Day 9

Author: tmoux

AdventOfCode23.cabal

@@ -83,3 +83,8 @@ executable Day6
  import: warnings, dependencies
  main-is: Main.hs
  hs-source-dirs: Day6
+
+executable Day9
+ import: warnings, dependencies
+ main-is: Main.hs
+ hs-source-dirs: Day9

Day9/Main.hs

@@ -0,0 +1,26 @@
+module Main where
+
+-- https://stackoverflow.com/questions/9512513/how-do-you-compute-the-difference-between-successive-elements-of-a-list-of-unkno
+diffs :: Num a => [a] -> [a]
+diffs [] = []
+diffs l = zipWith (-) (tail l) l
+
+part1 :: [[Int]] -> Int
+part1 = sum . map f
+ where
+ f [] = 0
+ f xs | all (0 ==) xs = 0
+ | otherwise = last xs + f (diffs xs)
+
+part2 :: [[Int]] -> Int
+part2 = sum . map f
+ where
+ f [] = 0
+ f xs | all (0 ==) xs = 0
+ | otherwise = head xs - f (diffs xs)
+
+main :: IO ()
+main = do
+ input <- map (map read . words) . lines <$> getContents
+ putStrLn $ "Part 1: " ++ show (part1 input)
+ putStrLn $ "Part 2: " ++ show (part2 input)

Day9/Makefile

@@ -0,0 +1,16 @@
+CC = g++-13
+CFLAGS = -O2 -std=c++20 -Wall
+
+all: sol
+
+sol: sol.cpp
+$(CC) $(CFLAGS) -o sol sol.cpp
+
+gen: gen.cpp
+$(CC) $(CFLAGS) -o gen gen.cpp
+
+brute: brute.cpp
+$(CC) $(CFLAGS) -o brute brute.cpp
+
+clean:
+$(RM) sol gen brute

Day9/old.cpp

@@ -0,0 +1,83 @@
+#include <bits/stdc++.h>
+using namespace std;
+using ll = long long;
+
+// Template
+#define REP(n) for (int _=0; _<(n); _++)
+#define FOR(i, a, b) for (int i=a; i<(b); i++)
+#define F0R(i, a) for (int i=0; i<(a); i++)
+#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
+#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
+ 
+#define sz(x) (int)(x).size()
+#define all(x) x.begin(), x.end()
+ 
+template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
+template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
+
+namespace std {
+ template<class Fun>
+ class y_combinator_result {
+ Fun fun_;
+ public:
+ template<class T>
+ explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
+ 
+ template<class ...Args>
+ decltype(auto) operator()(Args &&...args) {
+ return fun_(std::ref(*this), std::forward<Args>(args)...);
+ }
+ };
+ 
+ template<class Fun>
+ decltype(auto) y_combinator(Fun &&fun) {
+ return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun));
+ }
+} // namespace std
+
+#define DEBUG(x) cerr << #x << ": " << x << '\n'
+template<typename A, typename B> 
+ostream& operator<<(ostream &os, const pair<A, B> &p) { 
+ return os << '(' << p.first << ", " << p.second << ')'; 
+}
+template<typename T_container, 
+ typename T = typename enable_if<!is_same<T_container, string>::value, 
+ typename T_container::value_type>::type> 
+ostream& operator<<(ostream &os, const T_container &v) { 
+ os << '['; string sep; 
+ for (const T &x : v) 
+ os << sep << x, sep = ", "; 
+ return os << ']'; 
+}
+mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
+
+vector<ll> diff(vector<ll> A) {
+ vector<ll> B;
+ for (int i = 1; i < sz(A); i++) {
+ B.push_back(A[i] - A[i-1]);
+ }
+ return B;
+}
+
+ll solve(vector<ll> A) {
+ if (all_of(A.begin(), A.end(), [&](ll x) { return x == 0; })) return 0;
+ auto B = diff(A);
+ return A.back() + solve(B);
+}
+
+int main() {
+ ios_base::sync_with_stdio(false); cin.tie(NULL);
+ string line;
+
+ ll ans = 0;
+ while (getline(cin, line)) {
+ stringstream ss(line);
+ vector<ll> A;
+ ll x;
+ while (ss >> x) {
+ A.push_back(x);
+ }
+ ans += solve(A);
+ }
+ cout << ans << '\n';
+}

