๐จ๏ธ ๐ชจ ๐ชจ
#AoC 2022 | Day 17 | Solutions & Spoilers
gloomy hedge
hot mantle
hi
tough plaza
hi
tough plaza
lol
spiral cedar
These elephants need to calm down
left lotus
my part 1 works for example input but not actual input
hot mantle
i think i got it but i actually don't
tough plaza
oh
peak lance
n_repeats = (1000000000000 - dropped) // d_dropped
dropped += n_repeats * d_dropped
extra_rows += n_repeats * d_max_y
Accidentally divided by d_max_y so it found way too many cycles
Also explains why it didn't work for the test until I changed extra_rows to be += instead of =
I wonder why it works better for the test than for the real input though
ยฏ_(ใ)_/ยฏ
shrewd herald
In the part 2 example, no matter how I cut it, I'm consistently getting that every 15 rocks, the height increases by 23, so apart from an offset, shouldn't the answer be a fixed amount off of 1000000000000/15*23? I'm getting 1,533,333,333,338 when the real answer is 1,514,285,714,288... how am I off by billions and billions?
spiral cedar
I got every 35 rocks increase by 53 for the example, which gives 1,514,285,714,263 for 1,000,000,000,000//35*53
midnight marten
what are yall's cycle times
brazen fossil
i want to die
midnight marten
because for me
i got different
cycle ranges
*change amounts
depending on how many
rows i went down
brazen fossil
i just noticed that every time i looped back to the start of the jets
i had the same board loadout & etc
so i just simulated 1000000000000 % rocks_per_cycle -> rocks_per_cycle being how many rocks i get through every time i loop back to 0
and then added 1000000000000 // rocks_per_cycle * height_per_cycle to the resulting height
eager barn
ah p2 was fun
I started by trying to use regex on the differences in height to try to find repetition but then I just moved to using vscode's repetition highlighting
Here's my p2
https://paste.pythondiscord.com/ehusixofix
eager barn
what are yall's cycle times
mine is 1755
blissful bobcat
I see you people doing cycle detection in code while I output into a txt file and am did ctrl f
peak lance
Lol
blissful bobcat
I see you people doing cycle detection in code while I output into a txt file an...
I feel like a genius right now
2634 is my cycle length lul
eager barn
ah p2 was fun
I started by trying to use regex on the differences in height to t...
ah found a way to code it
https://paste.pythondiscord.com/hocawilequ
eager barn
ah found a way to code it
https://paste.pythondiscord.com/hocawilequ
this one has some fun control flow
shrewd herald
I got every 35 rocks increase by 53 for the example, which gives 1,514,285,714,2...
Thanks for the tip. I ended up just ditching all the changes I made to go to part 2 and started again with part 1 figuring that must have had the correct cycles for the example, and turns out that was right, so not even sure what mistake I made in that first attempt, but just threw it out.
hot mantle
i don't get why mine won't work
instead of doing |.####..| |....##.| |....##.| |....#..| |..#.#..| |..#.#..| |#####..| |..###..| |...#...| |..####.| +-------+ it does ```
|####...|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
I kinda liked these two lines of my code feature 3 enumerates:
directions = iter(itertools.cycle(enumerate(data)))
for rocknum, (rockstyle, rock) in enumerate(itertools.cycle(enumerate(rocks)), 1):```
I can't decide if that is ugly or elegant, but it worked well for me.If it tries to drop and can't its just done and comes to a rest, it doesn't continue to get pushed left and right, those left and right pushes will be used for the next piece. Make sure your doing left/right push (only if can) followed by drop (stop if can't) then back to the left/right push.
hot mantle
If it tries to drop and can't its just done and comes to a rest, it doesn't cont...
this is confusing
shrewd herald
this is confusing
The first one it falls from height 2 to 1, so that is clear. Then it sees if the left/right push works, and that is open. Finally, it tries to fall again, but this time something is blocking it, so it comes to a rest.
hot mantle
anyways here's the nice visualization of that part now working https://gyazo.com/44784bc26797b7a3216233636ca57ed4
atomic cedar
not sure if my brain is broken right now, but i can't think of a way to detect a cycle
left lotus
I would assume when the wind cycle repeats, but I'm not even done with part 1 so ๐คท
sullen tiger
Finally done ๐ https://paste.pythondiscord.com/vedoxorike.py
sullen tiger
not sure if my brain is broken right now, but i can't think of a way to detect a...
Essentially, a cycle is going to occur when the conditions are the same as they have been in the past. Spoiler: ||you can use a combination of the current position in the cycle of jet directions, the current position in the cycle of rock shapes, and the 'shape' of the top chunk of the tower||
atomic cedar
hmm, word, thanks, i'll try that
pallid epoch
hi people
i
i've manually checked that my code is correct for all of the sample
but it continues to fail
left lotus
i've manually checked that my code is correct for all of the sample
Same
pallid epoch
does anyone want to find the very easy bug
that i probably missed due to sleep deprive
(the result is half the expected)
wild rainBOT
Hey @fading sinew!
pallid epoch
the paste bin says my log file is too long
wild rainBOT
Hey @pallid epoch!
fading sinew
https://pastebin.com/m7WTg1Vj
Solution for 10**4299 rocks
pallid epoch
404 not found
fading sinew
wtf
silver light
that's what mine looks like too...
fading sinew
https://paste.pythondiscord.com/
doesn't work for me
The entire page freezes when I save
idk why
pallid epoch
The entire page freezes when I save
i get a "something went wrong"
silver light
How big are you pasting? Is there a limit?
I just tested it ok: https://paste.pythondiscord.com/efonegiboh
pallid epoch
i mean when i try to paste the log of all the starting block positions
pallid epoch
its like 18 million lines or so idk
fading sinew
its like 18 million lines or so idk
nah, it's tiny
pallid epoch
i meant my log file
fading sinew
340 ms for 10**4299
170 ms for the actual solution both parts
left lotus
Finally done with part 1
pallid epoch
what was the issue, out of curiosity
left lotus
I was replacing existing rock with air
I forgot you could slide rocks underneath other rocks
Smh haven't played tetirs for too long
fading sinew
Elephant + 1000000000000 rocks = A heck of a lot of dead elephants ๐
left lotus
what are y'all runtimes?
