Wait-Free Computability for General Tasks Companion slides for Distributed Computing Through Combinatorial Topology Maurice Herlihy & Dmitry Kozlov & Sergio Rajsbaum TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAA 1

Road Map Inherently colored tasks Solvability for colored tasks Protocol ) map Map ) protocol A Sufficient Topological Conditions

Star Review ¾ Star(¾,K) is the complex of facets of K containing ¾ Distributed Computing through Combinatorial Topology

Facet Review Facet not a facet A facet is a simplex of maximal dimension Distributed Computing through Combinatorial Topology

Review Open Star Staro(¾,K) union of interiors of simplexes containing ¾ Point Set Distributed Computing through Combinatorial Topology

Review Link Link(¾,K) is the complex of simplices of Star(¾,K) not containing ¾ Complex Distributed Computing through Combinatorial Topology

A simplicial map Á is rigid if Review A simplicial map Á is rigid if dim Á(¾) = dim ¾. 18-Nov-18

Road Map Inherently colored tasks Solvability for colored tasks Protocol ) map Map ) protocol A Sufficient Topological Conditions

The Hourglass task I O

Single-Process Executions ¢ R Q R,0 Q,0 I O

P and R only (P and Q Symmetric) P P ¢ R R I O

Q and R only ¢ I O

Hourglass satisfies conditions of fundamental theorem … Claim: Hourglass satisfies conditions of fundamental theorem … But has no wait-free immediate snapshot protocol! 18-Nov-18

Hourglass satisfies conditions of fundamental theorem … Claim: Hourglass satisfies conditions of fundamental theorem … But has no wait-free immediate snapshot protocol! 18-Nov-18

homotopy: |I |  |O| O I carried by ¢

Hourglass satisfies conditions of fundamental theorem … Claim: Hourglass satisfies conditions of fundamental theorem … But has no wait-free immediate snapshot protocol! 18-Nov-18

The Hourglass task solves 2-set agreement … Claim: The Hourglass task solves 2-set agreement … Which has no wait-free read-write protocol. 18-Nov-18

Write input value to announce array … Protocol: Write input value to announce array … Run Hourglass task … 18-Nov-18

Look in announce[] array … P P P Find non-null announce[] value P,1 R or Q R Q R Q Look in announce[] array …

What Went Wrong? f: |I|  |O|... carried by ¢ Theorem A colorless (I,O,¢) has a wait-free immediate snapshot protocol iff there is a continuous map … f: |I|  |O|... We will prove a simple theorem that completely characterizes when it is possible to solve a colorless task in wait-free read-write memory. The theorem states that a colorless task $(\cI^*,\cO^*,\Delta^*)$ has a wait-free read-write protocol if and only if there is a continuous map between their polyhedrons, carried by ¢ 18-Nov-18

One Direction is OK f: |I|  |O|... carried by ¢ Theorem If (I,O,¢) has a wait-free read-write protocol … then there is a continuous map … We will prove a simple theorem that completely characterizes when it is possible to solve a colorless task in wait-free read-write memory. The theorem states that a colorless task $(\cI^*,\cO^*,\Delta^*)$ has a wait-free read-write protocol if and only if there is a continuous map between their polyhedrons, f: |I|  |O|... carried by ¢ 18-Nov-18

The Other Direction Fails Theorem? If there is a continuous map … f: |I|  |O|... carried by ¢ … We will prove a simple theorem that completely characterizes when it is possible to solve a colorless task in wait-free read-write memory. The theorem states that a colorless task $(\cI^*,\cO^*,\Delta^*)$ has a wait-free read-write protocol if and only if there is a continuous map between their polyhedrons, then does (I,O,¢) have a wait-free IS protocol? 18-Nov-18

Simplicial approximation Review f: |I|  |O|... Simplicial approximation Á: BaryN I  O Repeated snapshot Protocol 18-Nov-18

Review f: |I|  |O|... Á: BaryN I  O Simplicial approximation Not color-preserving Á: BaryN I  O Repeated snapshot Another process’s output? Protocol 18-Nov-18

Road Map Inherently colored tasks Solvability for colored tasks Protocol ) map Map ) protocol A Sufficient Topological Conditions

