Day 13 Part 2

Author: monpie3

Day_13/README.md

@@ -54,3 +54,34 @@ Your puzzle answer was `[REDACTED]`.
 
 The first half of this puzzle is complete! It provides one gold star: ⭐
 
+## \--- Part Two ---
+
+As you go to win the first prize, you discover that the claw is nowhere near where you expected it would be. Due to a unit conversion error in your measurements, the position of every prize is actually `10000000000000` higher on both the `X` and `Y` axis!
+
+Add `10000000000000` to the `X` and `Y` position of every prize. After making this change, the example above would now look like this:
+
+```
+Button A: X+94, Y+34
+Button B: X+22, Y+67
+Prize: X=10000000008400, Y=10000000005400
+
+Button A: X+26, Y+66
+Button B: X+67, Y+21
+Prize: X=10000000012748, Y=10000000012176
+
+Button A: X+17, Y+86
+Button B: X+84, Y+37
+Prize: X=10000000007870, Y=10000000006450
+
+Button A: X+69, Y+23
+Button B: X+27, Y+71
+Prize: X=10000000018641, Y=10000000010279
+```
+
+Now, it is only possible to win a prize on the second and fourth claw machines. Unfortunately, it will take _many more than `100` presses_ to do so.
+
+Using the corrected prize coordinates, figure out how to win as many prizes as possible. _What is the fewest tokens you would have to spend to win all possible prizes?_
+
+Your puzzle answer was `[REDACTED]`.
+
+Both parts of this puzzle are complete! They provide two gold stars: ⭐⭐

Day_13/task_13b.py

@@ -0,0 +1,50 @@
+import re
+from sympy import symbols, Eq, solve
+
+A_PRICE = 3
+B_PRICE = 1
+
+
+def load_data(filename):
+    with open(filename, "r") as f:
+        data = f.read().split("\n\n")
+
+    machines = []
+    for machine in data:
+        machine = re.findall(r"\d+", machine)
+        machine = list(map(int, machine))
+        machines.append(machine)
+
+    return machines
+
+
+def find_way_to_win(machine):
+    a1, a2, b1, b2, r1, r2 = machine
+    r1 = r1 + 10000000000000  # fix for unit conversion
+    r2 = r2 + 10000000000000  # fix for unit conversion
+
+    x, y = symbols("x,y")
+    eq1 = Eq((a1 * x + b1 * y), r1)
+    eq2 = Eq((a2 * x + b2 * y), r2)
+
+    sol_dict = solve((eq1, eq2), (x, y))
+    x = sol_dict[x]
+    y = sol_dict[y]
+
+    if x == int(x) and y == int(y):
+        return x, y
+    else:
+        return 0, 0
+
+
+def calculate_tokens(machines):
+    token = 0
+    for machine in machines:
+        a, b = find_way_to_win(machine)
+        token += a * A_PRICE + b * B_PRICE
+    return token
+
+
+if "__main__" == __name__:
+    machines = load_data("Day_13/puzzle_input.txt")
+    print(calculate_tokens(machines))

Day_13/test_task_13b.py

@@ -0,0 +1,13 @@
+from task_13b import load_data, calculate_tokens
+import pytest
+
+
+@pytest.mark.parametrize(
+    "file_path, expected",
+    [
+        ("Day_13/example_13a.txt", 875318608908),
+    ],
+)
+def test_calculate_tokens(file_path, expected):
+    machines = load_data(file_path)
+    assert calculate_tokens(machines) == expected