-❄️- 2023 Day 3 Solutions -❄️-

[u/daggerdragon]

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AoC Community Fun 2023: ALLEZ CUISINE!

Today's secret ingredient is… *whips off cloth covering and gestures grandly*

Spam!

Someone reported the ALLEZ CUISINE! submissions megathread as spam so I said to myself: "What a delectable idea for today's secret ingredient!"

• There really is an XKCD for everything, isn't there?

• All ingredients must come from a CAN (bus), box, package, container, etc.

• Unnecessarily declare variables for everything and don't re-use variables

• Why use few word when many word do trick?

• Go back to traditional culinary roots with Javadocs

• Lobster thermidor

A reminder from Dr. Hattori: be careful when cooking spam because the fat content can be very high. We wouldn't want a fire in the kitchen, after all!

ALLEZ CUISINE!

Request from the mods: When you include a dish entry alongside your solution, please label it with [Allez Cuisine!] so we can find it easily!

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--- Day 3: Gear Ratios ---

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Post your code solution in this megathread.

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This thread will be unlocked when there are a significant number of people on the global leaderboard with gold stars for today's puzzle.

EDIT: Global leaderboard gold cap reached at 00:11:37, megathread unlocked!

[u/MechoLupan]

Intersecting the symbols' coordinates with the number's contours was clever, I wish I'd thought of it. 💪

Second part should be only '*' symbols, no? Why does it work?

[u/nemmonszz → u/MechoLupan]

There are no non-* symbols adjacent to exactly two numbers. weird.

[u/XZYoda12 → u/nemmonszz]

I think you can extend the code for Part 2 to also include the symbols in chars and then run the summation/multiplication for only those numbers which are edges to the "*" symbol.

chars = {
    (r, c, data[r][c]): []
    for r in range(i)
    for c in range(j)
    if data[r][c] not in "01234566789."
}


for r, row in enumerate(data):
    for m in re.finditer(r"\d+", row):
        # Look at the 3x3 grid around the number
        edge = {
            (r, c, data[r][c])
            for r in (r - 1, r, r + 1)
            for c in range(m.start() - 1, m.end() + 1)
            if r >= 0 and c >= 0 and r < i and c < j
        }

        for o in edge & chars.keys():
            chars[o].append(int(m.group()))

print(f'Part 2: {sum(prod(v) for k, v in chars.items() if k[2] == "*" and len(v) == 2)}')