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Understanding Distributed Programs and Correctness

Explore examples of distributed programs like maxima finding and token ring, and learn about correctness proofs. Includes design, invariant, and proof of correctness for mutual exclusion with Peterson’s algorithm.

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Understanding Distributed Programs and Correctness

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  1. Lecture 2:Distributed Programs and their Correctness Anish Arora CSE 6333

  2. Distributed programs: Example 0 maxima finding Given : Graph (V,E), V = {1,2,3} E = {(1,2), (1,3), (2,1), (3,1)} constant id.1, id.2, id.3 : integer var m.1, m.2, m.3 : integer Design : (j: jV : id.j=m.j)leads-to (j : jV : m.j=(max k : kV: id.k)) program max m.1 < m.2  m.1 = m.2 ▯ m.1 < m.3  m.1 = m.3 ▯m.2 < m.1  m.2 = m.1 ▯ m.3 < m.1  m.3 = m.1

  3. Distributed programs: Example 1 program max-abbreviated parameter j, k : V (▯ j, k : (j, k)  E : m.j < m.k  m.j = m.k) id.1 id.2 id.3 1 2 3

  4. Distributed programs: Example 1 program token ring Given : Graph (V,E), where V = {0,1,…,N-1} E = {Uj : jV : (j, j N 1)) ... vart.j : boolean Design : Initially, exactly one node j has the token (t.j is true) Infinitely often, every node receives that unique token programtring ▯j : j  V : t.j t.j, t.(j N 1):= false, true 1 N-2 0 N-1

  5. Proof of programs (distributed) 0. program max invariant: (j : jV : (k : kV : m.j = id.k))  (j : jV : m.j = (max k : kV : id.k)) variant function: (j : jV : (max k : kV : id.k)–m.j) 1. program tring invariant: (j : jV : t.j)  (j,k : j,kV : (t.j  t.k)  j = k) variant function for j: clockwise distance between node with token & j

  6. Mutual Exclusion program Peterson’s_Mutual_Exclusion var f.1, f.2, cs.1, cs.2, pc.1, pc.2 : Boolean ; turn : {1,2} process j : {1,2} begin  pc.j  f.j  pc.j, f.j := true, true ▯pc.j  pc.j, turn := false, 3-j ▯  pc.j  f.j (f.(3-j)  turn=(3-j))  cs.j := true ▯cs.j  cs.j, f.j := false, false end

  7. Proof of Correctness • An invariant S is (j : j  {1,2} : pc.j  f.j  cs.j  ( f.j  pc.j (f.(3-j)  pc.(3-j)  turn=(3-j)) Safety Proof: Observe that S  (cs.1  cs.2) • If the first two statements of process j are changed to : pc.j  f.j  pc.j, turn := true, 2 ▯ pc.j  pc.j, f.j := false, true there exists a state transition that violates S

  8. Proof of Correctness Liveness There exists a sequence of state transitions that yield a state where  cs.1  cs.2 holds Proof Consider a state where all boolean valued variables are false Now: • execute the first action of process 2; • execute the first, second & third action of process 1; • finally execute the second & third action of process 2

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