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copied!<p><code>^((.)(?1)\2|.?)$</code></p> <p>The <code>^</code> and <code>$</code> asserts the beginning and the end of the string respectively. Let us look at the content in between, which is more interesting:</p> <pre><code>((.)(?1)\2|.?) 1------------1 // Capturing group 1 2-2 // Capturing group 2 </code></pre> <p>Look at the first part <code>(.)(?1)\2</code>, we can see that it will try to match any character, and that same character at the end (back reference <code>\2</code>, which refers to the character matched by <code>(.)</code>). In the middle, it will recursively match for the whole capturing group 1. Note that there is an implicit assertion (caused by <code>(.)</code> matching one character at the beginning and <code>\2</code> matching the same character at the end) that requires the string to be at least 2 characters. The purpose of the first part is chopping the identical ends of the string, recursively.</p> <p>Look at second part <code>.?</code>, we can see that it will match one or 0 character. This will only be matched if the string initially has length 0 or 1, or after the leftover from the recursive match is 0 or 1 character. The purpose of the second part is to match the empty string or the single lonely character after the string is chopped from both ends.</p> <p>The recursive matching works:</p> <ul> <li>The whole string must be palindrome to pass, asserted by <code>^</code> and <code>$</code>. We cannot start matching from any random position.</li> <li>If the string is <= 1 character, it passes.</li> <li>If the string is > 2 characters, whether it is accepted is decided by the first part only. And it will be chopped by 2 ends if matches.</li> <li>The leftover if matches, can only be chopped by the 2 ends, or passes if its length is <= 1.</li> </ul>

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