# Is it known what input for sha256 would yield the same result as no input?

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

• The empty string ≠ no input. – fkraiem Aug 15 '17 at 9:29
• 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) – CodesInChaos Aug 15 '17 at 9:40

## 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.

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.
• I have seen the empty string denoted $\lambda$, $\epsilon$, or $\varepsilon$, but never $\sigma$... – fkraiem Aug 16 '17 at 7:48