new post: day 4

Author: thiagowfx

Diff for: content/posts/2024-12-06-advent-of-code-day-4.md

@@ -0,0 +1,208 @@
+---
+title: "Advent of Code: Day 4"
+date: 2024-12-06T11:44:39+01:00
+tags:
+  - dev
+  - devops
+---
+
+Link to [Day #4](https://adventofcode.com/2024/day/4) puzzle.
+
+<!--more-->
+
+It's a pretty typical 2D matrix search problem, or a graph search problem, if
+you will.
+
+The problem is naturally unraveled into the following searches:
+
+- horizontally
+- horizontally, reversed
+- vertically
+- vertically, reversed
+- diagonally, all 4 directions (NW, NE, SW, SE)
+
+It's possible to write a single pair of for loops that addresses the general
+case. The (classic) idea is to think of all 8 compass directions to move along
+the matrix:
+
+- (1, 0)
+- (-1, 0)
+- (0, 1)
+- (0, -1)
+- (1, 1)
+- (-1, -1)
+- (-1, 1)
+- (1, -1)
+
+Within the inner iteration, change `x += dx` and `y += dy` (or `i += di`, `j +=
+dj`, naming is hard). I did this many times in C++ though, and I want to write
+elegant Python code.
+
+Therefore I came up with the following solution instead, with nested list
+comprehensions:
+
+```python
+def search_horizontal(matrix, keyword):
+    return sum((True for row in matrix for i in range(len(row) - len(keyword) + 1) if "".join(row[i:i + len(keyword)]) in [keyword, keyword[::-1]]))
+```
+
+It follows the same principle as the original intent, however it leverages
+list slices so that we can omit the `dx/dy` step.
+
+The vertical search is pretty straightforward: it is just a matter of running
+the horizontal search in the transposed matrix (`zip(*matrix)`).
+
+I must confess that using `zip` to transpose matrices always felt magical and a
+mere coincidence that it just works™. Ruby has a `.transpose` method, which is
+more readable.
+
+For the diagonal search, I couldn't think of an elegant list comprehension
+manner to address it. Is it even possible to "2D slice" in Python?
+
+After-the-fact I decided to ask ChatGPT, and it is indeed possible, but it
+requires NumPy:
+
+> If a is 2-D, returns the diagonal of a with the given offset, i.e., the
+> collection of elements of the form a[i, i+offset]. If a has more than two
+> dimensions, then the axes specified by axis1 and axis2 are used to determine
+> the 2-D sub-array whose diagonal is returned. The shape of the resulting array
+> can be determined by removing axis1 and axis2 and appending an index to the
+> right equal to the size of the resulting diagonals.
+
+The method call resembles `numpy.array([[1, 2], [3, 4]]).diagonal(offset=1)`,
+perhaps with the aid of `.flip()` to account for the other direction.
+
+Anyway, my plain diagonal search is:
+
+```python
+def search_diagonal(matrix, keyword):
+    rows = len(matrix)
+    cols = len(matrix[0])
+
+    count = 0
+
+    for i in range(rows):
+        for j in range(cols):
+            if i + len(keyword) <= rows and j + len(keyword) <= cols:
+                if "".join(matrix[i + k][j + k] for k in range(len(keyword))) in [keyword, keyword[::-1]]:
+                    count += 1
+            if i + len(keyword) <= rows and j - len(keyword) >= -1:
+                if "".join(matrix[i + k][j - k] for k in range(len(keyword))) in [keyword, keyword[::-1]]:
+                    count += 1
+
+    return count
+```
+
+Part two is fundamentally a different problem.
+
+One way to address it is to search for all `'A'` characters, and then look
+around its "edges" to see if they contain exactly two `'M'` and two `'S'`, and
+that they are properly arranged:
+
+```python
+def search_double_mas(matrix):
+    rows = len(matrix)
+    cols = len(matrix[0])
+
+    count = 0
+
+    for i in range(1, rows - 1):
+        for j in range(1, cols - 1):
+            if matrix[i][j] != 'A':
+                continue
+
+            # look at a QWERTY keyboard to make sense of these variable names
+            q = matrix[i - 1][j - 1]
+            e = matrix[i - 1][j + 1]
+            z = matrix[i + 1][j - 1]
+            c = matrix[i + 1][j + 1]
+            edges = [q, e, z, c]
+
+            if edges.count('M') != 2 or edges.count('S') != 2:
+                continue
+
+            if q == e or q == z:
+                count += 1
+
+    return count
+```
+
+I couldn't find an opportunity for reuse of the solution from part one.
+
+The full solution:
+
+```python
+#!/usr/bin/env python3
+import sys
+
+def search_horizontal(matrix, keyword):
+    return sum((True for row in matrix for i in range(len(row) - len(keyword) + 1) if "".join(row[i:i + len(keyword)]) in [keyword, keyword[::-1]]))
+
+def search_vertical(matrix, keyword):
+    return search_horizontal(zip(*matrix), keyword)
+
+def search_diagonal(matrix, keyword):
+    rows = len(matrix)
+    cols = len(matrix[0])
+
+    count = 0
+
+    for i in range(rows):
+        for j in range(cols):
+            if i + len(keyword) <= rows and j + len(keyword) <= cols:
+                if "".join(matrix[i + k][j + k] for k in range(len(keyword))) in [keyword, keyword[::-1]]:
+                    count += 1
+            if i + len(keyword) <= rows and j - len(keyword) >= -1:
+                if "".join(matrix[i + k][j - k] for k in range(len(keyword))) in [keyword, keyword[::-1]]:
+                    count += 1
+
+    return count
+
+
+def search_double_mas(matrix):
+    rows = len(matrix)
+    cols = len(matrix[0])
+
+    count = 0
+
+    for i in range(1, rows - 1):
+        for j in range(1, cols - 1):
+            if matrix[i][j] != 'A':
+                continue
+
+            # look at a QWERTY keyboard to make sense of these variable names
+            q = matrix[i - 1][j - 1]
+            e = matrix[i - 1][j + 1]
+            z = matrix[i + 1][j - 1]
+            c = matrix[i + 1][j + 1]
+            edges = [q, e, z, c]
+
+            if edges.count('M') != 2 or edges.count('S') != 2:
+                continue
+
+            if q == e or q == z:
+                count += 1
+
+    return count
+
+
+
+def main():
+    with open(sys.argv[1]) as input:
+        lines = input.read().splitlines()
+
+    keyword = "XMAS"
+
+    # ['abcd', 'efgh', 'ijkl'] -> [['a', 'b', 'c', 'd'], ['e', 'f', 'g', 'h'], ['i', 'j', 'k', 'l']]
+    matrix = [list(line) for line in lines]
+
+    # part one
+    print(search_horizontal(matrix, keyword) + search_vertical(matrix, keyword) + search_diagonal(matrix, keyword))
+
+    # part two
+    print(search_double_mas(matrix))
+
+
+if __name__ == '__main__':
+    main()
+```