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;
}

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.

Rust

Part 1 Part 2

Fun with iterators! For the second part I went to regex for help.

I feel ok about part 1, and just terrible about part 2.

day01.factor on github (with comments and imports):

: part1 ( -- )
  "vocab:aoc-2023/day01/input.txt" utf8 file-lines
  [
    [ [ digit? ] find nip ]
    [ [ digit? ] find-last nip ] bi
    2array string>number
  ] map-sum .
;

MEMO: digit-words ( -- name-char-assoc )
  [ "123456789" [ dup char>name "-" split1 nip ,, ] each ] H{ } make
;

: first-digit-char ( str -- num-char/f i/f )
  [ digit? ] find swap
;

: last-digit-char ( str -- num-char/f i/f )
  [ digit? ] find-last swap
;

: first-digit-word ( str -- num-char/f )
  [
    digit-words keys [
      2dup subseq-index
      dup [
        [ digit-words at ] dip
        ,,
      ] [ 2drop ] if
    ] each drop                           !
  ] H{ } make
  [ f ] [
    sort-keys first last
  ] if-assoc-empty
;

: last-digit-word ( str -- num-char/f )
  reverse
  [
    digit-words keys [
      reverse
      2dup subseq-index
      dup [
        [ reverse digit-words at ] dip
        ,,
      ] [ 2drop ] if
    ] each drop                           !
  ] H{ } make
  [ f ] [
    sort-keys first last
  ] if-assoc-empty
;

: first-digit ( str -- num-char )
  dup first-digit-char dup [
    pick 2dup swap head nip
    first-digit-word dup [
      [ 2drop ] dip
    ] [ 2drop ] if
    nip
  ] [
    2drop first-digit-word
  ] if
;

: last-digit ( str -- num-char )
  dup last-digit-char dup [
    pick 2dup swap 1 + tail nip
    last-digit-word dup [
      [ 2drop ] dip
    ] [ 2drop ] if
    nip
  ] [
    2drop last-digit-word
  ] if
;

: part2 ( -- )
  "vocab:aoc-2023/day01/input.txt" utf8 file-lines
  [ [ first-digit ] [ last-digit ] bi 2array string>number ] map-sum .
;

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")]))

I’m a bit late to the party. I forgot about this.

Anyways, my (lazy) C solutions: https://git.sr.ht/~aidenisik/aoc23/tree/master/item/day1

[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

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);
};

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;
}

Part 02 in Rust 🦀 :

use std::{
    collections::HashMap,
    env, fs,
    io::{self, BufRead, BufReader},
};

fn main() -> io::Result<()> {
    let args: Vec = env::args().collect();
    let filename = &args[1];
    let file = fs::File::open(filename)?;
    let reader = BufReader::new(file);

    let number_map = HashMap::from([
        ("one", "1"),
        ("two", "2"),
        ("three", "3"),
        ("four", "4"),
        ("five", "5"),
        ("six", "6"),
        ("seven", "7"),
        ("eight", "8"),
        ("nine", "9"),
    ]);

    let mut total = 0;
    for _line in reader.lines() {
        let digits = get_text_numbers(_line.unwrap(), &number_map);
        if !digits.is_empty() {
            let digit_first = digits.first().unwrap();
            let digit_last = digits.last().unwrap();
            let mut cat = String::new();
            cat.push(*digit_first);
            cat.push(*digit_last);
            let cat: i32 = cat.parse().unwrap();
            total += cat;
        }
    }
    println!("{total}");
    Ok(())
}

fn get_text_numbers(text: String, number_map: &HashMap<&str, &str>) -> Vec {
    let mut digits: Vec = Vec::new();
    if text.is_empty() {
        return digits;
    }
    let mut sample = String::new();
    let chars: Vec = text.chars().collect();
    let mut ptr1: usize = 0;
    let mut ptr2: usize;
    while ptr1 < chars.len() {
        sample.clear();
        ptr2 = ptr1 + 1;
        if chars[ptr1].is_digit(10) {
            digits.push(chars[ptr1]);
            sample.clear();
            ptr1 += 1;
            continue;
        }
        sample.push(chars[ptr1]);
        while ptr2 < chars.len() {
            if chars[ptr2].is_digit(10) {
                sample.clear();
                break;
            }
            sample.push(chars[ptr2]);
            if number_map.contains_key(&sample.as_str()) {
                let str_digit: char = number_map.get(&sample.as_str()).unwrap().parse().unwrap();
                digits.push(str_digit);
                sample.clear();
                break;
            }
            ptr2 += 1;
        }
        ptr1 += 1;
    }

    digits
}

Ruby

https://github.com/snowe2010/advent-of-code/blob/master/ruby_aoc/2023/day01/day01.rb

Part 1

execute(1, test_file_suffix: "p1") do |lines|
  lines.inject(0) do |acc, line|
    d = line.gsub(/\D/,'')
    acc += (d[0] + d[-1]).to_i
  end
end

