7
$\begingroup$

Let us have any linear secret sharing scheme (LSSS) that works on some field $Z_{p}$, where p is some prime or a power of a prime e.g., Shamir Secret Sharing, Additive secret Sharing. The problem at hand is simple, for any secret shared value in $Z_{p}$, is it possible to convert it (and its shares) to elements on $Z_{q}$, where q could be either greater or smaller than p? Note that I do not need conversion from $Z_{2^n}$ to $Z_{2}$.

I am aware of the results on PRSS, and well as in this 2008 work, however the former does not address field conversion, and the latter, have explanations that are not clear nor it offers perfect security.

Any help is more than welcome!

$\endgroup$
  • $\begingroup$ There is a way to do share conversion from binary to arithmetic and vice-versa using this paper: thomaschneider.de/papers/DSZ15.pdf $\endgroup$ – Dragos May 19 '17 at 14:36
  • $\begingroup$ @Dragos, as I pointed out on the explanation of the problem I 'm not interested on $Z_{2}$ to $Z_{2^r}$ conversions. Thanks anyways! $\endgroup$ – DaWNFoRCe May 19 '17 at 14:48
  • $\begingroup$ For me, arithmetic means $Z_p$. You can use the method above by first converting from $Z_p \rightarrow Z_2$ and then from $Z_2 \rightarrow Z_q$. $\endgroup$ – Dragos May 19 '17 at 14:53
  • $\begingroup$ @Dragos. My inputs are bounded by the conversion target, if I do any transformation to a smaller field, let say $Z_{2}$ I would loose information (or be forced to use bit decomposition which I also want to avoid). This is why I cannot use $Z_{2}$, as I pointed out. Any idea is welcome! $\endgroup$ – DaWNFoRCe May 19 '17 at 14:58
  • $\begingroup$ I see..Even converting shares from $Z_p \rightarrow Z_2^{r}$ is a hard problem - at least I have encountered it several times and don't know how to solve it efficiently other than bit decompose and simulate the binary circuit in $Z_p$. $\endgroup$ – Dragos May 19 '17 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.