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copied!<p>Encrypting twice is more secure than encrypting once, even though this may not be clear at first. </p> <p>Intuitively, it appears that encrypting twice with the same algorithm gives no extra protection because an attacker might find a key which decrypts all the way from the final cyphertext back to the plaintext. ... But this is not the case. </p> <p>E.g. I start with plaintext <strong>A</strong> and encrypt with key <strong>K1</strong> it to get <strong>B</strong>. Then I encrypt <strong>B</strong> with key <strong>K2</strong> to get <strong>C</strong>. </p> <p>Intuitively, it seems reasonable to assume that there may well be a key, <strong>K3</strong>, which I could use to encrypt <strong>A</strong> and get <strong>C</strong> directly. If this is the case, then an attacker using brute force would eventually stumble upon <strong>K3</strong> and be able to decrypt <strong>C</strong>, with the result that the extra encryption step has not added any security. </p> <p>However, it is highly unlikely that such a key exists (for any modern encryption scheme). (When I say "highly unlikely" here, I mean what a normal person would express using the word "impossible"). </p> <p>Why?<br> Consider the keys as functions which provide a mapping from plaintext to cyphertext.<br> If our keys are all <strong>KL</strong> bits in length, then there are 2^KL such mappings.<br> However, if I use 2 keys of <strong>KL</strong> bits each, this gives me (2^KL)^2 mappings.<br> Not all of these can be equivalent to a single-stage encryption. </p> <p>Another advantage of encrypting twice, <em>if 2 different algorithms are used</em>, is that if a vulnerability is found in one of the algorithms, the other algorithm still provides some security. </p> <p>As others have noted, brute forcing the key is typically a last resort. An attacker will often try to break the process at some other point (e.g. using social engineering to discover the passphrase). </p> <p>Another way of increasing security is to simply use a longer key with one encryption algorithm. </p> <p>...Feel free to correct my maths!</p>

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