Part 2

map = {
  "one": 1,
  "two": 2,
  "three": 3,
  "four": 4,
  "five": 5,
  "six": 6,
  "seven": 7,
  "eight": 8,
  "nine": 9,
}

execute(2) do |lines|
  lines.inject(0) do |acc, line|
    first_num = line.sub(/(one|two|three|four|five|six|seven|eight|nine)/) do |key|
      map[key.to_sym]
    end
    last_num = line.reverse.sub(/(enin|thgie|neves|xis|evif|ruof|eerht|owt|eno)/) do |key|
      map[key.reverse.to_sym]
    end

    d = first_num.chars.select { |num| numeric?(num) }
    e = last_num.chars.select { |num| numeric?(num) }
    acc += (d[0] + e[0]).to_i
  end
end

Then of course I also code golfed it, but didn’t try very hard.

P1 Code Golf

execute(1, alternative_text: "Code Golf 60 bytes", test_file_suffix: "p1") do |lines|
  lines.inject(0){|a,l|d=l.gsub(/\D/,'');a+=(d[0]+d[-1]).to_i}
end

P2 Code Golf (ignore the formatting, I just didn’t want to reformat to remove all the spaces, and it’s easier to read this way.)

execute(1, alternative_text: "Code Golf 271 bytes", test_file_suffix: "p1") do |z|
  z.inject(0) { |a, l|
    w = %w(one two three four five six seven eight nine)
    x = w.join(?|)
    f = l.sub(/(#{x})/) { |k| map[k.to_sym] }
    g = l.reverse.sub(/(#{x.reverse})/) { |k| map[k.reverse.to_sym] }
    d = f.chars.select { |n| n.match?(/\d/) }
    e = g.chars.select { |n| n.match?(/\d/) }
    a += (d[0] + e[0]).to_i
  }
end

Java

My take on a modern Java solution (parts 1 & 2).

package thtroyer.day1;

import java.util.*;
import java.util.stream.IntStream;
import java.util.stream.Stream;


public class Day1 {
    record Match(int index, String name, int value) {
    }

    Map numbers = Map.of(
            "one", 1,
            "two", 2,
            "three", 3,
            "four", 4,
            "five", 5,
            "six", 6,
            "seven", 7,
            "eight", 8,
            "nine", 9);

    /**
     * Takes in all lines, returns summed answer
     */
    public int getCalibrationValue(String... lines) {
        return Arrays.stream(lines)
                .map(this::getCalibrationValue)
                .map(Integer::parseInt)
                .reduce(0, Integer::sum);
    }

    /**
     * Takes a single line and returns the value for that line,
     * which is the first and last number (numerical or text).
     */
    protected String getCalibrationValue(String line) {
        var matches = Stream.concat(
                        findAllNumberStrings(line).stream(),
                        findAllNumerics(line).stream()
                ).sorted(Comparator.comparingInt(Match::index))
                .toList();

        return "" + matches.getFirst().value() + matches.getLast().value();
    }

    /**
     * Find all the strings of written numbers (e.g. "one")
     *
     * @return List of Matches
     */
    private List findAllNumberStrings(String line) {
        return IntStream.range(0, line.length())
                .boxed()
                .map(i -> findAMatchAtIndex(line, i))
                .filter(Optional::isPresent)
                .map(Optional::get)
                .sorted(Comparator.comparingInt(Match::index))
                .toList();
    }


    private Optional findAMatchAtIndex(String line, int index) {
        return numbers.entrySet().stream()
                .filter(n -> line.indexOf(n.getKey(), index) == index)
                .map(n -> new Match(index, n.getKey(), n.getValue()))
                .findAny();
    }

    /**
     * Find all the strings of digits (e.g. "1")
     *
     * @return List of Matches
     */
    private List findAllNumerics(String line) {
        return IntStream.range(0, line.length())
                .boxed()
                .filter(i -> Character.isDigit(line.charAt(i)))
                .map(i -> new Match(i, null, Integer.parseInt(line.substring(i, i + 1))))
                .toList();
    }

    public static void main(String[] args) {
        System.out.println(new Day1().getCalibrationValue(args));
    }
}

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

I did this in C. First part was fairly trivial, iterate over the line, find first and last number, easy.

Second part had me a bit worried i would need a more string friendly library/language, until i worked out that i can just strstr to find “one”, and then in place switch that to “o1e”, and so on. Then run part1 code over the modified buffer. I originally did “1ne”, but overlaps such as “eightwo” meant that i got the 2, but missed the 8.