150ms for p1
fading sinew
150ms for p1
80 ms for p1 alone
toxic geyser
This was fun. I anticipated part two, spend way too much time on part one, and then had part two almost immediately.
fading sinew
Question:
why is
x=None
if x is None
like a billion times faster than
why is
x=1
if x is 1
toxic geyser
Is it that much faster?
!e
Ignoring the syntax warning:
import timeit
def check_none():
a = None
if a is None:
return 2
else:
return 3
def check_one():
a = 1
if a is 1:
return 2
else:
return 3
print(timeit.timeit(check_none))
print(timeit.timeit(check_one))
@toxic geyser :white_check_mark: Your 3.11 eval job has completed with return code 0.
001 | <string>:14: SyntaxWarning: "is" with a literal. Did you mean "=="?
002 | 0.08325721399160102
003 | 0.08740069600753486
They are the same when I test that too.
pallid epoch
maybe some impl about doing an actual identity check on 1
hot mantle
so 5 hours after this puzzle released i still don't know why my code is failing
no overlappings are occuring anymore
nvm i found the problem
fading sinew
!e
Ignoring the syntax warning:
```py
import timeit
def check_none():
a = ...
welp, wtf. my code just continues to run to with no end in sight with 1 but finishes in 170 ms if None
fucking weird
sullen tiger
welp, wtf. my code just continues to run to with no end in sight with 1 but fini...
Sounds more like a logic error 
fading sinew
ok now it returns 3127
sounds about in the range of the p1 answers
fading sinew
i'm testing the test example
ah ok
fading sinew
i'm testing the test example
misssing a condition check perhaps?
hot mantle
5 lines to be uncommented for visual debugging and 3 lines are currently uncommented to see if overlaps occured
fading sinew
5 lines to be uncommented for visual debugging and 3 lines are currently uncomme...
you code like you're golfing ๐ญ
fading sinew
5 lines to be uncommented for visual debugging and 3 lines are currently uncomme...
wherre are the lch & fall fncs?
and rch
hot mantle
wherre are the lch & fall fncs?
def fall(m,mx,my,x,y):
if y==0:return 0
for n in range(my):
ty=n-1
for i in range(mx):
if m[n,i] and (n==0 or not m[n-1,i]) and arr[y+ty,i+x]:return 0
return 1
def lch(m,my,x,y):
if x==0:return 0
x-=1
for n in range(my):
if m[n,0]and arr[y+n,x]:return 0
return 1
def rch(m,mx,my,x,y):
if (x:=x+mx)==7:return 0
for n in range(my):
if m[n,-1]and arr[y+n,x]:return 0
return 1
Erm, mine is 10092 characters.
fading sinew
thanks, golfing rn lol
hot mantle
sounds about in the range of the p1 answers
i gave up and wtf
3127 was the right answer to my input in part 1
the wrong answer to the example input was the right answer to my input
fading sinew
the wrong answer to the example input was the right answer to my input
LMAO ๐คฃ
sucks to be you, i guess ๐ญ
fading sinew
what are the chances of this
1/range(min(answer_list),max(answer_list))
hot mantle
now I have to figure out why I am off by 21 on both parts for some reason
i wanna figure out why i had the wrong answer to the example input which was the right answer to my input
left lotus
probably by chance I would imagine
left lotus
now I have to figure out why I am off by 21 on both parts for some reason
it doesn't happen on the puzzle input either
left lotus
fine
fluid siren
tfw
fluid siren
i blame pycharm
i selected the console and then when i cmd c it highlights the code itself
for some odd reason
fading sinew
i selected the console and then when i cmd c it highlights the code itself
Sublime >>> Pycharm
atleast for smaller codes. Pycharm over sublime for projects
fluid siren
is cycle detection with just the top five rows fine or do i need more rows
fluid siren
i think i might have a slight error in my code
peak lance
LMAO what on earth did you do
fluid siren
import antigravity
does a cycle always have to cycle back to the start here?
peak lance
No
peak lance
DW I made that wrong assumption too lol
fluid siren
hmm
peak lance
Actually you could probably make it work like that
fluid siren
i've managed to detect a cycle
peak lance
Might be shorter for the golf as well
fluid siren
well
if i knew it cycled at the start i could just divide stuff by two
but i know it doesn't
so there's an offset
which is super annoying
actually no i can't even do that
bruh
livid nexus
Completely hacked together my solution for today with cycles using just my height after each rock. Time to get back to yesterday's problem
frank scaffold
i still don't know how my part2 solution works tbh
I thought the cycle length must just be lcm(len(movements), 5)
peak lance
Completely hacked together my solution for today with cycles using just my heigh...
There's no way that should work
frank scaffold
but no, it's weirdly different; I ended up just writing a search checking different cycle lengths. for my real input, the lcm was ~50k and the actual cycle length was ~40k
livid nexus
There's no way that should work
Works for both test and puzzle data
peak lance
i still don't know how my part2 solution works tbh
ML be like
peak lance
Works for both test and puzzle data
I said 'should'
fluid siren
are you guys like shifting the board down or anything
livid nexus
Going to slightly clean up my code then I'll share it
peak lance
Does your input end on a piece lock-in?
That's the only way I see that approach working
peak lance
are you guys like shifting the board down or anything
No, just skipping some drops in the middle
livid nexus
What do you mean by piece lock-in?
peak lance
I just gave my program 5k rows to drop in lol
peak lance
What do you mean by piece lock-in?
When moving down fails so the piece locks in/stops falling/whatever
peak lance
For part 2 I switched to a defaultdict lol
livid nexus
Not sure what you mean, every piece locks in then?
peak lance
(technically I used aoc_helper.SparseGrid but I didn't actually use any of its methods lol)
fluid siren
hmm
peak lance
Not sure what you mean, every piece locks in then?
