Solved part one in about thirty seconds. But wow, either my brain is just tired at this hour or I’m lacking in skill, but part two is harder than any other year has been on the first day. Anyway, I managed to solve it, but I absolutely hate it, and will definitely be coming back to try to clean this one up.

https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day01.rs

impl Solver for Day01 {
    fn star_one(&self, input: &str) -> String {
        let mut result = 0;

        for line in input.lines() {
            let line = line
                .chars()
                .filter(|ch| ch.is_ascii_digit())
                .collect::>();
            let first = line.first().unwrap();
            let last = line.last().unwrap();
            let number = format!("{first}{last}").parse::().unwrap();
            result += number;
        }

        result.to_string()
    }

    fn star_two(&self, input: &str) -> String {
        let mut result = 0;

        for line in input.lines() {
            let mut first = None;
            let mut last = None;

            while first == None {
                for index in 0..line.len() {
                    let line_slice = &line[index..];
                    if line_slice.starts_with("one") || line_slice.starts_with("1") {
                        first = Some(1);
                    } else if line_slice.starts_with("two") || line_slice.starts_with("2") {
                        first = Some(2);
                    } else if line_slice.starts_with("three") || line_slice.starts_with("3") {
                        first = Some(3);
                    } else if line_slice.starts_with("four") || line_slice.starts_with("4") {
                        first = Some(4);
                    } else if line_slice.starts_with("five") || line_slice.starts_with("5") {
                        first = Some(5);
                    } else if line_slice.starts_with("six") || line_slice.starts_with("6") {
                        first = Some(6);
                    } else if line_slice.starts_with("seven") || line_slice.starts_with("7") {
                        first = Some(7);
                    } else if line_slice.starts_with("eight") || line_slice.starts_with("8") {
                        first = Some(8);
                    } else if line_slice.starts_with("nine") || line_slice.starts_with("9") {
                        first = Some(9);
                    }

                    if first.is_some() {
                        break;
                    }
                }
            }

            while last == None {
                for index in (0..line.len()).rev() {
                    let line_slice = &line[index..];
                    if line_slice.starts_with("one") || line_slice.starts_with("1") {
                        last = Some(1);
                    } else if line_slice.starts_with("two") || line_slice.starts_with("2") {
                        last = Some(2);
                    } else if line_slice.starts_with("three") || line_slice.starts_with("3") {
                        last = Some(3);
                    } else if line_slice.starts_with("four") || line_slice.starts_with("4") {
                        last = Some(4);
                    } else if line_slice.starts_with("five") || line_slice.starts_with("5") {
                        last = Some(5);
                    } else if line_slice.starts_with("six") || line_slice.starts_with("6") {
                        last = Some(6);
                    } else if line_slice.starts_with("seven") || line_slice.starts_with("7") {
                        last = Some(7);
                    } else if line_slice.starts_with("eight") || line_slice.starts_with("8") {
                        last = Some(8);
                    } else if line_slice.starts_with("nine") || line_slice.starts_with("9") {
                        last = Some(9);
                    }

                    if last.is_some() {
                        break;
                    }
                }
            }

            result += format!("{}{}", first.unwrap(), last.unwrap())
                .parse::()
                .unwrap();
        }

        result.to_string()
    }
}

My solutin in Elixir for both part 1 and part 2 is below. It does use regex and with that there are many different ways to accomplish the goal. I’m no regex master so I made it as simple as possible and relied on the language a bit more. I’m sure there are cooler solutions with no regex too, this is just what I settled on:

defmodule AdventOfCode.Day01 do
  def part1(args) do
    number_regex = ~r/([0-9])/

    args
    |> String.split(~r/\n/, trim: true)
    |> Enum.map(&first_and_last_number(&1, number_regex))
    |> Enum.map(&number_list_to_integer/1)
    |> Enum.sum()
  end

  def part2(args) do
    number_regex = ~r/(?=(one|two|three|four|five|six|seven|eight|nine|[0-9]))/

    args
    |> String.split(~r/\n/, trim: true)
    |> Enum.map(&first_and_last_number(&1, number_regex))
    |> Enum.map(fn number -> Enum.map(number, &replace_word_with_number/1) end)
    |> Enum.map(&number_list_to_integer/1)
    |> Enum.sum()
  end

  defp first_and_last_number(string, regex) do
    matches = Regex.scan(regex, string)
    [_, first] = List.first(matches)
    [_, last] = List.last(matches)

