2012 | OriginalPaper | Buchkapitel
When and How Process Groups Can Be Used to Reduce the Renaming Space
verfasst von : Armando Castañeda, Michel Raynal, Julien Stainer
Erschienen in: Principles of Distributed Systems
Verlag: Springer Berlin Heidelberg
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Considering the
M
-renaming problem and process groups, this paper investigates the following question: Is there a relation between the number of groups and the size of the new name space
M
? This question can be rephrased as follows: Can the initial partitioning of the processes into
m
groups allows the size of the renaming space
M
to be reduced, and if yes, how much?
This paper answers the previous questions. Let
n
denote the number of processes. Assuming that the processes are initially partitioned into
m
=
n
− ℓ non-empty groups, such that each process knows only its identity and its group number, the paper first presents a wait-free
M
-renaming algorithm whose size of the new name space is
M
=
n
+ 2ℓ − 1. For
$\frac{n}{2} < m \leq n-1$
(i.e.
$1\leq \ell < \frac{n}{2}$
), we have
M
< 2
n
− 1, which shows that, when the number of groups is greater than
$\frac{n}{2}$
, groups allow to circumvent the renaming lower bound in read/write systems. Then, on the lower bound size, the paper shows that there are pairs of values (
n
,
m
) such that there is no read/write wait-free
M
-renaming algorithm for which
M
≤ 2
n
− 2. This impossibility result breaks our hope to have a renaming algorithm providing a new name space whose size would decrease “regularly” as the number of groups increases from 1 to
n
. Finally, the paper considers the case where each group includes at least
s
processes. This algorithm shows that, when
m
is such that
$\frac{n}{s+1}< m < \frac{n}{s}$
, there is an
M
-renaming algorithm where
M
= 3
n
− (
s
+ 1)
m
− 1 =
n
(2 −
s
) + (
s
+ 1)ℓ − 1. Hence, the paper leaves open the following question: For any
n
and
s
= 1, does the predicate
$m > \frac{n}{2}$
define a threshold on the number of groups which allows the 2
n
− 2 lower bound on the renaming space size to be bypassed?