You might have a piece in the middle of a drop at the end of your input
E.g. <><>
livid nexus
I think you might misunderstand what I'm doing. Will share code in a moment
peak lance
<>< locks in the first piece, then > moves a second piece without locking it in
fluid siren
some people said that their data cycled after just one cycle of the moves
so
hmm
so if i know i have a cycle
then i can use the turtle bunny algorithm thing
to find the cycle length
i think
livid nexus
Here's my hacked together solution. I'm hoping it'll work with any input and I wasn't just lucky with my input.
def part_two():
rock_count = 10000
cave_height = rock_count // 5 * 13
cave = np.full((cave_height, 7), 0)
height = 0
heights = []
idx = 0
for i in range(rock_count):
rock = rocks[i % 5]
y = cave_height - height - rock.shape[0] - 3
x = 2
while (
np.all(cave[y : y + rock.shape[0], x : x + rock.shape[1]] + rock <= 1)
and y < cave_height
):
match gas[idx]:
case "<":
x = max(0, x - 1)
if np.any(
cave[y : y + rock.shape[0], x : x + rock.shape[1]] + rock > 1
):
x += 1
case ">":
x = min(7 - rock.shape[1], x + 1)
if np.any(
cave[y : y + rock.shape[0], x : x + rock.shape[1]] + rock > 1
):
x -= 1
idx = (idx + 1) % len(gas)
y += 1
y -= 1
height = max(height, (cave_height - y))
heights.append(height)
cave[y : y + rock.shape[0], x : x + rock.shape[1]] += rock
n = 1
repeat = 0
repeat_value = 0
for i in range(1, 2000):
if (
heights[(n + 1) * i] - heights[(n) * i]
== heights[(n + 2) * i] - heights[(n + 1) * i]
== heights[(n + 3) * i] - heights[(n + 2) * i]
):
repeat = i
repeat_value = heights[(n + 1) * i] - heights[(n) * i]
break
offset = 1000000000000 % repeat
return 1000000000000 // repeat * repeat_value + heights[offset - 1]
It was possible for this to be wrong, if, for example, every other time or every third that combination occurred it fell differently because of the state of the board, but such a state would be hard to propagate through so many pieces .
fluid siren
right as i manage to figure it out for the test data
i try to use the real data
and i get 54
wtf
how does it work so nicely for the example
and then break so horribly in the actual data
slow belfry
hard to play tetris with these pieces
fluid siren
no t spins sadly
midnight marten
hard to play tetris with these pieces
how do terminal animations work
does it like print things so fast
it seems like it's animating
slow belfry
how do terminal animations work
you find the parts of the terminal you need to change, and just update those parts by moving the cursor to a specific location and printing the character you need
midnight marten
how do you move the cursor?
slow belfry
with ansi escapes
slow belfry
yep
small current
you find the parts of the terminal you need to change, and just update those par...
so you can print("hello world") 10 times. and then move the cursor up 5 rows and chnage only one of those already printed hello world strings to asdfdasdfasdfa ?
small current
i alwys thought cursor location can only be changed in the last line printed ๐
peak lance
Won't work in conhost (CMD) or idle though
small current
whats conhost
small current
oh
peak lance
Use Windows Terminal
small current
powershell?
small current
i dont know the difference between command prompt and terminal tbh
when i enter terminal it only shows me the cmd.exe
peak lance
Windows Terminal is ANSI-compliant though
peak lance
i dont know the difference between command prompt and terminal tbh
You download Terminal from the windows store, it can run any of cmd, powershell, wsl
delicate marten
cmd is a shell, winterm is a good terminal emulator. when you open cmd.exe, however, it defaults to conhost, a really bad emulator.
peak lance
To conhost's credit it's more lightweight than Terminal
If I want to do something quick and dirty conhost is usually good enough
small current
`print('\x1b[4Ayourtexthere')` will do it I believe
ye it goes "up" 4 rows
but it will erase the lines below it
small current
Only if you print over them
well ye i thought thats the whole point if one wants to change only a thing in one of the already printed lines
need to print over the existing one i thought
peak lance
Yes but it doesn't erase the other lines unless you print over those too
FYI if you want to do cursor position stuff I've always used colorama
Includes a conhost compatibility thing too
small current
Yes but it doesn't erase the other lines unless you print over those too
for i in range(20):
print("Hello world",i)
print("\x1b[15Aasdfasfasdfasd")
it does for me somehow
peak lance
Might be your shell, try printing 10 newlines
for i in range(20):
print("Hello world",i)
print("\x1b[15Aasdfasfasdfasd")
print('\n' * 10)
If that doesn't work then I don't know
It should work
!e I'm curious, does it work in the bot?
for i in range(20):
print("Hello world",i)
print("\x1b[15Aasdfasfasdfasd")
print('\n' * 10)
@peak lance :white_check_mark: Your 3.11 eval job has completed with return code 0.
001 | Hello world 0
002 | Hello world 1
003 | Hello world 2
004 | Hello world 3
005 | Hello world 4
006 | Hello world 5
007 | Hello world 6
008 | Hello world 7
009 | Hello world 8
010 | Hello world 9
011 | Hello world 10
... (truncated - too many lines)
Full output: https://paste.pythondiscord.com/orokuvapih.txt?noredirect
peak lance
Hello world 0
Hello world 1
Hello world 2
Hello world 3
Hello world 4
Hello world 5
Hello world 6
Hello world 7
Hello world 8
Hello world 9
Hello world 10
Hello world 11
Hello world 12
Hello world 13
Hello world 14
Hello world 15
Hello world 16
Hello world 17
Hello world 18
Hello world 19
[43m15Aasdfasfasdfasd
I don't know if that got corrupted or just didn't work lol
Ok no it just didn't work
Wait ANSI doesn't render properly on mobile at all lol
keen frigate
not sure if my brain is broken right now, but i can't think of a way to detect a...
me last night
i was so braindead
atomic cedar
haha
slow belfry
i've been going to bed super early last few days, i'm not willing to give up more sleep for aoc
blissful phoenix
I was doing that for most days and today I just fell asleep in my chair in the middle of AoC at midnight.
