See the full AoC 2021 Day 14 puzzle at the URL: https://adventofcode.com/2021/day/14
- Use input file: day14-input.txt
- Fortran source code: aoc_day14_p1.f90
Task
The incredible pressures at this depth are starting to put a strain on your submarine. The submarine has polymerization equipment that would produce suitable materials to reinforce the submarine, and the nearby volcanically-active caves should even have the necessary input elements in sufficient quantities.
The submarine manual contains instructions for finding the optimal polymer formula; specifically, it offers a polymer template and a list of pair insertion rules (your puzzle input). You just need to work out what polymer would result after repeating the pair insertion process a few times.
For example:
NNCB
CH -> B
HH -> N
CB -> H
NH -> C
HB -> C
HC -> B
HN -> C
NN -> C
BH -> H
NC -> B
NB -> B
BN -> B
BB -> N
BC -> B
CC -> N
CN -> C
The first line is the polymer template - this is the starting point of the process.
The following section defines the pair insertion rules. A rule like AB -> C means that when elements A and B are immediately adjacent, element C should be inserted between them. These insertions all happen simultaneously.
So, starting with the polymer template NNCB, the first step simultaneously considers all three pairs:
- The first pair (NN) matches the rule NN -> C, so element C is inserted between the first N and the second N.
- The second pair (NC) matches the rule NC -> B, so element B is inserted between the N and the C.
- The third pair (CB) matches the rule CB -> H, so element H is inserted between the C and the B.
Note that these pairs overlap: the second element of one pair is the first element of the next pair. Also, because all pairs are considered simultaneously, inserted elements are not considered to be part of a pair until the next step.
After the first step of this process, the polymer becomes NCNBCHB.
Here are the results of a few steps using the above rules:
Template: NNCB
After step 1: NCNBCHB
After step 2: NBCCNBBBCBHCB
After step 3: NBBBCNCCNBBNBNBBCHBHHBCHB
After step 4: NBBNBNBBCCNBCNCCNBBNBBNBBBNBBNBBCBHCBHHNHCBBCBHCB
This polymer grows quickly. After step 5, it has length 97; After step 10, it has length 3073. After step 10, B occurs 1749 times, C occurs 298 times, H occurs 161 times, and N occurs 865 times; taking the quantity of the most common element (B, 1749) and subtracting the quantity of the least common element (H, 161) produces 1749 - 161 = 1588.
Apply 10 steps of pair insertion to the polymer template and find the most and least common elements in the result.
Answer needed : What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
Build the Fortran Source Code
gfortran -static-libgcc -o aoc_day14_p1 aoc_day14_p1.f90
or
fpm run aoc_day14_p1
- Use input file: day14-input.txt
- Fortran source code: aoc_day14_p2.f90
Task
Answer needed :
Build the Fortran Source Code
gfortran -static-libgcc -o aoc_day14_p2 aoc_day14_p2.f90
or
fpm run aoc_day14_p2