diff --git a/2021/src/bin/day08.rs b/2021/src/bin/day08.rs index dc0a44a..7820792 100644 --- a/2021/src/bin/day08.rs +++ b/2021/src/bin/day08.rs @@ -7,7 +7,7 @@ use itertools::Itertools; use std::array; const DAY: usize = 8; -type Parsed<'a> = Vec<([SSD; 10], [SSD; 4])>; +type Parsed = Vec<([SSD; 10], [SSD; 4])>; const VALID_DISPLAYS: [SSD; 10] = [119, 36, 93, 109, 46, 107, 123, 37, 127, 111]; @@ -16,8 +16,8 @@ type SSD = u32; struct Mapping([SSD; 7]); impl Mapping { - fn translate(&self, i: SSD) -> SSD { - 1 << self.0[i as usize] + fn translate(&self, i: usize) -> SSD { + 1 << self.0[i] } } @@ -25,7 +25,7 @@ fn parse(s: &str) -> SSD { ['g', 'f', 'e', 'd', 'c', 'b', 'a'].iter().map(|&c| s.contains(c)).fold(0, |acc, b| (acc | (b as SSD)) << 1) >> 1 } -fn bit_at(x: SSD, n: SSD) -> bool { +fn bit_at(x: SSD, n: usize) -> bool { (x >> n) & 1 != 0 } @@ -44,44 +44,35 @@ fn parse_input(raw: &str) -> Parsed { .collect() } -fn part1<'a>(parsed: &Parsed<'a>) -> usize { - parsed.iter().flat_map(|(_, output)| output).filter(|&&input| [2, 3, 4, 7].contains(&input.count_ones())).count() +fn part1(parsed: &Parsed) -> usize { + parsed.iter().flat_map(|(_, output)| output).filter(|input| [2, 3, 4, 7].contains(&input.count_ones())).count() } -fn part2<'a>(parsed: &Parsed<'a>) -> usize { +fn part2(parsed: &Parsed) -> usize { parsed .iter() - .map(|(input, raw_output)| { + .map(|(input, output)| { let [&one, &four, &seven] = [2, 4, 3].map(|n| input.iter().find(|s| s.count_ones() == n).unwrap()); // We know the position of a for sure because it’s the only difference between 7 and 1 let a = (0..7).position(|n| bit_at(difference(seven, one), n)).unwrap(); // And c and f are these two (both used in 1). - // Contrary to the name, these two values are both c_or_f, - // so we know c and f are these two, but we don’t know which is which. let (c, f) = (0..7).positions(|n| bit_at(one, n)).next_tuple().unwrap(); + // Determine which is which by their frequency in the input. + let (c, f) = if input.iter().filter(|&&i| bit_at(i, c)).count() == 8 { (c, f) } else { (f, c) }; // 4 uses b, c, d, f, but we already know c and f from 1, so this leaves b and d. let (b, d) = (0..7).positions(|n| bit_at(difference(four, one), n)).next_tuple().unwrap(); + let (b, d) = if input.iter().filter(|&&i| bit_at(i, b)).count() == 6 { (b, d) } else { (d, b) }; // Now e and g have to be in the remaining two positions. - let (e, g) = (0..7).filter(|n| ![a, b, c, d, f].contains(n)).next_tuple().unwrap(); - // Now there are 8 possible combinations from multiplying the 3 x_or_y we constructed above. - // This is a manual implementation of itertools::iproduct specialized for 3 small - // arrays because it’s much faster this way. - let mapping = [[c, f], [f, c]] - .into_iter() - .flat_map(|[c, f]| [[c, f, b, d], [c, f, d, b]]) - .flat_map(|[c, f, b, d]| [[c, f, b, d, e, g], [c, f, b, d, g, e]]) - .map(|[c, f, b, d, e, g]| { - let mut m = [0; 7]; - let mut cur = 0; - for i in [a, b, c, d, e, f, g] { - m[i] = cur; - cur += 1; - } - Mapping(m) - }) - .find(|m| input.iter().all(|&i| VALID_DISPLAYS.contains(&(0..7).map(|n| (bit_at(i, n) as SSD) * m.translate(n)).sum()))) - .unwrap(); - raw_output + let (e, g) = (0..7).filter(|n| ![a, b, c, d, f].contains(n)).map_into().next_tuple().unwrap(); + let (e, g) = if input.iter().filter(|&&i| bit_at(i, e)).count() == 4 { (e, g) } else { (g, e) }; + let mut m = [0; 7]; + let mut cur = 0; + for i in [a, b, c, d, e, f, g] { + m[i] = cur; + cur += 1; + } + let mapping = Mapping(m); + output .iter() .map(|&i| (0..7).map(|n| (bit_at(i, n) as SSD) * mapping.translate(n)).sum()) .map(|ssd: SSD| VALID_DISPLAYS.iter().position(|d| &ssd == d).unwrap())