Probably a good thing I did though. I've been stuck in cycle detection for P2 for god knows how long now. It's right on the test input but wrong on my actual input. Only guess I have is it might have to do with overhangs. I'm looking for same piece type, same heights of each column, and same index on the jet pattern. If anyone has any ideas, I'm all ears at this point
https://paste.pythondiscord.com/cirubifume
pallid epoch
finally found why my p1 is wrong
at some step the list just starts eating itself and i have no idea why
looks like a standard memory corruption but this shouldnโt happen in python?
like it seems like the list pointer is getting overwritten
brazen fossil
Probably a good thing I did though. I've been stuck in cycle detection for P2 fo...
if you want a "cheaty" solution try to store the max heights and the current piece every time your jets wrap back around to 0 for like 10 times and observe the difference in the max heights
(might just be my own input but it might work for you idk)
keen frigate
i've been going to bed super early last few days, i'm not willing to give up mor...
yeah, idk if i wanna give up my sleep anymore either
i also have a shuttle and flight to catch early tomorrow morning, so it's time to drop down the leaderboard tonight ๐
peak lance
Rip
plain gyro
i have this function that is supposed to shift the falling rock. it works fine for left and right, but gets kinda weird when moving down :/
def shift_block(grid:list[str], dir: tuple[int,int]):
dx, dy = dir
positions = []
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == '@':
positions.append((i, j))
step = 1 if dx < 0 else -1
for i, j in positions[::step]:
grid[i+dy][j+dx] = '@'
grid[i][j] = '.'
midnight marten
did you try using a debugger or anything of the sort
midnight marten
bro i'm so sorry
plain gyro
the step was to figure out in which order to remove and add the part of the blocks, but it only relevant if its an horizontal move
midnight marten
so in this case it's 1?
plain gyro
yeah
def shift_block(grid:list[str], dir: tuple[int,int]):
dx, dy = dir
positions = []
for y in range(len(grid)):
for x in range(len(grid[0])):
if grid[y][x] == '@':
positions.append((x, y))
new_positions = [(x, y+1) for x, y in positions]
for x, y in positions: # remove block
grid[y][x] = '.'
for x, y in new_positions: # build shifted block
grid[y][x] = '@'
...
shift_block(grid, (0,1))
...
this is what really blows my mind, how does this happen in one instruction grid[y][x] = '@'
it just added three @ out of nowhere
the length of the new_positions is 5, yet it places 7 @
how is that possible
midnight marten
def shift_block(grid: list[str], dir: tuple[int,int]):
dx, dy = dir
positions = []
for y in range(len(grid)):
for x in range(len(grid[0])):
if grid[y][x] == '@':
positions.append((x, y))
new_positions = [(x, y+1) for x, y in positions]
for x, y in positions: # remove block
grid[y][x] = '.'
for x, y in new_positions: # build shifted block
grid[y][x] = '@'
grid = [
list('...@...'),
list('..@@@..'),
list('...@...'),
list('.......'),
list('.......'),
list('.......')
]
shift_block(grid, (0, 1))
for i in grid:
print(''.join(i))
this seems to work fine for me?
plain gyro
what
plain gyro
[['.', '.', '.', '@', '.', '.', '.'], ['.', '.', '@', '@', '@', '.', '.'], ['.', '.', '.', '@', '.', '.', '.'], ['.', '.', '.', '.', '.', '.', '.'], ['.', '.', '.', '.', '.', '.', '.'], ['.', '.', '.', '.', '.', '.', '.']]
wait a sec
['.', '.', '.', '@', '.', '.', '.']
['.', '.', '@', '@', '@', '.', '.']
['.', '.', '.', '@', '.', '.', '.']
['.', '.', '.', '.', '.', '.', '.']
['.', '.', '.', '.', '.', '.', '.']
['.', '.', '.', '.', '.', '.', '.']
plain gyro
thing?
midnight marten
the grid you're testing w/
midnight marten
remove the shift_block
just wanna make sure it's due to the shift_block and not bc of anything else
plain gyro
midnight marten
try printing out new_positions?
midnight marten
try printing the grid after the loop with grid[y][x] = '.'
plain gyro
midnight marten
and after the
for x, y in new_positions: # build shifted block
grid[y][x] = '@'
```it's weird?print out y, x in each iteration
plain gyro
midnight marten
(gonna be honest i'm just throwing darts here, i've no clue what's wrong either)
plain gyro
yeah ive been like that for like two hours
midnight marten
and print out the grid after each individual iteration
plain gyro
1 3
-------
.......
...@...
.......
.......
.......
.......
2 2
-------
.......
...@...
..@....
.......
.......
.......
2 3
-------
.......
...@...
..@@...
.......
.......
.......
2 4
-------
.......
...@...
..@@@..
.......
.......
.......
3 3
-------
.......
...@...
..@@@..
...@...
...@...
...@...
-------
how??
midnight marten
what????
plain gyro
midnight marten
surprised pikachu face:
plain gyro
same
midnight marten
ok uh
just try setting
grid[3][3] = '@'
see what that does
OH WAIT
I THINK I KNOW
the last two elements
they're linked to the index 3 row
midnight marten
like yknow how
x = [0, 1, 2, 3]
y = x
x[0] = 1234
print(y) # [1234, 1, 2, 3]
huh?
midnight marten
i think that's the bug
plain gyro
i didnt get it
midnight marten
something similar to that is happening in your grid
because y didn't copy x
it's a reference to x
so if you change x, y will change along w/ it
plain gyro
but where is the bug
midnight marten
that's the bug
each row isn't independent
you might've screwed up grid initialization
plain gyro
okay, ill look into it, thanks a lot ^^
so the bugs here ๐ง
def insert_rock(rock, grid, width):
for i, line in enumerate(rock):
grid.insert(i, ['.'] * 2 + list(line) + ["."] * (width - len(line) - 2))
# resize_grid(grid, rock)
rock_height = len(rock)
space_between = ['.'] * width
for _ in range(3):
grid.insert(rock_height,space_between)
return grid
maybe?
plain gyro
oh
i see it
space_between = ['.'] * width
for _ in range(3):
grid.insert(rock_height,space_between)
the space between is the same in the last 3 rows
midnight marten
np!
warm zealot
This is driving me round the bend
For part 2, I've got a solution that works perfectly for the test case, agrees for every rock num I could be bothered to simulate, and yet doesn't give the correct answer for the main thing
quick meteor
I managed to complete part 1 with straightforward method which take roughly 20 seconds to execute. I see there is a repeating pattern on rock agencement but I'm too lazy to code this out...
