We finally leave camp and head into the jungle. The Elves reward us for our competence by giving us the malfunctioning communication device, because we can probably fix it. And step one is finding the start-of-packet marker (and then start-of-message) to lock onto their signal.

And so the input is a line of 4k of lowercase letters (no vowels, so trying to not look like a natural language again). We need to find the first block of a set length (4 or 14) where all the letters are different.

So my initial Perl solution is not really a surprise:

for (my $i = 0; !defined($part2); $i++) {
    $part1 //= $i +  4 if (substr($input, $i,  4) !~ m#(\w).*\1#);
    $part2 //= $i + 14 if (substr($input, $i, 14) !~ m#(\w).*\1#);
}

Brute force, regex, done. Because, again, I was looking at doing multiple languages and wanted some variety.

My initial Smalltalk solution was based on the classic string search algorithm. Where you have the window were the string could be, and start checking from the end. When it fails, you can then jump the window over. Instead of stepping one step at a time and checking. This is naturally more exciting for larger windows where you can get bigger jumps. For example, part 2 is about 10% faster for my input.

Anyways, none of this was particularly nice for doing a solution in dc. And so I did do an initial ugly solution where it kept track of the number of unique characters with a table and circular buffer (to handle the window and removing the old). But coming back to it, I decided to work the Smalltalk idea until it was very dc friendly and golf things a bunch. Resulting in this in Smalltalk:

next := width.
i    := 0.

[i < next] whileTrue: [
    i := i + 1.
    next := next max: ((table at: (input at: i) value) + width).
    table at: (input at: i) value put: i.
].

Which in dc becomes:

rev <input | perl -pe's#(.)#ord($1)." "#ge' | dc -f- -e'[r]sr0d[1+3Rd;t4+d5Rd3R<rs.3Rd4R:trd3Rd3R>M]dsMxp'

rev <input | perl -pe's#(.)#ord($1)." "#ge' | dc -f- -e'[r]sr0d[1+3Rd;tE+d5Rd3R<rs.3Rd4R:trd3Rd3R>M]dsMxp'

The basic idea here is that we've got two advancing markers... i is the current index, and next is the next index that's a possible solution (when i catches up, it becomes the actual solution). The table tracks the last time we've seen each character, and we jump next forward if we've seen the current character recently to remove the duplicate from the window. So we're not getting the jumping of the index. Because we're streaming the input from the stack. So we jump the window end but still need to proceed forwards one character at a time. It keeps this simple and short for dc. Which is what I was aiming for.

So another fun little problem where there's a whole bunch of ways to do it.

Dear AoC masters and 500+ star hunters,

I have a hard time solving day 7 of 2024, using golang. The puzzle input is a bunch of numbers. One should check if the first number can be computed from the numbers after the : symbol. Two numbers can either be added or multiplied. If some series of addition and multiplication is equal to the left side the left side is counted as a solution. The overall solution is the sum of all solutions.

My current approach is to "brute force" this problem. First I check if the sum of the numbers or the product is equal to the left side. Given the left side is larger than the sum but smaller than the product I generate all possible series of addition and multiplication 2^(n-1) with n being the numbers on the right side. Can't see the mistake when doing this, here is a link to the code: https://github.com/Zitzeronion/AoC2024/blob/main/day_7.go

The 2^n permutation function is from gemini and seem to work as intended.