Frequency analysis of the ciphertext in Edgar Allen Poe's "The Gold Bug" short story

goldbug.js

// https://poestories.com/read/goldbug
const cipherText = `53‡‡†305))6*;4826)4‡.)4‡);806*;48†8¶60))85;1‡(;:‡*8†83(88)5*†;46(;88*96*?;8)*‡(;485);5*†2:*‡(;4956*2(5*-4)8¶8*;4069285);)6†8)4‡‡;1(‡9;48081;8:8‡1;48†85;4)485†528806*81(‡9;48;(88;4(‡?34;48)4‡;161;:188;‡?;`;

const getFrequencyTable = (list) => {
  return list.reduce((table, element) => {
    if (!table[element]) {
      table[element] = {
        count: 0,
        percentage: 0,
      };
    }
    table[element].count++;
    const percentage = Number(
      ((table[element].count / list.length) * 100).toFixed(2)
    );
    table[element].percentage = percentage;
    return table;
  }, {});
};

const cipherLetters = cipherText.split("");

// Sort in descending order by letter count to get the most popular letters first
letterFrequencies = Object.entries(getFrequencyTable(cipherLetters)).sort(
  (entry1, entry2) => {
    const count1 = entry1[1].count;
    const count2 = entry2[1].count;
    return count2 - count1;
  }
);

/*
[
  [ '8', { count: 33, percentage: 16.26 } ],
  [ ';', { count: 26, percentage: 12.81 } ],  
  [ '4', { count: 19, percentage: 9.36 } ],    
  [ '‡', { count: 16, percentage: 7.88 } ],   
  [ ')', { count: 16, percentage: 7.88 } ],   
  [ '*', { count: 13, percentage: 6.4 } ],     
  [ '5', { count: 12, percentage: 5.91 } ],
  [ '6', { count: 11, percentage: 5.42 } ],
  [ '(', { count: 10, percentage: 4.93 } ],
  [ '1', { count: 8, percentage: 3.94 } ],     
  [ '†', { count: 8, percentage: 3.94 } ],
  [ '0', { count: 6, percentage: 2.96 } ],
  [ '2', { count: 5, percentage: 2.46 } ],
  [ '9', { count: 5, percentage: 2.46 } ],
  [ '3', { count: 4, percentage: 1.97 } ],
  [ ':', { count: 4, percentage: 1.97 } ],
  [ '?', { count: 3, percentage: 1.48 } ],
  [ '¶', { count: 2, percentage: 0.99 } ],
  [ '.', { count: 1, percentage: 0.49 } ],
  [ '-', { count: 1, percentage: 0.49 } ]
]
*/
console.log(letterFrequencies);

const decipheredMessage = cipherText
  // 8 is the most frequently occurring letter, so likely e
  .replace(/8/g, "e")
  // ;48 appears often, suggesting it is "the"
  .replace(/;/g, "t")
  .replace(/4/g, "h")
  // with the above transformations, we have:
  // 53‡‡†305))6*the26)h‡.)h‡)te06*the†e¶60))e5t1‡(t:‡*e†e3(ee)5*†th6(tee*96*?te)*‡(the5)t5*†2:*‡(th956*2(5*-h)e¶e*th0692e5)t)6†e)h‡‡t1(‡9the0e1te:e‡1the†e5th)he5†52ee06*e1(‡9thet(eeth(‡?3hthe)h‡t161t:1eet‡?t
  // ‡ appears twice in a row and also after h, suggesting it's o
  .replace(/‡/g, "o")
  // thet(eeth => the tree th...
  .replace(/\(/g, "r")
  // 9thetreethro?3hthe => ? = u and 3 = g
  .replace(/\?/g, "u")
  .replace(/3/g, "g")
  // 5goo†g05))6*the26)ho.)ho)te06*the†e¶60))e5t1ort:o*e†egree)5*†th6rtee*96*ute)*orthe5)t5*†2:*orth956*2r5*-h)e¶e*th0692e5)t)6†e)hoot1ro9the0e1te:eo1the†e5th)he5†52ee06*e1ro9thetreethroughthe)hot161t:1eetout
  // Since ) appears twice in a row multiple times, safe to assume it's 's' since we already found O and E, the two other letters that frequently repeat themselves.
  .replace(/\)/g, "s")
  // *orth = north or forth, north seems likely given it's a map
  .replace(/\*/g, "n")
  // th6rteen => 6 = i
  .replace(/6/g, "i")
  // thirteen9inutesnorthe => 9 = m
  .replace(/9/g, "m")
  // se¶enth => ¶ = v
  .replace(/¶/g, "v")
  // one†egree => † = d
  .replace(/†/g, "d")
  // inthedevi0ss => 0 = l
  .replace(/0/g, "l")
  // onedegrees5ndthirteenminutesnorthe => 5 = a
  .replace(/5/g, "a")
  // 1rom => 1 = f
  .replace(/1/g, "f")
  // atfort:onedegrees => : = y
  .replace(/\:/g, "y")
  // and2ynorth => 2 = b
  .replace(/2/g, "b")
  // bynorthmainbran-hseventhlimbeast => - = c
  .replace(/\-/g, "c")
  // agoodglassinthebisho.shostel => . = p
  .replace(/\./g, "p");

// a good glass in the bishops hostel in the devils seat forty one degrees
// and thirteen minutes northeast and by north main branch seventh limb
// east side shoot from the left eye of the deaths head a beeline from
// the tree through the shot fifty feet out
console.log(decipheredMessage);