Same person as @[email protected], different instance.
Lemmy is open source, so you can easily download its source code and open it in neovim :)
The article only summarizes it shortly, but the parallels to the Munich Agreement from 1938 are really scary.
Hitler’s aim was to take over all of Czechoslovakia by breaking it apart. The subject of the Munich Agreement was the Sudetenland, the region bordering Germany. Before there were some votes and local political forces expressing the wish of the German minority in the Sudetenland to create an independent state (See the parallels with DNR, LNR and Crimea). This was used by Hitler to justify taking over the region. Suddenly it wasn’t about independence anymore, but about inclusion into Germany.
The Czechoslovakian government in Prague obviously hated the idea, but they were not invited to the talks in Munich. Only afterwards were they made aware of the decision that would be imposed on their nation. Who was invited was fellow fascist Mussolini from Italy, as well as France and UK, who gave in and signed this agreement, giving international support to Germany just taking over parts of neighboring nations.
Their reasoning was, if they were to disagree, Hitler would assert his will by force and take Czechoslovakia militarily, starting a large European war (that is also the reason Prague was forced to accept the decision: the alternative was a war they could never win, they could not count on any outside help). This was the so-called appeasement policy by the UK. They bought “peace” in exchange for territories they didn’t own but felt the right to decide over. We all know how this heavily-priced peace turned out. At most it gave the allied forces one more year to prepare for WWII.
Reminds me of the beginning of Zelenskys presidency in 2020 when he refused the deal with Trump to investigate Hunter Biden. Trump really believes he can get Ukraine to do anything for him.
Just wait until they figure out how much carbon is captured by planting a tree.
Only if you convert rubles by “purchase power parity” as opposed to the market rate, which seems like a weird way to manipulate data to fit some narrative.
Funny idea, but good luck getting this leaky bucket from the Baltic sea all the way via Gibraltar to the Black sea.
Nah, this photo is definitely not from this year.
Pretty sure that’s correct. You can also see the elevated U2 in the background.
“Fraunce” is my favorite part of this map.
Nice ending for this year. Lock and key arrays are just added together and all elements must be <= 5. Merry Christmas!
fn flatten_block(block: Vec<Vec<bool>>) -> [u8; 5] {
let mut flat = [0; 5];
for row in &block[1..=5] {
for x in 0..5 {
if row[x] {
flat[x] += 1;
}
}
}
flat
}
fn parse(input: &str) -> (Vec<[u8; 5]>, Vec<[u8; 5]>) {
let mut locks = Vec::new();
let mut keys = Vec::new();
for block_s in input.split("\n\n") {
let block: Vec<Vec<bool>> = block_s
.lines()
.map(|l| l.bytes().map(|b| b == b'#').collect::<Vec<bool>>())
.collect();
assert_eq!(block.len(), 7);
// Lock
if block[0].iter().all(|e| *e) {
locks.push(flatten_block(block));
} else {
keys.push(flatten_block(block));
}
}
(locks, keys)
}
fn part1(input: String) {
let (locks, keys) = parse(&input);
let mut count = 0u32;
for l in locks {
for k in &keys {
if l.iter().zip(k).map(|(li, ki)| li + ki).all(|sum| sum <= 5) {
count += 1;
}
}
}
println!("{count}");
}
fn part2(_input: String) {
println!("⭐");
}
util::aoc_main!();
Also on github
Yikers. Part 2 took a while, staring at this diagram for hours. Eventually I noticed that each of these blocks has two pairs of (XOR, AND) gates sharing the same inputs (and inputs aren’t changed). So I matched up these pairs based on a distance metric of how much needs to be swapped to fit together. This helped me identify 4 blocks with errors, the rest was solved using pen and paper (one block is missing as it became apparent at that point):
There is also some code, but do yourself and me a favor and don’t look at it. While it does turn up the correct solution, it probably won’t with any other input, especially not the examples.
