thanks
I've probably learned that term at some point, so thanks for naming it. That made me realise my algorithm was too thicc and could just be greedy.
swlabr
joined 1 year ago
22
uh
pretty straightforward. At least it's not a grid!
21!
Finally managed to beat this one into submission.
P1
I created this disgusting mess of a recursive search that happened to work. This problem was really hard to think about due to the levels of indirection. It was also hard because of a bug I introduced into my code that would have been easy to debug with more print statements, but hubris.
P2
Recursive solution from P1 was too slow, once I was at 7 robots it was taking minutes to run the code. It didn't take long to realise that you don't really care about where the robots other than the keypad robot and the one controlling the keypad robot are since the boundary of each state needs all the previous robots to be on the A button. So with memoisation, you can calculate all the shortest paths for a given robot to each of the directional inputs in constant time, so O(kn) all up where n is the number of robots (25) and k is the complexity of searching for a path over 5 or 11 nodes.
What helped was looking at the penultimate robot's button choices when moving the keypad robot. After the first one or two levels, the transitions settle into the table in the appendix. I will not explain the code.
appendix
(P(0, 1), P(0, 1)): [],
(P(0, 1), P(0, 2)): [btn.r],
(P(0, 1), P(1, 0)): [btn.d, btn.l],
(P(0, 1), P(1, 1)): [btn.d],
(P(0, 1), P(1, 2)): [btn.d, btn.r],
(P(0, 2), P(0, 1)): [btn.l],
(P(0, 2), P(0, 2)): [],
(P(0, 2), P(1, 0)): [btn.d, btn.l, btn.l],
(P(0, 2), P(1, 1)): [btn.l, btn.d],
(P(0, 2), P(1, 2)): [btn.d],
(P(1, 0), P(0, 1)): [btn.r, btn.u],
(P(1, 0), P(0, 2)): [btn.r, btn.r, btn.u],
(P(1, 0), P(1, 0)): [],
(P(1, 0), P(1, 1)): [btn.r],
(P(1, 0), P(1, 2)): [btn.r, btn.r],
(P(1, 1), P(0, 1)): [btn.u],
(P(1, 1), P(0, 2)): [btn.u, btn.r],
(P(1, 1), P(1, 0)): [btn.l],
(P(1, 1), P(1, 1)): [],
(P(1, 1), P(1, 2)): [btn.r],
(P(1, 2), P(0, 1)): [btn.l, btn.u],
(P(1, 2), P(0, 2)): [btn.u],
(P(1, 2), P(1, 0)): [btn.l, btn.l],
(P(1, 2), P(1, 1)): [btn.l],
(P(1, 2), P(1, 2)): [],
1.1 trillion? Good luck, chuck.
21 (wip)
2 meme 2 memeious
Understandable, have a nice day
ok disco
It took me too long to read the prompt and see that without the shortcuts, it's a single path. I wasted too much time on search algorithms.
P1: Here's what I did: Walk the path. Every time you hit a new grid, check if the two shortcuts you can take will save you 100 ps.
To calculate the shortcut saving:
If you index every grid position on the main path from 0, then it takes X ps to reach position X, The time it takes to travel from start to X, then a shortcut to Y, then from Y to the end, is X + 1 + (main path length - Y). The time saved is then just Y - X - 1, modulo maybe like 5 fence post errors.
P2. The prompt wasn't really clear about whether or not cheating meant you can only travel through one set of walls before your cheat ends, or if it meant you could just move through walls for 20ps to wherever you could reach. Turns out, it's the latter.
The above formula is then a special case of Y - X - manhattan distance(X, Y).
20: currently a WIP but:
meme
Wait, so it’s all grids?
🧑🏿🚀🔫🧑🏿🚀
Day 19! (the cuervo gold....)
disc and code
Ok so my path to this answer was circuitous and I now hate myself a little.
P1: Ok, a repeated dfs on suffixes. that shouldn't be too hard. (it was not hard)
P2: Ok, a repeated dfs is a little too slow for me, I wonder how I can speed it up?
forgets about memoisation, a thing that you can do to speed this sort of thing up
I guess the problem is I'm doing an O(mn) match (where m is the number of towels, n is the max towel length) when I can do O(n). I'll build a prefix tree!
one prefix tree later
Ok that still seems to be quite slow. What am I doing wrong?
remembers that memoisation exists
Oh I just need to memoise my dp from part 1. Oops.
Anyway posting the code because I shrunk it down to like two semicolons worth of lines.
(
List<String> input = getLines();
Set<String> ts = Set.from(input.first.split(', '));
Map<String, int> dp = {};
int dpm(String s) => dp.putIfAbsent(
s,
() => s.isNotEmpty
? ts
.where((t) => t.matchAsPrefix(s) != null)
.map((t) => dpm(s.substring(t.length)))
.fold(0, (a, b) => a + b)
: 1);
void d19(bool sub) {
print(input
.skip(2)
.map((s) => dpm(s))
.map((i) => sub
? i
: i > 0
? 1
: 0)
.reduce((a, b) => a + b));
}
gilding the lily a bit but
yes
What is this, day 16?
Abstracted abstract:
Frontier models are increasingly trained and deployed as autonomous agents, which significantly increases their potential for risks. One particular safety concern is that AI agents might covertly pursue misaligned goals, hiding their true capabilities and objectives – also known as scheming. We study whether models have the capability to scheme in pursuit of a goal that we provide in-context and instruct the model to strongly follow. We evaluate frontier models on a suite of six agentic evaluations where models are instructed to pursue goals and are placed in environments that incentivize scheming.
I saw this posted here a moment ago and reported it*, and it looks to have been purged. I am reposting it to allow us to sneer at it.
*
43
Playing [Texas’s] energy market has become more profitable than mining bitcoin
(www.economist.com)
view more: next ›
I did part 2 manually! I will not bother writing a code solution unless I feel like it.
well well well
AoC, so you thought you could dredge up my trauma as an EE grad by making me debug a full-adder logic circuit? How dare you. You succeeded.