Solution to problem 3 in Python

src/Year_2016/P3.py

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+# --- Day 3: Squares With Three Sides ---
+
+# Now that you can think clearly, you move deeper into the labyrinth of
+# hallways and office furniture that makes up this part of Easter Bunny HQ.
+# This must be a graphic design department; the walls are covered in
+# specifications for triangles.
+
+# Or are they?
+
+# The design document gives the side lengths of each triangle it describes,
+# but... 5 10 25? Some of these aren't triangles. You can't help but mark the
+# impossible ones.
+
+# In a valid triangle, the sum of any two sides must be larger than the
+# remaining side. For example, the "triangle" given above is impossible,
+# because 5 + 10 is not larger than 25.
+
+# In your puzzle input, how many of the listed triangles are possible?
+
+from copy import copy
+from itertools import combinations, zip_longest
+from typing import Iterable, Iterator, Tuple
+
+with open("files/P3.txt") as f:
+    triangles = [line for line in f.read().strip().split("\n")]
+
+
+def part_1() -> None:
+    possible_triangles = 0
+    for triangle in triangles:
+        valid_triangles = 0
+        all_sides = triangle.split()
+        # make pairs of all combinations
+        two_sides = list(combinations(all_sides, 2))
+        for pair in two_sides:
+            # we need to remove from the list due to possible duplicates
+            all_sides_ = copy(all_sides)
+            all_sides_.remove(pair[0])
+            all_sides_.remove(pair[1])
+            missing = int(all_sides_[0])
+            sum_two_sides = sum(int(side) for side in pair)
+            if sum_two_sides > missing:
+                valid_triangles += 1
+        if valid_triangles == 3:
+            possible_triangles += 1
+
+    print(f"There are {possible_triangles} possible triangles")
+
+
+# --- Part Two ---
+
+# Now that you've helpfully marked up their design documents, it occurs to you
+# that triangles are specified in groups of three vertically. Each set of three
+# numbers in a column specifies a triangle. Rows are unrelated.
+
+# For example, given the following specification, numbers with the same
+# hundreds digit would be part of the same triangle:
+
+# 101 301 501
+# 102 302 502
+# 103 303 503
+# 201 401 601
+# 202 402 602
+# 203 403 603
+
+# In your puzzle input, and instead reading by columns, how many of the listed
+# triangles are possible?
+
+# split by spaces, instead of newlines
+with open("files/P3.txt") as f:
+    triangles_ = [line for line in f.read().strip().split()]
+
+
+def grouper(iterable: Iterable[str], n: int) -> Iterator[Tuple[int, int, int]]:
+    "Collect data into fixed-length chunks or blocks"
+    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
+    args = [iter(iterable)] * n
+    return zip_longest(*args, fillvalue=None)
+
+
+first_col = triangles_[::3]
+second_col = triangles_[1::3]
+third_col = triangles_[2::3]
+triangles_vert = list(grouper(first_col + second_col + third_col, 3))
+
+
+# almost identical to part 1
+def part_2() -> None:
+    possible_triangles = 0
+    for triangle in triangles_vert:
+        valid_triangles = 0
+        all_sides = list(triangle)
+        # make pairs of all combinations
+        two_sides = list(combinations(all_sides, 2))
+        for pair in two_sides:
+            # we need to remove from the list due to possible duplicates
+            all_sides_ = copy(all_sides)
+            all_sides_.remove(pair[0])
+            all_sides_.remove(pair[1])
+            missing = int(all_sides_[0])
+            sum_two_sides = sum(int(side) for side in pair)
+            if sum_two_sides > missing:
+                valid_triangles += 1
+        if valid_triangles == 3:
+            possible_triangles += 1
+
+    print(f"There are {possible_triangles} possible triangles")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()