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Manifold Protocols

Companion slides forDistributed Computing

Through Combinatorial TopologyMaurice Herlihy & Dmitry Kozlov & Sergio Rajsbaum

2

Kinds of Results

Impossibility

Separation

Task T cannot be solvedin model M

Task T can be solved by a protocol for T’, but not vice-versa

Distributed Computing Through Combinatorial Topology

3

Road MapManifolds

Sperner’s Lemma and k-Set Agreement

Immediate Snapshot Model

Weak Symmetry-Breaking

Separation results


4

Road MapManifolds




Separation results


5

Simplicial Complex


6

Manifolds

Every (n-1)-simplex aface of two n-simplexes


7

Manifolds

Every (n-1)-simplex aface of two n-simplexes

technically, a pseudo-manifold


8Distributed Computing Through Combinatorial Topology

Manifold with BoundaryInternal (n-1)-simplex aface of two n-simplexes

Boundary (n-1)-simplex aface of one n-simplex

Boundary C


9

Why Manifolds?

Nice combinatorial properties

Many useful theorems

Easy to prove certain claims

True, most complexes are not manifolds ….

Still a good place to start.

10

Road MapManifolds




Separation results



11

Immediate Snapshot Executions

Restricted form of Read-Write memory

Protocol complexes are manifolds

(Equivalent to regular R-W memory)But we will not prove it yet.


0 0 0

1

Write

Single-writer, multi-reader variables12


1 0 0

100Snapshot

Single-writer, multi-reader variables13


14

Immediate Snapshot Executions

Pick a set of processes

write together

snapshot together

Repeat with another set

Example Executions

P Q R

write

snap

write

snap

write

snap

P?? PQ? PQR

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Each process writes, then takes snapshot

Each process has a view

time


Example Executions

P Q R

write

snap

write write

snap snap

write

snap

P?? PQR PQR

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Moving last processone round earlier,

Changes this view from PQ? to PQR

P Q R

write

snap

write

snap

write

snap

P?? PQ? PQR


P Q R

write write write

snap snap snap

write write

snap snap

PQR PQR PQR

P Q R

write write write

snap snap snap

write write

snap snap

PQR PQR PQR

P Q R

write

snap

write write

snap snap

write

snap

P?? PQR PQR

Example Executions

17

Moving last processesone round earlier,

Changes this view from P?? to PQRDistributed Computing Through

Combinatorial Topology


18

Protocol ComplexProcess (color)

& view


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Protocol ComplexP Q R

write

snap

write

snap

write

snap

P Q R

write write write

snap snap snap

P Q R

write

snap

write write

snap snap

20

Standard Chromatic Subdivision



21

Vertex (Pi, ¾i)

Combinatorial Definition (I)

Process name view ¾i µ ¾

Input simplex ¾

Protocol complex Ch(¾)


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Combinatorial Definition (II)

Each process's write appears in its own view

Pi 2 names(¾i).

Snapshots are ordered

¾i µ ¾j or vice-versa

Each snapshot ordered immediately after writeif Pi 2 names(¾j), then ¾i µ ¾j.

If Pj saw Pi write … then Pj‘s snapshot not later than Pj‘s.


23

Manifold Theorem

If the input complex I is a manifold …


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Manifold Theorem

… so is the immediate snapshot protocol complex Ch(I).


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Proof of Manifold Theorem

Let ¿n-1 be an (n-1)-simplex of Ch(I).

Must prove that ¿n-1 is a face of exactly one or two n-simplexes of Ch(I).


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Without Loss of Generality …

¿n¡ 1 = fhP0;¾0i ; : : :;hPn¡ 1;¾n¡ 1ig

Simplex of I seen by last process

Where ¾i µ ¾i+1 for 0 · i < n

Re-index processes in execution order(ignoring ties)


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Proof Strategy

Count the ways we can extend…

Where ¾n is an n-simplex of Ch(¾)

