add day 13 part 2

day13/code.py

@@ -61,6 +61,7 @@ def parse_input(lines: list[str]) -> list[Machine]:
     return lst
 
 def _solve1(m: Machine) -> int:
+    # why not brute-force? :)
     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
 
@@ -77,6 +78,19 @@ def _solve1(m: Machine) -> int:
     return min(costs)
     
     
+def _solve2(m: Machine) -> int:
+    cost = None
+    times_b = (m.prize.y * m.a.x - m.prize.x * m.a.y) / (m.b.y * m.a.x - m.b.x * m.a.y)
+    times_a = (m.prize.x - m.b.x * times_b) / m.a.x
+
+    #if 100 >= times_a >= 0 and 100 >= times_b >= 0 and times_a.is_integer() and times_b.is_integer():
+    if times_a.is_integer() and times_b.is_integer():
+        cost = int(times_a) * 3 + int(times_b)
+
+    return cost
+
+
+
 def solve1(lines: list[str]):
     machines = parse_input(lines)
 
@@ -94,11 +108,10 @@ def solve2(lines: list[str]):
     machines = parse_input(lines)
 
     cost = 0
-    for m in machines[:1]:
+    for m in machines:
-        continue
         m.prize.x += 10000000000000
         m.prize.y += 10000000000000
-        c = _solve1(m)
+        c = _solve2(m)
         if not c:
             continue
         cost += c
@@ -114,4 +127,4 @@ if __name__ == '__main__':
     print('part 1:', solve1(lines)) # 27157
 
     #part 2
-    print('part 2:', solve2(lines)) # 862486
+    print('part 2:', solve2(lines)) # 104015411578548

day13/task.txt

@@ -73,4 +73,6 @@ Now, it is only possible to win a prize on the second and fourth claw machines.
 
 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 104015411578548.
 
+Both parts of this puzzle are complete! They provide two gold stars: **