How Shall I Name Thee? Let me count the ways...

How Shall I Name Thee?Let me count the ways... Hagit Attiya, Technion Armando Castañeda, Technion Maurice Herlihy, BrownAmi Paz, Technion

M-Renaming p1 p1 5 Renaming p2 p2 8 p3 p3 2 Unboundedidentifiers Unique namesin 1,…,M

M-Renaming Bounds n+1,... 2n-1,... 2n-1,... M 1,...,n 1,...,n 2n-2 n is a primepower ? n is nota primepower

Weak Symmetry Breaking (WSB) p1 p1 WSB p2 p2 p3 p3 0 / 1 outputs not all 0’s or all 1’s WSB  (2n-2)-renaming WSB solvable  n is a prime power

Model • ... • p1 • p2 • p3 • pn nasynchronousprocesses, communicating by atomic read / write operations Wait-freealgorithms: a process outputs after a finite number of steps, regardless of the others read write

Today • Impossibility of WSB when n is a prime power • n-process algorithm for WSB when n is not a prime power, with step complexity(is the largest prime factor of )  Same results for -renaming

Immediate Snapshots (IS) An execution is a sequence of blocks: sets of processes that • Write together and then • Scan (read everything) together

IS Executions • Indistinguishableexecutions: , if the process p has the same view in and in

IS Executions • Indistinguishableexecutions: , if the process p has the same view in and in • A process is seenin if it appears in some other process' view; otherwise, it is unseen

AR Lemma • Indistinguishableexecutions: , if the process p has the same view in and in • A process is seenin if it appears in some other process' view; otherwise, it is unseen • If is seen in an IS execution by P, then there is a unique IS execution by P s.t. • Also, is seen in

AR Lemma: Proof (Case 1) Consider the last seen round of • If is seen in an IS execution by P, then there is a unique IS execution by P s.t. • Also, is seen in

AR Lemma: Proof (Case 2) Consider the last seen round of • If is seen in an IS execution by P, then there is a unique IS execution by P s.t. • Also, is seen in

Pairing AR Lemma means that the IS executions by P in which is seen can be divided into pairs • If is seen in an IS execution by P, then there is a unique IS execution by P s.t. • Also, is seen in

The Lower Bound

Univalued Signed Count (USC) For an IS execution • Odd-sized blocks do not affect the sign • AR Lemma gives executions with opposite signs

Proof Strategy Prove that the USC of a WSB algorithm is • Trim the algorithm, preserving the USC • Prove the trimmed algorithm has USC The algorithm has a univalued IS execution!

: Trimming an Algorithm A When all processes show up, there is little uncertainty  Outcome is determined by the partial execution before all processes show up Write your input and scan If all processes have arrived, output 0 Repeat: Simulate a step of A, write and scan If all processes arrive, output 1 If A outputs, output the same

: Trimming an Algorithm A • A and have the same univalued signed count Write your input and scan If all processes have arrived, output 0 Repeat: Simulate a step of A, write and scan If all processes arrive, output 1 If A outputs, output the same

Proof of Trimming • A and have the same univalued signed count (0, … ,0,…,0) (1, … ,1,0,0,…,0) (1, … ,1,1,0,…,0) (1, … ,1,…,1)

Proof of Trimming • A and have the same univalued signed count are all tuples s.t. • output 1 in • output 0 in signed count of = (0, … ,0,…,0) (1, … ,1,?,0,…,0) (1, … ,1,…,1)

Signed Count = USC (part 1) {0,1} is output in Output vector = (1,…,1,0,..,0) • s.t. • Cancel out in the signed count

Signed Count = USC (part 2) {0} or {1} is output in • appears in the USC, and with the same sign Same holds for

Signed Count of and (seen) is seen in : Cancel out by AR Lemma. Same holds for

Signed Count of and (unseen) is unseenin : Let be an execution of induced by the same blocks (until runs alone) = , all but decide the same  • and have the same signed count • and have the same USC

What is the USC of ? No execution in which only 1 is output (last process to start always outputs 0) Write your input and scan If all processes have arrived, output 0 Repeat: Simulate a step of A, write and scan If all processes arrive, output 1 If A outputs, output the same

What is the USC of ? • All processes output 0 if they arrive together Write your input and scan If all processes have arrived, output 0 Repeat: Simulate a step of A, write and scan If all processes arrive, output 1 If A outputs, output the same

What is the USC of ? • All processes output 0 if they arrive together • But there could be other executions in which all processes output 0 • Need to assume the algorithm is symmetric

Symmetric Algorithms • p4 • p4 • p3 • p3 • p2 • p2 E.g., the same outputs in an execution by p1, p2, p3 and the equivalent execution with p1, p2, p4 Equivalent executions have the same sign • p1 • p1 • p1 • p1

What is the USC of ? One execution contributes 1 Take a univalued execution in which processes output 0 in A equivalent executions (with same sign) in which all processes output 0 When , they contribute 0 mod to the USC  USC mod  USC

The Upper Bound

Topology 101: Simplex • Set of vertexes Dimension: Number of vertexes -1 A faceis a sub-simplex • Chromatic: vertexes have different colors

Topology 101: Simplicial Complexes • Gluing simplexestogether • Some complexes are subdivisionsof others

Distributed Computing  Topology A vertex represents a local state of a process A simplex represents a global system state y x z x z a

Distributed Computing  Topology Glue together indistinguishabileexecutions y x z x z a a

Distributed Computing  Topology Glue together indistinguishabileexecutions • If is seen in an IS execution by P, then there is a unique IS execution by P s.t. • Also, is seen in • In the complex of IS executions, a simplex shares each face (of dimension n) with one simplex

Distributed Computing  Topology Glue together indistinguishabileexecutions Initial configurations, glued together, make the input complex The IIS executions correspond to the standard (chromatic) subdivision of the input complex • In the complex of IS executions, a simplex shares each face (of dimension n) with one simplex

IS Executions as the Standard Subdivision: 2 Processes • Execution:

IIS Executions as the Standard Subdivision: 2 Processes • Execution:

IIS Executions as the Standard Subdivision: 3 Processes • Execution:

Standard Subdivision (Std S) StdKS = apply the standard subdivision Ktimes

Iterated Immediate Snapshot • Each process starts on a corner of StdkS • After iterations of immediate snapshot end on a simplex, spanned by the participating processes steps per iteration [Borowsky Gafni] step complexity

Solving WSB Do iterated immediate snapshots & output the value associated with the vertex (its output function) • WSB is solved by finding and an output function from Stdk S

Output Function for WSB Binary values No univalued n-simplexes Symmetric onthe boundaries

Output Function for WSB Binary values No univalued n-simplexes Symmetric onthe boundaries An output function from StdK Sgives a WSB algorithm with O(Kn2) step complexity

Two-Step Plan (and a Detour) Step 1:findk and a (boundary symmetric) output function from StdKSwhose univalued signed count is 0 (only 0-univalued n-simplexes) Step 2:remove the univalued n-simplexes in StdK S, while preserving symmetry

Step One: Zero USC • Number of subsets with processes: • If n is not a prime power, these quantities are relatively prime •  there are small such that

Step One: Univalued Signed Count Std S

How Shall I Name Thee? Let me count the ways...