How can we adapt this theorem to colored tasks? A colorless (I,O,¢) has a wait-free immediate snapshot protocol iff there is a continuous map … f: |I|  |O|... carried by ¢ We will prove a simple theorem that completely characterizes when it is possible to solve a colorless task in wait-free read-write memory. The theorem states that a colorless task $(\cI^*,\cO^*,\Delta^*)$ has a wait-free read-write protocol if and only if there is a continuous map between their polyhedrons, How can we adapt this theorem to colored tasks? 18-Nov-18

Fundamental Theorem for Colored Tasks (I,O,¢) has a wait-free read-write protocol iff … I has a chromatic subdivision Div I … & color-preserving simplicial map Á: Div I  O… We will prove a simple theorem that completely characterizes when it is possible to solve a colorless task in wait-free read-write memory. The theorem states that a colorless task $(\cI^*,\cO^*,\Delta^*)$ has a wait-free read-write protocol if and only if there is a continuous map between their polyhedrons, carried by ¢ 18-Nov-18

¢ Quasi-Consensus Q,1 P,1 Q,1 P,1 P,0 Q,0 P,0 Q,0 I O Each of $P_0$ and $P_1$ is given a binary input. If both have input $v$, then both must decide $v$. If they have mixed inputs, then either they agree, or $P_1$ may decide $0$ and $P_0$ may decide 1 (but not vice-versa). P,0 Q,0 P,0 Q,0 I O

¢ Quasi-Consensus Q,1 P,1 Q,1 P,1 P,0 Q,0 P,0 Q,0 I O Each of $P_0$ and $P_1$ is given a binary input. If both have input $v$, then both must decide $v$. If they have mixed inputs, then either they agree, or $P_1$ may decide $0$ and $P_0$ may decide 1 (but not vice-versa). P,0 Q,0 P,0 Q,0 I O

¢ Quasi-Consensus P,0 Q,0 P,1 Q,1 Q,1 P,1 P,0 Q,0 I O Each of $P_0$ and $P_1$ is given a binary input. If both have input $v$, then both must decide $v$. If they have mixed inputs, then either they agree, or $P_1$ may decide $0$ and $P_0$ may decide 1 (but not vice-versa). P,0 Q,0 I O

¢ Not a colorless task! Quasi-Consensus P,0 Q,0 P,1 Q,1 Q,1 P,1 P,0

Quasi-Consensus P,0 Q,0 P,1 Q,1 Q,1 P,1 ¢ P,0 Q,0 I O

¢ No simplicial map: I  O carried by ¢ Quasi-Consensus P,0 Q,0 P,1 It is easily seen that there is no simplicial map carried by $\Delta$ from the input complex to the output complex. The vertexes of input simplex $\set{ang{P,0},\ang{Q,1}}$ $\ang{P,0}$ and $\ang{Q,1}$, but there is no single output simplex containing both vertexes. P,0 Q,0 No simplicial map: I  O carried by ¢ I O

Q,1 P,1 Q,1 P,1 Á Nevertheless, there is a map satisfying the conditions of the theorem from a subdivision of the input complex. If input simplex {<P,0>,<Q,1>} is subdivided as shown then it can be ``folded'' around the output complex, allowing input vertexes <P,0> and <Q,1> to be mapped to their counterparts in the output complex. P,0 Q,0 P,0 Q,0 Div I O

Code is asymmetric! // code for P T decide(T input) { announce[P] = input; if (input == 1) return 1; else if (announce[Q] != 1) return 0 else return 1 } Code is asymmetric! // code for Q T decide(T input) { announce[P] = input; if (input == 0) return 0; else if (announce[P] != 0) return 1 else return 0 } Here is a simple protocol for quasi-consensus. If P has input 0 and Q has input 1, then this protocol admits three distinct executions: one in which both decide 0, one in which both decide 1, and one in which Q decides 0 and P decides 1.These three executions correspond to the three simplexes in the subdivision of {<P,0>,<Q,1}>. 18-Nov-18

Road Map Inherently colored tasks Solvability for colored tasks Protocol ) map Map ) protocol A Sufficient Topological Conditions

Protocol ) Map I ¥(I) O protocol decision map ¥ δ input complex output complex carried by ¢ 18-Nov-18

Protocol ) Map ¥(I) O δ subdivision carried by of input ¢ complex 18-Nov-18

Road Map Inherently colored tasks Solvability for colored tasks Protocol ) map Map ) protocol A Sufficient Topological Conditions

