ROT13 JavaScript One-Liner Explained

The ROT13 JavaScript One-Liner

One of the most elegant ROT13 implementations is this JavaScript one-liner:

const rot13 = s => s.replace(/[a-zA-Z]/g, c =>
  String.fromCharCode(c.charCodeAt(0) + (c.toLowerCase() < 'n' ? 13 : -13))
);

Let's break it down piece by piece.

Step 1: The Regex /[a-zA-Z]/g

This regular expression matches every alphabetic character (both upper and lowercase) in the string. The g flag ensures all matches are processed, not just the first.

Non-alphabetic characters (digits, spaces, punctuation) are not matched and pass through unchanged.

Step 2: The Replace Callback

String.prototype.replace accepts a callback function that receives each matched character. The callback returns the replacement character.

Step 3: charCodeAt(0)

Returns the Unicode code point of the character:

• A = 65, B = 66, ..., Z = 90
• a = 97, b = 98, ..., z = 122

Step 4: The Ternary Trick

c.toLowerCase() < 'n' ? 13 : -13

This is the clever part. Instead of using modular arithmetic:

• If the letter is in the first half (A–M or a–m): add 13 to shift forward
• If the letter is in the second half (N–Z or n–z): subtract 13 to shift backward

This avoids the need for the modulo operator entirely.

Step 5: String.fromCharCode

Converts the new code point back to a character.

Trace Example

For the string "Hello":

H (72) → 72 + ('h' < 'n' ? 13 : -13) = 72 + 13 = 85 → U
e (101) → 101 + ('e' < 'n' ? 13 : -13) = 101 + 13 = 114 → r
l (108) → 108 + ('l' < 'n' ? 13 : -13) = 108 + 13 = 121 → y
l (108) → 121 → y
o (111) → 111 + ('o' < 'n' ? 13 : -13) = 111 - 13 = 98 → b

Result: "Uryyb"

Alternative: Using Modulo

const rot13 = s => s.replace(/[a-zA-Z]/g, c => {
  const base = c <= 'Z' ? 65 : 97;
  return String.fromCharCode(((c.charCodeAt(0) - base + 13) % 26) + base);
});

This version uses explicit modular arithmetic and is slightly more readable.