    [first, last]
  end

  defp number_list_to_integer(list) do
    list
    |> List.to_string()
    |> String.to_integer()
  end

  defp replace_word_with_number(string) do
    numbers = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]

    String.replace(string, numbers, fn x ->
      (Enum.find_index(numbers, &(&1 == x)) + 1)
      |> Integer.to_string()
    end)
  end
end

Trickier than expected! I ran into an issue with Lua patterns, so I had to revert to a more verbose solution, which I then used in Hare as well.

Lua:

-- SPDX-FileCopyrightText: 2023 Jummit
--
-- SPDX-License-Identifier: GPL-3.0-or-later

local sum = 0
for line in io.open("1.input"):lines() do
  local a, b = line:match("^.-(%d).*(%d).-$")
  if not a then
    a = line:match("%d+")
    b = a
  end
  if a and b then
    sum = sum + tonumber(a..b)
  end
end
print(sum)

local names = {
  ["one"] = 1,
  ["two"] = 2,
  ["three"] = 3,
  ["four"] = 4,
  ["five"] = 5,
  ["six"] = 6,
  ["seven"] = 7,
  ["eight"] = 8,
  ["nine"] = 9,
  ["1"] = 1,
  ["2"] = 2,
  ["3"] = 3,
  ["4"] = 4,
  ["5"] = 5,
  ["6"] = 6,
  ["7"] = 7,
  ["8"] = 8,
  ["9"] = 9,
}
sum = 0
for line in io.open("1.input"):lines() do
  local firstPos = math.huge
  local first
  for name, num in pairs(names) do
    local left = line:find(name)
    if left and left < firstPos then
      firstPos = left
      first = num
    end
  end
  local last
  for i = #line, 1, -1 do
    for name, num in pairs(names) do
      local right = line:find(name, i)
      if right then
        last = num
        goto found
      end
    end
  end
  ::found::
  sum = sum + tonumber(first * 10 + last)
end
print(sum)

Hare:

// SPDX-FileCopyrightText: 2023 Jummit
//
// SPDX-License-Identifier: GPL-3.0-or-later

use fmt;
use types;
use bufio;
use strings;
use io;
use os;

const numbers: [](str, int) = [
	("one", 1),
	("two", 2),
	("three", 3),
	("four", 4),
	("five", 5),
	("six", 6),
	("seven", 7),
	("eight", 8),
	("nine", 9),
	("1", 1),
	("2", 2),
	("3", 3),
	("4", 4),
	("5", 5),
	("6", 6),
	("7", 7),
	("8", 8),
	("9", 9),
];

fn solve(start: size) void = {
	const file = os::open("1.input")!;
	defer io::close(file)!;
	const scan = bufio::newscanner(file, types::SIZE_MAX);
	let sum = 0;
	for (let i = 1u; true; i += 1) {
		const line = match (bufio::scan_line(&scan)!) {
		case io::EOF =>
			break;
		case let line: const str =>
			yield line;
		};
		let first: (void | int) = void;
		let last: (void | int) = void;
		for (let i = 0z; i < len(line); i += 1) :found {
			for (let num = start; num < len(numbers); num += 1) {
				const start = strings::sub(line, i, strings::end);
				if (first is void && strings::hasprefix(start, numbers[num].0)) {
					first = numbers[num].1;
				};
				const end = strings::sub(line, len(line) - 1 - i, strings::end);
				if (last is void && strings::hasprefix(end, numbers[num].0)) {
					last = numbers[num].1;
				};
				if (first is int && last is int) {
					break :found;
				};
			};
		};
		sum += first as int * 10 + last as int;
	};
	fmt::printfln("{}", sum)!;
};

export fn main() void = {
	solve(9);
	solve(0);
};

import re
numbers = {
    "one" : 1,
    "two" : 2,
    "three" : 3,
    "four" : 4,
    "five" : 5,
    "six" : 6,
    "seven" : 7,
    "eight" : 8,
    "nine" : 9
    }
for digit in range(10):
    numbers[str(digit)] = digit
pattern = "(%s)" % "|".join(numbers.keys())
   
re1 = re.compile(".*?" + pattern)
re2 = re.compile(".*" + pattern)
total = 0
for line in open("input.txt"):
    m1 = re1.match(line)
    m2 = re2.match(line)
    num = (numbers[m1.group(1)] * 10) + numbers[m2.group(1)]
    total += num
print(total)

There weren’t any zeros in the training data I got - the text seems to suggest that “0” is allowed but “zero” isn’t.

