I’m way behind, but I’m trying to learn F#.

I’m using the library Combinatorics in dotnet, which I’ve used in the past, generate in this case every duplicating possibility of the operations. I the only optimization that I did was to use a function to concatenate numbers without converting to strings, but that didn’t actually help much.

I have parser helpers that use ReadOnlySpans over strings to prevent unnecessary allocations. However, here I’m adding to a C# mutable list and then converting to an FSharp (linked) list, which this language is more familiar with. Not optimal, but runtime was pretty good.

I’m not terribly good with F#, but I think I did ok for this challenge.

F#

// in another file:
let concatenateLong (a:Int64) (b:Int64) : Int64 =
    let rec countDigits (n:int64) =
        if n = 0 then 0
        else 1 + countDigits (n / (int64 10))   

    let bDigits = if b = 0 then 1 else countDigits b
    let multiplier = pown 10 bDigits |> int64
    a * multiplier + b

// challenge file
type Operation = {Total:Int64; Inputs:Int64 list }

let parse (s:ReadOnlySpan<char>) : Operation =
    let sep = s.IndexOf(':')
    let total = Int64.Parse(s.Slice(0, sep))
    let inputs = System.Collections.Generic.List<Int64>()
    let right:ReadOnlySpan<char> = s.Slice(sep + 1).Trim()

   // because the Split function on a span returns a SpanSplitEnumerator, which is a ref-struct and can only live on the stack, 
   // I can't use the F# list syntax here
    for range in right.Split(" ") do
        inputs.Add(Int64.Parse(sliceRange right range))
        
    {Total = total; Inputs = List.ofSeq(inputs) }

let part1Ops = [(+); (*)]

let execute ops input =
    input
    |> PSeq.choose (fun op ->
        let total = op.Total
        let inputs = op.Inputs
        let variations = Variations(ops, inputs.Length - 1, GenerateOption.WithRepetition)
        variations
        |> Seq.tryFind (fun v ->
            let calcTotal = (inputs[0], inputs[1..], List.ofSeq(v)) |||> List.fold2 (fun acc n f -> f acc n) 
            calcTotal = total
            )
        |> Option.map(fun _ -> total)
        )
    |> PSeq.fold (fun acc n -> acc + n) 0L

let part1 input =
    (read input parse)
    |> execute part1Ops

let part2Ops = [(+); (*); concatenateLong]

let part2 input = (read input parse) |> execute part2Ops

The Gen0 garbage collection looks absurd, but Gen0 is generally considered “free”.

V2 - concatenate numbers did little for the runtime, but did help with Gen1 garbage, but not the overall allocation.

Uiua

This turned out to be reasonably easy in Uiua, though this solution relies on macros which maybe slow it down.

(edit: removing one macro sped it up quite a bit)

(edit2: Letting Uiua build up an n-dimensional array turned out to be the solution, though sadly my mind only works in 3 dimensions. Now runs against the live data in around 0.3 seconds.)

Try it here

Data   ← ⊜(□⊜⋕⊸(¬∈": "))⊸≠@\n "190: 10 19\n3267: 81 40 27\n83: 17 5\n156: 15 6\n7290: 6 8 6 15\n161011: 16 10 13\n192: 17 8 14\n21037: 9 7 18 13\n292: 11 6 16 20"
Calib! ← ≡◇⊢▽⊸≡◇(∈♭/[^0]:°⊂) # Calibration targets which can be constructed from their values.
&p/+Calib!⊃(+|×)Data
&p/+Calib!⊃(+|×|+×ⁿ:10+1⌊ₙ₁₀,)Data

Python

It is a tree search

def parse_input(path):

  with path.open("r") as fp:
    lines = fp.read().splitlines()

  roots = [int(line.split(':')[0]) for line in lines]
  node_lists = [[int(x)  for x in line.split(':')[1][1:].split(' ')] for line in lines]

  return roots, node_lists

def construct_tree(root, nodes, include_concat):

  levels = [[] for _ in range(len(nodes)+1)]
  levels[0] = [(str(root), "")]
  # level nodes are tuples of the form (val, operation) where both are str
  # val can be numerical or empty string
  # operation can be *, +, || or empty string

  for indl, level in enumerate(levels[1:], start=1):

    node = nodes[indl-1]

    for elem in levels[indl-1]:

      if elem[0]=='':
        continue

      if elem[0][-len(str(node)):] == str(node) and include_concat:
        levels[indl].append((elem[0][:-len(str(node))], "||"))
      if (a:=int(elem[0]))%(b:=int(node))==0:
        levels[indl].append((str(int(a/b)), '*'))
      if (a:=int(elem[0])) - (b:=int(node))>0:
        levels[indl].append((str(a - b), "+"))

  return levels[-1]

def solve_problem(file_name, include_concat):

  roots, node_lists = parse_input(Path(cwd, file_name))
  valid_roots = []

  for root, nodes in zip(roots, node_lists):

    top = construct_tree(root, nodes[::-1], include_concat)

    if any((x[0]=='1' and x[1]=='*') or (x[0]=='0' and x[1]=='+') or
           (x[0]=='' and x[1]=='||') for x in top):

      valid_roots.append(root)

  return sum(valid_roots)