lucid gulch
test case works perfectly fine up to the point that i can track, but completely off on the result
def play(inputs, count):
types = [[(0, 0), (1, 0), (2, 0), (3, 0)],
[(1, 0), (0, 1), (1, 1), (2, 1), (1, 2)],
[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)],
[(0, 0), (0, 1), (0, 2), (0, 3)],
[(0, 0), (1, 0), (0, 1), (1, 1)]]
field = []
bottom = -1
i = 0
for r in range(count):
while len(field) < bottom+8: # field padding for next rock
field.append(['.']*7)
rock = types[r % 5]
x, y = 2, bottom+4
dropped = False
while not dropped:
m = '< >'.index(inputs[i])-1
# left/right
if all(0 <= x+a+m < 7 and field[y+b][x+a+m] == '.'
for a, b in rock):
x = x+m
i = (i+1) % len(inputs)
# down
if y > 0 and all(field[y+b-1][x+a] == '.' for a, b in rock):
y -= 1
else: # block done
for a, b in rock:
field[y+b][x+a] = '#'
dropped = True
bottom = max(y + b for _, b in rock)
while field[-1] == ['.']*7: # cut off empty rows
field.pop()
return field
print(len(play('>>><<><>><<<>><>>><<<>>><<<><<<>><>><<>>', 2022)))``` any help appreciated, like I said, seems to emulate the test case perfectly fine but the endresult is off by a massive 500
lilac torrent
1'000'000'000'000
that's a lot of fckin rocks
it takes me a bit less than a second to process 100'000 rocks
looks like I have the wrong approach here
lucid gulch
possibly pattern repetition?
lilac torrent
maybe
lucid gulch
probably after len(input) * 5
lilac torrent
there's a few leads to improve perf further but the order of magnitude isn't good enough
lucid gulch
which should only be 50k
lilac torrent
you've done part 2 already ?
lucid gulch
no i'm still failing the test case lmao
but after len(input)*5 you should start a new cycle since that's where input index 0 and rock index 0 will overlap again
just have to be careful about how the cycles connect
well that's my guess anyways
lilac torrent
you're right for the example
welp I won't finish yesterday's puzzle tonight after all ๐ข
lucid gulch
yea i'm also 2 behind now since i'll have to go to bed
lilac torrent
rip
oh
but
you actually don't have to handle the cycle connection do you
you can do 2 full circles and match the second with all of the other
lucid gulch
yea that works as well
@hot mantle what was your off by 500 error, I might be stuck at the same problem
hot mantle
<@310263589913100288> what was your off by 500 error, I might be stuck at the sa...
wdym?
โซthe final revision of the code i wrote was off by 59
lucid gulch
it's off by 499
here you said 499
hot mantle
oh that
hot mantle
here you said 499
that had overlaps
lucid gulch
so it went down where it shouldn't?
hot mantle
so it went down where it shouldn't?
it went left or right where it shouldn't
hot mantle
instead of doing ```
|.####..|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|...
@lucid gulch this?
lucid gulch
ahhh so it doesn't eat inputs if it won't move
ok thank you! that will hopefully be it
maybe not hmm
lucid gulch
test case works perfectly fine up to the point that i can track, but completely ...
well if anyone sees a problem on first glance lmk please
lucid gulch
congratz T.T
hot mantle
oh i got pretty close
14 off
68 off on actual input
anyways here's a visualization
jovial siren
i don't get the issue with my solution
welp
might be screwed today
i changed like nothing and now it works
that's pretty cool
hard totem
wait what. today was pretty simple. if i had started earlier i would have earned points. ๐ณ
tough plaza
๐ณ
hard totem
place 100 for part 2 is 40:48
that's lots of time.
not exactly sure when i started. 
small current
is todays second part another one of those
where a bad implementation iin part 1
leads to absurd amount of time necessary for computation?
hard totem
i don't think so.
i think part 2 should be easy independent of the specific part 1 implementation. 
okay, i just checked. apparently i spent about 45 minutes on part 1. ๐
lots of small errors here and there.
keen frigate
my part 2 was part 1 but extended
with cycle detection
but than that the code was basically the same
fluid siren
I rewrote my entire code for part 2
keen frigate
oh really? mine was adding a few variables and if statements here and there and it worked out
also i did some funny cursed shit with my Rock class
class Rock(namedtuple("Rock", "shape left_x bottom_y")):
def __lshift__(self, value):
return Rock(self.shape, self.left_x - value, self.bottom_y)
def __rshift__(self, value):
return Rock(self.shape, self.left_x + value, self.bottom_y)
def __sub__(self, value):
return Rock(self.shape, self.left_x, self.bottom_y - value)
def __add__(self, value):
return Rock(self.shape, self.left_x, self.bottom_y + value)
@property
def positions(self):
return [
(self.left_x + dx, self.bottom_y + dy)
for dy, row in enumerate(self.shape)
for dx, filled in enumerate(row)
if filled
]
def validate_move(self, grid):
return all(0 <= pos[0] < 7 and pos[1] >= 0 and pos not in grid for pos in self.positions)
so then i could do
match jet:
case ">": rock >>= 1
case "<": rock <<= 1
rock -= 1
I had a really bad hashing method for the board state
so I just changed my board to a 200k tall numpy array
keen frigate
huh really? interesting
fluid siren
an I piece slid down like 40 rows in my input
keen frigate
oh i saved state for the entire grid in a set
that contained positions that were filled
and then hashed the last 10 rows
if i'm understanding what you're saying
fluid siren
yeah
I had to hash the last 50 rows
last 25 rows wasn't enough
because the I piece would fall down 40 rows and I'd get a cycle of length 1
brazen fossil
so then i could do
```py
match jet:
case ">": rock >>= 1
case "<": rock ...
