Expand description
Advent of Code 2022 Day 10: https://adventofcode.com/2022/day/10
Because the number of numbers and total states is so small I decided to
simply compute the state of the register at each cycle and then analyze it
afterwards for each part of the challenge. We basically start with the
register x at 1 and then if we see a noop push it into the cycles
vector as-is, otherwise compute the change that’s being made and push it
once as-is, and then again with the adjustment (the instructions state
that the addx operation takes two cycles). This gives us a vector with
the state of the x register at each cycle during the “program”’s
execution.
In part one we then need to calculate a sum based on the state of the
register at various points (cycle 20 and then every 40 cycles after).
And in part two we instead need to draw the output based on the state of
the register. It actually took me a while to figure out what the prompt
was actually asking for so I’ll try to reproduce it here in a hopefully
easier way for me to understand.
The drawing operation works like this: we draw from left to right and from
top to bottom one pixel at a time. We know whether to draw a # or a .
depending on the state of the register for that pixel (the index on the
cycles vector is row * 40 + column. The “sprite” is the value of the
register plus or minus one, and so if the current column falls into the
range of x - 1 or x + 1 then we draw the # character, otherwise the
. character.
Functions§
- The solution for part one of the day ten challenge.
- The solution for part two of the day ten challenge.