this post was submitted on 12 Dec 2023
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Advent Of Code
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AoC 2023
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T counts the number of ways to place the blocks with lengths specified in b in the remaining a.size - ai slots. If there are no more slots left, there are two cases: Either there are also no more blocks left, then everything is fine, and the current situation is 1 way to place the blocks in the slots. Otherwise, there are still blocks left, and no more space to place them in. This means the current sitution is incorrect, so we contribute 0 ways to place the blocks. This is what the
if bi >= b.size then 1L else 0L{.scala} does.The start at
size + 1is necessary, as we need to compute every table entry before it may get looked up. When placing the last block, we may check the entry(ai + b(bi) + 1, bi + 1), whereai + b(bi)may already equala.size(in the case where the block ends exactly at the end ofa). The+ 1in the entry is necessary, as we need to skip a slot after every block: If we looked at(ai + b(bi), bi + 1), we could start ata.size, but then, for e.g.b = [2, 3], we would consider...#####.a valid placement.Let me know if there are still things unclear :)
Thanks for the detailed explanation. It helped a lot, especially what the
tblactually holds.I've read your code again and I get how it works, but it still feels kinda strange that we are considering values outside of range of
aandb, and that we are marking them as correct. Like in first row of the example???.### 1,1,3, there is no spring at8and no group at3but we are marking(8,3)and(7,3)as correct. In my mind, first position that should be marked as correct is4,2, because that's where group of 3 can fit.If you make the recurrent case a little more complicated, you can sidestep the weird base cases, but I like reducing the endpoints down to things like this that are easily implementable, even if they sound a little weird at first.
You are probably right. Just my rumblings. Thanks for the help.