Haskell

A surprisingly gentle one for the weekend! Avoiding string operations for concatenate got the runtime down below one second on my machine.

import Control.Arrow
import Control.Monad
import Data.List
import Data.Maybe

readInput :: String -> [(Int, [Int])]
readInput = lines >>> map (break (== ':') >>> (read *** map read . words . tail))

equatable :: [Int -> Int -> Int] -> (Int, [Int]) -> Bool
equatable ops (x, y : ys) = elem x $ foldM apply y ys
  where
    apply a y = (\op -> a `op` y) <$> ops

concatenate :: Int -> Int -> Int
concatenate x y = x * mag y + y
  where
    mag z = fromJust $ find (> z) $ iterate (* 10) 10

main = do
  input <- readInput <$> readFile "input07"
  mapM_
    (print . sum . map fst . (`filter` input) . equatable)
    [ [(+), (*)],
      [(+), (*), concatenate]
    ]

Haskell

import Control.Arrow
import Data.Char
import Text.ParserCombinators.ReadP

numP = read <$> munch1 isDigit
parse = endBy ((,) <$> (numP <* string ": ") <*> sepBy numP (char ' ')) (char '\n')

valid n [m] = m == n
valid n (x : xs) = n > 0 && valid (n - x) xs || (n `mod` x) == 0 && valid (n `div` x) xs

part1 = sum . fmap fst . filter (uncurry valid . second reverse)

concatNum r = (+r) . (* 10 ^ digits r)
    where
        digits = succ . floor . logBase 10 . fromIntegral

allPossible [n] = [n]
allPossible (x:xs) = ((x+) <$> rest) ++ ((x*) <$> rest) ++ (concatNum x <$> rest)
    where
        rest = allPossible xs

part2 = sum . fmap fst . filter (uncurry elem . second (allPossible . reverse))

main = getContents >>= print . (part1 &&& part2) . fst . last . readP_to_S parse

Haskell

I’m not very proud, I copied my code for part two.

import Control.Arrow hiding (first, second)

import qualified Data.List as List
import qualified Data.Char as Char

parseLine l = (n, os)
        where
                n = read . takeWhile (Char.isDigit) $ l
                os = map read . drop 1 . words $ l

parse :: String -> [(Int, [Int])]
parse s = map parseLine . takeWhile (/= "") . lines $ s

insertOperators target (r:rs) = any (target ==) (insertOperators' r rs)
insertOperators' :: Int -> [Int] -> [Int]
insertOperators' acc []     = [acc]
insertOperators' acc (r:rs) = insertOperators' (acc+r) rs ++ insertOperators' (acc*r) rs

insertOperators2 target (r:rs) = any (target ==) (insertOperators2' r rs)
insertOperators2' :: Int -> [Int] -> [Int]
insertOperators2' acc []     = [acc]
insertOperators2' acc (r:rs) = insertOperators2' (acc+r) rs ++ insertOperators2' (acc*r) rs ++ insertOperators2' concatN rs
        where
                concatN = read (show acc ++ show r)

part1 ls = filter (uncurry insertOperators)
        >>> map fst
        >>> sum
        $ ls
part2 ls = filter (uncurry insertOperators2)
        >>> map fst
        >>> sum
        $ ls

main = getContents >>= print . (part1 &&& part2) . parse

Factor

TUPLE: equation value numbers ;
C: <equation> equation

: get-input ( -- equations )
  "vocab:aoc-2024/07/input.txt" utf8 file-lines [
    split-words unclip but-last string>number
    swap [ string>number ] map <equation>
  ] map ;

: possible-quotations ( funcs numbers -- quots )
  dup length 1 -
  swapd all-selections
  [ unclip swap ] dip
  [ zip concat ] with map
  swap '[ _ prefix >quotation ] map ;

: possibly-true? ( funcs equation -- ? )
  [ numbers>> possible-quotations ] [ value>> ] bi
  '[ call( -- n ) _ = ] any? ;

