How Shall I Name Thee?

Let me count the ways...

Hagit Attiya, TechnionArmando Castañeda, TechnionMaurice Herlihy, BrownAmi Paz, Technion

2

M-Renaming

Unboundedidentifiers

Unique namesin 1,…,M

5

8

2

Renaming

p1

p2

p3

p1

p2

p3

3

M-Renaming Bounds

1,...,nM 2n-1,... 1,...,n 2n-1,... n+1,... 2n-2

n is a primepower

n is nota primepower ?

4

Weak Symmetry Breaking (WSB)

0 / 1 outputsnot all 0’s or all 1’s

WSB

p1

p2

p3

p1

p2

p3

WSB (2n-2)-renaming

WSB solvable n is a prime power

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Model

n asynchronous processes, communicating by atomic read / write operationsWait-free algorithms: a process outputs after a finite number of steps, regardless of the others

...p1 p2 p3 pn

read write

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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

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

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• Indistinguishable executions: , if the process p has the same view in and in

IS Executions

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• Indistinguishable executions: , if the process p has the same view in and in

• A process is seen in if it appears in some other process' view; otherwise, it is unseen

IS Executions

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AR Lemma

• Indistinguishable executions: , if the process p has the same view in and in

• A process is seen in 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

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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

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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

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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

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Univalued Signed Count (USC)For an IS execution

• Odd-sized blocks do not affect the sign• AR Lemma gives executions with opposite signs

∑¿ only   0   i s  output  in  α

𝒔𝒊𝒈𝒏 (𝜶 )+(−𝟏 )𝒏∗ ∑¿o nly  1   i s  output   in   α

𝒔𝒊𝒈𝒏 (𝜶 )

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Proof StrategyProve 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!

∑¿ only   0   i s  output  in  α

𝒔𝒊𝒈𝒏 (𝜶 )+(−𝟏 )𝒏∗ ∑¿o nly  1   i s  output   in   α

𝒔𝒊𝒈𝒏 (𝜶 )

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: Trimming an Algorithm A

Write your input and scanIf all processes have arrived, output 0Repeat:

Simulate a step of A, write and scanIf all processes arrive, output 1If A outputs, output the same

When all processes show up, there is little uncertainty Outcome is determined by the partial execution before all processes show up

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: Trimming an Algorithm A

Write your input and scanIf all processes have arrived, output 0Repeat:

Simulate a step of A, write and scanIf all processes arrive, output 1If A outputs, output the same

A and have the same univalued signed count

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Proof of TrimmingA and have the same univalued signed count

(1, … ,1,0,0,…,0)

(0, … ,0,…,0)

(1, … ,1,…,1)(1, … ,1,1,0,…,0)

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Proof of TrimmingA and have the same univalued signed count

are all tuples s.t. • output 1 in • output 0 in

signed count of =

(1, … ,1,?,0,…,0)

(0, … ,0,…,0)

(1, … ,1,…,1)

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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= ∑(𝜶 ,𝒑𝒊 )∈𝑿 𝑨

(−𝟏)𝒊 ∙ 𝒔𝒊𝒈𝒏(𝜶)

USC= ∑¿ only   0   i s  output   in   α

𝒔𝒊𝒈𝒏 (𝜶 )+(−𝟏 )𝒏∗ ∑¿ only  1  i s  output  in  α

𝒔𝒊𝒈𝒏 (𝜶)

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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= ∑(𝜶 ,𝒑𝒊 )∈𝑿 𝑨

(−𝟏)𝒊 ∙ 𝒔𝒊𝒈𝒏(𝜶)

USC= ∑¿ only   0   i s  output   in   α

𝒔𝒊𝒈𝒏 (𝜶 )+(−𝟏 )𝒏∗ ∑¿ only  1  i s  output  in  α

𝒔𝒊𝒈𝒏 (𝜶)

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Signed Count of and ( seen)

is seen in : Cancel out by AR Lemma.Same holds for

Signed  count= ∑(𝜶 ,𝒑𝒊 )∈𝑿 𝑨

(−𝟏)𝒊 ∙ 𝒔𝒊𝒈𝒏(𝜶)

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Signed Count of and ( unseen)

is unseen in : 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

Signed  count= ∑(𝜶 ,𝒑𝒊 )∈𝑿 𝑨

(−𝟏)𝒊 ∙ 𝒔𝒊𝒈𝒏(𝜶)

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No execution in which only 1 is output (last process to start always outputs 0)

What is the USC of ?

