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    <p>Try following decryption sequence</p> <pre><code>%s/\v(...)(...)/\2\1/g %s/\v(.)(.)(.)(.)/\2\4\1\3/g %s/\v(.)\1{2}(.)\1/\2\1\1\1\1/g %s/\v(.)\1{3}/\1\1/g </code></pre> <p><strong>Notes :</strong></p> <ol> <li><code>1,$</code> is well written as <code>%</code></li> <li>Use very-magic specifier <code>\v</code> to avoid abundant escaping using <code>\</code></li> </ol> <hr> <blockquote> <p>What does \1 or \2\2\2 mean before the forward slash?</p> </blockquote> <p>Since <code>\1,\2,..</code> captures what's have been matched within <code>()</code>. Hence when they are used in matching pattern itself, they are meant to match repeating characters.</p> <ul> <li><p><code>(.)</code> will match any character but <code>(.)\1</code> will match only a charater followed by itself, or in other terms a character repeated 2 times. E.g. in string <code>abbxyz</code>, it will match <code>bb</code></p></li> <li><p><code>(.)(.)\2\2\2</code> will match only a character followed by another character repeating 4 times. It's better written as <code>(.)(.)\2{3}</code>. E.g. in string <code>baabbbbaa</code>, it will match <code>abbbb</code></p></li> </ul> <hr> <blockquote> <p>And what does &amp; or &amp;&amp; mean?</p> </blockquote> <p><code>&amp;</code> contains a string with entire match. So <code>&amp;&amp;</code> is obviously 2 times repetition of matched string.</p> <p>E.g. input string <code>abbxyz</code> and matching pattern is <code>(.)\1</code> then it matches <code>bb</code> which is stored in <code>&amp;</code>. Hence value of <code>&amp;&amp;</code> will be <code>bbbb</code></p> <hr> <p><strong>P.S.</strong> I would strongly recommended you to go through <a href="http://www.regular-expressions.info/" rel="nofollow"><strong><code>The Premier website about Regular Expressions</code></strong></a> before jumping into such kind of problems.</p>
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    1. POVim: how to undo a regex with a new regex?
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    1. COhow does the \v option work? Will this /\v([2-9])/ be the same as this /\([2-9]\)/?
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    2. CO@Fizunik No. With `\v` you don't need to escape `()` for capturing. so `/\v([2-9])/` is same as `/\([2-9]\)/`. [Here](http://vim.wikia.com/wiki/Simplifying_regular_expressions_using_magic_and_no-magic) are more details
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