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6
Concurrent Systems/notes/6 - Atomicity.md
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@ -51,12 +51,12 @@ We now show that $\to$ is acyclic.
- by contradiction, consider a shortest cycle
- adjacent edges cannot belong to the same order (e.g. not both $\to_X$), otw. the cycle would be shortable, because of transitivity of the total order!
- adjacent edges cannot belong to orders on different objects
- this would mean that an operation is involved in both $\to_X$ and $\to_Y$ but it is not possible of course
- Hence, at least one $\to_X$ exists and it must be between two $\to_H$ i.e.: $$op1 \to_H op2 \to_X op3 \to_H op4$$, with op1 = op4
- this would mean that an operation is involved in both $\to_X$ and $\to_Y$ but it is not possible of course, so the cycle can only happen edges in $\to_X$ and $\to_H$.
- Hence, at least one $\to_X$ exists and it must be between two $\to_H$ i.e.: $$op1 \to_H op2 \to_X op3 \to_H op4$$, likely with op1 = op4
- can this be a cycle?
- $op1 \to_H op2$ means that $res(op1) <_H inv(op2)$
- $op2 \to_X op3$ entails that $inv(op2) <_H res(op3)$
- if not, as is a total order, we would have that $res(op3) <_H inv(op2)$, but we then would have a cycle of lenght 2...
- if not, as is a total order, we would have that $res(op3) <_H inv(op2)$, but we then would have $op3 \to_H op2$ a cycle of lenght 2...
- $op2 \to_H op3$ entails that $inv(op2) <_H res(op3)$