There are so many ways to learn about the Paxos protocol (see Lampson, Cachin, Howard Howard 2, Guerraoui, Kladov, Krzyzanowski, Lamport, Wikipedia and many more), this post is one more way. The emphasis of this post is on a decomposition of Paxos for omission failures that will later help when we do a similar decomposition in the Byzantine failure case (for PBFT and HotStuff).

This post has embedded a set of 11 simple exercises - try to go over them and post your answers.

The model is partial synchrony with $f<n/2$ omission failures and the goal is consensus (see below for exact details).

The Paxon parliament’s protocol provides a new way of implementing the state-machine approach to the design of distributed systems. Lamport, The Part-Time Parliament.

We approach Paxos by starting with two major simplifications:

1. Use a simple revolving primary strategy based on the assumptions of perfectly synchronized clocks. A later posts shows how to extend to a stable leader, how to rotate leaders with responsiveness, and how not to rely on clock synchronization.
2. Focus on a single-shot consensus. A later post shows how to extend to multi-shot consensus and state machine replication.

## View-based protocol with simple revolving primary

The protocol progresses in views, each view has a designated primary party. The role of the primary is rotated. For simplicity, the primary of view $v$ is party $v \bmod n$.

Clocks are perfectly synchronized, and $\Delta$ (the maximum message delay after GST) is known. View $v$ is set to be the time interval $[v(10 \Delta),(v+1)(10 \Delta))$. In other words, each $10\Delta$ clock ticks each party triggers a view change and increments the view by one. Clocks are assumed to be perfectly synchronized, so all parties move in and out of each view in complete synchrony (lock step).

## Single-shot consensus

In this setting, each party has some input value and the goal is to output a single value with the following three properties:

Uniform Agreement: if any two parties output $v$ and $v’$ then $v=v’$. Note that this is a strictly stronger property than Agreement which just requires that all non-faulty parties that output a value, output the same value.

Termination: all non-faulty parties eventually output a value and terminate. Note that this is a strictly stronger property than Liveness which just requires that all non-faulty parties eventually output a value.

Validity: the output is an input of one of the parties. Note that this is a strictly stronger property than Weak Validity which just requires that if all parties have the same input value then this is the output value.

Recoverable Broadcast is a pair of simple protocols: Broadcast and Recover. The Broadcast protocol has a designated leader party that has an input value val (we the term leader to indicate it does nor have to be the primary, but in our case it will be).

Recall that there are $n$ parties in total and at most $f<n/2$ can have omission corruptions.

Broadcast protocol:

leader sends <val> to all

send <echo, val> to all

Upon receiving n-f <echo, val>,
output val


Note that for simplicity, the leader also acts as a regular party. So it also sends <val> to itself and upon seeing its own message, it sends an <echo, val> message to all parties (again including sending it to itself).

Recover protocol. The output is either a value or a special $\bot$ value (which we write as bot in pseudo-code):

upon start,
if you sent <echo, val> in Broadcast,
then send <recover, val> to all
if you did not send <echo, val> in Broadcast yet,
then send <recover, bot> to all
Upon receiving n-f <recover, *>,
if one is <recover, val> then output val
otherwise (all are bot) then output bot



We will later detail what triggers starting the Recover protocol. Note that given <echo, val>, sending <recover, val> could be made redundant. We keep both for clarity for now.

### 4 properties of Recoverable Broadcast:

Validity: If a party outputs a value in Broadcast then it is the leader’s input value. If a party outputs a value in Recover then it is either the leader’s input value or $\bot$.

Weak Termination of Broadcast: If the leader is non-faulty then all non-faulty parties output a value and terminate.

Termination of Recover: If all parties start Recover, then all non-faulty parties output a value and terminate.

Recoverability: If all parties start Recover after some party outputs a value from Broadcast then all parties will output this value in the Recover.

#### Observe that:

2. Recover may return a non-$\bot$ value, even if no party has yet output a value during Broadcast!
3. Recover may return $\bot$, even if some party outputs the value from Broadcast!

Exercise 1: Write down three detailed executions, each one highlighting one of the observations above.

Exercise 2: Prove Validity and the two Termination properties. Explain where you used the assumption that there are at most $f$ failures.

Exercise 3: Prove recoverability and explain where you use (1) the assumption that $f<n/2$; (2) the Pigeonhole principle; (3) and where exactly you use the fact that the Recover is started after some party outputs a value from Broadcast.

