day13notes.txt

# Part 1

"The transparent paper is pretty big"… how big exactly? There are
735 points in my input, and the maximum coordinate is 1309.

Great, this doesn't fit in memory. I still have my page file experiment
from day 5 if necessary, but I'd rather find a more elegant solution.

One-row-at-a-time did work pretty well for day 5. This looks easy to do
for folds along x=…, but folding vertically will most likely require
multiple file reads.

But for now, we only need to realize a single fold.

Annoyingly, the sample has the first fold done vertically, but my input
has it horizontal.

But the text shows how the sample is supposed to look like after its first
(vertical) fold, so:

  awk '{for(i=1;i<=length($0);i++){if(substr($0,i,1)=="#") printf "%d,%d\n", i-1,NR-1}}'|sort -g

gives me the dot list for that output.

Curious, how big will my final output be? I have 5 horizontal folds to make,
and 7 vertical. That will divide the width by 32, and the height by 128,
bringing the size down to a much more manageable 40x10 grid…

  (((((40*2+1)*2+1)*2+1)*2+1)*2+1) = 1311
  (((((((10*2+1)*2+1)*2+1)*2+1)*2+1)*2+1)*2+1) = 1407

Something to watch out for: after a fold, some dots may end up with negative coordinates.

Wait a second, why on earth have I been thinking in terms of a matrix? I really
only need to handle the dot coordinate pairs!

Storing 735 short pairs will be easy. I just need to count how many distinct pairs remain
after the folds. I could try doing something smart… or just go for the n^2 algorithm, with
such a small n!

Parsing done, after a long hour…

Horizontal folding:

  fold along x=3

  f(x) = 3 - (x - 3) = 2*3 - x
  f(4) = 3 - (4 - 3) = 3 - 1 = 2
  f(6) = 3 - (6 - 3) = 3 - 3 = 0

After another hour, the first horizontal fold is executed.
Turns out there was no negative coordinate involved.

# Part 2

No big surprise here. I still need two things:
  - implement vertical folding
  - draw the resulting matrix