Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Functional programming, design and analysis of algorithms / 5 / a / i

Start with $(d[1],d[2],d[3],d[4])=(0,\infty ,\infty ,\infty )$. With in-place edge relaxation, the Bellman-Ford algorithm makes the following changes during its first lexicographic pass:The next pass makes no change; any remaining fixed-count passes also make none. Thus the final vector is $(0,4,2,3)$. These are sequential updates within a pass, rather than the simultaneous values of the Synchronous Bellman-Ford recurrence.