My solution is not pretty at all, but it does the job: Golang

Dart solution

This has got to be one of the biggest jumps in trickiness in a Day 1 puzzle. In the end I rolled my part 1 answer into the part 2 logic. [Edit: I’ve golfed it a bit since first posting it]

import 'package:collection/collection.dart';

var ds = '0123456789'.split('');
var wds = 'one two three four five six seven eight nine'.split(' ');

int s2d(String s) => s.length == 1 ? int.parse(s) : wds.indexOf(s) + 1;

int value(String s, List digits) {
  var firsts = {for (var e in digits) s.indexOf(e): e}..remove(-1);
  var lasts = {for (var e in digits) s.lastIndexOf(e): e}..remove(-1);
  return s2d(firsts[firsts.keys.min]) * 10 + s2d(lasts[lasts.keys.max]);
}

part1(List lines) => lines.map((e) => value(e, ds)).sum;

part2(List lines) => lines.map((e) => value(e, ds + wds)).sum;

Uiua solution

I may add solutions in Uiua depending on how easy I find them, so here’s today’s (also available to run online):

Inp ← {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"}
# if needle is longer than haystack, return zeros
SafeFind ← ((⌕|-.;)< ∩⧻ , ,)
FindDigits ← (× +1⇡9 ⊠(□SafeFind∩⊔) : Inp)
"123456789"
⊜□ ≠@\s . "one two three four five six seven eight nine"
∩FindDigits
BuildNum ← (/+∵(/+⊂⊃(×10↙ 1)(↙ 1⇌) ▽≠0.⊔) /↥)
∩BuildNum+,

or stripping away all the fluff:

Inp ← {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"}
⊜□ ≠@\s."one two three four five six seven eight nine" "123456789"
∩(×+1⇡9⊠(□(⌕|-.;)<⊙:∩(⧻.⊔)):Inp)
∩(/+∵(/+⊂⊃(×10↙1)(↙1⇌)▽≠0.⊔)/↥)+,

[Rust] 11157/6740

use std::fs;

const m: [(&str, u32); 10] = [
    ("zero", 0),
    ("one", 1),
    ("two", 2),
    ("three", 3),
    ("four", 4),
    ("five", 5),
    ("six", 6),
    ("seven", 7),
    ("eight", 8),
    ("nine", 9)
];

fn main() {
    let s = fs::read_to_string("data/input.txt").unwrap();

    let mut u = 0;

    for l in s.lines() {
        let mut h = l.chars();
        let mut f = 0;
        let mut a = 0;

        for n in 0..l.len() {
            let u = h.next().unwrap();

            match u.is_numeric() {
                true => {
                    let v = u.to_digit(10).unwrap();
                    if f == 0 {
                        f = v;
                    }
                    a = v;
                },
                _ => {
                    for (t, v) in m {
                        if l[n..].starts_with(t) {
                            if f == 0 {
                                f = v;
                            }
                            a = v;
                        }
                    }
                },
            }
        }

        u += f * 10 + a;
    }

    println!("Sum: {}", u);
}

Link

A new C solution: without lookahead or backtracking! I keep a running tally of how many letters of each digit word were matched so far: https://github.com/sjmulder/aoc/blob/master/2023/c/day01.c

int main(int argc, char **argv)
{
	static const char names[][8] = {"zero", "one", "two", "three",
	    "four", "five", "six", "seven", "eight", "nine"};
	int p1=0, p2=0, i,c;
	int p1_first = -1, p1_last = -1;
	int p2_first = -1, p2_last = -1;
	int nmatched[10] = {0};
	
