Not sure I get the rationality aspect of this—seems like a pretty basic high-school geometry problem. The guard is stationary on a line, and his visibility is a semi-circle. The line from the guard to you is perpendicular to the wall, so you can go a quarter-circle in either direction. Circumference is 2*Pi*r, so a quarter is ^{1}⁄_{2} * Pi * r. And you’ll cross the wall at the same distance from the guard you start at, r. Well, r plus epsilon (for both calculations), I guess, but that rounds to r for this calculation.

That’s not the answer you give, so maybe I don’t understand something. Or maybe it’s a formatting error.

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.

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.

Not sure I get the rationality aspect of this—seems like a pretty basic high-school geometry problem. The guard is stationary on a line, and his visibility is a semi-circle. The line from the guard to you is perpendicular to the wall, so you can go a quarter-circle in either direction. Circumference is 2*Pi*r, so a quarter is

^{1}⁄_{2}* Pi * r. And you’ll cross the wall at the same distance from the guard you start at, r. Well, r plus epsilon (for both calculations), I guess, but that rounds to r for this calculation.That’s not the answer you give, so maybe I don’t understand something. Or maybe it’s a formatting error.

Please clarify the problem statement (maybe include a diagram of the initial setup):

Does the night watch move? a. If so, what are our relative speeds? b. Does he need to catch me, or just see me and raise the alarm?

Do I understand correctly that the night watch’s field of view is a circle with some radius (blocked by the wall)?

Is the night watch right next to the wall, or some distance away?

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.