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Author SHA1 Message Date
1b6502df7a Solve 2024:9 p1-2 "Disk Fragmenter"
Hard one. Comparing to earlier years, the
difficulty level seems higher than usual. I blame
the LLM hipsters.

This one I asumed, but missed when I tried to
find it:
> "If a block contains free space, skip it instead."

Lucky me.
2025-01-05 00:06:18 +01:00
dc44be89bf Solve 2024:8 p1-2 "Resonant Collinearity"
As many have said, the puzzle text this day
was a bit challenging to parse and understand.

Especially this part of pt 2 was easy to miss:

> In fact, the three T-frequency *antennas* are all
> exactly in line with two antennas,
> so *they are all also antinodes*!

not including the antennas in pt 2 gave an slightly
too low answer.
2025-01-05 00:06:18 +01:00
81fe14a407 Solve 2024:7 p1-2 "Bridge Repair"
Got thrown of big time at pt 1 since there is a
duplicate test value in my puzzle input. The code
initially asumed that all test values should be
distinct. Biased from seeing "test" in the
termology, most likely.

Earlier editions of the code also tried to create
all combinations using binary strings, but it fell
short on pt 2 when a third operation was introduced.

From some inspiration in the solutions mega
thread on Reddit, the final code is queue-based
instead. Apparently, I also learned this kind of
problem is usually well suited for DFS search, and
apparently the final code very much is _in deed_
a DFS.
2025-01-05 00:06:18 +01:00
7b2b101b8b Solve 2024:6 p1-2 "Guard Gallivant"
Pt 1 was easy, Pt 2 was pure horror.

Earlier drafts of the code tried to be way too
smart. At one point, I just came to the conclusion
to place a new obstacle (#) on the grid and just
rerun the thing to look for loops.

2 things:

- the visited positions from pt 1 can be used as a
subset for positions to consider for the extra
"#".
- The track of loops can be optimized to look at
bounces on "#"s instead of each individual y,x pos,
given that the direction is remembered.

pt 2 is familiar, the last time a puzzle required
look detection the puzzle used lazer beams and
reflectors. Not sure what Event or day it was.
2025-01-05 00:06:18 +01:00
44f6aa0f53 Solve 2024:5 p1-2 "Print Queue"
In p2, it took some tries to find a correct way
to rearrange incorrect pagesets. The code ended
up just rearranging first found incorrect order
and readding the updated pageset to queue.
2025-01-05 00:06:18 +01:00
f751b3b8d5 Solve 2024:4 p1-2 "Ceres Search"
Not gonna lie, This is not my gig skillwise. I had
much fun though!

A matrix would most likely make the code more
readable.

I relied massively on test cases here, so I share
them here instead of in the source code:

assert 1 == solve("""
    X.....
    .M....
    ..A...
    ...S..
    ..X...
    """.strip())[0]

    assert 2 == solve("""
    XS....
    .MA...
    ..AM..
    ...SX.
    ..X...
    """.strip())[0]

    assert 2 == solve("""
    ..S......
    ...A.....
    ....M....
    .....X...
    ..X...M..
    .......A.
    ........S
    """.strip())[0]

    assert 4 == solve("""
    X.SS
    M.AA
    A.MM
    S.XX
    ...M
    ...A
    ...S
    """.strip())[0]

    assert 4 == solve("""
    ...X...
    ...M...
    .X.A...
    XMASAMX
    .A.....
    .S.....
    """.strip())[0]

    assert 1 == solve("""
    ....X
    ...M.
    ..A..
    .S...
    """.strip())[0]

    assert 2 == solve("""
    ...SX
    ..AM.
    .MA..
    XS...
    """.strip())[0]

    assert 2 == solve("""
    ......X
    .....M.
    ....A..
    ...S...
    ..A....
    .M.....
    X......
    """.strip())[0]

    assert 1 == solve("""
    M.S
    .A.
    M.S
    """.strip())[1]

    assert 1 == solve("""
    M.M
    .A.
    S.S
    """.strip())[1]

    assert 1 == solve("""
    S.M
    .A.
    S.M
    """.strip())[1]

    assert 1 == solve("""
    S.S
    .A.
    M.M
    """.strip())[1]

    assert 1 == solve("""
    S.S
    .A.
    M.M
    """.strip())[1]

    assert 1 == solve("""
    .A.
    M.M
    .A.
    S.S
    """.strip())[1]

    assert 1 == solve("""
    M.M.
    .A.A
    S.S.
    """.strip())[1]

    assert 1 == solve("""
    M.M
    .A.
    S.S
    .A.
    """.strip())[1]

    assert 1 == solve("""
    .M.M
    A.A.
    .S.S
    """.strip())[1]
2025-01-05 00:06:18 +01:00
cb622409f9 Solve 2024:3 p1-2 "Mull It Over"
Felt paranoid on this one, was expecting something
much worse for pt2.

The code that solved the puzzle was _extremely_
naive:

- It asumes puzzle input only contains mul() with
1-3 chars. The versioned code have `{1,3}` as safe
guard.
- p2 asumed initial chunk does not begin with
"n't". Would have been easy to handle though.
- p2 asumed junk strings as "don", "do", "don()t"
would not exist.

In other words, I was really lucky since I did not
look that closely on puzzle input beforehand.

Might have cut 60-90 seconds further if I had just
ran pt2 immidately instead of staring dumbly on the
test data (that changed for pt2).

Also, I got reminded that \d in regular expressions
is equal to `0-9`: it does not include commas and
punctations. The code that solved the puzzle was
paranoid and instead used `0-9`.

Managed to score ~1500th somehow despite this.
2025-01-05 00:06:18 +01:00
5deb351504 Solve 2024:2 p1-2 "Red-Nosed Reports"
DDOS of adventofcode.com, did not get to open the
puzzle until 06:05 (five minutes local time).

Screwed up the zip() and lost valuable minutes by
trying to replace it with itertools, only to find
I have made a typo (did AB CD EF instead of AB BC
CD).

I also lost valuable minutes by tring to solve p1
with one nested loop. Stupid.

For part 2, I created the issafe() helper to be
able to remove items from reports and test, one
at the time.
2025-01-05 00:06:18 +01:00
81741232af Solve 2024:1 p1-2 "Historian Hysteria"
Realized afterwards I got the ints() helper, so
original code before cleanup splitted the input and
mapped all words with int(). valuable seconds lost
there.

Also, collections.Counter() was used initially since
I was too tired to remember count() method in lists.

Line 8 took the longest time to figure out. A typo
took 1-3 minutes to find for part 2.

Form: Sleep deprived, felt slow.
2025-01-05 00:06:18 +01:00
cb50d62e3f Prepare for AOC 2024
Python, again.

10th year anniverary.
2025-01-05 00:06:18 +01:00