vault backup: 2025-03-03 19:21:59

This commit is contained in:
Marco Realacci 2025-03-03 19:21:59 +01:00
parent a5add556fb
commit 8991168716

View file

@ -217,6 +217,11 @@ Induction (true for , to be proved for +1):
- p at level can increase its level by writing its FLAG at +1 and its index in $A_Y[+1]$ - p at level can increase its level by writing its FLAG at +1 and its index in $A_Y[+1]$
- let $p_x$ be the last one that writes `A_Y[+1]`, so `A_Y[+1]=x` - let $p_x$ be the last one that writes `A_Y[+1]`, so `A_Y[+1]=x`
- for $p_x$ to pass at level +1, it must be that $∀k≠x. F[k] < +1$, then $p_x$ is the only proc at level +1 and the thesis holds, since 1<=n--1 - for $p_x$ to pass at level +1, it must be that $∀k≠x. F[k] < +1$, then $p_x$ is the only proc at level +1 and the thesis holds, since 1<=n--1
- otherwise, $p_x$ is blocked in the wait and so we have at most n--1 processes at level +1 (i.e., those at level , that by induction are at most n-, except for px that is blocked in its (+1)-th wait). - otherwise, $p_x$ is blocked in the wait and so we have at most n--1 processes at level +1: those at level , that by induction are at most n-, except for px that is blocked in its wait.
##### Starvation freedom proof ##### Starvation freedom proof
**Lemma:** every process at level ($\leq n-1$) eventually wins $\to$ starvation freedom holds by taking $=0$.
Reverse induction on
Base ($=n-1$): trivial