add day 13 part 1

day13/code.py

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+import sys
+sys.path.append('../aoclib')
+
+from aoclib import Input
+import re
+
+class Button:
+    def __init__(self, cost, x, y):
+        self.cost = cost
+        self.x = x
+        self.y = y
+
+    def __str__(self):
+        return f'x: {self.x}, y: {self.y}, cost: {self.cost}'
+
+
+class Prize:
+    def __init__(self, x, y):
+        self.x = x
+        self.y = y
+
+    def __str__(self):
+        return f'x: {self.x}, y: {self.y}'
+        
+
+class Machine:
+    def __init__(self, a, b: Button, p: Prize):
+        self.a = a
+        self.b = b
+        self.prize = p
+
+    def __str__(self):
+        s = ''
+        s += 'B_A: ' + self.a.__str__() + '\n'
+        s += 'B_B: ' + self.b.__str__() + '\n'
+        s += 'Prize: ' + self.prize.__str__() + '\n' 
+        return s
+        
+
+def parse_input(lines: list[str]) -> list[Machine]:
+    lst = list()
+    brx = re.compile(r'X\+(\d+), Y\+(\d+)')
+    prx = re.compile(r'X=(\d+), Y=(\d+)')
+
+    b_a = None
+    b_b = None
+    m = None
+    for l in lines:
+        if l.startswith('Button A'):
+            m = brx.search(l)
+            b_a = Button(3, int(m.group(1)), int(m.group(2)))
+        if l.startswith('Button B'):
+            m = brx.search(l)
+            b_b = Button(1, int(m.group(1)), int(m.group(2)))
+        if l.startswith('Prize'):
+            m = prx.search(l)
+            p = Prize(int(m.group(1)), int(m.group(2)))
+            machine = Machine(b_a, b_b, p)
+            lst.append(machine)
+
+    return lst
+
+def _solve1(m: Machine) -> int:
+    max_a = min(m.prize.x // m.a.x, m.prize.y // m.a.y)+1
+    max_b = min(m.prize.x // m.b.x, m.prize.y // m.b.y)+1
+
+    costs = []
+    for _a in range(max_a):
+        for _b in range(max_b):
+            x, y, cost = m.a.x*_a + m.b.x*_b, m.a.y*_a + m.b.y*_b, m.a.cost*_a + m.b.cost*_b 
+            if x == m.prize.x and y == m.prize.y:
+                costs.append(cost)
+
+    if not costs:
+        return None
+
+    return min(costs)
+    
+
+def solve1(lines: list[str]):
+    machines = parse_input(lines)
+
+    cost = 0
+    for m in machines:
+        c = _solve1(m)
+        if not c:
+            continue
+        cost += c
+        
+    return cost
+
+
+def solve2(lines: list[str]):
+    machines = parse_input(lines)
+
+    cost = 0
+    for m in machines[:1]:
+        continue
+        m.prize.x += 10000000000000
+        m.prize.y += 10000000000000
+        c = _solve1(m)
+        if not c:
+            continue
+        cost += c
+        
+    return cost
+
+
+if __name__ == '__main__':
+    #lines = Input('input_test.txt').lines()
+    lines = Input('input.txt').lines()
+
+    #part 1
+    print('part 1:', solve1(lines)) # 27157
+
+    #part 2
+    print('part 2:', solve2(lines)) # 862486

day13/input_test.txt

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+Button A: X+94, Y+34
+Button B: X+22, Y+67
+Prize: X=8400, Y=5400
+
+Button A: X+26, Y+66
+Button B: X+67, Y+21
+Prize: X=12748, Y=12176
+
+Button A: X+17, Y+86
+Button B: X+84, Y+37
+Prize: X=7870, Y=6450
+
+Button A: X+69, Y+23
+Button B: X+27, Y+71
+Prize: X=18641, Y=10279

day13/task.txt

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+--- Day 13: Claw Contraption ---
+Next up: the lobby of a resort on a tropical island. The Historians take a moment to admire the hexagonal floor tiles before spreading out.
+
+Fortunately, it looks like the resort has a new arcade! Maybe you can win some prizes from the claw machines?
+
+The claw machines here are a little unusual. Instead of a joystick or directional buttons to control the claw, these machines have two buttons labeled A and B. Worse, you can't just put in a token and play; it costs 3 tokens to push the A button and 1 token to push the B button.
+
+With a little experimentation, you figure out that each machine's buttons are configured to move the claw a specific amount to the right (along the X axis) and a specific amount forward (along the Y axis) each time that button is pressed.
+
+Each machine contains one prize; to win the prize, the claw must be positioned exactly above the prize on both the X and Y axes.
+
+You wonder: what is the smallest number of tokens you would have to spend to win as many prizes as possible? You assemble a list of every machine's button behavior and prize location (your puzzle input). For example:
+
+Button A: X+94, Y+34
+Button B: X+22, Y+67
+Prize: X=8400, Y=5400
+
+Button A: X+26, Y+66
+Button B: X+67, Y+21
+Prize: X=12748, Y=12176
+
+Button A: X+17, Y+86
+Button B: X+84, Y+37
+Prize: X=7870, Y=6450
+
+Button A: X+69, Y+23
+Button B: X+27, Y+71
+Prize: X=18641, Y=10279
+This list describes the button configuration and prize location of four different claw machines.
+
+For now, consider just the first claw machine in the list:
+
+Pushing the machine's A button would move the claw 94 units along the X axis and 34 units along the Y axis.
+Pushing the B button would move the claw 22 units along the X axis and 67 units along the Y axis.
+The prize is located at X=8400, Y=5400; this means that from the claw's initial position, it would need to move exactly 8400 units along the X axis and exactly 5400 units along the Y axis to be perfectly aligned with the prize in this machine.
+The cheapest way to win the prize is by pushing the A button 80 times and the B button 40 times. This would line up the claw along the X axis (because 80*94 + 40*22 = 8400) and along the Y axis (because 80*34 + 40*67 = 5400). Doing this would cost 80*3 tokens for the A presses and 40*1 for the B presses, a total of 280 tokens.
+
+For the second and fourth claw machines, there is no combination of A and B presses that will ever win a prize.
+
+For the third claw machine, the cheapest way to win the prize is by pushing the A button 38 times and the B button 86 times. Doing this would cost a total of 200 tokens.
+
+So, the most prizes you could possibly win is two; the minimum tokens you would have to spend to win all (two) prizes is 480.
+
+You estimate that each button would need to be pressed no more than 100 times to win a prize. How else would someone be expected to play?
+
+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 27157.
+
+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?
+
+