Added 2019 day 14 part 1
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2019/14/README.md
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# 2019 Day 14: Space Stoichiometry
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Copyright (c) Eric Wastl
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#### [Direct Link](https://adventofcode.com/2019/day/14)
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## Part 1
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As you approach the rings of Saturn, your ship's **low fuel** indicator turns on. There isn't any fuel here, but the rings have plenty of raw material. Perhaps your ship's Inter-Stellar Refinery Union brand **nanofactory** can turn these raw materials into fuel.
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You ask the nanofactory to produce a list of the **reactions** it can perform that are relevant to this process (your puzzle input). Every reaction turns some quantities of specific **input chemicals** into some quantity of an **output chemical**. Almost every **chemical** is produced by exactly one reaction; the only exception, `ORE`, is the raw material input to the entire process and is not produced by a reaction.
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You just need to know how much **`ORE`** you'll need to collect before you can produce one unit of **`FUEL`**.
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Each reaction gives specific quantities for its inputs and output; reactions cannot be partially run, so only whole integer multiples of these quantities can be used. (It's okay to have leftover chemicals when you're done, though.) For example, the reaction `1 A, 2 B, 3 C => 2 D` means that exactly 2 units of chemical `D` can be produced by consuming exactly 1 `A`, 2 `B` and 3 `C`. You can run the full reaction as many times as necessary; for example, you could produce 10 `D` by consuming 5 `A`, 10 `B`, and 15 `C`.
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Suppose your nanofactory produces the following list of reactions:
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```
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10 ORE => 10 A
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1 ORE => 1 B
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7 A, 1 B => 1 C
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7 A, 1 C => 1 D
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7 A, 1 D => 1 E
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7 A, 1 E => 1 FUEL
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```
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The first two reactions use only `ORE` as inputs; they indicate that you can produce as much of chemical `A` as you want (in increments of 10 units, each 10 costing 10 `ORE`) and as much of chemical `B` as you want (each costing 1 `ORE`). To produce 1 `FUEL`, a total of **31** `ORE` is required: 1 `ORE` to produce 1 `B`, then 30 more `ORE` to produce the 7 + 7 + 7 + 7 = 28 `A` (with 2 extra `A` wasted) required in the reactions to convert the `B` into `C`, `C` into `D`, `D` into `E`, and finally `E` into `FUEL`. (30 `A` is produced because its reaction requires that it is created in increments of 10.)
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Or, suppose you have the following list of reactions:
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```
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9 ORE => 2 A
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8 ORE => 3 B
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7 ORE => 5 C
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3 A, 4 B => 1 AB
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5 B, 7 C => 1 BC
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4 C, 1 A => 1 CA
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2 AB, 3 BC, 4 CA => 1 FUEL
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```
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The above list of reactions requires **165** `ORE` to produce 1 `FUEL`:
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- Consume 45 `ORE` to produce 10 `A`.
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- Consume 64 `ORE` to produce 24 `B`.
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- Consume 56 `ORE` to produce 40 `C`.
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- Consume 6 `A`, 8 `B` to produce 2 `AB`.
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- Consume 15 `B`, 21 `C` to produce 3 `BC`.
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- Consume 16 `C`, 4 `A` to produce 4 `CA`.
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- Consume 2 `AB`, 3 `BC`, 4 `CA` to produce 1 `FUEL`.
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Here are some larger examples:
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- **13312** `ORE` for 1 `FUEL`:
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```
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157 ORE => 5 NZVS
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165 ORE => 6 DCFZ
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44 XJWVT, 5 KHKGT, 1 QDVJ, 29 NZVS, 9 GPVTF, 48 HKGWZ => 1 FUEL
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12 HKGWZ, 1 GPVTF, 8 PSHF => 9 QDVJ
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179 ORE => 7 PSHF
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177 ORE => 5 HKGWZ
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7 DCFZ, 7 PSHF => 2 XJWVT
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165 ORE => 2 GPVTF
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3 DCFZ, 7 NZVS, 5 HKGWZ, 10 PSHF => 8 KHKGT
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```
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- **180697** ORE for 1 FUEL:
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```
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2 VPVL, 7 FWMGM, 2 CXFTF, 11 MNCFX => 1 STKFG
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17 NVRVD, 3 JNWZP => 8 VPVL
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53 STKFG, 6 MNCFX, 46 VJHF, 81 HVMC, 68 CXFTF, 25 GNMV => 1 FUEL
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22 VJHF, 37 MNCFX => 5 FWMGM
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139 ORE => 4 NVRVD
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144 ORE => 7 JNWZP
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5 MNCFX, 7 RFSQX, 2 FWMGM, 2 VPVL, 19 CXFTF => 3 HVMC
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5 VJHF, 7 MNCFX, 9 VPVL, 37 CXFTF => 6 GNMV
