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December 2021

10dec2021 · Surveying lava basins with BQN and fixpoints

Yesterday, Advent of Code had an interesting problem: given the heightmap of a lava cave, compute the lowest points and the size of their basins (connected regions).

Let’s do this in BQN again, as this problem teaches some good ways to think in array languages.

First, let’s load the input data into a matrix:

   d ← > '0' -˜ •FLines"day09"

We subtract the ASCII lines from the character 0 to get numerical rows. The merge function (>) then converts this list-of-lists into a 2-dimensional array. For the sample data, we get:

┌─
╵ 2 1 9 9 9 4 3 2 1 0
  3 9 8 7 8 9 4 9 2 1
  9 8 5 6 7 8 9 8 9 2
  8 7 6 7 8 9 6 7 8 9
  9 8 9 9 9 6 5 6 7 8
                      ┘

A low point is a point that is lower than every orthogonally adjacent point. Thanks to array progamming, we can solve this for the whole matrix at once without any loops!

The core idea is to shift the array into each cardinal direction, and then compute the minimum of these arrays. If the original array is smaller then the array of the minimums, it’s a low point.

By default, shifting («, ») in BQN inserts zeroes for numerical arrays. But since we are looking for the minimum, we need to shift in a value that is higher than any. We can simply use .

So, to shift in from the left, we use:

   ∞»˘d
┌─
╵ ∞ 2 1 9 9 9 4 3 2 1
  ∞ 3 9 8 7 8 9 4 9 2
  ∞ 9 8 5 6 7 8 9 8 9
  ∞ 8 7 6 7 8 9 6 7 8
  ∞ 9 8 9 9 9 6 5 6 7
                      ┘

Unfortunately, shifting from the top is not so easy:

   ∞»d
Error: shift: =𝕨 must be =𝕩 or ¯1+=𝕩 (0≡=𝕨, 2≡=𝕩)
at ∞»d

We need would need to make a list of long enough to agree with the array width:

   (∞¨⊏d)»d
┌─
╵ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞
  2 1 9 9 9 4 3 2 1 0
  3 9 8 7 8 9 4 9 2 1
  9 8 5 6 7 8 9 8 9 2
  8 7 6 7 8 9 6 7 8 9
                      ┘

However, since we need to do this on every side, we can also look at the problem differently: we shift in from the left under rotation by 0, 90, 180, 270 degrees.

How do we “rotate” a matrix? We reverse the rows () and then transpose () it.

   ⍉⌽d
┌─
╵ 9 8 9 3 2
  8 7 8 9 1
  9 6 5 8 9
  9 7 6 7 9
  9 8 7 8 9
  6 9 8 9 4
  5 6 9 4 3
  6 7 8 9 2
  7 8 9 2 1
  8 9 2 1 0
            ┘

By using the repeat modifier () we can easily rotate several times.

   (⍉∘⌽⍟(↕4)) d
┌─
· ┌─                      ┌─            ┌─                      ┌─
  ╵ 2 1 9 9 9 4 3 2 1 0   ╵ 9 8 9 3 2   ╵ 8 7 6 5 6 9 9 9 8 9   ╵ 0 1 2 9 8
    3 9 8 7 8 9 4 9 2 1     8 7 8 9 1     9 8 7 6 9 8 7 6 7 8     1 2 9 8 7
    9 8 5 6 7 8 9 8 9 2     9 6 5 8 9     2 9 8 9 8 7 6 5 8 9     2 9 8 7 6
    8 7 6 7 8 9 6 7 8 9     9 7 6 7 9     1 2 9 4 9 8 7 8 9 3     3 4 9 6 5
    9 8 9 9 9 6 5 6 7 8     9 8 7 8 9     0 1 2 3 4 9 9 9 1 2     4 9 8 9 6
                        ┘   6 9 8 9 4                         ┘   9 8 7 8 9
                            5 6 9 4 3                             9 7 6 7 9
                            6 7 8 9 2                             9 8 5 6 9
                            7 8 9 2 1                             1 9 8 7 8
                            8 9 2 1 0                             2 3 9 8 9
                                      ┘                                     ┘
                                                                              ┘

Finally, we perform the shift operation under () the rotation, that is, BQN rotates the array, does the shift, and knows how to undo the rotation!

┌─
· ┌─                      ┌─                      ┌─                      ┌─
  ╵ ∞ 2 1 9 9 9 4 3 2 1   ╵ 3 9 8 7 8 9 4 9 2 1   ╵ 1 9 9 9 4 3 2 1 0 ∞   ╵ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞
    ∞ 3 9 8 7 8 9 4 9 2     9 8 5 6 7 8 9 8 9 2     9 8 7 8 9 4 9 2 1 ∞     2 1 9 9 9 4 3 2 1 0
    ∞ 9 8 5 6 7 8 9 8 9     8 7 6 7 8 9 6 7 8 9     8 5 6 7 8 9 8 9 2 ∞     3 9 8 7 8 9 4 9 2 1
    ∞ 8 7 6 7 8 9 6 7 8     9 8 9 9 9 6 5 6 7 8     7 6 7 8 9 6 7 8 9 ∞     9 8 5 6 7 8 9 8 9 2
    ∞ 9 8 9 9 9 6 5 6 7     ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞     8 9 9 9 6 5 6 7 8 ∞     8 7 6 7 8 9 6 7 8 9
                        ┘                       ┘                       ┘                       ┘
                                                                                                  ┘