	while ((c = getchar()) != EOF)
		if (c == '\n') {
			p1 += p1_first*10 + p1_last;
			p2 += p2_first*10 + p2_last;
			p1_first = p1_last = p2_first = p2_last = -1;
			memset(nmatched, 0, sizeof(nmatched));
		} else if (c >= '0' && c <= '9') {
			if (p1_first == -1) p1_first = c-'0';
			if (p2_first == -1) p2_first = c-'0';
			p1_last = p2_last = c-'0';
			memset(nmatched, 0, sizeof(nmatched));
		} else for (i=0; i<10; i++)
			/* advance or reset no. matched digit chars */
			if (c != names[i][nmatched[i]++])
				nmatched[i] = c == names[i][0];
			/* matched to end? */
			else if (!names[i][nmatched[i]]) {
				if (p2_first == -1) p2_first = i;
				p2_last = i;
				nmatched[i] = 0;
			}

	printf("%d %d\n", p1, p2);
	return 0;
}

Did this in Odin (very hashed together, especially finding the last number in part 2):

package day1

import "core:fmt"
import "core:strings"
import "core:strconv"
import "core:unicode"

p1 :: proc(input: []string) {
    total := 0

    for line in input {
        firstNum := line[strings.index_proc(line, unicode.is_digit):][:1]
        lastNum := line[strings.last_index_proc(line, unicode.is_digit):][:1]

        calibrationValue := strings.concatenate({firstNum, lastNum})
        defer delete(calibrationValue)

        num, ok := strconv.parse_int(calibrationValue)

        total += num
    }

    // daggonit thought it was the whole numbers
    /*
    for line in input {
        firstNum := line

        fFrom := strings.index_proc(firstNum, unicode.is_digit)
        firstNum = firstNum[fFrom:]

        fTo := strings.index_proc(firstNum, proc(r:rune)->bool {return !unicode.is_digit(r)})
        if fTo == -1 do fTo = len(firstNum)
        firstNum = firstNum[:fTo]


        lastNum := line
        lastNum = lastNum[:strings.last_index_proc(lastNum, unicode.is_digit)+1]
        lastNum = lastNum[strings.last_index_proc(lastNum, proc(r:rune)->bool {return !unicode.is_digit(r)})+1:]

        calibrationValue := strings.concatenate({firstNum, lastNum})
        defer delete(calibrationValue)

        num, ok := strconv.parse_int(calibrationValue, 10)
        if !ok {
            fmt.eprintf("%s could not be parsed from %s", calibrationValue, line)
            return
        }

        total += num;
    }
    */

    fmt.println(total)
}

p2 :: proc(input: []string) {
    parse_wordable :: proc(s: string) -> int {
        if len(s) == 1 {
            num, ok := strconv.parse_int(s)
            return num
        } else do switch s {
            case "one"  : return 1
            case "two"  : return 2
            case "three": return 3
            case "four" : return 4
            case "five" : return 5
            case "six"  : return 6
            case "seven": return 7
            case "eight": return 8
            case "nine" : return 9
        }

        return -1
    }

    total := 0

    for line in input {
        firstNumI, firstNumW := strings.index_multi(line, {
            "one"  , "1",
            "two"  , "2",
            "three", "3",
            "four" , "4",
            "five" , "5",
            "six"  , "6",
            "seven", "7",
            "eight", "8",
            "nine" , "9",
        })
        firstNum := line[firstNumI:][:firstNumW]


        // last_index_multi doesn't seem to exist, doing this as backup
        lastNumI, lastNumW := -1, -1
        for {
            nLastNumI, nLastNumW := strings.index_multi(line[lastNumI+1:], {
                "one"  , "1",
                "two"  , "2",
                "three", "3",
                "four" , "4",
                "five" , "5",
                "six"  , "6",
                "seven", "7",
                "eight", "8",
                "nine" , "9",
            })

            if nLastNumI == -1 do break

            lastNumI += nLastNumI+1
            lastNumW  = nLastNumW
        }
        lastNum := line[lastNumI:][:lastNumW]

        total += parse_wordable(firstNum)*10 + parse_wordable(lastNum)
    }

    fmt.println(total)
}

Had a ton of trouble with part 1 until I realized I misinterpreted it. Especially annoying because the example was working fine. So paradoxically part 2 was easier than 1.