: solve ( funcs -- n )
  get-input
  [ possibly-true? ] with filter
  [ value>> ] map-sum ;

: part1 ( -- n )
  { + * } solve ;

: _|| ( m n -- mn )
  [ number>string ] bi@ append string>number ;

: part2 ( -- n )
  { + * _|| } solve ;

Uiua

Credits to @mykl@lemmy.world for the approach of using reduce and also how to split the input by multiple characters.
I can happily say that I learned quite a bit today, even though the first part made me frustrated enough that I went searching for other approaches ^^

Part two just needed a simple modification. Changing how the input is parsed and passed to the adapted function took longer than changing the function itself actually.

Run with example input here

PartOne ← (
  &rs ∞ &fo "input-7.txt"
  ⊜□≠@\n.
  ≡◇(⊜□≠@:.)
  ≡⍜⊡⋕0
  ≡⍜(°□⊡1)(⊜⋕≠@ .)
  ⟜(⊡0⍉)

  # own attempt, produces a too low number
  # ≡(:∩°□°⊟
  #   ⍣(⍤.◡⍣(1⍤.(≤/×)⍤.(≥/+),,)0
  #     ⊙¤⋯⇡ⁿ:2-1⊸⧻
  #     ⊞(⍥(⟜⍜(⊙(↙2))(⨬+×⊙°⊟⊡0)
  #         ↘1
  #       )⧻.
  #       ⍤.=0⧻.
  #     )
  #     ∈♭◌
  #   )0)

  # reduce approach found on the programming.dev AoC community by mykl@lemmy.world
  ≡(◇(∈/(◴♭[⊃(+|×)]))⊡0:°⊂)
  °□/+▽
)

PartTwo ← (
  &rs ∞ &fo "input-7.txt"
  ⊜(□⊜⋕¬∈": ".)≠@\n.
  ⟜≡◇⊢
  ≡◇(∈/(◴♭[≡⊃⊃(+|×|⋕$"__")]):°⊂)
  °□/+▽
)

&p "Day 7:"
&pf "Part 1: "
&p PartOne
&pf "Part 2: "
&p PartTwo

C

Big integers and overflow checking, what else is there to say 😅​

#include "common.h"

/* returns 1 on sucess, 0 on overflow */
static int
concat(uint64_t a, uint64_t b, uint64_t *out)
{
	uint64_t mul;

	for (mul=1; mul<=b; mul*=10) ;

	return 
	    !__builtin_mul_overflow( mul, a, out) &&
	    !__builtin_add_overflow(*out, b, out);
}

static int
recur(uint64_t expect, uint64_t acc, uint64_t arr[], int n, int p2)
{
	uint64_t imm;

	return
	    n < 1 ? acc == expect :
	    acc >= expect ? 0 :
	    recur(expect, acc + arr[0], arr+1, n-1, p2) ||
	    recur(expect, acc * arr[0], arr+1, n-1, p2) ||
	    (p2 && concat(acc, arr[0], &imm)
	        && recur(expect, imm, arr+1, n-1, p2));
}

int
main(int argc, char **argv)
{
	char buf[128], *tok, *rest;
	uint64_t p1=0, p2=0, arr[32], expect;
	int n;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));
	
	while ((rest = fgets(buf, sizeof(buf), stdin))) {
		assert(strchr(buf, '\n'));
		expect = atoll(strsep(&rest, ":"));

		for (n=0; (tok = strsep(&rest, " ")); n++) {
			assert(n < (int)LEN(arr));
			arr[n] = atoll(tok);
		}

		p1 += recur(expect, 0, arr, n, 0) * expect;
		p2 += recur(expect, 0, arr, n, 1) * expect;
	}

	printf("07: %"PRIu64" %"PRIu64"\n", p1, p2);
	return 0;
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day07.c

C#

public class Day07 : Solver
{
  private ImmutableList<(long, ImmutableList<long>)> equations;

  public void Presolve(string input) {
    equations = input.Trim().Split("\n")
      .Select(line => line.Split(": "))
      .Select(split => (long.Parse(split[0]), split[1].Split(" ").Select(long.Parse).ToImmutableList()))
      .ToImmutableList();
  }

  private bool TrySolveWithConcat(long lhs, long head, ImmutableList<long> tail) {
    var lhs_string = lhs.ToString();
    var head_string = head.ToString();
    return lhs_string.Length > head_string.Length &&
      lhs_string.EndsWith(head_string) &&
      SolveEquation(long.Parse(lhs_string.Substring(0, lhs_string.Length - head_string.Length)), tail, true);
  }

