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copied!<p>Putting this into another answer because it's just too much to type for a comment.</p> <p>The definition of 2nd-preimage-resistant is you have h(x) and x, and can't create x'. </p> <p>The definition of preimage-resistant (without second!) means you have only h(x), and can't create x.</p> <p>And the definition of collision resistant is you have nothing, and may choose any h(x), x and x'.</p> <p>If you use the hash to sign a plaintext message, you need 2nd-preimage-resistancy, but not collision resistancy. It doesn't matter to you if someone can find two colliding messages that produce a hash that is different from yours, but you want to make sure noone is able to craft a <em>different</em> message that has the your hash, even if they know your plaintext. </p> <p>If you use the hash to store hashed passwords, you don't care about collision resistance, and you don't care about 2nd-preimage-resistance, preimage-resistance is all you need. If an attacker knows one password, you don't really care if he can use that password to find a different one.</p> <p>So these were two examples where collision resistance is not required, but preimage-resistance or 2nd-preimage-resistance is.</p> <p>As to "Not collision resistant not necessarily means Not 2nd-preimage resistant", why is that? , consider the hash function <code>if x has less then 24 bits, then h(x)=0, else h(x)=sha256(x)</code>. This is very obviously not collision resistant (choose any 2 words that have less than 4 letters), but, as long as your text is longer, this function is preimage-resistant and 2nd-preimage-resistant (assuming sha256 hasn't been broken yet).</p>

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