Day 14 part 2

Author: jcisio

day14/d14.py

@@ -20,18 +20,35 @@
             continue
         distance[material] = max([distance[i] for i in recipes[material]['ingredients'].keys()]) + 1
 
-needed = {'FUEL': 1}
-while len(needed) > 1 or 'ORE' not in needed:
-    material = max(needed, key=lambda x: distance[x])
-    quantity = needed[material]
-    del needed[material]
-    if material == 'ORE':
-        needed[material] = quantity
-        continue
-    base_quantity, ingredients = recipes[material].values()
-    for a, b in ingredients.items():
-        if a not in needed:
-            needed[a] = 0
-        needed[a] += math.ceil(quantity/base_quantity)*b
 
-print('Part 1:', needed['ORE'])
+def required_ore(fuel):
+    global materials, recipes, distance
+    needed = {'FUEL': fuel}
+    while len(needed) > 1 or 'ORE' not in needed:
+        material = max(needed, key=lambda x: distance[x])
+        quantity = needed[material]
+        del needed[material]
+        if material == 'ORE':
+            needed[material] = quantity
+            continue
+        base_quantity, ingredients = recipes[material].values()
+        for a, b in ingredients.items():
+            if a not in needed:
+                needed[a] = 0
+            needed[a] += math.ceil(quantity/base_quantity)*b
+    return needed['ORE']
+
+def search_fuel_target(ore):
+    one_unit = required_ore(1)
+    target = ore//one_unit
+    used_ore = required_ore(target)
+    while True:
+        target += (ore-used_ore)//one_unit + 1
+        used_ore = required_ore(target)
+        if used_ore > ore:
+            break
+    return target - 1
+
+
+print('Part 1:', required_ore(1))
+print('Part 2:', search_fuel_target(1000000000000))

day14/problem.md

@@ -81,4 +81,14 @@ Here are some larger examples:
 121 ORE => 7 VRPVC
 7 XCVML => 6 RJRHP
 5 BHXH, 4 VRPVC => 5 LTCX
-Given the list of reactions in your puzzle input, what is the minimum amount of ORE required to produce exactly 1 FUEL?
+Given the list of reactions in your puzzle input, what is the minimum amount of ORE required to produce exactly 1 FUEL?
+
+--- Part Two ---
+After collecting ORE for a while, you check your cargo hold: 1 trillion (1000000000000) units of ORE.
+
+With that much ore, given the examples above:
+
+The 13312 ORE-per-FUEL example could produce 82892753 FUEL.
+The 180697 ORE-per-FUEL example could produce 5586022 FUEL.
+The 2210736 ORE-per-FUEL example could produce 460664 FUEL.
+Given 1 trillion ORE, what is the maximum amount of FUEL you can produce?