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Vector Clock Size Lower Bound: Proof and Implementation Repercussions

Prof. Jennifer Welch presents a theorem on vector clock size bounds using real number vectors in distributed systems. The proof highlights the implications for implementations and demonstrates contradictions that arise with smaller vector sizes.

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Vector Clock Size Lower Bound: Proof and Implementation Repercussions

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  1. Prof. Jennifer Welch

  2. Vector Clock Size Lower Bound Theorem: Any implementation of vector clocks using vectors of real numbers requires vectors of length n (number of processes). Proof: For any value of n, consider this execution: ai : first send event at process pi bi : last receive event at pi

  3. Example Bad Execution For n = 4:

  4. Vector Clock Size Lower Bound Claim 1:ai+1 || bi for all i (with wrap-around) Proof: Since each proc. does all sends before any receives, there is no transitivity. Also pi+1 does not send to pi. Claim 2:ai+1 bj for all j ≠ i. Proof: If j = i+1, obvious. If j ≠ i+1, then pi+1 sends to pj:

  5. Vector Clock Size Lower Bound • Suppose in contradiction, there is a way to implement vector clocks with k-vectors of reals, where k < n. • By Claim 1, ai+1 || bi => V(ai+1) and V(bi) are incomparable => V(ai+1) is larger than V(bi) in some coordinate h(i) => h : {0,…,n-1}  {0,…,k-1}

  6. Vector Clock Size Lower Bound • Since k < n, the function h is not 1-1. So there exist distinct i and j such that h(i) = h(j). Let r be this common value of h. V(a0) V(a1) … V(ai+1) … V(aj+1) … V(an-1) V(b0) … V(bi) … V(bj) … V(bn-2) V(bn-1) > in comp. h(0) > in comp. h(i) two of these components are the same, say h(i) = h(j) = r > in comp. h(j) > in comp. h(n-2) > in comp. h(n-1)

  7. Vector Clock Size Lower Bound V(bi) > in component r > in component r, contradicts aj+1 bi V(ai+1) ≤ in all components, since ai+1 bj V(bj) > in component r V(aj+1)

  8. Vector Clock Size Lower Bound • So V(ai+1) is larger than V(bi) in coordinate r and V(aj+1) is larger than V(bj) in coordinate r also. • V(aj+1)[r] > V(bj)[r] by def. of r ≥ V(ai+1)[r] by Claim 2 (ai+1bj) & correct. ≥ V(bi)[r] by def. of r • Thus V(aj+1) !< V(bi), contradicting Claim 2 (aj+1bi) and assumed correctness of V.

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