Introduction

    In the section CometBFT’s expected behaviour, we presented the most common behaviour, usually referred to as the good case. However, the grammar specified in the same section is more general and covers more scenarios that an Application designer needs to account for.

    In this section, we give more information about these possible scenarios. We focus on methods introduced by ABCI++: PrepareProposal and ProcessProposal. Specifically, we concentrate on the part of the grammar presented below.

    consensus-height    = *consensus-round decide commit
consensus-round     = proposer / non-proposer

proposer            = [prepare-proposal process-proposal]
non-proposer        = [process-proposal]

    We can see from the grammar that we can have several rounds before deciding a block. The reasons why one round may not be enough are:

    If we assume that the consensus algorithm decides on block $X$ in round $r$, in the rounds $r’ <= r$, CometBFT can exhibit any of the following behaviours:

    1. Call PrepareProposal and/or ProcessProposal for block $X$.
    2. Call PrepareProposal and/or ProcessProposal for block $Y \neq X$.
    3. Does not call PrepareProposal and/or ProcessProposal.

    In the rounds in which the process is the proposer, CometBFT’s PrepareProposal call is always followed by the ProcessProposal call. The reason is that the process also broadcasts the proposal to itself, which is locally delivered and triggers the ProcessProposal call. The proposal processed by ProcessProposal is the same as what was returned by any of the preceding PrepareProposal invoked for the same height and round. While in the absence of restarts there is only one such preceding invocations, if the proposer restarts there could have been one extra invocation to PrepareProposal for each restart.

    As the number of rounds the consensus algorithm needs to decide in a given run is a priori unknown, the application needs to account for any number of rounds, where each round can exhibit any of these three behaviours. Recall that the application is unaware of the internals of consensus and thus of the rounds.

    Possible scenarios

    The unknown number of rounds we can have when following the consensus algorithm yields a vast number of scenarios we can expect. Listing them all is unfeasible. However, here we give several of them and draw the main conclusions. Specifically, we will show that before block $X$ is decided:

    1. On a correct node, PrepareProposal may be called multiple times and for different blocks (Scenario 1).
    2. On a correct node, ProcessProposal may be called multiple times and for different blocks (Scenario 2).
    3. On a correct node, PrepareProposal and ProcessProposal for block $X$ may not be called (Scenario 3).
    4. On a correct node, PrepareProposal and ProcessProposal may not be called at all (Scenario 4).

    Basic information

    Each scenario is presented from the perspective of a process $p$. More precisely, we show what happens in each round’s $step$ of the Tendermint consensus algorithm. While in practice the consensus algorithm works with respect to voting power of the validators, in this document we refer to number of processes (e.g., $n$, $f+1$, $2f+1$) for simplicity. The legend is below:

    Round X

    1. Propose: Describes what happens while $step_p = propose$.
    2. Prevote: Describes what happens while $step_p = prevote$.
    3. Precommit: Describes what happens while $step_p = precommit$.

    Scenario 1

    $p$ calls ProcessProposal many times with different values.

    Round 0

    1. Propose: The proposer of this round is a Byzantine process, and it chooses not to send the proposal message. Therefore, $p$’s $timeoutPropose$ expires, it sends $Prevote$ for $nil$, and it does not call ProcessProposal. All correct processes do the same.
    2. Prevote: $p$ eventually receives $2f+1$ $Prevote$ messages for $nil$ and starts $timeoutPrevote$. When $timeoutPrevote$ expires it sends $Precommit$ for $nil$.
    3. Precommit: $p$ eventually receives $2f+1$ $Precommit$ messages for $nil$ and starts $timeoutPrecommit$. When it expires, it moves to the next round.

    Round 1

    1. Propose: A correct process is the proposer in this round. Its $validValue$ is $nil$, and it is free to generate and propose a new block $Y$. Process $p$ receives this proposal in time, calls ProcessProposal for block $Y$, and broadcasts a $Prevote$ message for it.
    2. Prevote: Due to network asynchrony less than $2f+1$ processes send $Prevote$ for this block. Therefore, $p$ does not update $validValue$ in this round.
    3. Precommit: Since less than $2f+1$ processes send $Prevote$, no correct process will lock on this block and send $Precommit$ message. As a consequence, $p$ does not decide on $Y$.

    Round 2

    1. Propose: Same as in Round 1, just another correct process is the proposer, and it proposes another value $Z$. Process $p$ receives the proposal on time, calls ProcessProposal for new block $Z$, and broadcasts a $Prevote$ message for it.
    2. Prevote: Same as in Round 1.
    3. Precommit: Same as in Round 1.

    Rounds like these can continue until we have a round in which process $p$ updates its $validValue$ or until we reach round $r$ where process $p$ decides on a block. After that, it will not call ProcessProposal anymore for this height.

    Scenario 2

    $p$ calls PrepareProposal many times with different values.

    Round 0

    1. Propose: Process $p$ is the proposer in this round. Its $validValue$ is $nil$, and it is free to generate and propose new block $Y$. Before proposing, it calls PrepareProposal for $Y$. After that, it broadcasts the proposal, delivers it to itself, calls ProcessProposal and broadcasts $Prevote$ for it.
    2. Prevote: Due to network asynchrony less than $2f+1$ processes receive the proposal on time and send $Prevote$ for it. Therefore, $p$ does not update $validValue$ in this round.
    3. Precommit: Since less than $2f+1$ processes send $Prevote$, no correct process will lock on this block and send non-$nil$ $Precommit$ message. As a consequence, $p$ does not decide on $Y$.

    After this round, we can have multiple rounds like those in Scenario 1. The important thing is that process $p$ should not update its $validValue$. Consequently, when process $p$ reaches the round when it is again the proposer, it will ask the mempool for the new block again, and the mempool may return a different block $Z$, and we can have the same round as Round 0 just for a different block. As a result, process $p$ calls PrepareProposal again but for a different value. When it reaches round $r$ some process will propose block $X$ and if $p$ receives $2f+1$ $Precommit$ messages, it will decide on this value.

    Scenario 3

    $p$ calls PrepareProposal and ProcessProposal for many values, but decides on a value for which it did not call PrepareProposal or ProcessProposal.

    In this scenario, in all rounds before $r$ we can have any round presented in Scenario 1 or Scenario 2. What is important is that:

    Round $r$

    1. Propose: A correct process is the proposer in this round, and it proposes block $X$. Due to asynchrony, the proposal message arrives to process $p$ after its $timeoutPropose$ expires and it sends $Prevote$ for $nil$. Consequently, process $p$ does not call ProcessProposal for block $X$. However, the same proposal arrives at other processes before their $timeoutPropose$ expires, and they send $Prevote$ for this proposal.
    2. Prevote: Process $p$ receives $2f+1$ $Prevote$ messages for proposal $X$, updates correspondingly its $validValue$ and $lockedValue$ and sends $Precommit$ message. All correct processes do the same.
    3. Precommit: Finally, process $p$ receives $2f+1$ $Precommit$ messages, and decides on block $X$.

    Scenario 4

    Scenario 3 can be translated into a scenario where $p$ does not call PrepareProposal and ProcessProposal at all. For this, it is necessary that process $p$ is not the proposer in any of the rounds $0 <= r’ <= r$ and that due to network asynchrony or Byzantine proposer, it does not receive the proposal before $timeoutPropose$ expires. As a result, it will enter round $r$ without calling PrepareProposal and ProcessProposal before it, and as shown in Round $r$ of Scenario 3 it will decide in this round. Again without calling any of these two calls.

    Decorative Orb