NotAwfulTech

  int? check3(String line, int index) {
    if (line.length < index + 3) {
      return null;
    }

    String sub = line.substring(index, index + 3);
    return sub == "one"
        ? 1
        : sub == "two"
            ? 2
            : sub == "six"
                ? 6
                : null;
  }

The approach this time was:

1. iterate through the characters to find a * symbol
2. Search the characters around it for a digit.
3. Get the value of the number associated with that digit by searching backwards until you find the start of a number
4. Use union-find to track whether or not you've seen this number before (because you can't assume that the same value is the same number)

A simpler approach would consider that you only have two numbers on the same line for the same gear if the character in the gear column is a non-digit; otherwise, if a number is adjacent to a gear, there is only one on that row. Union-find is completely overkill, but I like using it even when I don't need to.

Anyway, upon reflecting on this, while the general approach is fine, I didn't think too hard about the implementation and just ended up with globs of overly memoized spaghetti. I probably should check if Dart has a python-like tuple object or similar. Whatever. Behold!

void day3s() {
  List lines = [
    for (String? line = stdin.readLineSync();
        line != null;
        line = stdin.readLineSync())
      line
  ];

  List> digs = [for (int i = 0; i < lines.length; i++) Map()];
  int? isDigitMem(int r, int c) {
    return digs[r].putIfAbsent(c, () => isDigit(lines[r][c]));
  }

  // first entry is parentc, second is size
  List>> uf = List.generate(
      lines.length, (i) => List.generate(lines[0].length, (j) => [j, 1, -1]));

  int find(int r, int c) {
    if (uf[r][c][0] != c) {
      uf[r][c][0] = find(r, uf[r][c][0]);
      return uf[r][c][0];
    }
    return uf[r][c][0];
  }

  void union(int r, int cp, int c) {
    cp = find(r, cp);
    c = find(r, c);

    if (c == cp) {
      return;
    }

    if (uf[r][cp][1] >= uf[r][c][1]) {
      uf[r][c][0] = cp;
      uf[r][cp][1] += uf[r][c][1];
    } else {
      uf[r][cp][0] = c;
      uf[r][c][1] += uf[r][cp][1];
    }
  }

  int stoi(int row, int col) {
    int acc = 0;
    for (int i = col; i < lines[0].length; i++) {
      int? d = isDigitMem(row, i);
      if (d != null) {
        acc = (acc * 10) + d;
        union(row, col, i);
      } else {
        break;
      }
    }
    return acc;
  }

  int? stoiSearch(int row, int col) {
    assert(row >= 0 && col >= 0 && row < lines.length && col < lines[0].length);
    if (isDigitMem(row, col) == null) {
      return null;
    }
    int i = col - 1;
    while (i >= 0) {
      if (isDigitMem(row, i) == null) {
        return stoi(row, i + 1);
      }
      i--;
    }
    return stoi(row, 0);
  }

  List> s2i = [for (int i = 0; i < lines.length; i++) Map()];
  int? stoiSearchMem(int row, int col) {
    return s2i[row].putIfAbsent(col, () => stoiSearch(row, col));
  }

  int count = 0;
  for (int i = 0; i < lines.length; i++) {
    for (int j = 0; j < lines[0].length; j++) {
      if (lines[i][j] != "*") {
        continue;
      }

      List gearVals = List.empty(growable: true);
      for (int x = -1; x <= 1; x++) {
        if (i + x < 0 || i + x > lines.length) {
          continue;
        }

        Set parents = {};
        for (int y = -1; y <= 1; y++) {
          if (j + y < 0 || j + y > lines[0].length) {
            continue;
          }

          int? cur = stoiSearchMem(i + x, j + y);

          int parent = find(i + x, j + y);
          if (parents.contains(parent)) {
            continue;
          }

          parents.add(parent);

          if (cur != null) {
            gearVals.add(cur);
          }
        }
      }

      if (gearVals.length == 2) {
        count += gearVals[0] * gearVals[1];
      }
    }
  }

  print(count);
}

Only novel thing I did here was recognise that red, green and blue end in different letters, so no need to string match the whole world, just check the last letter of the colour.

Anyway, a) is a pretty straightforward job-interview style question, except AoC doesn't care about efficiency. Still, we all have our own versions of pride, so I did it with a set (Though whether or not dart's native Set is tree or hash based is not known to me right now).

int matches(String line) {
  ({List wn, List n}) card = getNumbers(line);
  Set wn = Set.from(card.wn);

  return card.n.fold(0, (prev, e) => prev + (wn.contains(e) ? 1 : 0));
}

void day4a(List lines) {
  print(lines.fold(0, (acc, line) {
    int m = matches(line);
    return acc + (m != 0 ? (1 << (m - 1)) : 0);
  }));
}

So the general approach is to formulate it like this:

T_n(total cards won by card n) = M_n(total matches for card n) + CW_n(total cards won by the copies that card n wins).

and

CW_n =

• 0 if M_n = 0
• sum of T_i, where i = n + 1 ... n + M_n

Caching T_n is the DP trick to making this performant (though once again, it does not need to be)

Anyway, the above approach is the top-down version of the DP; the bottom-up version is what I actually started with in my head. I gave up on that approach because I felt like the logic was too hard for me to figure out.

void day4b(List lines) {
  List totalMatches = lines.map((e) => matches(e)).toList();

  // total copies won, including copies won from copies.
  List cachedWins = List.filled(totalMatches.length, -1);
  int totalWins(int i) {
    // return cached result if it exists
    if (cachedWins[i] > 0) {
      return cachedWins[i];
    }

    int count = totalMatches[i];
    // count the copies won from the subsequent copied cards
    for (int j = 1; j <= totalMatches[i]; j++) {
      count += totalWins(i + j);
    }

    // cache the result
    cachedWins[i] = count;
    return count;
  }

  int totalCards = totalMatches.length; // count all the originals
  // count all the wins
  for (int i = 0; i < totalMatches.length; i++) {
    totalCards += totalWins(i);
  }
  print(totalCards);
}

Other than processing the input in the form of intervals, I don't think there were any notable tricks at play here, so this was more of an exercise in implementing code cleanly, which I struggle with.

So that being said, here's the mathematic solution:

The inequality to solve is just: x = time holding the button

x^2 - tx + d < 0

which occurs when x is between (t/2) +/- (t^2 - 4d). The total ways are the total integers between those two numbers, exclusive.

Writing that all in code is a little tricky thanks to int/double rounding/truncation intricacies, so the brute force solution is "better".

count = 0
while(all ghosts not on endpoint):
   for (all ghosts):
    move ghost to next node
  count++

print count

I put a timestamp for every 100mil iterations, which ticked once every two seconds.

So, how do we solve it? As mentioned in my hint, some fudged number theory is required.

Long story short, each ghost will eventually reach an end node. They could reach K endpoints, where K is the total number of end nodes available. They will follow the same path in a loop if they traverse infinitely.

The fudge is this: assume each ghost only hits one end node repeatedly. In this case, the solution is just the LCM of the number of steps it takes for each ghost to reach its end node from the initial state.

If given a case where a ghost hits multiple unique endpoints, you can probably find the configuration of counts and LCMs to yield the smallest one, but that proof is left to the reader.

The answer that I got was in the magnitude of 10 trillion, close to 20 trillion.

Interestingly, the problem itself is based on a method for finding closed-form general formulae for progressions of integers. You can use it to develop a formula for sums of squares, cubes, quartics etc. or any progression that doesn't seem to have an obvious closed form.

Anyway, there is probably a language where this problem can be solved with a one-liner, I got kinda close, see 9.dart. Check the commits if it's unreadable.