¾ ¢(¾) I O Task (I,O,¢) 18-Nov-18 Recall that a task is given by an input complex \cI, an output complex \cO, and a relation \Delta that maps each input simplex \sigma to a subcomplex of \cO. ¾ ¢(¾) I O 18-Nov-18

chromatic subdivision Div ¾ chromatic simplex ¾ Recall that a that a simplex $\sigma = \set{\vs_0, \ldots, \vs_n}$ is \emph{chromatic} if each vertex is labeled with a distinct color, and a \emph{chromatic subdivision} $\Div \sigma$ is a subdivision of $\sigma$ where (1) each simplex of the subdivision is chromatic, and (2) for each $\tau \subset \sigma$, each vertex in $\Div \tau$ is labeled with a color from $\tau$. chromatic subdivision Div ¾ 18-Nov-18

The theorem we want to prove says that if there exists a chromatic subdivision Div \cI of the input complex, then there is an algorithm. ¾ ¢(¾) I O 18-Nov-18

If there is a chromatic subdivision … Theorem says … If there is a chromatic subdivision … The theorem we want to prove says that if there exists a chromatic subdivision Div \cI of the input complex, then there is an algorithm. ¾ ¢(¾) I O 18-Nov-18

à ¾ ¢(¾) I O Theorem says … If there is a chromatic subdivision … and a simplicial map à carried by ¢ … à The theorem we want to prove says that if there exists a chromatic subdivision Div \cI of the input complex, then there is an algorithm. ¾ ¢(¾) I O 18-Nov-18

à ¾ ¢(¾) I O Theorem says … If there is a chromatic subdivision … and a simplicial map à carried by ¢ … à The theorem we want to prove says that if there exists a chromatic subdivision Div \cI of the input complex, then there is an algorithm. ¾ ¢(¾) I O … then there is a wait-free IS protocol! 18-Nov-18

Let’s start with something easier … 18-Nov-18

à ChN¾ ¢(¾) I O Let’s start with a special case … If there is a simplicial map Ã: ChN ¾  ¢(¾) … à ChN¾ ¢(¾) I O 18-Nov-18 18-Nov-18 47

Á ¢(¾) ChN¾ I O Let’s start with something easier … If there is a simplicial map Á: ChN ¾  ¢(¾) … Á ¢(¾) ChN¾ I O … then there is a wait-free IS protocol! 18-Nov-18 18-Nov-18 48

Iterated immediate snapshot Protocol Iterated immediate snapshot Informally, the processes start out on vertexes of a single simplex $\sigma$ of $\cI$, and they eventually halt on vertexes of a single simplex of $\Delta(\sigma) \subseteq \cI$. 18-Nov-18

For any chromatic subdivision Div ¾ … If there is a color and carrier-preserving simplicial map Á: ChN ¾  Div ¾ … Á Div ¾ ChN¾ 18-Nov-18 18-Nov-18 50

Geometric construction Inductively divide boundary 18-Nov-18

Geometric construction Displace vetexes from barycenter 18-Nov-18

Geometric construction 18-Nov-18

Geometric construction Mesh(Ch ¾) is max diameter of a simplex 18-Nov-18

Subdivision shrinks mesh mesh(Ch ¾) · c mesh(¾) for some 0 < c < 1 18-Nov-18

Open cover An \emph{open cover} $\bbU$ for a complex $\cK$ is a finite collection of open sets $U_0, \ldots, U_k$ such that $\cK \subseteq \cup_{i=0}^k U_i$. 18-Nov-18

Lesbesgue Number ¸ 18-Nov-18 If $U_0, \ldots, U_k$ an open cover for $\cK$, there exists a $\lambda > 0$ (called a \emph{Lebesgue number}) such that any set of diameter less than $\lambda$ lies in a single $U_i$. ¸ 18-Nov-18

Open stars form an open cover for a complex ostar(v) 18-Nov-18

Vertexes lie on a common simplex iff their open stars intersect Intersection Lemma Vertexes lie on a common simplex iff their open stars intersect 18-Nov-18

… each star of ChN ¾ lies in a open star of Div ¾ Pick N large enough that each (closed) star of ChN ¾ has diameter less than ¸ … … each star of ChN ¾ lies in a open star of Div ¾ 18-Nov-18

Simplicial by intersection lemma. Defines a vertex map …. Simplicial by intersection lemma. Á 18-Nov-18

Á Div ¾ ChN¾ Á: ChN ¾  Div ¾ … We have just proved the Simplicial Approximation Theorem Á Div ¾ ChN¾ There is a carrier-preserving simplicial map Á: ChN ¾  Div ¾ … Not necessarily color-preserving! 18-Nov-18 18-Nov-18 62

