bonanova

There is a third solution. It involves physics. [spoiler=Clue for the physics solution]What if the vectors pointed inward?

An arbitrary convex polygon has a vector drawn from the midpoint of each of its sides. The vectors are perpendicular to the sides, pointed outward, and have the same lengths as the sides themselves. What is the sum of these vectors?

One important assumption not to make is that the walls are all connected. There can be islands. That is, you won't necessarily get through the maze, nor explore all of its area, simply by keeping a wall on the robot's left or right.

No problem revisiting old problems, as far as I am concerned, although generally it's not desirable. For all practical purposes there is a new set of Denizens solving problems these days, and, as I have proven, the remaining solvers may have forgotten the old problems. In most cases if they remember, they just keep quiet and watch. No, it's not mine, and I don't know who to credit. I got this from a book that I had back then, and can't find now. It's not from Martin Gardner, my past muse, but possibly from Peter Winckler. I searched for it to revisit the answer, but could not find it. Finally, the site search found my old posting of it. Thankfully, I explained it then in a way that I can understand now. A tribute, I guess, to my former clarity. Also, it was not completely solved back then. Prof Templeton gave an example that semi-proved his answer, however, so I gave him credit.

Now I know that old age has caught up with me. The reason I had thought this one through a while ago is that in 2009. And now that I've re-read it, I understand the key thought.

This one has me stumped. The x and y views show a single gas cloud (blue/cyan) centered at the top of those fields of view. My thinking is that should create a central gas cloud in the z view, but none exists.

I meant to say his mean displacement from the origin (a vector quantity) is zero. But that's not what you asked.

Why is it unfortunate? What part of the solution are you challenging?

@gavinksong, I'm guessing the reply was a link to the paper.

I believe you, antel0pe, simply because you say so. Multiple choice is a fav of teachers. Thanks for contributing.

Yeah. Now we're at the point where the "obviously" impossible case is being asserted to be possible. Somewhere around 2008 I thought through the process to see it, but I can't do it again now. I love the clarity of the arguments being made here. I can't provide the answer, but I hope to understand it once again when it's explained.

Homework alert?

Why do I say that?

OK so I guess that approach is not helpful then ... I love puzzles of this type because when you see the solution it usually leads to ... .

It must mean that the now-circular 3 and 4 faces are the inscribed circles of the original square faces.

Define his walk. Is his azimuth not limited to vertical or horizontal steps? [spoiler='I think it does not matter']1. Since the direction of each step is random, the mean distance traveled is zero. 2. Since the mean is zero, the standard deviation and rms (root mean square) of the distance for large number of steps are the same, and equal to N0.5 yards, where N is the number of yard-steps taken. To travel a distance s yards requires s2 steps.