Day9/sol.cpp

@@ -0,0 +1,83 @@
+#include <bits/stdc++.h>
+using namespace std;
+using ll = long long;
+
+// Template
+#define REP(n) for (int _=0; _<(n); _++)
+#define FOR(i, a, b) for (int i=a; i<(b); i++)
+#define F0R(i, a) for (int i=0; i<(a); i++)
+#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
+#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
+ 
+#define sz(x) (int)(x).size()
+#define all(x) x.begin(), x.end()
+ 
+template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
+template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
+
+namespace std {
+ template<class Fun>
+ class y_combinator_result {
+ Fun fun_;
+ public:
+ template<class T>
+ explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
+ 
+ template<class ...Args>
+ decltype(auto) operator()(Args &&...args) {
+ return fun_(std::ref(*this), std::forward<Args>(args)...);
+ }
+ };
+ 
+ template<class Fun>
+ decltype(auto) y_combinator(Fun &&fun) {
+ return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun));
+ }
+} // namespace std
+
+#define DEBUG(x) cerr << #x << ": " << x << '\n'
+template<typename A, typename B> 
+ostream& operator<<(ostream &os, const pair<A, B> &p) { 
+ return os << '(' << p.first << ", " << p.second << ')'; 
+}
+template<typename T_container, 
+ typename T = typename enable_if<!is_same<T_container, string>::value, 
+ typename T_container::value_type>::type> 
+ostream& operator<<(ostream &os, const T_container &v) { 
+ os << '['; string sep; 
+ for (const T &x : v) 
+ os << sep << x, sep = ", "; 
+ return os << ']'; 
+}
+mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
+
+vector<ll> diff(vector<ll> A) {
+ vector<ll> B;
+ for (int i = 1; i < sz(A); i++) {
+ B.push_back(A[i] - A[i-1]);
+ }
+ return B;
+}
+
+ll solve(vector<ll> A) {
+ if (all_of(A.begin(), A.end(), [&](ll x) { return x == 0; })) return 0;
+ auto B = diff(A);
+ return A[0] - solve(B);
+}
+
+int main() {
+ ios_base::sync_with_stdio(false); cin.tie(NULL);
+ string line;
+
+ ll ans = 0;
+ while (getline(cin, line)) {
+ stringstream ss(line);
+ vector<ll> A;
+ ll x;
+ while (ss >> x) {
+ A.push_back(x);
+ }
+ ans += solve(A);
+ }
+ cout << ans << '\n';
+}

README.md

@@ -19,6 +19,7 @@ Unless otherwise specified, programs read from standard input and print to stand
 | 6 | 00:03:11 / 00:06:50 | 102 / 354 |
 | 7 | 00:14:08 / 00:28:48 | 333 / 726 |
 | 8 | 00:02:53 / 00:56:12 | 35 / 4397 |
+| 9 | 00:05:44 / 00:06:52 | 295 / 191 |
 
 ### Day 7 (Camel Cards)
 
@@ -38,3 +39,7 @@ Part 1 is a quick implementation exercise.
 Part 2 seems nearly intractable in the general case. After sitting for a while being confused why the problem was so hard, I looked at the inputs and discovered there were only 6 "A"-rooms, and their paths/cycles were disjoint, and only contained one "Z"-room, *and* the lengths of the cycles are equal to the distance to the "Z"-room!
 Therefore the problem reduces to a CRT problem, which itself degenerates to an LCM calculation.
 
+### Day 9 (Mirage Maintenance)
+
+This was a pretty straightforward problem. Unfortunately my performance was hindered by my slow typing today.
+I was also unsure how to handle a case where the list would become empty before we hit all zeros (e.g, `1 2`), but I eventually assumed this wouldn't be a problem and chugged on.