Leslie Lamport
Summary: Piras Maha

 History
 Problem
 Choosing A Value
 Learning
 Simple Example

 Leslie Lamport
 Presented in 1990
 Almost everyone dismissed it
 The first paper was a difficult read, hard to understand
 Took 10 years before publication, required almost a complete re-write
 Gained popularity really quickly in distributed system
 Lots of decomposition and optimizations, still being applied to date

 Assume a collection of processes
 Each process proposes a value
 Considered consensus algorithm( chooses a single value is chosen)
 If no value is proposed then no value should be chosen
 If a value has been chosen processes should be able to learn chosen value

 Safety requirement for consensus
 Only a value proposed may be chosen
 Only single value is chosen
 Process never learns value has been chosen unless it actually has been
 Three roles of algorithm preformed by three classes of agents
 Proposers
 Acceptors
 Learners

 Agents can communicate with each asynchronously
(through messaging)
 Agents operate at arbitrary speed
 May fail, may restart (information should be remembered)
 Messages may take arbitrary delivery times, can be duplicated, can be lost but not corrupted

 A simple method: have one acceptor
 A proposer sends a proposal to acceptor
 Acceptor chooses first value it receives
 Issue: if acceptor fails, make further progress not possible

 What if we used multiple acceptors?
 Proposer sends value to a set of acceptors
 An acceptor may accept proposed value
 Value chosen if majority (or large enough set of
acceptors) chooses it
 In absence of message loss or failure, want a value chosen even if only one value proposed by proposer
 P1: An acceptor must accept the first proposal that it receives

 P1 raises an issue: several values proposed by different proposers , where every accepted a value but no majority
 Or two values were proposed and half were chosen by one set of acceptors
 Failure of single acceptor could make it impossible to learn which value was chosen

 P1 and requirement considering value is chosen when accepted by a majority imply acceptor must be allowed to accept multiple proposals
 To track the proposals, each proposal consists of proposal number and value
 Different proposal would have different numbers
 A value is chosen when single proposal with that value is accepted by majority of acceptors

 Multiple proposals may be chosen but value must be same in all
 P2: If a proposal with value v is chosen, then every higher-number proposal that is chosen has value v
 This guarantees that only single value will be chosen
 Again, to be chosen, proposal must be accepted by at least one acceptor
 P2a: If a proposal with a value v is chosen, then every higher-numbered proposal accepted by any acceptor has value v

 Since communication is asynchronous, a proposal may be chosen with an acceptor c, not received any proposal
 At this point if a new proposer issues a higher-number proposal with a different value, P1 requires c to accept, thereby violating P2a
 P2b: IF a proposal with value v is chosen then every higher number proposal issued by any proposer has value v

 Since proposal must be issued before it is accepted,
P2b implies P2a which implies P2
 Prove P2
 Assume proposal with number m and value v chosen
 Any proposal issued with number n, n>m has value v
 Using induction on n, can prove n had value v, with additional assumption: every proposal number with a number in the range m to (n-1) has value v
 For proposal m to be chosen, a majority of acceptors C, where every acceptor in C has accepted it

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This m (Hypothesis ) is chosen and it implies:
 Every acceptor in C has accepted a proposal with number in the range m ...(n-1) and every proposal with a number in m...(n-1) accepted by any acceptor has value v
Since any set S, with majority acceptors, contains at least one C leads to proposal number n has value v by(leading to) :
P2c: For any v and n, if a proposal with value v and number n is issued, then there is a set S consisting of a majority of acceptors such that either
 (a) no acceptor in S has accepted any proposal number less than n

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 Or (b) v is the value of the highest numbered proposal among all proposals numbered less than n accepted by the acceptors in S
A proposer that wants to issue a proposal numbered n, must learn the highest numbered proposal with number less than n, if any, that has been accepted so far
To do this: a proposer chooses number n, sends request to acceptors asking for response containing
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Promise to not accept proposal numbered less than n
Proposal with highest number less than n, that has been accepted
Called: Prepare Request

 If proposer receives requested responses from majority, then issue proposal with number n and value v (where v is highest numbered proposal among responses or new number if no proposals reported)
 Proposer later sends a request that proposal has been accepted

 Acceptor can receive two kinds of requests from proposers
 Prepare request
 Accept request
 Can ignore requests
 P1a: an acceptor can accept a proposal number n if and only if it has not responded to a prepare request having a number greater than n
 With this, a complete algorithm for choosing value

