The shortest path to the wall avoiding the semicircular FOV is a quarter-circle with length pi/2 times r (arriving at a distance r from the night watch).
(I corrected the answer I gave for the first part, thank you) The answer needs a slight specification in that the r is not that of the night-watches observable field, but r+epsilon, indicating that the spy remains one point beyond the FOV of the night-watch.
You specify that the vision is a sharp cutoff, so it’s an infinitessimally small difference (will be referred to as “epsilon” when you get to calculus), which rounds to 0 in this problem, for any precision of r.
In that case, I get a different answer:
The shortest path to the wall avoiding the semicircular FOV is a quarter-circle with length pi/2 times r (arriving at a distance r from the night watch).
(I corrected the answer I gave for the first part, thank you) The answer needs a slight specification in that the r is not that of the night-watches observable field, but r+epsilon, indicating that the spy remains one point beyond the FOV of the night-watch.
You specify that the vision is a sharp cutoff, so it’s an infinitessimally small difference (will be referred to as “epsilon” when you get to calculus), which rounds to 0 in this problem, for any precision of r.