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If I created this algorithm:

Break the message $M$ up into $n$ blocks $m_1,\ldots,m_n$ of 512 bits each.

Compute the hash by alternating AND and OR operations on the blocks. More specifically:

  • The first block and the second block are combined using AND
  • The result is combined with the third block using OR
  • The result is combined with the fourth block using AND
  • The result is combined with the fifth block using OR • and so on…

In other words, $h(M) = (\ldots(((m_1 \wedge m_2) \vee m_3) \wedge m_4) \vee m_5) \ldots )$

Why would this be vulnerable to a preimage attack?

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  • $\begingroup$ This question was part of a facebook group on cyber security and nobody seems to know the correct answer $\endgroup$ – Brooney Apr 12 '20 at 14:20
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Firstly, remember what is the pre-image attack*;

  • The pre-image attack is that given a hash value $h$ one need to find an $x$ such that $h = \operatorname{Hash}(x)$. The founded value doesn't need to be the pre-image that is used for the given hash value rather it can be any input value that its hash matches the given hash value, i.e. $x' \neq x$ and $h = \operatorname{Hash}(x')$.

Let call $\operatorname{H}_\ell (M) = (\ldots(((m_1 \wedge m_2) \vee m_3) \wedge m_4) \vee m_5) \ldots [\vee|\wedge]\, m_{n-2})$ i.e words just before the last two operation. And

$$\operatorname{H}(M) = \big(\operatorname{H}_\ell (M) [\vee|\wedge]\, m_{n-1}\big) [\vee|\wedge]\, m_n$$

The number of the blocks determines the order of the last two operations, the $\vee$ is the last or $\wedge$.

Now the trick is that

  • by using $\wedge$ you can make any bit zero $x \wedge 0 =0$ and
  • by using $\vee$ you can make any bit one $x \vee 1 =1$.

Therefore only using the last two blocks - actually using only two blocks - the attacker can create any hash value.

There are much more simple attacks as a note by Fgriue:

  • $h = \operatorname{H}(h)$
  • $h = \operatorname{H}(h\mathbin\|h)$ since $h = h \wedge h $
  • $h = \operatorname{H}(1^n\mathbin\|h)$ since $h = 1^n \wedge h $
  • $h = \operatorname{H}(h\mathbin\|h\mathbin\|h)$ since $h = (h \wedge h) \vee h $
  • $h = \operatorname{H}(\bar h\mathbin\|h\mathbin\|h)$ since $h = (\bar h \wedge h) \vee h $

* The formal definition can be found in this seminal work Cryptographic Hash-Function Basics: Definitions, Implications, and Separations for Preimage Resistance, Second-Preimage Resistance, and Collision Resistance by P. Rogaway and T. Shrimpton.

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    $\begingroup$ Yes. The simplest preimage attack is that $H(h)=h$. Others are that $H(h\mathbin\|h)=h$, $H(1^n\mathbin\|h)=h$, $H(h\mathbin\|h\mathbin\|h)=h$, $H(\bar h\mathbin\|h\mathbin\|h)=h$. $\endgroup$ – fgrieu Apr 12 '20 at 20:03
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    $\begingroup$ @fgrieu that is brilliant. I've added them. Thanks. $\endgroup$ – kelalaka Apr 12 '20 at 20:15

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