AoC 2021 D8: Seven Segment Search

| Problem statement | Source code | Tags: Constraint satisfaction

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Part 1

Fun problem, but not much to say here. 1 uses 2 segments, 4 uses 4 segments, 7 uses 3 segments, and 8 uses all 7 segments. Just count how many times these lengths appear in the output values.

const count = data
  .map((line) => line.split(" | ")[1])
  .flatMap((output) => output.split(" "))
  .filter((digit) => [2, 3, 4, 7].includes(digit.length));

Part 2

If we use the following canonical segment labeling:

 aaaa
b    c
b    c
 dddd
e    f
e    f
 gggg

Then we have the following segment usage for each digit:

const digits = {
  abcefg: "0",
  cf: "1",
  acdeg: "2",
  acdfg: "3",
  bcdf: "4",
  abdfg: "5",
  abdefg: "6",
  acf: "7",
  abcdefg: "8",
  abcdfg: "9",
};

We can count the number of times each segment letter appears in all ten digits:

This means that if we count the letters in the 10 scrambled patterns, we can instantly determine b, e, and f, because they have unique counts. We then need to tell a from c (both appear 8 times), and d from g (both appear 7 times). We already know the patterns for 1, 4, 7, and 8 from part 1. Subtracting the pattern for 1 from that of 7 gives us a. Subtracting the pattern for 1 plus b from that of 4 gives us d.

Once we have the mapping from scrambled letters to actual segments, we can decode the output values by looking up the output patterns using the digits map above.

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