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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 everyS_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.
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’)
- 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