  private bool SolveEquation(long lhs, ImmutableList<long> rhs, bool with_concat = false) {
    if (rhs.Count == 1) return lhs == rhs[0];
    long head = rhs[rhs.Count - 1];
    var tail = rhs.GetRange(0, rhs.Count - 1);
    return (SolveEquation(lhs - head, tail, with_concat))
      || (lhs % head == 0) && SolveEquation(lhs / head, tail, with_concat)
      || with_concat && TrySolveWithConcat(lhs, head, tail);
  }

  public string SolveFirst() => equations
    .Where(eq => SolveEquation(eq.Item1, eq.Item2))
    .Select(eq => eq.Item1)
    .Sum().ToString();
  public string SolveSecond() => equations
    .Where(eq => SolveEquation(eq.Item1, eq.Item2, true))
    .Select(eq => eq.Item1)
    .Sum().ToString();
}

Made a couple of attempts to munge the input data into some kind of binary search tree, lost some time to that, then threw my hands into the air and did a more naïve sort-of breadth-first search instead. Which turned out to be better for part 2 anyway.
Also, maths. Runs in just over a hundred milliseconds when using AsParallel, around half a second without.

List<(long, int[])> data = new List<(long, int[])>();

public void Input(IEnumerable<string> lines)
{
  foreach (var line in lines)
  {
    var parts = line.Split(':', StringSplitOptions.TrimEntries);

    data.Add((long.Parse(parts.First()), parts.Last().Split(' ').Select(int.Parse).ToArray()));
  }
}

public void Part1()
{
  var correct = data.Where(kv => CalcPart(kv.Item1, kv.Item2)).Select(kv => kv.Item1).Sum();

  Console.WriteLine($"Correct: {correct}");
}
public void Part2()
{
  var correct = data.AsParallel().Where(kv => CalcPart2(kv.Item1, kv.Item2)).Select(kv => kv.Item1).Sum();

  Console.WriteLine($"Correct: {correct}");
}

public bool CalcPart(long res, Span<int> num, long carried = 0)
{
  var next = num[0];
  if (num.Length == 1)
    return res == carried + next || res == carried * next;
  return CalcPart(res, num.Slice(1), carried + next) || CalcPart(res, num.Slice(1), carried * next);
}

public bool CalcPart2(long res, Span<int> num, long carried = 0)
{
  var next = num[0];
  // Get the 10 logarithm for the next number, expand the carried value by 10^<next 10log + 1>, add the two together
  // For 123 || 45
  // 45 ⇒ 10log(45) + 1 == 2
  // 123 * 10^2 + 45 == 12345
  long combined = carried * (long)Math.Pow(10, Math.Floor(Math.Log10(next) + 1)) + next;
  if (num.Length == 1)
    return res == carried + next || res == carried * next || res == combined;
  return CalcPart2(res, num.Slice(1), carried + next) || CalcPart2(res, num.Slice(1), carried * next) || CalcPart2(res, num.Slice(1), combined);
}

python

45s on my machine for first shot, trying to break my will to brute force 😅. I’ll try improving on it in a bit after I smoke another bowl and grab another drink.

import itertools
import re
import aoc

def ltr(e):
    r = int(e[0])
    for i in range(1, len(e), 2):
        o = e[i]
        n = int(e[i + 1])
        if o == '+':
            r += n
        elif o == '*':
            r *= n
        elif o == '||':
            r = int(f"{r}{n}")
    return r

def pl(l, os):
    d = [int(x) for x in re.findall(r'\d+', l)]
    t, ns = d[0], d[1:]
    ops = list(itertools.product(os, repeat=len(ns) - 1))
    for o in ops:
        e = str(ns[0])
        for i, op in enumerate(o):
            e += f" {op} {ns[i + 1]}"
        r = ltr(e.split())
        if r == t:
            return r
    return 0

def one():
    lines = aoc.get_lines(7)
    acc = 0
    for l in lines:
        acc += pl(l, ['+', '*'])
    print(acc)

def two():
    lines = aoc.get_lines(7)
    acc = 0
    for l in lines:
        acc += pl(l, ['+', '*', '||'])
    print(acc)

one()
two()

Nim

Bruteforce, my beloved.

Wasted too much time on part 2 trying to make combinations iterator (it was very slow). In the end solved both parts with 3^n and toTernary.