An open-star cover is chromatic if every simplex ¿ of ChN ¾ is covered by open stars of of the same color. 18-Nov-18

If the open-star cover is chromatic …. Then the simplicial map …. Is color preserving! Á Must show that covering can be made chromatic … 18-Nov-18

Open Cover Fail Two simplexes conflict … If colors disjoint, but … polyhedrons overlap. cannot map to same color 18-Nov-18

We will show how to eliminate conflicting simplexes Open Cover Fail An open-star cover is chromatic iff there are no conflicting simplexes. We will show how to eliminate conflicting simplexes 18-Nov-18

Carriers Let $\Div \sigma$ be a subdivision of $\sigma$. For any simplex $\tau \in \Div \sigma$, the \emph{carrier} of $\tau$ in $\Div \sigma$ is the smallest face $\kappa$ of $\sigma$ such that $\tau$ is in $\Div \kappa$. 18-Nov-18

Perturbation 18-Nov-18

Star contains ² ball in carrier around vertex Room for perturbation Star contains ² ball in carrier around vertex 18-Nov-18

Room for perturbation Can perturb to any point within ² ball in carrier and still have subdivision 18-Nov-18

Open Cover Fail ½ has q+1 colors ¿ has p+1 colors 18-Nov-18

Simplexes lie in hyperplane of dimension p+q (because they overlap) 18-Nov-18

Some vertex has carrier of dimension p+q+1 (because there are p+q+2 colors) 18-Nov-18

Can perturb vertex within (p+q+1)-dimension ² ball … 18-Nov-18

Can perturb vertex within (p+q+1)-dimension ² ball … Out of the hyperplane 18-Nov-18

Repeat until star diameter < Lebesgue number: Construct Ch ChN-1* ¾ Perturb to ChN* ¾ So open-star cover is chromatic Construct color-preserving simplicial map 18-Nov-18

Div ¾ ¢(¾) Ã Given 18-Nov-18 18-Nov-18 77

ChN¾ Constructed Á Div ¾ ¢(¾) Ã Given 18-Nov-18 18-Nov-18 78

Iterated immediate snapshot here … ChN¾ Á Div ¾ ¢(¾) Ã Yields protocol here! 18-Nov-18 18-Nov-18 79

Road Map Inherently colored tasks Solvability for colored tasks Protocol ) map Map ) protocol A Sufficient Topological Condition

Link-Connected O is link-connected if for each ¿ 2 O, A geometric complex is subdivided by partitioning each of its simplexes into smaller simplexes without changing the complex's polyhedron. O is link-connected if for each ¿ 2 O, link(¿,O) is (n - 2 - dim ¿)-connected. 18-Nov-18

not a fan not link-connected. 18-Nov-18

Theorem If, for all ¾ 2 I, ¢(¾) is ((dim ¾)-1)-connected, and O is link-connected then (I,O,¢) has a wait-free IS protocol 18-Nov-18

Proof Strategy If, for all ¾ 2 I, ¢(¾) is ((dim ¾)-1)-connected, and O is link-connected, there exists subdivision Div & color-preserving simplicial map ¹: Div I ! O carried by ¢. 18-Nov-18

Proof Strategy If, for all ¾ 2 I, ¢(¾) is ((dim ¾)-1)-connected, and so protocol exists by colored theorem O is link-connected, there exists subdivision Div & color-preserving simplicial map ¹: Div I ! O carried by ¢. 18-Nov-18

Lemma rigid & color-preserving on boundary means color-preserving everywhere suppose we have a rigid simplicial map Á: Div ¾  O that is color-preserving on Div  ¾ then Á is color-preserving on Div ¾ 18-Nov-18

Lemma If O is link-connected … can extend rigid simplicial map Án-1: skeln-1 I  O to a rigid simplicial map Án: Div I  O where Div skeln-1 I = skeln-1 I 18-Nov-18

Á O I O Á’ div I Induction Base: n = 1 collapses hinge does not collapse0.eps Á’ div I does not collapse

Induction Step: Á does not collapse (n-1)-simplexes

Div I collapse2.eps O

Div I does not collapse Á’ O exploit connected link

Summary connectivity of ¢(¾) Inductively use … link-connectivity of O to construct a color-preserving simplicial map Á: Div I  Ocarried by ¢. protocol follows from main theorem 18-Nov-18

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