Rafbill_pc

98 posts

Rafbill_pc

Rafbill_pc

@Rafbill_pc

Katılım Aralık 2023
69 Takip Edilen294 Takipçiler
Rafbill_pc
Rafbill_pc@Rafbill_pc·
#AHC059, seed=0, moves=988 (Initialize with chokudai's solution, postprocess with my solution (which is similar to @wata_orz 's approach)).
GIF
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
I see that @terry_u16 's solution used randomized dfs, maybe this state representation is a way to turn randomized dfs into something suitable for simulated annealing?
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
The state representation is quite weird: the tree is determined by dfs with a priority stack (when popping an element, choose the one with the highest priority. In case of a tie, choose the one that was added last). The SA transition is to change the priorities of a few cells.
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
#AHC054 3rd place on the provisional leaderboard. If you do simulated annealing using simulations to estimate the expected score and let it run for several minutes, you can get a good score. But simulating is very slow (O(N^4) and you need multiple simulations per SA step).
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
The range ((L-2e12)/8, (U+2e12)/8) is used instead of (L/8,U/8), because otherwise elements close to L or U become too hard to create. The elements close to L or U remain harder to create, so it is better to consider them first (when we have 500 values available in A).
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
We can estimate how close we can get by looking at: log_2(Choose(500,8)) ~= 56 log_2(Choose(108,8)) ~= 38 log_2(4e12) ~= 42
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
I'm typing this on my phone while on a train, apologies if this is hard to understand.
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
A better strategy is to consider the whole list of vertices, view it as a list of 2d points (pairs of probabilities), and optimize the area of the region bounded by these points.
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Rafbill_pc
Rafbill_pc@Rafbill_pc·
#AHC051 Here is a short description of my approach, I'll write a more detailed description soon.
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