¿n-1 = {(P0, ¾0), …, (Pn-1, ¾n-1)}

¿n = {(P0, ¾0), …, (Pn-1, ¾n-1i, (Pn, ¾n)}


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Cases

¿n¡ 1 = fhP0;¾0i ; : : :;hPn¡ 1;¾n¡ 1ig

dim n-1 or dim n

Boundary or internal

3 cases


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¿ n-

1

¾n-

1

¿n¡ 1 = fhP0;¾0i ; : : :;hPn¡ 1;¾n¡ 1ig Boundary(n-1)-simplex

Case One


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Exactly one n-simplex ¾ of I contains ¾n-1

¾

¾n-

1

¿n-1

Case One


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Exactly one n-simplex ¿n of Ch(I) contains ¿n-1

¿n = ¿n-1 [ hPn, ¾i

¾

Case One


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¿n-1

¾n-1

¿n¡ 1 = fhP0;¾0i ; : : :;hPn¡ 1;¾n¡ 1ig Internal(n-1)-simplex

Case Two


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Exactly two n-simplexes ¾0 & ¾1 of I contain

¾n-1

¾0

¾1

Case Two


34

Exactly 2 n-simplexes ¿0n & ¿1

n of Ch(I) contain ¿n-1

¿n0 = ¿n-1 [ hPn, ¾0i

¿n1 = ¿n-1 [ hPn, ¾1i

¾0

¾1

Case Three


35

¿n-1

¾n-1

¿n¡ 1 = fhP0;¾0i ; : : :;hPn¡ 1;¾n¡ 1ig n-simplexCase Three

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Suppose

Pn 2 names(¾k)(Pk saw Pn’s write)

Pn names(¾k-1)(Pk-1 did not see Pn’s write)

Special case: k=0, ¾-1 = ;

¾n µ ¾k)

¾k-1 ½ ¾n)

Either ¾k-1 ½ ¾n ½ ¾k or ¾k-1 ½ ¾n = ¾k

Case Three


37

Pk-1 did not see Pn

Pk did see Pn

Pn between Pk-1 & Pk

Pn with Pk

This completes the proof.

Case Three


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Manifold Protocols

A protocol M(I) is a manifold protocol if

If I is a manifold, so is M(I)

M(I) = M(I).

Example: immediate snapshot

Important: closed under composition

39

Road MapManifolds




Separation results


40

k-set Agreement: before

32 19

21

40

19 19

21

41

k-set Agreement: after

k = 2Distributed Computing Through



42

Theorem

No Manifold Protocol can solve n-set agreement

Including Immediate SnapshotIncluding read-write memory

43

Sperner Coloring


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Sperner Coloring


“Corners” have distinct colors

45

Sperner Coloring


Edge vertexes have corner colors


46

Sperner Coloring


Edge vertexes have corner colors


Every vertex has face boundary colors

47

Sperner’s Lemma


In any Sperner coloring, at least one n-simplex has all n+1 colors

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Sperner Coloring for Manifolds with Boundary (base)

One color hereOther color here

Only those colors in interior


49

Sperner Coloring for Manifolds with Boundary (inductive)

M is n-manifold colored with ¦

M = [i Mi

Where Mi is non-empty (n-1)-manifold Sperner-colored with ¦ n {i }


50

Sperner’s Lemma for Manifolds


M

Odd number of n-simplexes have all

n+1 colors

Sperner Coloring on Manifold

M

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Proof of Sperner’s Lemma

Induction on nBase

One color hereOther color here

Odd number of changesDistributed Computing Through


52

Induction Step


53

Dual Graph

One vertex per n-simplex …


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Dual Graph

One vertex per n-simplex …

One “external” vertex


55

Discard One Color

n+1 colors

“restricted”n colors


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Edges

If two n-simplexes share an (n-1)-face …with restricted n colors …

add an edge.


57

Edges

Same for external vertex


58

Some Vertexes have Degree Zero

Even degreeDistributed Computing Through


59

Some Vertexes have Degree Two

Even degree

Only n restricted colors


60

All n+1 Colors ) Degree 1

Odd degreeDistributed Computing Through


61

Induction Hypothesis

Odd number of n-colored simplexes


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Induction Hypothesis

External vertex has odd degree


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OddDegrees

One external vertex

Even

n-simplexes with n+1 colors?

