Challenge #61: Save the Astronaut

[JosephSerpis]

Challenge Completed

[jamielaird]

This was great fun. Here's my solution - just a smidge under 40 minutes assuming a 1 second dwell time on each card and no final period. I'm making the assumption that longer dwell time for duplicate cards will be interpreted as multiple instances of that card.

EDIT - is it just me, or does anyone else find splitting a string to rows *incredibly* satisfying? It's such a simple thing, but it gets me every time.

Spoiler

[kelly_gilbert]

Pretty fun! My solution doesn't match the given solution due to one assumption (see detail in spoiler). Otherwise, my workflow is pretty similar to the given solution.

Spoiler

Note, my solution assumes the camera can make a complete rotation (i.e. through zero), so my result is slightly lower because it takes the most efficient path available.

The given solution assumes that the camera cannot make a complete rotation (e.g. rotating from 339 degrees, clockwise through zero, to 148 degrees = 21+148 = 169 degrees). In this case, the solution would take the longer route, counterclockwise, from 339 to 148 (passing the 180-degree point) = 191 degrees.

This occurs in the word "moon", since "on" = hex 6F6E.
When moving from the F (338.82 degrees) to 6 (148.23 degrees), the shortest path would be to rotate clockwise, from 338 --> past 0 --> 148 = 169.41 degrees.

The solution takes the longer way around (counterclockwise, from 338 --> past 180 --> 148 = 190.59 degrees).

[kat]

I let my camera go around in circles, back and forwards, which allowed it to be a little quicker. I tried to make it slower by adding end of transmission and moving it back to the last letter (that's how I understood that part). So interesting to see what different assumptions everyone makes :)

Spoiler

[danilang]

I assumed that the camera can pan both directions and paused for 1 second for each card.

Spoiler

i built a time between letters table and joined to the hex characters of the message assuming they started on the question.

Dan

[KOBoyle]

Solution attached. I calculated the shortest path to the next card. After reading others comments, I too went back and added a 1 second pause.

Spoiler

[Gauthier_B]

I got pretty close but could not manage to match your result.

I went on assuming that the rover could pan left or right based on the shorter distance.

Gonna have a look at the solution now

[OldDogNewTricks]

My solution is attached.

[ch12345]

I made a fundamental error in my processing: assuming that the message was never the target of the camera (so missing it from the 360 degree camera view, meaning that each letter was 22.5 degrees apart, simplifying finding the fastest route).

I also manually added in the start and end characters, although admittedly in the context they'd be fairly useless (unless you put a bell between 0 (message end) and 1 to indicate a new message had started).

Speculation aside, here is my inaccurate, and slower solution:

Spoiler