Avatar of brunnock

brunnock's solution

to Rotational Cipher in the JavaScript Track

Published at Aug 25 2018 · 0 comments
Test suite


This exercise has changed since this solution was written.

Create an implementation of the rotational cipher, also sometimes called the Caesar cipher.

The Caesar cipher is a simple shift cipher that relies on transposing all the letters in the alphabet using an integer key between 0 and 26. Using a key of 0 or 26 will always yield the same output due to modular arithmetic. The letter is shifted for as many values as the value of the key.

The general notation for rotational ciphers is ROT + <key>. The most commonly used rotational cipher is ROT13.

A ROT13 on the Latin alphabet would be as follows:

Plain:  abcdefghijklmnopqrstuvwxyz
Cipher: nopqrstuvwxyzabcdefghijklm

It is stronger than the Atbash cipher because it has 27 possible keys, and 25 usable keys.

Ciphertext is written out in the same formatting as the input including spaces and punctuation.


  • ROT5 omg gives trl
  • ROT0 c gives c
  • ROT26 Cool gives Cool
  • ROT13 The quick brown fox jumps over the lazy dog. gives Gur dhvpx oebja sbk whzcf bire gur ynml qbt.
  • ROT13 Gur dhvpx oebja sbk whzcf bire gur ynml qbt. gives The quick brown fox jumps over the lazy dog.


Go through the setup instructions for ECMAScript to install the necessary dependencies:



Install assignment dependencies:

$ npm install

Making the test suite pass

Execute the tests with:

$ npm test

In the test suites all tests but the first have been skipped.

Once you get a test passing, you can enable the next one by changing xtest to test.


Wikipedia https://en.wikipedia.org/wiki/Caesar_cipher

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.


import RotationalCipher from './rotational-cipher';

describe('Rotational cipher', () => {
  test('rotate a by 0, same output as input', () => {
    const expected = 'a';
    const actual = RotationalCipher.rotate('a', 0);


  xtest('rotate a by 1', () => {
    const expected = 'b';
    const actual = RotationalCipher.rotate('a', 1);


  xtest('rotate a by 26, same output as input', () => {
    const expected = 'a';
    const actual = RotationalCipher.rotate('a', 26);


  xtest('rotate m by 13', () => {
    const expected = 'z';
    const actual = RotationalCipher.rotate('m', 13);


  xtest('rotate n by 13 with wrap around alphabet', () => {
    const expected = 'a';
    const actual = RotationalCipher.rotate('n', 13);


  xtest('rotate capital letters', () => {
    const expected = 'TRL';
    const actual = RotationalCipher.rotate('OMG', 5);


  xtest('rotate spaces', () => {
    const expected = 'T R L';
    const actual = RotationalCipher.rotate('O M G', 5);


  xtest('rotate numbers', () => {
    const expected = 'Xiwxmrk 1 2 3 xiwxmrk';
    const actual = RotationalCipher.rotate('Testing 1 2 3 testing', 4);


  xtest('rotate punctuation', () => {
    const expected = 'Gzo\'n zvo, Bmviyhv!';
    const actual = RotationalCipher.rotate('Let\'s eat, Grandma!', 21);


  xtest('rotate all letters', () => {
    const expected = 'Gur dhvpx oebja sbk whzcf bire gur ynml qbt.';
    const actual = RotationalCipher.rotate('The quick brown fox jumps over the lazy dog.', 13);

function rotateChar(char,amount,min,max) {
  let charNum = char.charCodeAt(0);
  if (charNum>=min && charNum<=max) {
    charNum += amount;
    if (charNum > max) charNum = (min-1) + charNum%max;
class RotationalCipher {
  static rotate(input,amount) {
    let inputArr = input.split('');
    inputArr = inputArr.map(x=>rotateChar(x,amount,65,90));
    inputArr = inputArr.map(x=>rotateChar(x,amount,97,122));
    return inputArr.join('');

Community comments

Find this solution interesting? Ask the author a question to learn more.

What can you learn from this solution?

A huge amount can be learned from reading other people’s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

  • What compromises have been made?
  • Are there new concepts here that you could read more about to improve your understanding?