Now we insert (´) the minimum function () between these arrays and compute the minimum at each position:

   ⌊´{∞⊸»˘⌾(⍉∘⌽⍟𝕩)d}¨↕4
┌─
╵ 1 2 1 7 4 3 2 1 0 1
  2 1 5 6 7 4 3 2 1 0
  3 5 6 5 6 7 4 7 2 1
  7 6 5 6 7 6 5 6 7 2
  8 7 6 7 6 5 6 5 6 7
                      ┘

The positions where the original array d is still smaller are the low points, and we store them for part 2.

   l ← d < ⌊´{∞⊸»˘⌾(⍉∘⌽⍟𝕩)d}¨↕4
┌─
╵ 0 1 0 0 0 0 0 0 0 1
  0 0 0 0 0 0 0 0 0 0
  0 0 1 0 0 0 0 0 0 0
  0 0 0 0 0 0 0 0 0 0
  0 0 0 0 0 0 1 0 0 0
                      ┘

To finish part 1, we need to compute the risk level for each low point, which is 1 plus the height. So compute that:

   (1+d)×l
┌─
╵ 0 2 0 0 0 0 0 0 0 1
  0 0 0 0 0 0 0 0 0 0
  0 0 6 0 0 0 0 0 0 0
  0 0 0 0 0 0 0 0 0 0
  0 0 0 0 0 0 6 0 0 0
                      ┘

Finally, we compute the end result by deshaping () the array into a single long list and summing it up:

   +´⥊(1+d)×l
15

This concludes part 1.

Part 2 is less straight forward. We need to compute the basins around every low point, which is the area limited by the points of height 9.

Since we need to compute the size for each basin in the end, we need to know which point belongs to which basin. To get started, we first give each low point a unique number. One way to do this is to assign an index to each position and just use those that are used in the low point array.

In BQN, the shape () of an array is the list of sizes for each axis:

   ≢d
⟨ 5 10 ⟩

We count up to the product of these values and add one (to avoid numbering a basin with 0, which will be used for the basin limits). Then, we multiply this with the array of low points so all ones in it get turned into a unique basin index. We perform the multiplication under deshape, so we keep the shape of the input data:

   s ← (1+↕×´≢d) ×⌾⥊ l
┌─
╵ 0 2  0 0 0 0  0 0 0 10
  0 0  0 0 0 0  0 0 0  0
  0 0 23 0 0 0  0 0 0  0
  0 0  0 0 0 0  0 0 0  0
  0 0  0 0 0 0 47 0 0  0
                         ┘

Here’s the main idea how to solve part 2: we incrementally grow these areas by adding their neighbors until the whole array is filled. For one step, we do this with a function Rise:

   Rise ← (d≠9)⊸×(»⌈«⌈«˘⌈»˘⌈⊣)

Here, shifting in zeroes is good enough, so we can use the monadic shift functions. The train at the end computes the maximum () of the four shifted versions and the array itself (). We multiply it with a matrix that is 0 where the depth is 9, so the basin limits will be constantly zero.

Let’s run it once and twice to see how it works:

   Rise s
┌─
╵ 2  2  0  0 0  0  0  0 10 10
  0  0 23  0 0  0  0  0  0 10
  0 23 23 23 0  0  0  0  0  0
  0  0 23  0 0  0 47  0  0  0
  0  0  0  0 0 47 47 47  0  0
                              ┘
   Rise⍟2 s
┌─
╵ 2  2  0  0  0  0  0 10 10 10
  2  0 23 23  0  0  0  0 10 10
  0 23 23 23 23  0  0  0  0 10
  0 23 23 23  0  0 47 47  0  0
  0  0  0  0  0 47 47 47 47  0
                               ┘

As you can see, row 1 colum 3 does not get filled by basin #2 since it has height 9.

Now we could just iterate this step a often enough, or actually only until we reach a fixpoint; that is, applying Rise again doesn’t change the value anymore.

A simple way to implement a fixpoint operator is

_fix ← { 𝕩 ≡ 𝔽 𝕩 ? 𝕩 ; 𝕊 𝔽 𝕩 }

Let’s compute the filled map:

   Rise _fix s
┌─
╵  2  2  0  0  0 10 10 10 10 10
   2  0 23 23 23  0 10  0 10 10
   0 23 23 23 23 23  0 47  0 10
  23 23 23 23 23  0 47 47 47  0
   0 23  0  0  0 47 47 47 47 47
                                ┘

Now, we just need to compute the sizes of the basins, which means computing a histogram of the basin numbers. We deshape the array again, and only keep all values bigger than zero:

   m ← ⥊ Rise _fix s
   (m>0)/m
⟨ 2 2 10 10 10 10 10 2 23 23 23 10 10 10 23 23 23 23 23 47 10 23 23 23 23 23 47 47 47 23 47 47 47 47 47 ⟩

With the nice group indices function (), we can count how often each value appears:

   ≠¨⊔(m>0)/m
⟨ 0 0 3 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 ⟩

Finally, let’s compute the three largest values by sorting in descending order () and taking the first three entries ():

   3↑∨≠¨⊔(m>0)/m
⟨ 14 9 9 ⟩

We then multiply this together and get the result for part 2:

   ×´3↑∨≠¨⊔(m>0)/m
1134

NP: The New Basement Tapes—Quick Like a Flash

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