Most recent activity
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FWIW, I had no problems fitting new 25c and 25c gp5000s to the R (18mm internal) hooked rims.
My approach is not to try and use my thumbs to push the tyre bead over the rim but to sort of roll it on. Also after reading these horror stories tried getting them off and after releasing the bead from the rim on both sides (this was the hardest part) only needed one lever
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a meme
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been on the whole time. check if your stream is set to "ambient sound"
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And that's a wrap (modulo finishing some of the days in rust):
$ python main.py Day Part 1 Part 2 Time [ms] ---------------------------------------------------- Day 01 1374 1418 2.01 Day 02 1660158 1604592846 1.52 Day 03 3895776 7928162 4.52 Day 04 14093 17388 3.80 Day 05 6397 22335 177.87 Day 06 393019 1757714216975 0.79 Day 07 355989 102245489 1.30 Day 08 473 1097568 5.64 Day 09 600 987840 55.01 Day 10 321237 2360030859 2.38 Day 11 1637 242 28.11 Day 12 5252 147784 2.81 Day 13 724 CPJBERUL 2.63 Day 14 2233 2884513602164 3.94 Day 15 523 2876 772.99 Day 16 852 19348959966392 5.14 Day 17 14535 2270 8.71 Day 18 4435 4802 599.87 Day 19 396 11828 547.72 Day 20 5475 17548 65.86 Day 21 503478 716241959649754 50.66 Day 22 603661 1237264238382479 82.99 Day 23 19167 47665 153.68 Day 24 29599469991739 17153114691118 0.24 Day 25 516 319.61 ---------------------------------------------------- 2926.96Just snuck in under 3 seconds total solve time.
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Yep. Hopefully binary search or something is all you need but I'm skeptical.
See if you can spot a pattern in the calculations done on the input values. I have something which I suspect is right, but failed to collect the right information in my recursive call.
D23 done, A* too slow so I went with recursion and memoising which was shockingly fast by comparison
My basic approach using Dijkstra was ~4s, I did some mild perusal of other people's solutions and with an appropriate cost-function for A* got down to ~150ms.
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Today looks annoying as well
There must be a trick, because just naively trying the 9^14 combinations isn’t very feasible.
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Nearly caught up (still day 19 to do), but here was my part2 for today (~55ms)
from functools import cache def part2(inp): p1, p2 = inp p1 -= 1 p2 -= 1 moves = list(Counter(map(sum, product(*repeat(range(1, 4), 3)))).items()) @cache def winners(A, B, s1, s2): if s1 >= 21: return (1, 0) if s2 >= 21: return (0, 1) winA = 0 winB = 0 for move, freq in moves: newA = (A + move) % 10 news1 = s1 + newA + 1 x, y = winners(B, newA, s2, news1) winA += y * freq winB += x * freq return (winA, winB) return max(winners(p1, p2, 0, 0)) -
It definitely took me longer to parse the exercise than write the code...
https://gist.github.com/wence-/500e1bc4633192ac6fee67433aedd974#file-day16-py
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That's the naive implementation, if you use a heap, then insertion is O(log n)
Person, occasional wit.