We use a variation of Recoverable Broadcast to build a view-based consensus protocol. Recall that every $10 \Delta$ the parties change view and rotate the primary. Since clocks are perfectly synchronized this change of view is perfectly synchronized as well.

Here is a natural path: in view 1 the primary does a Recoverable Broadcast, with its input value. The output of the Broadcast is a consensus decision!

But there is a challenge: what if the first Primary is faulty and only some parties decide, but not all? For agreement to hold we must make sure that later primaries use the same value!

Exercise 4: If all that each primary does in its view is Broadcast its input value - show an execution that has a violation of the Agreement property.

A natural thing a primary of view >1 can do is call Recover (duh - that’s why we started with Recoverable Broadcast). In particular, if there was a decision in view 1 by some party, then we would like the Recover to notify the new primary.

Recall that for recoverability we need the output of the Broadcast to happen before the Recover starts! Solution: change the Broadcast so it stops when view $v$ ends:

Broadcast-in-view(v,p) has both a view $v$ and a proposal value $p$:

Primary of view v sends <propose(v,p)> to all

Upon receiving <propose(v,p)> from primary,
if myview = v then send <echo(v,p)> to all

Upon receiving n-f <echo(v,p)>,
if myview = v then output p



So we can run Recover when we enter view $v+1$ and thus guarantee the recoverability properties for all Broadcasts of previous views.

Another challenge: what if Recover returns different values from several previous views, which one should the primary use? For example, suppose that at view 10, we recover both a proposal $p’$ from view 4 and a proposal $p’’$ from view 6, which one should the primary choose?

The main Paxos algorithmic insight is:

Choose the recovered value associated with the most recent view you hear!

Recover-Max protocol for view $v$ that applies this insight:

upon start of view v
send <echoed-max(v,v',p)> of the highest view v' in which you sent <echo(v',p)>
or send <echoed-max(v,bot)> if you never sent echo yet

Primary waits for n-f responses <echoed-max(v,*)>
if all are bot then output bot
otherwise output the proposal p associated with the highest view v'



This Recover protocol needs only to send messages to the primary.

We are still not done. But let’s analyze the effect of Recover-Max. Assume that in each view, each primary just Broadcasts its own input.

Lemma 1: let $p$ be a value output by Broadcast in some view $v$, then any Recover-Max invocation in any view $>v$ will output some value $p’$ such that $p’$ was proposed at view $\geq v$.

Exercise 5: Prove Lemma 1. Show two examples where the Lemma is false: (1) If some party sends an echoed-max not of its highest echo but of a lower view; (2) or if the primary outputs not the proposal associated with the highest view but of a lower view.

Exercise 6: If each Primary just proposes its own input, show an example of why in the lemma above it may happen that $p \neq p’$.

So all that remains is to combine the two protocols to make sure the primary uses the result of Recover-Max if its output is not $\bot$. We are finally ready to define the consensus protocol:

The consensus protocol:

For view 1, the primary of view 1 with input $val$:

Broadcast-in-view (1,val)


For view $v>1$, the primary of view $v$ with input $val$:

p := Recover-Max(v)

if p = bot then
otherwise


In words: the primary will first try to recover the maximal echo. If no echo is seen, the primary is free to choose its own input. Otherwise, it proposes the value associated with the highest view in which it heard there was an echo.

This completes the description of the protocol. Let’s prove that the three properties of consensus hold.

### Agreement (Safety)

Lemma 2: Let $v^{\star}$ be the first view with $n-f$ echos of $(v^\star, x)$, then for any view $v>v^\star$ the proposal value of Broadcast-in-view (v,x) must be $x$.

Exercise 7: Prove the Agreement property follows from Lemma 2.

First step hints: Assume two parties decide on different values. Prove they could not have decided in the same view, so … apply Lemma 2.

We now prove Lemma 2, which is the essence of Paxos.

Proof of Lemma 2: consider the set $S$ (for Sentinels) consisting of the $n-f$ parties that sent <echo(v*,x)> in view $v^\star$. We will call them the Sentinels, because they guard safety.

We will prove by induction, that for any view $v\geq v^\star$:

1. The output of Recover-Max(v) is $x$.
2. For each party $P_i$ in $S$, the highest view $v’$ of <echoed-max(v,v',x)> it sends in Recover-Max(v) is such that: (1) $v’ \geq v^\star$; (2) and the value is $x$.

