Day 14 is all about stacking requirements and then working them to understand the inputs required to get the output desired. I’ll need to organize my list of reactions in such a way that I can work back from the desired end output to how much ore is required to get there.

## Challenge

The puzzle can be found here. I’m given a list of reactions that describe how various chemicals can be combined to make other chemicals. For example:

10 ORE => 10 A
1 ORE => 1 B
7 A, 1 B => 1 C
7 A, 1 C => 1 D
7 A, 1 D => 1 E
7 A, 1 E => 1 FUEL


The challenge is to determine how much ORE I need to start with to get 1 FUEL.

In part 2, I’ll be given a large amount of ORE, and asked to figure out how much FUEL it can produce. This isn’t just dividing by the amount of ORE required to make 1 FUEL, because there are leftover chemicals after 1 is created that can be used in the next.

## Solution

### Part 1

This challenge breaks into a couple parts. First, I need to read in the puzzle input and translate it into a structure to easily reference. I’m going to be looking at an output, and trying to determine how much input is needed, so I’ll use the output as the dictionary key. For each output, I’ll have an out which is the number that are produced, and an in, which is a dict of inputs as keys and amounts required as values:

with open(sys.argv[1], 'r') as f:

reactions = {}
for r in reaction_lines:
i, o = r.split(" => ")
o_num, o_chem = o.split(" ")
inputs = {}
for in_str in i.split(", "):
i_num, i_chem = in_str.split(" ")
inputs[i_chem] = int(i_num)
reactions[o_chem] = {"out": int(o_num), "in": inputs}


For example, the example about would produce:

{'A': {'out': 2, 'in': {'ORE': 9}}, 'B': {'out': 3, 'in': {'ORE': 8}}, 'C': {'out': 5, 'in': {'ORE': 7}}, 'AB': {'out': 1, 'in': {'A': 3, 'B': 4}}, 'BC': {'out': 1, 'in': {'B': 5, 'C': 7}}, 'CA': {'out': 1, 'in': {'C': 4, 'A': 1}}, 'FUEL': {'out': 1, 'in': {'AB': 2, 'BC': 3, 'CA': 4}}}


Next, I’ll create a dictionary of needs, which starts with the thing I’m trying to create, 1 FUEL:

chem_needs = defaultdict(int, {'FUEL': 1})


I’ll also initialize a list of leftover chemicals. This happens when I need three of something, but the recipe to make it creates four. I’ll make four, use three, and have one left. I’ll also create an ORE counter as well:

chem_have = defaultdict(int)
ore = 0


Now I loop as long as there are items in the needs list. For each item, I find out how to create it. If I have enough in the leftovers list (chem_have), I subtract that many from the have list, delete the item from the needs list, and continue the loop.

Otherwise, I’ll calculate the number more I need to create, and delete the balance in the have list and the need list. Then I’ll go about determining the inputs to make that many. First, I’ll find the number produced by one receipe, and then determine how many reactions I need to perform. I’ll add the extras to the have list. Then I’ll loop over the items required for that output and add them to the need list, unless they are ORE, in which case I add to the count needed.

while chem_needs:
item = list(chem_needs.keys())[0]
if chem_needs[item] <= chem_have[item]:
chem_have[item] -= chem_needs[item]
del chem_needs[item]
continue

num_needed = chem_needs[item] - chem_have[item]
del chem_have[item]
del chem_needs[item]
num_produced = reactions[item]["out"]

if (num_needed / num_produced) == int(num_needed / num_produced):
num_reactions = num_needed // num_produced
else:
num_reactions = (num_needed // num_produced) + 1

chem_have[item] += (num_reactions * num_produced) - num_needed
for chem in reactions[item]["in"]:
if chem == "ORE":
ore += reactions[item]["in"][chem] * num_reactions
else:
chem_needs[chem] += reactions[item]["in"][chem] * num_reactions

print(f"Part 1: {ore}")


When it’s done, it prints the amount of ORE needed:

$time ./day14.py 14-puzzle_input.txt Part 1: 136771 real 0m0.054s user 0m0.049s sys 0m0.004s  ### Part 2 For part 2, I need to think about how many FUEL I can make with 1000000000000 ORE (that’s 112). An initial guess might be that I could take the ORE required to make 1 FUEL and divide a trillion by that. But that’s not quite right, because as the lefover chemicals build, I’ll be able to start getting extra FUEL out. Rather than try to back calculate, I can use the work I’ve already done. I’ll convert the code from part 1 that finds how much ORE is required for one FUEL into a function, and have it take the number of FUEL as an input. I’ll think about a range of possible numbers of FUEL. It has to be at least a trillion divided by the number of ORE required to create one FUEL. I’ll use a loop to find a upper bound by multiplying that lower bound by 10 until it is over 112. low = 1e12 // ore_required(1) high = 10 * low while ore_required(high) < 1e12: low = high high = 10 * low  Now with an upper and lower bound in place, I’ll check how many ORE are required for the midpoint, and compare it to 112. Then I can adjust my range, until I know the most I can make without requiring more than 112. while low < high - 1: mid = (low + high) // 2 ore = ore_required(mid) if ore < 1e12: low = mid elif ore > 1e12: high = mid else: break print(f"Part 2: {int(mid)}")  This all still runs basically instantly: $ time ./day14.py 14-puzzle_input.txt
Part 1: 136771
Part 2: 8193614

real    0m0.059s
user    0m0.050s
sys     0m0.008s


## Final Code

#!/usr/bin/env python3

import sys
from collections import defaultdict

def ore_required(fuel=1):
chem_needs = defaultdict(int, {"FUEL": fuel})
chem_have = defaultdict(int)
ore = 0

while chem_needs:
item = list(chem_needs.keys())[0]
if chem_needs[item] <= chem_have[item]:
chem_have[item] -= chem_needs[item]
del chem_needs[item]
continue

num_needed = chem_needs[item] - chem_have[item]
del chem_have[item]
del chem_needs[item]
num_produced = reactions[item]["out"]

if (num_needed // num_produced) * num_produced == num_needed:
num_reactions = num_needed // num_produced
else:
num_reactions = (num_needed // num_produced) + 1
# print(num_needed, num_produced, num_reactions)

chem_have[item] += (num_reactions * num_produced) - num_needed
for chem in reactions[item]["in"]:
if chem == "ORE":
ore += reactions[item]["in"][chem] * num_reactions
else:
chem_needs[chem] += reactions[item]["in"][chem] * num_reactions
# print(f'{ore} {chem_needs} {chem_have}')

return ore

with open(sys.argv[1], "r") as f:

reactions = {}
for r in reaction_lines:
i, o = r.split(" => ")
o_num, o_chem = o.split(" ")
inputs = {}
for in_str in i.split(", "):
i_num, i_chem = in_str.split(" ")
inputs[i_chem] = int(i_num)
reactions[o_chem] = {"out": int(o_num), "in": inputs}

print(f"Part 1: {ore_required()}")

low = 1e12 // ore_required()
high = 10 * low

while ore_required(high) < 1e12:
low = high
high = 10 * low

while low < high - 1:
mid = (low + high) // 2
ore = ore_required(mid)
if ore < 1e12:
low = mid
elif ore > 1e12:
high = mid
else:
break

print(f"Part 2: {int(mid)}")