rock += ord(jet) - 61
trust
hot mantle
rock += ord(jet) - 61
rock+=1|-(jet<'>')
trust
keen frigate
rock += ord(jet) - 61
okay this is pretty smart
keen frigate
`rock+=1|-(jet<'>')`
trust
and i would say this is smart but why
brazen fossil
wtf does |- do
midnight marten
honestly today was kinda sus
because it relied on an i piece
not falling all the way to the bottom
unless someone can prove
that there's an upper bound on the maximum number of spots
a piece can fall down
fluid siren
I think someone on reddit said something about keeping track of the maximum number of drops of a piece
also all my days are kinda sus
well not all
a lot of my days
brazen fossil
honestly i'm still not really sure how to not "cheese" today's solution
not sure what to actually detect cycles in
tough plaza
i should think before posting, i forgot the thing was 7 wide
fluid siren
I mean you thought for quite a while before that already
tough plaza
i am bad at aoc, unfortunately
tough plaza
so what
brazen fossil
3 digit ranked wym
tough plaza
pretty much everyone here is 2 or 3 digit
fluid siren
3 digit ranked wym
leaderboard
brazen fossil
pydis lb or aoc lb
fluid siren
actual
tough plaza
i can semi-consistently hit 3 digits for problems
brazen fossil
sub 1k generally isn't hard
tough plaza
but that doesn't mean much
brazen fossil
sub 100 though ๐
brazen fossil
it's not hard to get sub 1k if you start when it releases
tough plaza
^
fluid siren
from my perspective you guys are like really good at aoc
tough plaza
this is my first year, i'm pretty inexperienced
fluid siren
I've been taking like 6 hours to solve these last few days
brazen fossil
my second year after not doing it in 2021 ๐คทโโ๏ธ
just remember this isn't like usaco
tough plaza
ye lol
brazen fossil
and the solution is what counts - not how you did it
tough plaza
sometimes i just waste 10 mins trying to optimize
only to realize that i can't
and it really was braindead simulation
brazen fossil
i.e. my today's solution looked like this
fluid siren
did you not need to simulate after all the cycles?
brazen fossil
wym
fluid siren
like
tough plaza
i.e. my today's solution looked like this
C=D=E=a=0;Z=x=1j
K=input()
F,G={0},{}
L=lambda x:all([(p:=B+x+A).imag>0<=p.real<7,0][p in F]for A in I)
while 1:
A=max(A.imag for A in F);Z=B=2+4j+A*x;I=((0,1,x,1+x),range(4),(1,x,2+x,1+2j),(0,1,2,2+x,2+2j),(0,x,2j,3j))[C:=-~C%5];E==2022>(a:=A)
if(C,D)in G:N,O=G[C,D];P,Q=divmod(1e12-E,N-E);Q>=0<a>print(a,A+(O-A)*P)
while Z:G[C,D]=E,A;B+=L(M:=ord(K[D])-61)*M;Z=L(-x);B-=x*Z;D=-~D%10091
F|={B+A for A in I};E+=1```๐
fluid siren
I had
(height before first cycle) + (cycle length)*(number of cycles) + (height of remaining partial cycle)
do you not need the remaining partial cycle?
brazen fossil
i basically noticed that every time jet loops around, a constant height is added, ergo that previous cycles don't matter
so i just simulated how much i needed to simulate (like 1000000000000 % rocks_per_cycle idk) how many rocks
and just added the height i get per cycle
very fun
hacky solution
tough plaza
i did every 5 times the jet looped
brazen fossil
wym
brazen fossil
it isn't-i made sure to check
brazen fossil
lmao
brazen fossil
wym
brazen fossil
if you wanted code that works for any input you probably should do that
brazen fossil
๐คทโโ๏ธ
midnight marten
who the heck pinged
lucid gulch
```python
def play(inputs, count):
types = [[(0, 0), (1, 0), (2, 0), (3, 0)]...
so after sleeping on it and watching the "game" run for a while i figured out my problem python bottom = max(y + b for _, b in rock)after setting the stone i used this to change the spawn point but didn't account for the fact that pieces could slot in below the current max
lucid gulch
am I right in the assumption that patterns only happen on a cycle basis i.e. len(input)*len(rock_shapes)?
plain gyro
my solution for p1 takes a couple minutes, pretty sure it would be an eternity to use it for part 2...
i have no idea how to optimize it
fluid siren
why does it take a couple minutes for your p1?
plain gyro
cause its not a very efficient solution, but its what i could get
wild rainBOT
Hey @plain gyro!
plain gyro
why does it take a couple minutes for your p1?
now it takes about 18 seconds
because i was using deepcopy and it is very slow
but still, its not good enough
fluid siren
try profiling to find the bad bits
fluid siren
you aren't meant to brute force 3 trillion rocks
plain gyro
so i should approach the problem in a smarter way?
and not place and move every rock?
fluid siren
no
find a way to reduce the number of rocks you need to place by taking advantage of repeating stuff
lilac torrent
I'm loosing my mind over part 2
past karma
Part 2 was weird for me in that I found it conceptually easy but really irritating to code
past karma
mathematical as in trying to find the height per repeating group of rocks?
lilac torrent
here's where I'm at
heigh grow as a repetitive sequence
for the example, the sequences are 7 cycles long, the first 2 are useless, and then it repeats every 35 cycles, which is 5 sequeences of 7 cycles
one cycle being one rock
and rn I'm trying to figure out where those magic numbers come from
lucid gulch
my pattern recognition works for the test case but doesn't produce anything on input
sad stuff
past karma
and rn I'm trying to figure out where those magic numbers come from
how did you figure out the magic number in the first place?
past karma
right
so what you need to do is programatically derive the 35 cycle number
and I agree, it's much more annoying to do programmatically than just looking at the output
lilac torrent
but it's not always 35 isn't it ?
lilac torrent
what I'm thinking of
it looks like the number of flow mov / the numbers of shapes
that's what I'm working out rn
lucid gulch
@past karma is it related to len(input)*5 ?
thought that should obviously be it but so far only produced results for the test for me
past karma
<@388062601281798144> is it related to len(input)*5 ?
not sure I'm following what you mean there
lucid gulch
well my thought was that obviously things should repeat an some multiple of len(input)*5 because thats where both stones and the input will repeat, starting a new cycle
which works for the test case
but i've realized the problem with that logic, still thought it might be related nonetheless
past karma
oh right. yeah that was my initial thought as well
seems not to be the case
but there is a repeating group
lucid gulch
so just brute force pattern recognition?
lilac torrent
there's repeating groups before nb_flow * nb_shapes
past karma
so just brute force pattern recognition?
i essentially did this
lucid gulch
alright, i'm gonna do it as well then
past karma
if you do it as dumbly as I did then it takes 2 mins to run
lucid gulch
i kinda don't wanna mess with my play algorithm but doing it afterwards requires knowledge of how long to run
lilac torrent
@past karma so people have been doing pattern recognition ?
past karma
<@388062601281798144> so people have been doing pattern recognition ?
