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Robert Mion
Robert Mion

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High-Entropy Passphrases

Advent of Code 2017 Day 4

Part 1

  1. Inventory Management System Redux!
  2. Writing my working algorithm

Inventory Management System Redux!

  • Remember 2018 Day 2?
  • It asked me to find exact counts of letters within strings
  • This feels similar, but easier, in my opinion
  • Because I don't need to build an object of letter counts for each string
  • I just need an Array and a Set, so I can compare length and size!

Writing my working algorithm

Split the input at each newline character into an array of strings

For each string, accumulate a tally, starting at 0
  Split the string at each space character into an array of strings
  If the length of the array is the same as the size of a Set generated from that array
    Increment the tally by 1, because there are no duplicate phrases
    Leave the tally unchanged

Return the tally
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In JavaScript, I use reduce(), split(), Array.length, Set.size and a ternary operator in a one long chained statement:

     .reduce((tally, current) => {
       return tally += current.split(' ').length 
                    == new Set(current.split(' ')).size 
                    ? 1 : 0
     }, 0)
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Part 2

  1. What a delightful twist!
  2. Three reduce()rs, that's how!
  3. Revealing my working reduce()rs

What a delightful twist!

  • Anagrams, eh? How fun!
  • And also...complicating should I attempt to solve this?

Three reduce()rs, that's how!

This is going to get complex really fast, so bear with me:

  1. An outer reduce() to track the tallies of valid passphrases
  2. An inner reduce() to collect all the stringified letter tallies for each word
  3. An inner-most reduce() to generate a stringified letter tally for one word

This animation illustrates each of the reduce()rs using the last example set of passphrases:
Animation of all three reducers

Revealing my working reduce()rs

This is the JavaScript I'm proud to have written in about 15 minutes:

return input
  .reduce((passphrases, passphrase) => {
    let letterCounts = passphrase
      .split(' ')
      .reduce((list, group) => {
        let counts = group
          .split(' ')
          .reduce((tallies, letter) => {
            tallies[letter] = (tallies[letter] || 0) + 1
            return tallies
          }, {})
            .map(el => el + '|' + counts[el]).join(''))
        return list
      }, [])
    return passphrases += letterCounts.length 
                       == new Set(letterCounts).size 
                       ? 1 : 0
  }, 0)
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I did it!!

  • I solved both parts!
  • I used a ton of reduce()!
  • I made a detailed GIF showing what each reduce() does!
  • I really enjoyed solving Part 2!
  • I'm now tied for my personal best star score in a given year: 40!

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