The latest articles on DEV Community by heavenhunter (@heavenhunter). https://dev.to/heavenhunter https://media2.dev.to/dynamic/image/width=90,height=90,fit=cover,gravity=auto,format=auto/https:%2F%2Fdev-to-uploads.s3.amazonaws.com%2Fuploads%2Fuser%2Fprofile_image%2F3499214%2F04371e37-4e40-4c84-93af-9ccc48378f55.png https://dev.to/heavenhunter en heavenhunter Sat, 13 Sep 2025 09:48:41 +0000 https://dev.to/heavenhunter/enigma-machine-how-a-step-on-its-rotor-change-the-mapping-24nf https://dev.to/heavenhunter/enigma-machine-how-a-step-on-its-rotor-change-the-mapping-24nf <p>I'm reading through publicly available e-book "Cryptography: An Introduction (3rd Edition)" by Nigel Smart, I'm confused when reading the following section on page 50.</p> <p><a href="https://media2.dev.to/dynamic/image/width=800%2Cheight=%2Cfit=scale-down%2Cgravity=auto%2Cformat=auto/https%3A%2F%2Fdev-to-uploads.s3.amazonaws.com%2Fuploads%2Farticles%2Fffsq2vghs4ix372d305b.png" class="article-body-image-wrapper"><img src="https://media2.dev.to/dynamic/image/width=800%2Cheight=%2Cfit=scale-down%2Cgravity=auto%2Cformat=auto/https%3A%2F%2Fdev-to-uploads.s3.amazonaws.com%2Fuploads%2Farticles%2Fffsq2vghs4ix372d305b.png" alt="Excerpt from the book" width="800" height="417"></a></p> <blockquote> <p>Now assume that rotor one moves on one step, so A now maps to D under rotor one, B to A, C to C and D to B.</p> </blockquote> <p>Feeling confused, I decided to dig deeper into that statement.</p> <p>As described within the book, the mapping under rotor 1 can be formulized as below:</p> <div class="table-wrapper-paragraph"><table> <tr> <th>Input</th> <td>A</td> <td>B</td> <td>C</td> <td>D</td> </tr> <tr> <th>Output (step=0)</th> <td>C</td> <td>A</td> <td>B</td> <td>D</td> </tr> <tr> <th>Output (step=1)</th> <td>D</td> <td>A</td> <td>C</td> <td>B</td> </tr> </table></div> <p>I was expecting the output under rotor 1 after it moves one step to be a shift from its initial output, from <code>CABD</code> becomes either <code>ABDC</code> or <code>DCAB</code>. But to my surprise, the book says <code>DACB</code> instead.</p> <p>The explanation available in Wikipedia is not sufficient for me, it's skipping a lot of detail. Then I see a related post in <a href="https://crypto.stackexchange.com/a/71233/52890" rel="noopener noreferrer">crypto.stackexchange.com</a> that can clear up my confusion.</p> <h2> The correct way </h2> <p>My mistake is I shift the output character, but I should shift the <strong>offsets</strong> instead. If the input is <code>A</code> and the output is <code>B</code> then the offset is <code>1</code>. If the input is <code>D</code> and the output is <code>C</code> then the offset is <code>-1</code> (or <code>3</code> in modulo 4, because we only have 4 alphabet characters <code>ABCD</code>).</p> <p>So we need to calculate the offsets, then shift the offset to determine the output for the following step.</p> <div class="table-wrapper-paragraph"><table> <tr> <th>Input</th> <td>A</td> <td>B</td> <td>C</td> <td>D</td> </tr> <tr> <th>Output (step=0)</th> <td>C</td> <td>A</td> <td>B</td> <td>D</td> </tr> <tr> <th>Offset (step=0)</th> <td>2</td> <td>-1</td> <td>-1</td> <td>0</td> </tr> <tr> <th>Offset (step=1)<br>(shift left)</th> <td>-1</td> <td>-1</td> <td>0</td> <td>2</td> </tr> <tr> <th>Output (step=1)</th> <td>D</td> <td>A</td> <td>C</td> <td>B</td> </tr> <tr> <th>Explanation<br>(input+offset)</th> <td>A-1</td> <td>B-1</td> <td>C+0</td> <td>D+2</td> </tr> </table></div> <p><code>Offset (step=1)</code> is obtained from shifting left (rotating) <code>Offset (step=0)</code>. The final <code>Output (step=1)</code> can be obtained from applying <code>Offset (step=1)</code> to <code>Input</code>.</p> <p>Now the final output is <code>DACB</code>, exactly matches what explained in the book 😄</p> cryptography encryption enigma