# Licht comments on [missing post]

• Thank you for the comment, I will try to incorporate some of the suggestions into the puzzle to help with clarity.

1) a. The night-watch is stationary, there are only two paths which you can take that will get you to the wall in the quickest possible way. b. Once in the field of view of the night-watch the alarm will raise and your cover blown.

2) You understand correctly.

3) They are situated on the wall.

• 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.

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