#include 
#include 
#include 
#include 
#include 

size_t readfile(char* fname, char* buffer, size_t buffer_len)
{

    int f = open(fname, 'r');
    assert(f >= 0);
    size_t total = 0;
    do {
        size_t nr = read(f, buffer + total, buffer_len - total);
        if (nr == 0) {
            return total;
        }
        total += nr;
    }
    while (buffer_len - total > 0);
    return -1;
}

int part1(const char* buffer, size_t buffer_len)
{
    int first = -1;
    int last = -1;
    int total = 0;
    for (int i = 0; i < buffer_len; i++)
    {
        char c = buffer[i];
        if (c == '\n')
        {
            if (first == -1) {
                continue;
            }
            total += (first*10 + last);
            first = last = -1;
            continue;
        }
        int val = c - '0';
        if (val > 9 || val < 0)
        {
            continue;
        }
        if (first == -1)
        {
            first = last = val;
        }
        else
        {
            last = val;
        }
    }
    return total;
}

void part2_sanitize(char* buffer, size_t len)
{
    char* p = NULL;
    while ((p = strnstr(buffer, "one", len)) != NULL)
    {
        p[1] = '1';
    }
    while ((p = strnstr(buffer, "two", len)) != NULL)
    {
        p[1] = '2';
    }
    while ((p = strnstr(buffer, "three", len)) != NULL)
    {
        p[1] = '3';
    }
    while ((p = strnstr(buffer, "four", len)) != NULL)
    {
        p[1] = '4';
    }
    while ((p = strnstr(buffer, "five", len)) != NULL)
    {
        p[1] = '5';
    }
    while ((p = strnstr(buffer, "six", len)) != NULL)
    {
        p[1] = '6';
    }
    while ((p = strnstr(buffer, "seven", len)) != NULL)
    {
        p[1] = '7';
    }
    while ((p = strnstr(buffer, "eight", len)) != NULL)
    {
        p[1] = '8';
    }
    while ((p = strnstr(buffer, "nine", len)) != NULL)
    {
        p[1] = '9';
    }
    while ((p = strnstr(buffer, "zero", len)) != NULL)
    {
        p[1] = '0';
    }
}

int main(int argc, char** argv)
{
    assert(argc == 2);
    char buffer[1000000];
    size_t len = readfile(argv[1], buffer, sizeof(buffer));
    {
        int total = part1(buffer, len);
        printf("Part 1 total: %i\n", total);
    }

    {
        part2_sanitize(buffer, len);
        int total = part1(buffer, len);
        printf("Part 2 total: %i\n", total);
    }
}

My solution in rust. I’m sure there’s a lot more clever ways to do it but this is what I came up with.

use std::{io::prelude::*, fs::File, path::Path, io };

fn main() 
{
    run_solution(false); 

    println!("\nPress enter to continue");
    let mut buffer = String::new();
    io::stdin().read_line(&mut buffer).unwrap();

    run_solution(true); 
}

fn run_solution(check_for_spelled: bool)
{
    let data = load_data("data/input");

    println!("\nProcessing Data...");

    let mut sum: u64 = 0;
    for line in data.lines()
    {
        // Doesn't seem like the to_ascii_lower call is needed but... just in case
        let first = get_digit(line.to_ascii_lowercase().as_bytes(), false, check_for_spelled);
        let last = get_digit(line.to_ascii_lowercase().as_bytes(), true, check_for_spelled);


        let num = (first * 10) + last;

        // println!("\nLine: {} -- First: {}, Second: {}, Num: {}", line, first, last, num);
        sum += num as u64;
    }

    println!("\nFinal Sum: {}", sum);
}

fn get_digit(line: &[u8], from_back: bool, check_for_spelled: bool) -> u8
{
    let mut range: Vec = (0..line.len()).collect();
    if from_back
    {
        range.reverse();
    }

    for i in range
    {
        if is_num(line[i])
        {
            return (line[i] - 48) as u8;
        }

        if check_for_spelled
        {
            if let Some(num) = is_spelled_num(line, i)
            {
                return num;
            }
        }
    }

    return 0;
}

fn is_num(c: u8) -> bool
{
    c >= 48 && c <= 57
}

fn is_spelled_num(line: &[u8], start: usize) -> Option
{
    let words = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"];

    for word_idx in 0..words.len()
    {
        let mut i = start;
        let mut found = true;
        for c in words[word_idx].as_bytes()
        {
            if i < line.len() && *c != line[i]
            {
                found = false;
                break;
            }
            i += 1;
        }

        if found && i <= line.len()
        {
            return Some(word_idx as u8 + 1);
        }
    }

    return None;
}

fn load_data(file_name: &str) -> String
{
    let mut file = match File::open(Path::new(file_name))
    {
        Ok(file) => file,
        Err(why) => panic!("Could not open file {}: {}", Path::new(file_name).display(), why),
    };

    let mut s = String::new();
    let file_contents = match file.read_to_string(&mut s) 
    {
        Err(why) => panic!("couldn't read {}: {}", Path::new(file_name).display(), why),
        Ok(_) => s,
    };
    
    return file_contents;
}