Toots for thinkb4coding@mastodon.social account

Written by Jérémie Chassaing on 2025-01-15 at 16:30

I finally closed my FB and Insta accounts that I didn't use for over 3 years now.

It feels good

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Written by Jérémie Chassaing on 2025-01-01 at 02:00

Happy new year 45², for those who celebrate!

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Written by Jérémie Chassaing on 2024-12-25 at 15:04

This is the first time I finish #AdventOfCode on then 25th of december (and I made this last one also in #Uiua ) !

All my code in #fsharp and #uiua is here:

https://github.com/thinkbeforecoding/advent-of-code-2024

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Written by Jérémie Chassaing on 2024-12-24 at 00:51

[#]bayrou c'est pire que le père fouettard.

Tu t'apprêtes à passer un bon Noël et lui:

Non, ce sera valls, darmanin et borne.

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Written by Jérémie Chassaing on 2024-12-22 at 12:06

Day 22 of #adventofcode wqs not too difficult.

For part 2, I brut forced find the total price for all present sequence of 4 changes, and took the min.

On 20 CPUs, it finished in a few minutes while I took a shower.

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Written by Jérémie Chassaing on 2024-12-21 at 12:02

Oh, I'm in the metro and just found why my solution is too low...

The arm should Never be on a gap...

Will fix this, but this evening as I won't have time before

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Written by Jérémie Chassaing on 2024-12-20 at 08:19

[#]AdventOfCode was quite easy today

Build a map(dictionay) of the track x,y -> distance

Part1: for each point in track check if point 2 moves in direction is in track, and d2-d1-2>=100

Part2: for all pairs of points in track, if Manhattan distance m is <= 20 it's a potential cheat, check cheat gain d2-d1-m >= 100

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Written by Jérémie Chassaing on 2024-12-17 at 17:24

It was also important to notice that all operations are bitwise operations.

X mod 8 is X & 0b111

X div 2^N is X shift right by N

The rest is just xor

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Written by Jérémie Chassaing on 2024-12-17 at 12:35

And the second number of the opcode that takes no argument also.

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Written by Jérémie Chassaing on 2024-12-17 at 12:32

The only difference is ptobably the 2 constants used to xor, and maybe which registers are used for comptation

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Written by Jérémie Chassaing on 2024-12-17 at 12:26

Day 17 part 2 was fun.

I guess all programs were similar

Taking bits from A 3by3

This bits are xor with 2 constants to get an index and a value

Bits from a at this index are xor with value before output

Solver for part 2 is trying numbers in depth first from 0 to 7.

Each value impose further bits, so keep a mask of fixed bits.

At each step validate value that is tried with already set bits, this cuts all invalid branches early and terminates in less than 100 microseconds

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Written by Jérémie Chassaing on 2024-12-15 at 12:33

Todays part2 was simpler when you always get back to boxes and not individual cells.

A few functions to go from cell to box and box to cells made the job.

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Written by Jérémie Chassaing on 2024-12-14 at 13:55

This was quite logical, with a width of 101, x coordinates are the same every 101. And every 103 for y. So every 101 or 103 you have x or y echo of the image. But it's both correct in x and y only every 101*103

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Written by Jérémie Chassaing on 2024-12-14 at 13:47

For part 2, we noticed an horizontal corelation at 10+101x and a vertical one on 70+103x

Solving the equation, both met at -3020, but due to modulo, repeated every 101*103 seconds.

Done

[#]AdventOfCode

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Written by Jérémie Chassaing on 2024-12-13 at 07:31

Today's advent of code was super easy if you solved the equation directly.

The longuest part was parsing, but it worked the first time with FParsec

https://github.com/thinkbeforecoding/advent-of-code-2024/blob/main/day13.fsx

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Written by Jérémie Chassaing on 2024-12-12 at 09:38

Today was not really easy.

My strategy:

de-duplicate area letters using a fill algorithm
finding areas is the easy
for each area, find all neighbors area
find top-left corner and follow to count border, remove external neighbors
for remaining neighbors (holes), find top left corner and follow inner border, remove internal neighbors until all internal neighbors have been seen

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