# Puzzle: 1

*(I have been wanting to try my hand at creating puzzles to be done quick for some time now and here is my first attempt. I don’t know yet how to rank its difficulty, so I will let you try and let me know what you think. I would also appreciate any feedback or suggestions on how I can improve, i.e. how to make it more challenging, where things were to obvious, how I might better word the puzzle to hide the logic, that sort of thing. Thank you, and have fun!)*

**The puzzle should be done as quickly as possible (and try your hand at doing it without pen and paper), so set a timer once you have red and understood the question to see how long it takes you, and have fun! **

You are a spy that is standing in front of a wall of infinite length. Your mission is to infiltrate the enemy, by getting past the wall, and put a stop to a life threatening event. Directly in front of you stands the night-watch, who although stationary, is able to see everything in all directions in front of the wall on which they stand, and right up to a point behind which you stand.

You have to infiltrate the enemy as fast as possible, for if you don’t, the population on earth will be dramatically altered. Once at the wall you have the espionage acumen to get past flawlessly and carry out the mission, but if caught by the night-watch at any point before reaching the wall, the alarm is raised immediately and you will fail the mission.

**The question is:** What is the distance you move to infiltrate the wall in the quickest time possible, without being detected by the night-watch?

*Bonus:* How far away from the night-watch are you when you have infiltrated the enemy?

*(The answers are below)*

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Main question: The distance that the spy will travel from their starting point, to arrive the quickest to the wall, and at which point they wont get the night-watches attention, is

Bonus question: They will arrive at the wall at a distance equal to the sight of the night watch plus their position point just beyond that. .

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.

(It was my first time trying tags, I had the problem set tag in mind. I also intended for it to be a timed test, but realized that it might be to on the nose. )It is a formatting error though and I’ve had to go back and edit a couple times to understand where I should make discrepancies in the text so that things aren’t left open ended. In certain cases defining certain aspects has changed the answer, but now I know the specificity needed for these types of exercises.

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.