# Paxos Algorithm Assignment Help

Discovered through Lamport in the late 80s (while trying to prove that a complex algorithm with regard to handling Byzantine failures utilized in a program had been impossible). Paxos is actually a good example of the consensus protocol.

Contains the first 3 phase commit algorithm that experienced the obviously stated correctness condition and a proof associated with correctness.

## Key Assumptions:

Group of procedures which run Paxos is well known a-priori. Procedures suffer crash failures as well as all procedures possess Greek names.

Three roles: proposer, acceptor as well as Learner (The node may act as a lot more than one client)

## Two phases-

Stage 1: Prepare request Response as well as

Stage 2: Accept request Response

Paxos algorithm family: basic paxos, multi paxos, cheap paxos, fast paxos, generalized paxos and Byzantine paxos.

Paxos Properties: Paxos can be an asynchronous consensus algorithm.

Asynchronous networks- No common clocks or even shared notion of time.

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Paxos is assured secure.

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### Paxos Conditions:

Proposer: Suggests values with regard to consideration through Acceptors. Advocates for a client.

Acceptor: Views the actual values proposed through proposers. Renders an accept or reject decision.

Learner: Learns the actual selected value. In practice, every node will often play all three functions.

Proposal: An alternative suggested with a proposer. Consists of a unique number as well as a proposed value.

### In Well-Behaved Runs:

#### Paxos is safe:

Intuition- If a proposal with value v is determined; next each higher-numbered proposal released through any proposer has value v.

#### Safety (proof):

Suppose (n, v) could be the first proposal in which approved. When probably none, basic safety holds.

Let (n’, v’) function as first given proposal after (n, v) using a different value v’!=v

Since (n’, v’) passed, it needs important regarding acceptors. Hence, several process approve equally (n, v) and (n’, v’), even though it's suggest value v with version number k= n.

As (n’, v’) passed, that need to be given a response (“ack”, n’, j, v’) to be able to it’s prepare request, with n j n Consider j v we get the contradiction

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