Write your input and scanIf all processes have arrived, output 0Repeat:

Simulate a step of A, write and scanIf all processes arrive, output 1If A outputs, output the same

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What is the USC of ?

No execution in which only 1 is output (last process to start always outputs 0)

Write your input and scanIf all processes have arrived, output 0Repeat:

Simulate a step of A, write and scanIf all processes arrive, output 1If A outputs, output the same

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What is the USC of ?

• All processes output 0 if they arrive together

Write your input and scanIf all processes have arrived, output 0Repeat:

Simulate a step of A, write and scanIf all processes arrive, output 1If A outputs, output the same

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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

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Symmetric Algorithms

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

p3

p1 p1

p2 p3

p1

p1

p2

p4

p4

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What is the USC of ? ∑

only  0  in  α𝒔𝒊𝒈𝒏 (𝜶 )+(−𝟏 )𝒏∗ ∑

only  1  in  α𝒔𝒊𝒈𝒏 (𝜶 )

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

USC mod USC

The Upper Bound

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Topology 101: Simplex

• Set of vertexes

Dimension: Number of vertexes -1 A face is a sub-simplex• Chromatic: vertexes have different colors

{𝑥 } {𝑥 , 𝑦 } {𝑥 , 𝑦 ,𝑧 } {𝑥 , 𝑦 ,𝑧 ,𝑤 }

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Topology 101: Simplicial Complexes

• Gluing simplexes together

• Some complexes are subdivisions of others

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A vertex represents a local state of a processA simplex represents a global system state

Distributed Computing Topology

zx

a

zx

y

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Distributed Computing Topology

Glue together indistinguishabile executions

zx

y

zx

aa

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Glue together indistinguishabile executions

Distributed Computing Topology

In the complex of IS executions, a simplex shares each face (of dimension n) with one simplex

If is seen in an IS execution by P, then there is a unique IS execution by P s.t.

Also, is seen in

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Distributed Computing Topology

Glue together indistinguishabile executionsInitial 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

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IS Executions as the Standard Subdivision: 2 Processes

• Execution:

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IIS Executions as the Standard Subdivision: 2 Processes

• Execution:

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IIS Executions as the Standard Subdivision: 3 Processes

• Execution:

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IIS Executions as the Standard Subdivision: 3 Processes

• Execution:

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IIS Executions as the Standard Subdivision: 3 Processes

• Execution:

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Standard Subdivision (Std S)

StdK S = apply the standard subdivision K times

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Iterated Immediate Snapshot

Each process starts on a corner of Stdk S After iterations of immediate snapshot end on a

simplex, spanned by the participating processes

step complexity

steps per iteration [Borowsky Gafni]

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Solving WSBDo 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

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Output Function for WSB

Binary values

No univalued n-simplexes

Symmetric onthe boundaries

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Binary values

No univalued n-simplexes

Symmetric onthe boundaries

Output Function for WSB

An output function from StdK S gives a WSB algorithm with O(Kn2) step complexity

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Two-Step Plan (and a Detour)

Step 1: find k and a (boundary symmetric) output function from StdK S whose univalued signed count is 0 (only 0-univalued n-simplexes)

Step 2: remove the univalued n-simplexes in StdK S, while preserving symmetry

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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

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Step One: Univalued Signed Count

Std S USC  = 1+∑𝑖=1

𝑛− 1

(𝑛𝑖 )𝑘𝑖=0

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Step One: Univalued Signed Count

