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I'm trying to understand the relationship between preimage resistance and second preimage resistance. Consider the following example.

Suppose there exists a hash function H(x). The function produces k-bit hash codes, and is collision resistant, preimage resistance, and second preimage resistance.

Then suppose there exists a hash function H'(x). The function produces (k+1) bit hash codes. When x is k bit long, the output is 0||x. When x is not k bit long the output is 1||H(x).

What resistances does H'(x) have and nor have, and why?

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    $\begingroup$ This is indistinguishable from homework; what have you tried? $\endgroup$ – fkraiem Nov 20 '16 at 6:37
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Well, it's obvious that H' is not preimage resistant for k bit values. If you have a hash that starts with 0, you will know that it's of the form 0||x and you got your x such as H'(x)=0||x.

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