m;"><span></span></span></span></span></span></span></span></span></span> simply by trying all possible keys - an approach called a <em>brute force attack</em>.</p>
<pre><code class="language-python">def BruteForce(ciphertext, plaintext1, plaintext2):
	for key in [0..2^n - 1]:
		if Enc(key, plaintext1) == ciphertext:
			return plaintext1
		if Enc(key, plaintext2) == ciphertext:
			return plaintext2
</code></pre>
<p>The reason we are not really worried about this attack, which works for <em>every</em> cipher