2
$\begingroup$

Since we know that Homomorphic encryption allow computation on encrypted data. Let two n-bit integers $(x,y)$ are encrypted using some LWE public key cryptography. For example - if $x'= HEnc(x,pk)$ and $y'= HEnc(y,pk)$, Finding $Max(x',y')$ or $Min(x',y')$ or $Equal(x',y')$ the above comparison is necessary.

Now, we need to perform comparison (>=,==) operation homomorphically. How can we do this? Can anyone describe this with example?

Improve this question
$\endgroup$
  • 1
    $\begingroup$ So both operands are encrypted and the result is an encrypted bit? Is that the scenario? $\endgroup$ – mikeazo Jan 26 '16 at 12:50
  • $\begingroup$ Yes, this scenario is right. Homomorphic encryption allow computation on encrypted data. For example if x'=HEnc(x,pk) and y'=HEnc(y,pk), Finding Max(x',y') or Min(x',y') the above comparison is necessary. $\endgroup$ – Tushar Saha Jan 27 '16 at 4:21
1
$\begingroup$

You'll have to write the function you are trying to calculate as a polynomial in the two inputs $x$ and $y$. If you are working over the field with $q$ elements as plaintexts, you have to calculate for equality the polynomial $(x-y)^{q-1}$. Greater-than-or-equal (however you define that for finite fields) will be even more complicated.

Improve this answer
$\endgroup$
0
$\begingroup$

Have a look at order revealing encryption (ORE) schemes:

https://eprint.iacr.org/2014/834.pdf

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Many thanks for suggesting this paper. The paper is about 'Order revealing encryption'. Is there any relation between 'Order revealing Encryption' and 'Homomorphic encryption'? I want to do this job homomorphically because I've some other job with this encryption. $\endgroup$ – Tushar Saha Jan 27 '16 at 7:18
  • 1
    $\begingroup$ No, ORE is rather unrelated to HOM. This "answer" is more of a comment. $\endgroup$ – Thomas M. DuBuisson Jan 27 '16 at 17:52

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.