GIAN Course on Distributed Network Algorithms. Distributed Objects
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1 GIAN Course on Distributed Network Algorithms Distributed Objects Stefan T-Labs, 2011
2 GIAN Course on Distributed Network Algorithms Distributed Objects How to access and manipulate data in an efficient and consistent manner? Yet another application of spanning trees! Stefan T-Labs, 2011
3 A fundamental challenge in distributed systems: coordinated access to a shared resource / object. read-write read-write Challenge: conditional modifications require mutual exclusion: avoid race conditions.
4 A fundamental challenge in distributed systems: coordinated access to a shared resource / object. Simple solution: get and lock file read-write read-write Challenge: conditional modifications require mutual exclusion: avoid race conditions.
5 How to implement consistent read/write access in a network?
6 Naive Solution: Idea: just keep a single copy at some leader node!
7 Naive Solution: Idea: just keep a single copy at some leader node! read/write remotely read/write remotely Access: 1. send message to leader, 2. leader processes request, 3. result sent back down the tree.
8 Problem 1: How to find leader? Idea: just keep a single copy at some leader node! read/write remotely read/write remotely Access: 1. send message to leader, 2. leader processes request, 3. result sent back down the tree.
9 Problem 1: How to find leader? Idea: Pre-define and search along a spanning tree: root! read/write remotely Idea: just keep a single copy at some leader node! read/write remotely Access: 1. send message to leader, 2. leader processes request, 3. result sent back down the tree.
10 Naive Solution: read/write remotely Problem 2: request costs Each request costs O(Diameter): O(Diameter): Inefficient if multiple Inefficient if multiple requests from requests from the same user! the same user! Better to move Better to move object close to object close to where it is needed! where it is needed! read/write remotely Access: 1. send message to leader, 2. leader processes request, 3. result sent back down the tree.
11 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree.
12 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree but allow users to get copy! Assume: Access by Bob
13 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree but allow users to get copy! Assume: Access by Bob
14 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree but allow users to get copy! What if Alice wants access?
15 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree but allow users to get copy! Home coordinates access: informs Alice when object free. What if Alice wants access?
16 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree but allow users to get copy! next in line: Home coordinates access: informs Alice when object free. What if Alice wants access?
17 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree. Once Bob releases access, Alice can get it! next in line: What if Alice wants access?
18 Better Solution: Idea: define home location of object: e.g., at root of a spanning tree. Once Bob releases access, Alice can get it! What if Alice wants access?
19 Everything great? Idea: define home location of object: e.g., at root of a spanning tree. Problem: home-based solution inefficient if Alice and Bob are close: triangular routing!
20 Everything great? Idea: define home location of object: e.g., at root of a spanning tree. Idea: make the root/home dynamic: home = current user! Problem: home-based solution inefficient if Alice and Bob are close: triangular routing!
21 Everything great? Idea: define home location of object: e.g., at root of a spanning tree. Idea: make the root/home dynamic: home = current user! Problem: home-based solution inefficient if Alice and Bob are close: triangular routing! The Arrow Protocol!
22 Everything great? Idea: define home location of object: e.g., at root of a spanning tree. Idea: make the root/home dynamic: home = current user! Problem: home-based solution inefficient if Alice and Bob are close: triangular routing! The Arrow Protocol! Based on link reversals: remember MST!
23 The Arrow Protocol Idea maintain invariant: directed spanning tree pointing to the object. In the Beginning 23
24 The Arrow Protocol Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! At Runtime 24
25 The Arrow Protocol Find object is easy due to invariant: simply route along directed spanning tree edges! Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! At Runtime 25
26 The Arrow Protocol Find object is easy due to invariant: simply route along directed spanning tree edges! Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! During lookup, invert directions of spanning tree edges! At Runtime 26
27 The Arrow Protocol Find object is easy due to invariant: simply route along directed spanning tree edges! Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! During lookup, invert directions of spanning tree edges! At Runtime 27
28 The Arrow Protocol Find object is easy due to invariant: simply route along directed spanning tree edges! Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! During lookup, invert directions of spanning tree edges! At Runtime 28
29 The Arrow Protocol Find object is easy due to invariant: simply route along directed spanning tree edges! Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! During lookup, invert directions of spanning tree edges! At Runtime 29
30 The Arrow Protocol Find object is easy due to invariant: simply route along directed spanning tree edges! Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! At Runtime Sweet! Spanning tree rooted at object location again. 30
31 The Arrow Protocol Find object is easy due to invariant: simply route along directed spanning tree edges! Idea maintain invariant: directed spanning tree pointing to the object. I need write lock! But what about concurrency!? At Runtime Sweet! Spanning tree rooted at object location again. 31
32 The Arrow Protocol: Concurrency Still in use by someone else! I want access! In the Beginning
33 The Arrow Protocol: Concurrency Still in use by someone else! I want access! In the Beginning
34 The Arrow Protocol: Concurrency However, like before: lookup, invert directions of spanning tree edges! Still in use by someone else! Phase 1
35 The Arrow Protocol: Concurrency However, like before: lookup, invert directions of spanning tree edges! Still in use by someone else! Next in line: Phase 1
36 The Arrow Protocol: Concurrency Still in use by someone else! Next in line: What if: I want access too! Phase 2
37 The Arrow Protocol: Concurrency Here the magic happens: gets routed to Bob! Which is before Alice in queue! Still in use by someone else! Next in line: What if: I want access too! Phase 2
38 The Arrow Protocol: Concurrency Here the magic happens: gets routed to Bob! Which is before Alice in queue! Still in use by someone else! Next in line: Invert pointers again! Phase 2
39 The Arrow Protocol: Concurrency Here the magic happens: gets routed to Bob! Which is before Alice in queue! Still in use by someone else! Next in line: Invert pointers again! Next in line: Phase 2
40 The Arrow Protocol: Concurrency Here the magic happens: gets routed to Bob! Which is before Alice in queue! Still in use by someone else! Next in line: And so on! Alice will be the next root, users will directed Invert pointers again! to her: A distributed queue implementation! Next in line: 40
41 Remark: Of course, once object becomes free, it can be sent to Bob directly. (No link reversals, no spanning tree.) The Arrow Protocol: Concurrency Still in use by someone else! Next in line: Next in line: 41
42 Arrow invert edge! wait myself? 42
43 Analysis Arrow - Message will only travel on static tree! - And can never traverse an edge twice (in opposite direction). Arrow is correct: find() terminates with message and time complexity D, where D is the diameter of the spanning tree. Completely asynchronous and concurrent environments! Proof. - Each edge {u,v} in the spanning tree is in one of four states: (A) u points to v, no message on the edge, v does not point to u (B) Message on the move from u to v (no pointer along edge) (C) v points to u, no message on edge, u does not point to v (D) Message on the move from v to u (no pointer along edge) QED 43
44 End of Lecture 44
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