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145 ORE => 6 MNCFX
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1 NVRVD => 8 CXFTF
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1 VJHF, 6 MNCFX => 4 RFSQX
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176 ORE => 6 VJHF
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```
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- **2210736** ORE for 1 FUEL:
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```
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171 ORE => 8 CNZTR
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7 ZLQW, 3 BMBT, 9 XCVML, 26 XMNCP, 1 WPTQ, 2 MZWV, 1 RJRHP => 4 PLWSL
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114 ORE => 4 BHXH
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14 VRPVC => 6 BMBT
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6 BHXH, 18 KTJDG, 12 WPTQ, 7 PLWSL, 31 FHTLT, 37 ZDVW => 1 FUEL
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6 WPTQ, 2 BMBT, 8 ZLQW, 18 KTJDG, 1 XMNCP, 6 MZWV, 1 RJRHP => 6 FHTLT
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15 XDBXC, 2 LTCX, 1 VRPVC => 6 ZLQW
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13 WPTQ, 10 LTCX, 3 RJRHP, 14 XMNCP, 2 MZWV, 1 ZLQW => 1 ZDVW
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5 BMBT => 4 WPTQ
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189 ORE => 9 KTJDG
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1 MZWV, 17 XDBXC, 3 XCVML => 2 XMNCP
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12 VRPVC, 27 CNZTR => 2 XDBXC
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15 KTJDG, 12 BHXH => 5 XCVML
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3 BHXH, 2 VRPVC => 7 MZWV
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121 ORE => 7 VRPVC
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7 XCVML => 6 RJRHP
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5 BHXH, 4 VRPVC => 5 LTCX
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```
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Given the list of reactions in your puzzle input, **what is the minimum amount of `ORE` required to produce exactly 1 `FUEL`**?
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## Part 2
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After collecting `ORE` for a while, you check your cargo hold: **1 trillion** (**1000000000000**) units of `ORE`.
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**With that much ore**, given the examples above:
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- The 13312 `ORE`-per-`FUEL` example could produce **82892753** `FUEL`.
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- The 180697 `ORE`-per-`FUEL` example could produce **5586022** `FUEL`.
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- The 2210736 `ORE`-per-`FUEL` example could produce **460664** `FUEL`.
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Given 1 trillion `ORE`, **what is the maximum amount of FUEL you can produce**?
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2019/14/code.py
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""" https://adventofcode.com/2019/day/14 """
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def readFile():
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with open(f"{__file__.rstrip('code.py')}input.txt", "r") as f:
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return [line[:-1] for line in f.readlines()]
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class Factory:
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def __init__(self):
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self.recipes = dict()
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self.storage = {"ORE": 0}
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self.ores = 0
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def addRecipe(self, recipe):
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left, right = recipe.split(" => ")
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left, right = left.split(", "), right.split()
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self.storage[right[1]] = 0
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self.recipes[right[1]] = {"amount": int(right[0]), "req": {}}
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for l in left:
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l = l.split()
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self.recipes[right[1]]["req"][l[1]] = int(l[0])
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def produce(self, chem="FUEL"):
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if "ORE" in self.recipes[chem]["req"]:
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cur = self.storage["ORE"]
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while self.storage["ORE"] < self.recipes[chem]["req"]["ORE"]:
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self.storage["ORE"] += self.recipes[chem]["req"]["ORE"]
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self.ores += self.recipes[chem]["req"]["ORE"]
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self.storage[chem] += self.recipes[chem]["amount"]
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self.storage["ORE"] -= self.recipes[chem]["req"]["ORE"]
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else:
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for r in self.recipes[chem]["req"]:
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while self.storage[r] < self.recipes[chem]["req"][r]:
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self.produce(r)
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self.storage[chem] += self.recipes[chem]["amount"]
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for r in self.recipes[chem]["req"]:
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self.storage[r] -= self.recipes[chem]["req"][r]
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while self.storage[r] < 0:
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self.produce(r)
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return
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def part1(vals):
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factory = Factory()
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for val in vals:
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factory.addRecipe(val)
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factory.produce()
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return factory.ores
