Skip to content

Latest commit

 

History

History
143 lines (114 loc) · 6.33 KB

day13.md

File metadata and controls

143 lines (114 loc) · 6.33 KB
Raw

Day 13

all / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13 / 14 / 15 / 16 / 17 / 18 / 19 / 20 / 21 / 22 / 23 / 24 / 25

Available as an RSS Feed

Prompt / Code / Rendered

Aw man, I feel like I would have leaderboarded today had I not been busy :'( These type of number theory problems are the ones I usually do well on.

Oh well! Silly internet points, right?

For part 1, you just need to minimize a function on each bus ID:

part1 :: Int -> [Int] -> (Int, Int)
part1 t0 xs = minimumBy (comparing snd)
    [ (x, waitTime)
    | x <- xs
    , let waitTime = x - (t0 `mod` x)
    ]

Part 2 is where things get interesting! Let's try to think of things inductively: start with small lists, and see how we would "add one more".

Let's say we had (offset, id) pairs (0,7) and (1,13), like in the example. This means that we want to find times where t `mod` 7 == 0 and (t + 1) `mod` 13 == 0.

We can sort of do a manual search by hand to get 14 as our lowest candidate. But also, note that 14 + (7*13)n for any integer n would preserve the offset property. 14, 14 + 91, 14 + 182, etc. So the family of all "valid" numbers are 14 + (7*13)n.

Next, what if we wanted to find the situation for pairs (0,7), (1,13), and (4,15)? Well, we already know that any solution that includes (0,7) and (1,13) will be of the form 14 + (7*13)n. So now we just need to find the first one of those that also matches (4,15)

-- 'until' repeatedly applies a function until it finds a value that matches a
-- predicate
ghci> until (\t -> (t + 4) `mod` 15 == 0) (+ (7*13)) 14
1106

Ah hah, good ol' 1106. Well, 1106 isn't the only number that works. We can see that 1106 + (7*13*15)n for any integer n would also work, since it preserves that mod property.

And so, we can repeat this process over and over again for each new number we see.

  1. Keep track of the current "lowest match" (14) and the current "search step" (7*13).
  2. When you see a number, search that family until you find a new lowest match that includes the new number.
  3. Use that new number as the next lowest match, and multiply it to get the new search step.
  4. Rinse and repeat.

Overall, this works pretty well as a foldl, where we keep this (lowest match, search step) pair as an accumulator, and update it as we see each new value in our list.

part2 :: [(Int, Int)] -> Int
part2 = fst . foldl' go (0, 1)
  where
    go (!base, !step) (offset, i) = (base', step * i)
      where
        base' = until (\n -> (n + offset) `mod` i == 0)
                      (+ step)
                      base

Back to all reflections for 2020

Day 13 Benchmarks

>> Day 13a
benchmarking...
time                 189.4 ns   (184.7 ns .. 198.3 ns)
                     0.992 R²   (0.985 R² .. 1.000 R²)
mean                 189.8 ns   (186.2 ns .. 199.6 ns)
std dev              19.00 ns   (7.817 ns .. 34.74 ns)
variance introduced by outliers: 90% (severely inflated)

* parsing and formatting times excluded

>> Day 13b
benchmarking...
time                 3.868 μs   (3.865 μs .. 3.872 μs)
                     1.000 R²   (1.000 R² .. 1.000 R²)
mean                 3.868 μs   (3.865 μs .. 3.876 μs)
std dev              14.47 ns   (9.762 ns .. 24.35 ns)

* parsing and formatting times excluded