"pattern recognition" seems rather fancy for what I did but I guess so
lilac torrent
I was ooking for a pure mathematical solution this whole time
lucid gulch
if find_pattern:
if pattern is None:
for k in range(highest-49):
if grid[highest-49:highest+1] == grid[k:k+50]:
pattern = grid[k:k+50]
print(f'repeat at r: {r} height: {highest} against {k+50}')
else:
if grid[k:k+50] == pattern:
print(f'repeat at r: {r} height: {highest} against {k+50}')```i get one immediate result but can't find anthing after that
past karma
not sure I follow what you did there, but I did a very dumb way of doing it
lucid gulch
i'm checking if the highest 50 lines can be found somewhere else in the grid, if so try to find it again after doing some more cycles so i know the pattern length
but trying to find it again doesn't produce anything
past karma
all I did was, after each rock fall, try and see if the grid could be split in 2 and have two halves exactly the same - but do that a number of times whilst progressively removing the bottom rows from the grid
lilac torrent
you're a genius
past karma
yeah well my genius solution takes two minutes to run lol
def check_for_repeating_groups(self, group_no):
for start, _ in enumerate(self._rocks_positional):
sublist = self._rocks_positional[start:]
if len(sublist) < 10:
# we are looking for large repeating groups
# use as quick and dirty way of ditching tiny repeating groups
return False, None, None
if len(sublist) % group_no != 0:
# looking for exact repeating groups so need a sublist divisible into equal parts
continue
groups = more_itertools.divide(group_no, sublist)
groups = iter(groups)
first = list(next(groups))
if all(first == list(x) for x in groups):
# found
group_len = len(first)
offset_len = len(self._rocks_positional) - group_len * group_no
return True, group_len, offset_len
return False, None, None
"rocks_positional" is a list[set[int]] where each set represents a row in the chamber, and each int in a set is the horizontal position of a rock in that row
I actually check to see if the grid can be split in 2, then again in 3 to work out the number of rocks for a repeating group
past karma
no problem! what was the solution runtime?
past karma
yh
lilac torrent
I figured something : if you take the diff (between current and last elevation), the current flow, and the current shape
you could use that as an id to match when the first repeating line was happening
past karma
sounds like a good optimisation :)
lilac torrent
just took me 2 days lol
and there's still a magic number : at one place I had to add a "-1)
but it's working I'll never modify it again
long long part2(std::vector<Shape> const& shapes, std::string const& gas_flow)
{
Tetris tetris(shapes, gas_flow);
std::vector<long long> cumulated;
std::unordered_map<RockInfos, long long, DfHash> rock_last_index;
int time_to_stabilize = 14;
int repeat_pattern_size = 35;
int i = 0;
while (true)
{
tetris.tick();
cumulated.push_back(tetris.max_y);
RockInfos rock{
tetris.max_y - cumulated.back(),
std::distance(tetris.gas_flow.begin(), tetris.current_flow_it),
std::distance(tetris.shapes.begin(), tetris.current_shape_it)
};
auto [l_it, l_inserted] = rock_last_index.insert({ rock, i});
if (!l_inserted)
{
repeat_pattern_size = i - l_it->second;
time_to_stabilize = l_it->second - 1;
break;
}
i++;
}
long long target = 1'000'000'000'000;
long long repeat = target / repeat_pattern_size;
long long remaining = target % repeat_pattern_size - time_to_stabilize;
long long result = cumulated[time_to_stabilize]
+ (cumulated[repeat_pattern_size + time_to_stabilize] - cumulated[time_to_stabilize]) * repeat
+ cumulated[time_to_stabilize + remaining - 1] - cumulated[time_to_stabilize];
return result;
}
for the curious
sorry it's in cpp
silver light
I'm pulling my hair out here. Part one works for the example, but it says the number is too low when using the puzzle input. I've gone over my code a bunch of times and even tried printing out the board after each step, but I can't find anything wrong.
past karma
want to post your code and see if we can figure out any issues?
wild rainBOT
Hey @silver light!
past karma
ah no worries :)
silver light
Just a hint please if you have one.
past karma
https://paste.pythondiscord.com/oloruwicav
your code is quite far off. For my main output, it's off by a factor of 50% larger that it should be
silver light
Yeah, but it gets the example right.
past karma
yes. same here. Very puzzling.
silver light
I thought the logic was good...
past karma
still looking into it
silver light
Thank you very much.
past karma
this is awkward lol - I ran the code on a different input which is why it crapped out
like, a different day's input
past karma
and running it on my actual day 17 input, it does give the correct answer
past karma
maybe redownload it
silver light
Just did that.
Trying now.
It's still giving me 3202, which is too low.
And never mind... It accepted the answer. I must have fat-fingered the entry...
silver light
I'm very sorry to have waited your time.
past karma
nah don't worry about it :)
silver light
Thank you very much!