All the rest


64

OddDegrees

One external vertex

Even


All the rest

Every graph

Even number of odd degree vertexes

sum of degrees is even…each edge contributes 2


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OddDegrees

One external vertex

Even


All the rest



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Parity

One external vertex


n-simplexes with n+1 colors?Even

Odd

Odd


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Parity

One external vertex


n-simplexes with n+1 colors?Even

Odd

Odd

QEDDistributed Computing Through


68

No Manifold Task can solve n-Set Agreement

Assume protocol exists: Run manifold task protocol

Choose value based on vertex

Idea:Color vertex with “winning”

process name …



Manifold Task for n-Set Agreement

manifold

Only P wins

Only Q and R win

Sperner coloring

70


manifold

Sperner coloring


71


manifold

Sperner coloring

n+1 colors

Contradiction: at most n can win

Execution with n+1 winners


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Road MapManifolds




Separation results



73


Group 0 Group 1

If all processes participate …

At least one process in each group


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Group 0 Group 1

If fewer participate …

we don’t care.

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Anonymous Protocols

Trivial solution: choose name parity

WSB protocol should be anonymous

Output depends on …Input …Interleaving …

But not name

Restriction on protocol, not task!

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Road MapManifolds




Separation results

77

Next Step

Construct manifold task that solves weak-symmetry-breaking

Because it is a manifold, it cannot solve n-set agreement

Separation: n-set agreement is harder than WSB


78

A Simplex


79

Standard Chromatic Subdivision


80

Glue Three Copies Together


81

Glue Opposite Edges


82

The Moebius Task


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Defines a Manifold Task


boundary boundary

boundary

84

Manifold Task


Note Sperner coloring on boundary

boundary boundary

boundary

85

Terminology


Each face has a central simplex

1-dim

2-dim

86

Subdivided Faces


external

internalinternal

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Black-and-White Coloring (I)


Central 2-simplex:

White near external face

Black elsewhere

88

Black-and-White Coloring (II)


Black

Central simplexof external face:

89

Black-and-White Coloring (III)


All others

White

90

Moebius Solves WSB


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Moebius Solves WSB


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Moebius Solves WSB


Every n-simplex has both black & white colors.

Boundary coloring is symmetric!

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General Construction

n+1 “copies” »0, … »n of Ch(¾)

»0 »n…

»1

(picture not exact!)

n = 2N


94


»0 »i-1…

n = 2N»i

»i+1

Face i is external

»k… …

»n




»0 »i-1…

n = 2N»i

»i+1

Face i is external

»k… …

»n

2N others



»0 »i-1…

n = 2N»i

»i+1 »k… …

»n

Identify Faces j

Rank j+N out of 2N

Face i is external



»0 »i-1…

n = 2N»i

»i+1 »k… …

»n

Symmetric!

Rank j+N out of 2N

Face i is external

Why it’s a(non-orientable)

manifold Why it works only for even dimensions

98

Open Problem

Generalize to odd dimensions …

or find counterexample.


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Black-and-White Coloring (I)


Central n-simplex:

White near external face

Black elsewhere

100

Black-and-White Coloring (II)


Black

Central m-simplexes

of external face for m > 0

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Black-and-White Coloring (III)


All others

White

102

No Monochromaticn-simplexes


Central n-simplexBy construction

103



Intersects internal face Black at n-center

White at edge

104



Intersects external face White at n-center

Black at m-center, 0 < m < n

105

Progress



Moebius Taskno

yes

Set Agreement

106

Next Step



Moebius Taskno

yes

Anonymous Set Agreement


107

Group 1

Choose name with anonymous

n-Set agreement

My namewritten? Group 0

no

· n+1

Write name

Read names

yes

Set Agreement WSB


108


n-Set agreement

Proof

Group 1


no

· n+1

Write name

Read names

yes

Set Agreement WSB

If all n+1 participate …First name written

joins Group 0Some name not chosen, it joins

Group 1


109


n-Set agreement

Proof

Group 1


no

· n+1

Write name

Read names

yes

Set Agreement WSB

Protocol is anonymous …Because we use anonymous

set agreement “black box”(Any set agreement protocol can be

made anonymous)

110

Conclusions

Some tasks harder than others …

n-set agreement solves weak-symmetry breaking

But not vice-versa


111

Remarks

Combinatorial and algorithmic arguments complement one

anotherCombinatorial: what we can’t do

Algorithmic: what we can do


112

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