Part 1 felt fairly pretty simple in Haskell:

import Data.Char (isDigit)

main = interact solve

solve :: String -> String
solve = show . sum . map (read . (\x -> [head x, last x]) . filter isDigit) . lines

Part 2 was more of a struggle, though I’m pretty happy with how it turned out. I ended up using concatMap inits . tails to generate all substrings, in order of appearance so one3m becomes ["","o","on","one","one3","one3m","","n","ne","ne3","ne3m","","e","e3","e3m","","3","3m","","m",""]. I then wrote a function stringToDigit :: String -> Maybe Char which simultaneously filtered out the digits and standardised them as Chars.

import Data.List (inits, tails)
import Data.Char (isDigit, digitToInt)
import Data.Maybe (mapMaybe)

main = interact solve

solve :: String -> String
solve = show . sum . map (read . (\x -> [head x, last x]) . mapMaybe stringToDigit . concatMap inits . tails) . lines
--                             |string of first&last digit| |find all the digits |   |all substrings of line|

stringToDigit "one"   = Just '1'
stringToDigit "two"   = Just '2'
stringToDigit "three" = Just '3'
stringToDigit "four"  = Just '4'
stringToDigit "five"  = Just '5'
stringToDigit "six"   = Just '6'
stringToDigit "seven" = Just '7'
stringToDigit "eight" = Just '8'
stringToDigit "nine"  = Just '9'
stringToDigit [x]
  | isDigit x         = Just x
  | otherwise         = Nothing
stringToDigit _       = Nothing

I went a bit excessively Haskell with it, but I had my fun!

Python

Questions and feedback welcome!

import re

from .solver import Solver

class Day01(Solver):
  def __init__(self):
    super().__init__(1)
    self.lines = []

  def presolve(self, input: str):
    self.lines = input.rstrip().split('\n')

  def solve_first_star(self):
    numbers = []
    for line in self.lines:
      digits = [ch for ch in line if ch.isdigit()]
      numbers.append(int(digits[0] + digits[-1]))
    return sum(numbers)

  def solve_second_star(self):
    numbers = []
    digit_map = {
      "one": 1, "two": 2, "three": 3, "four": 4, "five": 5,
      "six": 6, "seven": 7, "eight": 8, "nine": 9, "zero": 0,
      }
    for i in range(10):
      digit_map[str(i)] = i
    for line in self.lines:
      digits = [digit_map[digit] for digit in re.findall(
          "(?=(one|two|three|four|five|six|seven|eight|nine|[0-9]))", line)]
      numbers.append(digits[0]*10 + digits[-1])
    return sum(numbers)

Solution in C: https://github.com/sjmulder/aoc/blob/master/2023/c/day01-orig.c

Usually day 1 solutions are super short numeric things, this was a little more verbose. For part 2 I just loop over an array of digit names and use strncmp().

int main(int argc, char **argv)
{
	static const char * const nm[] = {"zero", "one", "two", "three",
	    "four", "five", "six", "seven", "eight", "nine"};
	char buf[64], *s;
	int p1=0,p2=0, p1f,p1l, p2f,p2l, d;
	
	while (fgets(buf, sizeof(buf), stdin)) {
		p1f = p1l = p2f = p2l = -1;

		for (s=buf; *s; s++)
			if (*s >= '0' && *s <= '9') {
				d = *s-'0';
				if (p1f == -1) p1f = d;
				if (p2f == -1) p2f = d;
				p1l = p2l = d;
			} else for (d=0; d<10; d++) {
				if (strncmp(s, nm[d], strlen(nm[d])))
					continue;
				if (p2f == -1) p2f = d;
				p2l = d;
				break;
			}

		p1 += p1f*10 + p1l;
		p2 += p2f*10 + p2l;
	}

	printf("%d %d\n", p1, p2);
	return 0;
}

Python 3

I’m trying to practice writing clear, commented, testable functions, so I added some things that are strictly unnecessary for the challenge (docstrings, error raising, type hints, tests…), but I think it’s a necessary exercise for me. If anyone has comments or criticism about my attempt at “best practices,” please let me know!

