I am looking for a one way function that can generate outputs deterministically. However, revealing any of the inputs dont allow the person to generate other inputs.

lets say we have a list of inputs: $i_{1},i_{2},i_{3},...$ that look completely random and a list of corresponding outputs: $o_{1},o_{2},o_{3},...$ that also look completely random which are each mapped by one way functions: $f_{1},f_{2},f_{3},...$
These outputs can be generated without knowing the inputs so anyone should be able to deterministically generate the outputs (perhaps by knowing a public shared data which is used in the function). But if one of these inputs is revealed, it should not reveal the other inputs.
If it is not mathematically possible, I need a proof for it.


"public shared data" $\:$ = $\:$ the ordered pair whose entries are
[a key for a full-domain statistically injective PRF] and [a signature verification key]

$\hspace{.04 in}f$ is an arbitrary one-way function.
For each positive integer $j$, [$o_j$ is the value of the PRF at $j\hspace{.03 in}$] and [$i_j$ is [[the signature for $o_j$] concatenated with [the string of zeros whose length equals the length of that signature]]]
and $\hspace{.04 in}f_j$ is given by
if $\hspace{.04 in}f_j$'s input has even length and its left half is a valid signature for $o_j$
and its right half is all-zero then output $o_j$ else output $\hspace{.04 in}f$ of $\hspace{.04 in}f_j$'s input.

  • $\begingroup$ Why should the inputs left half be all zero? You said the input is the signature for $j$? $\endgroup$ – abeikverdi Apr 23 '15 at 9:31
  • $\begingroup$ And how can someone generate the next output? He/she needs the signature of $j$. So if the key owner doesn't sign the next input $j$, the next output can not be calculated. $\endgroup$ – abeikverdi Apr 23 '15 at 9:34
  • $\begingroup$ The inputs' left half being all zero is a way to make sure that the $\hspace{.04 in}f_j$s are one-way. $\;$ $\endgroup$ – user991 Apr 23 '15 at 9:35
  • $\begingroup$ $o_j = j {}{}\;$ $\endgroup$ – user991 Apr 23 '15 at 9:35
  • $\begingroup$ umm.. interesting! but now the outputs are not random then. They are like 1,2,3,.. $\endgroup$ – abeikverdi Apr 23 '15 at 9:40

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