 Consider acceptor receives a prepare request with number n, but already responded to prepare request with number greater than n
 Therefore promised not to accept any proposal numbered n
 Which means no reason to respond to this request or respond to a request for a for a proposal that was already accepted

 Now, acceptor only needs to remember the highest – number proposal that was accepted and the number of highest-numbered prepare request which was responded to
 Since P2c kept invariant regardless of failures, acceptor must remember this information if failure or restart
 Note: proposer can abandon proposal and forget all about it, as long as it doesn’t issue another proposal with same number

 Phase 1
 Proposer sends prepare request with number n to majority of acceptors with proposal number n
 If an acceptor receives a prepare request with number n
(this n is greater than any prepare request which it has already responded up until now), then it responds to the request with a promise not to accept any more proposals numbered less than n and with the highest-numbered proposal (if any) that it has accepted

 Phase 2
 If proposer receives response to its prepare request (for example numbered n) from majority of acceptors
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Proposer sends accept request to all acceptors with proposal number n (with value v), where v is the value of the highestnumbered proposal from the responses OR is any value, if no responses were heard
If acceptor receives request for proposal number n, it will accept this proposal unless it already responded to a prepare with a number greater than n

 Multiple proposals can be made
 Follows algorithm each time
 Proposer can abandon in the middle of protocol
 Correctness should be maintained

 To learn a value is chosen, learner must first know the proposal has been accepted by majority
 Obvious solution to let each acceptor respond to all learners (sending proposal)
 Good: allows all learners find out about chosen value quickly
 Bad: requires each acceptor to respond to each learner
(large number of responses, number of learners x number of acceptors)

 Another option: have one distinguished learner
 Have this learner communicate to all other learners
(which value chosen)
 Bad: less reliable, if distinguished learner fails
 Requires extra round for all learners to discover chosen value
 Good: requires less responses( number of acceptors + number of learners)

 Next approach would be to have a set of distinguished learners
 Acceptors send acceptances to the set of distinguished learners
 Larger the set, more reliable but adds complexity/greater communication

 Consider: proposer p, completes phase 1, proposal number n1
 Another proposer q, then completes phase 1 for proposal number n2, n2>n1
 Proposer p’s accept request for n1 are ignored since acceptors promised not to accept proposal number< n2
 Then p, begins and completes phase 1 for n3, n3>n2
 Therefore q’s phase 2 accept requests ignored,
 ...

 Use distinguished proposer, only one to try issuing proposals
 If distinguished proposer communicates with majority of acceptors, using proposal number greater than any already used, then it can succeed issuing an accepted proposal
 Abandon proposal then trying again if it learns of new request with higher proposal number, this distinguished proposer will eventually choose high enough proposal number

 If enough of proposers, acceptors and communication network is working properly, electing single distinguished proposer can lead to liveness
 Reliable algorithm for electing distinguished proposer may require randomness or real-time (based on research results)

 Paxos algorithm assumes network of processes
 In its heart, considered consensus algorithm, with agents playing proposer, acceptor, learner
 Leader is chosen, plays distinguished proposer and learner
 Paxos consensus algorithm requests and responses sent as ordinary messages
 Stable storage (alive during failure) contains information acceptor must remember
 Acceptor records in stable storage before sending response

 Use collection of servers, each one implementing state machine independently
 Deterministic
 Guarantee execution osfsame sequence of state machine commands

 To put it simple terms: Paxos is used to have consistency (ordering of semantics) on a platform with multiple machines
 Why is this important?
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If ordering is jumbled, the meaning is lost
Unreliable
 Could cause safety issues

 Jane has $0 in her account
 Joe sends $100 to Jane
 When Jane is notified she received $100, she sends $50 cheque to CRA
 CRA cashs the cheque and spends it on something useless
 Without consistent ordering, each machine might view these in different order

 What would you like to convey to current and future researchers and practitioners?
Don’t pay attention to pedantic old farts like me telling you what to do.
 What are the most important lessons you’ve learned in your career?
My career has been idiosyncratic. Any meaningful lessons I could draw from it would, at best, apply to people starting their careers 30 years ago.
Ref: http://research.microsoft.com/en-us/um/people/lamport/pubs/ds-interview.pdf

 http://ayende.com/blog/4496/paxos-enlightment
 http://www.inf.usi.ch/faculty/pedone/MScThesis/mar co.pdf
 http://the-paper-trail.org/blog/?p=173
 Paxos Made Simple by Leslie Lamport (2001)