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Day 6: Guard Gallivant

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[-] CameronDev@programming.dev 3 points 1 week ago

How long did brute force take? Mine was 9s on an m1 with rust.

[-] Deebster@programming.dev 3 points 1 week ago* (last edited 1 week ago)

My rust code ran in 6s on my phone (Samsung A35 running under Termux). When I'm back at a computer it'd be interesting to compare times properly.

[-] Pyro@programming.dev 2 points 1 week ago* (last edited 1 week ago)

About 15-20 seconds, not too bad.

[-] CameronDev@programming.dev 1 points 1 week ago* (last edited 1 week ago)

I got mine down to 3s, but it wasn't a very smart loop detection. All I did was count steps and stop after 10000. The 9 second run was 100000 steps, which is obviously a bit excessive.

Does save iterating over the list of past visits, so probably a good shortcut.

[-] iAvicenna@lemmy.world 1 points 1 week ago* (last edited 1 week ago)

I did a similar approach (place obstacles on guards path). Takes about ~~80s~~ 10-15s in 11th Gen Intel(R) Core(TM) i7-11800H. Motivated by the code above, I also restricted the search to start right before the obstacle rather than the whole path which took it down from 80s to 10-15s

[-] TunaCowboy@lemmy.world 1 points 1 week ago

Mine was 9s

That's about how long it takes for my python solution to complete.

[-] CameronDev@programming.dev 2 points 1 week ago

How did you detect loops? I just ran for 100000 steps to see if I escaped, got my time down to 3s by doing only 10000 steps.

[-] Leavingoldhabits@lemmy.world 2 points 1 week ago

Not who you asked but: I save coordinates and direction into a vector each time the guard faces a #. Also every time the guard faces a #, I check if the position exists in the vector, if true, it’s an infinite loop. 78ms rust aolution.

[-] CameronDev@programming.dev 1 points 1 week ago

That's probably quite optimal, compared with checking every state in the path, or running off a fixed number of steps

[-] TunaCowboy@lemmy.world 2 points 1 week ago* (last edited 1 week ago)

I added each visited position/direction to a set, and when a 'state' is reached again you have entered a loop:

v = set()
while t[g.r][g.c] != 'X':
    state = (g.r, g.c, g.d)
    if state in v:
        acc += 1
        break
    v.add(state)
    g.move(t)

You can view my full solution here.

this post was submitted on 06 Dec 2024
26 points (96.4% liked)

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