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  1. PO
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    <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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    1. POWhat's the difference between NOT second preimage resistant and NOT collision resistant
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    1. USGuntram Blohm
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    1. COEm... Two points:1, again, this is a very good answer and examples but not quite related to the last question I asked in the comment. To answer my question, please suggest a way to prove a hash is "NOT 2nd-preimage resistant" with minimum effort. My answer to this is: We know there exists a certain input x in this hash which allows me to easily get its 2nd preimage(a x' where h(x) = h(x')). We don't necessarily to know which x is. Do you agree with me? If yes, I can ask my follow-up questions.
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    2. COAnd point 2: The last example hash function you gave, is actually not 2nd-preimage resistant. This property should not related to the input length. As long as we know there is some x which will lead to its 2nd-preimage, then we say this hash is Not-preimage resistant. (in this case, we know that for input "0", we can find one of its 2nd-preimage "00").
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    3. COIf you can find some kind of transformation on x, resulting in x', so that h(x)=h(x'), *for all possible values of x*, then the function isn't 2nd-preimage resistant, and of course not collision resistant. So, in your point 2, the hash function is 2nd-preimage-resistant, because the transformation only works on a very small subset of all possible x. As an example, h(x)="clear the 3rd bit on the 2nd byte of x, then calculate MD5". This isn't 2nd-preimage-resistant, with the transformation "toggle the 3rd bit on the 2nd byte".
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      1. USGuntram Blohm
 

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