Finding cliques in a graph, which is actually NP-comlete. For part two I did look up how to do it and implemented the Bron-Kerbosch algorithm. Adding the pivoting optimization improved the runtime from 134ms to 7.4ms, so that is definitely worth it (in some sense, of course I already had the correct answer without pivoting).
use rustc_hash::{FxHashMap, FxHashSet};
fn parse(input: &str) -> (Vec<Vec<usize>>, FxHashMap<&str, usize>) {
let mut graph = Vec::new();
let mut names: FxHashMap<&str, usize> = FxHashMap::default();
for l in input.lines() {
let (vs, ws) = l.split_once('-').unwrap();
let v = *names.entry(vs).or_insert_with(|| {
graph.push(vec![]);
graph.len() - 1
});
let w = *names.entry(ws).or_insert_with(|| {
graph.push(vec![]);
graph.len() - 1
});
graph[v].push(w);
graph[w].push(v);
}
(graph, names)
}
fn part1(input: String) {
let (graph, names) = parse(&input);
let mut triples: FxHashSet<[usize; 3]> = FxHashSet::default();
for (_, &v) in names.iter().filter(|(name, _)| name.starts_with('t')) {
for (i, &u) in graph[v].iter().enumerate().skip(1) {
for w in graph[v].iter().take(i) {
if graph[u].contains(w) {
let mut triple = [u, v, *w];
triple.sort();
triples.insert(triple);
}
}
}
}
println!("{}", triples.len());
}
// Bron-Kerbosch algorithm for finding all maximal cliques in a graph
fn bron_kerbosch(
graph: &[Vec<usize>],
r: &mut Vec<usize>,
mut p: FxHashSet<usize>,
mut x: FxHashSet<usize>,
) -> Vec<Vec<usize>> {
if p.is_empty() && x.is_empty() {
return vec![r.to_vec()];
}
let mut maximal_cliques = Vec::new();
let Some(&u) = p.iter().next() else {
return maximal_cliques;
};
let mut p_pivot = p.clone();
for w in &graph[u] {
p_pivot.remove(w);
}
for v in p_pivot {
let pn = graph[v].iter().filter(|w| p.contains(w)).copied().collect();
let xn = graph[v].iter().filter(|w| x.contains(w)).copied().collect();
r.push(v);
let new_cliques = bron_kerbosch(graph, r, pn, xn);
r.pop();
maximal_cliques.extend(new_cliques);
p.remove(&v);
x.insert(v);
}
maximal_cliques
}
fn part2(input: String) {
let (graph, names) = parse(&input);
let p = (0..graph.len()).collect();
let mut r = Vec::new();
let maximal_cliques = bron_kerbosch(&graph, &mut r, p, FxHashSet::default());
let maximum_clique = maximal_cliques
.iter()
.max_by_key(|clique| clique.len())
.unwrap();
let mut lan_names: Vec<&str> = names
.iter()
.filter(|(_, v)| maximum_clique.contains(v))
.map(|(name, _)| *name)
.collect();
lan_names.sort_unstable();
println!("{}", lan_names.join(","));
}
util::aoc_main!();
Also on github
Glad to be able to help. This one really was a doozy.
Nice breather today (still traumatized from the robots). At some point I thought you had to do some magic for predicting special properties of the pseudorandom function, but no, just collect all values, have a big table for all sequences and in the end take the maximum value in that table. Part 1 takes 6.7ms, part 2 19.2ms.
