Lecture 36: Consensus II
COSC 273: Parallel and Distributed Computing
Spring 2023
Annoucements
 Quiz due tonight
 Leaderboard 3 submissions, run Sunday morning
 Final Submissions: Tuesday, 05/09 by 5:00pm
 documentation = 40% of grade
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Primes Task (baseline 58810ms):
 The AMA (2070ms)
 Test Group Name Please Ignore (3530ms)
 Deadlock Dodgers (4467ms)
Sorting Task (baseline 8034ms):
 Deadlock Dodgers (1548ms)
 MRC (1550ms)
 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 (nonfaulty) processes decide on an output and terminate after a finite number of steps
Theorem (FLP, 1985). There is no algorithm that achieves consensus in the presence of even a single faulty process.
Today
We’ll prove:
Theorem. There is no waitfree protocol for consensus for shared memory with atomic read/write access.

waitfree means each process completes in a bounded number of steps, indpendent of actions of other threads
Game Plan
Theorem. There is no waitfree protocol for consensus for shared memory with atomic read/write registers.
 Define indistinguishability of executions

Lemma 1. indistinguishable executions have same results
 Define bivalent execution
 the final output (all 0s or all 1s) has not yet been determined

Lemma 2 Show there is a bivalent initial state
 Define critical executions: if any process takes a step, then output is determined

Lemma 3 Every protocol has a critical execution
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 $E’$:
 $P_i$ has same input
 sequence of read/write operations performed by $P_i$ are same
 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:
 If $P_i$ has not yet terminated, then $P_i$’s next step will be the same in any extension
 If $P_i$ has terminated, then $P_i$’s output is the same in $E$ and $E’$
Bivalent Executions
 Consider a (hypothetical) waitfree consensus protocol $A$
 Let $E$ be an execution of $A$
We say that $E$ is…

$0$valent: in every extension of $E$, all processes output $0$

$1$valent: in every extension of $E$, all processes output $1$

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 waitfree
 some process guaranteed to terminate even if one is not scheduled
Mere possibility of a crash together with waitfree 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:
 $E$ is bivalent
 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 $P_1$
 Start from a critical execution $E$
 $E$ is bivalent, but any extension is univalent
 Consider all possibilities for next step:
 both threads
read
for next step
 one thread
read
s, the other write
s
 both threads
write
 Show that in any case, we contradict either criticality of
E
or correctness of protocol
Assumptions
Without loss of generality:
 There are two processes $P_0$ and $P_1$
 $E$ 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:
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:
 Subcase a: write to different registers
 Subcase 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 waitfree consensus protocol using only read/write registers:
 Showed there is a bivalent initial state
 Showed there is a critical execution
 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