For the base case, $v=v^\star$ this follows from the definition of $S$. Now suppose the induction statement holds for all views $v^\star \leq v$ and consider view $v+1$:

Recover-Max(v+1) must get a response from $n-f$ parties, and that set must intersect with the set $S$ which is also of size $n-f$ by at least $n-2f>0$ one party. From $(2.)$ of the induction hypothesis on views $\leq v$ it follows that at least this one response will be of view $\geq v^\star$ and its value is $x$. From $(1.)$ it follows that any response from view $\geq v^\star$ will be of value $x$. Since Recover-Max(v+1) takes the value associated with the highest view, it must output $x$. This proves part $(1.)$ of the induction hypothesis for view $v+1$.

Observe that this is exactly the point in the proof where we used the main Paxos algorithmic insight:

Choose the recovered value associated with the most recent view you hear!

The reminder of the proof is to show that given the above, part $(2.)$ of the induction claim also holds:

Since the primary of view $v+1$ must propose $x$, then each party in $S$ either stays with its previous highest echo (from $(1.)$ of the induction hypothesis for view $\leq v$) or it updates it to the higher $(v+1,x)$. Clearly, in both cases, we proved that part $(2.)$ of the induction hypothesis holds for view $v+1$.

This concludes the proof of Lemma 2.

### Liveness

We proved Agreement, now let’s prove that eventually all non-faulty parties output a value.

Consider the view $v^+$ with the first non-faulty Primary that started after GST at time $T$. Since we are after GST, then on or before time $T+ \Delta$ the primary will receive <echoed-max(v+,*)> from all non-faulty parties (at least $n-f$). Hence will send a <propose(v+,p)> that will arrive at all non-faulty parties on or before time $T+2\Delta$. Hence all non-faulty parties will send <echo(v+,p)> (because they are still in view $v^+$). So all non-faulty parties will hear $n-f$ <echo(v+,p)> on or before time $T+3\Delta$. So all non-faulty will decide $p$ because they are still in view $v^+$.

This concludes the liveness proof.

Exercise 8: Show that there is no liveness (via an infinite execution) if view $v$ is set to be the time interval $[v(2 \Delta),(v+1)(2 \Delta))$. In other words, each $2\Delta$ clock ticks each party triggers an increment of the view by one.

Exercise 9: What is the minimal $\alpha$ such that the liveness property holds if view $v$ is set to be the time interval $[v(\alpha \Delta),(v+1)(\alpha \Delta))$. In other words, each $\alpha \Delta$ clock ticks each party triggers increments the view by one. What is the best time complexity you can get (see below)?

### Termination

We proved that all non-faulty parties output a value, but our protocol never terminates! For that we add the following termination gadget:

If the consensus protocol outputs p,
then send <decide, p> to all

Upon receiving <decide, p>
If you did not output yet,
Then output p and send <decide, p> to all

Upon receiving n-f <decide, p>
Terminate



Exercise 10: Prove termination - that the termination gadget causes all non-faulty parties to eventually terminate.

### Validity

Observe that Validity is trivial in these protocols. By induction, the only values used are the inputs of the parties.

What if a party has several different values (for example it received several different values from several different clients)? This is an example of how the standard validity of consensus does not address the challenges of MEV. More on that in later posts.

### Time and Message Complexity

Note that the time and number of messages before GST can be both unbounded. So for this post, we will measure the time and message complexity after GST.

Time complexity: since the Liveness proof waits for the first non-faulty primary that starts after GST this may take in the worst case: almost one “interrupted” view, then $f$ views of faulty primaries, then a view of a non-faulty party. So all parties will output a value in at most $(f+2)10 \Delta$ time after GST. A more careful analysis can improve the first and the last durations. We show here that $(f+1) \Delta$ is a worst case that cannot be avoided (but can have a small probability when using randomization).

Message Complexity: since each round has an all-to-all message exchange, the total number of messages sent after GST is $O((f+1) \times n^2) = O(n^3)$. We show here that $O(n^2)$ is the best you can hope for (for deterministic protocols or against strongly adaptive adversaries).

Exercise 11: Modify the protocol above to use just $O(n)$ messages per view (so total of $O(n^2)$ after GST). Explain why the proof still works, in particular, detail the Liveness proof and the Time complexity. Can you get the same bound as in Exercise 9?

## Acknowledgments

Many thanks to Kartik Nayak for insightful comments.