fn step(n: u32) -> u32 {
let a = (n ^ (n << 6)) % (1 << 24);
let b = a ^ (a >> 5);
(b ^ (b << 11)) % (1 << 24)
}
fn part1(input: String) {
let sum = input
.lines()
.map(|l| {
let n = l.parse().unwrap();
(0..2000).fold(n, |acc, _| step(acc)) as u64
})
// More than 2¹⁰ 24-bit numbers requires 35 bits
.sum::<u64>();
println!("{sum}");
}
const N_SEQUENCES: usize = 19usize.pow(4);
fn sequence_key(sequence: &[i8]) -> usize {
sequence
.iter()
.enumerate()
.map(|(i, x)| (x + 9) as usize * 19usize.pow(i as u32))
.sum()
}
fn part2(input: String) {
// Table for collecting the amount of bananas for every possible sequence
let mut table = vec![0; N_SEQUENCES];
// Mark the sequences we encountered in a round to ensure that only the first occurence is used
let mut seen = vec![false; N_SEQUENCES];
for l in input.lines() {
let n = l.parse().unwrap();
let (diffs, prices): (Vec<i8>, Vec<u8>) = (0..2000)
.scan(n, |acc, _| {
let next = step(*acc);
let diff = (next % 10) as i8 - (*acc % 10) as i8;
*acc = next;
Some((diff, (next % 10) as u8))
})
.unzip();
for (window, price) in diffs.windows(4).zip(prices.iter().skip(3)) {
let key = sequence_key(window);
if !seen[key] {
seen[key] = true;
table[key] += *price as u32;
}
}
// Reset seen sequences for next round
seen.fill(false);
}
let bananas = table.iter().max().unwrap();
println!("{bananas}");
}
util::aoc_main!();
Also on github
For me this was the hardest puzzle so far this year. First I did a bunch of things that turned out not to work properly. For example I tried to solve it with a greedy algorithm that always moved horizontally first then vertically, but that ignores the gaps that need to be avoided (what a sneaky requirement) and also somehow doesn’t guarantee the shortest sequence.
After reaching part 2 it became clear that a recursive function (with memoization) is needed again, and of course in the end it turned out a lot simpler than what I had cooked up before (you don’t want to see that). Now even part 2 takes just 1.6ms.
use euclid::{default::*, point2, vec2};
use rustc_hash::FxHashMap;
use std::iter;
type Move = Option<Vector2D<i32>>;
const KEYPAD_GAP: Point2D<i32> = point2(0, 3);
const DPAD_GAP: Point2D<i32> = point2(0, 0);
fn keypad_pos(n: u8) -> Point2D<i32> {
match n {
b'7' => point2(0, 0),
b'8' => point2(1, 0),
b'9' => point2(2, 0),
b'4' => point2(0, 1),
b'5' => point2(1, 1),
b'6' => point2(2, 1),
b'1' => point2(0, 2),
b'2' => point2(1, 2),
b'3' => point2(2, 2),
b'0' => point2(1, 3),
b'A' => point2(2, 3),
other => panic!("Invalid keypad symbol {other}"),
}
}
// `None` is used for A
fn dpad_pos(d: Move) -> Point2D<i32> {
match d {
Some(Vector2D { x: 0, y: -1, .. }) => point2(1, 0),
None => point2(2, 0),
Some(Vector2D { x: -1, y: 0, .. }) => point2(0, 1),
Some(Vector2D { x: 0, y: 1, .. }) => point2(1, 1),
Some(Vector2D { x: 1, y: 0, .. }) => point2(2, 1),
other => panic!("Invalid dpad symbol {other:?}"),
}
}
fn moves_for_diff(diff: Vector2D<i32>, pos: Point2D<i32>, gap: Point2D<i32>) -> Vec<Vec<Move>> {
let horizontal = iter::repeat_n(
Some(vec2(diff.x.signum(), 0)),
diff.x.unsigned_abs() as usize,
);
let vertical = iter::repeat_n(
Some(vec2(0, diff.y.signum())),
diff.y.unsigned_abs() as usize,
);
if pos + vec2(diff.x, 0) == gap {
// Must not move horizontal first, or we hit the gap
vec![vertical.chain(horizontal).chain(iter::once(None)).collect()]
} else if pos + vec2(0, diff.y) == gap {
vec![horizontal.chain(vertical).chain(iter::once(None)).collect()]
} else {
// Try both horizontal first and vertical first
vec![
horizontal
.clone()
