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If for example we would find a hash collision for SHA256 with $h(x) = h(x')$ for $x \ne x'$ and $x, \space x' < 128$ bit (meaning both $x$ and $x'$ are smaller than 128 bit), would SHA256 be considered broken?

Is it then always possible to create more collisions in a feasible time or could this hash collision just be considered an "anomaly"?

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    $\begingroup$ I think you mean $x,x' < 128$-bit $\endgroup$
    – kelalaka
    Feb 13, 2019 at 21:05

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If for example we would find a hash collision for SHA256 with $h(x) = h(x')$

Yes, pretty much. Now, I'm not certain what you meant by "$x \land x' < 128$ bit" (total bit length is less than 128 bits, or they disagree in at most 128 bits???), however, it doesn't really matter; finding any collision would demonstrate that SHA256 isn't quite as collision resistant as we had thought.

Is it then always possible to create more collisions in feasible time or could this hash collision just be considered an "anomaly"?

If $x, x'$ had the same length (or, more generally, used the same number of SHA-256 blocks when hashing), then it's easy to find more collisions; we would have $h(x || Pad_x || y) = h(x' || Pad_{x'} || y)$ for any string $y$ (where $Pad_x, Pad_{x'}$ is the padding that SHA-256 applies when hashing $x, x'$).

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