Day 17: Trick Shot
Haskell: Part 1 (02:15:58, rank 6580), Part 2 (03:02:25, rank 6672)
Not sure what happened here. Both parts were pretty simple in retrospect, but I got really confused at the beginning and went down the wrong path.
I basically built most of part 2 in my solution for part 1, then realised it didn't make sense and simplified it. I was trying to find the number of steps where the target area is reached on the x-axis, but only later realised that that's really an infinite number of steps when the x velocity reaches zero. I should have started with y, since I initially thought it could be arbitrarily high, but then realised (by looking at the output of scanl on the console) that the probe when fired upward will always pass back through y=0, at which time its velocity will be the negative of its initial velocity. This means that the maximum value is constrained to be smaller than the negation of the deepest point in the target range.
This part went ok after all the previous hassle: for each value of y and the number of steps that produced it, find the values of x that can reach the target area in the same number of steps.
Sounds simple when you say it.
As demonstrated in my summary of part 2, pausing and speaking out loud what I'm trying to achieve can be more effective than hacker typing.
I was explaining the puzzle to someone afterwards, and it got confusing pretty quickly, so I started dealing with the special case where the target is a 1x1 square at x=0,y=-2. Ignoring the role of x made it much easier to understand the behaviour for y, and we were able to draw the sequence of y positions over several timesteps for various initial velocity values.