.chain(vertical.clone())
.chain(iter::once(None))
.collect(),
vertical.chain(horizontal).chain(iter::once(None)).collect(),
]
}
}
fn dpad_sequence_len(
start: Move,
end: Move,
rounds: u32,
cache: &mut FxHashMap<(Move, Move, u32), u64>,
) -> u64 {
if rounds == 0 {
return 1;
}
if let Some(len) = cache.get(&(start, end, rounds)) {
return *len;
}
let start_pos = dpad_pos(start);
let end_pos = dpad_pos(end);
let diff = end_pos - start_pos;
let possible_paths = moves_for_diff(diff, start_pos, DPAD_GAP);
let shortest_sequence = possible_paths
.iter()
.map(|moves| {
moves
.iter()
.fold((0, None), |(cost, prev), &m| {
(cost + dpad_sequence_len(prev, m, rounds - 1, cache), m)
})
.0
})
.min()
.unwrap();
cache.insert((start, end, rounds), shortest_sequence);
shortest_sequence
}
fn keypad_sequence_len(start: u8, end: u8, rounds: u32) -> u64 {
let start_pos = keypad_pos(start);
let end_pos = keypad_pos(end);
let diff = end_pos - start_pos;
let possible_paths = moves_for_diff(diff, start_pos, KEYPAD_GAP);
let mut cache = FxHashMap::default();
possible_paths
.iter()
.map(|moves| {
moves
.iter()
.fold((0, None), |(cost, prev), &m| {
(cost + dpad_sequence_len(prev, m, rounds, &mut cache), m)
})
.0
})
.min()
.unwrap()
}
fn solve(input: &str, rounds: u32) -> u64 {
let mut sum: u64 = 0;
for l in input.lines() {
let mut prev = b'A';
let mut len = 0;
for b in l.bytes() {
len += keypad_sequence_len(prev, b, rounds);
prev = b;
}
let code_n: u64 = l.strip_suffix('A').unwrap().parse().unwrap();
sum += code_n * len;
}
sum
}
fn part1(input: String) {
println!("{}", solve(&input, 2));
}
fn part2(input: String) {
println!("{}", solve(&input, 25));
}
util::aoc_main!();
Also on github
There is exactly one path without cheating, so yes, the distance to one end is always the total distance minus the distance to the other end.
The important part here is to build two maps first that hold for every point on the path the distance to the start and the distance to the end respectively. Then calculating the path length for a cheat vector is a simple lookup. For part 2 I first generated all vectors with manhattan distance <= 20, or more specifically, exactly half of those vectors to avoid checking the same vector in both directions.
Part 2 takes 15ms. The code looks a bit unwieldy at parts and uses the pyramid of doom paradigm but works pretty well.
use euclid::{default::*, vec2};
const DIRS: [Vector2D<i32>; 4] = [vec2(1, 0), vec2(0, 1), vec2(-1, 0), vec2(0, -1)];
const MIN_SAVE: u32 = 100;
const MAX_DIST: i32 = 20;
fn parse(input: &str) -> (Vec<Vec<bool>>, Point2D<i32>, Point2D<i32>) {
let mut start = None;
let mut end = None;
let mut field = Vec::new();
for (y, line) in input.lines().enumerate() {
let mut row = Vec::new();
for (x, b) in line.bytes().enumerate() {
row.push(b == b'#');
if b == b'S' {
start = Some(Point2D::new(x, y).to_i32());
} else if b == b'E' {
end = Some(Point2D::new(x, y).to_i32());
}
}
field.push(row);
}
(field, start.unwrap(), end.unwrap())
}
fn distances(
field: &[Vec<bool>],
start: Point2D<i32>,
end: Point2D<i32>,
) -> (Vec<Vec<u32>>, Vec<Vec<u32>>) {
let width = field[0].len();
let height = field.len();
let mut dist_to_start = vec![vec![u32::MAX; width]; height];
let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height)).to_i32();
let mut cur = start;
let mut dist = 0;
dist_to_start[cur.y as usize][cur.x as usize] = dist;
while cur != end {
for dir in DIRS {
let next = cur + dir;
if bounds.contains(next)
&& !field[next.y as usize][next.x as usize]
&& dist_to_start[next.y as usize][next.x as usize] == u32::MAX
