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The answers to my HW say that a preimage of a single block is easily found. I do not understand how it is easily found. Please help.

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    $\begingroup$ Does this answer your question? AES-CBC Hash Function Collision Resistance. Usually the best practice is to ask the writer of the solution since they are paid for this and pretty sure they can explain better if one asks. $\endgroup$
    – kelalaka
    Nov 27, 2021 at 10:34

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A block cipher consists of an encryption function and a decryption function. In other words as well as $E_k$ there is also a function $D_k$ with the property that $D_k(E_k(m))=m$ and $E_k(D_k(c))=c$ for all $k$, $m$ and $c$.

Given a target hash value $v$, one can compute the $n$-long quantity $D_{IV}(v)$ and call this $m=m_1$. We then have that $H_1=f(H_0,m_1)=E_{IV}(m_1)=E_{IV}(D_{IV}(v))=v$. We conclude that $h(m)=v$.

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  • $\begingroup$ This was asked many times during the HW duration, and once answered here though I did not want to answer. I'm pretty sure that we can find more dupe if we search in older Q/A. A pretty basic question that circulates around. $\endgroup$
    – kelalaka
    Nov 27, 2021 at 10:55

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