Also, I thought it was odd that the correct answer to part 2 requires that you allow for overlapping letters such as “threeight”, but that doesn’t occur in the sample input. I imagine that many people will hit a wall wondering why their answer is rejected.

import re
from pathlib import Path


DIGITS = [
    "zero",
    "one",
    "two",
    "three",
    "four",
    "five",
    "six",
    "seven",
    "eight",
    "nine",
    r"\d",
]

PATTERN_PART_1 = r"\d"
PATTERN_PART_2 = f"(?=({'|'.join(DIGITS)}))"


def get_digit(s: str) -> int:
    """Return the digit in the input

    Args:
        s (str): one string containing a single digit represented by a single arabic numeral or spelled out in lower-case English

    Returns:
        int: the digit as an integer value
    """

    try:
        return int(s)
    except ValueError:
        return DIGITS.index(s)


def calibration_value(line: str, pattern: str) -> int:
    """Return the calibration value in the input

    Args:
        line (str): one line containing a calibration value
        pattern (str): the regular expression pattern to match

    Raises:
        ValueError: if no digits are found in the line

    Returns:
        int: the calibration value
    """

    digits = re.findall(pattern, line)

    if digits:
        return get_digit(digits[0]) * 10 + get_digit(digits[-1])

    raise ValueError(f"No digits found in: '{line}'")


def calibration_sum(lines: str, pattern: str) -> int:
    """Return the sum of the calibration values in the input

    Args:
        lines (str): one or more lines containing calibration values

    Returns:
        int: the sum of the calibration values
    """

    sum = 0

    for line in lines.split("\n"):
        sum += calibration_value(line, pattern)

    return sum


if __name__ == "__main__":
    path = Path(__file__).resolve().parent / "input" / "day01.txt"

    lines = path.read_text().strip()

    print("Sum of calibration values:")
    print(f"• Part 1: {calibration_sum(lines, PATTERN_PART_1)}")
    print(f"• Part 2: {calibration_sum(lines, PATTERN_PART_2)}")

import pytest
from advent_2023_python.day01 import (
    calibration_value,
    calibration_sum,
    PATTERN_PART_1,
    PATTERN_PART_2,
)


LINES_PART_1 = [
    ("1abc2", 12),
    ("pqr3stu8vwx", 38),
    ("a1b2c3d4e5f", 15),
    ("treb7uchet", 77),
]
BLOCK_PART_1 = (
    "\n".join([line[0] for line in LINES_PART_1]),
    sum(line[1] for line in LINES_PART_1),
)

LINES_PART_2 = [
    ("two1nine", 29),
    ("eightwothree", 83),
    ("abcone2threexyz", 13),
    ("xtwone3four", 24),
    ("4nineeightseven2", 42),
    ("zoneight234", 14),
    ("7pqrstsixteen", 76),
]
BLOCK_PART_2 = (
    "\n".join([line[0] for line in LINES_PART_2]),
    sum(line[1] for line in LINES_PART_2),
)


def test_part_1():
    for line in LINES_PART_1:
        assert calibration_value(line[0], PATTERN_PART_1) == line[1]

    assert calibration_sum(BLOCK_PART_1[0], PATTERN_PART_1) == BLOCK_PART_1[1]


def test_part_2_with_part_1_values():
    for line in LINES_PART_1:
        assert calibration_value(line[0], PATTERN_PART_2) == line[1]

    assert calibration_sum(BLOCK_PART_1[0], PATTERN_PART_2) == BLOCK_PART_1[1]


def test_part_2_with_part_2_values():
    for line in LINES_PART_2:
        assert calibration_value(line[0], PATTERN_PART_2) == line[1]

    assert calibration_sum(BLOCK_PART_2[0], PATTERN_PART_2) == BLOCK_PART_2[1]


def test_no_digits():
    with pytest.raises(ValueError):
        calibration_value("abc", PATTERN_PART_1)

    with pytest.raises(ValueError):
        calibration_value("abc", PATTERN_PART_2)

I wanted to see if it was possible to do part 1 in a single line of Python:

print(sum([(([int(i) for i in line if i.isdigit()][0]) * 10 + [int(i) for i in line if i.isdigit()][-1]) for line in open("input.txt")]))