From aada254e38c914a1d03e53edfb69af5951664768 Mon Sep 17 00:00:00 2001 From: Marco Realacci Date: Tue, 1 Apr 2025 09:09:49 +0200 Subject: [PATCH] vault backup: 2025-04-01 09:09:49 --- Concurrent Systems/notes/10 - Consensus Implementation.md | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/Concurrent Systems/notes/10 - Consensus Implementation.md b/Concurrent Systems/notes/10 - Consensus Implementation.md index f969a72..f30b265 100644 --- a/Concurrent Systems/notes/10 - Consensus Implementation.md +++ b/Concurrent Systems/notes/10 - Consensus Implementation.md @@ -51,7 +51,7 @@ p(C’) can be R1.read() or R1.write(v) and q(C’) can be R2.read() or R2.write - like point 1... We will again obtain a configuration that is both 0-valent and 1-valent 3. R1 = R2, with p that reads and q that writes (or viceversa) - - *Remark:* only p can distinguisc C' from p(C') (reads put the value read in a local variable, visible only by the process that performed the read) + - *Remark:* only p can distinguish C' from p(C') (reads put the value read in a local variable, visible only by the process that performed the read) - Let S' be the scheduling from C' where p stops and q decides: - S' starts with the write of q - S' leads q to decide 1, since q(C') is 1-valent @@ -59,4 +59,8 @@ p(C’) can be R1.read() or R1.write(v) and q(C’) can be R2.read() or R2.write - because of the initial remark, q decides 1 also here - Reactivate p - if p decides 0, then we would violate agreement - - if \ No newline at end of file + - if p decides 1, we contradict 0-valence of p(C') + +4. R1 = R2 and both operations are a write + - *Remark:* q(p(C)) = q(C) cannot be distinguished by q since the value written by p is lost after the write of q + - Then, work like in case (3). \ No newline at end of file