{
cur = next;
break;
}
}
dist += 1;
dist_to_start[cur.y as usize][cur.x as usize] = dist;
}
let total_dist = dist_to_start[end.y as usize][end.x as usize];
let dist_to_end = dist_to_start
.iter()
.map(|row| {
row.iter()
.map(|&d| {
if d == u32::MAX {
u32::MAX
} else {
total_dist - d
}
})
.collect()
})
.collect();
(dist_to_start, dist_to_end)
}
fn cheats(
field: &[Vec<bool>],
dist_to_start: &[Vec<u32>],
dist_to_end: &[Vec<u32>],
total_dist: u32,
) -> u32 {
let width = field[0].len();
let height = field.len();
let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height)).to_i32();
let mut count = 0;
for (y, row) in field.iter().enumerate() {
for (x, _w) in row.iter().enumerate().filter(|&(_i, w)| *w) {
let pos = Point2D::new(x, y).to_i32();
for (d0, &dir0) in DIRS.iter().enumerate().skip(1) {
for &dir1 in DIRS.iter().take(d0) {
let p0 = pos + dir0;
let p1 = pos + dir1;
if bounds.contains(p0) && bounds.contains(p1) {
let p0 = p0.to_usize();
let p1 = p1.to_usize();
if !field[p0.y][p0.x] && !field[p1.y][p1.x] {
let dist = dist_to_start[p0.y][p0.x].min(dist_to_start[p1.y][p1.x])
+ dist_to_end[p1.y][p1.x].min(dist_to_end[p0.y][p0.x])
+ 2; // Add 2 for cutting across the wall
if total_dist - dist >= MIN_SAVE {
count += 1;
}
}
}
}
}
}
}
count
}
fn part1(input: String) {
let (field, start, end) = parse(&input);
let (dist_to_start, dist_to_end) = distances(&field, start, end);
let total_dist = dist_to_start[end.y as usize][end.x as usize];
println!(
"{}",
cheats(&field, &dist_to_start, &dist_to_end, total_dist)
);
}
// Half of all vectors with manhattan distance <= MAX_DIST.
// Only vectors with positive x or going straight down are considered to avoid using the same
// vector twice in both directions.
fn cheat_vectors() -> Vec<Vector2D<i32>> {
let mut vectors = Vec::new();
for y in -MAX_DIST..=MAX_DIST {
let start = if y > 0 { 0 } else { 1 };
for x in start..=(MAX_DIST - y.abs()) {
assert!(x + y <= MAX_DIST);
vectors.push(vec2(x, y));
}
}
vectors
}
fn cheats20(
field: &[Vec<bool>],
dist_to_start: &[Vec<u32>],
dist_to_end: &[Vec<u32>],
total_dist: u32,
) -> u32 {
let vectors = cheat_vectors();
let width = field[0].len();
let height = field.len();
let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height)).to_i32();
let mut count = 0;
for (y, row) in field.iter().enumerate() {
for (x, _w) in row.iter().enumerate().filter(|&(_i, w)| !*w) {
let p0 = Point2D::new(x, y);
for &v in &vectors {
let pi1 = p0.to_i32() + v;
if bounds.contains(pi1) {
let p1 = pi1.to_usize();
if !field[p1.y][p1.x] {
let dist = dist_to_start[p0.y][p0.x].min(dist_to_start[p1.y][p1.x])
+ dist_to_end[p1.y][p1.x].min(dist_to_end[p0.y][p0.x])
+ v.x.unsigned_abs() // Manhattan distance of vector
+ v.y.unsigned_abs();
if total_dist - dist >= MIN_SAVE {
count += 1;
}
}
}
}
}
}
count
}
fn part2(input: String) {
let (field, start, end) = parse(&input);
let (dist_to_start, dist_to_end) = distances(&field, start, end);
let total_dist = dist_to_start[end.y as usize][end.x as usize];
println!(
"{}",
cheats20(&field, &dist_to_start, &dist_to_end, total_dist)
);
}
util::aoc_main!();
Also on github
Writing the same number a different way does not make it rational. There are no two natural numbers p and q so that p/q = 1 base pi.
The 3ms are for part 1 only, part 2 takes around 27ms. But I see that our approaches there are very similar. One difference that might make an impact is that you copy the substrings for inserting into the hashmap into String
s.
It’s the part in the video where you can see the Linux Desktop.