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As the title says: do we know already for sha256 what input would yield the same output as when using no input at all?

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    $\begingroup$ The empty string ≠ no input. $\endgroup$ – fkraiem Aug 15 '17 at 9:29
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    $\begingroup$ Two distinct inputs with the same output is a collision. We know of no collisions in SHA-2. (in practice it'd even amount to a second pre-image, which we can't even construct for MD5) $\endgroup$ – CodesInChaos Aug 15 '17 at 9:40
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Semantically, this makes no sense.

A function $\ f : X \to Y$, always takes an input from $X$ (by definition of a function).

In the case of $SHA256$ the set of possible inputs is the list of all possible bit strings, including the empty string denoted as $\epsilon$ :

$X = \{\\ \epsilon,\\ \texttt{0b0},\\ \texttt{0b1},\\ \texttt{0b00}\\ \texttt{0b01},\\ \texttt{0b10},\\ \ldots\}$

Remark: this is not a finite set.

As a result, there is no such thing as no input at all.


If your question is

Do we know $x$ such that $SHA256(x) = SHA256(\epsilon)$ where $\epsilon$ is the empty string ?

Then the answer is no, because this is called a collision ($x \neq y \land h(x) = h(y)$) and none has been found (yet) for SHA256.

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    $\begingroup$ I have seen the empty string denoted $\lambda$, $\epsilon$, or $\varepsilon$, but never $\sigma$... $\endgroup$ – fkraiem Aug 16 '17 at 7:48
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    $\begingroup$ SHA-2 is defined for all bit strings, the length doesn't need to be a multiple of 8 bits. $\endgroup$ – user31573 Aug 16 '17 at 16:48
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    $\begingroup$ Oh, sure, it's an unrelated thing, but I'd really like to see people get this correctly. $\endgroup$ – user31573 Aug 16 '17 at 17:14
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    $\begingroup$ By the way, strictly speaking, SHA-256 is limited to 2⁶⁴-1 bits, so the set of all inputs is finite. $\endgroup$ – user31573 Aug 16 '17 at 17:16
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    $\begingroup$ I won't bother with correcting that one :) $\endgroup$ – Biv Aug 16 '17 at 17:19

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