User:Rafat ahmad ali/sandbox

- In Lamport’s clock, C (e) <C (e’) does not imply e → e’; while in Vector timestamp, V (e)<V(e’) implies e → e’.

- Vector timestamps take up an amount of storage and message payload that is proportional to N,     the number of process; while Lamport’s clock does not.

- In Vector timestamp, let V(e) be the vector timestamp applied by the process at which e occurs. We can compare vector timestamps as follows:

- V = V’ iff V[j] = V’[j] for j = 1, 2, .. , N. - V ≤ V’ iff V[j] ≤ V’[j] for j = 1, 2, .. , N. - V < V’ iff V ≤ V’ ^ V ≠ V’.

- It is straightforward to show, by induction on the length of any sequence of events relating two events e and e’, that e → e’  V (e) <V (e’). However, in Lamport’s clock, the fact that from C (e) <C (e’) we cannot conclude that e → e’.

- Vector timestamps have the disadvantages, compared with Lamport timestamps, of taking up an amount of storage and message payload that is proportional to N,  the number of processes.