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For example:

Assuming someone has found a second preimage for SHA-256. In other words, for SHA256(M1)=H, someone has found M2 which SHA256(M2)=H

Now, is it possible to generate k-way second preimages faster than brute force? To illustrate my point:

For SHA256(M1)=SHA256(M2)=H, is there a way to find:

SHA256(M1)=SHA256(M2)=SHA256(M3)=SHA256(M4)=...=SHA256(Mk)=H

faster than brute force?

EDIT:

I did find some relevant information here: (https://pdfs.semanticscholar.org/bc44/a277e6e10ff318ea7ecb06af12a191d1193a.pdf)

According to the paper, by using an extension of Joux's multicollision technique, it is possible to generate multiple second preimages faster than brute force. However, this doesn't seem to match what I want to find, because the technique described in the paper creates multi-preimages with the same message block, but different initial values. I want a technique to find multi-preimages with different message blocks.

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  • $\begingroup$ I wonder if that is dependent on how the second pre-image was found. If you found it by sheer luck then I would guess the answer is no. Of course, if you'd find one by just trying then people would start scratching their heads and try to find out what had just happened. $\endgroup$ – Maarten Bodewes Apr 25 at 14:59

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