Lecture 36: Consensus II

Annoucements

1. Quiz due tonight
2. Leaderboard 3 submissions, run Sunday morning
3. Final Submissions: Tuesday, 05/09 by 5:00pm
   • documentation = 40% of grade

Leaderboard 2

Primes Task (baseline 58810ms):

1. The AMA (2070ms)
2. Test Group Name Please Ignore (3530ms)
3. Deadlock Dodgers (4467ms)

Sorting Task (baseline 8034ms):

1. Deadlock Dodgers (1548ms)
2. MRC (1550ms)
3. Benchmark (2013ms)

Binary Consensus

Setup: $n$ processes with binary inputs

• Agreement: all processes output the same value
• Validity: if all systems have the same input, they all output that valuex
• Termination: all (non-faulty) processes decide on an output and terminate after a finite number of steps

Today

We’ll prove:

Theorem. There is no wait-free protocol for consensus for shared memory with atomic read/write access.

• wait-free means each process completes in a bounded number of steps, indpendent of actions of other threads

Game Plan

Theorem. There is no wait-free protocol for consensus for shared memory with atomic read/write registers.

• Define indistinguishability of executions
  • Lemma 1. indistinguishable executions have same results

Executions

An execution $E$ of algorithm $A$ specifies

• Inputs of all processes
• Sequence of steps taken by processes
  • read, write, terminate, crash

Executions may be incomplete

• Not all nodes have terminated/crashed yet
  • encodes current state/history of execution

Executions may be extended by scheduling more steps

Example of Execution $E$

Indistinguishable Executions

• $E$ and $E’$ are executions
• they are indistinguishable at process $P_i$ if in $E$ and $E’$:
  1. $P_i$ has same input
  2. sequence of read/write operations performed by $P_i$ are same
  3. the sequence of values read and written by $P_i$ are the same

First Important Observation

Lemma 1. If executions $E$ and $E’$ are indistinguishable to process $P_i$ then:

1. If $P_i$ has not yet terminated, then $P_i$’s next step will be the same in any extension
2. If $P_i$ has terminated, then $P_i$’s output is the same in $E$ and $E’$

Bivalent Executions

• Consider a (hypothetical) wait-free consensus protocol $A$
• Let $E$ be an execution of $A$

We say that $E$ is…

1. $0$-valent: in every extension of $E$, all processes output $0$
2. $1$-valent: in every extension of $E$, all processes output $1$
3. bivalent if there exist
   • an extension $E’$ of $E$ in which all processes output $0$
   • an extension $E’’$ of $E$ in which all processes output $1$

Second Important Observation

Lemma 2. Suppose $A$ solves consensus. Then there is a bivalent initial state.

• Here an initial state is an execution in which no process has yet taken a step
  • the execution consists of only inputs for each process

Proof of Lemma 2

Must show: there is a bivalent initial state

Argument:

• by contradiction: suppose no bivalent initial state
• consider sequence of initial states
• show some are $0$-valent, some are $1$-valent
• show that some must be bivalent

$E_1$ is $0$-valent (Why?)

$E_5$ is $1$-valent

More Initial States

Assume: All Univalent

Adjacent Pair, Different Valency

All Extensions of $E_2$ Return $0$

All Extensions of $E_3$ Return $1$

$E_2’$ and $E_3’$ Indistinguishable

$E_2$ and $E_3$ Bivalent

Note

Don’t need to assume $P_2$ crashes

• just assume first step of $P_2$ is scheduled after some other thread outputs
• this is possible because we assume $A$ is wait-free
  • some process guaranteed to terminate even if one is not scheduled

Mere possibility of a crash together with wait-free assumption implies existence a bivalent initial state

• same holds if we require only termination with one fault

Critical Executions

An execution $E$ is critical if:

1. $E$ is bivalent
2. Extending $E$ by any single step of any process results in a univalent execution

Important Obvservation 3

Lemma 3. Every consensus protocol has a critical execution.

Proof of Lemma 3

Consider a bivalent initial state $E_0$

• Why does such a state exist?

Properties of Consensus

Lemma 2. Every consensus protocol has a bivalent initial state.

Lemma 3. Every consensus protocol has a citical execution $E$.

So far: Have not used any properties of atomic read/write registers

• These properties hold for all consensus protocols
  • even if other atomic operations are supported

Outline of Proof of FLP

Assume only 2 processes, $P_0$ and $P_1$

1. Start from a critical execution $E$
   • $E$ is bivalent, but any extension is univalent
2. Consider all possibilities for next step:
   • both threads read for next step
   • one thread reads, the other writes
   • both threads write
3. Show that in any case, we contradict either criticality of E or correctness of protocol

Assumptions

Without loss of generality:

1. There are two processes $P_0$ and $P_1$
2. $E$ is a critical state
   • if $P_0$ has next step, resulting execution is $0$-valent
   • if $P_1$ has next step, resulting execution is $1$-valent

Case 1: read/read

Assumption: next operations for both $P_0$ and $P_1$ are read

• Start from critical state $E$
  • if $P_0$ steps next, output is 0
  • if $P_1$ steps next, output is 1

read/read

read/read Next Step

read/read Problem

Case 2: read/write

Assumption:

• $P_0$’s next step is read

• $P_1$’s next step is write

• Start from critical state $E$
  • if $P_0$’s read step is next, output is 0
  • if $P_1$’s write step is next, output is 1

read/write Setup

read/write Next Step

read/write Indistinguishable

read/write $P_0$ Crashes

Case 3: write/write

Assumption: next operation for both $P_0$ and $P_1$ is write

Subcases:

• Sub-case a: write to different registers
• Sub-case b: write to same register

write/write Different Registers

write/write Next Step

write/write Indistinguishable

write/write Same Register

write/write Next Step

write/write Indistinguishable

Conclusion

In general:

• Indistinguishable executions produce same output

Assuming a wait-free consensus protocol using only read/write registers:

1. Showed there is a bivalent initial state
2. Showed there is a critical execution
3. Given a critical execution
   • found indistinguishable states that must give different outputs
   • this is a contradiction!

Remark. 1 and 2 hold for all protocols; 3 assumes only read/write registers