Runtime: 1.5s

func digits(n: int): int =
  result = 1; var n = n
  while (n = n div 10; n) > 0: inc result

func concat(a: var int, b: int) =
  a = a * (10 ^ b.digits) + b

func toTernary(n: int, len: int): seq[int] =
  result = newSeq[int](len)
  if n == 0: return
  var n = n
  for i in 0..<len:
    result[i] = n mod 3
    n = n div 3

proc solve(input: string): AOCSolution[int, int] =
  for line in input.splitLines():
    let parts = line.split({':',' '})
    let res = parts[0].parseInt
    var values: seq[int]
    for i in 2..parts.high:
      values.add parts[i].parseInt

    let opsCount = values.len - 1
    var solvable = (p1: false, p2: false)
    for s in 0 ..< 3^opsCount:
      var sum = values[0]
      let ternary = s.toTernary(opsCount)
      for i, c in ternary:
        case c
        of 0: sum *= values[i+1]
        of 1: sum += values[i+1]
        of 2: sum.concat values[i+1]
        else: raiseAssert"!!"
      if sum == res:
        if ternary.count(2) == 0:
          solvable.p1 = true
        solvable.p2 = true
        if solvable == (true, true): break
    if solvable.p1: result.part1 += res
    if solvable.p2: result.part2 += res

Codeberg repo

Java

Today was pretty easy one but for some reason I spent like 20 minutes overthinking part 2 when all it needed was one new else if. I initially through the concatenation operator would take precedence even tho it clearly says “All operators are still evaluated left-to-right” in the instructions…

I’m sure there are optimizations to do but using parallelStreams it only takes around 300ms total on my machine so there’s no point really

import java.io.IOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Path;
import java.util.Arrays;
import java.util.List;

public class Day7 {
    public static void main(final String[] _args) throws IOException {
        final List<Equation> equations = Files.readAllLines(Path.of("2024\\07\\input.txt"), StandardCharsets.UTF_8).stream()
            .map(line -> line.split(":\\s"))
            .map(line -> new Equation(
                    Long.parseLong(line[0]),
                    Arrays.stream(line[1].split("\\s"))
                        .map(Integer::parseInt)
                        .toArray(Integer[]::new)
                )
            ).toList();

        final char[] part1Operators = {'+', '*'};
        System.out.println("Part 1: " + equations.parallelStream()
            .mapToLong(equation -> getResultIfPossible(equation, part1Operators))
            .sum()
        );

        final char[] part2Operators = {'+', '*', '|'};
        System.out.println("Part 2: " + equations.parallelStream()
            .mapToLong(equation -> getResultIfPossible(equation, part2Operators))
            .sum()
        );
    }

    private static Long getResultIfPossible(final Equation equation, final char[] operators) {
        final var permutations = Math.pow(operators.length, equation.values.length - 1);
        for (int i = 0; i < permutations; i++) {
            long result = equation.values[0];
            int count = i;

            for (int j = 0; j < equation.values.length - 1; j++) {
                // If the result is already larger than the expected one, we can short circuit here to save some time
                if (result > equation.result) {
                    break;
                }

                final char operator = operators[count % operators.length];
                count /= operators.length;

                if (operator == '+') { result += equation.values[j + 1]; }
                else if (operator == '*') { result *= equation.values[j + 1]; }
                else if (operator == '|') { result = Long.parseLong(String.valueOf(result) + String.valueOf(equation.values[j + 1])); }
                else {
                    throw new RuntimeException("Unsupported operator " + operator);
                }
            }

            if (result == equation.result) {
                return result;
            }
        }

        return 0L;
    }

    private static record Equation(long result, Integer[] values) {}
}

Dart

Suspiciously easy, so let’s see how tomorrow goes… (edit: forgot to put the language! Dart for now, thinking about Uiua later)

import 'package:more/more.dart';

var ops = [(a, b) => a + b, (a, b) => a * b, (a, b) => int.parse('$a$b')];

bool canMake(int target, List<int> ns, int sofar, dynamic ops) {
  if (ns.isEmpty) return target == sofar;
  for (var op in ops) {
    if (canMake(target, ns.sublist(1), op(sofar, ns.first), ops)) return true;
  }
  return false;
}

solve(List<String> lines, dynamic ops) {
  var sum = 0;
  for (var line in lines.map((e) => e.split(' '))) {
    var target = int.parse(line.first.skipLast(1));
    var ns = line.skip(1).map(int.parse).toList();
    sum += (canMake(target, ns.sublist(1), ns.first, ops)) ? target : 0;
  }
  return sum;
}

part1(List<String> lines) => solve(lines, ops.sublist(0, 2));
part2(List<String> lines) => solve(lines, ops);