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def part2(vals):
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pass
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def test():
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assert part1(["10 ORE => 10 A","1 ORE => 1 B","7 A, 1 B => 1 C","7 A, 1 C => 1 D",
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"7 A, 1 D => 1 E","7 A, 1 E => 1 FUEL"]) == 31
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assert part1(["9 ORE => 2 A","8 ORE => 3 B","7 ORE => 5 C","3 A, 4 B => 1 AB",
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"5 B, 7 C => 1 BC","4 C, 1 A => 1 CA","2 AB, 3 BC, 4 CA => 1 FUEL"]) == 165
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assert part1(["157 ORE => 5 NZVS","165 ORE => 6 DCFZ","44 XJWVT, 5 KHKGT, 1 QDVJ, 29"
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" NZVS, 9 GPVTF, 48 HKGWZ => 1 FUEL","12 HKGWZ, 1 GPVTF, 8 PSHF => 9 QDVJ",
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"179 ORE => 7 PSHF","177 ORE => 5 HKGWZ","7 DCFZ, 7 PSHF => 2 XJWVT",
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"165 ORE => 2 GPVTF","3 DCFZ, 7 NZVS, 5 HKGWZ, 10 PSHF => 8 KHKGT"]) == 13312
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if __name__ == "__main__":
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test()
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vals = readFile()
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print(f"Part 1: {part1(vals)}")
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print(f"Part 2: {part2(vals)}")
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2019/14/input.txt
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1 JKXFH => 8 KTRZ
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11 TQGT, 9 NGFV, 4 QZBXB => 8 MPGLV
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8 NPDPH, 1 WMXZJ => 7 VCNSK
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1 MPGLV, 6 CWHX => 5 GDRZ
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16 JDFQZ => 2 CJTB
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1 GQNQF, 4 JDFQZ => 5 WJKDC
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2 TXBS, 4 SMGQW, 7 CJTB, 3 NTBQ, 13 CWHX, 25 FLPFX => 1 FUEL
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3 WMXZJ, 14 CJTB => 5 FLPFX
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7 HDCTQ, 1 MPGLV, 2 VFVC => 1 GSVSD
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1 WJKDC => 2 NZSQR
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1 RVKLC, 5 CMJSL, 16 DQTHS, 31 VCNSK, 1 RKBMX, 1 GDRZ => 8 SMGQW
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2 JDFQZ, 2 LGKHR, 2 NZSQR => 9 TSWN
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34 LPXW => 8 PWJFD
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2 HDCTQ, 2 VKWN => 8 ZVBRF
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2 XCTF => 3 QZBXB
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12 NGFV, 3 HTRWR => 5 HDCTQ
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1 TSWN, 2 WRSD, 1 ZVBRF, 1 KFRX, 5 BPVMR, 2 CLBG, 22 NPSLQ, 9 GSVSD => 5 NTBQ
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10 TSWN => 9 VFVC
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141 ORE => 6 MKJDZ
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4 NPSLQ, 43 VCNSK, 4 PSJL, 14 KTRZ, 3 KWCDP, 3 HKBS, 11 WRSD, 3 MXWHS => 8 TXBS
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8 VCNSK, 1 HDCTQ => 7 MXWHS
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3 JDFQZ, 2 GQNQF => 4 XJSQW
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18 NGFV, 4 GSWT => 5 KFRX
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2 CZSJ => 7 GMTW
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5 PHKL, 5 VCNSK, 25 GSVSD => 8 FRWC
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30 FRWC, 17 GKDK, 8 NPSLQ => 3 CLBG
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8 MXWHS, 3 SCKB, 2 NPSLQ => 1 JKXFH
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1 XJSQW, 7 QZBXB => 1 LGKHR
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115 ORE => 6 GQNQF
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12 HTRWR, 24 HDCTQ => 1 RKBMX
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1 DQTHS, 6 XDFWD, 1 MXWHS => 8 VKWN
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129 ORE => 3 XCTF
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6 GQNQF, 7 WJKDC => 5 PHKL
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3 NZSQR => 2 LPXW
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2 FLPFX, 1 MKLP, 4 XDFWD => 8 NPSLQ
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4 DQTHS, 1 VKWN => 1 BPVMR
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7 GMTW => 1 TXMVX
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152 ORE => 8 JDFQZ
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21 LGKHR => 9 NPDPH
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5 CJTB, 1 QZBXB, 3 KFRX => 1 GTPB
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1 MXWHS => 3 CWHX
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3 PHKL => 1 NGFV
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1 WMXZJ => 7 XDFWD
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3 TSWN, 1 VKWN => 8 GKDK
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1 ZVBRF, 16 PWJFD => 8 CMJSL
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3 VCNSK, 7 GDRZ => 4 HKBS
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20 XJSQW, 6 HTRWR, 7 CJTB => 5 WMXZJ
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12 ZVBRF, 10 FRWC, 12 TSWN => 4 WRSD
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16 HDCTQ, 3 GTPB, 10 NGFV => 4 KWCDP
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3 TXMVX, 1 NPDPH => 8 HTRWR
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9 NPDPH, 6 LPXW => 8 GSWT
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4 MKLP => 1 TQGT
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34 GTPB => 3 RVKLC
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25 VFVC, 5 RVKLC => 8 DQTHS
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7 KWCDP => 3 SCKB
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6 LGKHR => 8 MKLP
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39 MKJDZ => 9 CZSJ
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2 TSWN, 1 WMXZJ => 3 PSJL
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1
2019/14/solution.txt
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Part 1: 2486514
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