Std S

Symmetrically pick simplexes in every subdivision of

a face of dimension i

USC  = 1+∑𝑖=1

𝑛−1

(𝑛𝑖 )𝑘𝑖=0

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Step One: Univalued Signed Count

Std S

Divide and then extend

USC  = 1+∑𝑖=1

𝑛−1

(𝑛𝑖 )𝑘𝑖=0

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Step One: Univalued Signed Count

USC  = 1+∑𝑖=1

𝑛−1

(𝑛𝑖 )𝑘𝑖=0

Divide and then extend

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Step One: Univalued Signed Count

USC  = 1+∑𝑖=1

𝑛− 1

(𝑛𝑖 )𝑘𝑖=0

Divide and then extend Not

standard subdivision

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Step One: Univalued Signed Count

ki univalued simplexes from each face of

dimension i

USC  = 1+∑𝑖=1

𝑛−1

(𝑛𝑖 )𝑘𝑖=0

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Step One: Univalued Signed Count

ki univalued simplexes from each face of

dimension iColor the rest blue

USC  = 1+∑𝑖=1

𝑛−1

(𝑛𝑖 )𝑘𝑖=0

Some ki are negative some simplexes are counted as -1

Using small (k1, ... ,kn-1) Step one requires O(1) subdivisions

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Step Two: Remove Univalued Subdivide Std S further to eliminate all

univalued n-simplexes

Recursive algorithm that cancels a simplex counted +1 with a simplex counted -1

While preserving symmetry on the boundary

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Step Two: Adjacent Simplexes

• Eliminate univalued simplexes by subdivision, while preserving the boundary

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Step Two: Non-Adjacent Simplexes

• Eliminate univalued simplexes by subdivision, while preserving the boundary

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• Eliminate univalued simplexes by subdivision, while preserving the boundary

• Similar construction• subdivisions

for -length path.

Step Two: Longer Paths

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Step Two: Path Length

• Cannot eliminate simplexes with odd distance• Take univalued simplexes with opposite signsAR Lemma a path between them has even length

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Step Two: Connect

Find a path between univalued simplexes with opposite signs

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Step Two: Rinse

Subdivide until the path has no univalued simplexes

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Step Two: Rinse & Repeat

Subdivide until the path has no univalued simplexes

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Step Two: Rinse & Repeat

Until there are no univalued simplexes

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Step Two: Number of Subdivisions

Chop into paths with O(n) length, using a constant number of subdivisions

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Step Two: Number of Subdivisions

Chop into paths with O(n) length, using a constant number of subdivisions

O(n)O(n)

O(n)

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Step Two: Number of Subdivisions

Path of O(n) length is treated with O(n) iterationsEach iteration partitions a path P into paths of length |P| - 2, using O(1) subdivisions

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Step Two: Number of Subdivisions

Path of O(n) length is treated with O(n) iterationsEach iteration partitions a path P into paths of length |P| - 2, using O(1) subdivisions Two univalued simplexes are eliminated within O(n) subdivisions

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Step Two: Number of Subdivisions

Path of O(n) length is treated with O(n) iterationsEach iteration partitions a path P into paths of length |P| - 2, using O(1) subdivisions Two univalued simplexes are eliminated within O(n) subdivisions

In O(n) subdivisions, two univalued simplexes are eliminated, regardless of the path’s length

From (k1, ... ,kn-1) in step one: O(nq+2) paths in Std STotal number of subdivisions: O(nq+3)

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We Used Cone Subdivisions

How to get an algorithm?

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Getting an Algorithm

Map standard subdivision to cone subdivisionLift the output function

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Getting an Algorithm

Map simplexes to simplexes, preserving• process identifiers• structure of the subdivision

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Getting an Algorithm

Iterated immediate snapshots, for an appropriate number of times, then output the associated value

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Wrap Up

Step 1: symmetric subdivision, zero univalued signed count– subdivisions

Step 2: eliminate pairs of univalued simplexes, while preserving symmetry– subdivisions

Step 3: map from standard subdivision– No subdivision

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To be Continued…My course next semester: 236755

Thesis topics