decide 1, ma è l'ultima configurazione in cui è possibile. Quelle successive sono monovalenti.

p(C’) can be R1.read() or R1.write(v) and q(C’) can be R2.read() or R2.write(v’)

1. if R1 != R2

   • Whatever operations p and q issue, we have that q(p(C’)) = p(q(C’)) But q(p(C’)) is 0-val (because p(C’) is) whereas p(q(C’)) is 1-val
     • impossible case

2. R1 = R2 and both operations are a read

   • like point 1... We will again obtain a configuration that is both 0-valent and 1-valent

3. R1 = R2, with p that reads and q that writes (or viceversa)

   • Remark: only p can distinguish C' from p(C') (reads put the value read in a local variable, visible only by the process that performed the read)
   • Let S' be the scheduling from C' where p stops and q decides:
     • S' starts with the write of q
     • S' leads q to decide 1, since q(C') is 1-valent
   • Consider p(C') and apply S'
     • because of the initial remark, q decides 1 also here
   • Reactivate p
     • if p decides 0, then we would violate agreement
     • if p decides 1, we contradict 0-valence of p(C')

4. R1 = R2 and both operations are a write

   • Remark: q(p(C)) = q(C) cannot be distinguished by q since the value written by p is lost after the write of q
   • Then, work like in case (3).

CN(test&set) = 2

TS a test&set object init at 0
PROP array of proposals, init at whatever

propose(i, v) :=
	PROP[i] <- v
	if TS.test&set() = 0 then
		return PROP[i]
	else
		return PROP[1-i]

Wait-freedom, Validity and Integrity hold by construction.

Agreement: the first that performs test&set receives 0 and decides his proposal; the other one receives 1 and decides the other proposal.

Thm: there exists no A wait free that implements binary consensus for atomic R/W registers and test&set objects for 3 processes.

Proof: The structure is the same as the previous proof. Consider 3 proc.’s p, q and r. Let C be bivalent and S maximal s.t. S(C) (call it C’) is bivalent:

• p(C’) is 0-val, q(C’) is 1-val and r(C’) is monovalent (for example)

Let’s assume that:

• at C’ r stops for a long time
• op_{p} and op_{q} are the next operations that p and q issue from C’ by following A

1. op_p and op_q are both R/W operations on atomic registers
   • like in the previous proof
2. one is an operation on an atomic register and the other is a test&set but on different objects
   • like the first case of the previous proof, since p(q(C')) = q(p(C'))
3. they are both test&set on the same object
   • p(q(C’)) is 1-val whereas q(p(C’)) is 0-va

Let us now stop both p and q and resume r

• r cannot see any difference between p(q(C’)) and q(p(C’)) (the only diff.’s are the values locally stored by p and q as result of T&S)

Let S’ be a schedule of operations only from r that leads p(q(C’)) to a decision (that must be 1)

• Since r cannot see any difference between p(q(C’)) and q(p(C’)), if we run S’ from q(p(C’)) we must decide 1 as well
  • in contradiction with q(p(C')) be 0-val

CN(swap) = CN(fetch&add) = 2

S a swap object init at ⊥
PROP array of proposals, init at whatever

propose(i, v) :=
	PROP[i] <- v
	if S.swap(i) = ⊥ then
		return PROP[i]
	else
		return PROP[1-i]

FA a fetch&add object init at 0
PROP array of proposals, init at whatever

propose(i, v) :=
	PROP[i] <- v
	if FA.fetch&add(1) = 0 then
		return PROP[i]
	else
		return PROP[1-i]

CN(compare&swap) = ∞

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.:

CS a compare&swap object init at ⊥

propose(v) :=
	tmp <- CS.compare&swap(⊥, v)
	if tmp = ⊥ then
		return v
	else
		return tmp

Exercise: devise a consensus object with CN = ∞ by using the compare&swap that returns booleans.