[WIP] Solution to problem 3 in Python

src/Year_2019/P3.py

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+# --- Day 3: Crossed Wires ---
+
+# The gravity assist was successful, and you're well on your way to the Venus
+# refuelling station. During the rush back on Earth, the fuel management system
+# wasn't completely installed, so that's next on the priority list.
+
+# Opening the front panel reveals a jumble of wires. Specifically, two wires
+# are connected to a central port and extend outward on a grid. You trace the
+# path each wire takes as it leaves the central port, one wire per line of text
+# (your puzzle input).
+
+# The wires twist and turn, but the two wires occasionally cross paths. To fix
+# the circuit, you need to find the intersection point closest to the central
+# port. Because the wires are on a grid, use the Manhattan distance for this
+# measurement. While the wires do technically cross right at the central port
+# where they both start, this point does not count, nor does a wire count as
+# crossing with itself.
+
+# For example, if the first wire's path is R8,U5,L5,D3, then starting from the
+# central port (o), it goes right 8, up 5, left 5, and finally down 3:
+
+# ...........
+# ...........
+# ...........
+# ....+----+.
+# ....|....|.
+# ....|....|.
+# ....|....|.
+# .........|.
+# .o-------+.
+# ...........
+
+# Then, if the second wire's path is U7,R6,D4,L4, it goes up 7, right 6, down
+# 4, and left 4:
+
+# ...........
+# .+-----+...
+# .|.....|...
+# .|..+--X-+.
+# .|..|..|.|.
+# .|.-X--+.|.
+# .|..|....|.
+# .|.......|.
+# .o-------+.
+# ...........
+
+# These wires cross at two locations (marked X), but the lower-left one is
+# closer to the central port: its distance is 3 + 3 = 6.
+
+# Here are a few more examples:
+
+#     R75,D30,R83,U83,L12,D49,R71,U7,L72
+#     U62,R66,U55,R34,D71,R55,D58,R83 = distance 159
+#     R98,U47,R26,D63,R33,U87,L62,D20,R33,U53,R51
+#     U98,R91,D20,R16,D67,R40,U7,R15,U6,R7 = distance 135
+
+# What is the Manhattan distance from the central port to the closest
+# intersection?
+
+with open("files/test.txt") as f:
+    wire_a, wire_b = [path.split(",") for path in f.read().strip().split()]
+
+
+def get_final_point(init, direction, length):
+    if direction == "U":
+        init[0] += length
+    elif direction == "D":
+        init[0] -= length
+    elif direction == "R":
+        init[1] += length
+    elif direction == "L":
+        init[1] -= length
+    return init
+
+
+def get_path(wire):
+    init = [0, 0]
+    path = []
+    for step in wire:
+        direction, length = step[0], int(step[1:])
+        final_point = get_final_point(init, direction, length)
+        # make a copy of the final point to avoid reference to same object
+        path.append(final_point.copy())
+        init = final_point
+    # add starting point
+    return [[0, 0]] + path
+
+
+# print(len(get_path(wire_a)))
+# print(len(get_path(wire_b)))
+
+
+def line(p1, p2):
+    A = p1[1] - p2[1]
+    B = p2[0] - p1[0]
+    C = p1[0] * p2[1] - p2[0] * p1[1]
+    return A, B, -C
+
+
+def intersection(L1, L2):
+    D = L1[0] * L2[1] - L1[1] * L2[0]
+    Dx = L1[2] * L2[1] - L1[1] * L2[2]
+    Dy = L1[0] * L2[2] - L1[2] * L2[0]
+    if D != 0:
+        x = Dx // D
+        y = Dy // D
+        return x, y
+    else:
+        return False
+
+
+# L1 = line([10,1003], [476,1003])
+# L2 = line([250, 500], [250, 1200])
+
+# R = intersection(L1, L2)
+# if R:
+#     print(f"Intersection detected at {R}")
+# else:
+#     print("No single intersection point detected")
+
+path_a = get_path(wire_a)
+print(path_a[:5])
+path_b = get_path(wire_b)
+print(path_b[:5])
+
+
+for p1_a, p2_a, p1_b, p2_b in zip(
+    path_a[:-1], path_a[1:], path_b[:-1], path_b[1:]
+):
+    # for idx_b, (p1_b, p2_b) in enumerate(zip(path_b[:-1], path_b[1:])):
+    line_a = line(p2_a, p1_a)
+    print(line_a)
+    line_b = line(p2_b, p1_b)
+    print(line_b)
+    R = intersection(line_a, line_b)
+    if R:
+        print(p1_a, p2_a, p1_b, p2_b)
+        break