2014
DOI: 10.1007/978-3-319-14472-6_27
|View full text |Cite
|
Sign up to set email alerts
|

Space- and Time-Efficient Long-Lived Test-And-Set Objects

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
3
2

Citation Types

0
7
0

Year Published

2015
2015
2024
2024

Publication Types

Select...
3
1

Relationship

2
2

Authors

Journals

citations
Cited by 4 publications
(7 citation statements)
references
References 22 publications
0
7
0
Order By: Relevance
“…Applying this to the construction described in the proof of Theorem 2, yields a long-lived TAS object implemented from O(n log n) registers, where the expected step complexity of the test&set() operation is O(log * n) and worst-case step complexity of the reset() operation is O(1). Previously, the best space bound for achieving similar step complexities for test&set() and reset() was O(n 3/2 ) registers, achieved by, again, combining previous results [2,11]. The space lower bound for mutual exclusion [8] implies that any long-lived TAS implementation requires at least n registers.…”
Section: Introductionmentioning
confidence: 80%
See 3 more Smart Citations
“…Applying this to the construction described in the proof of Theorem 2, yields a long-lived TAS object implemented from O(n log n) registers, where the expected step complexity of the test&set() operation is O(log * n) and worst-case step complexity of the reset() operation is O(1). Previously, the best space bound for achieving similar step complexities for test&set() and reset() was O(n 3/2 ) registers, achieved by, again, combining previous results [2,11]. The space lower bound for mutual exclusion [8] implies that any long-lived TAS implementation requires at least n registers.…”
Section: Introductionmentioning
confidence: 80%
“…Since B has three registers, after any clean-sweep on B, a subsequent clean-sweep on B requires (at least) two processes to over-write the previous clean-sweep. These over-writers must have signature s, because, otherwise, an over-writer has a signature different from that of A and would signature-lose on B, implying that I already has an associated losing process via (2). If there are two processes with signature s then I can have at most one clean-sweep, and if there are three processes then I can have at most two clean-sweeps.…”
Section: Intuition For Correctness Proofmentioning
confidence: 99%
See 2 more Smart Citations
“…The reset() operation can only be executed by a process if its preceding operation on the object was a successful test&set(); in that case the reset() operation unconditionally resets the value of the TAS object to 0. Recently, Aghazadeh and Woelfel [3] showed that any TAS object implemented from m ℓ-bit registers can be transformed into a long-lived TAS object, using O(m • n) registers of size max{ℓ, log(n + m)} + O(1) bits. A reset() operation takes only constant time in the worstcase, and the step complexity of a test&set() operation of the long-lived object is the same (up to a constant additive term) as the one of the (one-shot) TAS object.…”
Section: Introductionmentioning
confidence: 99%