Holy Carp! 1000000000000 rocks? I'm going to need some optimizations...
silver light
I hope they don't forget what I'm doing for them here.
silver light
So I'm not seeing any loops in my input, so should I look for relationships at step intervals of the length of the input?
silver light
I've looked the bottom 10m records for any match to the first 10 and I haven't found a single match yet... does that sound right?
now i'm thinking that I can't use the bottom few rows.
lucid gulch
yea i think if you're gonna get a match it's gonna be xyzabcabcabc
lucid gulch
i think i'm on to something finally with my algorithm
lucid gulch
alright, thank you @past karma for the pointers you gave, i basically did what you said after converting my game to be set based instead of grid based python def find_pattern(grid, pattern=None): if pattern is None: for k in range(20, len(grid)//2): if grid[-k:] == grid[-k*2:-k]: pattern = grid[-k:] return pattern else: if grid[-len(pattern):] == pattern: return pattern and
if count > 1000000:
found = find_pattern(grid, repetition['pattern'])
if found:
if repetition['pattern'] is None:
repetition = {'pattern': found, 'cycle': r, 'height': len(grid)}
else:
cycles_rem, remainder = divmod(count - repetition['cycle'], r-repetition['cycle'])
count = r + remainder
magic = (cycles_rem-1)*(len(grid)-repetition['height'])``` magic gets initialized to 0 and will always be added to the final return, count is the number of cycles to be runah and you can speed up your solution to basically instant by setting the min length of blocks to 2000+ since that seems to be in the ballpark for input solutions
past karma
Good to know!
silver light
Finally got it. I feel kind of bad tho. I searched for the repeated blocks manually so it only works with my puzzle input.
pine kelp
Not sure if I'm being incredibly dumb, but I cannot find the cycle in the test input. I've completed Part 1 (which fully simulates tetris the rocks falling), so I assume my code is doing the right thing. But even visually inspecting the board, I can't find a pattern.
Am I missing something obvious or have I somehow cheesed Part 1 :/
past karma
Not sure if I'm being incredibly dumb, but I cannot find the cycle in the test i...
there is a pattern, but it doesn't start right at the bottom
there is an "offset", if you like, before the pattern starts
So, for the test input, the cycle only begins after you get to 25 height. And then the cycle is a repeating group of 53 height.
that corresponds to 15 rocks for the offset, and 35 rocks per repeating group.
hopefully that helps :)
past karma
Finally got it. I feel kind of bad tho. I searched for the repeated blocks manua...
try to see if you can do a general solution :) IMO for me the minimum bar is a solution that can work for both test and real input.
silver light
I will eventually do it, but I'm a little burnt out on that one. It took me hours to figure it out.
pine kelp
there is a pattern, but it doesn't start right at the bottom
Thanks for the tip. Stupidly I had changed how many rocks the simulation would run and didn't notice. So I never had quite enough of the chamber to find the pattern.
Annoyingly the maths I used to get the height for part 1 test using the repeating cycle doesn't work for part 2 test or actual input. As above, I've been at this on and off for days now. Gonna park it as I'm getting sick of rocks ๐
warm zealot
Thanks for the tip. Stupidly I had changed how many rocks the simulation would r...
A trick I used was not to store states at all, I just stored the heights so I didn't have to change my part1 solution much. Using the heights I calculated the deltas each time, and used a sliding window. So I iterated thru the windows (of arbitrary size like 150) until I found one that occurred later on. If a series of 150 deltas gets repeated you can be pretty sure it's a cycle
Idk if this helps 
blazing mortar
A trick I used was not to store states at all, I just stored the heights so I di...
oooooh, rocks cycle 5 and the jets cycle N, that would be slick
10,091 :/ hrmmm
<3% done on brute. force
blazing mortar
so it took ~10 minutes to do ~3% by brute force, and an hour to program something to skip to the end, if my computer didn't hang on the brute force one it would have taken ~5 hours
time saved!
spare raptor
Does anyone have the final state of the test input for part 1? I'm getting the wrong answer, but I can't find a flaw in my simulation logic either :/
past karma
Does anyone have the final state of the test input for part 1? I'm getting the w...
how exactly do you want the final state? I can provide an ascii printout if you desire?
spare raptor
how exactly do you want the final state? I can provide an ascii printout if you ...
Yeah, like what's in the problem but for the whole stack. I want to try to find where it's going wrong
past karma
ok. I'll try a printout for you
spare raptor
perfect, thanks!
past karma
!paste
wild rainBOT
spare raptor
Perfect! thank you!
spare raptor
And that's what I needed to find the bug! thanks, onto part 2 ๐
past karma
And that's what I needed to find the bug! thanks, onto part 2 ๐
what was the bug btw?
spare raptor
what was the bug btw?
The bit of code i has that checks for collisions between rocks was only looking at the bottom line of the falling rock for collisions, but for all lines of the rock. So the + ones would kind of get stuck on their corner. so instead of getting:
| + |
| +++|
| ####+ |
i'd get
| + |
| +++|
| + |
| #### |
does that make sense?
past karma
ah interesting
spare raptor
classic 'forgot to increase an iterator' bug :p But it didn't actually crop up until the 22nd rock, so it still seemed to be working for the test input and the simulation shown in the problem
past karma
yeah I just looked at the description example now - it's in none of the simulations
spare raptor
Hmm so now I'm thinking about part 2, seeing all the talk here about finding a repeating cycle. That seems pretty tough, but I have an idea to make it easier: instead of saving the lines as strings with one character per slot, i'll save them as integers, encoded as a bit mask (so the first line of the example would be 0011110. Then I could generate integer representations of a few hundred lines, and then just find the repeating cycle in that sequence of integers.
past karma
FYI I'm not saving the lines as string - internally they are saved in a structure similar to that (I use idx values from 0 - 6 but the concept is the same). I just print out when I need to (e.g. for easy unit testing)
spare raptor
Oh yeah I was only saving them as a string for easy printing out. But bit masks were the key for my day 16 solution so im looking for more ways to use that :p
Hmm i'm realizing that really only helps me to find the cycle in the output, but that's not enough to actually answer. I need to find the cycles in where the rocks end up
past karma
Hmm i'm realizing that really only helps me to find the cycle in the output, but...
not sure I follow the difference between the two there
spare raptor
well if I only look at the individual lines of the output, I can't easily distinguish between the individual rocks.
I also am assuming that the final cycle will be truncated, and I'll have to reconstruct that one to get the true final height
past karma
I also am assuming that the final cycle will be truncated, and I'll have to reco...
My final cycle was truncated yes. Of course the example input was not though ๐
spare raptor
naturally
past karma
well if I only look at the individual lines of the output, I can't easily distin...
Ah yes I gotcha. This is reliving PTSD for me ๐
spare raptor
Well, thank you for sharing your experience! saving me some stress for sure ๐
spare raptor
@past karma I finally got around to implementing a solution based on the cycles stuff, and it worked on the first try ๐
past karma
<@388062601281798144> I finally got around to implementing a solution based on t...
Ah awesome! Congrats :)